2023 CCEA GCSE Chemistry 1110 模拟试题及答案详解

(a)(i)–(ii) Standard definitions, as given in the specification. (iii) Mass number = protons + neutrons = 13 + 14 = 27. (iv) An aluminium atom (atomic number 13) has 13 protons and, since it forms Al³⁺ by losing 3 electrons, 13 − 3 = 10 electrons; the number of neutrons is unaffected by ion formation and is given as 14. (v) Relative atomic mass is the weighted mean of isotope masses by abundance: \( A_r = (0.20 \times 10) + (0.80 \times 11) = 2.0 + 8.8 = 10.8 \). (b) Calcium chloride, CaCl₂, requires one calcium atom to react with two chlorine atoms. Calcium (electronic structure 2,8,8,2) loses its 2 outer electrons to achieve the stable configuration 2,8,8, forming Ca²⁺; each chlorine atom (2,8,7) gains one of these electrons to complete its outer shell to 2,8,8, forming Cl⁻. In a dot-and-cross diagram, the 2 electrons lost by calcium (crosses) are shown redrawn into the outer shells of two separate chlorine atoms (as crosses among the chlorine atoms' own dots), with square brackets and charges around each resulting ion (Ca²⁺ and two Cl⁻). The oppositely charged ions are held together by strong electrostatic attraction, forming a giant ionic lattice structure. (c) Nitrogen (electronic structure 2,5) has 5 outer electrons and needs 3 more to complete its outer shell; each hydrogen atom has 1 outer electron and needs 1 more. Each hydrogen atom shares its single electron with one of nitrogen's unpaired electrons, forming a shared pair (a single covalent bond) between N and each H; this occurs three times, using 3 of nitrogen's 5 outer electrons, and leaves nitrogen's remaining 2 outer electrons as a non-bonding lone pair. In the dot-and-cross diagram, each N–H bond is shown as one dot (from H) and one cross (from N) shared between the two atoms, with the lone pair shown as two crosses on nitrogen alone. This gives a total of three bonding pairs of electrons (three N–H covalent bonds).
Final answer: (a) definitions as above; mass number 27; Al³⁺ has 13 protons, 14 neutrons, 10 electrons; Ar(B) = 10.8; (b) Ca loses 2e⁻ to form Ca²⁺, each of 2 Cl atoms gains 1e⁻ to form Cl⁻, giant ionic lattice; (c) N shares 3 bonding pairs with 3 H atoms, 1 lone pair remains on N.

(a)(i) 1 mark for correct definition. (ii) 1 mark for correct definition. (iii) 1 mark for 27. (iv) 1 mark each for 13 protons, 14 neutrons, 10 electrons — max 3. (v) 1 mark for 10.8 (with correct working shown; ECF applies for arithmetic slips using the correct method). (b) 1 mark for calcium losing 2 electrons; 1 mark for one Ca atom transferring one electron to each of two separate Cl atoms; 1 mark for both ions reaching a stable 2,8,8 configuration; 1 mark for describing the ionic bond as electrostatic attraction between oppositely charged ions; 1 mark for 'giant ionic lattice' — max 5. (c) 1 mark for identifying 3 shared/bonding pairs of electrons (one per N–H bond); 1 mark for correctly showing each bond as one electron from N and one from H; 1 mark for showing a lone pair remaining on nitrogen; 1 mark for correct total outer electron count on N (5) and H (1 each) before bonding; 1 mark for stating 'three' bonding pairs as the final answer — max 5.

(a) The mixture is first filtered: sand, being insoluble, is retained as the residue on the filter paper, while the dissolved salt passes through as the filtrate. The filtrate is then evaporated/crystallised (heated gently, e.g. in an evaporating basin, until crystals begin to form, then left to cool/evaporate further) to obtain pure, dry salt crystals. (b) The Rf value is defined as the distance travelled by the substance divided by the distance travelled by the solvent: \( R_f = \dfrac{\text{distance travelled by spot}}{\text{distance travelled by solvent}} = \dfrac{6.0}{8.0} = 0.75 \). (c) Sulfate ions are tested for by first acidifying the solution with dilute hydrochloric acid (to remove any carbonate ions that could give a false positive), then adding barium chloride solution; if sulfate ions are present, a white precipitate of insoluble barium sulfate forms immediately: \( \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s) \).
Final answer: (a) filtration (sand=residue, salt solution=filtrate) then evaporation/crystallisation of the filtrate; (b) Rf = 0.75; (c) add dilute HCl then BaCl₂ solution — white precipitate (BaSO₄) confirms sulfate; Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s).

(a) 1 mark for 'filtration' with sand correctly identified as residue and salt solution as filtrate; 1 mark for 'evaporation'/'crystallisation' applied to the filtrate; 1 mark for a valid procedural detail (e.g. heat gently, stop heating once crystals form); 1 mark for correctly obtaining 'dry salt crystals' as the final product — max 4. (b) 1 mark for correct formula (distance by spot ÷ distance by solvent); 1 mark for correct substitution (6.0/8.0); 1 mark for correct answer 0.75 — max 3. (c) 1 mark for 'dilute hydrochloric acid' (acidify first); 1 mark for 'barium chloride solution'; 1 mark for 'white precipitate' as the positive result; 1 mark for correct ionic equation with correct formulae; 1 mark for correct state symbols in the ionic equation — max 5.

(a) Copper(II) oxide (a base) reacts with sulfuric acid to form the salt copper(II) sulfate and water: \( \text{CuO(s)} + \text{H}_2\text{SO}_4(aq) \rightarrow \text{CuSO}_4(aq) + \text{H}_2\text{O(l)} \); checking atoms confirms this is already balanced (Cu 1=1, O 1+4=1+4, S 1=1, H 2=2). (b) Copper(II) oxide is a black solid; copper(II) sulfate crystals (hydrated) are blue. (c) Using excess solid copper(II) oxide means that once the reaction is complete, any unreacted acid has all been neutralised and the only substances left are dissolved copper(II) sulfate and unreacted solid copper(II) oxide — the excess solid can simply be filtered off, leaving a pure salt solution with no need for an indicator (which would contaminate the final product and require an extra removal step, unlike the excess-solid method). The student knows excess has been added because some black solid copper(II) oxide remains undissolved, visible at the bottom of the flask, even after thorough stirring. (d) The method follows the 'excess insoluble base added to acid' technique: warm a measured volume of dilute sulfuric acid gently, then add copper(II) oxide powder a little at a time, stirring continuously, continuing until the black copper(II) oxide is clearly in excess (i.e. some black solid remains undissolved even after further stirring), which ensures all of the acid has reacted. The mixture is then filtered to remove the excess, unreacted copper(II) oxide, leaving a clear blue filtrate containing dissolved copper(II) sulfate. This filtrate is gently heated/evaporated (for example in an evaporating basin) to reduce its volume and concentrate the solution, until the point of crystallisation is reached (often tested by dipping a glass rod into the solution and checking that crystals form on cooling on the rod). The concentrated solution is then left to cool undisturbed, allowing blue copper(II) sulfate crystals to form. Finally, the crystals are filtered from the remaining liquid and dried, for example by pressing gently between sheets of filter paper or leaving in a desiccator or low-temperature oven, avoiding strong heating that could remove the crystals' water of crystallisation.
Final answer: (a) CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l); (b) black; blue; (c) excess solid avoids needing an indicator and is removed by simple filtration; shown by undissolved solid remaining; (d) warm acid, add excess CuO with stirring, filter off excess solid, evaporate/concentrate the filtrate, cool to crystallise, filter and dry the crystals.

