Chemistry in the Earth System

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Define matter.
   Show the full solution

   Anything that has mass and occupies volume

2. State the difference between mass and weight.
   Show the full solution

   Mass is the quantity of matter and is independent of location; weight is the gravitational force on that mass and varies with location

3. State the law of conservation of mass, including the condition it requires.
   Show the full solution

   In a closed system, the total mass before a chemical change equals the total mass after it

4. A sealed flask contains 12.50 g of one reactant and 8.30 g of another. They react completely. What is the mass of the contents afterward?
   Show the full solution

   The flask is sealed, so nothing crosses the boundary. \( 12.50 + 8.30 \). 20.80 g

5. Give the usual reason an open reaction vessel appears to lose mass.
   Show the full solution

   A gaseous product escaped from the container

6. A student heats 5.00 g of a green solid in an open crucible and recovers 3.20 g of a black solid. Explain what happened to the missing 1.80 g and how you would test your explanation.
   Show the full solution

   Nothing is missing in the sense of destroyed. The decomposition released a gas that left the open crucible, so the 1.80 g is now in the room air. The test is to repeat the heating in a sealed vessel and weigh the whole vessel before and after: if the total is unchanged, the mass left the crucible rather than existence. A second test collects the gas over water and weighs it, which both accounts for the 1.80 g and identifies what was released. A gas escaped; confirm by repeating in a closed system and weighing the whole vessel

7. Wood burns to a small heap of ash that weighs far less than the log. Iron rusts and gains mass. Explain why these are the same phenomenon.
   Show the full solution

   Both are open systems exchanging gas with the air, and the direction of the apparent change depends only on whether the gas leaves or joins. Wood is largely carbon and hydrogen; burning converts it to carbon dioxide and water vapor, which leave, so the solid that stays behind is lighter. Iron is a solid metal; rusting combines it with oxygen from the air to make a solid oxide, which stays, so the solid gets heavier. Weigh either in a sealed container and the total is constant. Both exchange gas with the air, one losing it and one gaining it

8. Explain why "this substance is better for cleaning" is not a chemical claim, and rewrite it as one.
   Show the full solution

   Better is a judgment, and no measurement returns one, because better depends on what is being cleaned, at what cost, and with what risk. Chemistry can supply the facts the judgment uses but cannot make it. A checkable version fixes what is being measured and on what: "does a 5 percent solution of this substance remove a greater mass of a standard grease deposit from a glass slide in five minutes than a 5 percent solution of the alternative, at the same temperature?" Every term there can be measured and the result could come out either way. It is a value judgment; fix a measurable outcome, a quantity and a comparison

9. A student weighs a sealed bag of baking soda and vinegar, mixes them, observes vigorous bubbling and reweighs. The mass is unchanged, but the bag is now inflated. Explain both observations.
   Show the full solution

   The mass is unchanged because the bag is closed: the atoms present at the start are still present, rearranged into different substances. The inflation happens because one of those substances is a gas, and a gas occupies far more volume per unit mass than the solid and liquid it came from. Mass and volume are separate quantities, and a chemical change conserves the first while frequently altering the second dramatically. If the bag were opened, the gas would leave and the mass would then fall. Mass is conserved in the closed bag; volume is not conserved because a gas was produced

10. Lavoisier's contemporaries had the idea of conservation before him. Explain why he is nonetheless credited with establishing it.
    Show the full solution

    Stating a principle and establishing it are different achievements. Earlier writers offered conservation as a plausible philosophical position, with nothing that would show it false. Lavoisier made it a measured result by working in sealed vessels with a balance accurate enough to detect the differences at stake, and by doing so on reactions where the naive observation points the other way, such as metals gaining mass on heating. The principle became evidence-based rather than assumed, and it simultaneously destroyed the competing phlogiston account, which had to posit a substance with negative weight to survive. He measured it in closed systems rather than asserting it, on cases where the appearance contradicts it

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State the defining difference between a physical and a chemical change.
   Show the full solution

   A chemical change produces at least one new substance; a physical change does not

2. List the five common signs of a chemical change.
   Show the full solution

   Gas evolved, precipitate formed, color change, temperature change, light emitted

3. Classify density as intensive or extensive, and say why it matters.
   Show the full solution

   Intensive: it does not depend on how much you have, so it can be used to identify a substance

4. Classify these as physical or chemical changes: melting ice, souring milk, shredding paper, digesting food.
   Show the full solution

   Physical, chemical, physical, chemical

5. Is flammability a physical or a chemical property? Explain in one sentence.
   Show the full solution

   Chemical, because you can only observe it by changing the substance into something else

6. Bubbles form when hydrochloric acid is poured on limestone, and also when water is heated to boiling. Explain how you would tell the two situations apart experimentally.
   Show the full solution

   Identify the gas and check whether the starting material survives. Collect the gas from each: the gas from the limestone turns limewater cloudy and does not condense on cooling, so it is carbon dioxide, a substance that was not present before. The gas from the boiling water condenses back to liquid water on a cold surface, so it is the same substance in a different state. Then check the residue: the limestone has partly disappeared and a different solid remains, while cooling the steam returns all the water. One produced something new and the other did not. Identify the gas and test whether the original substance can be recovered by cooling

7. Dissolving sodium chloride in water absorbs a little heat, and dissolving it changes a clear liquid into a clear solution. Two of the five signs are arguably present. Explain why dissolving is nevertheless a physical change.
   Show the full solution

   Because the salt is still there and can be recovered unchanged. Evaporate the water and sodium chloride crystals come back with the same melting point, density and crystal form as the ones that went in, which would be impossible if they had been converted into something else. The temperature change comes from the energy needed to separate ions from the lattice against the energy released when water molecules surround them, and neither step forms new bonds between different elements. The signs are evidence that something happened, not evidence of what happened. The salt is recoverable unchanged by evaporation, so no new substance was formed

8. A student says mass is a good way to identify an unknown liquid. Explain the error and give a better property.
   Show the full solution

   Mass is extensive, so it depends entirely on how much of the liquid happens to be in the container and tells you nothing about what the liquid is. Any liquid can have a mass of 50 g. An identifying property must be intensive, so that the value is the same for a drop and for a barrel. Density works, because it is mass per unit volume and is fixed for a pure substance at a given temperature, as do boiling point and refractive index. In practice you would measure two or three intensive properties, since a single value can be shared by more than one substance. Mass is extensive; use an intensive property such as density or boiling point

9. A firework emits light and heat, and a glow stick emits light but gets slightly cooler. Are both chemical changes? Explain.
   Show the full solution

   Yes, both are. The direction of the energy flow does not decide the question; the production of new substances does. In the firework an exothermic reaction releases energy as heat and light, and the products are oxides quite different from the reactants. In the glow stick two substances are mixed and react, and the energy released by bond rearrangement leaves as light rather than heat, so the surroundings do not warm. Both produce new substances that cannot be separated back into the originals by physical means, which is the test. Both are chemical; energy direction does not decide it, new substances do

10. Explain why the candle in the worked example needs a wick, in terms of physical and chemical changes.
    Show the full solution

    Solid wax does not burn at a useful rate, because combustion happens where fuel molecules and oxygen molecules can mix freely, which is in the gas phase. A block of wax presents only its surface, so the reaction is slow and self-extinguishing. The wick performs the physical steps that solve this: heat from the flame melts wax at the base, capillary action carries liquid wax upward, and the heat at the tip vaporizes it, delivering a steady supply of wax vapor to the flame. The chemical change then has fuel in the phase it requires. The physical processes exist to feed the chemical one. The wick melts, transports and vaporizes the wax so the chemical change has gaseous fuel

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Name the two kinds of pure substance.
   Show the full solution

   Elements and compounds

2. State the difference between a compound and a mixture in terms of ratio.
   Show the full solution

   A compound has a fixed ratio of elements chemically bonded; a mixture can have any proportions and is not bonded

3. Classify each as homogeneous or heterogeneous: seawater, granite, air, sand in water.
   Show the full solution

   Homogeneous, heterogeneous, homogeneous, heterogeneous

4. Which property does distillation exploit?
   Show the full solution

   Difference in boiling point

5. Which property does filtration exploit?
   Show the full solution

   Difference in particle size, separating an insoluble solid from a liquid

6. Water is a compound of hydrogen and oxygen. Hydrogen burns explosively and oxygen supports combustion, yet water extinguishes fires. Explain what this shows about compounds.
   Show the full solution

   It shows that a compound's properties are not the sum or average of its elements' properties, because the atoms are chemically bonded and the resulting substance is genuinely new. The hydrogen in water is not present as hydrogen gas; it is bonded into molecules whose behavior has to be determined by experiment rather than predicted from the ingredients. This is the sharpest difference between a compound and a mixture: a mixture of hydrogen and oxygen gases is explosive, and the same atoms bonded as water are not. A compound's properties are new, not inherited from its elements

7. Explain how you would separate a mixture of two liquids that dissolve in each other, such as ethanol and water, and why filtration cannot work.
   Show the full solution

   Filtration separates by particle size and only works when one component is an undissolved solid. Two miscible liquids form a homogeneous mixture in which the particles are of molecular size and pass through any filter together, so no pore size helps. The property that does differ is boiling point: ethanol boils at about 78 degrees Celsius and water at 100. Fractional distillation exploits this, heating the mixture so the more volatile ethanol vaporizes preferentially, passing up a fractionating column where repeated condensation and vaporization improve the separation, then condensing in the condenser and collecting first. Fractional distillation, using the boiling point difference; filtration cannot separate molecular-scale mixtures

8. In the worked example, explain why the magnetic step must come before the water is added.
   Show the full solution

   Two reasons, one practical and one chemical. Practically, a magnet works poorly through a wet slurry: the filings clump with damp sand and are dragged along with material that should stay behind, so the recovered iron is contaminated and the recovered sand is depleted. Chemically, iron in contact with water and dissolved oxygen begins to rust, which converts some of the iron into iron oxide. That is a chemical change, so the iron can no longer be recovered unchanged and the mass balance at the end will not close. Ordering the steps to keep every stage physical is part of designing the separation. Wet filings separate badly, and iron in water begins to rust, which is a chemical change

9. Chromatography separates the pigments in a leaf extract. Explain what property difference it exploits and why the spots end up at different heights.
   Show the full solution

   It exploits the difference in how strongly each pigment is attracted to the stationary phase, the paper, compared with how readily it dissolves in the moving solvent. As the solvent rises through the paper it carries the pigments with it, but each one spends a different fraction of its time stuck to the paper rather than traveling in the solvent. A pigment with a strong attraction to the paper and low solubility in the solvent is held back and travels a short distance; one that dissolves readily and binds weakly travels nearly as far as the solvent front. The heights therefore report a ratio of two competing affinities, which is why the same pigment gives a reproducible position for a given paper and solvent. Competing affinity for the stationary paper against solubility in the moving solvent

10. Air is usually called a homogeneous mixture, but air also contains dust and water droplets. Explain how both statements can be defensible.
    Show the full solution

    The classification depends on the scale being described and on what is included in the sample. Treated as its gaseous components alone, air is genuinely homogeneous: nitrogen, oxygen, argon and carbon dioxide are mixed at the molecular level and no region differs from another, which is why a sample taken anywhere in a room has the same composition. Include the suspended solids and liquid droplets and the sample now has visibly distinct phases, which makes it heterogeneous, and fog and smoke are exactly that. Neither statement is careless; they answer different questions, and a good answer says which components it is counting. The gaseous components alone are homogeneous; including suspended particles and droplets makes it heterogeneous

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Convert 2.5 kg to milligrams.
   Show the full solution

   \( 2.5 \text{ kg} \times \frac{1000 \text{ g}}{1 \text{ kg}} \times \frac{1000 \text{ mg}}{1 \text{ g}} \). \( 2.5 \times 10^{6} \) mg

2. Convert 750 mL to liters.
   Show the full solution

   Divide by 1000. 0.750 L

3. State the two volume equalities linking milliliters, cubic centimeters and liters.
   Show the full solution

   \( 1 \text{ mL} = 1 \text{ cm}^3 \) and \( 1 \text{ L} = 1000 \text{ cm}^3 \)

4. Explain in one sentence why a conversion factor can be written either way up.
   Show the full solution

   Because the numerator and denominator are equal quantities, so the fraction equals one and multiplying by it cannot change the quantity

5. Convert 85 km/h to meters per second.
   Show the full solution

   \( 85 \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = \frac{85\,000}{3600} \). 23.6 m/s

6. A student converts 2.50 mL of mercury, density 13.53 g/cm³, to a mass in kilograms. Carry out the conversion and explain where the two traps are.
   Show the full solution

   First trap: the density is per cubic centimeter and the volume is in milliliters, which looks like a mismatch but is not, because \( 1 \text{ mL} = 1 \text{ cm}^3 \) exactly. Students often insert a spurious conversion here. So \( 2.50 \text{ cm}^3 \times 13.53 \; \frac{\text{g}}{\text{cm}^3} = 33.825 \text{ g} \). Second trap: the answer is wanted in kilograms, and stopping at grams is the commoner error than getting the arithmetic wrong. Divide by 1000 to get 0.033825 kg, then round to the three significant figures the data support. 0.0338 kg

7. Explain why checking that the units cancel is more useful than checking the arithmetic.
   Show the full solution

   Because the two failures are not equally likely or equally damaging. Arithmetic slips are caught by redoing the calculation, and a calculator rarely mismultiplies. Setting a conversion factor the wrong way up produces an answer that is wrong by a factor of the conversion squared, often by a thousand or a million, and it looks entirely plausible on the page because the numbers are the right numbers. The unit check catches exactly that error, before any arithmetic is done, because an inverted factor leaves an impossible surviving unit such as gallons squared per liter. It also catches a missing step, since the wanted unit simply fails to appear. It catches inverted or missing factors, which produce large plausible errors that rechecking arithmetic will not find

8. A laboratory procedure calls for 0.250 L of solution and a student measures 250 mL in a graduated cylinder. A second student says this is wrong because the units differ. Settle the disagreement.
   Show the full solution

   The measured quantity is correct, because 0.250 L and 250 mL are the same volume: multiplying 0.250 L by 1000 mL per liter gives 250 mL exactly. The second student has confused a difference in unit with a difference in quantity. There is a real point buried in the objection, though, and it concerns significant figures rather than size: 0.250 L is stated to three significant figures, so the measurement should be made in a vessel capable of supporting that precision. A 250 mL graduated cylinder typically reads to about one milliliter, which is adequate, while a 250 mL beaker's graduations are not. The volumes are identical; the defensible concern is precision, not magnitude

9. A carbon dioxide concentration is reported as 421 parts per million by volume. Express this as a percentage by volume, and comment on why the smaller unit is used.
   Show the full solution

   Parts per million means parts per \( 10^{6} \), and percent means parts per \( 10^{2} \), so divide by \( 10^{4} \): \( 421 \div 10^{4} = 0.0421 \) percent. The reason the atmospheric record is published in parts per million rather than percent is that the interesting changes are invisible at the coarser unit. A rise from 280 to 421 parts per million over two centuries appears as a change from 0.028 to 0.042 percent, which reads as almost nothing and requires three decimal places to express at all, whereas the same change in parts per million is a rise of 141 in a quantity near 300. The unit should be chosen so that ordinary variation lands in the ones or tens place. 0.0421 percent; parts per million is used so the variation is legible without leading zeros

10. Explain why every equation in this course that involves concentration or gas volume assumes liters, and what that implies for your working habits.
    Show the full solution

    The constants are defined that way. Molarity is defined as moles per liter, and the gas constant \( R \) has the value 0.0821 with units of liter atmospheres per mole kelvin, so a volume substituted in milliliters is wrong by a factor of a thousand and nothing in the equation reveals it. The result looks like a normal number and carries no warning. The working habit that prevents this is to convert units at the point of reading the question rather than at the point of substitution: write the volume down in liters, the temperature in kelvin and the mass in grams as you extract each one, so the quantities entering any formula are already in the units it expects. The constants are defined per liter, so convert at the moment you read the data, not at substitution

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. How many significant figures are in 0.00340?
   Show the full solution

   Leading zeros do not count; the trailing zero does because a decimal point is present. Three

2. How many significant figures are in 2050.0?
   Show the full solution

   All digits count here: the zero between non-zero digits, and the trailing zeros because the decimal point is shown. Five

3. State the rule for multiplication and division.
   Show the full solution

   The answer carries the fewest significant figures of any value used

4. Calculate \( 4.560 + 12.1 \) and report it correctly.
   Show the full solution

   Raw sum 16.66; the decimal-place rule allows one decimal place. 16.7

5. Calculate \( 3.20 \times 4.1 \) and report it correctly.
   Show the full solution

   Raw product 13.12; the smaller count is two significant figures. 13

6. Explain why 3400 is ambiguous and how scientific notation resolves it.
   Show the full solution

   Trailing zeros in a number with no decimal point may be measured digits or may only be holding the place value, and the notation cannot distinguish the two. A reading of 3400 g could mean the balance resolved the tens and ones digits and found both to be zero, or that it resolved only to the nearest hundred grams. Scientific notation makes the claim explicit because only the digits written in the coefficient are significant: \( 3.4 \times 10^{3} \) states two significant figures, \( 3.40 \times 10^{3} \) states three, and \( 3.400 \times 10^{3} \) states four, while all three represent the same magnitude. Trailing zeros without a decimal point may or may not be measured; scientific notation shows exactly which digits are claimed

7. A student measures a rectangle as 2.0 cm by 3.00 cm and reports an area of 6.00 cm². Find and explain the error.
   Show the full solution

   The multiplication rule takes the fewest significant figures of the values used, and 2.0 cm has only two, so the area may carry two: \( 6.0 \text{ cm}^2 \). Reporting 6.00 claims the area is known to the nearest hundredth of a square centimeter, which cannot be true when one side is known only to the nearest tenth of a centimeter. The underlying point is that a calculation cannot be more certain than the least certain measurement it used, and writing extra digits does not make the ruler better. 6.0 cm²; the least precise measurement limits the result

8. Explain why counting twelve test tubes does not limit the significant figures in a calculation that uses the number twelve.
   Show the full solution

   Because it is not a measurement and carries no uncertainty. Counting discrete objects gives an exact value: there are precisely twelve test tubes, not twelve give or take a tenth, so the number could be written as 12.000000 without making any false claim. Significant figure rules exist to propagate measurement uncertainty through a calculation, and a quantity with no uncertainty cannot contribute any. The same applies to defined conversions such as 1000 milliliters per liter and to the integers in a formula, which is why dividing a measured mass by an exact count leaves the significant figures set by the mass alone. It is an exact count with no uncertainty, so it contributes no limit; only measured values do

9. Three students weigh the same object on the same balance and get 4.512 g, 4.511 g and 4.513 g. The object's certified mass is 4.230 g. Describe the balance in terms of precision and accuracy, and say which problem is easier to fix.
   Show the full solution

   The readings agree with each other to within two thousandths of a gram, so the balance is highly precise: it returns the same value under the same conditions. It is badly inaccurate, since every reading is about 0.28 g above the certified value, an error more than a hundred times the scatter. The pattern of a constant offset with small scatter is the signature of a calibration fault rather than a random one. That is the easier problem to fix, because a systematic offset can be removed by recalibrating against a standard mass, whereas poor precision points to an instrument or a technique that would have to be replaced or retrained. Precise but inaccurate, with a systematic offset; recalibration fixes it easily

10. A student performs a four-step calculation, rounding to two significant figures after each step, and their answer differs from the correct one in the second digit. Explain the cause and the remedy.
    Show the full solution

    Rounding discards information, and rounding at every step discards it four times, so each subsequent step operates on a value that is already slightly wrong and the errors compound rather than cancel. Rounding to two significant figures can shift a value by nearly half a percent, and four such shifts can easily move the second digit of the final answer. The remedy is to carry the full precision of intermediate results through the whole calculation, either by keeping the value in the calculator or by writing down two or three extra guard digits, and to round only once when the final answer is reported. The significant figure rules describe what may be claimed at the end, not what must be written in the middle. Intermediate rounding compounds; carry full precision and round only the final answer

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Write 0.000 062 5 in scientific notation.
   Show the full solution

   Move the decimal point five places right. \( 6.25 \times 10^{-5} \)

2. Write \( 4.07 \times 10^{4} \) in ordinary decimal notation.
   Show the full solution

   40 700

3. Calculate \( (3.0 \times 10^{8}) \times (2.0 \times 10^{-5}) \).
   Show the full solution

   Coefficients \( 3.0 \times 2.0 = 6.0 \); exponents \( 8 + (-5) = 3 \). \( 6.0 \times 10^{3} \)

4. Calculate \( (6.0 \times 10^{5}) \div (2.0 \times 10^{8}) \).
   Show the full solution

   Coefficients \( 6.0 \div 2.0 = 3.0 \); exponents \( 5 - 8 = -3 \). \( 3.0 \times 10^{-3} \)

5. How many orders of magnitude separate \( 2 \times 10^{3} \) and \( 2 \times 10^{9} \)?
   Show the full solution

   The exponents differ by six. Six, a factor of a million

6. A student calculates \( (4.0 \times 10^{3}) + (2.0 \times 10^{2}) \) and answers \( 8.0 \times 10^{5} \). Find every error and give the correct result.
   Show the full solution

   They applied the multiplication rules to an addition, multiplying the coefficients and adding the exponents. Neither operation is permitted here. For addition the exponents must match first, so rewrite \( 2.0 \times 10^{2} \) as \( 0.20 \times 10^{3} \). Now the coefficients can be added: \( 4.0 + 0.20 = 4.2 \), and the shared exponent is unchanged, giving \( 4.2 \times 10^{3} \). A check in ordinary notation confirms it: \( 4000 + 200 = 4200 \). Their answer of 800 000 is nearly two hundred times too large, which an order-of-magnitude estimate would have caught immediately. \( 4.2 \times 10^{3} \)

7. Explain why scientific notation makes significant figures unambiguous, using 500 as the example.
   Show the full solution

   Written as 500, the two zeros might be measured digits or might be placeholders fixing the magnitude, and nothing in the notation says which, so the value could be claiming one, two or three significant figures. Scientific notation separates the two jobs completely: the power of ten carries all the magnitude information and the coefficient carries all the precision information, so every digit written in the coefficient is by definition significant. \( 5 \times 10^{2} \) claims one significant figure, \( 5.0 \times 10^{2} \) claims two and \( 5.00 \times 10^{2} \) claims three, and a reader knows exactly which measurement was made. Magnitude sits in the exponent and precision in the coefficient, so every written digit is significant

8. A calculation of the number of molecules in a drop of water returns \( 1.7 \times 10^{21} \). Another student's answer is \( 1.7 \times 10^{12} \). Explain how you could tell which is plausible without redoing the calculation.
   Show the full solution

   By comparing each against a fixed landmark. A mole is \( 6.02 \times 10^{23} \) particles and corresponds to about 18 g of water, so a drop of roughly 0.05 g is a few thousandths of a mole, which must therefore contain something in the region of \( 10^{21} \) molecules. The first answer sits exactly there. The second is nine orders of magnitude smaller, which would mean a drop of water contained about a millionth of a millionth of a mole, or a mass far below what any balance would register. Holding one or two anchor values in mind, such as Avogadro's number and the molar mass of water, lets you reject an answer by scale alone. The first; comparing with Avogadro's number shows the second is nine orders of magnitude too small

9. Explain why the answer to the worked example should be stated as "of the order of ten million" rather than as 7 600 000.
   Show the full solution

   Because the input did not support seven digits. The atomic diameter was given as about \( 1 \times 10^{-10} \) m, which is a single significant figure and is in any case a rough average, since atomic diameters range by roughly a factor of three from helium to cesium. An answer can never be more precise than the worst input, so at most one significant figure is defensible here, and even that digit depends on which element the atoms are. Writing 7 600 000 implies the count was determined to the nearest hundred, which no part of the calculation could justify. Stating an order of magnitude reports exactly what the estimate supports. The input had one significant figure and atomic sizes vary, so only the order of magnitude is supported

10. An atom is about \( 10^{-10} \) m across and a nucleus about \( 10^{-15} \) m. Express the relationship as an order of magnitude and explain why Rutherford found his result surprising.
    Show the full solution

    The diameters differ by five orders of magnitude, so the atom is about a hundred thousand times wider than its nucleus. Because volume scales with the cube of length, the nucleus occupies roughly \( 10^{15} \) times less volume than the atom, meaning the atom is overwhelmingly empty space. The surprise in Rutherford's gold foil experiment follows directly: on the prevailing model, with positive charge spread evenly through the atom, alpha particles should have passed through a thin foil with only slight deflections, and almost all of them did. The occasional particle returning nearly the way it came required all the positive charge and nearly all the mass to be concentrated in that tiny fraction of the volume, which lesson 2.2 takes up. Five orders of magnitude in diameter and about fifteen in volume, so the atom is nearly all empty space

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State the formula for density and a common unit for a solid.
   Show the full solution

   \( d = \frac{m}{V} \), in g/cm³

2. A sample has a mass of 22.5 g and a volume of 4.50 mL. Find its density.
   Show the full solution

   \( 22.5 \div 4.50 \). 5.00 g/mL

3. Find the mass of 15.0 mL of ethanol, density 0.789 g/mL.
   Show the full solution

   \( m = dV = 0.789 \times 15.0 = 11.835 \), to three significant figures. 11.8 g

4. Find the volume of 100.0 g of aluminum, density 2.70 g/cm³.
   Show the full solution

   \( V = \frac{m}{d} = \frac{100.0}{2.70} = 37.037 \), to three significant figures. 37.0 cm³

5. On a heating curve, what is happening during a flat section?
   Show the full solution

   A phase change: the supplied energy is overcoming attractions between particles rather than increasing their speed

6. Explain why a large steel ship floats while a small steel bolt sinks.
   Show the full solution

   Floating depends on the average density of the whole object including any space it encloses, not on the density of the material it is made from. The bolt is solid steel throughout, so its average density is steel's roughly 7.9 g/cm³, far above water's 1.00, and it sinks. The ship's hull encloses a very large volume of air, and the combined mass of steel and air divided by the total volume of the hull comes out below 1.00 g/cm³. Mass alone decides nothing, which is exactly why density rather than mass is the useful quantity. Average density including enclosed air, not the density of the material alone

7. A student reports the density of a liquid as 0.92 g/mL at 20 degrees Celsius and 0.89 g/mL at 60 degrees. Explain why density changes with temperature and whether this undermines its use as an identifying property.
   Show the full solution

   Heating gives the particles more kinetic energy, so they move further apart on average and the same mass occupies a larger volume. Mass is unchanged and volume has risen, so density falls. This does not undermine density as an identifying property, provided the temperature is stated: density is intensive, meaning it does not depend on sample size, but it is a function of temperature, so a reference value is always quoted at a specified temperature, usually 20 or 25 degrees Celsius. Comparing a measurement taken at 60 degrees against a table value for 20 degrees is the actual error to avoid. Volume expands on heating while mass is constant; the property is still identifying as long as the temperature is stated

8. Explain, in terms of particles, why the temperature stays constant while ice melts even though a burner is supplying heat throughout.
   Show the full solution

   Temperature is a measure of the average kinetic energy of the particles, so the temperature only rises when the energy supplied makes particles move faster. In a melting solid the energy is doing a different job: it is working against the attractive forces holding particles in fixed positions in the lattice, converting potential energy rather than kinetic energy. Until every particle has been freed from its lattice position the supplied energy continues to go into separation, so the average speed and therefore the temperature stay where they are. Once melting is complete the energy again goes into motion and the temperature resumes rising, which is why the curve turns upward at the end of the plateau. The energy overcomes attractions and raises potential energy, not average kinetic energy, so the temperature does not change

9. Ice floats on water. Explain what this shows about the arrangement of water molecules in the solid and why it matters for a lake in winter.
   Show the full solution

   Floating means ice is less dense than liquid water, so the same mass of water occupies more volume as a solid than as a liquid. That is the reverse of the usual pattern, and it means the molecules in ice must be held in an arrangement that is more open than the packing in the liquid. Hydrogen bonding is responsible: in the solid each molecule is locked into a fixed tetrahedral arrangement with gaps between, while in the liquid the bonds are constantly breaking and reforming and molecules can crowd closer together. For a lake this is decisive. Ice forms at the surface and stays there, insulating the water below, so the lake freezes downward slowly and liquid water persists at the bottom through winter. If ice sank, lakes would freeze solid from the bottom up. Hydrogen bonding holds the solid in a more open arrangement; the ice layer floats and insulates the liquid beneath

10. Two heating curves for equal masses of different substances both show a flat section at the melting point, but one plateau is three times longer than the other at the same heating rate. Explain what this tells you.
    Show the full solution

    At a constant heating rate, the length of the plateau measures the energy needed to melt the sample, because time multiplied by rate is energy supplied. A plateau three times as long therefore means three times as much energy is required to melt the same mass, so that substance has roughly three times the heat of fusion per gram. Physically, its particles are held together by stronger attractions that take more energy to overcome. This is also evidence you can use elsewhere: the same substance would be expected to have a higher melting point and a higher boiling point, because all three quantities depend on the strength of the forces between particles, which unit 3 takes up as intermolecular forces. It requires about three times the energy per gram to melt, so its interparticle attractions are considerably stronger

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State the law of definite proportions.
   Show the full solution

   A given compound always contains the same elements in the same proportion by mass, regardless of its source

2. State the law of multiple proportions.
   Show the full solution

   When two elements form more than one compound, the masses of one combining with a fixed mass of the other are in a ratio of small whole numbers

3. Name Dalton's two postulates that later proved false.
   Show the full solution

   That atoms are indivisible, and that all atoms of an element are identical in mass

4. Which law does the discovery of isotopes contradict, and which is unaffected?
   Show the full solution

   It contradicts Dalton's postulate of identical masses; the law of definite proportions is unaffected because isotope abundances are constant

5. Two compounds of nitrogen and oxygen contain 0.571 g and 1.142 g of oxygen per gram of nitrogen. State the ratio.
   Show the full solution

   \( 1.142 \div 0.571 = 2.00 \). 2 to 1

6. Explain why the law of multiple proportions is stronger evidence for atoms than the law of definite proportions.
   Show the full solution

   Definite proportions shows only that a given compound has a fixed recipe, which a continuous theory of matter can accommodate by saying that substances combine in some characteristic fixed ratio for reasons of affinity. It constrains the theory without forcing particles. Multiple proportions goes further, because it reports that the different ratios available to the same pair of elements are related to each other as small whole numbers. A continuous substance has no reason to prefer 2.000 over 1.837, whereas countable indivisible units can only combine one-to-one, two-to-one, three-to-two and so on, and those are exactly the ratios observed. Whole number ratios between compounds require discrete units; a fixed recipe alone does not

7. A student claims that water from a river and water from a glacier might have different compositions because their sources differ. Explain what the law of definite proportions does and does not say about this.
   Show the full solution

   The law applies to the compound water, and on that it is unambiguous: any sample of pure water is 11.2 percent hydrogen and 88.8 percent oxygen by mass, whatever its origin, because the ratio is fixed by the compound's formula and not by where it came from. What the student is actually describing is not water but a mixture. River water and glacier meltwater contain different dissolved minerals and gases in variable amounts, and mixtures have no fixed composition, which is precisely the distinction drawn in lesson 1.3. Separate the water from the dissolved solids by distillation and both samples give identical pure water. Pure water has a fixed composition; the samples differ because they are mixtures, not because the compound differs

8. Two oxides of sulfur contain 1.00 g of oxygen per gram of sulfur and 1.50 g of oxygen per gram of sulfur. Show that this fits the law and suggest formulas.
   Show the full solution

   The ratio is \( 1.50 \div 1.00 = 1.50 \), which is not a whole number as it stands, but the law requires a ratio of small whole numbers rather than an integer. Multiplying both sides by two gives 3 to 2, which qualifies. So for the same mass of sulfur the second compound contains three oxygen units wherever the first contains two. Formulas consistent with that are SO₂ and SO₃, and checking confirms it: sulfur dioxide has two oxygen atoms per sulfur and sulfur trioxide has three, a ratio of 3 to 2. The ratio 3 to 2 satisfies the law; SO₂ and SO₃

9. Explain why the existence of isotopes did not destroy Dalton's theory, given that it falsified one of his postulates.
   Show the full solution

   Because the falsified postulate was not the one doing the explanatory work. The laws of definite and multiple proportions follow from atoms being discrete countable units combining in fixed small ratios, and that requirement survives untouched if some atoms of an element happen to be heavier than others. What isotopes do change is the interpretation of atomic mass: the value on the periodic table becomes a weighted average of several isotope masses rather than the mass of a single kind of atom, which lesson 2.4 works through. Definite proportions still holds to the precision of ordinary measurement because isotope abundances are very nearly constant on Earth, so any sample contains the same mix. A theory can lose a postulate and keep its explanatory core. The core claim of discrete units in whole number ratios is untouched; only the interpretation of atomic mass changes

10. The worked example ended by saying the mass data alone cannot distinguish CO and CO₂ from C₂O₂ and C₂O₄. Explain why not, and what kind of evidence would settle it.
    Show the full solution

    Mass analysis reports ratios, and a ratio is unchanged by multiplying both terms by the same number. One carbon to one oxygen and two carbons to two oxygens give identical percentage compositions, so no amount of weighing can separate them. This is the same limitation that makes an empirical formula different from a molecular formula, which lesson 4.7 takes up. Settling it needs evidence about how many particles are present rather than how much they weigh, and that came from combining volumes of gases: when gases react, their volumes stand in small whole number ratios, and Avogadro's 1811 proposal that equal volumes contain equal numbers of particles converts those volume ratios directly into particle ratios. A ratio is preserved under multiplication, so mass alone cannot fix absolute numbers; combining gas volumes can

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Give the charge and relative mass of the proton, neutron and electron.
   Show the full solution

   Proton +1, mass 1; neutron 0, mass 1; electron -1, mass about 1/1836

2. Which particle did Thomson identify, and from what apparatus?
   Show the full solution

   The electron, from cathode rays deflected by electric and magnetic fields

3. What did Millikan's oil drop experiment measure?
   Show the full solution

   The charge on a single electron, showing charge comes in whole number multiples of a basic unit

4. State the observation in the gold foil experiment that required a nucleus.
   Show the full solution

   The small fraction of alpha particles deflected through more than ninety degrees

5. Why was the neutron the last of the three to be found?
   Show the full solution

   It has no charge, so it is not deflected by the electric and magnetic fields used to detect the others

