2025 CCEA GCSE Chemistry 1110 模擬試題連答案詳解

(a) A magnesium atom, electronic configuration (2,8,2), loses its 2 outer-shell electrons to form the Mg²⁺ ion, configuration (2,8). An oxygen atom, configuration (2,6), gains 2 electrons (the ones lost by magnesium) to form the O²⁻ ion, configuration (2,8). Both resulting ions have the stable electronic configuration of a noble gas (neon), and the oppositely charged ions are held together by strong electrostatic (ionic) attraction.
(b) Balancing: 2 Mg atoms are needed to supply the electrons for one O2 molecule (2 O atoms), giving 2Mg(s) + O2(g) → 2MgO(s). Final equation: 2Mg(s) + O2(g) → 2MgO(s).

(a) 1 mark: Mg loses 2 electrons to form Mg²⁺ (2,8); 1 mark: O gains 2 electrons to form O²⁻ (2,8). Accept 'transfers' for the pair as one combined idea only if both ions and charges correctly stated. (b) 1 mark: correct formulae of reactants (Mg, O2) and product (MgO); 1 mark: correctly balanced with state symbols 2Mg(s) + O2(g) → 2MgO(s). Reject unbalanced or missing state-symbol equations for the second mark.

(a) Each hydrogen atom has 1 outer electron and needs 1 more to fill its shell; oxygen has 6 outer electrons and needs 2 more. Oxygen shares one pair of electrons with each hydrogen atom, forming two single covalent bonds (O–H). After bonding, oxygen has a full outer shell of 8 electrons (2 from each shared pair, plus 2 non-bonding lone pairs of its own), and each hydrogen has a full outer shell of 2 electrons.
(b) Balancing hydrogen and oxygen atoms: 2 molecules of H2 react with 1 molecule of O2 to give 2 molecules of H2O, so 2H2(g) + O2(g) → 2H2O(l).

(a) 1 mark: correct description of two shared pairs of electrons between O and each H; 1 mark: oxygen has 8 electrons in outer shell after bonding. (b) 1 mark: correct formulae (H2, O2, H2O); 1 mark: correctly balanced with state symbols 2H2(g) + O2(g) → 2H2O(l). Accept H2O(g) only if steam is specified in the question; here reject as water is the liquid product expected.

(a) Group 1 metals react with water to produce a metal hydroxide and hydrogen gas. Balancing: 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g).
(b) A potassium atom loses one outer-shell electron to form a singly-charged positive ion with a stable noble-gas electronic configuration: K → K+ + e-.

(a) 1 mark: correct formulae of reactants and products; 1 mark: correctly balanced equation with state symbols. (b) 1 mark: correct species K and K+; 1 mark: correctly balanced half equation showing loss of one electron (K → K+ + e-). Reject equations showing gain of an electron.

(a) The hydrogen atom (1 outer electron) and the chlorine atom (7 outer electrons) each contribute one electron to a shared pair, forming a single covalent bond between them. Chlorine's outer shell then has 8 electrons in total: 2 in the shared (bonding) pair and 6 in three non-bonding lone pairs; hydrogen has a full outer shell of 2 electrons.
(b) Two moles of HCl are needed to supply both hydrogen ions that react with the carbonate ion, and calcium carbonate is a 1:1 reactant with CO2 and CaCl2: 2HCl(aq) + CaCO3(s) → CaCl2(aq) + H2O(l) + CO2(g).

(a) 1 mark: one shared pair of electrons between H and Cl described correctly; 1 mark: three lone pairs on chlorine identified. (b) 1 mark: correct formulae of all four species; 1 mark: correctly balanced equation with state symbols. Reject omission of state symbols for the second mark.