(a) 1 mark for correct formulae; 1 mark for correct balancing (already balanced as written); 1 mark for correct state symbols — max 3. (b) 1 mark for 'black'; 1 mark for 'blue' — max 2. (c) 1 mark for explaining excess solid is removed by simple filtration/no indicator needed; 1 mark for explaining an indicator would contaminate the product if used instead; 1 mark for 'solid remains undissolved' as the sign of excess — max 3. (d) Banded mark scheme (6 marks, QWC assessed). Indicative content: warm the dilute sulfuric acid; add copper(II) oxide gradually with stirring until in excess; filter to remove excess/unreacted copper(II) oxide; evaporate/heat the filtrate to concentrate it (to the point of crystallisation); cool to allow crystals to form; filter and dry the crystals (e.g. between filter paper or in a desiccator). Band A (5–6 marks): at least 5 of the indicative points, in a clear, logical sequence, accurate specialist vocabulary, few SPG errors. Band B (3–4 marks): 3–4 indicative points, reasonably clear, some SPG errors. Band C (1–2 marks): 1–2 indicative points, poorly organised, weak SPG.

(a) Mass of oxygen combined = mass of product − mass of magnesium = \( 4.03 - 2.43 = 1.60 \text{ g} \). (b) Using \( \text{moles} = \dfrac{\text{mass}}{A_r} \): moles Mg \( = \dfrac{2.43}{24} = 0.10125 \text{ mol} \); moles O \( = \dfrac{1.60}{16} = 0.10 \text{ mol} \). (c) Dividing both values by the smaller (0.10): Mg \( \frac{0.10125}{0.10} \approx 1.0 \), O \( \frac{0.10}{0.10} = 1.0 \); the simplest whole-number ratio of Mg:O is 1:1, giving the empirical formula MgO. (d) Small experimental errors are common in this classic experiment — for example, some of the hot magnesium can react with nitrogen gas in the air to form magnesium nitride as a side reaction, or a small amount of magnesium oxide 'smoke' can escape from the crucible when the lid is lifted, both of which would affect the measured masses slightly even though the calculated ratio still confirms the correct formula. (e)(i) Nanoparticles are defined as structures with a size of 1–100 nm. (ii) As the size of a particle decreases, its surface area to volume ratio increases sharply (for a cube, decreasing the side length by a factor of 10 increases the surface area to volume ratio by a factor of 10); this means a much larger proportion of a nanoparticle's atoms are located at its surface compared with the same substance in bulk, where most atoms are 'buried' inside; because surface atoms are more available to interact/react with their surroundings, nanoparticles can show different chemical and physical properties (e.g. reactivity) from the bulk material. (iii) Nanoparticles of substances such as titanium dioxide or zinc oxide are used in some sun creams because their small size gives more even, effective coverage of the skin and better absorption/reflection of the Sun's harmful ultraviolet rays; however, because nanoparticles are so small, there is a risk they could penetrate the skin (or be absorbed into the body) more easily than larger particles, raising concerns about potential cell damage, and there are also concerns about their environmental impact if washed into waterways.
Final answer: (a) 1.60 g; (b) 0.10125 mol Mg, 0.10 mol O; (c) empirical formula MgO; (d) e.g. side reaction with nitrogen/loss of oxide smoke; (e) 1–100 nm; higher surface area:volume ratio means more surface atoms able to react/interact; benefit — better UV protection/skin coverage; risk — potential cell damage/environmental harm.

(a) 1 mark for 1.60 g. (b) 1 mark for correct method for Mg; 1 mark for correct method for O; 1 mark for both correct values (0.10125 and 0.10) — max 3 (allow ECF from (a)). (c) 1 mark for correctly dividing by the smaller value and stating 'MgO' — accept if correct ratio (1:1) is shown even without explicit division shown, provided formula MgO given. (d) 1 mark for any valid, chemically sound source of experimental error (e.g. reaction with nitrogen, loss of product as smoke, incomplete reaction). (e)(i) 1 mark for '1–100 nm'. (ii) 1 mark for stating surface area to volume ratio increases as particle size decreases; 1 mark for linking this to a greater proportion of atoms being on the surface; 1 mark for linking this to different properties/reactivity from the bulk material — max 3. (iii) 1 mark for a valid benefit; 1 mark for a valid risk — max 2.

(a)(i) Since chlorine displaces the less reactive bromine from solution, the colourless potassium bromide solution turns orange/orange-brown as bromine is formed. (ii) Chlorine (as Cl₂) oxidises bromide ions to bromine, while itself being reduced to chloride ions: \( \text{Cl}_2(aq) + 2\text{Br}^-(aq) \rightarrow 2\text{Cl}^-(aq) + \text{Br}_2(aq) \). (iii) This displacement reaction occurs because chlorine, higher up Group 7, is more reactive than bromine, further down the group; a more reactive halogen can displace a less reactive halogen from a solution of its salt, so chlorine displaces bromine from potassium bromide solution. (b) Using the balanced equation \( \text{Cl}_2 + 2\text{KBr} \rightarrow 2\text{KCl} + \text{Br}_2 \): the relative formula mass of KBr is \( 39 + 80 = 119 \); moles of KBr \( = \dfrac{11.9}{119} = 0.1 \text{ mol} \); from the equation, 2 mol KBr produces 1 mol Br₂, so moles of Br₂ \( = \dfrac{0.1}{2} = 0.05 \text{ mol} \); the relative formula mass of Br₂ is \( 2 \times 80 = 160 \); mass of Br₂ \( = 0.05 \times 160 = 8.0 \text{ g} \). (c) Halogens react by gaining one electron into their outer shell to form a stable, noble-gas-like electronic configuration. Going down Group 7, each successive element has an additional electron shell, meaning its outer shell is further from the positively charged nucleus and is shielded from the nucleus's attractive pull by more inner, filled shells of electrons. This makes it progressively harder for the atom to attract and gain an extra electron, so the halogens become less reactive going down the group. (d) Chlorine gas is identified using damp litmus paper (or damp universal indicator paper): if chlorine is present, the paper first turns red (as chlorine dissolves in the water on the paper to form an acidic solution) and is then bleached white (as chlorine's bleaching action destroys the coloured dye).
Final answer: (a) turns orange/orange-brown; Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq); chlorine is more reactive, displaces bromine; (b) 8.0 g; (c) outer shell further from nucleus and more shielded down the group, so harder to gain an electron, reactivity decreases; (d) damp litmus/universal indicator paper turns red then is bleached white.

(a)(i) 1 mark for 'orange' or 'orange-brown'; 1 mark for correctly describing the colour change from colourless — max 2. (ii) 1 mark for correct species/formulae; 1 mark for correct balancing and state symbols — max 2. (iii) 1 mark for 'chlorine is more reactive than bromine'; 1 mark for reference to displacement of the less reactive halogen; 1 mark for reference to chlorine being reduced/bromide being oxidised — max 3. (b) 1 mark for correct Mr of KBr (119); 1 mark for correct moles of KBr (0.1); 1 mark for correct mole ratio applied (halving for Br₂); 1 mark for correct Mr of Br₂ (160); 1 mark for correct final answer (8.0 g) — max 5 (ECF applied throughout for an early slip). (c) 1 mark for 'outer shell further from the nucleus' down the group; 1 mark for 'more shielding by inner shells'; 1 mark for correctly concluding this makes it harder to gain an electron, so reactivity decreases — max 3. (d) 1 mark for 'damp litmus paper' or 'damp universal indicator paper'; 1 mark for 'turns red'; 1 mark for 'then bleached/turns white' — max 3.

