Explanations, Calculations and Representations: Exam Skills Guide (Chemistry)

How to use this guide

This guide is for students revising the whole course for the free-response section. It trains four habits that turn knowledge into credit: explaining with a named cause, setting out calculations, using representations, and justifying claims with data.

Every example and task here is original Marlbridge practice, not past exam material. The marks shown are a suggested Marlbridge rubric to help you judge your own answers; they are not official scoring and give no predicted score. For what each task verb asks for and how to plan your time, read the companion task verbs guide.

Try each worked example yourself before reading the strong answer.

The free-response section at a glance

Feature	Detail
Questions	7: three long questions (10 points each) and four short questions (4 points each)
Time	105 minutes
Share of the exam score	50%
Format	Questions are viewed in the Bluebook app; answers are handwritten in a paper booklet
Tools	A scientific or graphing calculator is recommended; a periodic table and formula sheet are provided

The course framework lists how the six science practices are weighted in this section. Mathematical routines (Practice 5) carry the largest share, 43–53%, and argumentation (Practice 6) carries 15–24%. Representing data and phenomena (Practice 3) carries 8–16%, and it is assessed only in the free-response section. Calculations, explanations and drawings earn most of the points.

Explanations that name the cause

An explanation has three parts: what happens, the particle-level cause, and the link between them. Weak answers restate the observation (“it boils higher because it is harder to boil”). Strong answers name the particles, the force or event, and the direction of the effect.

Four causes cover most of the course:

Cause	Use it for	What to say about the particles
Coulombic attraction	ionization energy, atomic radius, lattice strength, bond strength	size of the charges and the distance between them; more charge or less distance gives a stronger attraction
Intermolecular forces	boiling point, vapour pressure, solubility, chromatography	which forces act between molecules, which is strongest, and how structure (electron count, polarity, O–H or N–H) causes it
Collisions	rate, effect of concentration, temperature, catalyst	how often particles collide, what fraction have energy at least Eₐ, and orientation
Equilibrium shift	adding or removing a species, changing volume, comparing Q with K	how a change makes Q differ from K and which net reaction restores Q = K

Three rules make the cause visible:

• Compare both. If a question compares two substances, discuss each.
• Use the right words. Intermolecular forces are “overcome”, not “broken bonds”. Particles collide; substances do not.
• Close the loop. End with the observation: “…so more energy is needed to separate the molecules.”

Setting out a calculation

A reader should be able to follow every step without guessing. Use this layout:

1. Write the relationship in symbols (n = cV, q = mcΔT, rate = k[A]ᵐ).
2. Substitute the numbers with units.
3. Give the answer with a unit, a sign where it matters, and sensible significant figures.
4. State any assumption you used.

Significant figures. In multiplication and division, keep the number of significant figures of the least precise value. In addition and subtraction, keep the fewest decimal places; a difference can lose significant figures (26.7 − 21.3 = 5.4, two figures). Keep all digits in your calculator and round once, at the end.

Common assumptions to state:

Calculation	Assumption
Calorimetry	no energy exchanged with the surroundings; solution has c = 4.18 J g⁻¹ °C⁻¹ and density 1.00 g mL⁻¹
Titration	the end point is the equivalence point; the reaction goes to completion
Gases	the gas behaves ideally
Weak acid pH	x is small compared with the starting concentration (check: under about 5%)

Drawing and reading representations

Particulate diagrams. A good one passes four checks:

• Atoms are conserved: the same number of each kind before and after a reaction.
• The ratio of particles matches the formula or equation.
• Ions carry charges, and the charges in the box add to zero.
• In solution, water molecules point their δ− oxygen at cations and their δ+ hydrogens at anions. Gas particles are spread out; liquid particles touch but are disordered.

Graphs. Label both axes with quantity and unit. To use a graph as evidence, quote values read from it (“the plateau runs from 120 s to 420 s”). Know what a straight line means in the course: ln[A] against t linear means first order (slope = −k); 1/[A] against t linear means second order (slope = +k); absorbance against concentration linear means Beer–Lambert behaviour.