J. J. Thomson's Plum Pudding model described the atom as a sphere of positive charge with negatively charged electrons embedded throughout it, like plums in a pudding, with no central nucleus. Ernest Rutherford's gold-foil alpha-particle scattering experiment showed that most alpha particles passed straight through the foil, but a small number were deflected at large angles and a very few bounced almost straight back. This showed that an atom is mostly empty space, with almost all of its mass and all of its positive charge concentrated in a tiny central region – the nucleus – around which the electrons orbit. This nuclear model replaced the Plum Pudding model. James Chadwick later discovered the neutron, a neutral particle in the nucleus, which explained the mass of the atom that could not be accounted for by protons alone. This led to today's model: a central nucleus containing positively charged protons and neutral neutrons (which together account for almost all the mass of the atom), surrounded by negatively charged electrons occupying shells around the nucleus.

Marked using three levels of response. Level 3 (5–6 marks): a full, coherent account referencing the Plum Pudding model, Rutherford's scattering experiment and its conclusions (nucleus small, dense, positive; atom mostly empty space), and Chadwick's discovery of the neutron, using accurate specialist terminology (nucleus, proton, neutron, electron, electron shells) throughout with few or no errors of spelling, punctuation or grammar. Level 2 (3–4 marks): most of the above points made but with less detail or one omission (e.g. no explicit link to Chadwick or an incomplete description of the scattering experiment's conclusions); reasonably clear use of specialist terms. Level 1 (1–2 marks): basic, list-like statements, e.g. 'the model changed over time' or naming one scientist with little explanation; limited use of specialist terms. 0 marks: no relevant content.

(a) Down Group 7, the halogens change state from gas (chlorine) to liquid (bromine) to solid (iodine) at room temperature and pressure.
(b) A more reactive halogen displaces a less reactive halogen from a solution of its halide. Since reactivity decreases going down the group, astatine (below iodine) is less reactive than iodine, so astatine cannot displace iodine from potassium iodide solution – no reaction/displacement would occur.
(c) Halogen atoms react by gaining one electron into their outer shell to form a stable, singly-charged negative ion with a noble-gas electronic configuration. Going down the group, atoms have more electron shells, so the outer shell is further from the positively charged nucleus and is increasingly shielded from its attractive pull by the extra inner shells of electrons. This makes it harder for the nucleus to attract an additional electron into the outer shell, so the atoms become less reactive going down the group.

(a) 1 mark: correctly states gas → liquid → solid down the group. (b) 1 mark: correct prediction (no displacement); 1 mark: correct reason (astatine less reactive than iodine / reactivity decreases down group). (c) 1 mark: outer shell further from nucleus (more shells); 1 mark: greater shielding by inner shells; 1 mark: harder to attract/gain an electron, so reactivity decreases. All other valid answers will be credited.

(a) Sodium chloride has a giant ionic lattice structure: the Na+ and Cl– ions are held together by strong electrostatic forces of attraction that act in all directions throughout the giant lattice, so a great deal of thermal energy is required to overcome these forces and separate the ions – giving a high melting point. Iodine has a simple molecular covalent structure: strong covalent bonds hold the two iodine atoms together within each I2 molecule, but the forces of attraction between separate I2 molecules (van der Waals' forces) are weak. On melting, only these weak intermolecular forces need to be overcome, not the strong covalent bonds, so only a small amount of energy is needed and the melting point is much lower.
(b) In diamond, each carbon atom forms four strong covalent bonds to four neighbouring carbon atoms, creating a rigid, giant three-dimensional lattice that extends throughout the whole crystal. Because the whole structure is held together by a continuous network of strong covalent bonds, a very large amount of energy is needed to break enough of these bonds to melt the substance, giving diamond an extremely high melting point.

(a) 1 mark: NaCl giant ionic lattice with strong ionic attraction in all directions requiring much energy to break; 1 mark: iodine is molecular covalent, held together between molecules only by weak van der Waals' forces; 1 mark: correct comparison – little energy needed to overcome weak intermolecular forces vs much energy for strong ionic bonds. (b) 1 mark: diamond is giant covalent structure; 1 mark: each carbon bonded to four others by strong covalent bonds; 1 mark: large amount of energy needed to break many strong covalent bonds throughout the lattice.

