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Chemistry 30

Chemistry 30 · Diploma prep

Sample diploma exam

A full-length exam in the diploma format: 44 multiple-choice and 16 numerical-response questions of equal value, weighted to the same blueprint the real exam uses. Give yourself 3 hours, then read every solution — they are written out in full, starting on their own printed page.

These questions were written for this site. Alberta's own exams are secured; for real released items, use the links on the diploma prep page.

Sample Diploma Exam

Chemistry 30 · maddyhelps.com

Name
Date
Score
/ 60

Time: 3 hours, as on the real exam. Every student may take up to 6 hours if they need it.

You will need one approved calculator and the Chemistry Data Booklet. Any constant a question does not supply is in that booklet.

Multiple choice: choose the best of the four answers. Numerical response: first digit on the left, unused boxes blank, a 0 before the decimal point for answers between 0 and 1, and round only at the end, to the accuracy each question names.

Use the following information to answer questions 1 and 2 and numerical-response question 1.

A student burns a sample of ethanol under a metal can containing 250.0 g of water. The water warms from 21.0 °C to 33.5 °C. (c of water = 4.19 J/(g·°C))

1. The heat absorbed by the water is

  1. A. 13.1 kJ
  2. B. 1.31 kJ
  3. C. 131 kJ
  4. D. 34.9 kJ

Numerical Response

1. If 0.150 mol of ethanol burned, the molar enthalpy of combustion determined by this experiment is −______ kJ/mol.

(Record your answer to the nearest tenth in the numerical-response section on the answer sheet.)

2. The value obtained is much smaller in magnitude than the accepted molar enthalpy of combustion of ethanol. The most likely reason is that

  1. A. too much water was used
  2. B. heat was lost to the surroundings rather than absorbed by the water
  3. C. the ethanol was impure
  4. D. the thermometer read high

3. Which of the following processes is endothermic?

  1. A. The melting of ice
  2. B. The condensation of steam
  3. C. The combustion of methane
  4. D. The neutralization of HCl by NaOH

Use the following information to answer questions 4 and 5.

On a potential energy diagram for a reaction, the reactants sit at 150 kJ, the peak is at 260 kJ, and the products sit at 90 kJ.

4. The activation energy of the forward reaction and the ΔH of the reaction are

  1. A. 110 kJ and +60 kJ
  2. B. 170 kJ and −60 kJ
  3. C. 110 kJ and −60 kJ
  4. D. 260 kJ and −60 kJ

5. The activation energy of the reverse reaction is

  1. A. 170 kJ
  2. B. 110 kJ
  3. C. 60 kJ
  4. D. 260 kJ

6. Adding a catalyst to this reaction would

  1. A. lower both the activation energy and ΔH
  2. B. lower ΔH and leave the activation energy unchanged
  3. C. raise the activation energy of the reverse reaction only
  4. D. lower the activation energy and leave ΔH unchanged

7. For the reaction 2H₂(g) + O₂(g) → 2H₂O(g), ΔH = −484 kJ, the molar enthalpy of combustion of hydrogen is

  1. A. −242 kJ/mol
  2. B. −484 kJ/mol
  3. C. −968 kJ/mol
  4. D. +242 kJ/mol

8. Which statement about bond energies in an exothermic reaction is correct?

  1. A. No bonds are broken in an exothermic reaction
  2. B. The energy released forming bonds in the products exceeds the energy absorbed breaking bonds in the reactants
  3. C. The energy absorbed breaking bonds exceeds the energy released forming them
  4. D. Breaking bonds releases energy and forming them absorbs energy

Use the following information to answer numerical-response question 2.

Three thermochemical equations are given below.

  • 1 C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l) ΔH = −1411 kJ
  • 2 C₂H₆(g) + 3½O₂(g) → 2CO₂(g) + 3H₂O(l) ΔH = −1560 kJ
  • 3 H₂(g) + ½O₂(g) → H₂O(l) ΔH = −286 kJ

Numerical Response

2. For the hydrogenation reaction C₂H₄(g) + H₂(g) → C₂H₆(g), ΔH = −______ kJ.

(Record your answer in the numerical-response section on the answer sheet.)

