Section A — Short Structured Questions (Q1–5 · 50 marks)
Q1
(a) Write the electronic configuration of a nitrogen atom and state its group and period in the Periodic Table. [3]
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- Electronic configuration: 2,5 [1]
- Group: V (Group 5) — 5 outer shell electrons [1]
- Period: 2 — 2 occupied electron shells [1]
(b) Draw a dot-and-cross diagram for a molecule of ammonia (NH₃), showing outer shell electrons only. [2]
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- Central N atom with 3 N–H bonding pairs (each shown as one × from N and one • from H, shared) [1]
- One lone pair on N shown correctly; each H has no lone pairs [1]
N has 5 outer electrons (×): 3 shared in bonds + 1 lone pair (×× = 2 electrons). Each H has 1 outer electron (•): 1 shared in bond.
(c) Nitrogen forms N₂O₄ (dinitrogen tetroxide). State the type of bonding in N₂O₄ and predict two physical properties consistent with its structure. [3]
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- Bonding: covalent (non-metal compound, electron sharing) [1]
- N₂O₄ is a simple molecular covalent substance. Properties (any two): low melting/boiling point [1]; does not conduct electricity in any state [1]; soluble in non-polar solvents; exists as gas or low-boiling liquid at room temperature [1]
Simple molecular covalent: weak intermolecular forces between molecules → low m.p./b.p. No ions or free electrons → non-conductor.
⚠ Trap: Do not say "weak covalent bonds" — the bonds WITHIN the molecule are strong. What are weak are the intermolecular forces BETWEEN molecules.
(d) Silicon dioxide (SiO₂) has a giant covalent structure similar to diamond. Explain why SiO₂ has a much higher melting point than N₂O₄, even though both are covalent. [2]
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- SiO₂ has a giant covalent lattice — every Si atom is covalently bonded to 4 O atoms throughout the structure. To melt, ALL these strong covalent bonds must be broken, requiring enormous energy. [1]
- N₂O₄ is a simple molecular covalent substance. Melting only requires overcoming weak intermolecular forces between molecules — the strong covalent bonds within molecules are not broken. This requires far less energy. [1]
Q2
(a) Calculate the Mr of Fe₂O₃. [1]
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- Mr = (2 × 56) + (3 × 16) = 112 + 48 = 160 [1]
(b) Calculate the number of moles in 400 g of Fe₂O₃. [1]
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- moles = mass / Mr = 400 / 160 = 2.5 mol [1]
(c) Calculate the mass of iron produced from 400 g of Fe₂O₃ (assume excess CO). [2]
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- Mole ratio Fe₂O₃ : Fe = 1 : 2. Moles Fe = 2 × 2.5 = 5 mol [1]
- Mass Fe = 5 × 56 = 280 g [1]
(d) Calculate the volume of CO₂ gas produced at RTP from 400 g of Fe₂O₃. [2]
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- Mole ratio Fe₂O₃ : CO₂ = 1 : 3. Moles CO₂ = 3 × 2.5 = 7.5 mol [1]
- Volume = 7.5 × 24 = 180 dm³ [1]
(e) In practice, only 224 g of iron is obtained from 400 g of Fe₂O₃. Calculate the percentage yield. [2]
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- % yield = (actual / theoretical) × 100 = (224 / 280) × 100 [1]
- = 80% [1]
(f) State one reason why the yield is less than 100%. [1]
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- Any one of: the reaction is reversible / incomplete; some iron is lost during collection/processing; side reactions occur; impurities in the ore react instead. [1]
Q3
(a) Write the word equation and balanced symbol equation for the reaction of HCl with zinc. Include state symbols. [3]
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- Word equation: zinc + hydrochloric acid → zinc chloride + hydrogen [1]
- Symbol equation: Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g) [1 for correct formula/balancing, 1 for all state symbols correct]
(b) Describe a test to confirm that the gas produced in (a) is hydrogen. [2]
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- Hold a lit splint near the mouth of the test tube [1]
- A squeaky pop/small explosion is heard, confirming hydrogen [1]
(c) The student wants to prepare zinc chloride crystals from the reaction in (a). Describe the full procedure to obtain a pure, dry sample of zinc chloride crystals. [3]
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- Add excess zinc to the dilute HCl (to ensure all the acid reacts — excess zinc ensures no unreacted HCl remains in solution) [1]
- Filter the mixture to remove excess unreacted zinc [1]
- Gently evaporate the filtrate (ZnCl₂ solution) until a small volume remains / until crystals begin to form on the surface; allow to cool and crystallise; filter off crystals and dry between filter paper or in a warm oven [1]
Must not boil to dryness — crystals may decompose. "Evaporate to dryness" loses 1 mark for ZnCl₂ which is a hydrated salt and would spit.
