These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Where a question asks you to draw, sketch your diagram on paper with a key, then compare it with the description in the solution. Solubility data: AgCl is insoluble in water; CaCl₂, Ca(NO₃)₂ and all sodium, potassium and nitrate salts are soluble.
Question 1 (multiple choice · foundation)
Hydrogen sulfide burns in oxygen:
2 H₂S(g) + 3 O₂(g) → 2 SO₂(g) + 2 H₂O(g)
A container holds 4 H₂S molecules and 6 O₂ molecules before the reaction. The reaction goes to completion. Which describes the contents of the container afterwards?
- (A) 4 SO₂ molecules and 4 H₂O molecules
- (B) 2 SO₂ molecules and 2 H₂O molecules
- (C) 4 SO₂ molecules, 4 H₂O molecules and 3 O₂ molecules
- (D) 4 SO₂ molecules, 4 H₂ molecules and 2 O₂ molecules
Answer and explanation
Answer: (A). The ratio H₂S : O₂ is 2 : 3, so 4 H₂S need exactly 6 O₂. Nothing is left over. The products form in the ratio 2 : 2 for every 2 H₂S, so 4 H₂S give 4 SO₂ and 4 H₂O. Atom check: before, S 4, H 8, O 12; after, S 4, H 8, O 4 × 2 + 4 = 12.
- (B) copies the coefficients instead of scaling them to the box. It has only 2 S atoms after, so 2 S atoms have vanished.
- (C) leaves 3 O₂ unreacted, which would need 18 O atoms after. Only 12 are available.
- (D) conserves every atom, but it shows H₂ and O₂ instead of H₂O. A diagram must show the products in the equation, not just the right atom totals.
Question 2 (multiple choice · core)
Which description gives the best particle representation of a small sample of potassium sulfate solution, K₂SO₄(aq)? Water molecules are not shown.
- (A) Three K₂SO₄ units, each with two K atoms and one SO₄ group touching
- (B) Six separate K⁺ ions and three separate SO₄²⁻ ions, each sulfate drawn as one particle
- (C) Three separate K⁺ ions and three separate SO₄²⁻ ions
- (D) Six separate K⁺ ions, three separate S²⁻ ions and twelve separate O²⁻ ions
Answer and explanation
Answer: (B). Potassium sulfate is a soluble ionic compound, so in water it exists as separate ions. Each formula unit gives 2 K⁺ and 1 SO₄²⁻, so the ratio is 2 : 1. The sulfate ion is held together by covalent bonds and stays in one piece. Charge check: 6(+1) + 3(−2) = 0.
- (A) shows undissolved formula units, which is the solid, not the solution.
- (C) has the wrong ratio. Its charge would be 3(+1) + 3(−2) = −3, which is impossible for a neutral sample.
- (D) breaks the S–O covalent bonds. Dissolving does not split a polyatomic ion.
Question 3 (multiple choice · core)
Elements A and B are both made of diatomic molecules. A container holds 3 A₂ molecules and 6 B₂ molecules. After a reaction, it holds 4 AB₃ molecules and 1 A₂ molecule. Which balanced equation represents the reaction?
- (A) 3 A₂ + 6 B₂ → 4 AB₃ + A₂
- (B) A₂ + 3 B₂ → 2 AB₃
- (C) A₂ + 2 B₂ → AB₃
- (D) 3 A₂ + 6 B₂ → 4 AB₃
Answer and explanation
Answer: (B). One A₂ molecule is left over, so only 2 A₂ reacted. All 6 B₂ reacted and 4 AB₃ formed. The used-and-formed ratio 2 : 6 : 4 reduces to 1 : 3 : 2. Check: A 2 = 2, B 6 = 6.
- (A) balances, but it keeps the unreacted A₂ on both sides. An equation shows only what changes, in the smallest whole-number ratio.
- (C) does not balance: 2 A and 4 B on the left, 1 A and 3 B on the right.
- (D) ignores the leftover A₂ and so has 6 A atoms on the left but only 4 on the right.
Question 4 (constructed response · core)
A solution containing 3 Ag⁺ ions and 3 NO₃⁻ ions is mixed with a solution containing 2 Ca²⁺ ions and 4 Cl⁻ ions. Silver chloride precipitates.
(a) Write the net ionic equation for the reaction. (b) Describe a particle diagram of the mixture after the reaction. Give the number of each particle and say how each is arranged. (c) Show that your diagram in (b) is consistent with conservation of charge.
Worked solution and suggested Marlbridge rubric
(a) Ag⁺(aq) + Cl⁻(aq) → AgCl(s).
(b) The ratio is 1 : 1. There are 3 Ag⁺ and 4 Cl⁻, so all 3 Ag⁺ react with 3 Cl⁻; 1 Cl⁻ is left over.
- Solid: a packed cluster of 3 Ag⁺ and 3 Cl⁻ ions (3 AgCl units), at the bottom of the container.
- In solution, all separate and not touching: 2 Ca²⁺, 3 NO₃⁻ (each one particle) and 1 Cl⁻.
- No Ag⁺ remains in solution.
