AP Chemistry Practice Exam 2 - Build Your Test-Taking Confidence - 5 Steps to a 5: AP Chemistry 2017 (2016)

5 Steps to a 5: AP Chemistry 2017 (2016)

STEP 5

Build Your Test-Taking Confidence

AP Chemistry Practice Exam 2—Multiple Choice

ANSWER SHEET

The AP exam is a timed exam; keep this in mind as you prepare. When taking the various tests presented in this book, you should follow the AP exam rules as closely as possible. Anyone can improve his or her score by using notes, books, or an unlimited time. You will have none of these on the AP exam, so resist the temptation to use them on practice exams. Carefully time yourself, do not use other materials, and use a calculator only when expressly allowed to do so. After you have finished an exam, you may use other sources to go over questions you missed or skipped. We have seen many students get into trouble because the first time they attempted a test under “test conditions” was on the test itself.

AP Chemistry Practice Exam 2, Section I (Multiple Choice)

Time—1 hour and 30 minutes
NO CALCULATOR MAY BE USED WITH SECTION I

Which of the labeled arrows in the diagram above represents the strongest intermolecular force?

(A) A

(B) B

(D) D

The following information applies to questions 2–5.

A student builds a Galvanic cell to take advantage of the following reaction:

Zn(s) + 2 AgNO₃ (1 M ) → 2 Ag(s) + Zn(NO₃ )₂ (1 M )

The cell”s voltage is measured and found to be +1.56 volts.

What happens to the voltage when a small amount of silver nitrate is added to the silver compartment?

(A) There is no change in the voltage.

(B) The voltage becomes zero.

(D) The voltage decreases, but stays positive.

What happens to the cell voltage when the student replaces the Zn(NO₃)₂ solution with 1 M ZnCl₂ ?

(A) There is no change in the voltage.

(B) The voltage becomes zero.

(D) The voltage decreases, but stays positive.

If the standard reduction potential of silver is 0.80 V, what is the standard reduction potential of zinc?

(A) +0.76 V

(B) −0.76 V

(D) +0.04 V

What happens to the cell voltage after the cell has reached equilibrium?

(A) There is no change in the voltage.

(B) The voltage becomes zero.

(D) The voltage decreases, but stays positive.

Use the following information to answer questions 6–10.

Many metal salts crystallize from solution in the form of a hydrate. If the formula of the anhydrous salt is A_a X_x , then the generic formula of the hydrated form would be A_a X_x •xH₂ O. A student conducts an experiment to determine the formula of a metal oxide by collecting the following data:

The hydrated salt is finely powdered to ensure complete water loss to produce the anhydrous salt.

How many grams of water were in the hydrate?

(A) 5.557 g

(B) 4.737 g

(D) 0.820 g

What is the approximate percent water in the hydrated salt?

(A) 14%

(B) 86%

(D) 28%

The student did not have time to finish the experiment during the lab period and was forced to store the anhydrous salt and crucible in his lab drawer until the next day. At the beginning of the next lab period the student weighed the anhydrous salt and beaker. After calculating the percent water in the sample, the student found that the percentage was lower than predicted. Assuming all weighing were done correctly, what might be the cause of the lower than expected percentage?

(A) The sample dried further overnight.

(B) The student performed one or more weighings before the crucible had cooled to room temperature.

(D) The sample decomposed to another salt.

In another experiment on a different metal hydrate a student found that the salt was 62.9% water. In this case, the molar mass of the anhydrous salt was 106 g mol⁻¹. Which of the following general formulas gives the correct value of x ?

(A) A_a X_x •5H₂ O

(B) A_a X_x •10H₂ O

(D) A_a X_x •6H₂ O

Manganese normally forms one of four oxides. The oxides are MnO, Mn₂O₃ , MnO₂ , and Mn₃ O₄ . Which of the four oxides has the highest percentage of oxygen? (Molar masses: O = 16.0 g mol⁻¹ , Mn = 54.9 g mol⁻¹ , MnO = 70.9 g mol⁻¹ , Mn₂ O₃ = 157.8 g mol⁻¹ , MnO₂ = 86.9 g mol⁻¹ , and Mn₃ O₄ = 228.7 g mol⁻¹ .)

(A) MnO₂

(B) Mn₃ O₄

(D) MnO

For the following reaction, 2 NO(g) + O₂(g) → 2 NO₂ (g), the rate law is: Rate = k [NO]² [O₂ ]. In one experiment, the rate of appearance of NO₂ was determined to be 0.0138 M s⁻¹ when [NO] = 0.0125 M and [O₂ ] = 0.0125 M . What was the value of the rate constant?

(A) 7.1 × 10³ M ² s⁻¹ .

(B) 7.1 × 10³ M s⁻¹ .

(D) 7.1 × 10³ .

The table above gives the initial concentrations and rate for three experiments involving the decomposition of urea, CN₂H₄ O. The reaction is H⁺ (aq) + 2 H₂ O(l) + CN₂ H₄ O(aq) → 2 NH₄ ⁺ (aq) + HCO₃ ⁻ (aq). What is the rate law for this reaction?

(A) Rate = k [CN₂ H₄ O]

(B) Rate = k [CN₂ H₄ O]² [H⁺ ]²

(D) Rate = k [CN₂ H₄ O]² [H⁺ ]

Oxalic acid, H₂C₂ O₄ , is a useful chemical for rust removal. A student prepared five oxalic acid samples by dissolving 0.9000 grams of oxalic acid in 100.00 mL of water and pipetting 10.00 mL samples of this solution into five separate beakers. Each of the samples was diluted with deionized water, and an appropriate indicator was added as an indicator. The samples were then titrated with standard 0.05000 M sodium hydroxide, NaOH, until the appearance of a permanent color change of the indicator indicated the endpoint of the titration. The following volumes were obtained. Molar mass of H₂ C₂ O₄ = 90.04 g mol⁻¹ .

The student calculated a concentration of approximately 0.20 M in each case. This is not the correct value. What is the most likely mistake that the student made?

(A) The student used the total sample volume of the acid instead of the pipetted sample volume.

(B) The student did not use the correct mole ratio in the calculation.

(D) The student contaminated the samples during preparation.

Given the above standard reduction potentials, estimate the approximate value of the ΔG ° for the following reaction:

Zn + Co²⁺ → Zn²⁺ + Co

(A) −4.6 × 10⁴ J mol⁻¹

(B) −9.3 × 10⁴ J mol⁻¹

(D) +4.6 × 10⁴ J mol⁻¹

Use the following information to answer questions 15–19.

pH versus volume of titrant added

The diagram above represents the idealized titration curve for the reaction of pure sodium carbonate, Na₂ CO₃ , with a strong acid such as hydrochloric acid, HCl. E and F represent the pH at the endpoints corresponding to the formation of HCO₃ ⁻ and H₂ CO₃ , respectively. G and H correspond to the quantity of acid required to reach the endpoints.

A trial run used a sample of pure sodium carbonate. How does the volume of acid necessary to reach G from 0 compare to the volume of acid necessary to get from G to H?