💡 Key principle: For a soluble salt made from an insoluble reactant (metal/metal oxide/carbonate + acid): use EXCESS solid → filter → evaporate → crystallise.
(d) The student tests the final product with universal indicator and gets pH 6. Explain why the solution is not exactly pH 7 (neutral) and how the procedure could be improved. [2]
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- pH 6 is slightly acidic — a small amount of HCl may remain if insufficient zinc was added, OR ZnCl₂ solution is mildly acidic due to hydrolysis of the Zn²⁺ ion in water [1]
- Improvement: ensure a large excess of zinc is used so all acid reacts; recrystallise the product from distilled water to purify further [1]
Q4
(a) Explain why the rate of reaction decreases over time in Experiment 1. [2]
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- As the reaction proceeds, the concentration of HCl decreases (acid is consumed) [1]
- Fewer acid particles per unit volume → less frequent collisions between H⁺ ions and CaCO₃ surface → decreasing rate of reaction [1]
(b) Compare the two experiments. Explain why Experiment 2 has a faster initial rate but the same total volume of CO₂. [4]
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- Faster initial rate in Experiment 2: higher concentration of HCl (2 mol/dm³ vs 1 mol/dm³) means more H⁺ ions per unit volume → more frequent collisions with CaCO₃ surface → more successful collisions per second → faster rate [2]
- Same total volume of CO₂: the same mass of marble chips was used in both experiments, so the same number of moles of CaCO₃ reacts (assuming marble is limiting reagent). The total amount of CO₂ depends on the amount of CaCO₃, not the HCl concentration [1]. Both produce 75 cm³ total [1]
💡 Key distinction: Rate tells you HOW FAST — affected by concentration, temperature, surface area, catalyst. Total yield tells you HOW MUCH — determined by the limiting reagent and the mole ratio.
(c) Calculate the average rate of reaction (in cm³/s) during the first 20 s of Experiment 1. [1]
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- Rate = volume / time = 28 / 20 = 1.4 cm³/s [1]
(d) Describe ONE change that would increase the rate of Experiment 1 WITHOUT changing the total volume of CO₂ produced. Explain how this change increases rate using collision theory. [3]
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- Valid change (any one): use powdered CaCO₃ instead of chips (same mass) [1] / raise temperature of the HCl [1] / add a catalyst [1]
- Explanation for surface area increase: smaller particles have greater surface area exposed → more CaCO₃ particles can contact acid simultaneously → more frequent collisions per second → faster rate. Same mass of CaCO₃ → same total CO₂ [2]
Do NOT increase HCl concentration — this would increase total CO₂ if HCl is limiting. Do NOT increase the amount of CaCO₃ — this increases total CO₂.
Q5
(a) State what is observed at each electrode and write the electrode equations. [4]
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- Cathode (−): pink/red copper metal is deposited [1]. Equation: Cu²⁺(aq) + 2e⁻ → Cu(s) [1]
- Anode (+): copper anode dissolves/decreases in size [1]. Equation: Cu(s) → Cu²⁺(aq) + 2e⁻ [1]
With copper electrodes, the copper anode dissolves to replenish Cu²⁺ ions — the concentration of CuSO₄ stays approximately constant. This is the principle used in copper purification/electroplating.