6. Explain why the fact that most alpha particles passed straight through did not by itself refute the plum pudding model.
   Show the full solution

   Because plum pudding predicts exactly that. In that model the positive charge is spread thinly through the atom's whole volume and the electrons are far too light to obstruct a massive alpha particle, so the electric field encountered anywhere inside is weak and a fast particle should sail through with a small deflection at most. An observation that both competing models predict cannot distinguish between them. The decisive evidence has to be something one model forbids, and that is the large-angle scattering, which plum pudding cannot produce at any frequency. Plum pudding predicts easy passage too, so that observation discriminates nothing

7. Thomson found the same charge-to-mass ratio using cathodes made of different metals. Explain why that detail mattered more than the value itself.
   Show the full solution

   If the particles emitted by a zinc cathode had differed from those emitted by an aluminum one, the obvious reading would have been that each metal gives off fragments peculiar to itself, which says nothing general about atoms. Getting the identical ratio from every metal tested means the particle is not a property of zinc or aluminum but a component common to all of them. That is what makes it a constituent of matter in general, and it is what forces the conclusion that atoms have internal structure, since Dalton's atoms were indivisible and could not shed a universal fragment. The number was interesting; the universality was the argument. A universal particle implies internal structure common to all atoms, contradicting indivisibility

8. An alpha particle has about seven thousand times the mass of an electron. Explain why this means the electrons in the foil could not have caused the large deflections.
   Show the full solution

   In a collision the lighter object is the one that changes direction substantially. An alpha particle meeting an electron is like a bowling ball meeting a grain of sand: the electron is flung aside and the alpha particle continues almost undisturbed, its path bent by a negligible angle. To reverse the alpha particle's direction the target must be at least comparable to it in mass, so that it can absorb the momentum rather than being pushed out of the way, and it must also repel rather than attract it, which requires positive charge. Both requirements point to the same object, and together they mean nearly all the atom's mass as well as all its positive charge sits in that tiny volume. A target must be comparably massive to reverse the alpha particle; electrons are far too light

9. Explain why the rarity of large-angle deflection, rather than merely its existence, was the quantitative evidence for the nucleus being small.
   Show the full solution

   Existence establishes that a concentrated target is present somewhere; frequency establishes how big it is. Each alpha particle passing through the foil samples the cross-section it happens to cross, so the fraction turned back is a measure of how much of that cross-section the strongly repelling region occupies. A frequency of about one in eight thousand for large deflections means the effective target area is a minute fraction of the atom's area, and since area scales with the square of a radius, a ratio of that order in area corresponds to a radius ratio of roughly one in ten thousand to one in a hundred thousand. Had one particle in ten been turned back, the nucleus would have had to be a substantial fraction of the atom. The frequency measures the target's share of the cross-section, which converts to a size

10. The nuclear model created a problem it could not solve. State the problem and explain why it was serious rather than a detail.
    Show the full solution

    Electromagnetic theory, well established by 1911, requires that an accelerating charged particle radiates energy, and an electron moving in a curved orbit is accelerating. It should therefore lose energy continuously and spiral into the nucleus in a small fraction of a second. The problem is serious because it is not a matter of a missing detail but a flat contradiction with the most basic observation of all, namely that matter exists and is stable. A model that predicts every atom collapsing instantly cannot be accepted as it stands, so either electromagnetic theory or the model had to be modified. The resolution required electrons to occupy fixed allowed energies rather than arbitrary orbits, and the evidence for that came from line spectra, which lesson 2.5 takes up. An orbiting electron should radiate and spiral in, so the model predicts atoms cannot exist

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. What does the atomic number count, and what does it determine?
   Show the full solution

   Protons; it determines which element the atom is

2. Give the number of neutrons in uranium-235.
   Show the full solution

   Uranium is element 92, so \( 235 - 92 \). 143

3. How many protons, neutrons and electrons are in a neutral potassium-39 atom?
   Show the full solution

   Potassium is element 19. 19 protons, 20 neutrons, 19 electrons

4. Define isotopes.
   Show the full solution

   Atoms of the same element with the same proton count but different numbers of neutrons

5. How many electrons does Al³⁺ have?
   Show the full solution

   Aluminum is element 13; a 3+ charge means three electrons lost. 10

6. Explain why isotopes of an element have nearly identical chemical properties but measurably different physical ones.
   Show the full solution

   Chemical behavior is determined by electrons, specifically by how many are in the outer level and how strongly the nucleus holds them. Isotopes have the same proton count and therefore the same electron arrangement and the same nuclear charge, so they form the same bonds, the same ions and the same compounds. The neutrons contribute mass but no charge, so they are nearly invisible to chemistry. Physical properties that depend on mass do differ: heavier isotopes diffuse more slowly, have slightly higher boiling points and behave differently in a mass spectrometer, which is why isotope separation is done by physical methods such as diffusion or centrifuging rather than by any chemical reaction. Electrons govern chemistry and are unchanged; neutrons change mass, which governs the physical differences

7. A student says an atom that loses two electrons becomes a different element. Correct this and explain the underlying rule.
   Show the full solution

   Losing electrons makes an ion, not a new element. Elemental identity is fixed by the proton count and by nothing else, so a calcium atom that loses two electrons is still calcium, now written Ca²⁺, with its twenty protons intact. It happens to have the same number of electrons as an argon atom, and it is therefore isoelectronic with argon, but it is not argon and does not behave like it: argon is an unreactive gas while Ca²⁺ is an ion that sits in a lattice with anions. Changing the proton count requires a nuclear reaction, which is outside chemistry entirely. It becomes an ion of the same element; only a change in proton count changes the element

8. Explain why the mass number is always a whole number while the atomic mass on the periodic table usually is not.
   Show the full solution

   They are different quantities. The mass number counts particles, protons plus neutrons in one specific atom, and a count of objects cannot be fractional. The value printed on the periodic table is the weighted average mass of all the naturally occurring isotopes of that element, in atomic mass units, and an average over a mixture of different masses in unequal proportions will only be a whole number by coincidence. Chlorine's 35.45 is the clearest illustration: no single chlorine atom has that mass, because every one is either about 35 or about 37, and 35.45 reflects the roughly three-to-one abundance of the lighter isotope. Mass number counts particles in one atom; atomic mass is a weighted average over a mixture of isotopes

9. Carbon-14 is used for dating and carbon-12 is not. Explain what makes them different given that they have identical chemistry.
   Show the full solution

   The difference is nuclear stability rather than chemistry. Carbon-12 has six protons and six neutrons in a stable arrangement, while carbon-14 has six protons and eight neutrons and is unstable, decaying at a known and constant rate. Their identical chemistry is precisely what makes the method work: a living organism takes up carbon in whatever isotopic proportion the environment offers, making no distinction between the two, so the ratio inside it matches the atmosphere. When the organism dies it stops exchanging carbon, the carbon-14 continues to decay and is no longer replaced, and the falling ratio measures elapsed time. Chemistry provides the uniform starting point and the nucleus provides the clock. Carbon-14 is nuclearly unstable and decays at a known rate; identical chemistry is what gives a uniform starting ratio

10. An ion has 18 electrons, 16 protons and 16 neutrons. Identify it fully and explain how you know it is not argon.
    Show the full solution

    The proton count is sixteen, so the element is sulfur, and nothing else is relevant to that identification. Mass number is \( 16 + 16 = 32 \), so this is sulfur-32. With eighteen electrons and sixteen protons there are two more negative charges than positive, giving a charge of 2-, so the species is S^2-. It is not argon because argon is defined by having eighteen protons, and this species has sixteen. Having eighteen electrons makes it isoelectronic with argon, which means the electron arrangement is identical and explains why this is a stable ion to form, but the nuclear charge pulling on those eighteen electrons is smaller, so the ion is larger than an argon atom and behaves entirely differently. Sulfide, S^2-, from sulfur-32; the proton count identifies the element, not the electron count

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Write the formula for average atomic mass in words.
   Show the full solution

   The sum of each isotope's mass multiplied by its fractional abundance

2. Convert an abundance of 19.9 percent to a fractional abundance.
   Show the full solution

   0.199

3. How is the atomic mass unit defined?
   Show the full solution

   As one twelfth of the mass of a carbon-12 atom

4. Boron has isotopes of 10.013 u at 19.9 percent and 11.009 u at 80.1 percent. Find the average atomic mass.
   Show the full solution

   \( 10.013 \times 0.199 = 1.9926 \) and \( 11.009 \times 0.801 = 8.8182 \); the sum is 10.8108. 10.81 u

5. Copper has isotopes of 62.930 u at 69.17 percent and 64.928 u at 30.83 percent. Find the average atomic mass.
   Show the full solution

   \( 62.930 \times 0.6917 = 43.5287 \) and \( 64.928 \times 0.3083 = 20.0173 \); the sum is 63.5460. 63.55 u

6. A student calculates chlorine's average atomic mass as 36.0 by averaging 35 and 37. Explain the error and why the correct answer must be below 36.
   Show the full solution

   They took an unweighted mean, which would only be correct if the two isotopes were equally abundant. They are not: chlorine-35 makes up about three quarters of natural chlorine and chlorine-37 about one quarter, so the lighter isotope should count three times as heavily in the average. Any weighted average is pulled toward the more abundant value, so with the lighter isotope dominating the result must fall below the midpoint of 36, and it does, at 35.45. The sign of the error is predictable before any arithmetic is done, which is why the between-and-nearer check is worth applying every time. Unweighted averaging; the answer must lie below 36 because the lighter isotope is three times more abundant

7. An element has two isotopes and an average atomic mass almost exactly halfway between them. What does this tell you, and name an example.
   Show the full solution

   A weighted average sits at the midpoint only when the weights are equal, so this says the two isotopes are present in very nearly equal abundance, close to fifty percent each. Bromine is the standard example: bromine-79 and bromine-81 occur at roughly 50.7 and 49.3 percent, which is why bromine's average atomic mass of 79.90 sits almost exactly between 79 and 81. The reasoning runs in both directions, so an average close to one isotope's mass tells you that isotope dominates, and this is a useful qualitative read of the periodic table: chlorine at 35.45 signals a lopsided mixture, bromine at 79.90 signals a balanced one. The two isotopes are nearly equally abundant; bromine is the standard case

8. Explain why a chemist doing a molar mass calculation uses 35.45 for chlorine rather than 35 or 37.
   Show the full solution

   Because any real sample contains an enormous number of atoms in the natural isotopic proportions, and the mass of the sample is set by the average rather than by either extreme. A single mole of chlorine atoms is about \( 6.02 \times 10^{23} \) of them, roughly three quarters of which are the lighter isotope, so the mass of that mole is 35.45 g and not 35 or 37. Using 35 would underestimate the mass of every chlorine-containing sample by about one and a half percent, which propagates into every stoichiometry result. The individual mass numbers only matter when a single atom or a separated isotope is under discussion, which in this course means nuclear chemistry rather than reaction calculations. A weighable sample contains all isotopes in natural proportion, so its mass follows the weighted average

9. Silicon has three stable isotopes. Explain how the calculation changes and what check still applies.
   Show the full solution

   Nothing changes in principle, only in length: form one product of mass and fractional abundance for each isotope and add all three instead of two. The abundances must still sum to one, which for three isotopes is a more useful check than for two because an omitted or mistyped value is easier to hide. The between-and-nearer check still applies but in a weaker form: the average must lie between the lightest and heaviest isotope masses, and must be nearest the most abundant, though with three values you can no longer read off the abundance ratio from the position alone. Working backward also becomes harder, since one equation cannot determine two unknown abundances without a further constraint. Add one product per isotope; the abundances must still sum to one and the average must lie between the extremes

10. Two samples of chlorine are collected, one from seawater and one from a volcanic gas, and both give an average atomic mass of 35.45. Explain why this consistency is necessary for the law of definite proportions to hold.
    Show the full solution

    The law of definite proportions says a compound has the same composition by mass whatever its source. Composition by mass depends on the masses of the atoms involved, so if the isotopic mixture of chlorine varied appreciably from place to place, sodium chloride made from seawater chlorine and sodium chloride made from volcanic chlorine would have measurably different mass percentages of chlorine, and the law would fail at the precision of a good balance. It does not fail, because isotopic abundances on Earth are very nearly uniform, having been set when the material of the planet was mixed. This is also why the periodic table can print a single value at all: a genuinely variable abundance would make atomic mass a property of the sample rather than of the element. Uniform isotopic abundance is what makes composition by mass independent of source

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State the difference between a continuous and a line spectrum.
   Show the full solution

   A continuous spectrum contains all wavelengths; a line spectrum contains only certain specific wavelengths

2. When is light emitted by an atom?
   Show the full solution

   When an electron falls from a higher energy level to a lower one

3. Which carries more energy per photon, red or violet light?
   Show the full solution

   Violet, because it has the shorter wavelength

4. What does "quantized" mean?
   Show the full solution

   Restricted to certain specific values rather than able to take any value continuously

5. Why is a line spectrum useful for identifying an element?
   Show the full solution

   The pattern of lines is unique to each element and reproducible for every sample of it

6. Explain why the gaps between the lines are better evidence for quantization than the lines themselves.
   Show the full solution

   The lines establish that certain energies are emitted, which any model allowing those particular transitions could accommodate. The gaps establish something stronger, namely that the wavelengths in between are never emitted by any atom in the sample, no matter how many atoms are present or how much energy is supplied. A model permitting continuous electron energies cannot produce a gap, because for any wavelength you name there would be some pair of energies differing by exactly that amount, and with billions of atoms available some of them would make that transition. Absence across a whole range is only explicable if the energies themselves are restricted. Absent wavelengths cannot occur under continuous energies; the gaps are what a continuous model forbids

7. A sample of sodium vapor emits a strong yellow line at 589 nm. Explain what this tells you and why the same wavelength appears as a dark line in the sun's spectrum.
   Show the full solution

   The yellow line means sodium has two energy levels separated by exactly the energy carried by 589 nm light, and that this transition is a common one, since the line is intense. The same energy gap works in both directions, so a sodium atom in a lower level will readily absorb a photon of exactly 589 nm to climb to the higher level. In the sun, white light from the hot interior passes outward through cooler gas containing sodium atoms, which remove photons at precisely that wavelength and re-emit them in random directions. Along the line of sight to Earth there is a deficit at 589 nm, which appears as a dark absorption line. Emission and absorption lines therefore occur at identical wavelengths and both identify the element. Two levels differ by that energy; the same gap absorbs the same wavelength, producing a dark line

8. Explain why line spectra resolved the stability problem left by Rutherford's model.
   Show the full solution

   Rutherford's difficulty was that an orbiting electron should radiate continuously and spiral into the nucleus, and a continuous loss of energy would show up as a continuous spectrum as the electron passed smoothly through every energy on the way in. The spectra show the opposite: energy leaves the atom only in fixed amounts, at a few specific values, which is incompatible with a smooth spiral. Quantized levels solve the problem because there is a lowest allowed level with nothing below it, so an electron in the ground state has no lower state to fall to and cannot radiate at all. The atom is stable not because the electron stops moving but because there is nowhere further down to go. Quantized levels give a lowest allowed state, so a ground-state electron has nothing to fall to and cannot radiate

9. The lines in hydrogen's spectrum crowd closer together toward the violet end. Explain what this shows about the spacing of the energy levels.
   Show the full solution

   Each line corresponds to a drop from some higher level to a common lower one, and the lines toward the violet end come from drops that start further up. That those lines crowd together means the energies of successive transitions differ by less and less as the starting level rises, which can only happen if the levels themselves are packed more closely the higher you go. So the energy levels are not evenly spaced like the rungs of a ladder; the gaps shrink as the levels rise and converge toward a limit. Physically that limit is the energy at which the electron is no longer bound to the atom at all, and supplying more than that ionizes the atom, which connects directly to ionization energy in lesson 2.7. The levels converge as they rise, approaching the ionization limit

10. Fireworks produce colors by including salts of different metals. Explain the mechanism and why the color is characteristic of the metal rather than of the explosive.
    Show the full solution

    The explosive supplies heat, which excites electrons in the metal ions present, promoting them to higher energy levels. As those electrons fall back they emit light at the wavelengths fixed by that metal's energy level spacings, and it is those emitted wavelengths that the eye sees as color. The explosive determines how much energy is available and therefore how bright and how complete the excitation is, but it cannot alter the spacing of energy levels inside a strontium or barium ion, so it cannot change which wavelengths come out. Strontium salts give red and barium salts give green in any composition, which is exactly the reproducibility that makes a line spectrum an identification, and the same principle underlies the flame tests used to identify unknown ionic compounds. Heat excites the metal's electrons and the emitted wavelengths are fixed by that metal's level spacings, not by the heat source

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. How many electrons can an s, a p and a d sublevel hold?
   Show the full solution

   2, 6 and 10

2. How many electrons does a single orbital hold?
   Show the full solution

   Two

3. Write the full electron configuration of oxygen (Z = 8).
   Show the full solution

   1s² 2s² 2p⁴

4. Write the noble gas shorthand for potassium (Z = 19).
   Show the full solution

   Argon accounts for 18, and the nineteenth goes into 4s. [Ar]4s¹

5. How many valence electrons does phosphorus have?
   Show the full solution

   Group 15, configuration [Ne]3s² 3p³. Five

6. Write the configuration of iron (Z = 26) and explain the ordering decision you had to make.
   Show the full solution

   Argon accounts for the first 18 electrons. The decision is whether the next electrons enter 3d or 4s, and the answer is 4s, because at this point in the table the 4s sublevel lies slightly below 3d in energy and electrons fill from the lowest energy upward. Placing two in 4s brings the total to 20, and the remaining six go into 3d. Writing 3d before 4s in the sequence is common and is not wrong as a way of grouping by level, but the filling order is 4s first and that is what determines which electrons are removed first when iron forms ions. [Ar]4s² 3d⁶; 4s fills before 3d because it is lower in energy

7. Explain why the periodic table can be used to write a configuration without memorizing the filling order.
   Show the full solution

   Because the table was arranged by chemical behavior and chemical behavior is determined by electron arrangement, so the table turns out to be a direct map of the filling sequence. Groups 1 and 2 are where an s sublevel is being filled, groups 13 to 18 where a p sublevel is, and the ten transition columns where a d sublevel is, with the row number giving the level for s and p blocks and one less than the row number for the d block. Tracing from hydrogen along each row to the element of interest therefore writes the configuration out in order, and it automatically handles the 4s before 3d ordering because the fourth row reaches potassium and calcium before it reaches scandium. Each block corresponds to a sublevel being filled, so reading across the rows reproduces the filling order

8. Explain why elements in the same group have similar chemical properties, in terms of configuration.
   Show the full solution

   Chemical behavior is governed by the outermost electrons, because those are the ones that can be lost, gained or shared when atoms meet. Elements in the same group have the same number of electrons in the same kind of sublevel in their highest occupied level, so lithium is [He]2s¹ and sodium is [Ne]3s¹ and potassium is [Ar]4s¹, all with a single outer s electron. Faced with the same reaction they do the same thing, losing that one electron to form a 1+ ion. The inner electrons differ in number but are held in filled levels and do not participate. The similarity is therefore a consequence of the configuration, not a coincidence that the table records. Same number of valence electrons in the same sublevel type, so the same bonding behavior

9. A student writes bromine's valence electron count as seventeen. Diagnose the error and give the correct count.
   Show the full solution

   They have counted every electron outside the argon core, taking 4s² 3d¹⁰ 4p⁵ as all valence. Valence electrons are defined as those in the highest occupied energy level, which for bromine is level 4, containing 4s² and 4p⁵ for a total of seven. The ten 3d electrons sit in level 3 and are inner electrons even though they were filled later than the 4s pair, which is the genuinely confusing part: filling order and level number do not run in lockstep across the d block. The check is that seven matches bromine's position in group 17 and explains its formation of Br^- by gaining one electron. Seven; the 3d electrons are in level 3 and are not valence electrons

10. Explain why the noble gases are unreactive in terms of their configurations, and why helium belongs with them despite having only two valence electrons.
    Show the full solution

    Every noble gas from neon onward has a completely filled outer s and p sublevel, giving eight valence electrons, and a full outer level is a low energy arrangement with no strong tendency to gain, lose or share. There is no vacancy to fill and removing an electron from a filled level costs a great deal of energy, so these atoms have little reason to react. Helium has only two electrons and so cannot have eight, but its outer level is level 1, which contains only the 1s orbital and is therefore completely full at two. The criterion that matters is a full outer level rather than the number eight, and helium satisfies it, which is why it is grouped with the others and is the least reactive element of all. A filled outer level, not the number eight; level 1 is full at two electrons

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Define effective nuclear charge.
   Show the full solution

   The net attraction an outer electron feels, the nuclear charge reduced by shielding from inner electrons

2. State what happens to atomic radius across a period and down a group.
   Show the full solution

   It decreases across a period and increases down a group

3. Define ionization energy.
   Show the full solution

   The energy required to remove the outermost electron from a gaseous atom

4. Which element has the highest electronegativity?
   Show the full solution

   Fluorine

5. Is a sodium ion larger or smaller than a sodium atom?
   Show the full solution

   It has lost its only third-level electron, so the outer level is now the second. Much smaller

6. Explain why atomic radius decreases across a period even though electrons are being added.
   Show the full solution

   Because the electrons being added go into the same energy level as the ones already there, and electrons in the same level shield each other very poorly. Each step across the period adds one proton to the nucleus and one electron to the outer level, so the nuclear charge rises by one while the shielding barely changes, and the net pull on every outer electron increases. That stronger pull draws the whole outer level closer to the nucleus, so the atom contracts. The contrast with moving down a group is the useful one: there, the added electron goes into a new level further out and is shielded by all the filled levels beneath it, so the pull weakens and the atom expands. Added electrons enter the same level and shield poorly, so effective nuclear charge rises and the level contracts

7. Explain why ionization energy and atomic radius run in opposite directions.
   Show the full solution

   Both are consequences of the same underlying quantity, the strength of the pull on the outermost electron, and they simply report it in opposite senses. A large effective nuclear charge pulls the outer electron in close, which makes the atom small, and the same strong attraction is what must be overcome to detach that electron, which makes the ionization energy high. A weakly held electron sits further out, giving a large atom, and comes away easily, giving a low ionization energy. The inverse relationship is therefore not a separate fact to learn but the same fact stated twice, which is why predicting one trend from the other is legitimate. Both follow from the pull on the outer electron: strong pull means small atom and high ionization energy

8. Explain why a cation is smaller than its parent atom while an anion is larger.
   Show the full solution

   For a cation the usual case is that the atom has lost every electron in its outermost level, so the ion's outer level is the next one in, which is genuinely closer to the nucleus. On top of that, the unchanged number of protons is now pulling on fewer electrons, so each remaining one is held more tightly. Both effects shrink the ion, and the loss of a whole level makes the change large. For an anion no level is added, but electrons are, so the same nuclear charge is now shared among more electrons and the mutual repulsion between them increases. The outer level expands, and the anion is larger than the atom it came from. Neither change involves any alteration to the nucleus. Cations usually lose a whole level and hold the rest more tightly; anions gain repulsion without gaining protons

9. Fluorine is the most electronegative element, yet helium and neon are not assigned electronegativity values at all. Explain.
   Show the full solution

   Electronegativity is defined as an atom's tendency to attract a shared pair of electrons within a bond, so the quantity only has meaning for an atom that forms bonds. Helium and neon have completely filled outer levels, form no compounds under ordinary conditions, and therefore never have a shared pair to attract, which leaves nothing to measure. This is not a claim that they attract electrons weakly; by the related measure of ionization energy they hold their own electrons more tightly than fluorine does. It is a claim that the concept does not apply. Some tables do assign values to the heavier noble gases such as xenon, which does form compounds with fluorine and oxygen. The quantity is defined for bonded atoms, and these form no bonds, so there is nothing to measure

10. Na⁺, Mg²⁺, F^- and O^2- all have ten electrons. Arrange them by size and explain the ordering.
    Show the full solution

    All four are isoelectronic, so shielding and electron count are identical and the only variable left is nuclear charge. The more protons pulling on the same ten electrons, the tighter they are held and the smaller the ion. The proton counts are oxygen 8, fluorine 9, sodium 11 and magnesium 12, so size decreases in that order of increasing charge, giving \( \text{O}^{2-} \gt \text{F}^{-} \gt \text{Na}^{+} \gt \text{Mg}^{2+} \) from largest to smallest. The general rule for an isoelectronic series is therefore that size falls as atomic number rises, which is the reverse of what a student expecting anions to be large and cations small might guess from position alone. Largest to smallest: O^2-, F^-, Na⁺, Mg²⁺, ordered by increasing nuclear charge on the same ten electrons

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State the octet rule and its exception for hydrogen.
   Show the full solution

   Atoms tend toward eight outer electrons; hydrogen and helium are full at two

2. Name the three types of chemical bond.
   Show the full solution

   Ionic, covalent and metallic

3. Which bond type forms between a metal and a nonmetal?
   Show the full solution

   Ionic

4. An electronegativity difference of 0.96 predicts which bond type?
   Show the full solution

   It falls between 0.4 and 1.7. Polar covalent

5. Is energy released or absorbed when a bond forms?
   Show the full solution

   Released

6. Explain why a bond forms at all, without using the words "want" or "need".
   Show the full solution

   Two separate atoms have a certain total energy, and in some cases the arrangement in which their outer electrons are shared or transferred has a lower total energy than the separate atoms do. Systems are found in lower energy arrangements because the excess energy is released to the surroundings when the change happens and is not spontaneously reabsorbed. The full outer level is significant only because it is the configuration for which that energy drop is largest, since it places electrons where the attraction to the nuclei is strong and the repulsion between electrons is comparatively small. No intention is involved anywhere. The bonded arrangement is lower in energy, and the excess is released to the surroundings

7. Give two exceptions to the octet rule and explain why the rule is still worth teaching.
   Show the full solution

   Boron trifluoride leaves boron with only six electrons and is a stable compound; sulfur hexafluoride places twelve around sulfur, which is possible because sulfur is in period 3 and has empty d orbitals available; and nitrogen dioxide has an odd number of valence electrons, so no arrangement can pair them all. The rule remains worth teaching because it is correct for the great majority of compounds made from period 2 elements, which is most of organic chemistry and most of what a first course covers, and because it gives a procedure for drawing structures that otherwise could only be memorized. A model is judged on whether it predicts well enough for its purpose, and this one does, provided the exceptions are named rather than hidden. Boron trifluoride with six and sulfur hexafluoride with twelve; the rule still predicts most period 2 compounds correctly

8. Explain why the ionic and covalent categories are better described as a continuum than as two boxes.
   Show the full solution

   Because the physical quantity underlying the classification, the unevenness of electron sharing, varies smoothly and has no natural break in it. At zero electronegativity difference the pair is shared equally; as the difference grows the electron density shifts steadily toward the more electronegative atom, giving partial charges that get larger; and at a large difference the shift is so nearly complete that describing the result as separate ions is the more useful account. Nothing discontinuous happens at 1.7, which is a convention chosen for convenience. Real compounds sit at every point along the range, and some, such as aluminum chloride, genuinely behave as intermediate cases that neither label describes well. Electron sharing varies smoothly with electronegativity difference, and the cut-offs are conventions

9. Predict the bond type in potassium bromide and in nitrogen trihydride, using the values K 0.82, Br 2.96, N 3.04, H 2.20, and state a property that follows from each answer.
   Show the full solution

   Potassium and bromine differ by \( 2.96 - 0.82 = 2.14 \), above 1.7, so the bonding is ionic and the compound is a lattice of K⁺ and Br^-. A property that follows is a high melting point, because melting requires overcoming strong attractions throughout a three-dimensional lattice, and conduction when molten, because the ions become free to move. Nitrogen and hydrogen differ by \( 3.04 - 2.20 = 0.84 \), inside the middle band, so ammonia is polar covalent and exists as discrete molecules. A property that follows is a low boiling point compared with an ionic solid, because only the attractions between separate molecules need to be overcome, not bonds within them. KBr ionic with a high melting point; NH₃ polar covalent and molecular with a low boiling point

10. Metallic bonding does not fit the electronegativity rule, since two metals have a small difference yet do not form covalent molecules. Explain what makes metals different.
    Show the full solution

    The rule compares how strongly two atoms compete for a shared pair, and it assumes there is a pair to localize between them. Metals have few valence electrons and low ionization energies, so no atom holds its outer electrons tightly and there is nothing to compete over. Instead of one pair sitting between two nuclei, the valence electrons leave their parent atoms entirely and move freely through the whole lattice of positive ions, which is why the model is described as a sea of delocalized electrons. The low electronegativity difference correctly predicts that the sharing is even; what it cannot capture is that the sharing extends over the entire crystal rather than over one bond, and that delocalization is what produces electrical conductivity, malleability and metallic luster. Both atoms hold electrons weakly, so the valence electrons delocalize across the whole lattice instead of localizing in a pair

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. What charge does a group 2 metal form?
   Show the full solution

   2+

2. Write the formula of the compound formed by magnesium and chlorine.
   Show the full solution

   Mg²⁺ with Cl^-, so two chlorides per magnesium. MgCl₂

3. Write the formula of sodium carbonate, given the carbonate ion CO₃^2-.
   Show the full solution

   Two Na⁺ are needed for one carbonate. Na₂CO₃

4. Write the formula of iron(III) hydroxide.
   Show the full solution

   Fe³⁺ with OH^-, three hydroxides needed, and brackets are required. Fe(OH)₃

5. Why does a solid ionic compound not conduct electricity?
   Show the full solution

   Its ions are fixed in the lattice and cannot move, so no charge can flow

6. Explain why ionic compounds have much higher melting points than covalent molecular substances.
   Show the full solution

   The comparison is between two different things being overcome. Melting an ionic solid means breaking the electrostatic attraction between every ion and all its neighbors, and those attractions are strong, act in all directions and extend through the entire crystal, so a great deal of energy is needed and the melting point is high. Melting a covalent molecular solid does not break any covalent bonds at all: the molecules stay intact and only the comparatively weak attractions between separate molecules have to be overcome. The covalent bonds inside the molecules may well be stronger than the ionic attractions, which is why the comparison has to be made between the forces actually broken during melting rather than between bond strengths in general. Melting an ionic solid breaks strong lattice-wide attractions; melting a molecular solid only separates molecules

7. Explain why the formula MgO is written rather than Mg₂O₂, and what the formula is actually claiming.
   Show the full solution

   Because an ionic formula reports the simplest whole number ratio of ions in the lattice, not a count of particles in a molecule. There are no MgO molecules and no Mg₂O₂ molecules either; there is a lattice containing enormous and equal numbers of Mg²⁺ and O^2- ions. Since the ratio is one to one, MgO expresses it and Mg₂O₂ merely expresses the same ratio redundantly. The formula claims that the compound is electrically neutral with one magnesium ion for every oxide ion, and the smallest group that satisfies that description is called a formula unit rather than a molecule. An ionic formula gives the simplest ratio in the lattice; the unit is a formula unit, not a molecule

8. A student predicts that sodium chloride will bend when struck because the ionic attraction is strong. Explain why it shatters instead.
   Show the full solution

   Strength and flexibility are different properties, and the student has assumed a strong attraction implies a tough material. What matters is what happens when layers are forced to slide. In an ionic lattice the arrangement alternates in charge, so sliding one layer by a single ion width replaces every attraction across that plane with a repulsion between like charges. The crystal then pushes itself apart along the plane, and it does so suddenly, which is exactly what shattering looks like. The strength of the attraction makes the crystal hard, meaning difficult to scratch or compress, while the alternating arrangement makes it brittle. A metal lattice has no such alternation, so its layers slide without any repulsion appearing and the metal deforms instead. Sliding brings like charges opposite each other and the plane repels; hard and brittle are compatible

9. Predict whether calcium oxide or sodium chloride has the higher melting point, and justify the prediction.
   Show the full solution

   Calcium oxide. The strength of the attraction between two ions increases with the product of their charges and decreases as they get further apart. Sodium chloride is built from singly charged ions, Na⁺ and Cl^-, giving a charge product of one. Calcium oxide is built from Ca²⁺ and O^2-, a charge product of four, so each attraction in the lattice is substantially stronger. The ions are also somewhat smaller, especially the oxide ion compared with chloride, which brings the centers closer and strengthens the attraction further. Both factors point the same way, so calcium oxide should melt far higher, and it does, at about 2600 degrees Celsius against about 800 for sodium chloride. Calcium oxide, because doubly charged smaller ions attract far more strongly

10. Explain why an ionic compound dissolved in water conducts electricity, and what this shows about what dissolving does.
    Show the full solution

    Conduction requires mobile charge carriers, and a solution of an ionic compound conducts, so mobile ions must be present. That is direct evidence about the dissolving process: the water has not merely dispersed intact formula units, it has separated the lattice into individual hydrated Na⁺ and Cl^- ions free to move independently through the liquid. The test discriminates, because a solution of sugar in water does not conduct at all, showing that sugar dissolves as whole neutral molecules with no charge separation. Conductivity therefore distinguishes ionic solutes from molecular ones experimentally, which is the basis of the strong electrolyte idea used in lesson 5.6 for net ionic equations. Dissolving separates the lattice into free ions, which carry charge; sugar solutions do not conduct because no ions form

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. What is a covalent bond?
   Show the full solution

   A pair of electrons shared between two atoms and attracted to both nuclei

2. How many valence electrons are used to draw CH₄?
   Show the full solution

   Carbon 4 plus four hydrogens at 1 each. 8

3. How many valence electrons are used to draw NH₃?
   Show the full solution

   Nitrogen 5 plus three hydrogens at 1 each. 8

4. How many valence electrons are used to draw the nitrate ion, NO₃^-?
   Show the full solution

   \( 5 + 18 + 1 = 24 \), the extra one for the negative charge. 24

5. How many lone pairs are on the oxygen in a water molecule?
   Show the full solution

   Oxygen has six valence electrons, two used in bonds. Two

6. Explain why the electron count must be adjusted for charge, using NO₃^- as the example.
   Show the full solution

   Because the count is a count of the electrons actually present in the species, and an ion has a different number from the neutral atoms it was built from. Nitrogen contributes 5 and three oxygens contribute 18, giving 23 for the neutral collection, which is an odd number and cannot be arranged in pairs at all. The single negative charge means the ion has gained one electron from somewhere, bringing the total to 24, which pairs properly and gives every atom an octet. Skipping the adjustment produces a structure that either fails the octet rule or is drawn with an impossible unpaired electron, and since every later step uses the total, the error propagates through the whole drawing. An ion has gained or lost electrons, so add one per negative charge and subtract one per positive; nitrate needs 24, not 23