9. A reaction is written with the energy term included: N₂(g) + O₂(g) + 180 kJ → 2NO(g). This reaction is

  1. A. exothermic, with ΔH = −180 kJ
  2. B. exothermic, with ΔH = +180 kJ
  3. C. endothermic, with ΔH = +180 kJ
  4. D. endothermic, with ΔH = −180 kJ

Use the following information to answer numerical-response question 3.

A 50.0 g piece of metal at 95.0 °C is placed in 150.0 g of water at 18.0 °C. The final temperature of both is 21.2 °C.

Numerical Response

3. The specific heat capacity of the metal is ______ J/(g·°C).

(Record your answer to the nearest hundredth in the numerical-response section on the answer sheet.)

10. The oxidation number of chromium in Cr₂O₇²⁻ is

  1. A. +6
  2. B. +3
  3. C. +7
  4. D. +12

11. In which compound does oxygen have an oxidation number of −1?

  1. A. H₂O
  2. B. CO₂
  3. C. Na₂O
  4. D. H₂O₂

Use the following information to answer questions 12 and 13 and numerical-response question 4.

Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)

12. In this reaction, the reducing agent is

  1. A. Zn(s)
  2. B. Ag⁺(aq)
  3. C. Zn²⁺(aq)
  4. D. Ag(s)

Numerical Response

4. Using the standard reduction potentials Ag⁺ + e⁻ → Ag (E° = +0.80 V) and Zn²⁺ + 2e⁻ → Zn (E° = −0.76 V), the standard cell potential for this reaction is ______ V.

(Record your answer to the nearest hundredth in the numerical-response section on the answer sheet.)

13. If this reaction is used in a voltaic cell, the mass of the zinc electrode will

  1. A. decrease, because zinc is reduced
  2. B. decrease, because zinc is oxidized into solution
  3. C. increase, because silver plates onto it
  4. D. stay constant, because only electrons move

14. In a voltaic cell, the salt bridge

  1. A. prevents any contact between the two solutions
  2. B. supplies the energy that drives the reaction
  3. C. allows ions to migrate so that neither solution builds up a net charge
  4. D. carries electrons between the half-cells

15. Which statement correctly describes an electrolytic cell?

  1. A. It requires a power supply, and its anode is negative
  2. B. It operates spontaneously, and its anode is positive
  3. C. It operates spontaneously, and its cathode is negative
  4. D. It requires a power supply, and its anode is positive

Use the following information to answer question 16.

An unbalanced half-reaction is shown below.

  • Cr₂O₇²⁻(aq) → Cr³⁺(aq)

16. When this half-reaction is balanced in acidic solution, the number of electrons transferred is

  1. A. 6
  2. B. 3
  3. C. 2
  4. D. 14

17. When balancing a redox equation in basic solution, the final step is to

  1. A. double every coefficient
  2. B. add OH⁻ to both sides to neutralize every H⁺, then simplify the water
  3. C. add H⁺ to both sides
  4. D. remove all water molecules

Use the following information to answer question 18 and numerical-response question 5.

A cell is constructed from a copper electrode in Cu²⁺(aq) (E° = +0.34 V) and an iron electrode in Fe²⁺(aq) (E° = −0.45 V).

18. The cell notation for this cell is

  1. A. Fe²⁺(aq) | Fe(s) || Cu(s) | Cu²⁺(aq)
  2. B. Cu²⁺(aq) | Cu(s) || Fe(s) | Fe²⁺(aq)
  3. C. Fe(s) | Fe²⁺(aq) || Cu²⁺(aq) | Cu(s)
  4. D. Cu(s) | Cu²⁺(aq) || Fe²⁺(aq) | Fe(s)

Numerical Response

5. The standard potential of this cell is ______ V.

(Record your answer to the nearest hundredth in the numerical-response section on the answer sheet.)