(b) Calculate the charge passed during the electrolysis. [1]
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- Q = It = 2 × 965 = 1930 C [1]
(c) Calculate the moles of electrons transferred and hence the mass of copper deposited at the cathode. [3]
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- Moles e⁻ = Q / F = 1930 / 96 500 = 0.02 mol [1]
- From equation: 2 mol e⁻ → 1 mol Cu. So moles Cu = 0.02 / 2 = 0.01 mol [1]
- Mass Cu = 0.01 × 64 = 0.64 g [1]
(d) Identify the oxidation and reduction processes in this electrolysis, and state which electrode is the oxidising agent and which is the reducing agent. [2]
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- Cathode: Cu²⁺ gains electrons → REDUCTION. The cathode is the reducing agent (it reduces Cu²⁺) [1]
- Anode: Cu loses electrons → OXIDATION. The anode is the oxidising agent (it oxidises Cu → Cu²⁺) [1]
OIL RIG: Oxidation at anode (loss of electrons). Reduction at cathode (gain of electrons). The electrode that causes reduction is the reducing agent; the electrode that causes oxidation is the oxidising agent.
Section B — Longer Structured Questions (Q6–8 · 30 marks)
Q6
(a) Draw the full structural formula of propane (C₃H₈) and butene (C₄H₈). [2]
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- Propane: H₃C–CH₂–CH₃ (three C atoms in chain, each C has correct number of H atoms, all single bonds) [1]
- But-1-ene: H₂C=CH–CH₂–CH₃ (four C atoms, one C=C double bond shown, correct H atoms) [1]
(b) Propane undergoes complete combustion. Write the balanced equation. Then explain why incomplete combustion is dangerous. [3]
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- C₃H₈ + 5O₂ → 3CO₂ + 4H₂O [1 for correct balancing]
- Incomplete combustion produces carbon monoxide (CO) [1]
- CO is toxic — it binds to haemoglobin in red blood cells more strongly than oxygen, preventing oxygen transport and causing asphyxiation (carbon monoxide poisoning) [1]
(c) Cracking of long-chain alkanes produces shorter alkanes and alkenes. Write the equation for the cracking of decane (C₁₀H₂₂) to produce octane (C₈H₁₈) and one other product. State the conditions required. [3]
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- C₁₀H₂₂ → C₈H₁₈ + C₂H₄ (ethene) [1 for correct equation — check: 10C = 8+2 ✓; 22H = 18+4 ✓]
- Other product: ethene (C₂H₄) — an alkene [1]
- Conditions: high temperature (approximately 500–700°C) AND a catalyst (silica/alumina/zeolite) [1]
Accept other valid crack products (e.g. C₁₀H₂₂ → C₅H₁₂ + C₅H₁₀) as long as the equation balances and an alkene is produced.
(d) The ethene produced in cracking is used to make poly(ethene). Explain, using diagrams or equations, how addition polymerisation works. [2]
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- The C=C double bond in each ethene monomer opens up [1]
- Many ethene monomers join to form a long polymer chain: n CH₂=CH₂ → (–CH₂–CH₂–)ₙ. No atoms are lost — all atoms of each monomer are incorporated into the polymer. [1]
Accept a diagram showing: 3–5 ethene molecules with C=C, then arrow to polymer chain with C–C (single bonds). The repeat unit (–CH₂–CH₂–) in brackets with n is acceptable.
Q7
(a) Identify solutions A, B, C and D, giving reasons for each identification. [4]
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- A = Cu(NO₃)₂: blue precipitate with NaOH = Cu(OH)₂ (Cu²⁺ ions give blue precipitate) [1]
- B = FeSO₄: green precipitate with NaOH = Fe(OH)₂ (Fe²⁺ ions give green precipitate) [1]
- C = FeCl₃: brown/rust precipitate with NaOH = Fe(OH)₃ (Fe³⁺ ions give brown precipitate) [1]
- D = ZnSO₄: white precipitate with NaOH = Zn(OH)₂ which dissolves in excess NaOH (amphoteric) [1]
(b) Describe how the student could distinguish between solutions A (copper nitrate) and C (iron(III) chloride) using anion tests, without using NaOH. [3]
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- To test for Cl⁻ (in FeCl₃): add dilute nitric acid then silver nitrate solution [1]
- FeCl₃ gives a white precipitate of AgCl — confirms Cl⁻ ions present [1]
- Cu(NO₃)₂ gives no precipitate (NO₃⁻ does not react with AgNO₃) [1]
Always acidify with dilute HNO₃ (NOT HCl — would introduce Cl⁻ ions) before adding AgNO₃ to remove any CO₃²⁻ that might give a false positive.