7. A student draws CO₂ with two single bonds and gives carbon two lone pairs. Find the error using the electron count.
   Show the full solution

   Count first: carbon contributes 4 and two oxygens contribute 12, giving 16 electrons to place. Their structure uses 4 in the two single bonds and 4 in carbon's two lone pairs, leaving 8 for the oxygens, which is only two lone pairs each, so each oxygen has 6 electrons rather than 8. Carbon also has only 8 counting its lone pairs, but has placed electrons on itself while leaving outer atoms short, which the procedure forbids: outer atoms are completed first. The correct structure has two double bonds, giving carbon 8 from four bonding pairs and each oxygen 8 from two bonding pairs and two lone pairs. The count checks: \( 8 \text{ bonding} + 8 \text{ lone pair} = 16 \). Two double bonds are required; their version leaves both oxygens with only six electrons

8. Explain what resonance means and what evidence shows it is an averaging rather than a flickering.
   Show the full solution

   Resonance is the situation where the Lewis procedure yields two or more equally valid structures differing only in the placement of a multiple bond. The obvious reading is that the molecule alternates rapidly between them, but bond length measurements rule that out. In carbonate, a genuine flicker would mean each bond is sometimes a short double bond and sometimes a longer single bond, and techniques that measure bond lengths would show a distribution of two values. What is found instead is three identical bonds of one intermediate length, at all times. The ion therefore has a single unchanging structure in which the extra bonding pair is delocalized over all three positions, and the multiple drawings are an artifact of a notation that can only place a pair between two specific atoms. All bonds are measured to be identical and intermediate in length, which a rapid alternation would not produce

9. Explain why a triple bond is both shorter and stronger than a single bond between the same two elements.
   Show the full solution

   A covalent bond exists because shared electrons sit between the two nuclei and attract both, so the amount of shared electron density in that region determines how strongly the nuclei are pulled together. A triple bond shares three pairs there rather than one, so the attraction is roughly three times greater, and the nuclei are drawn closer until the repulsion between the two positive nuclei balances it, giving a shorter bond. Breaking that bond means separating atoms held by three times the shared density, so it requires considerably more energy. The pattern is consistent across nitrogen, where the triple bond of N₂ is both very short and very strong, which is exactly why atmospheric nitrogen is chemically unreactive. More shared pairs between the nuclei means stronger attraction, so the bond is both shorter and harder to break

10. Explain why the final electron count is a worthwhile check even when the structure looks right.
    Show the full solution

    Because the commonest Lewis structure errors produce drawings that look entirely plausible. Adding a lone pair too many, forgetting to subtract the electrons used in a bond, or drawing a double bond without removing the lone pair it came from all yield pictures in which every atom appears to have an octet, and visual inspection cannot distinguish them from a correct structure. The total is an independent constraint that none of those errors can satisfy: if the finished drawing contains twenty-six electrons when twenty-four were available, two were invented somewhere, and the check says so without needing to find where. It takes a few seconds and it is the only step that verifies the structure against the starting data rather than against expectation. Errors such as an invented lone pair still look like valid octets; only the total detects them

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. What counts as one electron domain?
   Show the full solution

   Any single bond, double bond, triple bond or lone pair

2. Give the shape and bond angle for four domains with no lone pairs.
   Show the full solution

   Tetrahedral, 109.5 degrees

3. Give the shape of a molecule with four domains and one lone pair.
   Show the full solution

   Trigonal pyramidal

4. State the bond angle in water and in ammonia.
   Show the full solution

   About 104.5 degrees and about 107 degrees

5. Is CCl₄ polar or nonpolar?
   Show the full solution

   Tetrahedral and symmetrical, so the four polar bonds cancel. Nonpolar

6. Explain the sequence 109.5, 107, 104.5 degrees for methane, ammonia and water.
   Show the full solution

   All three have four electron domains around the central atom, so all three start from a tetrahedral arrangement with an ideal angle of 109.5 degrees. The difference is how many of those domains are lone pairs. A lone pair is held by only one nucleus rather than shared between two, so its electron density is closer to the central atom and spread more widely, and it repels neighboring domains more strongly than a bonding pair does. Methane has no lone pairs and keeps the ideal angle. Ammonia has one, which pushes the three bonding pairs slightly closer together, reducing the angle to about 107. Water has two, which push twice as hard and bring the angle down to about 104.5. The steady decrease is a direct count of lone pairs. Each lone pair repels more strongly than a bonding pair and compresses the remaining bond angles

7. Explain why CO₂ is nonpolar but SO₂ is polar, given that both contain a central atom double bonded to two oxygens.
   Show the full solution

   The difference is the lone pair on sulfur, which changes the shape. Carbon in carbon dioxide has two domains and no lone pairs, so the molecule is linear and the two identical bond polarities point in exactly opposite directions and cancel completely. Sulfur in sulfur dioxide has three domains, two bonds and one lone pair, so the molecule is bent at about 120 degrees. The two bond polarities now point in directions that are not opposite, so they add rather than cancel and leave a net polarity toward the oxygen side. The bonds are similarly polar in both molecules; only the geometry differs, which is exactly the point that shape determines molecular polarity. Sulfur's lone pair makes SO₂ bent, so its bond polarities do not cancel; CO₂ is linear and they do

8. CH₄ and CH₃Cl are both tetrahedral. Explain why one is nonpolar and the other polar.
   Show the full solution

   Cancellation requires the pulls to be equal as well as symmetrically arranged. In methane the four bonds are identical C-H bonds, each barely polar and each pulling with the same small magnitude toward carbon, and the tetrahedral arrangement places them so that the four equal pulls sum to zero. Replacing one hydrogen with chlorine leaves the geometry tetrahedral but destroys the equality: the C-Cl bond is considerably more polar than a C-H bond, with a difference of 0.61 against 0.35, and it pulls electron density strongly toward the chlorine. That larger pull is no longer balanced by the three smaller ones opposite it, so a net polarity remains along the C-Cl direction. Symmetry of shape is necessary for cancellation but not sufficient; the bonds must also be alike. Tetrahedral symmetry cancels only identical bonds; the more polar C-Cl bond is unbalanced

9. Explain why the shape name describes only the atom positions and not the lone pairs, using water as the example.
   Show the full solution

   Because shape names were established to describe what can be observed, and experimental methods locate nuclei rather than lone pairs. Techniques such as X-ray diffraction determine where the atoms sit and therefore give bond lengths and bond angles directly; a lone pair has no nucleus and does not appear. Water has four electron domains arranged tetrahedrally, but only three of the positions are occupied by atoms, and the three atoms lie in a line that is bent, so the molecule is called bent rather than tetrahedral. The electron domain arrangement is still doing the explanatory work, since it is what fixes the angle at close to the tetrahedral value, and that is why the domain count is determined first and the shape name assigned afterward. Shape names report observable atom positions; the tetrahedral domain arrangement explains the angle but is not the name

10. Water's bent shape is a consequence of two lone pairs on oxygen. Explain how different the world would be if water were linear.
    Show the full solution

    A linear water molecule would have its two O-H bond polarities pointing in exactly opposite directions, so they would cancel and the molecule would be nonpolar, like carbon dioxide. Almost everything water does depends on it being polar. It would no longer dissolve ionic compounds, because there would be no partial charges to surround and separate the ions, so seawater and the ionic chemistry of cells could not exist. Hydrogen bonding between molecules would be absent, so water's boiling point would fall to something like that of hydrogen sulfide, around minus sixty degrees Celsius, and there would be no liquid water on Earth's surface at all. Its unusually high specific heat capacity and its expansion on freezing both follow from the same hydrogen bonding, so the climate moderation described in lesson 7.2 and the floating ice of lesson 1.7 would also disappear. Two lone pairs on one atom account for a remarkable amount. It would be nonpolar, dissolve no ionic compounds, form no hydrogen bonds, and be a gas at ordinary temperatures

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Name the three kinds of intermolecular force in order of strength.
   Show the full solution

   Hydrogen bonding, dipole-dipole, London dispersion

2. Which force acts between all molecules?
   Show the full solution

   London dispersion

3. State the condition for hydrogen bonding.
   Show the full solution

   Hydrogen must be bonded directly to nitrogen, oxygen or fluorine

4. What is broken when a molecular substance boils?
   Show the full solution

   Intermolecular forces between molecules, not the covalent bonds within them

5. Why does dispersion strengthen down the halogen group?
   Show the full solution

   More electrons in a larger, more easily distorted electron cloud

6. Explain why ammonia boils at -33 degrees Celsius while phosphine, PH₃, boils at -88 despite being heavier.
   Show the full solution

   On molar mass alone phosphine should boil higher, since it has more electrons and therefore stronger dispersion forces. The prediction fails because ammonia has hydrogen bonded directly to nitrogen, one of the three elements small enough and electronegative enough to support hydrogen bonding, while phosphorus is neither sufficiently electronegative nor sufficiently small. Ammonia molecules therefore attract one another through hydrogen bonds that phosphine entirely lacks, and that additional attraction outweighs phosphine's advantage in dispersion by about fifty degrees. It is the same pattern as water against hydrogen sulfide, one group to the left, which is why both anomalies appear in the same corner of the periodic table. Ammonia hydrogen bonds and phosphine cannot, which outweighs phosphine's greater dispersion

7. A student says steam is a mixture of hydrogen and oxygen gas. Correct this and explain what boiling actually does.
   Show the full solution

   Steam is water in the gas phase: the molecules are still H₂O, with their covalent O-H bonds entirely intact. Boiling is a physical change in the sense of lesson 1.2, and the evidence is that cooling steam gives back water rather than anything else, which would be impossible if the molecules had been taken apart. What boiling overcomes is the hydrogen bonding between separate molecules, allowing them to move apart into the gas phase. Splitting water into hydrogen and oxygen is a chemical change requiring far more energy, and it is achieved by electrolysis or by temperatures of thousands of degrees, not by a kettle. Steam is intact water molecules; boiling separates molecules without breaking covalent bonds

8. Explain why dispersion forces should not be described as negligible, using the halogen data.
   Show the full solution

   Because the per-interaction weakness of dispersion is offset by how strongly it scales with molecular size, and the halogens isolate the effect cleanly since they are all nonpolar and have no other force acting. Across chlorine, bromine and iodine the only variable is the number of electrons, and the boiling point rises from -34 to 59 to 184 degrees Celsius, a range of more than two hundred degrees produced by dispersion alone. Iodine is a solid at room temperature entirely because of it. The accurate statement is that dispersion is the weakest force between two small molecules and can become the dominant force between large ones, which is why very large nonpolar molecules such as waxes and polymers are solids. It scales with electron count, and alone it spans more than two hundred degrees across the halogens

9. Predict which of propane, C₃H₈, and ethanol, C₂H₅OH, has the higher boiling point. Both have a molar mass near 45 g/mol.
   Show the full solution

   Ethanol, by a wide margin. With the molar masses nearly equal, dispersion forces are comparable and cannot be the deciding factor, so the difference has to come from the other categories. Propane is a hydrocarbon: its C-H bonds are barely polar and the molecule is effectively nonpolar, so dispersion is all it has. Ethanol contains an O-H group, which means hydrogen bonded directly to oxygen and therefore hydrogen bonding between molecules, together with the dipole-dipole attraction from its polarity. Separating ethanol molecules requires overcoming those hydrogen bonds as well, so its boiling point is far higher. The measured values are about -42 degrees Celsius for propane and 78 for ethanol, a gap of a hundred and twenty degrees from one functional group. Ethanol, because its O-H group allows hydrogen bonding while propane has only dispersion

10. Explain why intermolecular forces, rather than bond strength, determine whether a substance is a gas, liquid or solid at room temperature.
    Show the full solution

    Because the state of a molecular substance is decided by whether thermal energy at that temperature is enough to pull whole molecules away from each other, and that contest involves only the attractions between molecules. The covalent bonds inside each molecule are not being tested at all, since they remain intact through melting and boiling alike. Nitrogen illustrates the point sharply: the triple bond in N₂ is among the strongest covalent bonds known, yet nitrogen is a gas down to -196 degrees Celsius, because the nonpolar molecules attract each other only through weak dispersion. Water has much weaker covalent bonds and is a liquid, because hydrogen bonding between its molecules is comparatively strong. Bond strength governs chemical reactivity; intermolecular forces govern physical state. State depends on separating whole molecules, which tests only the forces between them; nitrogen's strong triple bond still leaves it a gas

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Name NaBr.
   Show the full solution

   Sodium bromide

2. Give the formula and charge of the sulfate ion.
   Show the full solution

   SO₄^2-

3. Name CaCO₃.
   Show the full solution

   Calcium carbonate

4. Write the formula of potassium nitrate.
   Show the full solution

   K⁺ with NO₃^-, one each. KNO₃

5. Which metals in a first course require no roman numeral?
   Show the full solution

   Group 1, group 2, aluminum, zinc and silver

6. Name CuCl₂ and explain how you determined the numeral.
   Show the full solution

   Start from the anion, whose charge is known and fixed: chloride is Cl^- and there are two of them, giving a total of -2. The compound must be electrically neutral, so the single copper ion has to supply +2, making it Cu²⁺ and the name copper(II) chloride. The numeral is therefore derived from the anion charges and the requirement of neutrality, never read off a subscript. Copper is one of the metals that needs a numeral precisely because it can also form Cu⁺, and copper(I) chloride is the quite different compound CuCl. Copper(II) chloride; two chlorides give -2, so the one copper must be 2+

7. Explain why calcium chloride is never called calcium dichloride.
   Show the full solution

   Because the prefix would carry no information. Calcium is in group 2 and forms only Ca²⁺, and chloride is always Cl^-, so the requirement of neutrality fixes the ratio at one calcium to two chlorides and no other formula is possible. The name calcium chloride already specifies the compound completely, and adding di- would be redundant. Prefixes are used in covalent naming precisely because there the ratio is not fixed by charge: nitrogen and oxygen form NO, N₂O, NO₂ and N₂O₄ among others, so without prefixes the name could not distinguish them. The presence or absence of prefixes is therefore a signal about which kind of compound is being named. The charges fix the ratio uniquely, so a prefix would add nothing; prefixes exist for covalent compounds where the ratio varies

8. Write the formulas of iron(II) sulfate and iron(III) sulfate, and comment on why the numeral is essential here.
   Show the full solution

   Iron(II) is Fe²⁺ and sulfate is SO₄^2-, so they balance one to one: FeSO₄. Iron(III) is Fe³⁺ against the same 2- sulfate, so the lowest common multiple is 6 and the formula is two irons to three sulfates: Fe₂(SO₄)₃, with brackets because more than one polyatomic ion is present. The numeral is essential because without it the name iron sulfate would be ambiguous between two genuinely different compounds with different formulas, different molar masses, different colors and different chemistry. This is exactly the situation roman numerals exist to resolve, and it does not arise for sodium or magnesium, which have only one available charge. FeSO₄ and Fe₂(SO₄)₃; without the numeral the name would not specify which compound

9. A student writes the formula of sodium nitrate as NaNO₃₂. Diagnose the error and explain the bracket rule.
   Show the full solution

   They have attached the subscript from the polyatomic ion to the wrong place, or have written a subscript of thirty-two, and in either reading the formula is meaningless. Sodium nitrate needs only one nitrate ion, since Na⁺ and NO₃^- balance one to one, so the correct formula is simply NaNO₃ with no brackets needed at all. The bracket rule applies when more than one polyatomic ion is required: the whole ion goes inside brackets and the number of them is written outside, as in Ca(NO₃)₂. Without the brackets that formula would read CaNO₃₂ and lose the information that nitrate travels as an intact unit of one nitrogen and three oxygens. NaNO₃; brackets are needed only when more than one polyatomic ion is required, and the count goes outside them

10. Given that sulfate is SO₄^2-, predict the formula and charge of sulfite, and explain the reasoning pattern.
    Show the full solution

    Sulfite is SO₃^2-. The -ate and -ite endings distinguish two oxyanions of the same central element, and the rule is that the -ite form has one fewer oxygen atom than the -ate form while carrying the same charge. So dropping one oxygen from sulfate gives sulfite with the charge unchanged at 2-. The same pattern holds across the set: nitrate NO₃^- gives nitrite NO₂^-, and phosphate PO₄^3- gives phosphite PO₃^3-. Knowing the -ate form of each family therefore yields the -ite form without separate memorization, which halves the list that has to be learned. SO₃^2-; the -ite form has one fewer oxygen and the same charge as the -ate form

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Name CO.
   Show the full solution

   Carbon monoxide

2. Name CCl₄.
   Show the full solution

   Carbon tetrachloride

3. Write the formula of diphosphorus pentoxide.
   Show the full solution

   P₂O₅

4. Name HNO₃.
   Show the full solution

   Nitrate gives an -ic acid. Nitric acid

5. Name HCl in aqueous solution.
   Show the full solution

   Binary acid, so hydro- and -ic. Hydrochloric acid

6. Explain why covalent compounds need prefixes when ionic compounds do not.
   Show the full solution

   In an ionic compound the ratio of ions is determined entirely by their charges and the requirement that the compound be neutral, so for any given pair of ions only one formula is possible and the name alone specifies it. Covalent compounds have no such constraint, because atoms share electrons in whatever combination is energetically available rather than balancing fixed charges, so the same two elements commonly form several different compounds. Nitrogen and oxygen give NO, N₂O, NO₂ and N₂O₄, all real substances with quite different properties, and a name of nitrogen oxide would be useless. The prefixes supply the information that charge balance supplies on the ionic side. Ionic ratios are fixed by charge, so a name determines the formula; covalent ratios vary, so the count must be stated

7. Name H₂SO₃ and explain how its name relates to sulfuric acid.
   Show the full solution

   Remove the two hydrogens and the remaining ion is SO₃^2-, which is sulfite rather than sulfate, since sulfite has one fewer oxygen than sulfate at the same charge. An -ite ion produces an -ous acid, so the compound is sulfurous acid. Its relationship to sulfuric acid, H₂SO₄, is exactly the relationship between sulfite and sulfate: one oxygen fewer, same charge, and the suffix change from -ic to -ous records it. The same pairing runs through nitric and nitrous acids from nitrate and nitrite, so learning one family gives the pattern for all of them. Sulfurous acid; the -ous ending corresponds to the -ite ion, which has one fewer oxygen than the -ate

8. A student names MgCl₂ as magnesium dichloride and N₂O as nitrogen(I) oxide. Diagnose both errors.
   Show the full solution

   Both errors are the same error, made in opposite directions: the wrong naming system was selected. Magnesium is a metal, so magnesium chloride is ionic and takes no prefixes, because the two-to-one ratio is already forced by Mg²⁺ against Cl^- and di- adds nothing; the correct name is magnesium chloride. Nitrogen and oxygen are both nonmetals, so N₂O is covalent and takes prefixes rather than a roman numeral, which is reserved for variable-charge metals in ionic compounds; the correct name is dinitrogen monoxide. Checking whether the first element is a metal before writing anything would have prevented both. Magnesium chloride and dinitrogen monoxide; each used the other compound's system

9. Explain why HCl is called hydrogen chloride in one situation and hydrochloric acid in another.
   Show the full solution

   Because the two names describe the same formula in two different physical states, and the chemistry genuinely differs between them. As a pure substance HCl is a covalent molecular gas, named by the covalent system as hydrogen chloride, and in that state it does not behave as an acid at all: dry hydrogen chloride will not turn dry litmus red. Dissolved in water the molecules ionize, releasing H⁺ into solution and producing the properties that define an acid, so the aqueous solution takes the acid name hydrochloric acid. The distinction is not a bookkeeping quirk but a record of the fact that acidic behavior requires water, which lesson 11.1 develops through the Arrhenius definition. The pure gas is a covalent compound; only the aqueous solution ionizes and behaves as an acid

10. Given only the formula, describe the procedure you would use to decide which naming system applies, and identify the one case where the answer is ambiguous.
    Show the full solution

    Look at the first element in the formula. If it is a metal, the compound is ionic: name the cation, then the anion with -ide or its polyatomic name, adding a roman numeral only if the metal has more than one possible charge. If both elements are nonmetals, the compound is covalent and takes prefixes on both elements, with mono- dropped from the first. If the formula begins with hydrogen and the compound is described as aqueous, it is an acid: use hydro-(root)-ic if there is no oxygen, and otherwise name from the polyatomic ion with -ate going to -ic and -ite to -ous. The ambiguous case is a hydrogen compound with no state specified, since hydrogen is itself a nonmetal and H₂S is hydrogen sulfide as a gas and hydrosulfuric acid in water. The formula alone does not settle it and the question has to say which is meant. Test the first element for metal, nonmetal or hydrogen; hydrogen compounds with no stated state are the ambiguous case

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State the value of Avogadro's number.
   Show the full solution

   \( 6.022 \times 10^{23} \) particles per mole

2. How many atoms are in 2.00 mol of iron?
   Show the full solution

   \( 2.00 \times 6.022 \times 10^{23} \). \( 1.20 \times 10^{24} \) atoms

3. What is the mass of one mole of carbon atoms?
   Show the full solution

   12.01 g

4. Which contains more particles, one mole of lead or one mole of helium?
   Show the full solution

   Neither; both contain \( 6.022 \times 10^{23} \) particles

5. How many moles is \( 3.011 \times 10^{23} \) molecules?
   Show the full solution

   That is half of Avogadro's number. 0.500 mol

6. Explain why the mole exists at all, rather than chemists simply using grams.
   Show the full solution

   Because reactions are governed by particle counts and not by masses. Two hydrogen molecules react with one oxygen molecule to make two water molecules, a fixed relationship between numbers of particles, and the corresponding mass relationship is nothing so simple, since 4 g of hydrogen reacts with 32 g of oxygen. A chemist who mixes equal masses of two reactants has almost certainly mixed them in the wrong ratio. Since particles cannot be counted directly, the mole provides a fixed package size that converts a mass on a balance into a particle count, so that the whole number ratio from the equation can actually be used in the laboratory. Reactions occur in fixed particle ratios, not mass ratios, and the mole converts a measurable mass into a count

7. Explain why "one mole of oxygen" is an ambiguous statement, and what it should say.
   Show the full solution

   Because oxygen exists as individual atoms in compounds and as diatomic O₂ molecules as a free element, and a mole of each is a different quantity of matter. One mole of oxygen atoms is \( 6.022 \times 10^{23} \) atoms with a mass of 16.00 g, while one mole of oxygen molecules is the same number of molecules, therefore twice as many atoms, with a mass of 32.00 g. Any calculation that uses the wrong one is out by a factor of two. The statement should specify the particle: one mole of O atoms, or one mole of O₂ molecules. The same ambiguity affects every diatomic element, which in a first course means hydrogen, nitrogen, oxygen and the halogens. A mole of O atoms is 16.00 g and a mole of O₂ molecules is 32.00 g; the particle must be named

8. Explain why Avogadro's number has the particular value it has rather than being a round number.
   Show the full solution

   Because the number was not chosen; the convenience was chosen and the number followed. Chemists wanted a package whose mass in grams would be numerically equal to the atomic mass in atomic mass units, so that the periodic table could be read directly as a table of molar masses. Since the atomic mass unit was already defined as one twelfth of the mass of a carbon-12 atom, the required package is whatever count of carbon-12 atoms weighs exactly 12 g, and measurement shows that count to be \( 6.022 \times 10^{23} \). Had a round number such as \( 10^{23} \) been chosen instead, the convenient correspondence with atomic masses would have been lost and every molar mass would carry an awkward conversion factor. It is whatever count makes 12 g of carbon-12, which is what preserves the link to atomic masses

9. A student says a mole of lead is "more" than a mole of carbon because lead is heavier. Diagnose the error precisely.
   Show the full solution

   The word "more" is being applied to two different quantities without distinguishing them. By particle count the two are identical: both contain \( 6.022 \times 10^{23} \) atoms, because that is what a mole means. By mass they differ substantially, since a mole of lead is about 207 g and a mole of carbon about 12 g, reflecting that a lead atom is roughly seventeen times as massive as a carbon atom. The correct statements are that a mole of lead has a greater mass than a mole of carbon, and that both contain the same number of atoms. The underlying confusion is treating the mole as a unit of amount of stuff rather than as a count. Same number of atoms, greater mass; the mole counts particles and says nothing about mass

10. A mole of water occupies about 18 mL and a mole of carbon dioxide gas occupies about 22 400 mL at STP. Explain why the same number of molecules takes up such different volumes.
    Show the full solution

    Because in a liquid the molecules are in contact and in a gas they are overwhelmingly separated by empty space, so the volume occupied is set almost entirely by the spacing rather than by the size of the molecules themselves. In liquid water the hydrogen bonding of lesson 3.5 holds molecules touching one another, so the volume is roughly the combined volume of the molecules. In a gas at ordinary pressure the molecules are perhaps ten molecular diameters apart on average, which is about a thousand times the volume per molecule, and that ratio matches the observed factor of roughly 1200 between the two figures. It also explains why the molar volume of a gas is nearly the same for every gas, as lesson 4.4 shows, while molar volumes of liquids differ substantially. Gas volume is set by the spacing between molecules, not their size, so the same count occupies about a thousand times more space

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Calculate the molar mass of H₂O.
   Show the full solution

   \( 2 \times 1.008 + 16.00 \). 18.02 g/mol

2. Calculate the molar mass of CO₂.
   Show the full solution

   \( 12.01 + 2 \times 16.00 \). 44.01 g/mol

3. Calculate the molar mass of NaCl.
   Show the full solution

   \( 22.99 + 35.45 \). 58.44 g/mol

4. Calculate the molar mass of Mg(OH)₂.
   Show the full solution

   1 Mg, 2 O, 2 H: \( 24.31 + 32.00 + 2.016 \). 58.33 g/mol

5. Calculate the molar mass of glucose, C₆H₁₂O₆.
   Show the full solution

   \( 6(12.01) + 12(1.008) + 6(16.00) = 72.06 + 12.10 + 96.00 \). 180.16 g/mol

6. Calculate the molar mass of calcium phosphate, Ca₃(PO₄)₂, showing the atom count. Use P 30.97.
   Show the full solution

   Count first: the bracket holds PO₄ with a subscript 2, so there are 3 calcium, 2 phosphorus and \( 4 \times 2 = 8 \) oxygen. Then \( 3 \times 40.08 = 120.24 \), \( 2 \times 30.97 = 61.94 \) and \( 8 \times 16.00 = 128.00 \). Adding gives 310.18 g/mol. The error to avoid is taking four oxygens instead of eight, which would give 246.18 and is off by more than twenty percent with nothing on the page to reveal it. 310.18 g/mol

7. Explain why the bracket subscript error is harder to catch than most arithmetic mistakes.
   Show the full solution

   Because it produces an answer that is entirely plausible in isolation. A mistyped digit often gives an absurd result that the eye rejects, and a dropped decimal place shifts the magnitude by a factor of ten, which a rough estimate catches. Missing a bracket subscript changes the answer by a factor somewhere between one and a half and three, which is neither absurd nor obviously wrong, and the resulting number sits comfortably in the range a molar mass ought to occupy. Nothing downstream flags it either, since the wrong molar mass simply produces a wrong number of moles that then propagates quietly through the rest of the calculation. The only reliable defense is procedural: write out the atom count before any multiplication. It gives a plausible number in the right range, so neither estimation nor inspection detects it

8. A student calculates the molar mass of Fe₂O₃ as 71.85 g/mol. Diagnose the error without being told what they did.
   Show the full solution

   Apply the sanity check first: iron alone has an atomic mass of 55.85, so a compound containing two iron atoms must have a molar mass above 111.70 whatever else is in it. An answer of 71.85 is therefore impossible and can be rejected before the source of the error is found. Locating it, 71.85 is exactly \( 55.85 + 16.00 \), so the student used one iron and one oxygen and ignored both subscripts. The correct calculation is \( 2 \times 55.85 + 3 \times 16.00 = 111.70 + 48.00 = 159.70 \) g/mol. The check that the total exceeds the mass of the heaviest atom present, multiplied by its subscript, is worth applying to every molar mass. 159.70 g/mol; they ignored both subscripts, and the answer is below the mass of the iron alone

9. Explain why rounding chlorine's atomic mass to 35 is a worse decision than it looks.
   Show the full solution

   The true value is 35.45, so rounding to 35 introduces an error of 0.45 in 35.45, which is about 1.3 percent. That sounds small, but molar mass sits in the denominator of every conversion from mass to moles, so the error transfers directly into the mole count and then into every quantity derived from it, including the mass of product predicted by a stoichiometry calculation. In a compound with several chlorines the error multiplies: in CCl₄ the four chlorines contribute 141.8 correctly against 140 rounded, and the whole molar mass shifts by more than one percent. Since experimental percent yields are often reported to three significant figures, an avoidable systematic error of this size can be larger than the effect being measured. A 1.3 percent error enters the denominator of every mole calculation and multiplies with the number of chlorine atoms

10. Explain why a hydrate's molar mass includes the water, and what goes wrong if a student weighs a hydrate and uses the anhydrous molar mass.
    Show the full solution

    The water of crystallization is physically present in the solid being weighed. It occupies fixed positions in the crystal lattice in a definite ratio to the rest of the formula unit, so it contributes to the mass on the balance just as every other atom does, and excluding it would describe a different substance. If a student weighs 25.0 g of CuSO₄·5H₂O and divides by the anhydrous molar mass of 159.62 rather than the correct 249.70, they obtain 0.157 mol instead of 0.100 mol, overstating the amount by more than fifty percent. Every quantity derived from that figure is then too large by the same factor, and since blue copper sulfate crystals are the hydrate while the white powder is anhydrous, the decision has to be made by looking at what is actually on the balance. The water is part of the solid being weighed; using 159.62 instead of 249.70 overstates the moles by more than half

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. What do you divide by to convert mass to moles?
   Show the full solution

   The molar mass

2. Find the mass of 3.00 mol of CO₂.
   Show the full solution

   \( 3.00 \times 44.01 \). 132 g

3. How many moles is \( 1.204 \times 10^{24} \) molecules?
   Show the full solution

   \( 1.204 \times 10^{24} \div 6.022 \times 10^{23} \). 2.00 mol

4. Find the number of moles in 58.44 g of NaCl.
   Show the full solution

   That is exactly one molar mass. 1.00 mol

5. Why can mass never be converted to particles in a single step?
   Show the full solution

   Because no conversion factor links them directly; both are connected only through moles

6. Find the number of molecules in 8.00 g of oxygen gas, showing the units canceling.
   Show the full solution

   Oxygen gas is O₂, with a molar mass of 32.00 g/mol, and using 16.00 here is the standard error. Setting up the chain: \( 8.00 \text{ g} \times \frac{1 \text{ mol}}{32.00 \text{ g}} \times \frac{6.022 \times 10^{23}}{1 \text{ mol}} \). Grams cancel against grams and moles against moles, leaving molecules. The arithmetic is \( 8.00 \div 32.00 = 0.250 \) mol, then \( 0.250 \times 6.022 \times 10^{23} = 1.5055 \times 10^{23} \). \( 1.51 \times 10^{23} \) molecules

7. Find the mass of \( 3.011 \times 10^{23} \) molecules of ammonia, NH₃.
   Show the full solution

   Travel the map from particles to moles to mass. First \( 3.011 \times 10^{23} \div 6.022 \times 10^{23} = 0.5000 \) mol, which is a recognizable half of Avogadro's number and worth spotting. The molar mass of ammonia is \( 14.01 + 3 \times 1.008 = 17.03 \) g/mol. Then \( 0.5000 \times 17.03 = 8.515 \). To the three significant figures the data support, this is 8.52 g. The check on magnitude is that half a mole should weigh about half the molar mass, and it does. 8.52 g

8. Explain why 1.00 mol of CaCl₂ contains \( 6.022 \times 10^{23} \) formula units but \( 1.81 \times 10^{24} \) ions.
   Show the full solution

   A mole always contains Avogadro's number of whatever particle is named, and the particle named in the first figure is the formula unit. One formula unit of CaCl₂ is not a molecule, since the compound is ionic, but it does represent one Ca²⁺ together with two Cl^-, which is three ions in total. So one mole of formula units contains three moles of ions, and \( 3 \times 6.022 \times 10^{23} = 1.807 \times 10^{24} \). The factor of three comes from reading the formula and is a separate step from any mole conversion, which is why the answer must always state what is being counted rather than reporting a bare number of particles. Each formula unit contains three ions, so the ion count is three times the formula unit count

9. A student converts 10.0 g of helium to atoms and gets \( 6.02 \times 10^{24} \). Check the answer and explain the significance of the result either way.
   Show the full solution

   Helium has a molar mass of 4.00 g/mol, so \( 10.0 \div 4.00 = 2.50 \) mol, and \( 2.50 \times 6.022 \times 10^{23} = 1.51 \times 10^{24} \) atoms. The student's answer is exactly ten times Avogadro's number, which corresponds to 10.0 mol rather than 2.50 mol, so they have treated the mass in grams as a number of moles and skipped the division by molar mass entirely. The error is caught quickly by asking whether 10.0 g of helium is plausibly ten moles: ten moles would weigh 40.0 g. Helium's low molar mass makes this slip especially easy, because the numbers are small and the missing step is not obvious in the arithmetic. \( 1.51 \times 10^{24} \) atoms; they used the mass as a mole count without dividing by 4.00

10. Explain why setting conversions up as fractions with visible units is worth the extra writing, even for a one-step problem.
    Show the full solution

    Because the habit is what makes multi-step problems reliable, and it cannot be acquired at the moment it becomes necessary. In a one-step conversion a student can usually guess correctly whether to multiply or divide, so the fraction seems like overhead. In a three-step chain, where a mass becomes moles becomes a mole ratio becomes another mass, guessing fails often, and an inverted factor produces an answer wrong by the square of the molar mass with nothing on the page to reveal it. Written as fractions, the units either cancel to leave the quantity asked for or they do not, and the check happens before any arithmetic. The practice also makes an answer auditable by someone else, which matters when the working is what earns the marks. The habit transfers to multi-step chains where guessing fails, and the unit check verifies the setup before calculating

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State the molar volume of a gas at STP.
   Show the full solution

   22.4 L/mol

2. What temperature and pressure does STP specify?
   Show the full solution

   0 degrees Celsius, which is 273 K, and 1 atmosphere

3. How many moles is 11.2 L of any gas at STP?
   Show the full solution

   \( 11.2 \div 22.4 \). 0.500 mol

4. Find the volume of 44.0 g of CO₂ at STP.
   Show the full solution

   \( 44.0 \div 44.01 = 1.00 \) mol, then \( \times 22.4 \). 22.4 L

5. State Avogadro's law.
   Show the full solution

   Equal volumes of gases at the same temperature and pressure contain equal numbers of particles