(A) They are the same.

(B) It takes more to reach point G.

(D) It is impossible to determine.

The analysis of a sample contaminated with NaHCO₃gave slightly different results. How does the volume of acid necessary to reach G from 0 compare to the volume of acid necessary to get from G to H for the second sample?

(A) It takes more to get from G to H.

(B) It takes more to reach point G.

(D) It is impossible to determine.

How could a student determine if there was a strong acid or a strong base contaminant in the original sample?

(A) The presence of an acid contaminant would require less acid to reach H from G than to reach G from 0.

(B) The presence of a base contaminant would require less acid to reach G from 0 than to reach F from G.

(D) It is impossible to determine.

In addition to water, what are the predominant species in solution at F?

(A) Na₂ CO₃ and HCl

(B) Na⁺ , Cl⁻ , and H₂ CO₃

(D) Na⁺ , Cl⁻ , H⁺ , and CO₃ ²⁻

At what point on the graph for the titration of pure sodium carbonate is the pH = pK_a2 for carbonic acid?

(A) At point G

(B) Halfway between the start and point G

(D) Halfway between points G and H

Three steel containers hold gas samples. The containers are all the same size and at the same temperature. One container has 4.0 g of helium, another has 28.0 g of nitrogen, and the third has 44.0 g of carbon dioxide. Pick the FALSE statement from the following list:

(A) The densities increase in the order helium < nitrogen < carbon dioxide.

(B) The number of molecules in all the containers is the same.

(D) The average speed of all the molecules is the same.

When potassium perchlorate, KClO₄, dissolves in water, the temperature of the resultant solution is lower than the initial temperature of the components. Which of the following conclusions may be related to this?

(A) This is a spontaneous process because it is exothermic.

(B) This is a spontaneous process because of an entropy increase.

(D) This is a spontaneous process because it is exothermic.

What is the reason that the lightest member of Group 15 does not follow the trend of the other members, which show that the boiling point decreases with decreasing atomic mass of the Group 15 element?

The graph shows the variation of boiling point with Group number for the hydrogen compounds of the four lightest members of Group 15 on the periodic table (NH₃ , PH₃ , AsH₃ and SbH₃ ).

(A) Ionic bonds

(B) Hybrid orbitals

(D) Hydrogen bonding

A dimer consists of two closely associated molecules. In the gas phase, acetic acid tends to form dimers as illustrated on the left in the above diagram. Acetyl chloride, on the right in the above diagram, is not very efficient in forming dimers. Why is acetic acid better able to form dimers than acetyl chloride?

(A) The molecular mass of acetyl chloride is higher than that of acetic acid making it harder for the acetyl chloride to form dimers.

(B) It is easier to form a covalent bond between acetic acid molecules than between acetyl chloride molecules.

(D) Acetic acid is an acidic compound but acetyl chloride is a neutral compound.

Two compounds with the formula C₂H₂ Cl₂ appear in the above diagram. These two compounds are isomers. The molecules are planar and have the approximate structures shown in the diagram. The boiling point of trans-1, 2-dichloroethene is 47.5°C and the boiling point of cis-1,2-dichloroethene is 60.3°C. Which of the following best explains why cis-1,2-dichloroethene has a higher boiling point than its isomer, trans-1, 2-dichloroethene?

(A) The higher boiling isomer is more polar than the other isomer.

(B) The higher boiling isomer is better able to form hydrogen bonds than the other isomer.

(D) The higher boiling isomer has greater London dispersion forces than the other isomer.

Use the information on the following proposed mechanism to answer questions 25–27.

The above represents a proposed mechanism for the gaseous dinitrogen pentoxide, N₂O₅ . What are the overall products of the reaction?

(A) N₂ O₄ (g) + O₂ (g)

(B) 2 NO₂ (g) + O₂ (g)

(D) 4 NO₂ (g) + O₂ (g)

Choose the energy profile that best describes this mechanism.

(A)

(B)

(C)

(D)

What is the rate law for the reaction?

(A) Rate = k [N₂ O₅ ]²

(B) k = [N₂ O₅ ]

(D)

Use the information on the containers in the following diagram to answer questions 28–30.

Approximate molar masses:

If a sample of Kr effuses at a rate of 35 mL per second at 298 K, which of the gases below will effuse at approximately double the rate under the same conditions?

(A) Ar

(B) He

(D) Impossible to determine

Assuming all four gases are behaving ideally, which of the following is the same for all the gas samples?

(A) Average kinetic energy of the atoms

(B) Average speed of the atoms

(D) All properties are the same for gases behaving ideally

Which of the gases will show the greatest deviation from ideal behavior?

(A) He

(B) Ne

(D) Kr

Use the information on the containers in the following diagram to answer questions 31–34 concerning the following equilibrium.

An equilibrium mixture (container D) is at 75°C. Which of the following changes may increase the amount of the HCl in the container?

(A) Increasing the volume of the container

(B) Decreasing the volume of the container

(D) Adding 1 mole of He(g) to the container

Containers A, B, and C are not at equilibrium. If each container begins with equal amounts of the indicated substances present at the same temperature, which of the three will reach equilibrium first?

(A) A because producing products from reactants is faster than the reverse.

(B) B because producing reactants from products is faster than the reverse.

(D) It is impossible to determine which will reach equilibrium first.

If the initial partial pressure of Cl₂in container A is 1.0 atm and the initial partial pressure of H₂ O is also 1.0 atm, what will be the pressure at equilibrium?

(A) > 2.0 atm

(B) = 2.0 atm

(D) Impossible to determine

If the moles of HCl in container B are equal to five times the O₂present, what can be said about the moles of O₂ present at equilibrium?

(A) It will be zero because it is the limiting reagent

(B) It will remain the same

(D) It is impossible to determine

Use the following information on the bases in the following diagram to answer questions 35–37.

Ammonia is only present as a reference. Questions 35–37 only refer to the other three bases.

Solutions of methylamine, dimethylamine, and hydroxylamine are titrated. The base concentrations were 0.1000M , and 0.1000 M hydrochloric acid, HCl, was used for the titrations. Which of the three bases will yield the highest pH at the equivalence point?

(A) Hydroxylamine

(B) Methylamine

(D) The concentrations of all the bases were the same; therefore, the pH at the equivalence point will be the same.

All the bases in the diagram behave as Brønsted-Lowry bases in the same way; in each case, they accept a hydrogen ion to the same atom. How is this acceptance of a hydrogen ion accomplished?

(A) A hydrogen ion attaches to the lone pair on the nitrogen atom.

(B) The hydroxide ion reacts with the hydrogen ion to form water.

(D) The hydrogen ion combines with a hydrogen atom from the base to form H₂ gas.

Which of the following explains why the pH of a hydroxylamine solution is lower than any of the other solutions?

(A) The −OH is capable of donating a hydrogen ion, which will lower the pH.

(B) The presence of carbon makes the bases less stable.

(D) There is insufficient information to explain this observation.

Oxidation of which of the following substances will yield a stronger acid?