(c) When the student mixes iron(II) sulfate and copper(II) nitrate in equal volumes, a reaction occurs. Write the ionic equation for this reaction and identify what is oxidised and what is reduced. [3]
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- No reaction between FeSO₄ and Cu(NO₃)₂ solutions occurs — Fe is less reactive than Cu, so Fe²⁺ cannot displace Cu²⁺. However, if the question intends Fe metal + Cu²⁺: Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s). Ionic equation: Fe + Cu²⁺ → Fe²⁺ + Cu [1]
- Fe is oxidised: Fe → Fe²⁺ + 2e⁻ (loses electrons, oxidation state increases from 0 to +2) [1]
- Cu²⁺ is reduced: Cu²⁺ + 2e⁻ → Cu (gains electrons, oxidation state decreases from +2 to 0) [1]
Accept if student notes that between the two solutions (both aqueous) no visible reaction occurs and explains why. Award marks for correct oxidation/reduction identification in context.
Q8
(a) Explain why aluminium is extracted by electrolysis but iron is extracted by reduction with carbon. [2]
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- Aluminium is above carbon in the reactivity series — carbon cannot displace Al from its oxide (Al is more reactive than C). The more powerful method of electrolysis is required. [1]
- Iron is below carbon in the reactivity series — carbon CAN displace Fe from its oxide (C is more reactive than Fe). Reduction by carbon (coke) in the blast furnace is cheaper and more practical. [1]
(b) Write the equation for the reduction of iron(III) oxide by carbon monoxide in the blast furnace. [2]
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- Fe₂O₃ + 3CO → 2Fe + 3CO₂ [1 correct formula, 1 correctly balanced]
CO acts as the reducing agent — it removes oxygen from Fe₂O₃. CO is oxidised to CO₂ in the process. Fe₂O₃ is reduced to Fe.
(c) Aluminium is more reactive than iron, yet aluminium resists corrosion better in air. Explain why. [2]
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- Aluminium reacts rapidly with oxygen in air to form a thin, dense layer of aluminium oxide (Al₂O₃) on its surface [1]
- This oxide layer is strongly adherent and impermeable — it prevents further oxygen and water from reaching the metal underneath, protecting it from further corrosion [1]
Iron's rust (Fe₂O₃·xH₂O) is flaky and porous — it does NOT protect the iron underneath, allowing continuous corrosion. This is the key difference.
(d) The extraction of aluminium requires large amounts of electricity. State two environmental concerns associated with aluminium production and suggest how one of these concerns could be reduced. [3]
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- Environmental concern 1: large amounts of electricity needed → burning fossil fuels to generate electricity releases CO₂ → contributes to greenhouse effect and climate change [1]
- Environmental concern 2: mining bauxite destroys habitats/ecosystems; produces toxic red mud waste; deforestation of mining sites [1]
- Reduction measure: recycle aluminium (recycling uses only ~5% of the energy needed for primary extraction from ore) [1] / use renewable energy sources to power the electrolysis / restore mining sites after use [1]
(e) Copper is used for electrical wiring rather than aluminium, even though aluminium is cheaper and more abundant. Suggest one reason, using data from your chemistry knowledge. [1]
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- Any one of: copper has higher electrical conductivity than aluminium (lower resistance per unit length) [1] / copper is more ductile (can be drawn into thin wires more easily) [1] / copper is denser and stronger for a given wire diameter, preventing breakage at connections [1] / copper does not form an insulating oxide layer at connections that would interrupt current flow (aluminium oxide is a non-conductor and forms on Al wire joints, causing resistance) [1]
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