6. Find the mass of 3.36 L of nitrogen gas at STP.
   Show the full solution

   Convert to moles first: \( 3.36 \div 22.4 = 0.150 \) mol. Nitrogen gas is N₂, so the molar mass is \( 2 \times 14.01 = 28.02 \) g/mol, and using 14.01 would halve the answer. Then \( 0.150 \times 28.02 = 4.203 \), which to three significant figures is 4.20 g. The magnitude check is that 3.36 L is well under a mole, so the mass should be well under 28 g. 4.20 g

7. Explain why there is a single molar volume for all gases but no single molar volume for liquids.
   Show the full solution

   In a gas the particles are separated by distances far larger than their own diameters, so the volume is almost entirely empty space and is set by how far apart the particles sit rather than by how big they are. That spacing depends only on temperature and pressure, so at fixed conditions every gas has the same volume per mole. In a liquid the particles are in contact, so the volume is essentially the sum of the particle volumes together with however efficiently they pack, and both of those vary enormously between substances. A mole of water occupies about 18 mL and a mole of mercury about 15 mL despite mercury being eleven times denser, because the two quantities depend on molecular size and packing in ways that no single constant can capture. Gas volume comes from spacing, which is identical at fixed conditions; liquid volume comes from particle size and packing, which differ

8. A question gives a gas volume at 25 degrees Celsius and a student uses 22.4 L/mol. Explain what goes wrong and in which direction the answer errs.
   Show the full solution

   The constant 22.4 L/mol applies only at 0 degrees Celsius and 1 atmosphere. Raising the temperature to 25 degrees gives the particles more kinetic energy, so at constant pressure they push further apart and the same number of moles occupies a larger volume, about 24.5 L/mol. Using 22.4 therefore divides the volume by a number that is too small, which overstates the number of moles by roughly nine percent, and every quantity derived from that mole count, including any mass of product, is correspondingly too large. The lesson is to check the stated conditions before reaching for the constant, since the phrase "at STP" is doing real work in the question and is not decoration. The molar volume at 25 degrees is about 24.5 L/mol, so using 22.4 overstates the moles by about nine percent

9. Two identical balloons at the same temperature and pressure are filled, one with helium and one with argon. Compare the number of atoms and the masses.
   Show the full solution

   The atom counts are equal. By Avogadro's law equal volumes at the same temperature and pressure contain equal numbers of particles, and since both balloons have the same volume, both contain the same number of moles and therefore the same number of atoms. Both gases are monatomic, so no diatomic correction is needed. The masses are very different: helium has a molar mass of 4.00 g/mol and argon 39.95, so the argon balloon is almost exactly ten times heavier. This is precisely why the helium balloon rises and the argon one does not, since buoyancy depends on the density of the gas compared with the surrounding air rather than on how many particles are inside. Equal numbers of atoms; the argon is about ten times the mass

10. Explain how Avogadro's law let chemists determine molecular formulas that mass measurements alone could not, referring back to lesson 2.1.
    Show the full solution

    Mass analysis gives ratios, and a ratio is unchanged by multiplying both terms by the same number, so weighing could establish that a carbon oxide had twice as much oxygen per carbon as another without distinguishing CO and CO₂ from C₂O₂ and C₂O₄. Avogadro's law breaks the deadlock because it converts a measured volume directly into a particle count. When gases react, their volumes are found to stand in small whole number ratios, and if equal volumes contain equal numbers of particles, those volume ratios are particle ratios. Observing that two volumes of hydrogen combine with one volume of oxygen to give two volumes of water vapor fixes the numbers of molecules involved, not merely their masses, which is exactly the evidence mass measurements could not supply. Volume ratios become particle ratios, giving absolute molecule counts rather than mass ratios

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Write the formula for percent by mass of an element in a compound.
   Show the full solution

   Mass of that element in one mole, divided by the molar mass, times 100

2. Find the percent oxygen in H₂O.
   Show the full solution

   \( \frac{16.00}{18.02} \times 100 \). 88.8 percent

3. Find the percent carbon in CO₂.
   Show the full solution

   \( \frac{12.01}{44.01} \times 100 \). 27.3 percent

4. What check should every percent composition answer pass?
   Show the full solution

   The percentages must add to 100 within rounding

5. Find the mass of carbon in 200. g of CO₂.
   Show the full solution

   \( 200. \times 0.2729 \). 54.6 g

6. Find the percent composition of calcium carbonate, CaCO₃, and check your answer.
   Show the full solution

   The molar mass is \( 40.08 + 12.01 + 3 \times 16.00 = 100.09 \) g/mol, which is conveniently close to 100 and makes the percentages almost readable off the masses. Calcium: \( \frac{40.08}{100.09} \times 100 = 40.04 \) percent. Carbon: \( \frac{12.01}{100.09} \times 100 = 12.00 \) percent. Oxygen: \( \frac{48.00}{100.09} \times 100 = 47.96 \) percent. The check: \( 40.04 + 12.00 + 47.96 = 100.00 \). 40.04 percent Ca, 12.00 percent C, 47.96 percent O

7. Explain why the denominator is the molar mass of the compound rather than the atomic mass of the element.
   Show the full solution

   Because the question being answered is what fraction of the compound's total mass a given element accounts for, and a fraction requires the whole in the denominator. Dividing the mass of an element by its own atomic mass answers a different question entirely, namely how many moles of that element are present, and the result is a mole count rather than a fraction. The error is easy to spot in the units: mass divided by mass is dimensionless and can sensibly be multiplied by 100 to give a percentage, whereas grams divided by grams per mole gives moles, and multiplying moles by 100 produces nothing meaningful. A percentage is a part over a whole, so the whole compound's molar mass must be the denominator

8. CH₂O and C₆H₁₂O₆ have identical percent compositions. Explain why, and what this means for chemical analysis.
   Show the full solution

   Because the second formula is exactly six times the first, so every element's mass contribution and the total molar mass are all multiplied by six, and a fraction is unchanged when numerator and denominator are multiplied by the same number. Percent composition therefore reports only the ratio of atoms, not their absolute numbers, and every compound sharing that ratio gives the same analysis. For a laboratory this means an elemental analysis alone can never determine a molecular formula; it can determine the simplest ratio, called the empirical formula, and no more. A second and independent measurement of the molar mass is required to find the multiplier, which is the procedure of lesson 4.7. This is the same limitation that lesson 2.1 identified in Dalton's mass ratios. One formula is a whole multiple of the other, so the ratios are identical; analysis gives the empirical formula only

9. Two nitrogen fertilizers are offered: ammonium nitrate, NH₄NO₃, at 35.00 percent nitrogen, and urea, CO(NH₂)₂. Calculate urea's nitrogen percentage and say which delivers more nitrogen per kilogram.
   Show the full solution

   Count urea's atoms carefully, since the bracket subscript applies to both the nitrogen and the two hydrogens: 1 C, 1 O, 2 N and 4 H. The molar mass is \( 12.01 + 16.00 + 2 \times 14.01 + 4 \times 1.008 = 12.01 + 16.00 + 28.02 + 4.03 = 60.06 \) g/mol. The nitrogen percentage is \( \frac{28.02}{60.06} \times 100 = 46.65 \) percent. Urea therefore delivers more nitrogen per kilogram than ammonium nitrate, 466 g against 350 g, which is why it is the more concentrated product and is cheaper to transport per unit of nitrogen supplied. 46.65 percent; urea delivers more nitrogen per kilogram

10. Explain why percent composition being fixed for a compound is the same statement as the law of definite proportions.
    Show the full solution

    The law of definite proportions says a given compound always contains the same elements in the same proportion by mass regardless of its source, and percent composition is precisely that proportion expressed as a number out of a hundred. If water is always 11.19 percent hydrogen and 88.81 percent oxygen, then the law holds for water, and if a sample gave different percentages it would be either a different compound or a mixture. The connection also explains why percent composition is a useful identification tool: since the value is fixed by the formula, a measured analysis can be compared against calculated values for candidate compounds, and one that matches identifies the substance in the same way that density did for the metal in lesson 1.7. Percent composition is the fixed proportion by mass that the law asserts, stated numerically

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Define an empirical formula.
   Show the full solution

   The simplest whole number ratio of atoms in a compound

2. State the four steps of the procedure in order.
   Show the full solution

   Percent to mass, mass to moles, divide by the smallest, multiply to whole numbers

3. Why is it legitimate to assume a 100 g sample?
   Show the full solution

   Because the answer is a ratio, which is the same for any sample size

4. A ratio comes out as 1 : 1.33. What should you multiply by?
   Show the full solution

   1.33 is four thirds. 3, giving 3 : 4

5. Give the empirical formula of C₄H₈.
   Show the full solution

   Divide both subscripts by 4. CH₂

6. A compound is 52.2 percent carbon, 13.0 percent hydrogen and 34.8 percent oxygen. Find its empirical formula.
   Show the full solution

   Take 100 g, so 52.2 g C, 13.0 g H and 34.8 g O. Converting to moles: \( 52.2 \div 12.01 = 4.346 \), \( 13.0 \div 1.008 = 12.90 \) and \( 34.8 \div 16.00 = 2.175 \). The smallest is 2.175, so dividing through gives carbon \( 4.346 \div 2.175 = 1.998 \), hydrogen \( 12.90 \div 2.175 = 5.931 \) and oxygen 1.000. These round to 2, 6 and 1 within experimental error, so the empirical formula is C₂H₆O. That is ethanol, and note that no further multiplication is needed because the values are already close to whole numbers. C₂H₆O

7. Explain why the masses cannot simply be compared as a ratio without converting to moles.
   Show the full solution

   Because a formula states a ratio of atoms and a mass states a ratio of weights, and the two differ by however much the atoms differ in mass. In the first worked example the sample contained 40.0 g of carbon and 6.7 g of hydrogen, a mass ratio of about six to one, yet the formula has twice as many hydrogen atoms as carbon atoms. The discrepancy is entirely accounted for by a carbon atom being about twelve times heavier than a hydrogen atom, so a small mass of hydrogen represents a large number of atoms. Dividing each mass by its own molar mass removes the weighting and converts a mass ratio into a count ratio, which is exactly the job the mole was introduced to do. Atoms differ in mass, so a mass ratio is not a count ratio; dividing by molar mass removes the weighting

8. A student obtains a ratio of 1 : 2.5 and rounds it to 1 : 3. Explain the error and give the correct formula ratio.
   Show the full solution

   A value of 2.5 is not close to either 2 or 3, so it cannot be a rounding artifact; it is a genuine half-integer ratio and rounding it in either direction describes a different compound. The correct treatment is to multiply both values by 2, giving 2 : 5, which is a whole number ratio and therefore a valid empirical formula. The distinction that matters is between small deviations and clean fractions: a value of 2.97 or 3.02 should be rounded to 3, because experimental error of one or two percent is expected, whereas 2.5, 2.33 and 2.25 are exact fractions that should be cleared by multiplying by 2, 3 and 4 respectively. Multiply by 2 to give 2 : 5; a half-integer is a real fraction, not a rounding error

9. A 2.50 g sample of a hydrocarbon burns completely to give 7.86 g of CO₂ and 3.21 g of H₂O. Find the empirical formula.
   Show the full solution

   All the carbon in the sample ends up in the carbon dioxide and all the hydrogen in the water, so those product masses give the mole counts. Carbon: \( 7.86 \div 44.01 = 0.1786 \) mol CO₂, and each contains one carbon, so 0.1786 mol C. Hydrogen: \( 3.21 \div 18.02 = 0.1782 \) mol H₂O, and each contains two hydrogens, so \( 2 \times 0.1782 = 0.3563 \) mol H. Dividing both by the smaller gives carbon 1.000 and hydrogen 1.995, which is 2 within error, so the empirical formula is CH₂. As a check, the masses of carbon and hydrogen are \( 0.1786 \times 12.01 = 2.145 \) g and \( 0.3563 \times 1.008 = 0.359 \) g, which sum to 2.504 g and account for the 2.50 g sample, confirming that it contained no oxygen. CH₂

10. Explain why combustion analysis requires the mass check in the previous question when the compound might contain oxygen.
    Show the full solution

    Because the oxygen in the combustion products comes from two sources and the analysis cannot distinguish them. The carbon dioxide and water contain oxygen that was supplied by the air as well as any oxygen that was in the original compound, so unlike carbon and hydrogen, the oxygen content cannot be read off a product mass. It has to be obtained by difference: add the masses of carbon and hydrogen determined from the products, subtract that total from the original sample mass, and whatever remains was oxygen in the compound. In the previous question the carbon and hydrogen accounted for the whole 2.50 g, so the remainder was zero and the compound was a pure hydrocarbon. Had they summed to 1.80 g, the missing 0.70 g would be oxygen and would be converted to moles and included in the ratio. Product oxygen comes partly from the air, so compound oxygen must be found by subtracting the carbon and hydrogen masses from the sample mass

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State the relationship between an empirical and a molecular formula.
   Show the full solution

   The molecular formula is a whole number multiple of the empirical formula

2. How is the multiplier found?
   Show the full solution

   Divide the molar mass by the empirical formula mass

3. A compound has empirical formula NO₂ and molar mass 92.0 g/mol. Find the molecular formula.
   Show the full solution

   Empirical mass \( 14.01 + 32.00 = 46.01 \); \( 92.0 \div 46.01 = 2.00 \). N₂O₄

4. A compound has empirical formula CH₂ and molar mass 42.0 g/mol. Find the molecular formula.
   Show the full solution

   Empirical mass \( 12.01 + 2.016 = 14.03 \); \( 42.0 \div 14.03 = 2.99 \), so 3. C₃H₆

5. Why do ionic compounds have no molecular formula?
   Show the full solution

   They form lattices rather than molecules, so the formula is a ratio by nature

6. A student finds a multiplier of 3.4 and rounds it to 3. Explain why this is not acceptable and what they should do.
   Show the full solution

   The multiplier has to be a whole number because a molecule contains a whole number of each kind of atom, so a value of 3.4 is not a molecule that has been slightly mismeasured but a signal that one of the inputs is wrong. Deviations of one or two percent are expected from experimental error and 2.98 would be safely rounded to 3, but 3.4 is thirteen percent from the nearest integer, which is far too large to attribute to measurement. The correct response is to look for the error: recheck the empirical formula, since a ratio that was rounded rather than multiplied in lesson 4.6 would produce exactly this symptom, and recheck the molar mass measurement. Rounding hides the fault and produces a formula for a compound that was never analyzed. A multiplier that far from an integer indicates an error in the empirical formula or molar mass, which should be rechecked rather than rounded

7. Explain why the molar mass has to be measured separately rather than calculated from the percent composition.
   Show the full solution

   Because percent composition is a ratio and is unchanged by multiplying every subscript by the same number, so it carries no information about the absolute size of the molecule. CH₂O, C₂H₄O₂ and C₆H₁₂O₆ all give exactly 40.0 percent carbon, 6.7 percent hydrogen and 53.3 percent oxygen, so no amount of care in the elemental analysis can distinguish them. The molar mass is an independent quantity that does scale with molecular size, which is precisely why it supplies the missing information. In practice it is determined by a physical method such as mass spectrometry, or from the density of the vapor using the gas laws of unit 8, rather than by any further chemical analysis. Percent composition is scale-free, so it cannot reveal size; the molar mass is an independent physical measurement

8. Glucose and fructose both have the molecular formula C₆H₁₂O₆ yet differ in taste and in how the body processes them. Explain what this shows about the limits of a formula.
   Show the full solution

   A molecular formula states which atoms are present and how many of each, and nothing at all about how they are joined. Glucose and fructose contain identical atoms in identical numbers but arranged differently, with the carbon skeleton and the position of the carbonyl group differing, and such compounds are called isomers. Because chemical behavior depends on structure rather than on composition, isomers can have quite different properties, as lesson 3.4 showed for shape and polarity. This means the sequence from analysis to empirical formula to molecular formula, complete as it is, still stops short of identifying a substance, and determining the structure requires further evidence from methods such as spectroscopy. A formula says nothing about arrangement; isomers share a formula and differ in structure and therefore in behavior

9. A compound is 92.3 percent carbon and 7.7 percent hydrogen, with a molar mass of 78.1 g/mol. Find both formulas.
   Show the full solution

   Take 100 g, giving 92.3 g C and 7.7 g H. Moles: \( 92.3 \div 12.01 = 7.685 \) and \( 7.7 \div 1.008 = 7.639 \). Dividing both by 7.639 gives carbon 1.006 and hydrogen 1.000, so the ratio is 1 : 1 and the empirical formula is CH. The empirical formula mass is \( 12.01 + 1.008 = 13.02 \), so the multiplier is \( 78.1 \div 13.02 = 6.00 \). Multiplying both subscripts by 6 gives the molecular formula C₆H₆, which is benzene. Note that the empirical formula CH would be a molecule with one carbon and one hydrogen, which is not a stable substance, so in this case the empirical formula does not correspond to any real compound at all. Empirical CH, molecular C₆H₆

10. Explain the full chain of reasoning from a laboratory analysis to a molecular formula, naming what each step contributes.
    Show the full solution

    The laboratory measures masses, either as percentages from an elemental analysis or as the masses of combustion products, and masses alone cannot give a formula because atoms of different elements have different masses. Dividing each mass by the appropriate molar mass converts the mass ratio into a mole ratio, which is a ratio of atom counts, and this is the step the mole exists to make possible. Dividing through by the smallest value and clearing any fraction turns that ratio into whole numbers, giving the empirical formula, which is the simplest ratio consistent with the analysis. That ratio is scale-free, so an independent measurement of the molar mass is required, and dividing it by the empirical formula mass gives the whole number multiplier. Applying that multiplier to every subscript gives the molecular formula. Even then the structure remains undetermined, since isomers share a molecular formula. Masses give mole ratios, mole ratios give the empirical formula, the molar mass gives the multiplier, and structure still needs separate evidence

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. List the five common signs of a chemical reaction.
   Show the full solution

   Gas evolution, precipitate formation, color change, temperature change, light emission

2. Define a precipitate.
   Show the full solution

   An insoluble solid formed when two solutions are mixed

3. What test identifies carbon dioxide?
   Show the full solution

   It turns limewater cloudy

4. Why must a conservation of mass experiment use a closed system?
   Show the full solution

   So that no gas can enter or escape, which would change the measured mass without any matter being created or destroyed

5. Give a physical process that mimics gas evolution.
   Show the full solution

   Boiling, or a dissolved gas coming out of solution when warmed

6. A student warms a blue solution and it turns green. Explain why this is not sufficient evidence of a reaction and what would settle it.
   Show the full solution

   Color change is the weakest of the five signs because several physical processes produce it. Warming can shift the equilibrium between two differently colored forms of the same dissolved species, and evaporation of solvent concentrates the solution, which deepens and can apparently alter the color without any new substance existing. The test is reversibility together with identification: cool the solution back to its original temperature and see whether the blue returns, since a physical shift reverses and a reaction generally does not. If the color persists, evaporate a sample and examine the residue to determine whether it is the original compound or a different one. Color changes can be physical and reversible; cool it back and identify the dissolved substance

7. Explain why identifying a product is stronger evidence than any of the five signs.
   Show the full solution

   Because the five signs are symptoms that a chemical change could produce, while a new substance is the definition of a chemical change having occurred. Each sign is consistent with a physical process as well, so observing one narrows the possibilities without deciding between them, and a list of signs can never be made long enough to close that gap. Determining that a substance is present which was not present before, and which cannot be obtained from the reactants by any physical means, establishes the conclusion directly rather than by inference. This is why laboratory work moves as quickly as possible from watching to testing, using limewater, splint tests, flame tests and melting points. The signs are symptoms consistent with physical change; a new substance is what chemical change means

8. A reaction in an open beaker shows a mass decrease of 0.44 g. Explain what most likely happened and how to confirm it.
   Show the full solution

   A gaseous product has almost certainly escaped from the open beaker into the room. Nothing has been destroyed; the mass has simply left the system being weighed, which is the standard explanation for an apparent loss and is exactly the wood-burning case from lesson 1.1. Confirming it has two parts. First, repeat the reaction in a sealed flask and weigh the whole vessel before and after: an unchanged total shows the mass left the beaker rather than existence. Second, collect the gas, either over water or in a gas syringe, weigh it and identify it, which should account for the missing 0.44 g and name the substance responsible. The figure itself is suggestive, since 0.44 g is one hundredth of a mole of carbon dioxide. A gas escaped; repeat in a sealed vessel and collect, weigh and identify the gas

9. Explain why dissolving sodium hydroxide in water releases heat but is still not a chemical reaction, while neutralizing it with acid is.
   Show the full solution

   Dissolving separates the ions of the lattice and surrounds each with water molecules, and the energy released when water molecules arrange around the ions exceeds the energy needed to break the lattice apart, so the solution warms. No new substance results: the sodium and hydroxide ions are the same ions that were in the solid, now dispersed, and evaporating the water returns solid sodium hydroxide unchanged. Neutralization is different because hydroxide ions and hydrogen ions combine to form water molecules, which is a new substance with entirely different properties, and the solution afterward contains a salt rather than the original base. The recovery test discriminates: the dissolved base comes back, the neutralized one does not. Dissolving disperses existing ions recoverably; neutralization forms water, a new substance

10. Explain why the conclusion in the worked example is phrased as "consistent with conservation of mass" rather than "proves conservation of mass".
    Show the full solution

    Because a single experiment with a balance of finite precision cannot prove a general law, and claiming more than the evidence supports is the specific error lesson 1.5 identified. The measurement shows that the mass did not change by more than the balance could detect, which might be a hundredth of a gram, so it rules out any change larger than that and says nothing about smaller ones. It also concerns one reaction in one vessel, while the law is a claim about every chemical change everywhere. What gives the law its standing is not any single result but the accumulation of many such measurements, at steadily improving precision, none of which has found a discrepancy, together with the atomic explanation of why a discrepancy should not occur. One measurement of finite precision on one reaction cannot establish a universal law, only fail to contradict it

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. What may be changed when balancing an equation, and what may not?
   Show the full solution

   Coefficients may be changed; subscripts may not

2. Give the state symbol for a substance dissolved in water.
   Show the full solution

   (aq)

3. Balance: H₂ + O₂ gives H₂O.
   Show the full solution

   \( 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \)

4. Balance: Fe + O₂ gives Fe₂O₃.
   Show the full solution

   Iron 4 and 4, oxygen 6 and 6. \( 4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 \)

5. Balance: Zn + HCl gives ZnCl₂ + H₂.
   Show the full solution

   \( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \)

6. Balance the combustion of ethanol, C₂H₆O, and show your check.
   Show the full solution

   Carbon first: two carbons give \( 2\text{CO}_2 \). Hydrogen next: six hydrogens give \( 3\text{H}_2\text{O} \). Now count oxygen on the right: \( 2 \times 2 + 3 = 7 \) atoms. The left already has one oxygen inside the ethanol molecule, which is the step most often missed, so the O₂ needs to supply only \( 7 - 1 = 6 \) atoms, which is \( 3\text{O}_2 \). The balanced equation is \( \text{C}_2\text{H}_6\text{O} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \). Check: carbon 2 and 2, hydrogen 6 and 6, oxygen \( 1 + 6 = 7 \) and \( 4 + 3 = 7 \). \( \text{C}_2\text{H}_6\text{O} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \)

7. A student balances H₂ + O₂ to H₂O by writing H₂O₂. Explain precisely what is wrong.
   Show the full solution

   They have altered a subscript, which is not a bookkeeping adjustment but a change of substance. H₂O₂ is hydrogen peroxide, a corrosive liquid that decomposes readily and is used as a bleach, and it is not what forms when hydrogen burns in oxygen. The atoms now balance, but the equation is a true statement about a different reaction and a false statement about the one asked for. Coefficients may be changed because they count how many of each molecule take part, which is genuinely adjustable; subscripts describe the internal composition of a molecule, which is fixed by the compound's identity. The correct balance is \( 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \). Changing a subscript changes the compound to hydrogen peroxide; only coefficients may be adjusted

8. Explain why leaving oxygen until last makes combustion equations easier.
   Show the full solution

   Because balancing is easiest when each adjustment affects only one element, and an element appearing in several formulas violates that. In a combustion equation oxygen appears in the O₂ on the left and in both the carbon dioxide and the water on the right, so any change to an oxygen-containing coefficient disturbs at least two counts at once. Carbon appears in only two places and hydrogen in only two, so fixing them is unambiguous and, once done, the right-hand side is completely determined. Oxygen can then be counted rather than balanced: total the oxygens on the right and halve to get the O₂ coefficient. Leaving the most widely distributed element until last is a general strategy, and in combustion that element is reliably oxygen. Oxygen appears in three formulas, so adjusting it disturbs several counts; fixing the others first determines it

9. Balance the reaction of aluminum with copper(II) chloride, and explain how treating chloride as a unit helps.
   Show the full solution

   The unbalanced equation is \( \text{Al} + \text{CuCl}_2 \rightarrow \text{AlCl}_3 + \text{Cu} \). Chlorine does not change partners in a way that breaks it up, so count chlorines directly: the left has them in twos and the right in threes, and the lowest common multiple is 6. That fixes \( 3\text{CuCl}_2 \) and \( 2\text{AlCl}_3 \). Aluminum then needs a coefficient of 2 and copper a coefficient of 3, giving \( 2\text{Al} + 3\text{CuCl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{Cu} \). Checking: aluminum 2 and 2, copper 3 and 3, chlorine 6 and 6. Going straight to the element that constrains the answer, rather than starting at the left and working across, turns a trial-and-error problem into one step. \( 2\text{Al} + 3\text{CuCl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{Cu} \)

10. Explain why a balanced equation with coefficients 4, 2, 4 is considered wrong even though the atoms balance.
    Show the full solution

    Because the convention is that coefficients are the smallest whole numbers that achieve the balance, and 4, 2, 4 shares a common factor of 2 with a simpler set of 2, 1, 2. The atoms do balance, so the equation is not false, but it fails to state the reaction in its simplest terms and can suggest a relationship that is not there. The coefficients are read as a mole ratio in unit 6, and while 4 : 2 : 4 and 2 : 1 : 2 are the same ratio, only the reduced form makes that ratio immediately legible. The convention parallels the one for ionic formulas in lesson 3.2, where Mg₂O₂ is reduced to MgO for the same reason. Coefficients must be the smallest whole number set; 4, 2, 4 reduces to 2, 1, 2

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Name the five reaction types.
   Show the full solution

   Synthesis, decomposition, single replacement, double replacement, combustion

2. Classify: \( 2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2 \).
   Show the full solution

   One reactant, two products. Decomposition

3. Classify: \( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \).
   Show the full solution

   Synthesis

4. Classify: \( \text{BaCl}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{NaCl} \).
   Show the full solution

   Two compounds exchanging partners. Double replacement

5. What are the products when a hydrocarbon burns completely?
   Show the full solution

   Carbon dioxide and water

6. Predict whether iron will displace silver from silver nitrate solution, and write the equation if it does.
   Show the full solution

   Iron sits above silver in the activity series, so iron is the more reactive metal and the displacement occurs. Iron forms Fe²⁺ in this reaction and nitrate carries a single negative charge, so iron(II) nitrate is Fe(NO₃)₂ and two silver nitrates are consumed per iron atom, releasing two silver atoms: \( \text{Fe} + 2\text{AgNO}_3 \rightarrow \text{Fe(NO}_3)_2 + 2\text{Ag} \). The observable result is silver depositing as a gray coating on the iron while the solution takes on the pale green of iron(II). This is the basis of the traditional silver tree demonstration. It occurs: \( \text{Fe} + 2\text{AgNO}_3 \rightarrow \text{Fe(NO}_3)_2 + 2\text{Ag} \)

7. Explain why predicting products for a reaction that does not occur is a worse error than writing "no reaction".
   Show the full solution

   Because the two answers claim different things about the chemistry. Writing "no reaction" is a substantive prediction that can be tested and that happens to be correct, and it demonstrates that the activity series was consulted and understood. Writing plausible products asserts that a chemical change happens when it does not, which would be discovered immediately in the laboratory by nothing at all occurring. It also indicates that the student treated the question as a pattern-matching exercise, swapping symbols according to the shape of the equation without checking whether the swap is energetically possible. The activity series exists precisely to rule such predictions out, so ignoring it defeats the purpose of having asked. "No reaction" is a correct testable prediction; inventing products asserts a change that does not happen

8. Burning magnesium is both a synthesis and a combustion. Explain why this is not a flaw in the classification scheme.
   Show the full solution

   Because the five categories are not defined by a single criterion and were never intended to partition reactions into exclusive boxes. Four of them classify by the pattern of the equation, counting reactants and products and looking for free elements, while combustion classifies by what is reacting and what happens energetically, namely rapid reaction with oxygen releasing heat and light. A reaction can satisfy both a structural description and an energetic one at once, and burning magnesium does: two substances combine into one, which is synthesis, and the combination is with oxygen and releases a great deal of energy, which is combustion. Categories are models in the sense of lesson 1.2, judged by usefulness rather than by whether they carve reality at exclusive joints. Combustion is defined by what reacts and the energy released, the others by equation pattern, so overlap is expected

9. Explain what drives a double replacement reaction, given that ions in solution are already free to move.
   Show the full solution

   Simply mixing two solutions of soluble salts produces a mixture of four kinds of ion all moving independently, and if all four possible combinations are soluble then nothing changes and no reaction has occurred, however the equation is written. A double replacement happens only when one of the new pairings removes ions from solution, which it can do in three ways: forming an insoluble solid that precipitates out, forming a gas that escapes, or forming water, which is a molecular compound that does not dissociate. In each case the ions concerned are no longer free in the solution, so the change is real and irreversible under those conditions. Predicting these reactions therefore requires the solubility rules, which lesson 5.5 sets out, and the point is made explicit by the net ionic equations of lesson 5.6. Only the formation of a precipitate, a gas or water removes ions from solution; otherwise nothing changes

10. Explain how the activity series was established and why it has to be experimental rather than derived.
    Show the full solution

    It was built by testing pairs: put each metal into solutions of the others' salts and record which displacements occur, then arrange the metals so that every observed displacement runs from higher to lower in the list. Consistency across all the pairs is what makes a single ordering possible, and the position of hydrogen was fixed by testing which metals release hydrogen from acids. The ranking is experimental because it reflects the overall energy change of a real process, which depends on ionization energy, the energy needed to break up the metal lattice and the energy released when the resulting ion is surrounded by water molecules, all combining in a way that no single periodic trend predicts. Reactivity does correlate loosely with low ionization energy, so the alkali metals sit at the top, but the ordering of zinc, iron and lead cannot be read off the periodic table. By systematic pairwise displacement tests; the ordering depends on several competing energy terms that no single trend predicts

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Give the products of complete combustion of a hydrocarbon.
   Show the full solution

   Carbon dioxide and water

2. Give the additional products possible in incomplete combustion.
   Show the full solution

   Carbon monoxide and carbon as soot

3. What flame color indicates incomplete combustion?
   Show the full solution

   Yellow and sooty, rather than blue

4. Balance the complete combustion of methane.
   Show the full solution

   \( \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \)

5. Is combustion exothermic or endothermic?
   Show the full solution

   Exothermic

6. Explain why carbon monoxide is more dangerous than the carbon dioxide produced by complete combustion.
   Show the full solution

   Carbon dioxide is not especially toxic and becomes dangerous only at concentrations high enough to displace oxygen from the air, and it is also detectable indirectly because rising levels make a room feel stuffy. Carbon monoxide is toxic at concentrations far too low to affect the oxygen content of the air, because it binds to hemoglobin in the blood in place of oxygen and does so far more strongly, so the blood loses its capacity to carry oxygen even while the lungs are working normally. It is colorless, odorless and tasteless, so there is no sensory warning at all, and the early symptoms of headache and drowsiness resemble ordinary tiredness and discourage the victim from leaving. That combination of potency and undetectability is why detectors are required near fuel-burning appliances. It binds hemoglobin far more strongly than oxygen, is toxic at low concentrations, and gives no sensory warning

7. A gas burner burns with a yellow sooty flame and blackens the base of a beaker. Explain what is happening and how to correct it.
   Show the full solution

   The flame is burning with insufficient oxygen, so combustion is incomplete. Instead of every carbon atom reaching carbon dioxide, some form carbon monoxide and some remain as solid carbon particles, which glow yellow as they are heated in the flame and deposit as soot on any cool surface they meet. The black deposit is that unreacted carbon, which represents fuel whose energy was never released, so the flame is also cooler and less efficient. The correction is to increase the air supply by opening the air hole at the base of the burner, which premixes air with the gas before it reaches the flame. The flame should turn blue and stop depositing soot, which is the visible test that combustion is now complete. Incomplete combustion from insufficient oxygen; open the air hole to premix more air and the flame turns blue

8. Balance the complete combustion of butane, C₄H₁₀, showing where the odd-even trick is needed.
   Show the full solution

   Carbon first: four carbons give \( 4\text{CO}_2 \). Hydrogen next: ten hydrogens give \( 5\text{H}_2\text{O} \). Now count oxygen on the right: \( 4 \times 2 + 5 = 13 \) atoms, which is odd. Since the only source on the left is the diatomic O₂, an odd atom count cannot be supplied without a fractional coefficient, so double every coefficient. Two butanes give \( 8\text{CO}_2 \) and \( 10\text{H}_2\text{O} \), so the right side now holds \( 16 + 10 = 26 \) oxygen atoms, requiring \( 13\text{O}_2 \). The balanced equation is \( 2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O} \). Check: carbon 8 and 8, hydrogen 20 and 20, oxygen 26 and 26. \( 2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O} \)

9. Explain why every carbon atom in a fuel ends up as one carbon dioxide molecule in complete combustion, and why this matters beyond the equation.
   Show the full solution

   Conservation of mass requires that atoms are neither created nor destroyed, and in complete combustion the only carbon-containing product is carbon dioxide, each molecule of which contains exactly one carbon atom. Every carbon that went in as fuel must therefore come out as one molecule of carbon dioxide, with no route to anywhere else. This matters because it makes emissions calculable from composition alone: the mass of carbon dioxide produced by burning a fuel can be worked out from the fuel's formula and mass without any measurement of the exhaust, which is the calculation lesson 6.7 carries out. It also means emissions cannot be reduced by adjusting how a fuel is burned, only by burning less of it or by burning a fuel with less carbon per unit of energy released. Conservation of atoms and the single carbon in CO₂ fix the ratio, so emissions follow from fuel composition alone