(a) Solubility is defined as the mass of a solid that is required to saturate 100 g of water at a given temperature. (b) At 60 °C, a saturated solution in 50 g of water contains \( \dfrac{110}{100} \times 50 = 55 \text{ g} \) of dissolved potassium nitrate. At 20 °C, the maximum mass that can remain dissolved in 50 g of water is \( \dfrac{32}{100} \times 50 = 16 \text{ g} \). Since the solution originally contained 55 g of dissolved potassium nitrate, but only 16 g can remain dissolved at the lower temperature, the mass that crystallises out is \( 55 - 16 = 39 \text{ g} \). (c) Solubility describes the maximum amount of solute that can dissolve in a given amount of water at a particular temperature. As temperature decreases, the solubility of most solids (including potassium nitrate) decreases, meaning the water can hold less dissolved solute at the lower temperature; since the solution was already holding the maximum amount possible at 60 °C, cooling it means it can no longer hold all of the dissolved potassium nitrate, so the excess amount separates back out of solution as solid crystals.
Final answer: (a) mass of solid needed to saturate 100 g water at a given temperature; (b) 39 g; (c) solubility decreases on cooling, so the water can no longer hold all the dissolved solute, and the excess crystallises out.

(a) 1 mark for a correct definition matching the specification. (b) 1 mark for correctly calculating mass dissolved at 60 °C (55 g); 1 mark for correctly calculating maximum mass soluble at 20 °C (16 g); 1 mark for correct method (subtracting the two values); 1 mark for correct final answer (39 g) — max 4 (ECF applied). (c) 1 mark for 'solubility decreases as temperature decreases'; 1 mark for correctly linking this to the excess solute separating out/crystallising once the solution can no longer hold it all dissolved — max 2.

(a)(i) The vigour of the reaction with dilute hydrochloric acid reflects each metal's position in the reactivity series: magnesium, being highly reactive, reacts vigorously, producing rapid bubbling of hydrogen gas and a noticeable rise in temperature; zinc, being less reactive than magnesium, reacts more slowly, with a slower, steadier stream of bubbles; copper, being below hydrogen in the reactivity series, does not react with dilute hydrochloric acid at all, so no bubbles are seen. (ii) A metal's reactivity depends on how readily it loses electrons to form a positive ion — the more easily a metal atom loses electrons (the greater its tendency to form a positive ion), the more reactive it is. (b) In phytomining, specially chosen plants are grown on land where the soil contains low concentrations of copper compounds, too low for conventional mining to be economical; as the plants grow, they absorb copper compounds from the soil through their roots. Once grown, the plants are harvested and burned, producing an ash that contains a concentrated amount of the copper compounds originally spread through the soil. Acid is then added to this ash, dissolving the copper compounds to produce a solution known as a leachate. Finally, copper metal is obtained from this leachate by displacement: scrap iron, being more reactive than copper, is added to the leachate, and iron displaces copper from solution (Fe + Cu²⁺ → Fe²⁺ + Cu), depositing copper metal, which can then be collected. This method avoids the environmental damage of digging, moving and disposing of large amounts of rock associated with traditional mining. (c) Using the 'loss/gain of oxygen' definition of oxidation and reduction: magnesium starts as an element (Mg) and ends as magnesium oxide (MgO), meaning it has gained oxygen — this is oxidation. Copper(II) oxide (CuO) starts containing oxygen and ends as copper metal (Cu), meaning it has lost oxygen — this is reduction. (d)(i) To show both air and water are needed for rusting, three test tubes are set up as a controlled comparison: Tube 1 contains an iron nail with a drying agent such as anhydrous calcium chloride and is sealed, so air is present but water is absent; Tube 2 contains an iron nail fully submerged in water that has been boiled (to remove dissolved oxygen/air) and is then covered with a layer of oil to prevent air from re-dissolving, so water is present but air is (largely) absent; Tube 3 contains an iron nail in ordinary water, open to the air, so both air and water are present, acting as the control/comparison. After leaving the tubes for several days, only the nail in Tube 3 (both air and water present) shows visible rusting, showing that both are needed for rusting to occur. (ii) The rusting of iron is summarised by the word equation: iron + oxygen + water → hydrated iron(III) oxide. (e) Zinc lies above iron in the reactivity series, meaning it is more reactive and loses electrons more readily than iron. When galvanised steel is scratched, exposing the iron underneath, the zinc coating still protects the iron because, being more reactive, the zinc corrodes/reacts (loses electrons) in preference to the iron, even though the iron itself is directly exposed to air and water. Because the zinc is corroded instead of the iron, this method of protection is called sacrificial protection — the zinc acts as a 'sacrificial' metal that is used up so that the more important iron structure is preserved.
Final answer: (a) Mg reacts fastest/vigorously, Zn more slowly, Cu not at all; more reactive metals lose electrons more readily; (b) grow plants to absorb Cu compounds → burn to ash → add acid to form leachate → displace Cu using scrap iron; (c) Mg oxidised (gains oxygen), CuO reduced (loses oxygen); (d) three-tube comparison (air only/water only/air+water) shows rusting needs both; iron + oxygen + water → hydrated iron(III) oxide; (e) zinc is more reactive, corrodes preferentially, sacrificially protecting the iron even where scratched.

(a)(i) 1 mark for Mg reacting vigorously/fastest with bubbling; 1 mark for Zn reacting more slowly/steadily; 1 mark for Cu showing no reaction; 1 mark for correctly linking the differences to relative reactivity — max 4 (accept marks for correct observations even if the explicit reactivity link is given as part of (ii)). (ii) 1 mark for 'more reactive metals lose electrons more easily'; 1 mark for correctly linking this to forming a positive ion — max 2. (b) 1 mark each for: plants absorb copper compounds from soil; plants harvested and burned to ash; acid added to ash to form a leachate; copper obtained from leachate by displacement using scrap iron; correct explanation that this avoids traditional mining of rock — max 5 (award any 5 valid points). (c) 1 mark for 'magnesium is oxidised'; 1 mark for correct reason (gains oxygen); 1 mark for 'copper(II) oxide is reduced'; 1 mark for correct reason (loses oxygen) — max 4. (d)(i) 1 mark for a tube with air only (drying agent, no water); 1 mark for a tube with water only (boiled/deoxygenated water, oil layer excluding air); 1 mark for a tube with both air and water present (open to air, in water) as the comparison; 1 mark for correctly stating rusting only occurs in the tube with both present; 1 mark for a valid additional experimental detail (e.g. leaving for several days, comparing nails) — max 5. (ii) 1 mark for 'iron + oxygen + water → hydrated iron(III) oxide' (or equivalent naming 'rust'). (e) 1 mark for 'zinc is more reactive than iron'; 1 mark for zinc corroding/reacting in preference to iron; 1 mark for this occurring even where the coating is scratched/iron exposed; 1 mark for naming this 'sacrificial protection' — max 4.

(a)(i) In a closed system at dynamic equilibrium, the forward and reverse reactions are both still occurring, but they happen at exactly the same rate, so there is no overall change in the concentrations/amounts of reactants and products over time. (ii) Le Châtelier's Principle states that if a system at equilibrium is subjected to a change in conditions (such as temperature, pressure or concentration), the position of equilibrium will shift in the direction that acts to oppose/minimise the effect of that change. (b) In the Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), there are 4 moles of gas on the reactant side but only 2 moles of gas on the product side. According to Le Châtelier's Principle, increasing the pressure causes the equilibrium to shift in the direction that reduces the number of gas molecules present, since this partially opposes the increase in pressure — this is the product (ammonia) side, so a high pressure increases the equilibrium yield of ammonia. Because the forward reaction is exothermic, Le Châtelier's Principle also predicts that a lower temperature would shift equilibrium further towards the exothermic (product) side, giving a higher yield of ammonia. However, at a low temperature, the rate of reaction would be very slow, meaning it would take an impractically long time to reach equilibrium, even though the eventual yield would be higher. A temperature of around 450 °C is therefore chosen as a compromise (trade-off): it is high enough to give a reasonably fast rate of reaction (so ammonia is produced quickly enough to be economical), while still giving an acceptable, though not maximum, equilibrium yield, since a much higher temperature would reduce yield further even though it would increase rate even more. (c) Collision theory states that for a reaction to occur, particles must collide with each other with sufficient energy (at least the activation energy). Increasing the concentration of hydrochloric acid means there are more acid particles (H⁺ ions) present in the same volume of solution; this increases the frequency of collisions between acid particles and the surface of the magnesium per second. Since the proportion of collisions with enough energy to react stays the same, but the total number of collisions per second increases, the number of successful collisions per second also increases, so the rate of reaction increases. (d)(i) A catalyst is defined as a substance that increases the rate of a chemical reaction without itself being permanently chemically changed or used up in the reaction. (ii) A catalyst works by providing an alternative reaction pathway that has a lower activation energy than the uncatalysed reaction; since a greater proportion of particle collisions now have enough energy to exceed this lower activation energy, a greater proportion of collisions are successful, increasing the rate of reaction.
Final answer: (a) rate forward = rate reverse, amounts constant; equilibrium shifts to oppose an imposed change; (b) high pressure shifts equilibrium towards fewer gas molecules (ammonia side), increasing yield; 450°C is a trade-off between a faster rate (favoured by higher T) and a higher yield (favoured by lower T, since forward reaction is exothermic); (c) higher concentration increases collision frequency, increasing successful collisions per second and rate; (d) catalyst increases rate without being used up, by providing a lower-activation-energy pathway.