(A) HNO₂

(B) HNO₃

(D) H₃ PO₄

Which of the compounds in the above diagram is capable of participating in hydrogen bonding?

(A) C₃ H₉ N

(B) CH₃ F

(D) C₄ H₁₁ N

Choose the reaction expected to have the greatest decrease in entropy.

(A) C(s) + CO₂ (g) → 2 CO(g)

(B) 2 Na(s) + O₂ (g) → Na₂ O₂ (s)

(D) 2 NI₃ (s) → 3 I₂ (s) + N₂ (g)

A certain reaction is nonspontaneous under standard conditions, but becomes spontaneous at higher temperatures. What conclusions may be drawn under standard conditions?

(A) ΔH < 0, ΔS < 0 and ΔG = 0

(B) ΔH > 0, ΔS < 0 and ΔG > 0

(D) ΔH > 0, ΔS > 0 and ΔG > 0

Which of the following is the correct order of increasing acid strength?

(A) H₂ SeO₃ < H₂ SO₃ < HClO < HBrO

(B) HClO < H₂ SeO₃ < HBrO < H₂ SO₃

(D) H₂ SO₃ < H₂ SeO₃ < HClO < HBrO

Use the following information on the bases in the following diagram to answer questions 43–44.

A large number of compounds adopt the sodium chloride structure. The following table contains some examples of the compounds in this group with their respective melting points and the sum of the cation-anion radii.

Why are the melting points of the alkaline earth oxides (CaO, SrO, and BaO) so much higher than those of the other two compounds in the table?

(A) Smaller ions have lower lattice energy.

(B) Smaller ionic charges lead to higher lattice energy.

(D) Higher ionic charges lead to higher lattice energy.

Why do the melting points of the alkaline metal oxides decrease in the order CaO > SrO > BaO?

(A) Larger ions have higher lattice energies, which leads to lower melting points.

(B) Smaller ions have lower lattice energies, which leads to higher melting points.

(D) Smaller ions have a greater affinity for oxygen, which leads to a higher melting point.

Joseph Priestly discovered oxygen gas by the decomposition of solid mercury(II) oxide, HgO, to oxygen gas, O₂, and liquid mercury metal, Hg. How many moles of oxygen gas will form when 4.32 g of solid mercury(II) oxide decomposes? The formula mass of mercury(II) oxide is 216 g mol⁻¹ .

(A) 0.100 mol

(B) 0.0100 mol

(D) 0.0150 mol

According to the data in the table above, which of the following best explains the trend in increasing melting points?

(A) All the molecules are nonpolar and, for such molecules, intermolecular forces increase with increasing molar mass.

(B) All the molecules are polar and, for such molecules, intermolecular forces increase with increasing molar mass.

(D) The sequence is a coincidence since all the molecules have the same intermolecular forces.

Determine the H⁺(aq) concentration in 1.0 M phenol, C₆ H₅ OH, solution. (The K _a for phenol is 1 × 10⁻¹⁰ .)

(A) 1 × 10⁻¹⁰ M

(B) 1 × 10⁻⁹ M

(D) 1 × 10⁻⁵ M

Ammonia is the best-known nitrogen-hydrogen compound; however, there are a number of other nitrogen-hydrogen compounds, three of which are in the above diagram. Which of these has the longest average N-N bond length?

(A) N₂ H₂

(B) N₃ H₃

(D) They are all the same.

The above equilibrium was established in a 1.00 L container at a certain temperature. Once the system came to equilibrium, it was found that the following amounts of materials were present in the container:

(NH₄ )₂ CO₃ = 8.00 moles, NH₃ = 4.00 moles, CO₂ = 2.00 moles, and H₂ O = 2.00 moles.

Determine the value of K _c at this temperature.

(A) 8.00

(B) 64.0

(D) 32.0

A solution of a weak base is titrated with a solution of a standard strong acid. The progress of the titration is followed with a pH meter. Which of the following observations would occur?

(A) Initially, the pH slowly decreases, then there is a rapid decrease to give a pH below 7 at the equivalence point.

(B) The pH of the solution gradually decreases throughout the experiment and the pH at the equivalence point is below 7.

(D) Initially, the pH quickly decrease, then there is a gradual decrease to the equivalence point where the pOH equals the pK _b of the base.

A student mixes 100.0 mL of a 0.10M hydrofluoric acid, HF, solution with 50.0 mL of a 0.10 M potassium hydroxide, KOH, solution. The K _a for hydrofluoric acid is 6.8 × 10⁻⁴ . Which of the diagrams below best represents the species, other than H₂ O, in the solution after the acid reacts with the base?

(A)

(B)

(C)

(D)

Which of the following best represents the result for the reaction of 50.0 mL of a 0.20M barium hydroxide, Ba(OH)₂ , solution with 50.0 mL of a 0.10 M sulfuric acid, H₂ SO₄ , solution to form a precipitate of BaSO₄ ?

(A)

(B)

(C)

(D)

The container in the above diagram had an initial volume of 10.0 L. The container was heated until the temperature was double the original temperature. After heating, the pressure was measured and the piston moved down until the pressure was four times the original value. What was the final volume of the container?

(A) 10.0 L

(B) 1.25 L

(D) 20.0 L

The diagram above shows the structure of molecules of CCl₄and CBr₄ and the above table gives the boiling points and molar masses of the compounds. Which of the compounds is nonpolar?

(A) Dichlorodifluoromethane

(B) Dibromodichloromethane

(D) Tetrafluoromethane

2 SO₂ (g) + O₂ (g) 2 SO₃ (g)

The reaction above is allowed to continue until equilibrium is established. After equilibrium is established a catalyst is added to the system. How does the rate of the forward reaction compare to the rate of the reverse reaction after the addition of the catalyst?

(A) The forward rate is faster than the reverse rate.

(B) The forward and reverse rates are the same.

(D) There is insufficient information to determine the relative rates.

A reaction has the following suggested mechanism:

NH₃(aq) + OCl⁻(aq) → NH₂ Cl(aq) + OH⁻ (aq)
NH₃(aq) + NH₂Cl(aq) → Cl⁻ (aq) + N₂ H₅ ⁺ (aq)
OH⁻(aq) + N₂H₅ ⁺ (aq) → N₂ H₄ (aq) + H₂ O(l)
Referring to the above mechanism, which of the following would support the suggested mechanism?

(A) Heating increases the rate of the reaction.

(B) Spectroscopy shows that N₂ H₅ ⁺ (aq) is present in trace amounts.

(D) The first step is the rate-determining step.

Determine the final temperature, in °C, of a sample of helium gas. The sample initially occupied a volume of 5.00 L at 127°C and 875 mm Hg. The sample was heated, at constant pressure, until it occupied a volume of 10.00 L.

(A) 454°C

(B) 527°C

(D) 181°C

A solution contains 2.00 mole of acetic acid, CH₃COOH, and 1.00 mole of calcium acetate, Ca(CH₃ COO)₂ . The solution is able to resist the addition of a small amount of strong acid or strong base with only minor changes in the pH of the solution. Larger quantities of strong acid or strong base can cause a significant change in pH. How many moles of nitric acid, HNO₃ , may be added before the pH begins to change significantly?