19. A steel pipeline is protected from corrosion by attaching blocks of magnesium. The magnesium protects the steel because it

  1. A. is harder than steel and shields it physically
  2. B. reduces any Fe²⁺ back to iron metal
  3. C. prevents oxygen from dissolving in the surrounding water
  4. D. is more easily oxidized than iron, so it is oxidized instead

Use the following information to answer numerical-response question 6.

A current of 1.50 A is passed through a solution of AgNO₃ for 20.0 minutes.

Numerical Response

6. The mass of silver deposited is ______ g.

(Record your answer to the nearest hundredth in the numerical-response section on the answer sheet.)

20. Producing 1 mol of aluminum from Al³⁺ requires how many times as much charge as producing 1 mol of silver from Ag⁺?

  1. A. 3
  2. B. 1
  3. C. 2
  4. D. 27

21. Which of the following is not a redox reaction?

  1. A. 2H₂O(l) → 2H₂(g) + O₂(g)
  2. B. AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
  3. C. 2Mg(s) + O₂(g) → 2MgO(s)
  4. D. Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

Use the following information to answer numerical-response question 7.

Four processes are listed below.

  • 1 Electrons leave the anode
  • 2 Ions migrate through the salt bridge
  • 3 Metal ions are reduced at the cathode
  • 4 A metal atom is oxidized at the anode surface

Numerical Response

7. Beginning with the oxidation, the order in which these occur is ___, ___, ___, and ___.

(Record all four digits of your answer in the numerical-response section on the answer sheet.)

22. During the electroplating of a spoon with silver, the spoon is

  1. A. the cathode, where it is oxidized
  2. B. the anode, where Ag⁺ is reduced onto its surface
  3. C. the cathode, where Ag⁺ is reduced onto its surface
  4. D. the anode, where it is oxidized

23. During the electrolysis of aqueous sodium bromide, hydrogen gas bubbles from one electrode. The statement “hydrogen gas was produced by the reduction of water” is

  1. A. an observation, because the bubbles were seen
  2. B. a hypothesis, because it was written before the experiment
  3. C. a prediction, because it states what will happen
  4. D. an inference, because it explains an observation rather than reporting it

24. The general formula CₙH₂ₙ₋₂ describes

  1. A. alkynes
  2. B. alkanes
  3. C. alkenes
  4. D. alcohols

Use the following information to answer question 25.

CH₃CH(CH₃)CH₂CH₂CH₃

25. The IUPAC name of the compound above is

  1. A. hexane
  2. B. 2-methylpentane
  3. C. 4-methylpentane
  4. D. 2-methylbutane

26. Which pair of compounds are isomers?

  1. A. propane and propanol
  2. B. methane and ethane
  3. C. butane and methylpropane
  4. D. ethane and ethene

27. A compound whose name ends in -oic acid contains which functional group?

  1. A. −COOH
  2. B. −OH
  3. C. −COO−
  4. D. −CHO

Use the following information to answer questions 28 and 29.

Ethanoic acid is warmed with methanol and a few drops of concentrated sulfuric acid.

28. The organic product of this reaction is

  1. A. an aldehyde
  2. B. a ketone
  3. C. an ether
  4. D. an ester

29. The other product of this reaction is

  1. A. no other product forms
  2. B. water
  3. C. hydrogen
  4. D. carbon dioxide

30. Ethene decolourizes bromine water rapidly, but ethane does not react with it in the dark. This is because ethene

  1. A. is more soluble in water than ethane
  2. B. contains more hydrogen than ethane
  3. C. has a double bond that can open and add bromine atoms
  4. D. is a larger molecule than ethane

Use the following information to answer numerical-response question 8.

Four organic compounds are listed below.

  • 1 CH₃CH₂OH
  • 2 CH₃COOH
  • 3 CH₃CH₂CH₃
  • 4 CH₃COOCH₃

Numerical Response

8. The compounds above that contain a carbonyl group, C=O, are numbered ______ and ______.

(Record both digits of your answer in any order in the numerical-response section on the answer sheet.)

Use the following information to answer numerical-response question 9.