(a)(i) 1 mark for 'rate of forward reaction = rate of reverse reaction'; 1 mark for 'amounts of reactants/products remain constant' — max 2. (ii) 1 mark for 'equilibrium shifts to oppose a change in conditions'; 1 mark for correctly referencing conditions (temperature/pressure/concentration) — max 2. (b) 1 mark for identifying fewer gas moles on the product side; 1 mark for correctly applying Le Châtelier to explain high pressure favours the product side/higher yield; 1 mark for identifying the forward reaction as exothermic; 1 mark for explaining a lower temperature would favour a higher yield; 1 mark for explaining a lower temperature would give too slow a rate; 1 mark for correctly concluding 450 °C is a compromise/trade-off between rate and yield — max 6. (c) 1 mark for 'more particles/higher concentration'; 1 mark for 'more frequent collisions'; 1 mark for 'more successful collisions per second' (or equivalent); 1 mark for correctly concluding rate increases — max 4. (d)(i) 1 mark for correct definition. (ii) 1 mark for 'lower activation energy pathway' correctly linked to increased rate.

(a)(i) The alkanes have the general formula CₙH₂ₙ₊₂. (ii) A homologous series is a group of organic compounds that share the same general formula, have similar chemical properties (since they share the same functional group, or lack one, as with alkanes), show a gradual change (gradation) in their physical properties (such as boiling point) as chain length increases, and differ from one member to the next by a CH₂ unit. (iii) With 3 carbon atoms, propane's formula follows CₙH₂ₙ₊₂: \( C_3H_{2(3)+2} = C_3H_8 \). (b)(i) Cracking breaks down large, saturated hydroccarbon molecules (alkanes), which are less useful/in lower demand, into smaller, more useful molecules; because the products contain fewer hydrogen atoms relative to carbon than the original alkane, some of the products are unsaturated alkenes. It requires a high temperature (with a catalyst commonly used to allow a lower temperature to be used than would otherwise be needed). (ii) Balancing carbon and hydrogen atoms: C₁₀H₂₂ has 10 C and 22 H; C₈H₁₈ has 8 C and 18 H; the remaining product must have \( 10-8=2 \) carbon atoms and \( 22-18=4 \) hydrogen atoms, giving C₂H₄ (ethene). (c)(i) The alkenes' functional group is the C=C double covalent bond. (ii) Bromine water (an orange/brown solution of bromine) is added to a sample of the hydrocarbon and shaken; if the hydrocarbon is an alkene, the C=C double bond reacts with the bromine in an addition reaction, and the bromine water is decolourised (turns from orange/brown to colourless); an alkane, having no C=C double bond, does not react, so the bromine water stays orange/brown. (iii) Many ethene molecules join together, with each C=C double bond opening to form single bonds linking adjacent monomers, in a reaction called addition polymerisation; the resulting repeat unit, poly(ethene)'s basic repeating structure, is –CH₂–CH₂– repeated many times to form a very long chain molecule. (d)(i) To produce ethanol by fermentation, a sugar solution is mixed with yeast, which contains enzymes that catalyse the conversion of sugar into ethanol and carbon dioxide; the mixture is kept at a warm temperature (roughly 25–35 °C, warm enough for the enzymes to work efficiently but not so hot that they are denatured) and, crucially, in the absence of oxygen (anaerobic conditions), since oxygen would allow the ethanol to be oxidised further/allow unwanted microbial growth. (ii) Ethanol, when oxidised (by exposure to air over time, or more rapidly by acidified potassium dichromate solution), is converted into a carboxylic acid — specifically ethanoic acid. (e)(i) With sufficient oxygen, complete combustion of an alkane such as methane produces carbon dioxide and water: methane + oxygen → carbon dioxide + water. With insufficient oxygen, incomplete combustion instead produces carbon monoxide and water (and/or carbon/soot and water): methane + oxygen → carbon monoxide + water. (ii) Incomplete combustion is dangerous because it produces carbon monoxide, a toxic gas that binds to haemoglobin in red blood cells more strongly than oxygen does, reducing the blood's capacity to carry oxygen around the body, which can lead to illness or death, particularly since carbon monoxide is colourless and odourless and so hard to detect without a alarm. Fuels containing sulfur impurities release sulfur dioxide gas when burned; this sulfur dioxide dissolves in water vapour/rain in the atmosphere to form dilute sulfurous/sulfuric acid, which falls as acid rain, damaging stone buildings and statues, harming or killing vegetation, and lowering the pH of lakes and rivers, harming aquatic life. (f)(i) In a reaction profile diagram, the y-axis represents energy and the x-axis represents the progress of the reaction; the curve starts at the energy level of the reactants, rises to a peak, then falls again. The height from the reactants' energy level up to the peak represents the activation energy — the minimum energy that colliding particles need for a reaction to occur. Because the reaction is exothermic, the curve falls to a products energy level that is lower than the reactants' starting energy level, with the vertical difference between the reactants' and products' energy levels representing the overall energy released. (ii) Using \( \Delta H = \text{(energy to break bonds in reactants)} - \text{(energy released forming bonds in products)} \): bonds broken (reactants) = 1 mol H–H + 1 mol Cl–Cl \( = 436 + 242 = 678 \text{ kJ/mol} \); bonds formed (products) = 2 mol H–Cl \( = 2 \times 431 = 862 \text{ kJ/mol} \); overall energy change \( = 678 - 862 = -184 \text{ kJ/mol} \). Since the value is negative (more energy is released forming the new bonds than is needed to break the old ones), the reaction is exothermic.
Final answer: (a) CₙH₂ₙ₊₂; homologous series definition as above; C₃H₈; (b) cracking breaks large alkanes into smaller/more useful molecules incl. alkenes, needs high temperature; C₂H₄; (c) C=C double bond; bromine water decolourised by an alkene; addition polymerisation, repeat unit –CH₂–CH₂–; (d) sugar + yeast, warm, anaerobic → ethanol + CO₂; oxidised to a carboxylic acid; (e) complete combustion → CO₂+H₂O, incomplete → CO(+ soot)+H₂O; CO reduces blood's oxygen-carrying capacity; sulfur impurities → SO₂ → acid rain; (f) exothermic profile rises to activation energy peak then falls below the reactants' level; ΔH = −184 kJ/mol, exothermic.