(A) 0.500 mole

(B) 1.00 mole

(D) 3.00 mole

Use the following information in the following thermochemical equations to answer questions 59–60.

Determine ΔHfor the combustion of ethanol, C₂ H₅ OH, if H₂ O(g) formed in the above reaction instead of H₂ O(l).

(A) +1280 kJ

(B) −1280 kJ

(D) −1100 kJ

What is the energy change when 72.0 g of water vapor decomposes to the elements at constant pressure? The molar mass of H₂O is 18.0 g mol⁻¹ .

(A) −1144 kJ

(B) −572 kJ

(D) +1144 kJ

STOP: End of AP Chemistry Practice Exam 2, Section I (Multiple Choice).

Answers and Explanations for Exam 2, Section 1 (Multiple Choice)

1 . B —This is a dipole-dipole force, which is stronger than a dipole-induced dipole (A and C) or a London dispersion force (D).

2 . C —This will increase the concentration of Ag⁺ , causing a shift to the right, which will lead to an increase in the cell voltage.

3 . A —The zinc ion concentration remains the same (1 M); therefore, the voltage will remain the same. The identity of the anion associated with the zinc is not important unless the compound is not soluble. Zinc chloride must be soluble; otherwise the student could not make a 1 M solution.

4 . B —The cell voltage (1.56 V) is the sum of the standard reduction potential for silver (0.80 V) and the reverse of the standard reduction potential for zinc. Therefore, (1.56 − 0.80) V = 0.76 V = the reverse of the standard reduction potential for zinc. Reversing the value gives −0.76 V as the standard reduction potential for zinc.

5 . B —As the cell begins to run, the voltage decreases until the system reaches equilibrium where the voltage is zero.

6 . D —The mass of water is the difference in the masses of the hydrated salt and the anhydrous salt. In this case, (mass of crucible plus hydrated salt) − (mass of crucible plus anhydrous salt) = 58.677 g − 57.857 g = 0.820 g. This assumes the weight of the crucible is constant.

7 . A —The percent water is 100% times the grams of water divided by the mass of the hydrated salt. The mass of water is:

(mass of crucible plus hydrated salt) − (mass of crucible plus anhydrous salt) = (58.677 − 57.857) g = 0.820 g water.

The mass of the hydrated salt is:

(mass of crucible plus hydrated salt) − (mass of crucible) = (58.677 − 53.120) g = 5.557 g hydrated salt.

Finally, percent water 14.8%. The simplified calculation would be: Percent water

8 . C —Leaving the sample overnight in the lab drawer would cause the sample to be no longer anhydrous. The mass of the “anhydrous” salt would now be higher indicating a smaller amount of water loss, which would lead to a lower percentage of water in the sample.

9 . B —It is necessary to determine the empirical formula of the compound. If the sample is 63% water, then it is 37% anhydrous salt. Assuming 100 grams of compound, the masses of water and anhydrous salt are 63 g and 37 g, respectively. Converting each of these to moles gives:

Since there are ten times as many moles of water as moles of the anhydrous salt, the formula must be A_a X_x •10H₂ O.

10 . A —The general equation to determine the percent oxygen in the sample is: Percent oxygen =

However, no percent calculation is necessary. The sample with the highest Mn:O ratio will have the highest percentage. The ratios are: MnO = 1:1, Mn₂ O₃ = 1:15, MnO₂ = 1:2, and Mn₃ O₄ = 1:1.3, which means that MnO₂ will have the highest percentage.

11 . C —No calculations are necessary as this is the only answer with the correct units.

12 . A —Comparing experiments 1 and 3 shows that changing the hydrogen ion concentration has no effect upon the rate; therefore the reaction is zero order in hydrogen ion. Comparing experiments 2 and 3 shows that doubling the urea concentration doubles the rate; therefore, the reaction is first order with respect to urea.

13 . B —The first step is to calculate the true concentration of the sample to see how the student results compare. The balanced chemical equation is H₂ C₂ O₄ (aq) + 2 NaOH(aq) → Na₂ C₂ O₄ (aq) + 2 H₂ O(l)

All the volumes are similar; therefore, it is possible to use any one of them and calculate an approximate molarity using rounded numbers for simplicity.

The calculated value should be about half what the student reported. This indicates that the student did not include the 1:2 mole ratio relating the acid to the base or that the student incorrectly used a relationship such as M₁ V₁ = M₂ V₂ .

14 . B —The relationship is: ΔG = −nFE ° (given on the exam)

The number of electrons transferred = 2 = n

F = 96,500 coulombs mol⁻¹ and volt = 1 joule coulomb⁻¹ (both given on the exam), these two relationships lead to F = 96,500 J V⁻¹ mol⁻¹ .

E ° = (0.76 − 0.28) V = 0.48 V

Entering this information in the equation gives ΔG = −(2)(96,500 J V⁻¹ mol⁻¹ )(0.48 V) ≈

−(2)(100,000)(0.50) ≈ −100,000 J mol⁻¹ (actual value = −9.26 × 10⁴ J mol⁻¹ )

15 . A —At point G, all the CO₃ ²⁻ has been converted to HCO₃ ⁻ and the moles of HCO₃ ⁻ will equal the moles of CO₃ ²⁻ originally present. It will require an equal volume of acid to titrate an equal number of moles of HCO₃ ⁻ as required for the CO₃ ²⁻ . For pure sodium carbonate, F will always be 2G.

16 . A —At point G, all the CO₃ ²⁻ has been converted to HCO₃ ⁻ and the moles of HCO₃ ⁻ will equal the moles of CO₃ ²⁻ originally present plus the quantity of HCO₃ ⁻ originally present. It will require a greater volume of acid to titrate a greater number of moles of HCO₃ ⁻ as required for the CO₃ ²⁻ .

17 . C —It would be necessary to titrate the strong base and the CO₃ ²⁻ to reach G. However, it is only necessary to titrate the HCO₃ ⁻ to reach H, which means less acid is necessary.

18 . B —At G the CO₃ ²⁻ is now HCO₃ ⁻ , so no CO₃ ²⁻ remains. The Na⁺ did not react, so it is still present as ions. The Cl⁻ is from the HCl and remains as separate ions in solution. After G, the H⁺ from the acid begins to convert HCO₃ ⁻ to form H₂ CO₃ , which is complete at point F leaving no HCO₃ ⁻ in the solution. Other than water, all species are strong electrolytes and exist as ions in solution. The H₂ CO₃ will be decomposing to H₂ O and CO₂ (g).

19 . D —The pH will equal the pK _a2 when the concentration of HCO₃ ⁻ equals the concentration of H₂ CO₃ . This occurs when one-half of the HCO₃ ⁻ has been converted to H₂ CO₃ .

20 . D —The average kinetic energy, not the average speed, is the same if the temperatures are the same. Each container has one mole of gas, which means that at the same volume and temperature they will have the same pressure. The greater the molar mass, divided by a constant volume, the greater the density. One mole of gas will have Avogadro”s number of molecules.