C₄H₁₀(g) + ___O₂(g) → ___CO₂(g) + ___H₂O(g)

Numerical Response

9. When the equation above is balanced using the lowest whole-number coefficients with a coefficient of 2 in front of C₄H₁₀, the coefficient of O₂ is ______.

(Record your answer in the numerical-response section on the answer sheet.)

31. Polyethene is produced from ethene by

  1. A. condensation polymerization, releasing water
  2. B. substitution, releasing hydrogen
  3. C. cracking
  4. D. addition polymerization, with no by-product

Use the following information to answer questions 32 and 33.

Butan-1-ol boils at 118 °C, while butane boils at −1 °C, although the two molecules are of similar size.

32. The best explanation for the difference is that butan-1-ol

  1. A. is an ionic compound
  2. B. forms hydrogen bonds between its molecules
  3. C. has stronger covalent bonds within its molecules
  4. D. has a larger molar mass

33. In a fractional distillation tower, the fractions collected near the top of the tower have

  1. A. shorter chains and higher boiling points
  2. B. the same chain length as those at the bottom
  3. C. shorter chains and lower boiling points
  4. D. longer chains and higher boiling points

34. A system at chemical equilibrium is best described as one in which

  1. A. both reactions have stopped
  2. B. the concentrations of reactants and products are equal
  3. C. all of the reactants have been used up
  4. D. the forward and reverse reactions continue at equal rates

Use the following information to answer questions 35 to 37 and numerical-response question 10.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −198 kJ

35. The equilibrium constant expression for this system is

  1. A. 2[SO₃] ÷ (2[SO₂] × [O₂])
  2. B. [SO₃]² ÷ ([SO₂]²[O₂])
  3. C. ([SO₂]²[O₂]) ÷ [SO₃]²
  4. D. [SO₃]² ÷ ([SO₂]² + [O₂])

Numerical Response

10. At equilibrium, [SO₂] = 0.10 mol/L, [O₂] = 0.05 mol/L and [SO₃] = 0.30 mol/L. The value of K is ______.

(Record your answer in the numerical-response section on the answer sheet.)

36. Raising the temperature of this system would

  1. A. shift the equilibrium left and increase K
  2. B. have no effect on the position of equilibrium
  3. C. shift the equilibrium left and decrease K
  4. D. shift the equilibrium right and increase K

37. Decreasing the volume of the container would shift this equilibrium

  1. A. left, because there are fewer moles of gas on the reactant side
  2. B. neither way, since all species are gases
  3. C. right, because K increases with pressure
  4. D. right, because there are fewer moles of gas on the product side

Use the following information to answer numerical-response questions 11 and 12.

H₂(g) + I₂(g) ⇌ 2HI(g)

  • A 1.00 L flask initially holds 0.80 mol of H₂ and 0.80 mol of I₂. At equilibrium, 0.60 mol/L of H₂ remains.

Numerical Response

11. The equilibrium concentration of HI is ______ mol/L.

(Record your answer to the nearest hundredth in the numerical-response section on the answer sheet.)

Numerical Response

12. The value of K for this system is ______.

(Record your answer to the nearest tenth in the numerical-response section on the answer sheet.)

38. Adding a catalyst to a system at equilibrium

  1. A. shifts the equilibrium toward whichever side has fewer moles
  2. B. changes neither the position of equilibrium nor K, but equilibrium is reached sooner
  3. C. shifts the equilibrium toward the products
  4. D. increases K

Use the following information to answer question 39.

Ag⁺(aq) + Cl⁻(aq) ⇌ AgCl(s)

  • A few drops of silver nitrate solution are added to a saturated solution of sodium chloride.

39. Adding the silver ions shifts this equilibrium

  1. A. neither way, since AgCl is a solid
  2. B. right, which increases the value of K
  3. C. right, removing chloride ions from solution
  4. D. left, releasing chloride ions into solution

40. Which statement about a 0.10 mol/L solution of ethanoic acid is correct?

  1. A. It is a concentrated solution of a strong acid
  2. B. It ionizes completely, like hydrochloric acid
  3. C. Its pH is 1.0
  4. D. It is a dilute solution of a weak acid, and most of the acid remains as molecules

Use the following information to answer numerical-response question 13.