(a)(i) 1 mark for CₙH₂ₙ₊₂. (ii) 1 mark for 'same general formula/similar chemical properties'; 1 mark for 'gradation in physical properties' and/or 'differ by CH₂' — max 2. (iii) 1 mark for C₃H₈. (b)(i) 1 mark for 'breakdown of larger/saturated alkanes into smaller, more useful molecules, some unsaturated/alkenes'; 1 mark for a valid condition (high temperature, and/or catalyst) — max 2 (question allows up to 3 — award a 3rd mark for full, precise explanation covering both the products' size and unsaturation). (ii) 1 mark for C₂H₄. (c)(i) 1 mark for 'C=C double (covalent) bond'. (ii) 1 mark for 'add bromine water'; 1 mark for 'decolourised/turns colourless' as the positive (alkene) result — max 2. (iii) 1 mark for 'addition polymerisation'; 1 mark for correct repeat unit –CH₂–CH₂–; 1 mark for a valid description of how the double bond opens to link monomers — max 3. (d)(i) 1 mark for 'sugar + yeast'; 1 mark for 'warm temperature' (approx. 25–35 °C or equivalent); 1 mark for 'anaerobic/absence of oxygen' — max 3. (ii) 1 mark for 'carboxylic acid' (accept 'ethanoic acid'). (e)(i) 1 mark for correct complete combustion word equation; 1 mark for correct incomplete combustion word equation — max 2. (ii) 1 mark for CO binding to haemoglobin, reducing oxygen-carrying capacity; 1 mark for sulfur impurities forming sulfur dioxide; 1 mark for SO₂ dissolving in atmospheric water to form acid rain (with a named effect, e.g. damaging buildings/vegetation/aquatic life) — max 3. (f)(i) 1 mark for correctly describing the shape (rises to a peak then falls below the start for an exothermic reaction); 1 mark for correctly identifying activation energy as the rise from reactants to the peak — max 2. (ii) 1 mark for correct bonds-broken total (678); 1 mark for correct bonds-formed total (862); 1 mark for correct method (broken − formed); 1 mark for correct answer −184 kJ/mol; 1 mark for correctly stating 'exothermic' — max 5 (ECF applied).

(a)(i) Electrolysis is defined as the process by which an ionic compound, when molten or in aqueous solution (so its ions are free to move), is broken down (decomposed) into its constituent elements by passing an electric current through it. (ii) In electrolysis, the anode is the positive electrode, and the cathode is the negative electrode; positively charged ions (cations) move towards, and are discharged at, the cathode, while negatively charged ions (anions) move towards, and are discharged at, the anode. (b)(i) At the cathode (negative electrode), positively charged lead ions, Pb²⁺, gain electrons (are reduced) to form lead metal; a bead/pool of grey molten lead is observed forming at the cathode. At the anode (positive electrode), negatively charged bromide ions, Br⁻, lose electrons (are oxidised) to form bromine; an orange/brown gas (bromine vapour) is observed forming/bubbling at the anode. (ii) At the cathode: \( \text{Pb}^{2+}(l) + 2e^- \rightarrow \text{Pb}(l) \); at the anode: \( 2\text{Br}^-(l) \rightarrow \text{Br}_2(g) + 2e^- \). (c)(i) For an ionic compound to conduct electricity and undergo electrolysis, its ions must be free to move so they can carry charge to the electrodes. In solid aluminium oxide, the Al³⁺ and O²⁻ ions are held rigidly in a fixed giant ionic lattice and cannot move, so the solid cannot conduct electricity; melting (or dissolving) the aluminium oxide frees the ions to move throughout the liquid, allowing it to conduct electricity and be electrolysed. (ii) During the process, oxygen ions are discharged at the carbon anodes; at the high operating temperature used, this oxygen reacts with the carbon of the anodes themselves, gradually burning them away as carbon dioxide gas, so the anodes must be periodically replaced as they are used up.
Final answer: (a) electrolysis breaks down an ionic compound (molten/dissolved) into elements using electricity; anode = positive electrode, cathode = negative electrode; (b) cathode produces lead metal, anode produces bromine gas; Pb²⁺(l)+2e⁻→Pb(l), 2Br⁻(l)→Br₂(g)+2e⁻; (c) molten/dissolved state frees the ions to move and carry charge; anodes are replaced because they react with the oxygen produced, burning away as CO₂.

(a)(i) 1 mark for 'breaking down/decomposing an ionic compound'; 1 mark for 'molten or dissolved, using an electric current' — max 2. (ii) 1 mark for 'anode = positive electrode'; 1 mark for 'cathode = negative electrode' — max 2. (b)(i) 1 mark for 'lead (metal) at the cathode'; 1 mark for 'bromine (gas) at the anode'; 1 mark for a valid observation at either electrode (grey molten lead bead, or orange/brown gas) — max 3. (ii) 1 mark for correct cathode half-equation; 1 mark for correct anode half-equation (both must balance charge and atoms) — max 2. (c)(i) 1 mark for 'ions must be free to move to conduct/carry charge'; 1 mark for 'solid ions are fixed in a lattice and cannot move' — max 2. (ii) 1 mark for 'anodes react with/burn in the oxygen produced, forming carbon dioxide'.

(a) To obtain an accurate and reliable titre, the student should first rinse a pipette with a small amount of the sodium hydroxide solution (to avoid diluting it with any water left in the pipette), then use the pipette to measure exactly 25.0 cm³ of sodium hydroxide solution into a clean conical flask, adding a few drops of a suitable indicator such as phenolphthalein. A burette is similarly rinsed with the hydrochloric acid before being filled with it, and the initial burette reading is recorded. The hydrochloric acid is then added from the burette into the flask, swirling the flask constantly to mix the solutions, adding the acid more slowly (dropwise) as the colour change of the indicator is approached, until the indicator just permanently changes colour, marking the end point of the reaction; the final burette reading is then recorded, and the volume of acid added (the titre) is calculated by subtracting the initial reading from the final reading. This process is repeated, using the volume from the first (rough) titration as a guide to add acid more quickly at first, then more slowly near the expected end point, until at least two titres agree closely (are 'concordant', usually within 0.1 cm³ of each other); an average titre is then calculated using only the concordant results, to give a reliable value for use in further calculations. (b) The moles of hydrochloric acid used are found first: \( \text{moles HCl} = \text{concentration} \times \text{volume (in dm}^3) = 0.100 \times \dfrac{22.5}{1000} = 0.00225 \text{ mol} \). Since the equation shows a 1:1 mole ratio between HCl and NaOH, moles of NaOH = moles of HCl = 0.00225 mol. The concentration of the sodium hydroxide solution is then found using \( \text{concentration} = \dfrac{\text{moles}}{\text{volume (in dm}^3)} = \dfrac{0.00225}{25.0/1000} = \dfrac{0.00225}{0.0250} = 0.0900 \text{ mol/dm}^3 \). (c) First, the moles of calcium carbonate are calculated: \( \text{moles CaCO}_3 = \dfrac{\text{mass}}{M_r} = \dfrac{5.0}{100} = 0.05 \text{ mol} \). From the balanced equation, the mole ratio of CaCO₃ to CO₂ is 1:1, so moles of CO₂ produced = 0.05 mol. Using the fact that 1 mole of any gas occupies 24 dm³ at room temperature and pressure: \( \text{volume of CO}_2 = 0.05 \times 24 = 1.2 \text{ dm}^3 \). (d) Carbon dioxide gas is confirmed by bubbling it through limewater (a solution of calcium hydroxide); if carbon dioxide is present, the limewater turns from colourless to milky/cloudy, due to the formation of insoluble calcium carbonate. (e) The Earth's atmosphere is made up of approximately 21% oxygen by volume (alongside approximately 78% nitrogen and small amounts of other gases).
Final answer: (a) rinse pipette/burette, measure 25.0 cm³ NaOH with indicator, titrate to the end-point colour change, repeat to obtain concordant titres, average them; (b) 0.0900 mol/dm³; (c) 1.2 dm³; (d) bubble through limewater — turns milky/cloudy; (e) about 21%.