21 . B —The decrease in temperature indicates that the system absorbed heat, meaning that this is an endothermic process. For an endothermic process to be spontaneous, the entropy must increase.

22 . D —Hydrogen bonding may occur when hydrogen is attached directly to N, O, or F.

23 . C —The two molecules are hydrogen bonded together. Hydrogen bonding is a relatively strong intermolecular force. Acetyl chloride cannot exhibit anything stronger than dipole-dipole forces, which are, in general, weaker than hydrogen bonds.

24 . A —The higher boiling isomer is more polar than the other isomer because the two very electronegative chlorine atoms are on one side, which leads to their polar bonds working together. When the chlorine atoms are on opposite sides, their polar bonds work against each other.

25 . D —Add the equations together and cancel any species that appear on both sides (intermediates).

26 . A —The first step in the mechanism is the slow (rate-determining) step. It is the slowest because it has the highest activation energy. For the first step to be the slow step, the first peak must be the highest.

27 . C —The rate law always considers the slowest step in a mechanism. There is one molecule of N₂ O₅ as the reactant in the slow step; therefore, the rate law will only use the concentration of this reactant raised to a power equal to the number of molecules in the slow step.

28 . C —Use Graham”s law; a molecule with one-fourth the molar mass will diffuse at double the rate. Neon is the nearest to one-fourth the molar mass of krypton.

29 . A —This is a consequence of kinetic molecular theory. The average kinetic energy depends only on the absolute temperature.

30 . D —Heavier nonpolar species exhibit greater London dispersion forces, and stronger attractive forces lead to greater deviation from ideal behavior under a given set of conditions.

31 . A —An increase in volume will cause an equilibrium to shift toward the side with more gas. There are four gas molecules on the left side and five gas molecules on the right side; therefore, an increase in volume will result in a shift to the right, which increases the amount of HCl (and O₂ ). B will have the opposite effect. Cooling an endothermic equilibrium will cause a shift to the left, which will decrease the amount of HCl. D will yield no change because helium is not part of the equilibrium.

32 . D —The amount of time necessary is a kinetics problem. There is not kinetic data presented to make the determination possible.

33 . A —The ICE table for this equilibrium is:

From the equilibrium line on the table, the equilibrium pressure should be:

(1.0 − 2x ) + (1.0 − 2x ) + (+4x ) + (+x ) = 2.0 + x

Therefore, the equilibrium pressure will be greater than 2.0 atm by an amount equal to x .

34 . D —To determine the amount, it would be necessary to know the value of the equilibrium constant. A cannot be correct because it is not possible for the amount of any of the materials to be zero at equilibrium. B and C cannot be correct because at least some of the O₂ would be converted to Cl₂ and H₂ O leading to a decrease in the amount.

35 . B —The pH of a 1.0 M methylamine solution is the highest; therefore, it is the strongest of the bases. For this reason, the pH at the equivalence point of the methylamine titration will be the highest.

36 . A —As in ammonia, all these compounds behave as Brønsted-Lowry bases by accepting a hydrogen ion. The reaction involves the hydrogen ion attaching to the lone pair on the nitrogen atoms.

37 . C —These are all bases because the nitrogen atom is capable of reacting with a hydrogen ion by donating its lone pair to the hydrogen ion. The oxygen atom pulls electron density away from the nitrogen atom causing the nitrogen atom to attract the lone pair more strongly making it less able to donate the pair to a hydrogen ion.

38 . A —Only A can undergo oxidation, as there are higher oxidation states of nitrogen. For example, HNO₃ is the most likely oxidation product and, unlike HNO₂ , it is a strong acid.

39 . D —Hydrogen bonding is possible when hydrogen is attached to N, O, and F. D is the only compound in the diagram where this is true. The simple presence of hydrogen and N, O, or F is insufficient.

40 . B —In general, gases have much higher entropy than either liquids or solids. For this reason, the predictions depends primarily upon which reaction results in the greatest decrease in the number of moles of gas. A and D both result in an increase in the number of moles of gas, so there is an increase in entropy. C shows no change in the number of moles of gas; therefore, the change in entropy will be small.

41 . D —The relationship, given on the equations page of the exam is ΔG = ΔH − T ΔS . Nonspontaneous under standard conditions means: ΔG > 0. To become spontaneous, ΔG must be less than zero. Increasing the temperature will change the T ΔS term (entropy). If ΔH is greater than zero and ΔS is also greater than zero, the combination will be positive as long as the ΔH is greater than T ΔS . As the temperature increases, T ΔS will eventually become larger than ΔH making the process spontaneous.

42 . C —In general, the more oxygen atoms present not attached to hydrogen atoms, the stronger the oxyacid. If two oxyacids have the same number of oxygen atoms not attached to hydrogen atoms, the acid with the more electronegative central atom is the stronger acid. The number of oxygen atoms without hydrogen atoms attached are HClO = 0, HBrO = 0, H₂ SeO₃ = 1, and H₂ SO₃ = 1. The electronegativities increase in the order Se < S and Br < Cl.

43 . D —The melting points of ionic compounds increase with increasing lattice energy. Lattice energy increases with increasing ionic charge and with decreasing sum of ionic radii. It is apparent from comparing NaF to CaO that charge is more important than small changes in radii. The charges are Na⁺ , Ca²⁺ , Sr²⁺ , Ba²⁺ , F⁻ , Cl⁻ , and O²⁻ .

44 . C —The melting points of ionic compounds increase with increasing lattice energy. Lattice energy increases with increasing ionic charge and with decreasing sum of ionic radii. The oxide ion radius is a constant, while the metal radii decrease in the order Ba²⁺ > Sr²⁺ > Ca²⁺ . The decrease in metal radii is due to the smaller ions having fewer electron shells.

45 . B —The balanced chemical equation is: 2 HgO(s) → 2 Hg(l) + O₂ (g)

The calculation is:

46 . A —The substance with the highest melting point has the strongest intermolecular forces. All four molecules are nonpolar; therefore, the intermolecular forces are London dispersion forces. In general, London dispersion forces, for molecules with similar structures, London dispersion forces increase with increasing molar mass.

47 . D —Since phenol has a K _a values given, it is a weak acid, as such the equilibrium expression is:

C₆ H₅ OH(aq) H⁺ (aq) + C₆ H₅ O⁻ (aq)

Use the K _a expression:

This leads to:

[H⁺ ] = (1.0 × 1 × 10⁻¹⁰ )^1/2 = (1 × 10⁻¹⁰ )^1/2 = 1 × 10⁻⁵ M

48 . C —The higher the average number of bonds between the nitrogen atoms, the shorter the bond is. For diazene there are two bonds, for triazene the average is 1.5 bonds, and for tetrazene the average is 1.33 bonds. The length of the average bond length increases in the order 2 < 1.5 < 1.33.

49 . D —The carbon is a solid and the water is a liquid; therefore, neither of these will be in the calculation. Since the volume of the container is 1.00 L, the molarities of the other two substances are 4.00 M NH₃ and 2.00 M CO₂ .