Data tables. To find the effect of one variable, choose two rows in which only that variable changes. Say which rows you used.

Justifying a claim with data

When a question says “justify”, give three things in order:

Part	What to write	Example
Claim	a direct answer	“HX is the stronger acid.”
Evidence	specific values from the question	“At the same concentration, 0.10 M, HX has pH 1.9 and HY has pH 2.9.”
Reasoning	the principle that links evidence to claim	“Lower pH means higher [H₃O⁺], so a larger fraction of HX molecules has ionized and its Ka is larger.”

Evidence without numbers rarely earns the point.

Worked example 1: Coulombic attraction in ionic solids (Unit 2)

Question. Strontium oxide, SrO, has a much higher melting point than rubidium fluoride, RbF. Both have the same lattice type. Using the ionic radii in the calculator note, explain the difference in terms of Coulombic attraction. (2 points)

Weak answer. “SrO has stronger ionic bonds because Sr is 2+, so it is harder to melt.”

What it misses. It ignores the oxide ion’s charge, the distance between ions and RbF, and never says what melting does to the ions.

Strong answer. Melting separates ions held by Coulombic attraction, which is proportional to |q₊q₋| ÷ d. In SrO, |q₊q₋| = 2 × 2 = 4 and d = 118 + 140 = 258 pm. In RbF, |q₊q₋| = 1 × 1 = 1 and d = 152 + 133 = 285 pm. SrO has four times the charge product and a shorter distance, so its ratio is about 4.4 times larger (0.0155 against 0.00351). The attraction between ions in SrO is much stronger, so more energy is needed to separate them, and SrO melts at a much higher temperature.

Suggested Marlbridge rubric	Point
Uses both charges for each compound (4 compared with 1) and the smaller distance in SrO	1
Links stronger attraction between ions to more energy needed to melt	1

Worked example 2: a titration calculation (Unit 4)

Question. 20.00 mL of sulfuric acid, H₂SO₄(aq), is titrated with 0.1250 M NaOH(aq). The end point is reached after 31.48 mL. Calculate [H₂SO₄]. (3 points)

Weak answer. “0.03148 × 0.125 = 0.003935. 0.003935 ÷ 0.02 = 0.197.”

What it misses. No equation, so the 2 : 1 mole ratio is missed and the answer is twice too large. No units, no assumption.

Strong answer.

1. Equation: H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l).
2. n(NaOH) = cV = 0.1250 mol L⁻¹ × 0.03148 L = 3.935 × 10⁻³ mol.
3. n(H₂SO₄) = ½ × 3.935 × 10⁻³ mol = 1.968 × 10⁻³ mol (1.9675 × 10⁻³ unrounded).
4. [H₂SO₄] = n ÷ V = 1.9675 × 10⁻³ mol ÷ 0.02000 L = 0.09838 M (4 significant figures, matching the data).

Assumption: the end point is the equivalence point, and both protons of H₂SO₄ react.

Suggested Marlbridge rubric	Point
Moles of NaOH from cV, with volume in litres	1
Uses the 2 : 1 mole ratio from a balanced equation	1
[H₂SO₄] = 0.0984 M (or 0.09838 M) with unit	1

Worked example 3: justifying rate orders from data (Unit 5)

Question. For the fictional reaction P + Q → products at constant temperature:

Trial	[P] (M)	[Q] (M)	Initial rate (M s⁻¹)
1	0.050	0.10	3.0 × 10⁻⁵
2	0.10	0.10	1.2 × 10⁻⁴
3	0.050	0.20	6.0 × 10⁻⁵

A student claims that doubling [P] and doubling [Q] have the same effect on the rate. (a) Evaluate the claim using the data. (b) Write the rate law and calculate k. (c) Explain, in terms of collisions, why the rate in Trial 3 is greater than in Trial 1. (4 points)

Weak answer. “(a) Wrong, P has more effect. (b) rate = k[P]²[Q], k = 0.12. (c) More particles so more collisions.”