10. Faraday chose a candle as the subject of six lectures. Using lessons 1.2 and 5.4, explain what makes a candle a good vehicle for teaching chemistry.
    Show the full solution

    A candle contains an unusual density of separable phenomena in one familiar object. It shows three distinct physical changes, melting, capillary rise and vaporization, feeding one chemical change, so it forces the distinction of lesson 1.2 rather than merely illustrating it. It demonstrates combustion of a hydrocarbon with visible evidence of both products, since water condenses on a cold surface held in the flame and carbon dioxide turns limewater cloudy. It shows complete and incomplete combustion simultaneously, since the yellow luminous region contains glowing carbon particles while the blue base burns cleanly, which is why a cold spoon held in the yellow region collects soot. It shows that a fuel must reach the gas phase to burn rapidly, which is the point of the wick. Every one of those can be observed without apparatus, which is what Faraday meant by an open door. It combines several physical changes feeding one chemical change, with both complete and incomplete combustion visible and both products testable

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Which compounds are always soluble?
   Show the full solution

   All group 1 and ammonium compounds, and all nitrates

2. Name the two common insoluble chlorides.
   Show the full solution

   Silver chloride and lead(II) chloride

3. Will copper displace hydrogen from hydrochloric acid?
   Show the full solution

   Copper is below hydrogen in the activity series. No

4. Are most carbonates soluble or insoluble?
   Show the full solution

   Insoluble, except group 1 and ammonium carbonates

5. What must be true for a double replacement reaction to actually occur?
   Show the full solution

   One product must be a precipitate, a gas or water

6. Predict what happens when barium chloride solution is mixed with sodium sulfate solution, and write the equation.
   Show the full solution

   Swap the partners and build the formulas from the charges: barium is Ba²⁺ and sulfate is SO₄^2-, giving BaSO₄, while sodium is Na⁺ and chloride is Cl^-, giving NaCl. Checking the rules, sodium chloride is a group 1 compound and therefore soluble, but barium sulfate is one of the named exceptions among sulfates and is insoluble. A precipitate forms, so the reaction occurs: \( \text{BaCl}_2(aq) + \text{Na}_2\text{SO}_4(aq) \rightarrow \text{BaSO}_4(s) + 2\text{NaCl}(aq) \). The observable result is a dense white precipitate, and this reaction is the standard laboratory test for sulfate ions. A white precipitate of barium sulfate forms

7. A student predicts that mixing silver nitrate with calcium chloride gives AgCl₂ and CaNO₃. Find both errors.
   Show the full solution

   They have carried the subscripts across from the reactants instead of rebuilding the formulas from the charges. Silver forms only Ag⁺, so pairing it with Cl^- gives AgCl in a one-to-one ratio, and AgCl₂ would require a 2+ silver ion that does not occur here. Calcium forms Ca²⁺, so pairing it with the singly charged NO₃^- requires two nitrates, giving Ca(NO₃)₂ with brackets. The correct equation is \( 2\text{AgNO}_3(aq) + \text{CaCl}_2(aq) \rightarrow 2\text{AgCl}(s) + \text{Ca(NO}_3)_2(aq) \), with silver chloride precipitating as a white solid. The general rule is that a double replacement produces genuinely new compounds whose formulas must be derived afresh. AgCl and Ca(NO₃)₂; formulas must be rebuilt from charges, not copied

8. Explain why the activity series predicts that potassium cannot be stored in water but gold can be left exposed for centuries.
   Show the full solution

   The series ranks how readily a metal gives up electrons to form a positive ion, and the two metals sit at opposite ends of it. Potassium is at the very top, losing its single outer electron extremely readily, so it reacts vigorously even with the weak supply of hydrogen ions in water, displacing hydrogen gas and releasing enough heat to ignite it. That is why potassium is stored under oil, which excludes both water and air. Gold is at the very bottom, meaning the energy required to remove its electrons is not repaid by the compound that would form, so it does not react with water, oxygen or ordinary acids. Gold artifacts recovered from ancient burials are untarnished for exactly this reason, and it is also why gold occurs in nature as the free metal while potassium never does. Potassium gives up electrons extremely readily and reacts with water; gold does not, so it survives unreacted

9. Explain why sodium chloride and potassium nitrate solutions can be mixed with no reaction, even though both would give a precipitate with silver nitrate.
   Show the full solution

   Whether a reaction occurs depends on the particular pairings available, not on whether the ions are capable of forming a precipitate with something. Mixing sodium chloride with potassium nitrate puts Na⁺, Cl^-, K⁺ and NO₃^- into one solution, and every pairing of those four is soluble: sodium and potassium compounds are soluble by the group 1 rule and nitrates are soluble without exception. No combination removes anything from solution, so all four ions remain free and the mixture is simply a solution of four ions. Adding silver nitrate introduces Ag⁺, which does form an insoluble compound with chloride, so a precipitate appears. The chloride ion was present and available the whole time; it had no partner that could take it out of solution until silver arrived. All four possible pairings are soluble, so nothing leaves solution; the chloride needs a silver ion to precipitate

10. Explain why the solubility rules are stated as a list of exceptions rather than derived from the bonding ideas of unit 3.
    Show the full solution

    Because whether an ionic compound dissolves depends on a close competition between two large energy terms that unit 3 treats separately. Breaking up the lattice requires energy, and the amount depends on the charges and sizes of the ions, while surrounding the separated ions with water molecules releases energy, and that amount also depends on charge and size. Both terms are large and they are of similar magnitude, so solubility is decided by a small difference between two big numbers, and the sign of that difference cannot be predicted from ionic radius or charge alone. Compounds with similar-looking formulas therefore behave differently, as sodium chloride and silver chloride do. The rules are an empirical summary of measured behavior, which is honest, and lesson 9.1 sets out the energy account they summarize. Solubility is a small difference between two large competing energy terms, so it is measured rather than predicted

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Define a spectator ion.
   Show the full solution

   An ion that appears unchanged on both sides of the complete ionic equation and takes no part in the reaction

2. Which species are split into ions when writing a complete ionic equation?
   Show the full solution

   Only soluble ionic compounds labeled (aq)

3. Should a precipitate be split into ions? Explain in one sentence.
   Show the full solution

   No, because its ions are locked in a solid lattice and are not free in solution

4. What must balance in a net ionic equation besides atoms?
   Show the full solution

   Total charge on each side

5. Write the net ionic equation for barium ions reacting with sulfate ions.
   Show the full solution

   \( \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s) \)

6. Write all three equations for lead(II) nitrate reacting with potassium iodide.
   Show the full solution

   Molecular: \( \text{Pb(NO}_3)_2(aq) + 2\text{KI}(aq) \rightarrow \text{PbI}_2(s) + 2\text{KNO}_3(aq) \). Complete ionic, splitting every aqueous ionic compound and keeping the solid intact, and noting that the coefficient of 2 means two potassium ions and two iodide ions: \( \text{Pb}^{2+}(aq) + 2\text{NO}_3^{-}(aq) + 2\text{K}^{+}(aq) + 2\text{I}^{-}(aq) \rightarrow \text{PbI}_2(s) + 2\text{K}^{+}(aq) + 2\text{NO}_3^{-}(aq) \). Potassium and nitrate are spectators, so the net ionic equation is \( \text{Pb}^{2+}(aq) + 2\text{I}^{-}(aq) \rightarrow \text{PbI}_2(s) \). Charge check: \( (+2) + 2(-1) = 0 \) on the left and 0 on the right. \( \text{Pb}^{2+}(aq) + 2\text{I}^{-}(aq) \rightarrow \text{PbI}_2(s) \)

7. Explain why the net ionic equation for mixing sodium chloride and potassium nitrate solutions cannot be written.
   Show the full solution

   Because there is nothing left after the cancellation. Writing the complete ionic equation puts Na⁺, Cl^-, K⁺ and NO₃^- on the left, and since every possible pairing is soluble nothing precipitates, so exactly the same four ions appear on the right. Every one of them is therefore a spectator, and canceling them all leaves an empty equation. That emptiness is the formal statement that no reaction occurred, and it is a more informative result than it looks: the molecular equation can be written out and balanced and looks like a reaction, whereas the net ionic form makes the absence of any change explicit and unmistakable. All four ions are spectators, so nothing remains; the empty equation states that no reaction occurred

8. Explain why one net ionic equation can represent many different molecular equations.
   Show the full solution

   Because the net form records only the ions that changed, and the identity of the spectators is irrelevant to that change. When silver ions meet chloride ions they form solid silver chloride regardless of what the silver arrived with or what the chloride arrived with, so silver nitrate with sodium chloride, silver acetate with potassium chloride and silver fluoride with hydrochloric acid all reduce to the same net equation. The molecular equations look entirely different because they name the spectator partners, which the chemist happened to choose when selecting reagents. The net form therefore captures the chemistry, and the molecular form captures the recipe, which is why the net form is the more useful statement of what a reaction is. Only the reacting ions appear, and spectator identity does not affect the change, so many recipes share one reaction

9. Explain why writing the complete ionic equation is a better description of the beaker than the molecular equation, even though both are correct.
   Show the full solution

   Because the molecular equation implies particles that are not present. Writing NaCl(aq) suggests intact sodium chloride units floating in the water, whereas the conductivity evidence of lesson 3.2 shows the solution contains separated hydrated Na⁺ and Cl^- ions moving independently, with no association between a particular sodium and a particular chloride. The complete ionic equation represents that accurately. The molecular equation is not false, since it correctly reports what was weighed out and what the overall stoichiometry is, and it is the right form for calculating quantities. But as a picture of the mixture it is misleading, and the difference matters as soon as a student is asked why mixing two solutions that contain no solid at all can suddenly produce one. Soluble ionic compounds exist as free ions, not formula units, and the ionic equation shows what is actually present

10. Explain why charge balance is a useful check that the molecular equation does not provide.
    Show the full solution

    A molecular equation contains only neutral formulas, so both sides are automatically zero charge and the check is vacuous: it can never fail and therefore never detects anything. Once the compounds are split into ions, each side carries an explicit sum of positive and negative charges, and that sum must be equal on both sides because electrons are neither created nor destroyed. This gives an independent test that catches errors the atom count misses, such as omitting an ion during cancellation, writing a charge with the wrong sign, or forgetting that a coefficient of 2 applies to the ion charge as well as to the count. In the lead iodide example the left sums to zero from \( (+2) + 2(-1) \), and a student who wrote a single iodide would get a nonzero total and know immediately that something was dropped. Molecular equations are all neutral so the check is vacuous; ionic equations expose omitted ions and sign errors

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. What do the coefficients in a balanced equation represent?
   Show the full solution

   A ratio of moles, or equivalently of particles

2. For \( 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \), how many moles of water come from 4 mol of hydrogen?
   Show the full solution

   The ratio is 2 to 2, so one to one. 4 mol

3. State the three stages of the stoichiometry road map.
   Show the full solution

   Given quantity to moles, mole ratio, moles to wanted quantity

4. Which stage uses the balanced equation?
   Show the full solution

   Only the middle one, the mole ratio

5. Why must the equation be balanced before any calculation?
   Show the full solution

   Because the coefficients supply the ratio, and wrong coefficients give a wrong ratio

6. Explain why mixing reactants in the mass ratio given by the coefficients wastes one of them.
   Show the full solution

   Because the coefficients count particles and the masses of those particles differ, often by a large factor. In the ammonia synthesis the equation calls for three hydrogen molecules per nitrogen molecule, but a hydrogen molecule has a mass of 2.016 atomic mass units against nitrogen's 28.02, so three moles of hydrogen weigh only 6.05 g while one mole of nitrogen weighs 28.02 g. Measuring out 1 g of nitrogen and 3 g of hydrogen supplies about 0.036 mol of nitrogen and 1.49 mol of hydrogen, which is a ratio of roughly 1 to 41 rather than 1 to 3. Almost all the hydrogen would be left over, and the reaction would stop when the nitrogen ran out. Masses per mole differ, so equal coefficient masses give wildly wrong mole ratios and leave one reactant unused

7. For \( \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \), find the moles of oxygen needed for 0.250 mol of propane and the moles of water produced.
   Show the full solution

   Take each ratio from the coefficients with the wanted substance on top so the given moles cancel. For oxygen the ratio is five moles of O₂ per mole of propane, so \( 0.250 \times \frac{5}{1} = 1.25 \) mol of oxygen. For water the ratio is four moles per mole of propane, so \( 0.250 \times \frac{4}{1} = 1.00 \) mol of water. Both came from the same given quantity without any intermediate step, because the question asked in moles and answered in moles, so no molar mass was needed at all. 1.25 mol of oxygen and 1.00 mol of water

8. Explain why any two substances in an equation can be related directly, including two reactants.
   Show the full solution

   Because a single balanced equation fixes the proportions of everything in it simultaneously, so the coefficients form one consistent set rather than a series of separate pairwise statements. If one nitrogen goes with three hydrogens and one nitrogen goes with two ammonias, then three hydrogens go with two ammonias, and the ratio between any two species can be read off directly without passing through a third. This includes two reactants, which is what makes limiting reactant problems possible in lesson 6.4, and two products, which allows one product to be measured in order to determine another. The equation is a set of simultaneous proportions, not a recipe to be read from left to right. The coefficients form one consistent set of proportions, so any pair relates directly

9. A student calculates moles of product without balancing the equation first. Explain what happens and why the error is invisible in the answer.
   Show the full solution

   An unbalanced equation still has coefficients, implicitly all equal to one, so the student takes a one-to-one ratio and the calculation proceeds without any obstacle. The arithmetic is valid, the units cancel correctly and the answer emerges as a plausible mass in a plausible range, so nothing in the working signals a problem. The answer is nonetheless wrong by whatever factor the true coefficients would have supplied, which for \( 4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 \) would be a factor of two. The only defense is procedural: balance the equation and check every element before extracting any ratio, since no later step will catch the omission. An unbalanced equation supplies a one-to-one ratio that calculates cleanly and gives a plausible but wrong answer

10. Explain why the mole is indispensable to stoichiometry, referring to what a balance can and cannot measure.
    Show the full solution

    A balanced equation states a relationship between numbers of particles, and a balance measures mass, so the two are expressed in incompatible quantities and neither can be used directly to get the other. The mole resolves the incompatibility because it is defined as a count and yet has a known mass for every substance, which means it can be reached from either side. A mass is converted into moles by dividing by molar mass, the equation's ratio is applied to those moles because the coefficients are a particle ratio and therefore a mole ratio, and the resulting moles are converted back into a mass. Without that middle unit there is no route from what the equation says to what the laboratory can weigh, which is why every problem in this unit converts to moles first even when both the question and the answer are in grams. Equations relate particle counts and balances measure mass; the mole is the only quantity that is both a count and weighable

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State the three steps of a mass to mass calculation.
   Show the full solution

   Mass to moles, mole ratio, moles to mass

2. Which molar mass is used in the final step?
   Show the full solution

   That of the substance being asked for, not the given one

3. For \( 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \), find the mass of water from 10.0 g of hydrogen with oxygen in excess.
   Show the full solution

   \( 10.0 \div 2.016 = 4.960 \) mol H₂, ratio 2 to 2 so 4.960 mol H₂O, then \( \times 18.02 \). 89.4 g

4. For \( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \), find the mass of ammonia from 28.0 g of nitrogen with hydrogen in excess.
   Show the full solution

   \( 28.0 \div 28.02 = 0.9993 \) mol, \( \times 2 = 1.9986 \) mol NH₃, \( \times 17.03 \). 34.0 g

5. What free check is available when a problem asks for every product?
   Show the full solution

   The product masses must sum to the mass of reactant consumed

6. Explain why using the given substance's molar mass in the last step is a silent error.
   Show the full solution

   Because the calculation still produces a number in grams with all the units canceling correctly, so nothing in the setup reveals the mistake. The molar mass simply appears in the right place with the right units, and only its value is wrong. The size of the resulting error depends on how different the two molar masses are: in the limestone example, using 100.09 instead of 44.01 would report 50.0 g of carbon dioxide from 50.0 g of limestone, which happens to be exactly the starting mass and could easily be mistaken for a conservation check passing. The defense is to write the substance name next to each molar mass as it is used, so that a mismatch between the label and the step is visible. The units still cancel and the answer looks reasonable; only labeling each molar mass with its substance exposes it

7. Zinc reacts with hydrochloric acid: \( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \). Find the mass of hydrogen from 6.54 g of zinc.
   Show the full solution

   Zinc's molar mass is 65.38 g/mol, so \( 6.54 \div 65.38 = 0.1000 \) mol, a convenient round figure. The mole ratio from the coefficients is one hydrogen molecule per zinc atom, so 0.1000 mol of H₂ is produced. Hydrogen gas is diatomic with a molar mass of \( 2 \times 1.008 = 2.016 \) g/mol, and using 1.008 here would halve the answer. So \( 0.1000 \times 2.016 = 0.2016 \) g. The answer is small because hydrogen is the lightest substance there is, which is a useful sanity check on its magnitude. 0.202 g

8. Explain why the conservation check in the worked example is strong evidence rather than a coincidence.
   Show the full solution

   Because the two product masses were obtained by separate calculations that used different molar masses, 44.01 for carbon dioxide and 56.08 for calcium oxide, and there is no arithmetic reason for two independently wrong answers to sum to the starting mass. An error in either mole conversion, either molar mass or either ratio would shift one of the answers without shifting the other, and the sum would miss 50.0 g. The check is therefore testing the whole chain rather than repeating it. It would fail to catch one class of error, namely a mistake in the initial mole calculation, since that figure feeds both answers equally and would scale both by the same factor, which is why it is a strong check rather than a complete one. Two independent calculations with different molar masses have no reason to agree unless both are right

9. Magnesium burns in oxygen: \( 2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO} \). Find the mass of magnesium oxide from 2.43 g of magnesium, and relate the answer to lesson 1.1.
   Show the full solution

   Magnesium's molar mass is 24.31 g/mol, so \( 2.43 \div 24.31 = 0.1000 \) mol. The ratio is two to two, so one to one, giving 0.1000 mol of magnesium oxide. Its molar mass is \( 24.31 + 16.00 = 40.31 \) g/mol, so the mass is \( 0.1000 \times 40.31 = 4.031 \) g, or 4.03 g. This is precisely the worked example of lesson 1.1, where the apparent mass gain of 1.60 g was attributed to oxygen taken in from the air. The stoichiometry now supplies the number independently: 0.0500 mol of O₂ is consumed, which is \( 0.0500 \times 32.00 = 1.60 \) g, matching exactly. 4.03 g, and the 1.60 g gain is the mass of oxygen the calculation predicts

10. Explain why a chemist planning an industrial process would run this calculation before building anything.
    Show the full solution

    Because the calculation converts a required output into every input and byproduct, all of which have to be arranged for physically and financially before a plant exists. Working backward from a target tonnage of quicklime gives the tonnage of limestone that must be quarried and transported, which fixes the raw material cost and the size of the supply contract. It simultaneously gives the mass of carbon dioxide released, which determines the emissions permit required and, in many jurisdictions, a carbon cost. It also identifies which costs cannot be engineered away: the carbon dioxide from the limestone is fixed by the equation, so no improvement in the kiln's efficiency reduces it. Getting a factor of two wrong here means a plant sized wrongly, which is discovered only after it is built. It converts a target output into raw material needs, byproduct masses and emissions, all of which must be committed to in advance

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Name the four quantities that connect directly to moles.
   Show the full solution

   Mass, gas volume at STP, number of particles, and solution volume through molarity

2. Which step of the road map changes when the question asks for a gas volume rather than a mass?
   Show the full solution

   Only the final step; the mole ratio is unchanged

3. For \( 2\text{CO}(g) + \text{O}_2(g) \rightarrow 2\text{CO}_2(g) \), what volume of oxygen reacts with 10.0 L of carbon monoxide at the same conditions?
   Show the full solution

   Ratio one oxygen per two carbon monoxide. 5.00 L

4. When can the volume ratio shortcut be used?
   Show the full solution

   Only when both substances are gases at the same temperature and pressure

5. Find the volume at STP of 0.250 mol of any gas.
   Show the full solution

   \( 0.250 \times 22.4 \). 5.60 L

6. Zinc reacts with excess hydrochloric acid. Find the volume of hydrogen at STP from 6.54 g of zinc.
   Show the full solution

   The equation is \( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \). Zinc's molar mass is 65.38 g/mol, so \( 6.54 \div 65.38 = 0.1000 \) mol. The mole ratio is one hydrogen per zinc, giving 0.1000 mol of H₂. Converting to volume at STP: \( 0.1000 \times 22.4 = 2.24 \) L. Note that the volume shortcut is unavailable here because zinc is a solid, so the full road map is required, entering through mass and leaving through volume. 2.24 L

7. Explain why the volume ratio shortcut works for gases but would be badly wrong for solids.
   Show the full solution

   For gases, volume is set by the spacing between particles rather than by their size, and at a given temperature and pressure that spacing is the same for every gas, so volume is directly proportional to the number of moles with one universal constant. That proportionality is what allows moles to be skipped entirely: multiplying and dividing by the same constant cancels. For solids and liquids the particles are in contact, so the volume depends on how big the particles are and how efficiently they pack, and the volume per mole varies enormously between substances. Applying the shortcut to the limestone reaction would claim that a given volume of rock yields the same volume of carbon dioxide, when in fact 50 g of limestone occupies about 18 mL and produces 11.2 L of gas, an error of nearly a factor of six hundred. Gas volume per mole is universal because it comes from spacing; solid volume per mole depends on particle size and packing

8. A student uses 22.4 L/mol for a reaction carried out at 25 degrees Celsius. Explain the consequence and how to do it properly.
   Show the full solution

   The constant 22.4 L/mol holds only at 0 degrees Celsius and 1 atmosphere. At 25 degrees the particles have more kinetic energy and spread further apart at constant pressure, so a mole occupies about 24.5 L. Using 22.4 to convert a measured volume into moles therefore divides by a value that is too small and overstates the mole count by roughly nine percent, and that error carries into every subsequent quantity. The proper approach is to use the ideal gas law from lesson 8.4, \( PV = nRT \), with the actual temperature in kelvin, which handles any conditions rather than one special case. The molar volume shortcut is a special case of that equation evaluated at STP, and it should only be reached for when the question says STP. It overstates the moles by about nine percent; use \( PV = nRT \) with the actual temperature instead

9. For \( 2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l) \), explain why the volume ratio shortcut can be used for the two reactants but not to find the volume of water.
   Show the full solution

   Hydrogen and oxygen are both gases at the stated conditions, so their volumes are both proportional to moles with the same constant and the coefficient ratio of two to one transfers directly to volumes. Water in this equation is labeled as a liquid, and a liquid's volume per mole is governed by molecular size and packing rather than by the universal gas spacing, so no shared constant exists and the ratio does not carry across. Finding the volume of water requires the full road map: convert to moles of water, multiply by the molar mass of 18.02 g/mol to get a mass, then divide by the density of about 1.00 g/mL to get a volume. The state symbols in the equation are therefore not decoration; they determine which method is available. Both reactants are gases sharing one molar volume; liquid water does not, so its volume needs mass and density

10. Explain why the hub and spoke picture is a better way to learn this unit than memorizing separate procedures for mass, volume and particle problems.
    Show the full solution

    Because there is only one procedure, and treating it as several invites the student to memorize three sets of steps that differ in one place and to choose wrongly under pressure. Every problem in the unit does the same thing: convert whatever is given into moles, apply the mole ratio from the balanced equation, and convert the resulting moles into whatever is wanted. What varies is only which spoke is used at each end, and those are four independent conversions from unit 4 that were learned before stoichiometry began. Seeing it this way also makes mixed problems unremarkable, since entering by mass and leaving by volume is just a different pair of spokes, and it explains why the unit's genuinely new content is a single idea, the mole ratio, rather than a catalog of cases. All problems share one structure, with only the entry and exit conversions differing; mixed cases then need nothing new

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Define the limiting reactant.
   Show the full solution

   The reactant that runs out first and therefore limits how much product can form

2. State the method for identifying it.
   Show the full solution

   Convert both reactants to moles and divide each by its coefficient; the smallest result is limiting

3. Which reactant is used to calculate the product?
   Show the full solution

   The limiting reactant only

4. What signals that a question is a limiting reactant problem?
   Show the full solution

   Quantities are given for two or more reactants

5. Is the reactant with the smaller mass always limiting?
   Show the full solution

   No; molar masses and coefficients both differ, so mass alone decides nothing

6. Explain why dividing by the coefficient is necessary rather than just comparing moles.
   Show the full solution

   Because the equation does not require the reactants in equal amounts, so an equal number of moles does not mean an equal supply. Dividing each mole count by its coefficient converts both into the same unit of comparison, namely how many times the reaction as written could proceed on that reactant alone, and only then can the two be put side by side. In the ammonia synthesis a mixture containing two moles of nitrogen and three of hydrogen has more hydrogen by mole count, yet hydrogen is limiting: dividing gives two for nitrogen and one for hydrogen, and the reaction can run only once. Comparing raw moles would have picked the wrong reactant whenever the coefficients differ, which is most of the time. Coefficients set how much of each is needed per run, so dividing converts both to a comparable number of possible runs

7. For \( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \), 28.0 g of nitrogen is mixed with 6.00 g of hydrogen. Identify the limiting reactant and find the mass of ammonia.
   Show the full solution

   Convert both to moles: \( 28.0 \div 28.02 = 0.9993 \) mol of N₂ and \( 6.00 \div 2.016 = 2.976 \) mol of H₂. Divide each by its coefficient: nitrogen gives \( 0.9993 \div 1 = 0.9993 \) and hydrogen gives \( 2.976 \div 3 = 0.9921 \). Hydrogen is very slightly smaller, so hydrogen is limiting, and the mixture is close to the ideal ratio. Calculate from hydrogen only: \( 2.976 \times \frac{2}{3} = 1.984 \) mol of NH₃, and \( 1.984 \times 17.03 = 33.79 \) g. The closeness of the two figures shows why the method has to be applied rather than eyeballed. Hydrogen is limiting; 33.8 g of ammonia

8. Explain why calculating from the excess reactant gives an answer that is not merely wrong but physically impossible.
   Show the full solution

   Because it predicts an amount of product that would require more of the limiting reactant than exists. In the worked example, calculating from aluminum predicts 35.3 g of copper, which would need \( 0.5559 \) mol of copper(II) chloride, or 74.7 g, when only 50.0 g was supplied. The prediction therefore asserts that atoms appeared from nowhere, which conservation of mass forbids. This is a useful way to check an answer: take the predicted product back through the equation to the reactant not used in the calculation, and confirm that the amount required is actually available. A limiting reactant answer always passes this test and an excess reactant answer never does. It requires more of the limiting reactant than was supplied, so it violates conservation of mass

9. A mixture contains 4.00 mol of A and 4.00 mol of B for the reaction \( \text{A} + 2\text{B} \rightarrow \text{C} \). Identify the limiting reactant and explain the result in words.
   Show the full solution

   Divide each by its coefficient: A gives \( 4.00 \div 1 = 4.00 \) and B gives \( 4.00 \div 2 = 2.00 \). B is smaller, so B is limiting despite the two being present in identical amounts. In words, each run of the reaction consumes one A and two B, so the four moles of A could support four runs while the four moles of B can support only two. After two runs the B is exhausted, and 2.00 mol of A remains untouched, which is half of what was supplied. The example shows clearly that equal mole quantities do not imply a balanced mixture; the mixture is balanced only when the mole ratio matches the coefficient ratio, which here would require twice as much B as A. B is limiting: it supports two runs against A's four, leaving 2.00 mol of A unreacted

10. Explain why industrial processes often deliberately supply one reactant in excess, given that the excess is not consumed.
    Show the full solution

    Because the goal is usually to consume the expensive or difficult reactant completely rather than to avoid leftovers. Supplying a cheap reactant in excess drives the limiting reactant as close to full conversion as possible, which raises the yield relative to the costly input and reduces the amount of unreacted valuable material that has to be separated and recycled. The excess is often easy to recover and return to the process, as unreacted nitrogen and hydrogen are in ammonia synthesis, so it is not wasted in any real sense. There is also a rate consideration from unit 10: a higher concentration of one reactant speeds the reaction up, so an excess can shorten the time a batch occupies expensive equipment. The excess is chosen by cost, not by the equation. An excess of the cheap reactant maximizes conversion of the expensive one, and is usually recovered and recycled

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. How is the mass of excess remaining found?
   Show the full solution

   Mass supplied minus mass consumed

2. Which reactant is used to calculate the amount consumed?
   Show the full solution

   The limiting reactant, through the mole ratio between the two reactants

3. What must be done before subtracting?
   Show the full solution

   Convert the consumed moles to a mass, so both quantities are in the same unit

4. State the check on the subtraction.
   Show the full solution

   Consumed plus remaining must equal supplied

5. What does a full mass balance compare?
   Show the full solution

   The total mass supplied against the products plus the leftover excess

6. Explain why the mole ratio must be applied before the subtraction, rather than after.
   Show the full solution

   Because the two reactants are consumed in the proportion the equation specifies, not one for one, so the moles of limiting reactant are not the moles of excess reactant used up. In the worked example, 0.3719 mol of copper(II) chloride consumes only 0.2479 mol of aluminum, because the equation requires two aluminums for every three copper(II) chlorides. Subtracting 0.3719 from the supplied 0.3706 mol of aluminum would give a negative number and suggest aluminum was limiting, contradicting the analysis already done. The ratio converts a quantity of one substance into the corresponding quantity of another, and until that conversion is made the two numbers refer to different things and cannot be combined. The reactants are consumed in the coefficient ratio, so the limiting moles must be converted before they can be subtracted

7. Explain why subtracting moles from grams produces a meaningless result.
   Show the full solution

   Because addition and subtraction require the quantities to be the same kind of thing, and moles and grams are not. Grams measure mass and moles count particles, and the conversion between them differs for every substance, so there is no fixed relationship that would make the subtraction sensible. The result carries no unit that can be written down, since gram minus mole is not a quantity, and it cannot be interpreted as either a mass or a count. This is the same requirement that governs significant figures in lesson 1.5, where the addition rule applies to decimal places precisely because the quantities being added share a unit, and it is worth treating any subtraction in a chemistry problem as a prompt to check the units match. Addition requires matching units; moles and grams are different quantities with a substance-dependent conversion

8. Explain why the full mass balance verifies the entire problem rather than just the last step.
   Show the full solution

   Because every quantity in the problem appears in it. The masses of both products came from the limiting reactant through two different mole ratios and two different molar masses, and the leftover excess came from a third ratio and a subtraction, so the balance simultaneously tests the identification of the limiting reactant, three mole ratios, four molar masses and the subtraction. An error anywhere would shift one term without shifting the others and the total would miss 60.0 g. The check is powerful precisely because the terms were calculated independently, and it rests on conservation of mass from lesson 1.1, which guarantees that whatever was put into the sealed flask is still there in some form afterward. Every calculated quantity appears in the balance, so an error in any of them breaks the total

9. In the worked example, suppose 60.0 g of copper(II) chloride had been used instead. Determine what changes.
   Show the full solution

   Recheck the limiting reactant first, since changing a quantity can change which one it is. Now \( 60.0 \div 134.45 = 0.4463 \) mol of copper(II) chloride, and dividing by the coefficient gives \( 0.4463 \div 3 = 0.1488 \). Aluminum is unchanged at \( 0.3706 \div 2 = 0.1853 \). Copper(II) chloride is still the smaller, so it remains limiting, and every product simply scales up: copper becomes \( 0.4463 \times 63.55 = 28.4 \) g. The aluminum consumed rises to \( 0.4463 \times \frac{2}{3} = 0.2975 \) mol, or 8.03 g, so the leftover aluminum falls to \( 10.0 - 8.03 = 1.97 \) g. Adding more of the limiting reactant increases the product and reduces the excess, as expected. Copper(II) chloride is still limiting; copper rises to 28.4 g and leftover aluminum falls to 1.97 g

10. Explain why knowing the leftover excess matters in a real laboratory or plant, given that it does not affect the yield.
    Show the full solution

    Because the leftover is physically present in the vessel mixed with the products, and everything downstream has to deal with it. It sets the purification problem: the product cannot be used until the unreacted material is removed, and the choice of method depends on how much there is and what state it is in. It sets the recovery question, since a valuable excess is worth separating and returning to the process while a cheap one may not be. It sets a disposal question if the excess is hazardous, because the waste stream's composition must be known before it can be treated or discharged legally. It also affects the apparent yield if the product is weighed without purification, since contamination by leftover reactant inflates the mass and can produce a percent yield above one hundred, which lesson 6.6 treats as a diagnostic. It determines purification, recovery, waste handling, and whether an unpurified product mass can be trusted

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Define theoretical yield.
   Show the full solution

   The mass of product predicted by stoichiometry from the limiting reactant, assuming complete conversion

2. Write the formula for percent yield.
   Show the full solution

   Actual divided by theoretical, times 100

3. An experiment predicts 8.00 g and obtains 6.40 g. Find the percent yield.
   Show the full solution

   \( \frac{6.40}{8.00} \times 100 \). 80.0 percent

4. Which reactant is used to find the theoretical yield?
   Show the full solution

   The limiting reactant

5. What does a percent yield above 100 indicate?
   Show the full solution

   An error, most often a product weighed before it was fully dry

6. Give three physical and two chemical reasons a yield falls below 100 percent.
   Show the full solution

   Physical: product is lost during transfer between vessels, since some always adheres to glassware; product passes through or remains on the filter paper during separation; and some product stays dissolved in the solution rather than precipitating, since no solid is completely insoluble. Chemical: a side reaction consumes some of the limiting reactant into a different product, so those atoms never reach the intended one; and the starting material may be impure, so the mass weighed out contained less reactant than assumed and the true theoretical yield was lower than calculated. A sixth cause applies to reversible reactions, where equilibrium is reached before conversion is complete, which unit 10 develops. Transfer losses, filtration losses and residual solubility; side reactions and impure reactants

7. Explain why a shortfall in yield does not contradict conservation of mass.
   Show the full solution

   Because conservation of mass says atoms are neither created nor destroyed in a closed system, not that all of them end up in the vessel where the chemist is looking. A yield of 89.7 percent means that about ten percent of the copper atoms are somewhere other than the weighing boat: adhering to the beaker, trapped in the filter paper, clinging to the unreacted foil, or still dissolved in the discarded filtrate. Every one of them could in principle be recovered and weighed, and doing so would close the balance exactly. The apparatus is an open system, so the shortfall is a bookkeeping statement about where the product went, which is exactly the reasoning pattern of lesson 1.1. The missing atoms are elsewhere in the apparatus, not destroyed; the system is open