K _c = [NH₃ ]² [CO₂ ] = (4.00)² (2.00) = 32.0

50 . A —Strong acids and strong bases have pH = 7 at the equivalence point. The presence of a weak base with a strong acid lowers this value.

51 . C —The reaction is:

HF(aq) + KOH(aq) → KF(aq) + H₂ O(l)

The potassium hydroxide is the limiting reagent, and all of the hydroxide ions from the base combine with one-half of the hydrogen ions produced from the acid. The potassium fluoride is a strong electrolyte and is present in solution as fluoride ions and potassium ions. The hydrofluoric acid is a weak acid, as indicated by the K _a , and it will partially ionize in solution to form hydrogen ions and fluoride ions with the remaining acid being undissociated.

52 . B —The reaction is Ba(OH)₂ (aq) + H₂ SO₄ (aq) → BaSO₄ (s) + 2 H₂ O(l)

There are 0.010 moles of barium hydroxide and 0.0050 moles of sulfuric acid. The barium hydroxide is in excess and the sulfuric acid is limiting. All the sulfate ions combine with barium ions to form the BaSO₄ precipitate with the excess barium ions remaining in solution. All the hydrogen ions from the acid react with the hydroxide ions from the base to produce water and leave the excess hydroxide ions in solution.

53 . C —There are several ways of solving this problem. One way is to use the combined gas equation (P ₁ V ₁ )/T ₁ = (P ₂ V ₂ )/T ₂ . In this problem, V ₁ = 10.0 L, T ₂ = 2 T ₁ , and P ₂ = 4 P ₁ . Rearranging the combined gas equation and entering the values gives:

54 . D —Tetrafluoromethane is the only nonpolar molecule in the diagram. All the other compounds are polar.

55 . B —The catalyst will increase the rate of both the forward and reverse reactions. The rates will still remain the same as the presence of a catalyst does not alter the position of the equilibrium, just the time necessary to reach equilibrium.

56 . B —The presence of N₂ H₅ ⁺ (aq) supports this mechanism because the detection of an intermediate supports the mechanism. None of the other choices support or refute the overall mechanism.

57 . B —It is necessary to convert the temperature to kelvin and back again.

T ₂ = (V ₂ T ₁ )/V ₁ = [(10.00 L × (127 + 273) K)/(5.00 L)] − 273 = 527°C

Simplified by (10.00/5.0) = 2.00; therefore, 400 K × 2 = 800 − 273 = 527°C.

58 . C —The original solution is a buffer, which will resist changes in pH until all the acetic acid or acetate ions are reacted. The reaction of nitric acid with the acetate ion is HNO₃ (aq) + CH₃ COO^- (aq) → CH₃ COOH(aq) + NO₃ ⁻(aq)

One mole releases 2.00 mole of acetate ion into the solution. As long as any of the acetate ion remains, the solution will have some buffering ability and the pH will remain about the same. Once sufficient nitric acid has been added to react with all the acetate ion, it is possible to drastically lower the pH by adding more acid. The reaction will require 2.00 moles of nitric acid to completely react with the acetate ion.

59 . D —Using Hess”s law:

It is possible to simplify the problem by determining the heat of vaporization of water (H₂ O(l) → H₂ O(g)), which is 90 kJ.

60 . C —The decomposition of water vapor is the reverse of the last reaction shown; therefore, the enthalpy change is positive instead of negative. The amount of water decomposing is 4.00 mole, which is double the amount of water in the reaction. Double the water will require double the energy.

AP Chemistry Practice Exam 2, Section II (Free Response)

Question 1

A simplified diagram of one cell of a lead-acid battery as used in most automobiles is shown above. The half-reactions are:

PbO₂ (s) + HSO₄ ⁻ (aq) + 3 H⁺ (aq) + 2 e⁻ → PbSO₄ (s) + 2 H₂ O(l)
Pb(s) + HSO₄ ⁻ (aq) → PbSO₄ (aq) + H⁺ (aq) + 2 e-

The standard cell potential is +2.04 V at 25°C. Initially the electrode on the left is pure lead and the electrode on the right is pure lead coated with lead(IV) oxide.

(a) The standard reduction potential for the PbO₂ half-reaction is 1.68 V.

(i) Write the overall reaction for the lead-acid battery.

(ii) Calculate the value of the standard reduction potential for the Pb half-reaction.

(b) A student constructs a brand-new cell and it is fully charged.

(i) Calculate the initial value of ΔG ° for the cell.

(ii) Calculate the value of K for the cell.

(iii) The cell is allowed to operate until it reaches equilibrium. What is the value of ΔG ° for the cell at equilibrium?

(d) Why is the PbO₂ electrode made of Pb coated with PbO₂ and not pure PbO₂ ?

Question 2

2 NO(g) + Cl₂ (g) → 2 NOCl(g)

Thermodynamic values related to the above reaction are given in the table below.

(a) Determine the enthalpy change for the above reaction.

(b) Assuming that the NO bond in NOCl is a double bond, estimate the NO bond energy in NO.

(c) Does the NO bond energy in NO agree with any of the values in the table? What is there about the electronic structure of NO that might be the cause of this observation?

(d) Calculate the entropy change for the reaction.

(e) Is this reaction spontaneous or nonspontaneous at 25°C? Justify your prediction.

Question 3

5 Fe²⁺ (aq) + MnO₄ ⁻ (aq) + 8 H⁺ (aq) → Mn²⁺ (aq) + 5 Fe³⁺ (aq) + 4 H₂ O(l)

The above reaction is to be used in the analysis of a sample containing iron metal. A sample is dissolved in acid and the iron reduced to iron(II) ions with a little solid zinc. Excess zinc is filtered from the solution and a titration with standard potassium permanganate is performed immediately before air has had time to oxidize any of the iron(II) ions. The presence of a permanent pink color of excess permanganate ion indicates the endpoint of the titration.

(a) It is possible to standardize the potassium permanganate solution using solid iron(II) ammonium sulfate hexahydrate, Fe(NH₄ )₂ (SO₄ )₂ •6H₂ O. You are given a potassium permanganate solution. Outline the general experimental procedure (not the calculations) for standardizing this solution.

(b) Show how to calculate the concentration of the potassium permanganate solution.

(d) If some of the excess zinc was not filtered from the solution, would the reported percentage of iron in the sample be higher, lower, or the same as when the zinc is removed? Explain.

(e) Standardization with iron(II) ammonium sulfate hexahydrate has the advantage of employing the same reaction as in the iron determination. However, there is a problem with using this compound to standardize the potassium permanganate. What might this problem be?

(f) Give the electron configuration of iron(II) ions, Fe²⁺ .

Question 4

Ca(CN)₂ (s) + H₂ SO₄ (aq) + 2 H₂ O(l) → 2 HCN(g) + CaSO₄ •2H₂ O(s)

It is possible to generate HCN gas by the above reaction. The gas is extremely toxic and great care is necessary when using this compound or related cyanides in any form.