(a) Banded mark scheme (6 marks, QWC assessed). Indicative content: rinse pipette with NaOH solution; measure 25.0 cm³ NaOH into a conical flask using the pipette; add indicator (e.g. phenolphthalein); rinse and fill burette with HCl, record initial reading; add acid to flask, swirling, until indicator changes colour (end point); record final reading/calculate titre; repeat and obtain concordant results (within 0.1 cm³); calculate average titre from concordant results only. Band A (5–6 marks): at least 5 of the indicative points, in a clear, logical sequence, accurate specialist vocabulary, few SPG errors. Band B (3–4 marks): 3–4 indicative points, reasonably clear, some SPG errors. Band C (1–2 marks): 1–2 indicative points, poorly organised, weak SPG. (b) 1 mark for correct moles of HCl (0.00225); 1 mark for correctly applying the 1:1 mole ratio; 1 mark for correct method for concentration (moles ÷ volume in dm³); 1 mark for correct final answer 0.0900 mol/dm³ — max 4 (ECF applied). (c) 1 mark for correct moles of CaCO₃ (0.05); 1 mark for correctly applying the 1:1 mole ratio to CO₂; 1 mark for correct method (moles × 24); 1 mark for correct final answer 1.2 dm³ — max 4 (ECF applied). (d) 1 mark for 'limewater'; 1 mark for 'turns milky/cloudy' — max 2. (e) 1 mark for a value in the range 20–21%; 1 mark for 'oxygen' correctly identified/percentage correctly attributed — max 2 (accept '21%' alone for full marks).

(a) To compare the reactivity of the three metals fairly, the student should use the same form of each metal (e.g. a strip of ribbon of the same length, or the same mass of powder, giving the same surface area for each) and add each to a separate test tube containing the same volume and the same concentration of dilute hydrochloric acid, at the same starting temperature. A suitable measurement is then taken for each metal — for example, timing (using a stopwatch) how long it takes for the metal to completely disappear/react, or measuring the volume of hydrogen gas collected in a fixed time using a gas syringe connected to the test tube by a delivery tube. Keeping the acid's volume and concentration, the temperature, and the metal's form/surface area the same for each test ensures reactivity is the only variable affecting the result (a fair test). As a safety precaution, the student should wear eye protection throughout, since dilute hydrochloric acid can cause irritation if splashed in the eyes; working in a well-ventilated area is also sensible, as hydrogen gas is flammable. (b) The student can measure pH accurately using a pH meter (or a pH probe/sensor connected to a data logger); the probe is dipped into the first solution and the reading is recorded to at least one decimal place, then the probe is rinsed thoroughly with distilled water before being dipped into the next solution, to avoid any of the previous solution contaminating the next reading; this is repeated for all four solutions. (c) A suitable results table would have a first column headed 'Metal' listing magnesium, zinc and iron in separate rows, and further columns headed with the quantity measured and its unit — for example 'Time taken for metal to react / s' (or 'Volume of gas collected in 60 s / cm³') — ideally with additional columns for a repeat reading and the calculated average, so that reliability can be assessed and an anomalous reading identified if necessary.
Final answer: (a) same form/mass of each metal in equal volume/concentration of acid, time reaction or measure gas volume, wear eye protection; (b) use a pH meter/probe, rinsed between solutions, recording to at least 1 dp; (c) table with metal names in one column and the measured quantity (with unit) in further columns, including repeats/average.

(a) 1 mark for using the same volume of acid for each metal; 1 mark for using the same concentration of acid; 1 mark for using the same form/mass/surface area of metal for each; 1 mark for a valid measurement method (timing to react, or gas volume in fixed time); 1 mark for correct apparatus named (e.g. stopwatch, gas syringe); 1 mark for identifying reactivity/rate as the variable being compared, with other variables controlled; 1 mark for a valid safety precaution (eye protection); 1 mark for a further valid detail (e.g. same starting temperature, or working in a well-ventilated area) — max 8. (b) 1 mark for 'pH meter' or 'pH probe/sensor'; 1 mark for dipping the probe into each solution; 1 mark for rinsing with distilled water between solutions; 1 mark for recording the reading; 1 mark for 'to at least one decimal place' — max 5. (c) 1 mark for a column identifying the metal tested; 1 mark for a column for the measured quantity with a correct unit; 1 mark for a sensible measured quantity matching (a) (e.g. time in seconds, or volume in cm³); 1 mark for including a repeat column; 1 mark for including an average column — max 5.

(a) The procedure begins by using a pipette, together with a pipette filler (rather than mouth suction), to accurately measure exactly 25.0 cm³ of the sodium hydroxide solution into a clean conical flask. A few drops of phenolphthalein indicator are then added to the flask, which will appear pink in the alkaline solution. A burette, held securely in a burette holder/clamp on a stand, is filled with the dilute hydrochloric acid (having first been rinsed with the acid), and the initial volume reading on the burette is recorded, looking at the meniscus at eye level. The conical flask is placed on a white tile beneath the burette, which makes it easier to see the colour change clearly against a plain background. The acid is then run from the burette into the flask a little at a time, with the flask swirled continuously to mix the solutions thoroughly; as the pink colour begins to fade more slowly (indicating the end point is near), the acid is added drop by drop until the solution just turns from pink to colourless, marking the end point of the neutralisation. The final burette reading is recorded, and the titre (volume of acid used) is calculated by subtracting the initial reading from the final reading. (b) A single titration result could be affected by a small error (for example, in judging exactly when the colour changes, or in reading the burette), so repeating the titration allows the student to check whether their results are consistent/reproducible, and to spot and disregard any anomalous (clearly inconsistent) result. By calculating the average of titres that are concordant (in close agreement, usually within 0.1 cm³ of each other), the effect of small random errors in any one titration is reduced, giving a more accurate and reliable final titre to use in further calculations. (c) Since both the sodium hydroxide solution (an alkali) and the hydrochloric acid can be corrosive/irritant to skin and especially eyes, the student should wear eye protection (safety goggles) throughout the experiment.
Final answer: (a) pipette 25.0 cm³ NaOH into a flask with indicator, titrate acid from a burette (on a stand, over a white tile) until the colour just changes at the end point, record the titre; (b) repeating and averaging concordant results reduces random error and checks reliability; (c) wear eye protection.

(a) 1 mark for 'pipette' (with pipette filler) used to measure 25.0 cm³ NaOH; 1 mark for adding indicator (phenolphthalein) to the flask; 1 mark for 'burette' (in a stand/clamp) filled with the acid, with initial reading recorded; 1 mark for use of a white tile under the flask; 1 mark for adding acid while swirling the flask; 1 mark for adding acid dropwise near the end point; 1 mark for correctly identifying the end point (colour just changes, pink to colourless) and recording the final reading/titre — max 7. (b) 1 mark for checking reliability/reproducibility, or identifying an anomalous result; 1 mark for correctly defining/identifying 'concordant' results (in close agreement); 1 mark for explaining averaging concordant results reduces the effect of random error, giving a more accurate result — max 3. (c) 1 mark for 'eye protection/safety goggles'; 1 mark for a valid reason (acid/alkali can be corrosive/irritant) — max 2.