8. A student's theoretical yield is calculated from the excess reactant. Describe the effect on the percent yield and how you would spot it.
   Show the full solution

   Calculating from the excess reactant always overstates the theoretical yield, because the excess could support more reaction than the limiting reactant permits. Since the theoretical yield is the denominator of the percent yield, an inflated denominator produces a percent yield that is too low, and in the worked example using aluminum would give \( \frac{21.2}{35.3} \times 100 = 60.1 \) percent instead of 89.7. The symptom is a percent yield that is surprisingly poor for a reaction with no obvious difficulty, and the diagnostic is to check whether the limiting reactant was identified at all. A second check is to take the theoretical yield back through the equation and confirm the required amount of every reactant was actually available, which the inflated figure fails. The theoretical yield is too large, so the percent yield is too small; check that the limiting reactant was identified

9. A reaction has a theoretical yield of 45.0 g and a percent yield of 72.0 percent. Find the actual yield, and explain how to rearrange the formula reliably.
   Show the full solution

   Start from the definition rather than a memorized rearrangement: percent yield equals actual divided by theoretical times 100, so \( 72.0 = \frac{\text{actual}}{45.0} \times 100 \). Divide both sides by 100 to get the fraction, \( 0.720 = \frac{\text{actual}}{45.0} \), then multiply both sides by 45.0 to get \( \text{actual} = 0.720 \times 45.0 = 32.4 \) g. The reliable habit is to convert the percentage to a decimal first and then read the remaining equation as a simple proportion, which avoids the common slip of multiplying by 100 in the wrong place. The sanity check is that the actual yield must be smaller than the theoretical yield whenever the percentage is below one hundred, and 32.4 is indeed less than 45.0. 32.4 g

10. Explain why a chemical process with a yield of 40 percent might be preferred over one with a yield of 90 percent.
    Show the full solution

    Because percent yield measures only the efficiency of conversion and says nothing about cost, safety, speed or what else is produced. The forty percent route might start from a raw material that is a tenth of the price, so that the wasted sixty percent costs less than the ninety percent route's input. It might run at room temperature and atmospheric pressure while the high-yield route requires high temperature and pressure, which means expensive equipment and large energy costs. It might avoid a toxic reagent or a hazardous intermediate, reducing the cost of handling and waste treatment. It might produce a purer product that needs no further separation, or produce a valuable byproduct. It might also simply be much faster, so more batches run in the same plant. Yield is one input to the decision, not the decision. Yield ignores cost, energy, safety, speed and byproducts, any of which can dominate the choice

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. What happens to each carbon atom in a fuel during complete combustion?
   Show the full solution

   It becomes one molecule of carbon dioxide

2. Roughly what mass of carbon dioxide comes from 1 kg of a typical hydrocarbon fuel?
   Show the full solution

   About 3 kg

3. Where does the extra mass come from?
   Show the full solution

   Oxygen taken from the atmosphere

4. Give the molar mass of octane, C₈H₁₈.
   Show the full solution

   \( 8 \times 12.01 + 18 \times 1.008 \). 114.22 g/mol

5. Name one source of carbon dioxide in industry that is not combustion.
   Show the full solution

   Decomposition of limestone in cement manufacture

6. Find the mass of carbon dioxide released by burning 1.00 kg of methane, CH₄, and compare it with octane.
   Show the full solution

   The equation is \( \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \). Methane's molar mass is \( 12.01 + 4 \times 1.008 = 16.04 \) g/mol, so \( 1000 \div 16.04 = 62.34 \) mol. The ratio is one to one, giving 62.34 mol of carbon dioxide, and \( 62.34 \times 44.01 = 2744 \) g, or 2.74 kg. That is less than octane's 3.08 kg from the same fuel mass, because methane has a higher proportion of hydrogen and hydrogen burns to water rather than carbon dioxide. This is part of why natural gas is described as a lower-carbon fossil fuel, though a full comparison would also have to account for the energy released per kilogram. 2.74 kg, less than octane's 3.08 kg because methane is richer in hydrogen

7. Explain why improving an engine's efficiency reduces fuel consumption but not the carbon dioxide released per kilogram of fuel burned.
   Show the full solution

   Because efficiency and stoichiometry answer different questions. Efficiency describes how much of the energy released by the reaction is converted into useful work rather than lost as waste heat, so a more efficient engine travels further on each kilogram of fuel and therefore burns fewer kilograms for a given journey. The carbon dioxide produced per kilogram burned is fixed by conservation of atoms: every carbon atom in the fuel must leave as carbon dioxide in complete combustion, and no arrangement of pistons changes that. So efficiency reduces total emissions by reducing the amount of fuel burned, which is a real and substantial effect, while leaving the emission factor per kilogram exactly where it was. The only ways to change the factor itself are to burn a fuel with a different carbon to hydrogen ratio or not to burn one at all. Efficiency reduces how much fuel is burned; the carbon per kilogram is fixed by the fuel's formula

8. Explain why cement manufacture is described as having emissions that are hard to eliminate, referring to lesson 6.2.
   Show the full solution

   Cement production involves two separate carbon dioxide sources. The kiln has to be heated to around fourteen hundred degrees Celsius, which is normally done by burning fuel and produces combustion emissions, and those could in principle be eliminated by heating with electricity from a non-combustion source. The second source is the chemistry itself: the process requires calcium oxide, which is made by decomposing calcium carbonate, and lesson 6.2 showed that producing 28.0 kg of quicklime necessarily releases 22.0 kg of carbon dioxide from the limestone. Those process emissions come from the raw material rather than from any fuel, so they persist no matter how the kiln is heated. Removing them requires either capturing the gas or finding a different chemistry for cement, both of which are much harder problems than changing a heat source. The carbon dioxide comes from decomposing the limestone itself, so changing the fuel or heat source cannot remove it

9. A student says producing 3.08 kg of carbon dioxide from 1.00 kg of fuel violates conservation of mass. Refute this with a full mass balance.
   Show the full solution

   It violates nothing, because the fuel is not the only reactant. Burning 1.00 kg of octane is 8.755 mol, and the equation requires 12.5 moles of oxygen per mole of octane, so \( 8.755 \times 12.5 = 109.4 \) mol of O₂ is consumed, which at 32.00 g/mol is 3502 g, or 3.50 kg. The total reactant mass is therefore \( 1.00 + 3.50 = 4.50 \) kg. On the product side, the carbon dioxide is 3.08 kg and the water is \( 8.755 \times 9 = 78.80 \) mol at 18.02 g/mol, which is 1420 g or 1.42 kg. The products total \( 3.08 + 1.42 = 4.50 \) kg, matching exactly. The oxygen simply arrived from the atmosphere rather than from the tank, which is why the fuel mass alone does not bound the product mass. 3.50 kg of oxygen is consumed from the air, so 4.50 kg of reactants give 4.50 kg of products

10. Explain what this calculation does and does not establish about what should be done regarding fuel use.
    Show the full solution

    It establishes several things firmly. Burning a given mass of a given fuel releases a fixed and calculable mass of carbon dioxide; that mass is about three times the fuel mass for a hydrocarbon; the figure is determined by the fuel's formula and cannot be reduced by burning it more cleanly or efficiently per kilogram; and different fuels have different emission factors that can be compared quantitatively. Those are measurable facts and they constrain any sensible discussion. What the calculation cannot supply is what anyone should do, because that requires weighing the benefits obtained from the energy against the costs of the emissions, comparing alternatives that carry their own costs and risks, and deciding whose interests count and over what time horizon. None of those is a measurement, and lesson 1.1 established that science informs such decisions without making them. A student who presents the number as settling the policy has overstepped exactly the boundary this course opened with. It fixes the quantity of emissions and rules out some proposed remedies, but weighing costs against benefits is not a measurable question

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Define temperature in terms of particles.
   Show the full solution

   The average kinetic energy of the particles in a substance

2. Define heat.
   Show the full solution

   Energy transferred between objects because of a temperature difference

3. Convert 25 degrees Celsius to kelvin.
   Show the full solution

   \( 25 + 273 \). 298 K

4. In which direction does heat always flow?
   Show the full solution

   From higher temperature to lower temperature

5. Classify temperature and heat as intensive or extensive.
   Show the full solution

   Temperature is intensive; heat is extensive

6. Explain why two beakers of water at the same temperature, one holding 50 g and one holding 500 g, differ in how much energy they can transfer.
   Show the full solution

   Temperature reports the average kinetic energy of each particle, so at the same temperature the particles in both beakers are moving with the same average energy and neither is hotter than the other. Total thermal energy, however, is that average multiplied by the number of particles, and the larger beaker contains ten times as many. Cooling both to room temperature therefore releases ten times as much energy from the 500 g beaker. The larger beaker also stays hot far longer, because removing a given quantity of energy lowers its temperature by only a tenth as much. This is the intensive against extensive distinction from lesson 1.2 applied to energy. Same average energy per particle, but ten times as many particles, so ten times the transferable energy

7. Explain why it is wrong to say an object "contains heat".
   Show the full solution

   Because heat is defined as energy in transit, not as a property an object holds. What an object possesses is internal energy, which is the total kinetic and potential energy of its particles, and that quantity exists whether or not anything is happening. Heat is the name for internal energy crossing a boundary because of a temperature difference, so it exists only during a transfer and stops existing when equilibrium is reached. The distinction matters because saying an object contains heat suggests a fixed store that could be identified and measured inside it, when in fact the same object transfers different amounts depending entirely on what it is placed in contact with. The parallel is that a bank account contains money but not payments. Heat is energy in transit; what an object holds is internal energy

8. Explain why the kelvin scale rather than Celsius is required whenever temperature appears in a formula.
   Show the full solution

   Because the kelvin scale is proportional to the average kinetic energy of the particles, starting from zero where that energy is at its minimum, while the Celsius scale has its zero at an arbitrary point, the freezing temperature of water. Ratios are meaningless on a scale with an arbitrary zero: 40 degrees Celsius is not twice as energetic as 20 degrees Celsius, whereas 400 K genuinely does correspond to twice the average kinetic energy of 200 K. Any formula that multiplies or divides by temperature is therefore using a ratio and requires kelvin, which is why the gas laws of unit 8 fail dramatically if Celsius is substituted, and why a Celsius value that happens to be negative would produce an impossible negative volume. Kelvin is proportional to kinetic energy from a true zero, so ratios are meaningful; Celsius has an arbitrary zero

9. A student says a bucket of ice at 0 degrees Celsius has no thermal energy. Correct this.
   Show the full solution

   Zero degrees Celsius is not zero energy; it is the freezing temperature of water, which is an arbitrary reference point on that scale and corresponds to 273 K. The particles in ice at that temperature are still vibrating vigorously about their lattice positions and possess a substantial quantity of kinetic energy, which is why ice can melt and why it will warm something colder than itself placed against it. The temperature at which particle motion reaches its minimum is absolute zero, which is 0 K or about minus 273 degrees Celsius, and even there a residual zero-point energy remains. The error is treating a scale's zero as though it marked the absence of the quantity being measured. 0 degrees Celsius is 273 K, an arbitrary reference; particle motion and energy continue

10. Explain what happens at the particle level when a hot metal block is placed in cold water, and why the process stops.
    Show the full solution

    At the surface where the two touch, the metal's particles are vibrating with a much higher average kinetic energy than the water molecules against them. Collisions at that boundary transfer energy from the faster particles to the slower ones, on average, so the metal's particles lose kinetic energy and the water's gain it. That transfer is heat, and it is not directed by anything; it is the statistical result of energetic particles colliding with less energetic ones. As the exchange continues the metal cools and the water warms, so the difference in average energies shrinks. When the two averages become equal, collisions still occur and energy still crosses the boundary in both directions, but at equal rates, so there is no net transfer. That is thermal equilibrium, and it is a dynamic balance rather than a cessation of motion, which is the same structure as the chemical equilibrium of lesson 10.5. Collisions transfer energy from faster to slower particles until the average energies match, after which transfer continues equally in both directions

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Write the equation relating heat, mass, specific heat capacity and temperature change.
   Show the full solution

   \( q = mc\Delta T \)

2. State the specific heat capacity of water with its units.
   Show the full solution

   4.18 J/(g·°C)

3. Find the energy needed to heat 500.0 g of water by 25.0 degrees Celsius.
   Show the full solution

   \( 500.0 \times 4.18 \times 25.0 \). 52 250 J, or 52.2 kJ

4. How is \( \Delta T \) calculated?
   Show the full solution

   Final temperature minus initial temperature

5. What does a negative value of \( q \) mean?
   Show the full solution

   The substance cooled, so energy was released

6. Find the temperature rise when 5000 J is supplied to 200.0 g of aluminum, specific heat capacity 0.897 J/(g·°C).
   Show the full solution

   Rearrange to \( \Delta T = \frac{q}{mc} \), which is the form to derive rather than memorize, since the original equation is easy to recall and dividing both sides by \( mc \) takes a moment. Substituting, \( \Delta T = \frac{5000}{200.0 \times 0.897} = \frac{5000}{179.4} = 27.87 \). To three significant figures the rise is 27.9 degrees Celsius. A useful check is that the same energy into 200.0 g of water would give only \( \frac{5000}{836} = 5.98 \) degrees, about a fifth as much, which is consistent with aluminum's specific heat capacity being about a fifth of water's. 27.9 degrees Celsius

7. Explain why water's specific heat capacity is so much higher than that of a metal, referring to lesson 3.5.
   Show the full solution

   Temperature rises only when the energy supplied increases the average kinetic energy of the particles, so anything else that energy does suppresses the temperature rise. In a metal there is very little else available: the atoms sit in a lattice with delocalized electrons and energy supplied goes almost entirely into vibration, so a small input produces a large temperature rise. In water, molecules are held to one another by hydrogen bonds, which are comparatively strong intermolecular attractions, and a substantial fraction of the energy supplied goes into disrupting and rearranging those bonds rather than into speeding molecules up. That energy raises potential energy instead of kinetic energy, so it does not register as temperature, and far more joules are required per degree. Much of the energy goes into disrupting hydrogen bonds, raising potential rather than kinetic energy

8. A car radiator uses water rather than oil, which has a specific heat capacity of about 2 J/(g·°C). Explain the choice and one drawback.
   Show the full solution

   A radiator's job is to carry energy away from the engine, so the coolant should absorb as much energy as possible for a given temperature rise and a given mass circulated. Water's specific heat capacity of 4.18 is roughly twice that of oil, so the same mass of water removes about twice the energy for the same temperature rise, allowing a smaller pump and a smaller radiator. Water is also cheap, non-flammable and non-toxic. The drawbacks follow from its other properties: it freezes at 0 degrees Celsius and expands as it does so, as lesson 1.7 explained, which can crack an engine block, so antifreeze must be added. It also boils at 100 degrees Celsius unless the system is pressurized, and it promotes corrosion of metal parts. About twice the energy per gram per degree; but it freezes and expands, and boils at 100 degrees unless pressurized

9. Explain why a sea breeze blows from the sea toward the land during the day and often reverses at night.
   Show the full solution

   During the day the sun delivers roughly equal energy per square meter to land and sea, but the land has a much lower specific heat capacity, so its surface warms several times faster. Air in contact with the warm land is heated, expands, becomes less dense and rises, which lowers the pressure there, and cooler denser air from over the sea flows in to replace it. That inflow is the sea breeze. At night the process reverses because the land also loses its energy much faster, so the surface cools below the sea's temperature within a few hours while the sea, holding an enormous store of energy, has barely changed. Now the air over the sea is the warmer and rising one, and the breeze blows from land to sea. The same specific heat difference drives both directions. Land warms and cools much faster than water, so the rising air is over the land by day and over the sea at night

10. Explain why scientists studying the energy balance of the climate regard ocean heat content as a more reliable measure than surface air temperature.
    Show the full solution

    Because the ocean is where nearly all of the retained energy actually goes. It covers about seventy percent of the surface, extends kilometers deep and has the highest specific heat capacity of any common substance, so its thermal mass dwarfs that of the atmosphere by a factor of roughly a thousand. An energy imbalance therefore shows up in ocean heat content as a steady accumulation, whereas surface air temperature is a measurement on a thin, low-capacity layer that is stirred by weather and by exchanges with the ocean itself. Air temperature consequently fluctuates by large amounts from year to year for reasons that involve moving energy around rather than adding it, as when a warm ocean current releases stored energy to the atmosphere. Measuring the reservoir rather than the surface gives a signal with far less noise relative to the trend, which is the same reasoning as preferring a mean with a stated spread in lesson 1.5. The ocean holds the overwhelming majority of the energy, so it integrates the imbalance while air temperature fluctuates with exchanges

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State the central assumption of calorimetry.
   Show the full solution

   The energy lost by one substance equals the energy gained by the other

2. What is a coffee-cup calorimeter made from?
   Show the full solution

   Two nested polystyrene cups with a lid and a thermometer

3. At the end of a calorimetry experiment, how do the two temperatures compare?
   Show the full solution

   They are equal, at thermal equilibrium

4. Find the specific heat capacity of a 50.0 g metal that releases 460 J while cooling by 20.0 degrees Celsius.
   Show the full solution

   \( c = \frac{460}{50.0 \times 20.0} \). 0.460 J/(g·°C)

5. In which direction does a simple calorimeter's error usually run?
   Show the full solution

   Energy is lost to the surroundings, so measured values come out low in magnitude

6. Explain why the metal and the water have very different temperature changes even though the energy transferred is the same.
   Show the full solution

   Because temperature change is energy divided by the product of mass and specific heat capacity, and both of those factors differ between the two substances. In the worked example the water has twice the mass of the metal and about nine and a half times its specific heat capacity, so the product \( mc \) is roughly nineteen times larger for the water. The same 1630 J therefore produces a temperature change about nineteen times smaller in the water, which is why the water rose by 3.9 degrees while the metal fell by 74.1. The quantity \( mc \), sometimes called the heat capacity of the object, is what determines how much a given energy transfer moves the temperature. The same energy divided by very different values of \( mc \) gives very different temperature changes

7. A student uses the water's temperature change for the metal as well. Explain what happens to the answer.
   Show the full solution

   Using 3.9 instead of 74.1 for the metal's temperature change divides by a number about nineteen times too small, so the calculated specific heat capacity comes out about nineteen times too large: \( \frac{1630}{50.0 \times 3.9} = 8.36 \) J/(g·°C). That value is impossible for a metal, being twice water's and far above any known substance, so the error is detectable by inspection against the reference table, where no candidate is within an order of magnitude. The underlying mistake is conceptual rather than arithmetic: the two substances share a final temperature but not a temperature change, because they started from very different points. It gives about 8.4 J/(g·°C), roughly nineteen times too large and impossible for a metal

8. Explain why the calorimeter's error has a predictable direction rather than being random.
   Show the full solution

   Because the physical processes that violate the assumption all run the same way. The assumption is that every joule leaving the metal enters the water, and the ways it fails are conduction through the cup walls and lid, energy absorbed by the cup and the thermometer, and loss to the room air during transfer and while equilibrium is reached. There is no corresponding mechanism by which extra energy enters the water, since the room is cooler than the system for most of the experiment. Every failure therefore reduces the water's temperature rise, reduces the calculated \( q \), and reduces the calculated specific heat capacity. This makes the error systematic in the sense of lesson 1.5 rather than random: repeating the experiment gives the same low bias each time, so averaging more trials improves precision without improving accuracy. Every loss mechanism removes energy from the water and none adds any, so the bias is systematic and always low

9. A neutralization reaction in a coffee-cup calorimeter raises 100.0 g of solution by 6.8 degrees Celsius. Find the energy released, and state two assumptions you made.
   Show the full solution

   Treating the solution as water, \( q = 100.0 \times 4.18 \times 6.8 = 2842 \) J, which to two significant figures is about 2.8 kJ released by the reaction. The first assumption is that the dilute solution has the same specific heat capacity and density as pure water, which is reasonable for dilute solutions but introduces a small error. The second is that no energy was absorbed by the cup, lid or thermometer and none escaped to the room, which is the standard calorimetry assumption and means the true value is somewhat larger than 2.8 kJ. A third, implicit in taking the mass as 100.0 g, is that the mass of the combined solutions is the sum of the masses mixed. About 2.8 kJ released, assuming the solution behaves as water and that no energy was lost to the calorimeter or surroundings

10. Explain why precise measurements use a bomb calorimeter whose heat capacity is calibrated in advance.
    Show the full solution

    Because calibration removes the need for the assumption that fails. A simple calorimeter assumes the apparatus absorbs no energy, which is false, and the size of the error is unknown. A bomb calorimeter is instead characterized in advance by releasing a precisely known quantity of energy inside it, usually by burning a standard substance, and measuring the resulting temperature rise. That gives the heat capacity of the entire assembly, including the vessel, the water and the thermometer, as a single calibrated number in joules per degree. Subsequent measurements multiply the observed temperature rise by that constant, so the energy absorbed by the apparatus is accounted for rather than ignored. The sealed steel vessel also allows combustion at constant volume under high oxygen pressure, so the reaction goes to completion and no gaseous products escape. Calibrating the whole apparatus accounts for the energy it absorbs instead of assuming it absorbs none

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Is breaking a bond exothermic or endothermic?
   Show the full solution

   Endothermic; energy must be supplied

2. Write the bond energy formula for enthalpy change.
   Show the full solution

   Bonds broken minus bonds formed

3. What does a larger bond energy value indicate?
   Show the full solution

   A stronger bond, requiring more energy to break

4. A reaction breaks 1200 kJ of bonds and forms 1500 kJ. Find \( \Delta H \) and classify it.
   Show the full solution

   \( 1200 - 1500 = -300 \). -300 kJ, exothermic

5. Why is the bond energy method only an estimate?
   Show the full solution

   The values are averages across many compounds and apply only to gas phase species

6. Estimate \( \Delta H \) for \( \text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl} \), given H-H 436, Cl-Cl 243 and H-Cl 431 kJ/mol.
   Show the full solution

   Bonds broken: one H-H at 436 and one Cl-Cl at 243, a total of 679 kJ. Bonds formed: two H-Cl bonds, since the equation produces two molecules each with one bond, giving \( 2 \times 431 = 862 \) kJ. Then \( \Delta H = 679 - 862 = -183 \) kJ/mol, so the reaction is exothermic. The accepted value is about -184.6 kJ/mol, so the estimate is within one percent. Note that the coefficient of 2 in the equation had to be applied to the H-Cl bonds; forgetting it would give \( 679 - 431 = +248 \) kJ and reverse the classification entirely. -183 kJ/mol, exothermic

7. Explain why combustion reactions are reliably exothermic, in terms of specific bonds.
   Show the full solution

   Because the bonds that oxygen forms in the products are unusually strong compared with the bonds present in the fuel and in oxygen itself. A hydrocarbon contains C-H bonds at about 413 kJ/mol and C-C bonds at about 347, and molecular oxygen contains an O=O bond at 498, all moderate values. The products contain C=O bonds at about 799 kJ/mol and O-H bonds at about 463, and every carbon ends up with two C=O bonds while every pair of hydrogens ends up in a water with two O-H bonds. The total released on forming those product bonds therefore exceeds the total required to break the reactant bonds, generally by a wide margin, and the surplus appears as heat. The pattern holds across hydrocarbons because the bonds involved are the same ones in every case. C=O and O-H bonds in the products are much stronger than the C-H, C-C and O=O bonds broken

8. Explain why the activation energy of a reaction is consistent with it being exothermic overall.
   Show the full solution

   Because the two quantities describe different stages of the same process. Before any new bond can form, existing bonds must be broken or at least substantially stretched, and that requires an energy input regardless of what happens afterward. Activation energy is the size of that initial investment, the barrier that must be climbed before the reaction can proceed. The overall enthalpy change compares the starting and finishing states only, and if the product bonds are stronger, the system ends lower in energy than it began and the reaction is exothermic. A reaction can therefore need a spark to start and then release far more than the spark supplied, which is exactly what happens when a match lights a gas burner. Lesson 7.5 draws this as an energy diagram and lesson 10.2 uses the barrier to explain reaction rates. Activation energy is the initial cost of breaking bonds; the enthalpy change compares only the start and end states

9. Nitrogen gas is very unreactive despite nitrogen being an essential element. Explain using bond energy, given that the nitrogen triple bond is about 941 kJ/mol.
   Show the full solution

   At 941 kJ/mol the triple bond in N₂ is among the strongest bonds known, roughly twice the O=O bond and more than twice a typical single bond. Any reaction of nitrogen gas must begin by breaking or weakening that bond, so the energy barrier is enormous and the reaction is extremely slow at ordinary temperatures even when the overall enthalpy change is favorable. The consequence is that the atmosphere is seventy-eight percent nitrogen which is chemically almost inert, and that converting it into usable compounds requires either the high temperature and pressure of the industrial ammonia process with a catalyst, the energy of a lightning strike, or the specialized enzymes of nitrogen-fixing bacteria. A very strong bond makes a substance unreactive without making it low in energy. The triple bond is exceptionally strong, so the barrier to any reaction is very large and nitrogen is kinetically inert

10. Explain why a bond energy calculation cannot account for the energy released when the water produced in combustion condenses.
    Show the full solution

    Because bond energies describe only the making and breaking of covalent bonds within molecules, and condensation breaks no bonds and forms none. When water vapor condenses to liquid, the H-O covalent bonds inside each molecule are entirely unchanged; what happens is that hydrogen bonds form between separate molecules, which are intermolecular attractions of the kind in lesson 3.5. The bond energy method has no term for those, so it necessarily gives the enthalpy change with water as a gas. The difference is not small: condensing the water releases a further amount of energy, which is why the heat obtainable from burning methane is quoted as two different values depending on whether the exhaust water is allowed to condense, and why a condensing boiler is more efficient than a conventional one. Condensation forms intermolecular hydrogen bonds, not covalent bonds, so the method gives the gas-phase value only

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. What sign does \( \Delta H \) take for an exothermic reaction?
   Show the full solution

   Negative

2. Define the system and the surroundings.
   Show the full solution

   The system is the reaction itself; the surroundings are everything else

3. A flask feels cold during a reaction. Classify the reaction.
   Show the full solution

   Endothermic, with a positive \( \Delta H \)

4. On an energy diagram, where are the products for an exothermic reaction?
   Show the full solution

   Below the reactants

5. From where to where is activation energy measured?
   Show the full solution

   From the reactants up to the peak

6. Explain why a warm beaker indicates a negative \( \Delta H \), which seems backward.
   Show the full solution

   Because \( \Delta H \) reports the energy change of the system, and the beaker is part of the surroundings rather than part of the system. The warmth you feel is energy that has arrived in the surroundings, and by conservation of energy it must have come from somewhere, namely the reaction. The system therefore ended with less energy than it started with, which is a decrease, and a decrease is written with a negative sign. The convention only seems backward because the observation is made on the surroundings while the quantity is defined for the system, so the two always carry opposite signs. The reliable habit is to translate the observation into a sentence about the system before writing anything: "the surroundings warmed, so the system lost energy, therefore \( \Delta H \) is negative." \( \Delta H \) describes the system, which lost the energy the surroundings gained

7. Explain why an endothermic reaction can occur at all, given that systems tend toward lower energy.
   Show the full solution

   Because energy is not the only factor determining whether a change happens. Many endothermic processes proceed because they produce a large increase in disorder, and the tendency toward greater disorder can outweigh an unfavorable energy change, particularly at higher temperatures. Dissolving ammonium nitrate is the standard example: the process absorbs energy and cools the solution, yet it happens readily because an ordered crystal lattice becomes ions dispersed randomly through the solvent, which is a very large increase in disorder. The energy required is drawn from the thermal energy of the surroundings, which is why the solution cools. Full treatment of this requires the concept of entropy and free energy, which is beyond a first course, but the qualitative point matters: "lower energy is favored" is a tendency and not a law. An increase in disorder can outweigh an unfavorable energy change, with the energy drawn from the surroundings

8. Two reactions have the same \( \Delta H \) of -200 kJ/mol but very different activation energies. Describe how their diagrams and their behavior differ.
   Show the full solution

   Both diagrams start and finish at the same levels, with the products 200 kJ below the reactants, so the vertical distance across the diagram is identical and both release the same energy per mole. What differs is the height of the peak between them. The reaction with the low activation energy has a small hump, so a large fraction of collisions have enough energy to get over it and the reaction proceeds quickly at ordinary temperature. The reaction with the high activation energy has a tall hump, so very few collisions succeed and the reaction is extremely slow unless it is heated or a catalyst is provided. This is why energy released and speed of release are independent properties: petrol and wood both burn exothermically, but petrol vapor ignites from a spark while a log needs prolonged heating. Lesson 10.2 develops the collision picture behind this. Same start and end levels, different peak heights; the higher barrier makes the reaction much slower without changing the energy released

9. Photosynthesis is endothermic and respiration is exothermic. Explain the signs and the relationship between them.
   Show the full solution

   Photosynthesis converts carbon dioxide and water into glucose and oxygen, forming products whose bonds are collectively weaker than those of the reactants, so the system ends higher in energy and \( \Delta H \) is positive. The energy required is supplied as light, which is why the process stops in the dark. Respiration runs the same conversion in reverse, breaking down glucose with oxygen to give carbon dioxide and water, so the system ends lower in energy and \( \Delta H \) is negative, with the released energy captured by the organism. Because the two are reverse processes, their enthalpy changes are equal in magnitude and opposite in sign, which is an instance of the state function property that lesson 7.6 uses in Hess's law. Together they form the energy cycle that supports nearly all life, storing solar energy in chemical bonds and releasing it on demand. Photosynthesis is positive and respiration negative, equal in magnitude and opposite in sign because they are reverse processes

10. Explain why the same reaction can be reported with two different \( \Delta H \) values in different textbooks without either being wrong.
    Show the full solution

    Because \( \Delta H \) depends on how the equation is written and on the states of the substances, both of which must be specified for the value to mean anything. Doubling every coefficient doubles the enthalpy change, since twice as much reacts, so \( \text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O} \) and \( 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \) legitimately carry values differing by a factor of two. The state of the product matters too: producing liquid water releases more energy than producing steam, because condensation releases further energy as hydrogen bonds form, so the two values differ by that amount. This is why published values always accompany a specific balanced equation with state symbols, and why comparing two figures requires checking that both refer to the same equation. The value scales with the coefficients and depends on the physical states, so the equation and state symbols must be given with it

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State Hess's law.
   Show the full solution

   The total enthalpy change is the same whether a reaction occurs in one step or several

2. What happens to \( \Delta H \) when an equation is reversed?
   Show the full solution

   Its sign changes

3. What happens to \( \Delta H \) when an equation is doubled?
   Show the full solution

   It doubles

4. Define a state function.
   Show the full solution

   A quantity whose change depends only on the initial and final states, not on the route

5. A reaction has \( \Delta H = -250 \) kJ. Give \( \Delta H \) for the reverse reaction.
   Show the full solution

   +250 kJ

6. Explain why enthalpy being a state function is what makes Hess's law possible.
   Show the full solution

   A state function's change depends only on where the system starts and where it finishes, so any two routes between the same pair of states must give the same total. That is precisely what allows a real reaction to be replaced by an imaginary sequence of other reactions for the purpose of calculation: the sequence need not be one that anyone could actually carry out, as long as it begins and ends at the right places. If enthalpy were not a state function, the answer would depend on which route was chosen and the method would be meaningless. The physical guarantee comes from conservation of energy: if two routes gave different totals, running the cheaper one forward and the more expensive one backward would create energy from nothing. Any route between the same start and end gives the same total, so a convenient imaginary route may replace the real one

7. Given \( \text{A} \rightarrow \text{B} \) with \( \Delta H = +40 \) kJ and \( \text{B} \rightarrow \text{C} \) with \( \Delta H = -70 \) kJ, find \( \Delta H \) for \( \text{C} \rightarrow \text{A} \).
   Show the full solution

   First find \( \text{A} \rightarrow \text{C} \) by adding the two given steps as written, since B appears as a product of the first and a reactant of the second and cancels: \( \Delta H = +40 + (-70) = -30 \) kJ. The target runs the other way, from C to A, which is the reverse of what was just found, so the sign changes and \( \Delta H = +30 \) kJ. An alternative route reaches the same place: reverse both given steps individually, giving \( \text{C} \rightarrow \text{B} \) at +70 and \( \text{B} \rightarrow \text{A} \) at -40, then add to get +30. The two methods agreeing is itself an illustration of the state function property. +30 kJ

8. Explain why the method requires checking that unwanted substances cancel completely.
   Show the full solution

   Because the cancellation is the verification that the manipulation was correct. Reversing or scaling an equation is a choice, and there is no guarantee that a particular set of choices produces the target; the check is that adding the manipulated equations leaves exactly the target equation and nothing else. If a substance fails to cancel, or cancels only partially in a way the target does not contain, then the combination describes a different overall reaction and the enthalpy total belongs to that reaction rather than the one asked for. The arithmetic would still produce a number, and it would look entirely respectable, so the cancellation check is the only step that catches a wrong choice of manipulation. Complete cancellation to exactly the target equation is the only confirmation that the manipulations were the right ones

9. Explain why a student who reverses an equation but forgets the sign change will get an answer that looks reasonable.
   Show the full solution

   Because the error changes only one number in a sum of similar-sized numbers, and the result remains in a plausible range for an enthalpy change. In the worked example, forgetting the sign change gives \( -393.5 + (-283.0) = -676.5 \) kJ instead of -110.5, which is still negative, still an exothermic combustion value and still of a magnitude that chemical reactions commonly have, so nothing about it looks absurd. The check that catches it is physical rather than arithmetic: forming carbon monoxide cannot release more energy than forming carbon dioxide from the same carbon, because the carbon monoxide can still be burned further to release 283.0 kJ more. Any answer larger in magnitude than -393.5 is therefore impossible, and asking whether the result makes sense against the other values given is the habit that catches this. The wrong answer is still negative and of a normal magnitude; only a physical plausibility check detects it

10. Explain how Hess's law is used to obtain enthalpy values for reactions that have never been carried out.
    Show the full solution