(a) Calculate the number of moles of HCN present in 1.00 L of this gas at 273 K and 1.00 atm.

(b) What volume of HCN gas would the reaction of 1.00 g of Ca(CN)₂ with excess H₂ SO₄ form? The volume is measured at 298 K and 1.00 atm. The molar mass of Ca(CN)₂ is 92.1 g mol⁻¹ .

(c) The reaction in (b) went to completion; however, the volume of gas generated was lower than expected. Explain why the volume was low.

(d) In the box below, draw the Lewis electron-dot structure for HCN.

Question 5

Answer each of the following with respect to the plot of the boiling points of the hydrogen compounds in the above graph. Group 15 is the nitrogen family, Group 16 is the oxygen family, and Group 17 is the halogen family.

(a) On the above plot draw a line to indicate the boiling points of the hydrogen compounds in Group 14 (carbon family).

(b) What is the cause of the upward turn in the three lines shown on the graph?

Question 6

Five beakers (A–E) are on a countertop. Each contains 200 mL of a 0.10 M solution. Beaker A contains SrCl₂ ; beaker B contains NH₄ Cl; beaker C contains CH₃ CH₂ OH; beaker D contains Na₂ C₂ O₄ ; and beaker E contains Ni(NO₃ )₂ .

(a) Which beaker has the lowest pH and why?

(b) Which solution is most likely to be colored?

(d) Which solution has the lowest electrical conductivity and why?

Question 7

The solubility product constants for some insoluble sulfates are given in the table above.

(a) Write a balanced chemical equation for the dissolution equilibrium of silver sulfate, Ag₂ SO₄ , in water.

(b) A sodium sulfate solution is slowly titrated into a solution that is 0.10 M in Ba(NO₃ )₂ and 0.10 M in Sr(NO₃ )₂ .

(i) Show, with calculations, which salt will precipitate first, the barium salt or the strontium salt.

(ii) If the titration is continued, eventually the second salt will begin to precipitate. What is the concentration of the other ion (Ba²⁺ or Sr²⁺ ) remaining in solution when the second salt begins to precipitate?

STOP: End of AP Chemistry Practice Exam 2, Section II (Free Response).

Answers and Explanations for Exam 2, Section II (Free Response)

Question 1

(a) (i) PbO₂ (s) + 2 HSO₄ ⁻ (aq) + 2 H⁺ (aq) + Pb(s) → 2 PbSO₄ (aq) + 2 H₂ O(l)

You get 1 point for this answer.

If the cell voltage is 2.04 V then the two half-reaction voltages must add to this. Therefore, 2.04 V = (1.68 + ?) V making the voltage for the second half-reaction equal to 0.36 V. This is the voltage for the oxidation half-reaction; to change this to a standard reduction potential it is necessary to reverse the half-reaction, which means it is necessary to reverse the sign on the voltage. Thus, the standard reduction potential for the second half-reaction is −0.36 V.

You get 1 point for calculating 0.36 V and 1 point for changing the sign to a negative value.

(b) It is necessary to use two of the equations provided on the exam to answer this problem. (If you already know the equations, it will save you the time necessary to look them up).

(i) The exam (and the Appendixes at the back of this book) provides the equation ΔG ° = −nFE ° plus the values of the Faraday constant, F = 96,485 coulombs per mole of electrons, and 1 volt = . For this problem, n = 2 moles of electrons and E ° = + 2.04 V. Entering the values into the given equation: = 393659 = 3.94 × 10⁵ J or 394 kJ.

Give yourself 1 point for the correct setup of the equation including the conversion and 1 more point for the correct answer.

(ii) It is possible to get the answer by two different means by using the equations given on the test (in the back of this book). The values of the constants are also here. The important equations are ΔG ° = −nFE ° and ΔG ° = −RTln K . It is possible to combine these equations to . The first equation was used in part b(ii) and it is acceptable to use the answer from there and go directly to the second equation without recalculating the result. Using the third equation gives: (which is the same answer you get from using the answer to b(ii) and the second equation. Finally, if ln K = 159, then K = e¹⁵⁹ .

Give yourself 1 point for the correct setup of the equation including the conversion and 1 more point for the correct answer.

(iii) At equilibrium ΔG ° is always 0.

Give yourself 1 point for this answer.

(c) As the water evaporates, the concentration of sulfuric acid will increase, which leads to an increase in the hydrogen ion concentration. An increase in the hydrogen ion concentration will promote the PbO₂ half-reaction more than it will change the Pb half-reaction. The result is a shift toward the left, which will increase the voltage.

Give yourself 1 point for this answer.

(c) It is necessary to have a conductor present for the transfer of electrons. Lead(IV) oxide is an ionic compound, which does not conduct electricity. Lead is a metal, which does conduct electricity.

Give yourself 1 point for this answer. Subtract 1 point if any of the answers has the wrong number of significant figures.

Total your points for the various parts. There are 10 possible points.

Question 2

(a) ΔH _rxn ° = [2(50)] − [2(90) + 1(0)] = −80 kJ

The answer with appropriate units is worth 1 point. You do not need to get the exact answer, but your answer should round to this one.

(b) The answer from part (a) equals the bonds broken minus the bonds formed.

[(2 NO) + (Cl–Cl)] − 2 [(N=O) + (NCl)] = −80 kJ

[(2 NO) + (240)] − 2 [(610) + (200)] = −80 kJ

NO = 650 kJ

The setup (broken − formed) is worth 1 point, and the answer is worth 1 point. You do not need to get the exact answer, but your answer should round to this one. If your answer from (a) was wrong, but you use it correctly in this part, you still get your setup and answer points.

(c) The value is higher than the value given for an N=O bond. The NO molecule has an odd number of electrons and this will alter its bonding.

You get 1 point for the correct observation, and 1 point for any explanation involving an odd number of electrons. If you got the wrong answer for part (b), you can still get 1 or 2 points if you used the answer correctly on this part.

(d) ΔS _rxn ° = [2(260)] − [2(210) + 1(220)] = −120 J/K

The setup (products − reactants) is worth 1 point, and the answer is worth 1 point. You do not need to get the exact answer, but your answer should round to this one.

(e) It is necessary to calculate the free-energy change.

ΔG _rxn ° = ΔH _rxn ° − T ΔS _rxn ° = −80 kJ − (298 K)(1 kJ/1000 J)(−120 J/K) = −44 kJ

The negative value means the reaction is spontaneous.

You get 1 point for the prediction that the reaction is spontaneous. The setup (plugging into the equation) is worth 1 point if you remember to change the temperature to kelvin and the joule to kilojoule conversion (or vice versa for an answer in J). An additional 1 point comes from the answer. If you got the wrong value in either part (a) or (b), but used it correctly, you will still get the point for the answer. The free-energy equation is part of the material supplied in the exam booklet.

Total your points for the different parts. There is a maximum of 10 points possible. Subtract 1 point if all your answers do not have the correct number of significant figures.