(a) Sodium hydroxide solution is used to identify certain metal ions by the colour and behaviour of the precipitate formed. A green precipitate that does not redissolve in excess sodium hydroxide solution is characteristic of iron(II) ions, Fe²⁺, which form insoluble iron(II) hydroxide: \( \text{Fe}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Fe(OH)}_2(s) \). (b) To identify the halide ion present, the student should take a fresh sample of the solution and first add a few drops of dilute nitric acid; this acidifies the solution and removes/prevents interference from any carbonate ions that might otherwise also form a precipitate with the next reagent, which could be mistaken for a halide test result. Silver nitrate solution is then added dropwise to the acidified solution; a precipitate forms as the silver ions react with whichever halide ion is present, and the colour of this precipitate identifies which halide is present: a white precipitate indicates chloride ions are present (forming silver chloride), a cream/pale yellow precipitate indicates bromide ions are present (forming silver bromide), and a (darker) yellow precipitate indicates iodide ions are present (forming silver iodide). Observing and correctly distinguishing between these similar-looking precipitate colours (ideally in good light, against a dark background) allows the student to confidently identify which halide ion the original compound contains. (c) A cream precipitate identifies bromide ions, Br⁻, as being present (forming silver bromide, AgBr). If iodide ions, I⁻, had instead been present, the precipitate formed with silver nitrate solution would have been yellow (silver iodide, AgI), a distinctly different, more strongly yellow colour than the cream precipitate given by bromide.
Final answer: (a) iron(II), Fe²⁺; Fe²⁺(aq) + 2OH⁻(aq) → Fe(OH)₂(s); (b) acidify with dilute nitric acid, then add silver nitrate solution — white=chloride, cream=bromide, yellow=iodide; (c) bromide; iodide would give yellow.

(a) 1 mark for 'iron(II)'/'Fe²⁺'; 1 mark for correct ionic equation species; 1 mark for correct balancing/formula (Fe(OH)₂); 1 mark for correct state symbols — max 4. (b) Banded mark scheme (6 marks, QWC assessed). Indicative content: take a fresh sample of the solution; add dilute nitric acid (to remove carbonate interference); add silver nitrate solution; observe the colour of the precipitate formed; white precipitate = chloride; cream precipitate = bromide; yellow precipitate = iodide. Band A (5–6 marks): at least 5 of the indicative points, in a clear, logical method, accurate specialist vocabulary, few SPG errors. Band B (3–4 marks): 3–4 indicative points, reasonably clear, some SPG errors. Band C (1–2 marks): 1–2 indicative points, poorly organised, weak SPG. (c) 1 mark for 'bromide'/'Br⁻'; 1 mark for 'yellow' for iodide; 1 mark for correctly distinguishing this from the cream colour given by bromide — max 3.

(a) A suitable set-up connects a conical flask (containing the hydrogen peroxide solution, into which the manganese(IV) oxide catalyst is added) to a gas syringe via a bung fitted with a delivery tube; as oxygen gas is produced, it collects in the gas syringe, pushing the plunger along a graduated scale, allowing the volume of gas produced to be read off directly at regular time intervals (measured using a stopwatch), without gas escaping. (b) Using \( \text{rate} = \dfrac{\text{change in volume}}{\text{change in time}} \): over the first 20 seconds, the volume changes from 0 to 24 cm³, so \( \text{rate} = \dfrac{24 - 0}{20 - 0} = 1.2 \text{ cm}^3\text{/s} \). (c) The rate of this reaction depends on the frequency of successful collisions between hydrogen peroxide molecules and the surface of the (solid) manganese(IV) oxide catalyst. As the reaction proceeds, hydrogen peroxide is continuously being used up (converted into water and oxygen), so its concentration in the remaining solution steadily decreases; a lower concentration means there are fewer hydrogen peroxide particles present in a given volume of solution, so they collide with the catalyst surface less frequently, and the rate of reaction (shown by the gradient/steepness of the volume–time graph) decreases as a result. Eventually, once all of the hydrogen peroxide originally present has fully decomposed, no further oxygen gas can be produced, so the volume of gas collected stops increasing and the graph becomes a horizontal (flat) line. (d) To investigate the effect of concentration specifically, the student should repeat the same experiment (same volume of hydrogen peroxide solution, same mass/form of catalyst, same temperature, and the same method of collecting gas) several times, changing only the concentration of the hydrogen peroxide solution used between each repeat (for example, by diluting the hydrogen peroxide with different volumes of water), and then compare either the resulting volume–time graphs directly, or the initial rate of reaction calculated for each concentration (for example, from the gradient over the first 20 seconds, as in part (b)). (e) A result that does not fit the otherwise consistent pattern shown by the rest of the data is called an anomalous result; when analysing the data (for example, when drawing a line or curve of best fit through the plotted points), this anomalous reading should be identified and disregarded/ignored (it should not be used to influence the position of the best-fit line), and, where practical, the measurement at that time should be repeated to obtain a more reliable value.
Final answer: (a) conical flask with catalyst connected via delivery tube to a gas syringe, reading volume at set time intervals; (b) 1.2 cm³/s; (c) concentration of H₂O₂ falls as it is used up, so collision frequency with the catalyst falls, decreasing rate; graph flattens once all H₂O₂ has reacted; (d) repeat with only concentration changed, comparing graphs/initial rates; (e) an anomalous result — ignore it when drawing the best-fit line, and repeat if possible.

(a) 1 mark for 'gas syringe' correctly used to collect/measure the gas; 1 mark for correct connection (conical flask + delivery tube/bung to the syringe); 1 mark for reading the volume at set time intervals (with a stopwatch) — max 3. (b) 1 mark for correct method (change in volume ÷ change in time); 1 mark for correct substitution (24/20); 1 mark for correct answer 1.2 cm³/s — max 3. (c) 1 mark for 'hydrogen peroxide is used up/its concentration decreases'; 1 mark for 'fewer particles/lower concentration means less frequent collisions' with the catalyst; 1 mark for correctly linking this to a decreasing rate/gradient; 1 mark for correctly explaining the graph becomes horizontal once all hydrogen peroxide has reacted — max 4. (d) 1 mark for 'repeat with only concentration changed, all other variables controlled'; 1 mark for comparing resulting graphs/initial rates — max 2. (e) 1 mark for 'anomalous result'; 1 mark for 'ignore/exclude it when drawing the best-fit line' (accept also 'repeat the reading') — max 2.

(a) The student sets up the electrolysis cell by placing two inert graphite electrodes into a container of dilute sulfuric acid, connecting each electrode by a wire (via a crocodile clip) to opposite terminals of a d.c. power supply. Once the power supply is switched on, an electric current passes through the acid, and gas bubbles begin to form at the surface of each electrode as electrolysis takes place. To collect each gas separately, a test tube completely filled with the dilute sulfuric acid is inverted over each electrode (for example, using a small stand or clamp to hold it in place); as gas is produced at that electrode, it rises and collects in the upper part of the inverted test tube, displacing the acid downward and out, allowing the volume of gas collected to build up for testing. (b) To test the gas collected at the cathode, the student removes the test tube from over the electrode (keeping it inverted to avoid losing the gas) and, working quickly, brings a lighted splint to the mouth of the tube; if the gas is hydrogen, a squeaky 'pop' sound is heard as the hydrogen burns rapidly. To test the gas collected at the anode, the student instead inserts a glowing (not flaming) wooden splint into the mouth of that test tube; if the gas is oxygen, the splint relights, bursting back into flame (or glowing noticeably more brightly), because oxygen supports combustion. (c) At the cathode, hydrogen ions from the water/acid gain electrons (are reduced) to form hydrogen gas: \( 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) \). At the anode, hydroxide ions lose electrons (are oxidised) to form oxygen gas and water: \( 4\text{OH}^-(aq) \rightarrow \text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^- \).
Final answer: (a) inert electrodes in dilute H₂SO₄ connected to a d.c. supply, each gas collected in an inverted, acid-filled test tube over its electrode; (b) cathode gas (H₂): lit splint gives a pop; anode gas (O₂): glowing splint relights; (c) cathode: 2H⁺(aq)+2e⁻→H₂(g); anode: 4OH⁻(aq)→O₂(g)+2H₂O(l)+4e⁻.