What it misses. Part (a) quotes no trials or ratios; k has no unit; part (c) names no particles and ignores collision frequency, energy and orientation.

Strong answer.

(a) The claim is incorrect. Trials 1 and 2: [Q] is constant and [P] doubles; the rate rises from 3.0 × 10⁻⁵ to 1.2 × 10⁻⁴ M s⁻¹, a factor of 4 = 2², so the reaction is second order in P. Trials 1 and 3: [P] is constant and [Q] doubles; the rate doubles, so it is first order in Q. Doubling [P] has twice the effect of doubling [Q].

(b) rate = k[P]²[Q]. k = 3.0 × 10⁻⁵ M s⁻¹ ÷ ((0.050 M)² × 0.10 M) = 0.12 M⁻² s⁻¹.

(c) In Trial 3 there are twice as many Q particles in the same volume, so collisions between P and Q happen twice as often. The temperature is unchanged, so the same fraction of collisions has energy at least Eₐ and a suitable orientation. Twice as many successful collisions per second give twice the rate.

Suggested Marlbridge rubric	Point
Rejects the claim, citing trials and the rate ratios 4 and 2	1
Correct rate law	1
k = 0.12 with unit M⁻² s⁻¹	1
Collision frequency increases; fraction with enough energy unchanged	1

Worked example 4: which acid is stronger? (Unit 8)

Question. A student measures the pH of 0.100 M solutions at 25 °C: acetic acid, CH₃COOH, pH 2.87; fluoroacetic acid, FCH₂COOH, pH 1.83. (a) Make a claim about which acid is stronger and justify it with the data. (b) Calculate Ka for fluoroacetic acid. (c) Explain the difference in strength in terms of molecular structure. (4 points)

Weak answer. “(a) Fluoroacetic acid because its pH is lower. (b) Ka = (10⁻¹·⁸³)² ÷ 0.100 = 2.2 × 10⁻³. (c) Fluorine makes it more acidic.”

What it misses. (a) never links pH to ionization. (b) assumes x is small, but x is 15% of 0.100 M. (c) names an atom, not a cause.

Strong answer.

(a) Fluoroacetic acid is stronger. At equal concentration its pH is lower (1.83 against 2.87), so [H₃O⁺] is higher: 1.48 × 10⁻² M against 1.35 × 10⁻³ M. About 15% of fluoroacetic acid molecules ionize compared with about 1.3% of acetic acid molecules, so its Ka is larger.

(b) [H₃O⁺] = [FCH₂COO⁻] = 10⁻¹·⁸³ = 1.48 × 10⁻² M. [FCH₂COOH] = 0.100 − 0.0148 = 0.0852 M. Ka = (1.48 × 10⁻²)² ÷ 0.0852 = 2.6 × 10⁻³. The 5% check fails (15%), so the full expression is needed; the short-cut gives 2.2 × 10⁻³.

(c) Fluorine is very electronegative. It pulls electron density through the bonds towards itself, which spreads out the negative charge on the conjugate base, FCH₂COO⁻. A more stable conjugate base means the proton is given up more readily.

Suggested Marlbridge rubric	Point
Claim with both pH values or [H₃O⁺] values quoted	1
Links lower pH to a greater fraction ionized or larger Ka	1
Ka = 2.6 × 10⁻³ (2.2 × 10⁻³ accepted if the assumption is stated)	1
Electron withdrawal by F stabilizes the conjugate base	1

Common mistakes

• Calling intermolecular forces “bonds”, or saying covalent bonds break when a molecular liquid boils.
• Describing only one of the two substances in a comparison.
• Skipping the balanced equation, so the mole ratio is lost.
• Leaving units off k, ΔH or concentration, or dropping the minus sign on an exothermic ΔH.
• Rounding in the middle of a calculation, or quoting eight digits at the end.
• Using the “x is small” short-cut without checking it.
• Saying “the data show” without quoting a value, a trial number or a ratio.
• Particulate diagrams that lose atoms, give the wrong ratio, or point water’s hydrogen atoms at a cation.