A solution of hydrochloric acid has a concentration of 0.025 mol/L.

Numerical Response

13. The pH of this solution is ______.

(Record your answer to the nearest hundredth in the numerical-response section on the answer sheet.)

Use the following information to answer numerical-response question 14.

A solution of sodium hydroxide has a concentration of 1.0 × 10⁻³ mol/L.

Numerical Response

14. The pH of this solution at 25 °C is ______.

(Record your answer to the nearest tenth in the numerical-response section on the answer sheet.)

41. The conjugate base of HSO₄⁻ is

  1. A. SO₄²⁻
  2. B. H₂SO₄
  3. C. H₃O⁺
  4. D. OH⁻

Use the following information to answer numerical-response question 15.

A 0.200 mol/L solution of a weak monoprotic acid has a pH of 2.72.

Numerical Response

15. The Ka of this acid is ______ × 10⁻⁵.

(Record your answer to the nearest tenth in the numerical-response section on the answer sheet.)

Use the following information to answer question 42 and numerical-response question 16.

A 25.0 mL sample of hydrochloric acid is titrated to the equivalence point with 22.6 mL of 0.150 mol/L NaOH(aq).

Numerical Response

16. The concentration of the hydrochloric acid is ______ mol/L.

(Record your answer to the nearest hundredth in the numerical-response section on the answer sheet.)

42. If the acid had been ethanoic acid rather than hydrochloric acid, the pH at the equivalence point would be

  1. A. above 7, because excess base has been added
  2. B. above 7, because the ethanoate ion reacts with water to produce OH⁻
  3. C. exactly 7, because the acid is fully neutralized
  4. D. below 7, because some acid always remains

43. An indicator is best described as

  1. A. a salt that dissolves only at the equivalence point
  2. B. a buffer that resists pH change
  3. C. a weak acid whose conjugate base is a different colour
  4. D. a strong acid that changes colour at pH 7

44. A buffer solution resists a change in pH because it contains

  1. A. a strong acid and a strong base in equal amounts
  2. B. only a weak acid
  3. C. a saturated salt solution
  4. D. appreciable amounts of both a weak acid and its conjugate base

Answer key · Sample diploma exam

Chemistry 30 · maddyhelps.com

Multiple choice
1. A · 2. B · 3. A · 4. C · 5. A · 6. D · 7. A · 8. B · 9. C · 10. A · 11. D · 12. A · 13. B · 14. C · 15. D · 16. A · 17. B · 18. C · 19. D · 20. A · 21. B · 22. C · 23. D · 24. A · 25. B · 26. C · 27. A · 28. D · 29. B · 30. C · 31. D · 32. B · 33. C · 34. D · 35. B · 36. C · 37. D · 38. B · 39. C · 40. D · 41. A · 42. B · 43. C · 44. D

Numerical response
1. 87.3 · 2. 137 · 3. 0.55 · 4. 1.56 · 5. 0.79 · 6. 2.01 · 7. 4132 · 8. 24 · 9. 13 · 10. 180 · 11. 0.40 · 12. 0.4 · 13. 1.60 · 14. 11.0 · 15. 1.8 · 16. 0.14

Question 1 — A. 13.1 kJ

q = mcΔT = (250.0)(4.19)(12.5) = 13 094 J = 13.1 kJ. ΔT is the change, 33.5 − 21.0 = 12.5 °C, not the final temperature.

Numerical Response 1 — 87.3

ΔH = −13.09 kJ ÷ 0.150 mol = −87.3 kJ/mol. The accepted value is far larger because most of the energy heats the air and the can rather than the water — which is the point of the next question.

Question 2 — B. heat was lost to the surroundings rather than absorbed by the water

Simple calorimeters lose a great deal of energy to the air, the can and the stand. Everything measured is a lower bound, which is why experimental values come out low rather than randomly scattered.