(a) 1 mark for 'inert/graphite electrodes' in dilute sulfuric acid; 1 mark for correct connection to a d.c. power supply (via wires/crocodile clips); 1 mark for gas bubbles forming at each electrode on switching on; 1 mark for using an inverted test tube (initially filled with acid) over each electrode to collect gas; 1 mark for correctly explaining gas displaces the liquid downward as it collects; 1 mark for a further valid procedural detail — max 6. (b) 1 mark for 'lighted splint' at the cathode gas; 1 mark for 'pop' as the positive result; 1 mark for 'glowing splint' at the anode gas; 1 mark for 'relights/bursts into flame' as the positive result — max 4. (c) 1 mark for correct cathode half-equation (species, balancing, charge); 1 mark for correct anode half-equation (species, balancing, charge); 1 mark for both half-equations correctly balanced for charge as well as atoms — max 3.

(a) The student places the crucible, containing the weighed 2.50 g sample of hydrated copper(II) sulfate, on a pipeclay triangle resting on a tripod, and heats it using a Bunsen burner, gently at first and then more strongly, with the crucible lid tilted slightly ajar to let steam/water vapour escape while still reducing the chance of any solid spitting out. After a period of heating, the crucible is allowed to cool (for example on a heatproof mat) before being reweighed on a balance. This cycle of heating, cooling and reweighing is repeated, and the mass recorded each time; once two consecutive reweighings give the same mass, the sample has been heated 'to constant mass', meaning no further water is being lost. Heating for a fixed time instead would not reliably indicate that all of the water of crystallisation has actually been removed, since different samples or heating rates could mean the reaction finishes earlier or later than any single fixed time allowed; heating to constant mass instead directly confirms, from the evidence of the mass itself, that the process (in this case, loss of water of crystallisation) is fully complete, giving a reliable final mass for the anhydrous salt. (b) The mass of water lost on heating is \( 2.50 - 1.60 = 0.90 \text{ g} \); the number of moles of water lost is \( \dfrac{0.90}{18} = 0.05 \text{ mol} \). The number of moles of anhydrous copper(II) sulfate remaining is \( \dfrac{1.60}{160} = 0.01 \text{ mol} \). The value of x is the ratio of moles of water to moles of CuSO₄: \( x = \dfrac{0.05}{0.01} = 5 \). (This confirms the well-known formula of hydrated copper(II) sulfate, CuSO₄·5H₂O.) (c) As it is heated, the blue crystals of hydrated copper(II) sulfate gradually turn white (or pale grey/off-white) as they lose their water of crystallisation and become anhydrous copper(II) sulfate; water vapour (steam) can also be seen escaping from the crucible during heating.
Final answer: (a) heat in a crucible on a pipeclay triangle/tripod, cool and reweigh repeatedly until mass is unchanged, because this confirms all water has been lost; (b) x = 5; (c) blue crystals turn white, with steam/water vapour escaping.

(a) 1 mark for 'crucible on a pipeclay triangle/tripod'; 1 mark for heating with a Bunsen burner (gently, then more strongly); 1 mark for cooling then reweighing; 1 mark for repeating the heat–cool–reweigh cycle; 1 mark for 'until mass is unchanged/constant' as the end point; 1 mark for a correct explanation of why constant mass (rather than fixed time) is used, i.e. it directly confirms all the water has been removed — max 6. (b) 1 mark for correct mass of water lost (0.90 g); 1 mark for correct moles of water (0.05); 1 mark for correct moles of CuSO₄ (0.01); 1 mark for correct method (dividing moles of water by moles of CuSO₄); 1 mark for correct final answer x = 5 — max 5 (ECF applied). (c) 1 mark for 'blue crystals turn white/pale' (colour change); 1 mark for 'steam/water vapour given off' — max 2.

(a) The experiment uses three test tubes as a controlled comparison, each designed to isolate whether air, water, or both, are needed for rusting. Tube 1 contains an iron nail together with a drying agent such as anhydrous calcium chloride, and is sealed, so that air is present in the tube but water/moisture is not. Tube 2 contains an iron nail fully submerged in water that has first been boiled (to drive off any dissolved oxygen/air), with a layer of oil floated on top of the water to stop air redissolving into it from the atmosphere, so that water is present but air/oxygen is largely excluded. Tube 3 contains an iron nail simply left in ordinary (non-boiled) water in a tube open to the air, so that both air and water are present together, acting as the comparison/control condition. After leaving all three tubes undisturbed for several days, the nails are removed and compared by eye for the presence of an orange-brown rust coating; only the nail from Tube 3, where both air and water were present, shows visible rusting, showing both are required together for rusting to occur. (b)(i) The rusting of iron can be summarised as: iron + oxygen + water → hydrated iron(III) oxide. (ii) Balancing the equation for iron reacting with oxygen and water to form hydrated iron(III) oxide (using 4 iron atoms to match 2 formula units of Fe₂O₃, which contains 4 Fe and 6 O in total, requiring 3 O₂ molecules, with water included on both sides to represent the 'hydrated' water of crystallisation, unbalanced with respect to x): \( 4\text{Fe(s)} + 3\text{O}_2(g) + x\text{H}_2\text{O(l)} \rightarrow 2\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O(s)} \). (c) Zinc lies above iron in the reactivity series, meaning zinc atoms lose electrons (are oxidised/corrode) more readily than iron atoms. Even where a scratch in the zinc coating exposes the underlying iron directly to air and moisture, the zinc immediately surrounding the scratch continues to corrode in preference to the iron, because it is more reactive and so more willing to lose electrons/be oxidised; the iron is therefore still protected from rusting even though it is physically exposed. Because the zinc is deliberately used up ('sacrificed') to protect the more structurally important iron, this method is known as sacrificial protection. (d) During rusting, iron atoms are oxidised, losing electrons to form iron(II) ions: \( \text{Fe(s)} \rightarrow \text{Fe}^{2+}(aq) + 2e^- \). At the same time, oxygen (dissolved in the surrounding water) is reduced, gaining these electrons, along with water, to form hydroxide ions: \( \text{O}_2(g) + 2\text{H}_2\text{O(l)} + 4e^- \rightarrow 4\text{OH}^-(aq) \); the iron(II) ions and hydroxide ions formed subsequently react further and are oxidised to form the hydrated iron(III) oxide that makes up rust.
Final answer: (a) 3-tube comparison (air only / water only / air+water) shows both are needed; only the air+water tube rusts; (b) iron + oxygen + water → hydrated iron(III) oxide; 4Fe(s) + 3O₂(g) + xH₂O(l) → 2Fe₂O₃·xH₂O(s); (c) zinc is more reactive, corrodes preferentially even where scratched — sacrificial protection; (d) Fe(s)→Fe²⁺(aq)+2e⁻ (oxidation); O₂(g)+2H₂O(l)+4e⁻→4OH⁻(aq) (reduction).

(a) 1 mark for a tube with air only (drying agent, sealed, no water); 1 mark for a tube with water only (boiled water, oil layer excluding air); 1 mark for a tube with both air and water present (open, in ordinary water); 1 mark for leaving for a suitable period (e.g. several days); 1 mark for comparing the nails for visible rust; 1 mark for correctly concluding rusting only occurs where both air and water are present — max 6. (b)(i) 1 mark for 'iron + oxygen + water'; 1 mark for '→ hydrated iron(III) oxide' — max 2. (ii) 1 mark for correct formulae (Fe, O₂, H₂O, Fe₂O₃·xH₂O); 1 mark for correct balancing of Fe and O — max 2. (c) 1 mark for 'zinc is more reactive than iron'; 1 mark for zinc losing electrons/corroding more readily; 1 mark for this occurring even where the coating is scratched/iron is exposed; 1 mark for naming 'sacrificial protection' — max 4. (d) 1 mark for correct oxidation half-equation for iron; 1 mark for correct reduction half-equation for oxygen; 1 mark for both correctly balanced for charge as well as atoms — max 3.