Question 3

(a) 1. Samples of iron(II) ammonium sulfate hexahydrate are weighed into beakers, and deionized water is added to dissolve each of the samples.

A buret is rinsed with the potassium permanganate solution and then filled.
The potassium permanganate solution is titrated into the iron(II) standard until a permanent pink color from the permanganate solution appears.
Repeat these steps with the other samples.

Listing these will get you 2 points. Missing one or more of these will get you 1 point. Other items are not necessary and will not change the scoring.

(b) [(Mass of Fe(NH₄ )₂ (SO₄ )₂ •6H₂ O)/(Molar mass of Fe(NH₄ )₂ (SO₄ )₂ •6H₂ O)] × (1 mol MnO₄ ⁻ /5 mol Fe²⁺ )/(Volume of potassium permanganate solution)

You get 2 point for this complete answer. You get 1 point if a step is missing and 0 points if two or more steps are missing.

(c) (Volume permanganate solution) × (Concentration permanganate solution) × (5 mol Fe/1 mol MnO₄ ⁻ ) × (Molar mass of iron) = grams iron

You get 1 point for this answer.

Percent iron = (grams iron/grams sample) × 100%

You get 1 point for this answer.

(d) The percent iron would be higher.

You get 1 point for this answer.

The titration would require more potassium permanganate solution to reach the endpoint because it would be necessary to titrate not only the iron but also the zinc.

You get 1 point for this answer or by saying the zinc would reduce some of the iron back to iron(II), which would require it to be re-titrated.

(e) As noted in the initial comments for this question, air oxidizes iron(II) ions. For this reason, the iron(II) ions in iron(II) ammonium sulfate hexahydrate might undergo air oxidation before being used to standardize the potassium permanganate.

This answer or simply stating that the compound is unstable is worth 1 point.

(f) The electron configuration may be written as [Ar] 3d⁶ or 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ . Recall that for the transition metals the s-electrons, in this case 4s² , are the first electrons lost in oxidation.

This answer is worth 1 point.

Total your points for the various parts. There are 10 possible points.

Question 4

(a) The Ideal Gas Equation is one way of calculating the number of moles, n , of a gas.

HCN (three significant figures) As an alternative, if you recognize the 273 K and 1.00 atm is STP, you can use the molar volume of a gas at STP to calculate the number of moles.

This answer is worth 1 point.

(b) Again it is possible to use the Ideal Gas Equation to solve for the volume, V , of a gas.

This answer is worth 1 point. Unlike part (a) you cannot use the molar volume of a gas because the conditions are not STP.

(c) Based upon its structure, HCN is polar and, therefore, must be soluble in water. For this reason, some will not be in the gas phase but in the solution, which means the volume of the gas will be less than expected.

This answer, including the explanation, is worth 1 point.

(d)

There must be a single bond between the hydrogen atom and the carbon atom and a triple bond between the carbon atom and the nitrogen atom. The lone pair on the nitrogen atom can be anywhere as long as it is obvious that it is only associated with the nitrogen atom. Both the carbon and the nitrogen atoms have an octet of electrons. This answer is worth 1 point.

Total your points for each part. There are 4 possible points. Subtract 1 point if all reported answers did not have the correct number of significant figures.

Question 5

(a) The Group 14 line is the dotted line. The dotted line should always be below the other lines because the hydrogen compounds of the Group 14 elements are all nonpolar, while all the other hydrogen compounds on the graph are polar. The dotted line should not go up at the end because CH₄ is the only Period 2 hydrogen compound on the graph that is not capable of hydrogen bonding.

Give yourself 1 point if your line remains below all the other lines. Give yourself a second point if your line does not move up at the left end.

(b) The Period 2 compounds on the graph are, from top to bottom, H₂ O, HF, NH₃ , and CH₄ . The first three of these compounds are the only compounds plotted that are capable of hydrogen bonding. Hydrogen bonding in stronger than any of the other intermolecular forces present in the compounds on the graph, which means their boiling points should be higher than those of compounds without hydrogen bonding.

Give yourself 1 point for this explanation.

(c) Li is a Period 2 element; therefore, it would appear on the left side of the graph and it would have a much higher boiling point than the other compounds because it is an ionic compound and ionic compounds, because of their high lattice energies, have very high boiling points.

You get 1 point if you give the correct approximate position and mention ionic bonding.

A total of 4 points is possible.

Question 6

The compounds are supplied as solutions; therefore, they must be soluble. This means any argument stating that one or more of the original substances is not soluble in water is invalid.

(a) The lowest pH will be for the solution containing an acid. Beaker B has the lowest pH. The NH₄ ⁺ ion is the conjugate acid of a weak base, and as an acid (weak), it will lower the pH. There are no other acids present.

You get 1 point for choosing B with the correct explanation.

(b) Beaker E, because transition metal ions are often colored in solution.

You get 1 point for the correct choice.

(c) Solution D—solution D contains the conjugate base, C₂ O₄ ²⁻ , of a weak acid. The conjugate bases of weak acids undergo hydrolysis to produce basic solutions.

You get 1 point for the correct choice.

(d) Solution C contains a nonelectrolyte; therefore, it does not conduct. All the remaining solutions contain electrolytes, which are conductors.

You get 1 point for choosing solution C with the explanation.

Total your points for the different parts. There are 4 possible points.

Question 7

(a) Ag₂ SO₄ (s) 2 Ag⁺ (aq) + SO₄ ²⁻ (aq)

You get 1 point for this equation.

(b) (i) Since the stoichiometry of SrSO₄ and BaSO₄ are the same, the one with the lower K _sp will precipitate first (BaSO₄ ). However, the problem states that it is necessary to show this with calculations.

Since the stoichiometries are the same, it is possible to use the generic equation and generic mass action expression with M²⁺ = Ba²⁺ or Sr²⁺ .

Since both metals have the same concentration, it is only necessary to determine which requires the lower sulfate ion concentration. This is done as:

For BaSO₄ = 1.1 × 10⁻⁹ M SO₄ ²⁻ and for SrSO₄ = 3.2 × 10⁻⁶ M SO₄ ²⁻

You get 1 point for determining that the sulfate ion concentration shows that BaSO₄ will precipitate first.

(ii) From part (i), the second ion (Sr²⁺ ) begins to precipitate at 3.2 × 10⁻⁶ M SO₄ ²⁻ . Using this value with the BaSO₄ K _sp allows the calculation of the barium ion concentration when the SrSO₄ first begins to precipitate. This calculation is:

You get 1 point for this answer. You also get 1 point if you if you did this calculation correctly but made the wrong prediction in part (i)

(c) For the compounds with the same stoichiometry (BaSO₄ , CaSO₄ , and SrSO₄ ) the one with the largest K _sp is the most soluble; therefore, it is only necessary to compare the solubilities of Ag₂ SO₄ and CaSO₄ .

For CaSO₄

For Ag₂ SO₄

The value for x from Ag₂ SO₄ is larger; therefore it is more soluble.

Give yourself 1 point for this answer.

There is a total of 4 points possible. Subtract 1 point if any answer has the wrong number of significant figures.