Question 3 — A. The melting of ice

Melting absorbs energy to overcome the attractions holding the solid together. The other three all release energy to the surroundings.

Question 4 — C. 110 kJ and −60 kJ

Ea is measured from the reactants to the peak: 260 − 150 = 110 kJ. ΔH is products minus reactants: 90 − 150 = −60 kJ, so the reaction is exothermic.

Question 5 — A. 170 kJ

The reverse reaction starts at the products, 90 kJ, and must still reach the same peak at 260 kJ: 260 − 90 = 170 kJ. It equals the forward Ea plus |ΔH|, which is a useful check.

Question 6 — D. lower the activation energy and leave ΔH unchanged

A catalyst provides a lower-energy pathway, so the peak drops for both directions. The reactants and products are untouched, so their difference — ΔH — cannot change.

Question 7 — A. −242 kJ/mol

The equation releases 484 kJ for 2 mol of H₂, so each mole releases 242 kJ. Using the enthalpy of the equation where a molar enthalpy is required is one of the errors examiners report most often.

Question 8 — B. The energy released forming bonds in the products exceeds the energy absorbed breaking bonds in the reactants

Breaking always costs energy and forming always releases it. Exothermic means the forming side wins, so there is surplus energy to give to the surroundings.

Numerical Response 2 — 137

Take equation 1 as written, equation 3 as written, and reverse equation 2: −1411 + (−286) + 1560 = −137 kJ. Check by cancelling — the CO₂ and H₂O must all disappear.

Question 9 — C. endothermic, with ΔH = +180 kJ

Energy on the reactant side is energy being consumed, so the reaction absorbs it: endothermic, ΔH positive. Energy written on the product side would mean the opposite.

Numerical Response 3 — 0.55

Water gains (150.0)(4.19)(3.2) = 2011 J. The metal cooled by 95.0 − 21.2 = 73.8 °C, so c = 2011 ÷ (50.0 × 73.8) = 0.545 ≈ 0.55 J/(g·°C). Using 95.0 °C as the metal's ΔT is the usual error.

Question 10 — A. +6

Seven oxygens at −2 give −14. The ion is −2 overall, so the two chromiums share +12 and each is +6.

Question 11 — D. H₂O₂

Peroxides are the exception to the −2 rule, and examiners single this out as the one place oxidation numbers go wrong. In H₂O₂, two hydrogens at +1 require the two oxygens to total −2, so each is −1.

Question 12 — A. Zn(s)

Zinc goes from 0 to +2, losing electrons, so it is oxidized — which makes it the reducing agent. Ag⁺ gains electrons and is the oxidizing agent.

Numerical Response 4 — 1.56

E°cell = E°(cathode) − E°(anode) = 0.80 − (−0.76) = 1.56 V. Silver is reduced, so it is the cathode. Never double the 0.80 V to match the 2Ag⁺ — potential does not scale with amount.

Question 13 — B. decrease, because zinc is oxidized into solution

Zinc is the anode, and oxidation turns solid zinc into ions that dissolve. The silver cathode gains mass as Ag⁺ is reduced onto it.

Question 14 — C. allows ions to migrate so that neither solution builds up a net charge

Without ion migration, charge builds up in both half-cells within moments and the reaction stops. Electrons travel the external wire, never the bridge.

Question 15 — D. It requires a power supply, and its anode is positive

Electrolysis forces a non-spontaneous reaction, so an external supply is needed and it makes the anode positive. Oxidation is still at the anode — only the sign differs from a voltaic cell.

Question 16 — A. 6

Chromium falls from +6 to +3 for each of two atoms, so six electrons are gained: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.

Question 17 — B. add OH⁻ to both sides to neutralize every H⁺, then simplify the water

Balance in acid first, then neutralize: each H⁺ plus an OH⁻ becomes H₂O, and identical waters on both sides cancel.

Question 18 — C. Fe(s) | Fe²⁺(aq) || Cu²⁺(aq) | Cu(s)

Copper has the higher reduction potential, so it is the cathode and iron is the anode. Notation runs anode on the left, cathode on the right, with the double line for the salt bridge.