Practise it

Task 1 (Unit 1). Explain why the first ionization energy of potassium is lower than that of sodium.

Model answer

Potassium’s valence electron is in the 4s subshell (n = 4); sodium’s is in 3s (n = 3). The potassium electron is on average further from the nucleus and has an extra inner shell of core electrons shielding it, so the effective nuclear charge it feels is similar but the distance is larger. By Coulomb’s law the attraction is weaker, so less energy is needed to remove the electron.

Task 2 (Unit 3). Describe a particulate diagram for two formula units of MgBr₂ dissolved in water, showing water molecules around one cation and one anion.

Model answer

Two Mg²⁺ and four Br⁻ ions, separated (ratio 1 : 2; total charge 2 × (+2) + 4 × (−1) = 0). Water molecules around Mg²⁺ point their oxygen atoms (δ−) at the ion; those around Br⁻ point a hydrogen atom (δ+) at it. Other water molecules fill the space, touching but disordered. A key labels each symbol.

Task 3 (Unit 6). An electric heater supplying 40.0 J s⁻¹ is placed in a fictional pure liquid (M = 58.0 g mol⁻¹) that is already boiling. In 250 s the mass of liquid falls by 14.5 g. (a) Calculate the molar enthalpy of vaporization and state the assumption you used. (b) Explain why the temperature stays constant while the liquid boils.

Model answer

(a) q = 40.0 J s⁻¹ × 250 s = 10 000 J. n = 14.5 g ÷ 58.0 g mol⁻¹ = 0.250 mol. ΔH_vap = 10 000 J ÷ 0.250 mol = 40.0 kJ mol⁻¹. Assumption: all the heater’s energy goes into vaporizing the liquid, with none lost to the surroundings. If energy is lost, the true value is lower.

(b) During boiling, the energy supplied is used to overcome the attractions between particles as they separate into the gas, increasing their potential energy. The average kinetic energy of the particles does not change, so the temperature stays constant while liquid remains.

Task 4 (Unit 7). For the fictional equilibrium A₂(g) ⇌ 2A(g) in a 1.00 L container, an equilibrium particulate diagram shows 3 A₂ molecules and 4 A atoms. Each particle represents 0.10 mol. (a) Calculate Kc. (b) Two more A₂ molecules are added. Predict the direction of net reaction and justify it.

Model answer

(a) [A₂] = 0.30 M and [A] = 0.40 M. Kc = [A]² ÷ [A₂] = (0.40)² ÷ 0.30 = 0.53.

(b) Now [A₂] = 0.50 M, so Q = (0.40)² ÷ 0.50 = 0.32. Q (0.32) < K (0.53), so the net reaction goes forward, forming A, until Q = K again.

Task 5 (Unit 9). A cell is built from Ag⁺/Ag and Cr³⁺/Cr half-cells under standard conditions (potentials in the calculator note). (a) Identify the anode. (b) Calculate E°cell and ΔG° for 3Ag⁺(aq) + Cr(s) → 3Ag(s) + Cr³⁺(aq).

Model answer

(a) Chromium has the more negative reduction potential, so Cr is oxidized; Cr is the anode.

(b) E°cell = E°(cathode) − E°(anode) = +0.80 V − (−0.74 V) = +1.54 V. n = 3 mol electrons. ΔG° = −nFE° = −3 × 96 485 C mol⁻¹ × 1.54 V = −445 761 J = −446 kJ mol⁻¹ (per mole of reaction as written). Negative ΔG° agrees with positive E°cell: the reaction is thermodynamically favourable.

Where to practise next

• Intermolecular forces practice: comparison explanations.
• Titration practice: multi-step calculations with mole ratios.
• Rate law practice: orders and k from data tables.
• Reaction quotient and Le Châtelier practice: justifying shifts with Q and K.
• Molecular structure of acids and bases practice: structure-based acid strength.
• Galvanic and electrolytic cells practice: cell potentials and ΔG°.