    By building the target reaction out of standard enthalpies of formation, which are the enthalpy changes for forming one mole of each compound from its elements in their standard states. These have been measured or derived for thousands of compounds and tabulated. Any reaction can then be treated as an imaginary two-stage route: decompose all the reactants back into their elements, which costs the reverse of their formation enthalpies, then assemble the products from those elements, which releases their formation enthalpies. Adding gives the familiar rule that the enthalpy change equals the sum of the products' formation enthalpies minus the sum of the reactants'. Neither stage is a reaction anyone performs, and it does not matter, because enthalpy is a state function and only the endpoints count. This is how the energy change of a proposed industrial process can be evaluated before anyone attempts it. Tabulated formation enthalpies let any reaction be routed through the elements, since only the endpoints matter

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Name the three mechanisms of energy transfer.
   Show the full solution

   Conduction, convection and radiation

2. Which mechanism can cross empty space?
   Show the full solution

   Radiation

3. State the condition for a molecule to absorb infrared radiation.
   Show the full solution

   The vibration must change the molecule's dipole moment

4. Why are nitrogen and oxygen transparent to infrared?
   Show the full solution

   They are made of two identical atoms, so stretching never produces a dipole

5. Which reservoir holds most of the energy retained in the Earth system?
   Show the full solution

   The ocean

6. Explain how carbon dioxide can absorb infrared even though it is a nonpolar molecule.
   Show the full solution

   Because the absorption condition concerns the dipole moment during a vibration, not the dipole moment of the molecule sitting still. Carbon dioxide is linear and symmetrical at rest, so its two polar C=O bonds point in exactly opposite directions and cancel, giving zero net dipole. When the molecule bends, however, the two oxygens move to the same side and the cancellation fails, so the molecule becomes momentarily polar, and the same happens during an asymmetric stretch where one bond lengthens as the other shortens. Those vibrations therefore change the dipole moment and can interact with the oscillating electric field of infrared light. Only the symmetric stretch, which preserves the symmetry throughout, remains inactive. The resting shape tells you about the molecule's behavior in a solvent; the vibrational shapes tell you about its behavior in radiation. Bending and asymmetric stretching break the symmetry, creating a changing dipole even though the resting molecule has none

7. Explain why a gas present at 0.04 percent can matter when one at 78 percent does not.
   Show the full solution

   Because the contribution of each gas is its abundance multiplied by its capacity to absorb infrared, and nitrogen's capacity is exactly zero rather than merely small. Any quantity multiplied by zero is zero, so seventy-eight percent of the atmosphere contributes nothing whatsoever to the greenhouse effect however much of it there is, and the same applies to oxygen and argon. The entire effect therefore comes from the remaining fraction of a percent, within which water vapor and carbon dioxide are the principal absorbers. This also means that proportional reasoning about the atmosphere as a whole is misleading: increasing carbon dioxide from 0.028 to 0.042 percent sounds negligible as a fraction of the atmosphere, but it is a fifty percent increase in one of the few constituents doing any absorbing at all. Nitrogen's absorbing capacity is zero, so the whole effect comes from the trace gases and their proportional changes are what count

8. Explain why the greenhouse effect depends on the sun and the Earth emitting at different wavelengths.
   Show the full solution

   Because the effect is an asymmetry between incoming and outgoing radiation, and that asymmetry only exists if the two are at different wavelengths. The sun is extremely hot and emits mostly visible and near-infrared light, which greenhouse gases barely absorb, so sunlight passes through the atmosphere and reaches the surface. The Earth's surface is far cooler and therefore radiates at much longer wavelengths, in the thermal infrared, which is exactly where carbon dioxide and water vapor absorb strongly. Energy consequently enters easily and leaves with difficulty. Were both streams at the same wavelength, any gas that blocked the outgoing radiation would block the incoming radiation equally and there would be no net effect. The mechanism is a one-way filter, and it is one-way only because the two bodies are at very different temperatures. Incoming visible light passes through while outgoing infrared is absorbed; equal wavelengths would block both and produce no effect

9. Explain why Arrhenius stated his result in terms of doublings rather than absolute amounts.
   Show the full solution

   Because the relationship he found is logarithmic rather than linear, which is what his quoted rule says: carbon dioxide rising in geometric progression produces temperature rising in arithmetic progression. In plain terms, each successive doubling adds roughly the same amount of warming, so the first doubling and the second doubling have similar effects even though the second involves twice as much gas. The physical reason is saturation within the strongest absorption bands: once those wavelengths are almost completely absorbed by the gas already present, adding more can only capture radiation at the weaker edges of the bands, so each additional molecule contributes less than the one before. Quoting a sensitivity per doubling therefore states the result in the units in which it is approximately constant, which is more useful than a figure per part per million that would change as the concentration rose. The response is logarithmic, so each doubling adds about the same warming; band saturation is the physical reason

10. Explain the full path of energy through the Earth system, from the sun to space, in terms of the three transfer mechanisms.
    Show the full solution

    Energy arrives from the sun entirely by radiation, since nothing else crosses the vacuum, and most of it is visible light that passes through the atmosphere without being absorbed. At the surface it is absorbed by land and ocean, warming them. From there it moves in several ways. Conduction passes energy from the surface into the air and water immediately in contact with it, but both are poor conductors so this acts only over a short distance. Convection then does most of the work of redistribution: warmed air and water become less dense and rise, carrying energy upward and, through winds and ocean currents, from the equator toward the poles. Evaporation carries energy as well, released again when water vapor condenses at altitude. Finally the surface and the atmosphere radiate infrared, some of which is absorbed by greenhouse gases and re-emitted in all directions, slowing the loss, and the remainder escapes to space. The planet's temperature settles where the outgoing radiation matches the incoming. Radiation in from the sun, conduction and convection redistributing energy at the surface, and radiation out to space, with greenhouse gases slowing the last step

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State three assumptions of the kinetic molecular theory.
   Show the full solution

   Constant random motion, negligible particle volume, elastic collisions, no attractions, average kinetic energy proportional to absolute temperature (any three)

2. What causes the pressure of a gas?
   Show the full solution

   Collisions of the particles with the walls of the container

3. Convert 380 mmHg to atmospheres.
   Show the full solution

   \( 380 \div 760 \). 0.500 atm

4. Give the value of one atmosphere in kilopascals.
   Show the full solution

   101.325 kPa

5. Which quantity is proportional to average kinetic energy?
   Show the full solution

   Absolute temperature, in kelvin

6. Explain from the model why a gas fills its container completely while a liquid does not.
   Show the full solution

   The model assumes gas particles are in constant random motion and that there are no significant attractions between them, and those two assumptions together mean nothing holds a gas particle near any other. A particle travels in a straight line until it collides with something, and since the motion is random the particles spread until they are distributed throughout whatever space is available, filling the container and taking its shape. In a liquid the particles are close enough for intermolecular forces of the kind in lesson 3.5 to matter, and those attractions are strong enough to hold the particles together as a body while still allowing them to slide past one another. A liquid therefore takes the shape of its container but keeps its own volume, because it is held together but not held in place. No attractions and random motion, so gas particles disperse; in a liquid intermolecular forces hold the particles together

7. Explain why doubling the temperature raises the pressure more than merely doubling the collision frequency would.
   Show the full solution

   Because raising the temperature changes both of the factors that determine pressure, whereas compressing the gas changes only one. Faster particles reach the walls more often, which increases the collision frequency, and they also arrive carrying more momentum, so each impact delivers a larger force. Pressure is the product of how often and how hard, so both contributions act in the same direction and reinforce each other. By contrast, halving the volume increases the frequency of collisions but leaves the speed of each particle exactly as it was, since temperature is unchanged, so only one factor rises. This is why a complete answer to any temperature question names both effects; naming one finds half the explanation. Higher temperature raises both collision frequency and force per collision, while compression raises only frequency

8. Explain why helium leaks out of a balloon faster than the nitrogen and oxygen of air leak in.
   Show the full solution

   At a given temperature all gas particles have the same average kinetic energy, and kinetic energy depends on both mass and speed. A helium atom has a molar mass of about 4 g/mol while nitrogen and oxygen molecules are around 28 and 32, so to carry the same energy the much lighter helium atom must move considerably faster, by roughly a factor of two and a half to three. Escape through the microscopic pores in the balloon wall depends on how often a particle arrives at a pore and how readily it passes through, and faster smaller particles do both more often. The helium therefore leaves at a greater rate than air enters, so the balloon shrinks rather than simply exchanging its contents. Equal average kinetic energy means the lighter helium atoms move faster and reach the pores more often

9. Explain why the kinetic model requires collisions to be elastic, and what would follow if they were not.
   Show the full solution

   An elastic collision is one in which no kinetic energy is lost, so the particles rebound with the same total energy they arrived with. The assumption is needed because gas particles collide constantly, of the order of billions of times per second each, and if even a tiny fraction of the energy were lost in each collision the total kinetic energy would decay extremely rapidly. Since temperature is a measure of average kinetic energy, the gas would cool spontaneously to a standstill within a fraction of a second, and its pressure would fall to zero as the particles stopped striking the walls. A sealed flask of gas left alone does neither: its temperature and pressure stay constant indefinitely. That observation is the evidence for the assumption, and it also means any energy a gas loses must go somewhere identifiable, such as through the walls to a cooler surrounding. Any energy loss per collision would cool the gas to a standstill almost instantly; sealed gases keep constant temperature and pressure

10. A student says gases are compressible because the particles themselves squash. Correct this using the model.
    Show the full solution

    The particles do not change size at all; what changes is the space between them. The model assumes the volume of the particles is negligible compared with the volume of the container, which means a gas at ordinary pressure is overwhelmingly empty space, as the comparison of molar volumes in lesson 4.1 showed, where a mole of water occupies 18 mL as a liquid and 22 400 mL as a gas. Compressing a gas pushes the particles closer together, reducing that empty space, and the particles themselves are untouched. This also explains why liquids and solids are nearly incompressible: their particles are already in contact, so there is almost no empty space left to remove, and compressing further would require deforming the particles, which takes enormous force. The empty space between particles is reduced; the particles are unchanged, which is why liquids and solids barely compress

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State Boyle's law as an equation.
   Show the full solution

   \( P_1V_1 = P_2V_2 \) at constant temperature

2. State Charles's law as an equation.
   Show the full solution

   \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) at constant pressure

3. Convert 27 degrees Celsius to kelvin.
   Show the full solution

   300 K

4. A gas at 2.00 atm occupies 5.00 L. Find the volume at 1.00 atm.
   Show the full solution

   \( \frac{2.00 \times 5.00}{1.00} \). 10.0 L

5. What value of Celsius temperature corresponds to absolute zero?
   Show the full solution

   About -273 degrees Celsius

6. A gas at 300 K occupies 4.00 L. Find the volume at 450 K at constant pressure, and check the answer.
   Show the full solution

   Both temperatures are already in kelvin, so no conversion is needed, though confirming that is worth a moment. The gas is heated at constant pressure so the volume must increase, which sets the expectation. Substituting, \( V_2 = \frac{V_1 T_2}{T_1} = \frac{4.00 \times 450}{300} = 6.00 \) L. The check is proportional: the absolute temperature rose by a factor of 1.5, so the volume should rise by the same factor, and \( 4.00 \times 1.5 = 6.00 \) confirms it. Doing the proportional check separately from the formula catches a substitution made upside down. 6.00 L

7. Explain why the units of pressure and volume do not need converting but the units of temperature do.
   Show the full solution

   Because pressure and volume each appear on both sides of the equation, so any consistent unit cancels in the ratio. Expressing both volumes in liters or both in milliliters gives the same numerical answer, since only the ratio of the two matters, and the same is true of pressures in atmospheres or kilopascals. Temperature is different because the requirement is not merely consistency but that the scale has a true zero, so that ratios correspond to ratios of the underlying physical quantity. Converting 25 and 100 degrees Celsius to kelvin does not simply rescale them, it shifts them by 273, and a shift changes a ratio while a rescale does not. That is why consistency is sufficient for the other two and insufficient for temperature. P and V appear as ratios so units cancel; a Celsius-to-kelvin change is an offset, which alters a ratio

8. A sealed syringe of gas is pushed to half its volume at constant temperature. Explain what happens to the pressure using the kinetic model.
   Show the full solution

   The number of particles is unchanged because the syringe is sealed, and their average speed is unchanged because the temperature is constant, so each individual collision with the wall delivers exactly the same force as before. What changes is how far a particle travels between wall collisions, which is now half as far, so each particle strikes a wall twice as often. The total number of collisions per second therefore doubles while the force per collision is the same, and since pressure is the total force per unit area, the pressure doubles. This is Boyle's law derived from the model, and it also explains why the relationship is inverse rather than some other shape: collision frequency scales directly with how little distance there is to cross. Collision frequency doubles while force per collision is unchanged, so the pressure doubles

9. Explain how plotting Charles's law data located absolute zero before anyone could reach that temperature.
   Show the full solution

   By extrapolation from measurements taken at ordinary temperatures. Measuring the volume of a fixed quantity of gas at several accessible temperatures and plotting volume against Celsius temperature gives a straight line, and the line can be extended backward beyond the measured range. Every gas tested gives a line that, when extended, reaches zero volume at the same temperature of about -273 degrees Celsius, and that common intercept is the evidence. A volume cannot be negative, so no lower temperature is possible, which identifies -273 as an absolute minimum. The fact that gases of quite different molar masses and properties all extrapolate to the same intercept is what makes the result a property of temperature itself rather than an accident of one substance. No gas actually reaches zero volume, since every gas liquefies first, so the value is obtained entirely from where the line points. Volume against Celsius temperature is linear and extrapolates to zero volume at -273 for every gas, which is a common intercept no substance could reach

10. A student calculates that heating a gas from 20 degrees Celsius to 40 degrees Celsius doubles its volume. Diagnose the error and give the correct factor.
    Show the full solution

    They have taken the ratio of the Celsius values, 40 divided by 20, and concluded the volume doubles. Charles's law is a proportion in absolute temperature, so the temperatures must be converted first: 20 degrees Celsius is 293 K and 40 degrees is 313 K. The ratio is \( \frac{313}{293} = 1.068 \), so the volume increases by about seven percent, not by one hundred percent. The error is a factor of nearly twenty in the size of the change. It is worth noting how misleading the Celsius arithmetic is near room temperature: because 273 is large compared with ordinary Celsius readings, modest Celsius changes correspond to small absolute changes, and doubling a Celsius temperature almost never doubles anything. Convert first: 293 K to 313 K is a factor of 1.068, about a seven percent increase

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State Gay-Lussac's law as an equation.
   Show the full solution

   \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \) at constant volume

2. Write the combined gas law.
   Show the full solution

   \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \)

3. Which quantity is held constant in Gay-Lussac's law?
   Show the full solution

   Volume

4. What assumption do all three simple gas laws share?
   Show the full solution

   The amount of gas is fixed

5. A gas at 1.50 atm and 300 K is heated to 400 K at constant volume. Find the new pressure.
   Show the full solution

   \( \frac{1.50 \times 400}{300} \). 2.00 atm

6. A gas occupies 5.00 L at 2.00 atm and 250 K. Find the volume at 1.00 atm and 300 K.
   Show the full solution

   All three quantities change, so the full combined law is needed. Rearranging, \( V_2 = \frac{P_1V_1T_2}{T_1P_2} \). Before calculating, predict the direction of each effect separately: halving the pressure should roughly double the volume, and raising the temperature from 250 to 300 K should increase it by a further twenty percent, so the answer should be around 12 L. Substituting, \( V_2 = \frac{2.00 \times 5.00 \times 300}{250 \times 1.00} = \frac{3000}{250} = 12.0 \) L. Both effects acted in the same direction here, which is why the volume more than doubled. 12.0 L

7. Explain why learning the combined law is more reliable than memorizing three separate equations.
   Show the full solution

   Because it removes a decision that is easy to get wrong under pressure. With three equations a student must first identify which quantity is held constant, then recall which of three similar-looking formulas corresponds to that case, and the formulas differ only in which symbols appear. With the combined law there is one equation to recall, and the constant quantity simply appears on both sides and cancels, which is a mechanical step rather than a recall step. It also handles the case where all three quantities change, which none of the simple laws can, so there is no separate procedure for it. Deriving the three special cases once, as the worked example does, is what makes the single equation feel sufficient rather than like one more thing to remember. One equation replaces a choice among three, and it also covers the case where all three quantities change

8. Explain why an aerosol can carries a warning against heating, using both the gas law and the kinetic model.
   Show the full solution

   The can is rigid and sealed, so volume and amount of gas are fixed and Gay-Lussac's law applies: pressure rises in direct proportion to absolute temperature. The worked example showed a forty degree warming raising the pressure by about fourteen percent, and since aerosol cans are already pressurized to several atmospheres, fourteen percent of a large number is a substantial absolute increase. In kinetic terms, heating raises the average kinetic energy of the particles, so they strike the walls both more often and with greater force, and both effects raise the pressure. The can is designed with a margin above its filling pressure, but that margin is finite, and a fire or a closed car in summer can push the internal pressure past the point where the seam fails, which is why the warning specifies both direct heat and incineration. Rigid and sealed means Gay-Lussac applies, and faster particles strike more often and harder, raising pressure past the can's margin

9. Explain why the combined gas law cannot be used when a gas leaks from a container.
   Show the full solution

   Because every form of the law assumes the quantity of gas is unchanged between the two states, and that assumption is built into the derivation. The quantity \( \frac{PV}{T} \) is constant for a fixed amount of gas, and its value is proportional to the number of moles present, so comparing a before state and an after state with different amounts compares two different constants. If gas escapes, the pressure falls for a reason the equation has no term for, and applying it anyway would attribute that fall to a volume or temperature change that did not happen. The correct tool is the ideal gas law of lesson 8.4, which includes \( n \) explicitly and can therefore handle a change in the amount of gas as well as in the other three quantities. The law assumes a fixed amount of gas; use \( PV = nRT \), which includes \( n \) explicitly

10. A weather balloon rises, so both pressure and temperature fall. Explain why it expands rather than contracting.
    Show the full solution

    The two changes push the volume in opposite directions and the question is which wins. Falling external pressure allows the gas to expand, which increases the volume, while falling temperature reduces the particles' kinetic energy and tends to reduce the volume. Their relative sizes settle it. In the worked example the pressure halved, a factor of 2.00 in favor of expansion, while the temperature fell from 300 to 250 K, a factor of 0.833 in favor of contraction, and the product is 1.67 so the volume rose from 2.00 to 3.33 L. The general reason expansion usually wins is that pressure falls very steeply with altitude, dropping to a small fraction of its surface value, whereas absolute temperature falls only modestly because the kelvin scale starts so far below ordinary temperatures. This is why high-altitude balloons are launched slack and can eventually burst. Pressure falls by a much larger factor than absolute temperature, so the expansion outweighs the contraction

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Write the ideal gas law.
   Show the full solution

   \( PV = nRT \)

2. Give the value and units of R for pressure in atmospheres.
   Show the full solution

   0.0821 L·atm/(mol·K)

3. What does n represent?
   Show the full solution

   The number of moles of gas

4. Find the volume of 0.500 mol of gas at 1.50 atm and 350 K.
   Show the full solution

   \( V = \frac{nRT}{P} = \frac{0.500 \times 0.0821 \times 350}{1.50} \). 9.58 L

5. How does the ideal gas law differ from the combined gas law in what it describes?
   Show the full solution

   It describes a single state including the amount of gas, rather than comparing two states of a fixed amount

6. Show that the ideal gas law reproduces the molar volume of 22.4 L/mol at STP.
   Show the full solution

   Molar volume is the volume occupied by exactly one mole, so set \( n = 1.00 \), and STP specifies \( P = 1.00 \) atm and \( T = 273 \) K. Rearranging for volume, \( V = \frac{nRT}{P} = \frac{1.00 \times 0.0821 \times 273}{1.00} = 22.41 \) L. This reproduces the 22.4 L/mol used throughout unit 4, which confirms that the molar volume is not an independent fact to be memorized but a single evaluation of the ideal gas law at one particular set of conditions. It also explains immediately why the value changes at other conditions, since altering P or T alters the result, which is the reason lesson 4.4 insisted that 22.4 applies at STP only. 22.4 L, so the molar volume is the ideal gas law evaluated at STP

7. A student uses a pressure of 202.65 kPa with R = 0.0821. Explain what goes wrong and give the fix.
   Show the full solution

   The numerical value 0.0821 is inseparable from its units of liter atmospheres per mole per kelvin, so it is only correct when the pressure is expressed in atmospheres. Substituting 202.65 instead of the equivalent 2.00 atm inflates the pressure term by a factor of about 101, so a calculation of the number of moles comes out about a hundred times too large and a calculation of volume about a hundred times too small. Nothing in the arithmetic signals this, and the units check would catch it only if the student wrote kPa explicitly and noticed it failing to cancel against the atm in R. The fix is either to convert the pressure by dividing by 101.325 to get 2.00 atm, or to use R = 8.314 with pressure in kilopascals and volume in liters. Convert to 2.00 atm, or use R = 8.314 for kPa; otherwise the answer is off by about a hundredfold

8. Explain why the ideal gas law connects the gas laws to stoichiometry in a way the combined law cannot.
   Show the full solution

   Because it contains \( n \), the number of moles, and moles are the currency of every stoichiometry calculation. The combined gas law relates pressure, volume and temperature only, and the amount of gas cancels out of it entirely, so it can predict how a sample behaves when conditions change but can never say how much substance is present. The ideal gas law makes the amount explicit, so a measurement of pressure, volume and temperature in the laboratory converts directly into a mole count, which then enters the road map of lesson 6.1 exactly as a mass would after division by molar mass. This is what allows a reaction that produces a gas to be quantified by collecting it and measuring its volume, which is often far easier than isolating and weighing a product. It contains n, so measured P, V and T give a mole count that feeds directly into the stoichiometry road map

9. Derive the expression for molar mass from gas density, and explain why it was historically important.
   Show the full solution

   Start from \( PV = nRT \) and substitute \( n = \frac{m}{M} \), where m is the sample mass and M the molar mass, giving \( PV = \frac{m}{M}RT \). Multiply both sides by M and divide by PV to isolate it: \( M = \frac{mRT}{PV} \). Since the mass divided by the volume is the density d, this becomes \( M = \frac{dRT}{P} \). The importance is that it determines a molar mass from three quantities that are all straightforward to measure for a gas, namely its density, its temperature and its pressure, with no chemical analysis required at all. Before mass spectrometry this was the principal route to a molar mass, and lesson 4.7 showed that a molar mass is exactly the independent measurement needed to convert an empirical formula into a molecular formula. Combining an elemental analysis with a gas density measurement therefore determined a molecular formula completely. \( M = \frac{dRT}{P} \); it gave molar masses from purely physical measurements, supplying what empirical formulas lacked

10. Explain why the ideal gas law is called "ideal" and what that word is warning about.
    Show the full solution

    Because it describes the behavior of a hypothetical gas that exactly satisfies the assumptions of the kinetic model from lesson 8.1, in particular that the particles have no volume of their own and exert no attractions on one another. No real gas satisfies either assumption: real molecules occupy space and do attract each other, as the intermolecular forces of lesson 3.5 establish, which is precisely why real gases can be condensed into liquids while an ideal gas could not. The word is therefore a warning that the equation is an approximation whose accuracy depends on how nearly those assumptions hold. At ordinary temperatures and pressures the approximation is very good, often within a percent, because the particles are far apart and moving fast enough for attractions to be negligible. Lesson 8.6 sets out the two conditions under which it stops being good. It assumes zero particle volume and no attractions, which no real gas satisfies; the approximation is good only when particles are far apart and fast

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State Dalton's law of partial pressures.
   Show the full solution

   The total pressure of a mixture equals the sum of the partial pressures of its components

2. Define partial pressure.
   Show the full solution

   The pressure a gas would exert if it alone occupied the container

3. A mixture at 2.00 atm contains 0.300 mol of A in 1.20 mol total. Find the partial pressure of A.
   Show the full solution

   Mole fraction \( \frac{0.300}{1.20} = 0.250 \), times 2.00 atm. 0.500 atm

4. A gas is collected over water at a total pressure of 750 mmHg where the water vapor pressure is 20 mmHg. Find the dry gas pressure.
   Show the full solution

   730 mmHg

5. Does partial pressure depend on mole fraction or mass fraction?
   Show the full solution

   Mole fraction

6. Explain why Dalton's law follows from the kinetic model.
   Show the full solution

   The model assumes gas particles exert no attractions on one another and that their own volume is negligible, which together mean that a particle's behavior is unaffected by what other particles happen to be present. A nitrogen molecule in a mixture travels and collides with the walls exactly as it would if the oxygen were removed, because nothing about the oxygen influences it. Each gas therefore delivers to the walls precisely the collisions it would deliver alone, and its contribution to the pressure is its partial pressure. Pressure is total force per unit area, and forces add, so the total is the sum of the contributions. The law is not an independent discovery so much as an immediate consequence of the non-interaction assumption, which is also why it fails for real gases under the same conditions that make the ideal gas law fail. Non-interacting particles each behave as if alone, so their wall collisions and therefore their pressures simply add

7. Explain why partial pressure depends on the number of particles rather than their masses.
   Show the full solution

   Because pressure is produced by collisions with the walls, and what matters is how many collisions occur and how much momentum each transfers. At a given temperature all gas particles have the same average kinetic energy regardless of mass, so a heavy slow particle and a light fast one deliver the same average momentum per collision and arrive with frequencies that compensate for their speeds. The net effect is that each particle contributes the same amount to the pressure whatever it weighs. Consequently a mixture containing equal numbers of hydrogen and carbon dioxide molecules has equal partial pressures from each, even though the carbon dioxide accounts for about ninety-six percent of the mass. Using mass fraction instead of mole fraction would therefore be badly wrong for mixtures of gases with different molar masses. Equal average kinetic energy means each particle contributes equally to pressure regardless of mass, so counts are what matter

8. Explain why the water vapor correction is always in the same direction and what that makes it.
   Show the full solution

   Water vapor is always present above water and always adds to the total pressure, so the measured total always exceeds the pressure of the collected gas alone. Omitting the correction therefore always overstates the partial pressure of the gas, always overstates the number of moles calculated from it, and always overstates the yield of the reaction that produced it. Because the sign of the error never varies, this is a systematic error in the sense of lesson 1.5 rather than a random one, which has a practical consequence: repeating the experiment many times and averaging improves precision but does nothing whatever to the accuracy, since every trial is biased the same way. The only remedy is to apply the correction, which is why the vapor pressure tables exist. Water vapor always adds to the total, so omitting it always overstates the gas; it is a systematic error that averaging cannot remove

9. Air is about 78 percent nitrogen and 21 percent oxygen by volume. Find the partial pressure of oxygen at 1.00 atm, and explain why altitude sickness occurs.
   Show the full solution

   For gases, percentage by volume equals percentage by moles, since Avogadro's law makes volume proportional to particle count, so oxygen's mole fraction is 0.21 and its partial pressure at sea level is \( 0.21 \times 1.00 = 0.21 \) atm. At altitude the composition of the air is essentially unchanged, so oxygen remains 21 percent, but the total pressure falls substantially, to roughly 0.5 atm at about 5500 m. The partial pressure of oxygen therefore falls to around 0.11 atm. Since the uptake of oxygen into the blood depends on its partial pressure rather than on its percentage, less oxygen is absorbed per breath even though the air is just as oxygen-rich by composition. That deficit is what produces altitude sickness, and it explains why supplemental oxygen works: raising the fraction restores the partial pressure. 0.21 atm at sea level; at altitude the fraction is unchanged but total pressure falls, so oxygen's partial pressure and uptake fall

10. Explain why a gas cannot usefully be collected over water at 100 degrees Celsius.
    Show the full solution

    Because the vapor pressure of water at 100 degrees Celsius is 760 mmHg, which is exactly one atmosphere, and that is the definition of the normal boiling point. The water vapor alone would therefore account for the entire atmospheric pressure, leaving nothing for the collected gas: applying Dalton's law would give a partial pressure of zero or a negative value for the gas being measured. Physically, the water would be boiling vigorously and filling the collection vessel with steam, so the measured volume would be almost entirely vapor and no meaningful reading could be taken. The correction grows steeply with temperature for this reason, from about 17.5 mmHg at 20 degrees to 92 mmHg at 50 degrees, so gas collection over water is done cool and the temperature is recorded precisely. Water's vapor pressure reaches 760 mmHg at 100 degrees, so the vapor accounts for the whole pressure and nothing is left to attribute to the gas

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Under what two conditions do real gases deviate most from ideal behavior?
   Show the full solution

   High pressure and low temperature

2. Which assumption fails at high pressure?
   Show the full solution

   That the particles' own volume is negligible

3. Which assumption fails at low temperature?
   Show the full solution

   That there are no attractions between particles

4. When should \( PV = nRT \) be used instead of 22.4 L/mol?
   Show the full solution

   At any conditions other than STP

5. Which behaves more ideally, helium or water vapor?
   Show the full solution

   Helium, because its intermolecular forces are far weaker

6. Find the volume of hydrogen at 20.0 degrees Celsius and 1.00 atm produced when 6.54 g of zinc reacts with excess hydrochloric acid.
   Show the full solution

   The equation is \( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \). Zinc's molar mass is 65.38 g/mol, so \( 6.54 \div 65.38 = 0.100 \) mol, and the one-to-one ratio gives 0.100 mol of hydrogen. The conditions are not STP, so 22.4 L/mol does not apply and the ideal gas law is needed with \( T = 20.0 + 273 = 293 \) K: \( V = \frac{nRT}{P} = \frac{0.100 \times 0.0821 \times 293}{1.00} = 2.41 \) L. For comparison the STP volume would be 2.24 L, so the room-temperature answer is about seven percent larger, consistent with the ratio \( \frac{293}{273} \). 2.41 L

7. Explain why the two deviations from ideality push in opposite directions.
   Show the full solution

   Because they violate different assumptions with opposite consequences. Particle volume means that some of the measured container volume is occupied by the particles themselves and is unavailable for movement, so the space the gas can actually move in is smaller than the measured volume. The gas therefore resists compression more than predicted and its measured volume exceeds the ideal prediction. Intermolecular attractions work the other way: a particle approaching the wall is pulled backward by the particles behind it, so it arrives with less momentum than it otherwise would and each collision delivers less force, making the measured pressure lower than predicted. One effect makes a gas seem larger than ideal and the other makes it seem to push less hard, and which dominates depends on the conditions, with attractions dominating at low temperature and particle volume at very high pressure. Particle volume reduces the available space and raises measured volume; attractions soften wall collisions and lower measured pressure

8. Explain why the fact that gases can be liquefied is itself proof that they are not ideal.
   Show the full solution

   Because an ideal gas could never condense. The model assumes there are no attractions whatever between particles, and condensation is precisely the process of particles being held together by attractions strongly enough to form a liquid with a definite volume. If the assumption were literally true, cooling a gas would slow its particles indefinitely but nothing would ever pull them together, and compressing it would force them close but nothing would hold them there once the pressure was released. Every real gas can be liquefied given a low enough temperature and a high enough pressure, so every real gas must have attractions between its particles. The temperature at which this becomes possible tracks the strength of those forces, which is why helium must be cooled to about four kelvin while water condenses at 373 kelvin, exactly the ranking of lesson 3.5. Condensation requires attractions, which an ideal gas is defined not to have; the condensation temperature ranks their strength

9. A student uses 22.4 L/mol to find the volume of gas produced in a reaction run at 80 degrees Celsius. Estimate the size and direction of the error.
   Show the full solution

   The molar volume is proportional to absolute temperature at constant pressure, so the correct value at 80 degrees Celsius is \( 22.4 \times \frac{353}{273} = 29.0 \) L/mol. Using 22.4 therefore understates the volume by \( \frac{29.0 - 22.4}{29.0} \), which is about twenty-three percent, and the error is always in that direction because any temperature above 0 degrees Celsius gives a larger molar volume. If instead the student were working backward from a measured volume to find moles, the same error would overstate the mole count by nearly thirty percent, which would then propagate into every mass calculated from it. At room temperature the discrepancy is around nine percent and easy to overlook; at 80 degrees it is large enough to invalidate a result outright. About twenty-three percent too small, and always in that direction since the correct molar volume at 353 K is 29.0 L/mol

10. Explain what conditions make a gas behave most ideally, and connect the answer to the assumptions of the model.
    Show the full solution

    Low pressure and high temperature. Low pressure spreads the particles far apart, so the volume they themselves occupy becomes a negligible fraction of the container, which is exactly the first assumption the model makes. Being far apart also weakens the attractions between them, since intermolecular forces fall off steeply with distance, which supports the second assumption. High temperature gives the particles a large average kinetic energy, so any attraction they do experience is small compared with the energy of their motion and deflects them negligibly, which supports the same assumption from the other side. The model is therefore most accurate precisely when the particles are far apart and moving fast, which is the condition that makes both idealizations nearly true at once, and it is why ordinary laboratory conditions of about one atmosphere and room temperature give agreement to within a percent or so for most gases. Low pressure and high temperature, because both make the particles far apart and fast enough for zero volume and no attractions to be good approximations

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Define solute and solvent.
   Show the full solution

   The solute is the substance dissolved; the solvent is what dissolves it

2. Name the three energy steps of dissolving and say which releases energy.
   Show the full solution

   Separating solute, separating solvent, and hydration; only hydration releases energy

3. Does sodium chloride dissolve as ions or as molecules?
   Show the full solution

   As separate hydrated ions

4. What experimental test distinguishes an ionic solution from a molecular one?
   Show the full solution

   Electrical conductivity: ionic solutions conduct, molecular ones generally do not

5. Define hydration.
   Show the full solution

   The surrounding of a dissolved particle by a shell of water molecules

6. Explain why a solution of sugar does not conduct electricity while a solution of salt does, even though both look identical.
   Show the full solution

   Conduction requires mobile charged particles, and the two solutions differ in whether any are produced. Sodium chloride is ionic, so dissolving separates the lattice into free Na⁺ and Cl^- ions that move independently through the water and carry charge from one electrode to the other. Sugar is a molecular compound held together by covalent bonds, and dissolving merely disperses intact neutral sucrose molecules among the water molecules; no charges are separated and nothing can carry current. The appearance of the two solutions is identical because both are homogeneous mixtures of colorless particles far too small to scatter light, so the conductivity test is the one that discriminates, which is why it is used to classify electrolytes. Salt produces free ions that carry charge; sugar dissolves as neutral molecules with no charges to move