Numerical Response 5 — 0.79

0.34 − (−0.45) = 0.79 V. The positive value confirms the reaction is spontaneous as written, which is what makes it a voltaic cell.

Question 19 — D. is more easily oxidized than iron, so it is oxidized instead

Cathodic protection with a sacrificial anode: magnesium's reduction potential is lower, so it gives up electrons in preference to iron and is consumed.

Numerical Response 6 — 2.01

q = (1.50 A)(1200 s) = 1800 C. Moles of electrons = 1800 ÷ 96 500 = 0.01865 mol. Ag⁺ needs one electron each, so 0.01865 mol × 107.87 g/mol = 2.01 g.

Question 20 — A. 3

Al³⁺ + 3e⁻ → Al needs three electrons per atom against silver's one. Examiners note students handle Faraday calculations at a 1:1 ratio and stumble when the ratio changes.

Question 21 — B. AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

In the precipitation, every element keeps its oxidation number — the ions simply change partners. The other three all involve electron transfer.

Numerical Response 7 — 4132

A metal atom is oxidized (4), releasing electrons that leave the anode (1) and travel to the cathode, where ions are reduced (3); ions then migrate through the bridge (2) to balance the charge that transfer created.

Question 22 — C. the cathode, where Ag⁺ is reduced onto its surface

Plating deposits metal, which requires reduction, and reduction happens at the cathode. The silver bar is the anode, dissolving to replace what plates out.

Question 23 — D. an inference, because it explains an observation rather than reporting it

Bubbles at the electrode are the observation. Identifying the gas and attributing it to reduction is an interpretation of that observation — a distinction Alberta's bulletin says students routinely miss.

Question 24 — A. alkynes

A triple bond costs four hydrogens relative to an alkane, giving CₙH₂ₙ₋₂. Alkanes are CₙH₂ₙ₊₂ and alkenes CₙH₂ₙ.

Question 25 — B. 2-methylpentane

The longest chain is five carbons — pentane — with a methyl branch. Numbering from the end nearer the branch puts it on carbon 2; numbering the other way gives 4 and breaks the lowest-locant rule.

Question 26 — C. butane and methylpropane

Isomers share a molecular formula — both are C₄H₁₀ — with different structures, which is why they boil at different temperatures. The other pairs differ in formula.

Question 27 — A. −COOH

The carboxyl group defines a carboxylic acid. −OH alone is an alcohol, −COO− between carbons is an ester, and −CHO at the end of a chain is an aldehyde.

Question 28 — D. an ester

A carboxylic acid plus an alcohol gives an ester and water — esterification, a condensation reaction. The sulfuric acid is a catalyst, not a reactant.

Question 29 — B. water

Condensation reactions release a small molecule as the two pieces join, and here it is water. That is exactly what distinguishes condensation from addition.

Question 30 — C. has a double bond that can open and add bromine atoms

The exposed second bond of C=C is the reactive site, so alkenes undergo addition readily. Alkanes have only strong single bonds and need UV light or heat for substitution.

Numerical Response 8 — 24

A carboxylic acid (2) and an ester (4) both contain C=O. An alcohol (1) has only −OH, and propane (3) has no functional group at all.

Numerical Response 9 — 13

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O. The products hold 16 + 10 = 26 oxygen atoms, which is 13 O₂. Doubling the hydrocarbon is the standard way to clear the half-coefficient.

Question 31 — D. addition polymerization, with no by-product

The double bonds open and the monomers join end to end with nothing left over. Condensation polymerization — nylon, polyesters — releases a small molecule at every link.

Question 32 — B. forms hydrogen bonds between its molecules

The −OH group hydrogen bonds to neighbouring molecules, and boiling must overcome those attractions. Covalent bonds inside a molecule are not broken by boiling — a distinction examiners report students blur.

Question 33 — C. shorter chains and lower boiling points

Short-chain hydrocarbons have weaker intermolecular forces, boil at lower temperatures, and rise furthest before condensing. Bitumen and heavy oils condense low down.