(c) Solution: 2(+2) + 3(−1) + 1(−1) = 0. Solid: 3(+1) + 3(−1) = 0. Total charge is zero, as it was before mixing.
| Point | What earns it |
|---|---|
| 1 | Correct net ionic equation with state symbols, no spectator ions |
| 1 | Precipitate shown as 3 Ag⁺ and 3 Cl⁻ packed together |
| 1 | Spectators (2 Ca²⁺, 3 NO₃⁻) and the 1 leftover Cl⁻ shown separate in solution |
| 1 | Charge sum shown to be zero for the solution (or for the whole diagram) |
Do not award the third point if Ca²⁺ is drawn paired with NO₃⁻ or Cl⁻, because those salts are soluble.
Question 5 (constructed response · core)
A sealed syringe contains 4 water molecules in the gas phase at 110 °C. The syringe is cooled to 25 °C and the water condenses.
(a) Describe a particle diagram of the syringe contents after cooling. (b) Explain, using your diagram, why condensation is represented as a physical process. (c) A student draws the “after” box with 4 H₂ molecules and 2 O₂ molecules packed closely together. Explain what is wrong.
Worked solution and suggested Marlbridge rubric
(a) 4 H₂O molecules, each an O atom joined to two H atoms, close together and touching but in no regular pattern (liquid), collected at the bottom of the syringe.
(b) The same 4 H₂O molecules appear before and after. No O–H bonds break or form; only the distance between molecules changes, because attractions between molecules pull them close together. The equation is H₂O(g) → H₂O(l), with the same formula on both sides.
(c) The student has drawn new substances. 4 H₂ and 2 O₂ conserve the atoms (H 8, O 4), but they represent the decomposition 2 H₂O → 2 H₂ + O₂, which is a chemical change. Cooling does not break covalent bonds. Also, H₂ and O₂ are gases at 25 °C, so packing them closely is not consistent with their state.
| Point | What earns it |
|---|---|
| 1 | Four intact H₂O molecules, close together and disordered |
| 1 | Explains that the particles are the same before and after; only spacing changes |
| 1 | Identifies that the student’s diagram shows a different substance (a chemical change), with a reason |
Question 6 (constructed response · core)
Hydrogen reacts with fluorine: H₂(g) + F₂(g) → 2 HF(g). A “before” box contains 2 H₂ molecules and 3 F₂ molecules. A student draws the “after” box with 4 HF molecules and 2 F₂ molecules.
(a) Use atom counts to show that the student’s diagram is not consistent. (b) Describe the correct “after” box. (c) Explain how your corrected diagram shows the 1 : 1 : 2 ratio of the equation.
Worked solution and suggested Marlbridge rubric
(a) Before: H = 2 × 2 = 4, F = 3 × 2 = 6. Student’s after: H = 4, F = 4 + 2 × 2 = 8. Two fluorine atoms have appeared from nowhere, so atoms are not conserved.
(b) 2 H₂ react with 2 F₂ to give 4 HF. One F₂ is left over. Correct after box: 4 HF molecules and 1 F₂ molecule, all far apart (gases). Check: H 4, F 4 + 2 = 6.
(c) Used and formed: 2 H₂, 2 F₂ and 4 HF. The ratio 2 : 2 : 4 reduces to 1 : 1 : 2, which matches the coefficients. The leftover F₂ does not count towards the ratio.
| Point | What earns it |
|---|---|
| 1 | Shows F is 6 before and 8 in the student’s box |
| 1 | Correct after box: 4 HF and 1 F₂ |
| 1 | Shows that the particles used and formed are in the ratio 1 : 1 : 2, excluding the leftover |
Question 7 (constructed response · stretch)
A small flask contains 6 molecules of liquid hydrogen peroxide, H₂O₂. A catalyst is added. Afterwards the flask contains 6 molecules of liquid water and 3 molecules of oxygen gas.
(a) Write the balanced equation shown by these particle counts, with state symbols. (b) Describe how a consistent “after” diagram should show the different states of the two products. (c) A student says: “The number of molecules went up from 6 to 9, so the total mass must have increased.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) 6 H₂O₂ → 6 H₂O + 3 O₂ reduces to 2 H₂O₂(l) → 2 H₂O(l) + O₂(g). Check: H 4 = 4; O 4 = 2 + 2.
(b) The 6 H₂O molecules are close together and disordered at the bottom of the flask (liquid). The 3 O₂ molecules are far apart from each other and from the liquid, filling the space above it (gas). Each O₂ is drawn as two joined O atoms, and each H₂O as one O joined to two H.
(c) The claim is incorrect. Mass depends on the atoms, not on the number of molecules. Before: 12 H and 12 O atoms. After: 12 H and 6 + 6 = 12 O atoms. The same atoms are present, so the total mass is unchanged (if the O₂ stays in a closed flask). The number of molecules rose only because the atoms were rearranged into smaller molecules.
| Point | What earns it |
|---|---|
| 1 | Balanced equation in lowest whole-number ratio |
| 1 | Correct state symbols, (l), (l) and (g) |
| 1 | Liquid water shown close together; O₂ shown far apart |
| 1 | Rejects the claim, linking mass to conserved atoms (with an atom count or clear statement) |
An answer that also notes that an open flask would lose mass as O₂ escapes shows good reasoning, but the last point needs the link between conserved atoms and mass.
How did you do?
- Q1 or Q3 wrong: practise “subtract the leftovers, then reduce the ratio” in Worked example 1 of the study guide.
- Q2 or Q4 wrong: re-read “Representing reactions in solution” and Worked example 2.
- Q5 wrong: re-read “Representing physical processes”.
- Q6 or Q7 incomplete: your answer needs an atom count as evidence, not only a statement.
Then tick off the topic checklist.