7. Explain why the temperature change on dissolving can be in either direction.
   Show the full solution

   Because the observed change is the net of three separate energy terms, two of which cost energy and one of which releases it. Breaking the solute's lattice apart is endothermic and so is pushing solvent molecules aside to make room, while surrounding the separated particles with solvent is exothermic. Whether the sum is positive or negative depends on the particular solute, and specifically on how strongly its particles are held in the lattice against how strongly they are hydrated once free. Small highly charged ions are strongly hydrated and tend to give exothermic dissolving, while large singly charged ions are weakly hydrated and tend to give endothermic dissolving. Neither outcome tells you whether the substance dissolves, only what it does to the temperature while doing so. Two endothermic steps compete with one exothermic step, and which dominates depends on lattice strength against hydration strength

8. Explain why an endothermic dissolving happens at all, given that it makes the surroundings colder.
   Show the full solution

   Because the tendency toward lower energy is not the only factor determining whether a change occurs; the tendency toward greater disorder also matters, and at ordinary temperatures it can be decisive. A crystal of ammonium nitrate is an extremely ordered arrangement in which every ion occupies a fixed lattice position, while the dissolved state has those same ions dispersed randomly throughout the solvent with vastly more possible arrangements available. That increase in disorder is large, and it outweighs the unfavorable energy change. The energy required does not come from nowhere: it is drawn from the thermal energy of the water and its surroundings, which is exactly why the solution cools and why the process works as a cold pack. A full account of the competition requires entropy and free energy, which lies beyond a first course. The large increase in disorder outweighs the unfavorable energy change, with the energy drawn from the surroundings

9. Explain what determines whether a water molecule turns its oxygen or its hydrogens toward a dissolved ion.
   Show the full solution

   The charge on the ion, since the orientation is simply opposite charges attracting. A water molecule is polar in the sense of lesson 3.4, with a partial negative charge concentrated on the oxygen where the lone pairs are and partial positive charges on the two hydrogens. Facing a cation such as Na⁺, the water turns its negative oxygen end toward the ion, and several water molecules arrange themselves around it that way. Facing an anion such as Cl^-, the water turns a positive hydrogen end toward the ion instead. The hydration shells around a cation and an anion are therefore oriented oppositely, and this is only possible because the molecule is bent: a linear water molecule would be nonpolar and could not orient toward anything, which is the point lesson 3.4 closed on. Opposite charges attract: oxygen faces cations and hydrogens face anions, which requires water to be polar

10. Explain why water is called the universal solvent and why the name overstates the case.
    Show the full solution

    The name records that water dissolves a wider range of substances than almost any other common liquid, which follows from its polarity and its capacity for hydrogen bonding. Its partial charges let it surround and stabilize ions, so it dissolves most ionic compounds, and its hydrogen bonding lets it interact strongly with any molecule carrying an O-H or N-H group, so it dissolves sugars, alcohols and many biological molecules. The name overstates the case because those same properties make water a poor solvent for anything nonpolar: oils, fats, waxes and hydrocarbons are essentially insoluble in it, as lesson 9.2 explains, and many ionic compounds such as silver chloride and barium sulfate are insoluble too. A solvent that dissolved everything would also be impossible to store, since it would dissolve its container. Its polarity and hydrogen bonding give it exceptional range, but nonpolar substances and many ionic compounds do not dissolve in it

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State the like dissolves like rule.
   Show the full solution

   Polar solvents dissolve polar and ionic solutes; nonpolar solvents dissolve nonpolar solutes

2. Is hexane polar or nonpolar?
   Show the full solution

   Nonpolar

3. Why does ethanol mix with water?
   Show the full solution

   Its O-H group allows hydrogen bonding with water

4. Name the two parts of a soap molecule.
   Show the full solution

   A nonpolar hydrocarbon tail and an ionic head

5. Is carbon tetrachloride a polar or nonpolar solvent?
   Show the full solution

   Tetrahedral and symmetrical, so the bond polarities cancel. Nonpolar

6. Explain why water and oil separate, without saying that they repel each other.
   Show the full solution

   There is no repulsion between them; both are attracted to everything to some degree. The separation is a competition that the oil loses. Water molecules attract one another through hydrogen bonding, which is comparatively strong, while they can attract an oil molecule only through weak dispersion forces. Dissolving the oil would require separating water molecules from each other to make room, which costs a large amount of energy, and the surrounding of the oil molecule would release only a small amount in return. The net cost is unfavorable, so the water molecules preferentially stay associated with each other and exclude the oil. The oil molecules are pushed together into a separate layer not because they attract one another strongly but because the water will not accommodate them. Water's hydrogen bonding to itself is far stronger than anything it could offer oil, so the water excludes it

7. Explain why methanol mixes with water in all proportions but octanol, C₈H₁₇OH, is only slightly soluble.
   Show the full solution

   Both molecules have the same polar O-H group and can therefore hydrogen bond with water, but they differ in how much nonpolar hydrocarbon is attached to it. Methanol has a single carbon, so the polar group dominates the molecule and its behavior is essentially that of a polar substance, giving complete miscibility. Octanol carries an eight-carbon chain, which is a substantial nonpolar region that water cannot accommodate, and dissolving it would require the water to open up a large cavity for the chain at considerable energy cost. The single O-H group cannot pay for that, so solubility is low. The general pattern along a series of alcohols is that solubility falls steadily as the chain lengthens, because the ratio of nonpolar to polar character rises. The same polar group is attached to a much larger nonpolar chain, which water cannot accommodate

8. Iodine dissolves poorly in water but readily in hexane. Explain both observations with one principle.
   Show the full solution

   Iodine is a diatomic molecule of two identical atoms, so the bond is perfectly nonpolar and the molecule has no permanent dipole; its only intermolecular attraction is dispersion. In water, dissolving it would mean breaking hydrogen bonds between water molecules and replacing them with weak dispersion attractions to iodine, which is a poor exchange and does not happen appreciably. In hexane, which is also nonpolar and also held together only by dispersion, the attractions being broken and the attractions being formed are of the same kind and similar strength, so the exchange costs almost nothing and the iodine dissolves freely. One principle covers both: a substance dissolves when the new attractions can adequately replace the old ones, and that is most easily satisfied when solute and solvent are alike. Dispersion attractions to iodine cannot replace water's hydrogen bonds, but they match hexane's own dispersion forces exactly

9. Explain why dry cleaning uses a nonpolar solvent rather than water.
   Show the full solution

   Because the stains that resist water are the nonpolar ones, principally oils, greases and waxes, and by the like dissolves like rule those dissolve in a nonpolar solvent directly rather than needing to be packaged as soap does. A nonpolar solvent offers the grease molecules dispersion attractions comparable to the ones they have for each other, so they disperse readily. There is a second reason concerning the fabric itself: water penetrates natural fibers such as wool and silk, swelling them and disrupting the hydrogen bonds within the protein structure, which causes shrinkage and distortion. A nonpolar solvent interacts only weakly with those polar fibers and therefore leaves them dimensionally stable. The method removes the nonpolar soiling while leaving the polar fabric untouched, which is the rule used twice. Nonpolar solvents dissolve greasy stains directly and do not swell or distort polar natural fibers as water does

10. Silver chloride is ionic yet insoluble in water. Explain why the rule fails here and what the failure tells you.
    Show the full solution

    Because the rule is a summary of which attractions usually win, not a statement about a single quantity, and in this case the balance falls the other way. Dissolving any ionic solid requires the energy released on hydrating the separated ions to compete with the energy needed to break up the lattice. For sodium chloride hydration is sufficient and the compound dissolves. Silver chloride has a substantially larger lattice energy, partly because the bonding has significant covalent character rather than being purely ionic, so the ions are held more tightly than a simple charge argument suggests, and hydration cannot pay the cost. The failure tells you that solubility is decided by a small difference between two large competing energy terms, which is exactly why lesson 5.5 presented the solubility rules as an empirical list rather than deriving them. Its lattice energy exceeds what hydration can repay; solubility is a close competition between two large energies, so it must be measured

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Define molarity.
   Show the full solution

   Moles of solute per liter of solution

2. Convert 400.0 mL to liters.
   Show the full solution

   0.4000 L

3. Find the molarity of 0.500 mol in 2.00 L.
   Show the full solution

   0.250 M

4. Find the moles in 250.0 mL of 0.400 M solution.
   Show the full solution

   \( n = MV = 0.400 \times 0.2500 \). 0.100 mol

5. Is molarity per liter of solvent or per liter of solution?
   Show the full solution

   Per liter of solution

6. 58.44 g of sodium chloride is made up to 500.0 mL. Find the molarity.
   Show the full solution

   The molar mass of sodium chloride is \( 22.99 + 35.45 = 58.44 \) g/mol, so the mass given is exactly one molar mass and \( 58.44 \div 58.44 = 1.000 \) mol. The volume is \( 500.0 \text{ mL} = 0.5000 \) L, converted as its own step. Then \( M = \frac{1.000}{0.5000} = 2.000 \) M. The result is worth a moment's thought: one mole in half a liter is two moles per liter, so the concentration is double what one mole in a full liter would give, which is the check that the division went the right way. 2.00 M

7. Explain why a 1.00 M solution is not made by adding 1.00 L of water to one mole of solute.
   Show the full solution

   Because molarity is defined per liter of solution, and adding a liter of water to a solid gives a total volume larger than a liter. The dissolved particles occupy space among the water molecules, so the solution's volume is the water's volume plus some contribution from the solute, and for a concentrated solution that addition is not negligible. The resulting concentration would therefore be less than 1.00 M by an amount that depends on the solute and could not be predicted without measurement. The correct procedure dissolves the solute in less water than needed and then adds water up to a calibrated mark on a volumetric flask, which fixes the final volume of the solution at exactly the intended value whatever the solute contributed. The solute adds volume, so water plus solute exceeds a liter; make up to the mark instead

8. Describe how to prepare 500.0 mL of 0.100 M potassium nitrate solution. The molar mass is 101.10 g/mol.
   Show the full solution

   Find the moles first: \( n = MV = 0.100 \times 0.5000 = 0.0500 \) mol, taking care to use 0.5000 L rather than 500.0. Convert to a mass: \( 0.0500 \times 101.10 = 5.055 \) g, so weigh 5.06 g. Transfer it to a 500.0 mL volumetric flask, rinsing the weighing boat into the flask so that none of the solute is left behind, since any loss lowers the concentration. Add distilled water to about two thirds full and swirl until the solid has completely dissolved. Then add water carefully to the graduation mark, using a dropper for the last few drops so the meniscus sits on the line, and stopper and invert the flask several times to mix thoroughly. Weigh 5.06 g, dissolve in part of the water, then make up to the 500.0 mL mark

9. Explain why molarity is used in chemistry rather than a concentration in grams per liter.
   Show the full solution

   Because reactions occur in fixed ratios of particles, and molarity reports particles directly while grams per liter does not. A balanced equation says that one mole of one reactant reacts with two of another, so a chemist who knows the molarity and volume of a solution can obtain the moles present in a single multiplication, \( n = MV \), and feed that straight into the road map of lesson 6.1. A concentration in grams per liter would require an extra division by molar mass every time, and comparing two solutions would be misleading, since 10 g/L of sodium hydroxide and 10 g/L of potassium hydroxide contain quite different numbers of moles and therefore react with different amounts of acid. Molarity is chosen to make the quantity that matters immediately available. Reactions go by particle ratio, and molarity gives moles in one step while mass concentration hides them

10. A student weighs the solute into a beaker, dissolves it, and pours the solution into a volumetric flask without rinsing the beaker. Explain the effect on the concentration.
    Show the full solution

    Some solution inevitably clings to the beaker's walls and is left behind, and that residue contains solute. The flask therefore receives less than the intended number of moles while still being filled to the calibrated mark, so the volume is correct and the moles are low, and the concentration comes out below the target. The magnitude depends on how much is left but a few percent is easy to lose this way, which is significant for a solution intended as a standard. The error is also systematic rather than random: it always reduces the concentration, never raises it, so repeating the preparation does not average it out. The remedy is to rinse the beaker two or three times with small volumes of distilled water and add every rinsing to the flask before making up to the mark. Solute left in the beaker lowers the moles while the volume stays correct, so the concentration is systematically low

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Write the dilution equation.
   Show the full solution

   \( M_1V_1 = M_2V_2 \)

2. What quantity is unchanged during a dilution?
   Show the full solution

   The number of moles of solute

3. 100.0 mL of 2.00 M solution is diluted to 500.0 mL. Find the new concentration.
   Show the full solution

   \( M_2 = \frac{2.00 \times 100.0}{500.0} \). 0.400 M

4. Does \( V_2 \) mean the final volume or the water added?
   Show the full solution

   The final total volume

5. State the safety rule for diluting concentrated acid.
   Show the full solution

   Add the acid to the water, never water to the acid

6. How much water must be added to 25.0 mL of 12.0 M acid to make it 3.00 M?
   Show the full solution

   Find the final volume first: \( V_2 = \frac{M_1V_1}{M_2} = \frac{12.0 \times 25.0}{3.00} = 100.0 \) mL. The concentration was reduced by a factor of four, so the volume rose by a factor of four, which checks. The question asks for water added, not final volume, so subtract: \( 100.0 - 25.0 = 75.0 \) mL of water. Answering 100.0 mL would be giving the right calculation to the wrong question, and in the laboratory it would produce a solution of the wrong concentration, since adding 100 mL of water to 25 mL of acid gives 125 mL in total and a concentration of 2.40 M. 75.0 mL of water

7. Explain why the dilution equation follows from the definition of molarity.
   Show the full solution

   Molarity is moles per liter, so the moles present are the concentration multiplied by the volume, \( n = MV \). Dilution consists of adding more solvent and nothing else, so no solute enters or leaves and the number of moles after is identical to the number before. Writing that equality out gives \( M_1V_1 = M_2V_2 \), since each side is just \( n \) expressed through the concentration and volume at that moment. The equation is therefore not a separate rule to memorize but a statement that the solute is conserved, and recognizing this is useful because it makes the verification in the worked example obvious: calculating the moles at each stage must give the same answer, and if it does not, something other than dilution has been assumed. \( n = MV \) and the moles are unchanged, so the two products must be equal

8. Explain in energy terms why acid is added to water rather than the reverse.
   Show the full solution

   Mixing concentrated acid with water is strongly exothermic, because the acid molecules ionize and the resulting ions are hydrated, releasing a large amount of energy in the sense of lesson 9.1. The danger depends on where that energy goes. Adding acid slowly to a large volume of water spreads the released energy through a large mass of liquid with water's high specific heat capacity, so the temperature rise is small and controlled. Adding water to concentrated acid concentrates the same release into the small volume of water that has just arrived, typically at the surface, and that small thermal mass can be driven past its boiling point almost instantly. The resulting flash of steam ejects droplets of concentrated acid upward and outward, which is why the order is a rule rather than a preference. Acid into water spreads the released energy through a large thermal mass; water into acid boils locally and spatters concentrated acid

9. A solution is diluted by a factor of ten three times in succession. Find the overall dilution factor and explain why this is used in practice.
   Show the full solution

   Each step reduces the concentration to a tenth, and the steps multiply, so the overall factor is \( 10 \times 10 \times 10 = 1000 \) and the final solution is one thousandth of the original concentration. The technique is called serial dilution and it is used because measuring a very small volume accurately is difficult: preparing a thousandfold dilution in one step from a 1000 mL flask would require measuring 1.0 mL of stock, and the percentage error in that small measurement would dominate the result. Three successive tenfold dilutions instead measure comfortable volumes such as 10.0 mL into 100.0 mL each time, and although each step contributes its own small error, the total is far smaller than the single-step error would be. This is how the parts per million and parts per billion standards of lesson 9.6 are prepared. A factor of 1000; it avoids measuring one very small volume, which would dominate the error

10. Explain why the dilution equation allows milliliters while the molarity equation requires liters.
    Show the full solution

    Because of where the volume sits in each expression. In the dilution equation volume appears on both sides, multiplied by a concentration, so the equation is effectively a ratio and any unit used consistently cancels between the two sides; expressing both volumes in milliliters gives the same numerical answer as expressing both in liters. In the definition of molarity the volume appears once, in the denominator, and the unit of the result is determined by it, so a volume in milliliters produces a quantity in moles per milliliter rather than the moles per liter that the unit M means. Nothing cancels it and nothing signals the mismatch, which is why that calculation is a trap and this one is not. The general principle is that consistency suffices where a unit cancels and conversion is required where it does not. Volume cancels between the two sides in the dilution equation but sets the unit of the answer in the molarity definition

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Define a saturated solution.
   Show the full solution

   One holding the maximum solute that will dissolve at that temperature

2. What happens to the solubility of most solids as temperature rises?
   Show the full solution

   It increases

3. What happens to the solubility of gases as temperature rises?
   Show the full solution

   It decreases

4. What effect does increasing pressure have on gas solubility?
   Show the full solution

   It increases it

5. A point plotted above a solubility curve represents what kind of solution?
   Show the full solution

   Supersaturated, and therefore unstable

6. Explain why a carbonated drink fizzes when opened and goes flat when left out.
   Show the full solution

   The drink is bottled under a carbon dioxide pressure well above atmospheric, and gas solubility rises with pressure, so a large quantity dissolves. Opening the bottle drops the pressure above the liquid to atmospheric almost instantly, so the solubility falls sharply and the liquid now holds far more carbon dioxide than it can retain. The excess comes out of solution as bubbles, which is the fizz. Leaving the drink open lets the process continue more slowly, because carbon dioxide keeps escaping into the room and never builds up a pressure above the liquid, so dissolved gas continues to leave until almost none remains and the drink is flat. Warming the drink accelerates both stages, since gas solubility also falls with temperature. Opening drops the pressure so solubility falls and gas escapes; it continues until nearly all has left

7. Explain why potassium nitrate can be purified by recrystallization but sodium chloride cannot.
   Show the full solution

   Recrystallization works by dissolving the impure solid in hot solvent and cooling, so that the substance itself crystallizes out while the impurities, present in much smaller quantities, remain below their own saturation and stay dissolved. The method therefore depends on a large difference in solubility between the hot and cold temperatures. Potassium nitrate has a steep solubility curve, from 110 g per 100 g of water at 60 degrees down to 32 g at 20 degrees, so cooling recovers 78 g of every 110 g dissolved, about seventy percent, in purified form. Sodium chloride's curve is nearly flat, 37 g against 36 g across the same range, so cooling recovers about 1 g in 37 and the method is useless. Sodium chloride is therefore obtained by evaporating the water instead, which recovers all of it but also deposits the impurities. Recrystallization needs a steep solubility curve; sodium chloride's is nearly flat so cooling recovers almost nothing

8. Explain why warm water discharged into a river can kill fish without any pollutant being released.
   Show the full solution

   Because it attacks the oxygen supply from two directions at once. Gas solubility falls as temperature rises, so warmer river water simply cannot hold as much dissolved oxygen, and the concentration available to aquatic organisms drops. At the same time fish are ectothermic, meaning their body temperature follows the water's, so warming raises their metabolic rate and therefore their oxygen demand. Supply falls while demand rises, and the gap can be enough to cause death by suffocation even though the water is chemically unchanged in every other respect. Warmer water also accelerates bacterial decomposition of any organic matter present, which consumes still more oxygen, so the effect compounds. This is why thermal discharges are regulated as a form of pollution. Warmer water holds less dissolved oxygen while raising the fish's metabolic demand, so supply falls as demand rises

9. Explain how a supersaturated solution can exist and why it crystallizes suddenly when disturbed.
   Show the full solution

   A supersaturated solution holds more solute than its solubility permits, which is possible because crystallization requires a starting point. Crystals grow on a nucleus, which may be a dust particle, a scratch on the glass or a small existing crystal, and in a solution that has been cooled slowly in a clean smooth container no such nucleus is present. The dissolved particles have no site on which to assemble into an ordered lattice, so they remain in solution in an unstable condition, holding more than they should. Introducing a seed crystal or scratching the glass provides the missing nucleus, and crystallization then proceeds extremely rapidly because every excess particle has somewhere to go, releasing the energy of crystallization as it does so. Reusable hand warmers work exactly this way, using a supersaturated sodium acetate solution triggered by a metal clicker. No nucleus is available for crystals to grow on; supplying one lets the excess solute deposit all at once

10. Explain why solids and gases respond oppositely to temperature, in terms of what has to happen for each to dissolve.
    Show the full solution

    The two cases differ in which step is the obstacle. Dissolving a solid requires its lattice to be broken apart, which costs energy, and raising the temperature supplies more thermal energy and makes that separation easier, so solubility generally rises. A gas has no lattice to break: its particles are already separate and widely spaced, so nothing about dissolving it requires an energy input for separation. What holds a dissolved gas molecule in the liquid is its attraction to the solvent, and the only relevant effect of raising the temperature is to give it more kinetic energy with which to escape. Heating therefore helps a solid dissolve and helps a gas leave. Stated generally, dissolving a gas is an exothermic process overall while dissolving most solids is endothermic, and heating opposes an exothermic process, which is the principle lesson 10.6 develops as Le Chatelier's. Heating helps break a solid's lattice but only helps a dissolved gas escape, since a gas has no lattice to break

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. What does ppm mean as a fraction?
   Show the full solution

   Parts per \( 10^{6} \)

2. For a dilute aqueous solution, 1 ppm equals what concentration in mg/L?
   Show the full solution

   1 mg/L

3. Convert 2.5 ppm to ppb.
   Show the full solution

   Multiply by 1000. 2500 ppb

4. Convert 0.0050 percent to ppm.
   Show the full solution

   Multiply by \( 10^{4} \). 50 ppm

5. Find the mass of solute in 500.0 mL of a 2.5 ppm aqueous solution.
   Show the full solution

   2.5 ppm is 2.5 mg/L, and 500.0 mL is 0.5000 L. 1.25 mg

6. Explain why 1 ppm equals 1 mg/L for water but not for other solvents.
   Show the full solution

   The equivalence depends entirely on water's density being close to 1.00 g/mL. One liter of dilute aqueous solution therefore has a mass of about 1000 g, which is \( 10^{6} \) milligrams, so one milligram of solute in that liter is one part in \( 10^{6} \) by mass, which is one ppm. For a solvent of different density the arithmetic changes: one liter of hexane has a mass of about 655 g, so 1 mg in a liter is one part in 655 000, or about 1.5 ppm. The shortcut is a convenience of the particular solvent rather than a definition, and it also requires the solution to be dilute enough that the solute does not appreciably change the density. Quoting mg/L avoids the issue entirely, which is why regulatory limits are often given that way. It relies on water's density of 1 g/mL making a liter weigh \( 10^{6} \) mg; other solvents have different densities

7. A sample is reported as 0.8 ppm and a limit as 900 ppb. Determine whether the sample exceeds the limit.
   Show the full solution

   The two figures are in different units, so they cannot be compared until one is converted, and comparing 0.8 against 900 without converting would suggest the sample is far below the limit. Convert the sample to ppb by multiplying by 1000: \( 0.8 \text{ ppm} = 800 \text{ ppb} \). Now the comparison is legitimate, and 800 ppb is below the limit of 900 ppb, so the sample passes, though only by about eleven percent. Alternatively convert the limit to ppm, giving 0.9 ppm against the sample's 0.8 ppm, which reaches the same conclusion. Either direction works; what is not optional is converting before comparing. 0.8 ppm is 800 ppb, which is below the 900 ppb limit

8. Explain what the choice of unit in a regulation tells you about the substance.
   Show the full solution

   Units are chosen so that the values of interest are legible, so the unit encodes the scale at which the substance matters. A contaminant regulated in parts per million becomes a concern at concentrations around one in a million, while one regulated in parts per billion is a concern at concentrations a thousand times smaller, which indicates far greater toxicity or a much lower threshold for harm. Lead's action level of 15 ppb, against limits for substances like chloride that are quoted in hundreds of ppm, records a difference of roughly four orders of magnitude in how dangerous the two are per unit mass. The unit therefore carries information beyond arithmetic convenience, which is also why a report that mixes units without saying so is genuinely misleading rather than merely untidy. It reflects the concentration at which the substance matters, so ppb indicates roughly a thousandfold greater potency than ppm

9. Explain why atmospheric carbon dioxide is reported by volume rather than by mass, and what difference it makes.
   Show the full solution

   Because the quantity that matters physically is the number of molecules, and for a gas a fraction by volume is the same as a fraction by moles, since Avogadro's law makes volume proportional to particle count at fixed temperature and pressure. Infrared absorption in lesson 7.7 depends on how many absorbing molecules the radiation encounters, not on their combined mass, so 421 ppm by volume states directly that 421 molecules in every million are carbon dioxide. A mass fraction would be a larger number for the same air, because a carbon dioxide molecule at 44 g/mol is heavier than the average air molecule at about 29 g/mol, so the same molecular fraction corresponds to a mass fraction of roughly 640 parts per million. The two figures describe the same air, and reporting one as though it were the other would overstate or understate the absorbing molecules by about fifty percent. Volume fraction equals mole fraction for gases, which is what absorption depends on; by mass the same air would read about 640 ppm

10. Explain how you would prepare 1.00 L of a 5.00 ppm standard solution from a 1000 ppm stock, and why the method matters.
    Show the full solution

    The dilution equation of lesson 9.4 applies with concentrations in ppm, since the units cancel between the two sides: \( V_1 = \frac{C_2V_2}{C_1} = \frac{5.00 \times 1000}{1000} = 5.00 \) mL of stock, made up to 1.00 L. In practice that means pipetting 5.00 mL of the stock into a 1.00 L volumetric flask and adding water to the mark. The method matters because measuring 5.00 mL accurately requires a volumetric pipette rather than a measuring cylinder, and because any error in that small volume translates directly into the standard's concentration. For much larger dilution factors a single step would require an impractically small volume, so serial dilution is used instead: preparing a 10.0 ppm intermediate first and then diluting that. Standards prepared this way are what instruments are calibrated against, so their accuracy limits every subsequent measurement. Dilute 5.00 mL of stock to 1.00 L with a volumetric pipette and flask; for larger factors use serial dilution

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. Define reaction rate and give its usual units.
   Show the full solution

   Change in concentration per unit time, in mol/(L·s)

2. Name three properties that could be tracked to follow a reaction.
   Show the full solution

   Gas volume, mass, color, pH or conductivity (any three)

3. When is a reaction usually fastest?
   Show the full solution

   At the start, when reactant concentrations are highest

4. A concentration falls from 0.50 M to 0.35 M in 30 s. Find the average rate.
   Show the full solution

   \( \frac{0.15}{30} \). 0.0050 mol/(L·s)

5. What does the gradient of a concentration-time graph represent?
   Show the full solution

   The instantaneous rate of reaction

6. Explain why the rate of a reaction falls as it proceeds.
   Show the full solution

   Because the reactants are being used up, so their concentrations fall. A reaction occurs when reactant particles collide with sufficient energy, and the frequency of those collisions depends on how many reactant particles are present per unit volume. As the reaction consumes them there are fewer particles in the same space, collisions between them become less frequent, and the rate drops accordingly. Where a solid reactant is involved the effect is compounded, because the solid is being eaten away and its exposed surface area shrinks. The consequence is that a concentration-time graph curves rather than running straight, steep at the start and flattening toward the end as the reaction approaches completion. Falling reactant concentration means fewer collisions per second, and a consumed solid also loses surface area

7. Explain why a rate must state which substance was measured.
   Show the full solution

   Because different substances in the same reaction change at different rates, in the ratio of their coefficients. In \( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \), hydrogen is consumed three times as fast as nitrogen and ammonia is produced twice as fast as nitrogen is consumed, so three different numerical rates describe the same reaction perfectly correctly. Reporting a bare number therefore leaves the reader unable to reproduce or compare the result, and comparing two experiments that happened to measure different substances would give a meaningless ratio. The convention is either to name the substance explicitly or to divide each rate by the corresponding coefficient, which produces a single rate for the reaction as a whole. Substances change at rates in the ratio of their coefficients, so the number is ambiguous without naming one

8. Diamond converting to graphite releases energy, yet diamonds do not change. Explain what this shows.
   Show the full solution

   It shows that the energy change of a reaction and its rate are separate properties determined by different things. Graphite is slightly lower in energy than diamond, so the conversion is energetically favorable and would release a small amount of energy if it occurred. Whether it occurs at a measurable speed depends instead on the height of the energy barrier between the two, and converting diamond to graphite requires breaking and rearranging an enormous number of strong covalent carbon bonds in a rigid three-dimensional network. That barrier is so high that at room temperature essentially no carbon atoms have enough energy to surmount it, and the conversion proceeds at a rate that would take vastly longer than the age of the universe to notice. Diamonds are therefore not stable in the strict sense; they are simply unable to change. Energy released and reaction rate are independent; the barrier can be high enough to prevent a favorable change entirely

9. Explain why collecting carbon dioxide over water understates the early rate, and what to use instead.
   Show the full solution

   Carbon dioxide is appreciably soluble in water, unlike hydrogen or oxygen, so some of the gas produced dissolves into the water it is being collected over rather than appearing in the measuring vessel. The effect is largest at the start, when the water is not yet saturated with carbon dioxide and absorbs it readily, so the measured volume in the early seconds is lower than the volume actually produced and the calculated initial rate is too low. As the water approaches saturation the loss diminishes, so the error changes through the experiment rather than being a constant offset, which distorts the shape of the curve as well as its values. Using a gas syringe avoids water entirely and collects everything produced, which is why it is the standard apparatus for reactions generating soluble gases. Some carbon dioxide dissolves, most in the early stages; use a gas syringe instead of collection over water

10. A student reports "the rate was 0.015". Explain what is missing and why each omission matters.
    Show the full solution

    Three things are missing. There are no units, so the number could be moles per liter per second, grams per minute or centimeters cubed per second, differing by many orders of magnitude, and a reader cannot reproduce or compare it. The substance is not named, so it is unclear whether the figure refers to a reactant disappearing or a product appearing, and those differ by the ratio of the coefficients. The interval is not stated, and since the rate changes throughout a reaction, an average over the first ten seconds and an average over the whole run are different quantities. A complete report would read: the average rate of consumption of hydrochloric acid over the first 40 seconds was 0.015 mol/(L·s). Each omission makes the figure less usable, and together they make it useless. Units, the substance measured and the time interval; without all three the number cannot be reproduced or compared

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.

1. State the two conditions for a successful collision.
   Show the full solution

   Sufficient energy, and correct orientation

2. Define activation energy.
   Show the full solution

   The minimum energy colliding particles must have for a reaction to occur

3. What happens in a collision with insufficient energy?
   Show the full solution

   The particles bounce apart unchanged; no bonds break

4. Do all particles at a given temperature have the same energy?
   Show the full solution

   No; they have a range of energies around an average

5. What two quantities multiply to give the rate?
   Show the full solution

   The total collision frequency and the fraction of collisions that are successful

6. Explain why the orientation condition matters more for large molecules than for small ones.
   Show the full solution

   Because the fraction of approaches that count as correctly aligned falls as the molecule grows. Two atoms colliding have no orientation requirement worth speaking of, since a sphere looks the same from every direction and any approach delivers the energy where it is needed. A large organic molecule may have dozens of atoms with only one reactive site, so a collision arriving anywhere else deposits its energy in the wrong part of the molecule and achieves nothing, however energetic it was. The proportion of successful approaches can fall by several orders of magnitude for this reason alone, independently of any change in energy. It is also why enzymes are such effective catalysts: they bind their substrate in a fixed alignment, so that the orientation requirement is satisfied before the reacting species meet. A large molecule has one reactive site among many atoms, so most approaches deliver energy to the wrong place

7. Explain why a collision below the activation energy achieves nothing at all, rather than achieving something partial.
   Show the full solution

   Because reaching the products requires passing over an energy barrier, and a system that does not reach the top of the barrier simply returns to where it started. Partway up the barrier the existing bonds are stretched and distorted but not broken, and that arrangement is higher in energy and therefore unstable, so it relaxes back to the original molecules as soon as the particles separate. There is no intermediate product to accumulate. The consequence for kinetics is that low-energy collisions contribute nothing whatever to the rate, so increasing their number changes nothing, and any factor that raises the rate must either supply more energy or otherwise increase the fraction of collisions that clear the barrier. This is exactly why lowering the barrier with a catalyst is so effective, as lesson 10.4 shows. Without clearing the barrier the distorted arrangement relaxes back unchanged, so no partial product accumulates

8. Explain why solid reactants generally react more slowly than solutions of the same substances.
   Show the full solution

   Because collisions can only occur where the reactants can actually meet, and in two solids that is only at the small area of contact between the surfaces. The great majority of particles in each solid are buried in the interior, held in fixed lattice positions with no access to the other reactant at all, so they cannot participate. In solution the lattices have been broken up and the particles are dispersed throughout the liquid and free to move, so every particle of one reactant can encounter particles of the other, and the collision frequency is enormously higher. This is why so many laboratory reactions specify aqueous solutions, why grinding two solids together speeds a reaction a little by increasing contact area, and why melting them speeds it a great deal by mobilizing every particle. In solids only the surface particles can meet; dissolving frees every particle to collide

9. A reaction has an activation energy so low that almost every collision has enough energy. Predict what limits its rate.
   Show the full solution

   If the energy condition is satisfied by nearly every collision, it can no longer be the bottleneck, so the rate is limited by how often the particles meet at all and by whether they are correctly oriented when they do. For small simple particles the orientation requirement is also weak, so the limit becomes the collision frequency itself, which in solution is set by how fast the particles can diffuse through the solvent to find each other. Such reactions are called diffusion-controlled and they are about as fast as chemistry in solution can be. A practical consequence is that raising the temperature barely speeds them up, since the fraction clearing the barrier was already near one and only the modest increase in movement remains, which is a useful diagnostic: a reaction insensitive to temperature probably has a very low barrier. The collision frequency itself, set by diffusion; such reactions are barely sped up by heating

10. Explain why particles collide billions of times a second yet reactions can take hours.
    Show the full solution

    Because the fraction of collisions that succeed can be extraordinarily small, and the rate is the product of the two quantities rather than the frequency alone. A particle in solution undergoes of the order of \( 10^{12} \) collisions per second, but if only one collision in a million carries enough energy to clear the activation barrier and only one in a hundred of those is correctly oriented, then successful collisions occur roughly once in \( 10^{8} \), leaving about \( 10^{4} \) productive events per second per particle. Since a mole contains \( 6 \times 10^{23} \) particles and a substantial fraction of them must react for the change to be complete, a reaction with a high barrier can easily take hours. The enormous collision frequency and the enormous rarity of success very nearly cancel, and what is left over is the observable rate. The success fraction can be one in \( 10^{8} \) or smaller, so an enormous collision frequency still leaves a slow reaction