Question 34 — D. the forward and reverse reactions continue at equal rates

Equilibrium is dynamic: both reactions keep running, at matching rates, so concentrations hold steady — steady, not equal.

Question 35 — B. [SO₃]² ÷ ([SO₂]²[O₂])

Products over reactants, each raised to the power of its coefficient. Coefficients become exponents, never multipliers.

Numerical Response 10 — 180

K = (0.30)² ÷ [(0.10)²(0.05)] = 0.09 ÷ 0.0005 = 180. A value well above 1 says products are favoured at this temperature.

Question 36 — C. shift the equilibrium left and decrease K

The forward reaction is exothermic, so heat behaves as a product: adding heat pushes the system back toward reactants. Temperature is the only change that alters the value of K.

Question 37 — D. right, because there are fewer moles of gas on the product side

Three moles of gas on the left against two on the right, so compressing the system favours the side that occupies fewer moles. K itself does not change.

Numerical Response 11 — 0.40

H₂ fell by 0.20 mol/L, so x = 0.20. HI forms at twice that rate: 2x = 0.40 mol/L. The change row follows the coefficients — that is the whole ICE table.

Numerical Response 12 — 0.4

I₂ also fell by 0.20 to 0.60 mol/L, so K = (0.40)² ÷ [(0.60)(0.60)] = 0.16 ÷ 0.36 = 0.44 ≈ 0.4. A K below 1 says this mixture favours reactants.

Question 38 — B. changes neither the position of equilibrium nor K, but equilibrium is reached sooner

A catalyst speeds the forward and reverse reactions equally. It changes how quickly you arrive, not where you arrive.

Question 39 — C. right, removing chloride ions from solution

Adding a reactant drives the system toward products, precipitating AgCl and pulling Cl⁻ out of solution. Alberta's bulletin notes students miss stresses stated indirectly like this one.

Question 40 — D. It is a dilute solution of a weak acid, and most of the acid remains as molecules

Strength is how completely an acid ionizes; concentration is how much is dissolved. Ethanoic acid ionizes only slightly, so the solution is mostly intact molecules and its pH is near 2.9.

Numerical Response 13 — 1.60

HCl is a strong acid, so [H₃O⁺] = 0.025 mol/L and pH = −log(0.025) = 1.60. Only the decimal places in a pH count as significant figures.

Numerical Response 14 — 11.0

pOH = −log(1.0 × 10⁻³) = 3.00, so pH = 14.00 − 3.00 = 11.0. Answering 3.0 means stopping at the pOH.

Question 41 — A. SO₄²⁻

A conjugate base is what remains after the acid donates one proton: HSO₄⁻ − H⁺ = SO₄²⁻. Adding a proton instead gives H₂SO₄, its conjugate acid.

Numerical Response 15 — 1.8

[H₃O⁺] = 10⁻²·⁷² = 1.9 × 10⁻³ mol/L. Ka = (1.9 × 10⁻³)² ÷ 0.200 = 1.8 × 10⁻⁵. Using the ICE table backwards from pH is exactly the step examiners say students skip.

Numerical Response 16 — 0.14

Moles of NaOH = (0.0226)(0.150) = 3.39 × 10⁻³ mol. The ratio is 1:1, so the acid holds the same, and 3.39 × 10⁻³ ÷ 0.0250 = 0.1356 ≈ 0.14 mol/L.

Question 42 — B. above 7, because the ethanoate ion reacts with water to produce OH⁻

At equivalence no base is in excess. What remains is the conjugate base of a weak acid, which takes protons from water and leaves hydroxide behind — a prediction the bulletin flags as difficult.

Question 43 — C. a weak acid whose conjugate base is a different colour

Because the two forms differ in colour and their ratio depends on pH, an indicator changes over roughly two pH units. Choose one whose range falls on the steep part of the curve.

Question 44 — D. appreciable amounts of both a weak acid and its conjugate base

The weak acid neutralizes added base and the conjugate base neutralizes added acid, so the ratio shifts rather than the pH. Examiners note students can pick a buffer from a list but struggle to say what it then does.