(a) On the universal indicator colour scale, a solution of pH 2 (strongly acidic) turns the indicator red or orange-red; a solution of pH 10 (weakly alkaline) turns it blue or dark blue.
(b) Hydrochloric acid is a strong acid, which means it is completely ionised in aqueous solution: every HCl molecule splits fully into H+(aq) and Cl–(aq) ions. Ethanoic acid is a weak acid, which means it is only partially ionised in solution: only a small proportion of the CH3COOH molecules split into ions at any time, with most remaining as un-ionised molecules. Even though both solutions have the same starting concentration of acid, the hydrochloric acid solution therefore contains a much higher concentration of H+(aq) ions than the ethanoic acid solution. Since a higher concentration of H+(aq) ions corresponds to a lower pH, the hydrochloric acid has the lower pH (1) and the ethanoic acid, with fewer H+(aq) ions, has the higher pH (3).

(a) 1 mark: pH 2 red/orange-red; 1 mark: pH 10 blue/dark blue. (b) 1 mark: HCl is a strong acid, completely ionised; 1 mark: ethanoic acid is a weak acid, partially ionised; 1 mark: at same concentration, ethanoic acid produces fewer H+(aq) ions; 1 mark: lower [H+(aq)] gives higher pH, correctly linking to the observed pH values. All other valid answers will be credited.

(a) From the standard flame-test colours: a lilac flame indicates potassium ions (K+), so X = potassium. A brick-red flame indicates calcium ions (Ca2+), so Y = calcium. A yellow/orange flame indicates sodium ions (Na+), so Z = sodium.
(b) The nichrome wire loop is first cleaned by dipping it into concentrated hydrochloric acid (to remove any contamination from previous tests) and then heated in a blue Bunsen flame until it produces no colour. The clean loop is dipped into the concentrated hydrochloric acid again and then into the solid sample (or a small amount is picked up on the moistened loop), before being held in the non-luminous (roaring, blue) part of the Bunsen flame. The student observes and records the colour of the flame produced, wearing eye protection throughout and holding the wire so that hands are kept a safe distance from the flame.

(a) 1 mark each for correct identification of X (potassium), Y (calcium), Z (sodium). (b) 1 mark: clean the nichrome wire loop with concentrated hydrochloric acid; 1 mark: dip clean loop in acid then in the sample and hold in a non-luminous/blue Bunsen flame; 1 mark: relevant safety precaution stated (eye protection / safe handling distance from flame). All other valid answers will be credited.

(a) At 40°C, 100 g of water dissolves 64 g of potassium nitrate. For 250 g of water: mass dissolved = 64 × (250/100) = 64 × 2.5 = 160 g.
(b) At 60°C, 100 g of water holds 110 g of dissolved potassium nitrate (a saturated solution). On cooling to 20°C, 100 g of water can only hold 32 g dissolved. The mass of crystals deposited = mass dissolved at 60°C − mass that remains dissolved at 20°C = 110 − 32 = 78 g.
(c) A saturated solution is a solution in which no more solute (at a given temperature) will dissolve – the solution is in equilibrium with any undissolved solute present.

(a) 1 mark: correct method (scaling 64 g by 250/100); 1 mark: correct answer 160 g. (b) 1 mark: correct method (110 − 32); 1 mark: correct answer 78 g. (c) 2 marks: full correct definition referencing 'no more solute will dissolve' at 'that (particular) temperature'; 1 mark for a partial/incomplete definition. Accept ECF from an incorrect but consistent answer in (a).

(a) The relative atomic mass (Ar) of an element is the mass of an average atom of that element (a weighted mean of the mass numbers of its naturally occurring isotopes, taking into account their relative abundance) compared with the mass of an atom of the carbon-12 isotope, which is defined as having a mass of exactly 12.
(b) The empirical formula of a compound is the simplest whole-number ratio of the atoms of each element present in the compound.
(c) First find the empirical formula mass: Mr(CH2O) = 12 + (2 × 1) + 16 = 30. To find how many CH2O units make up the molecular formula, divide the given relative formula mass by the empirical formula mass: 180 ÷ 30 = 6. Multiplying every subscript in CH2O by 6 gives the molecular formula: C6H12O6.

(a) 1 mark: mass of atom compared with carbon-12 (=12 exactly); 1 mark: correct reference to weighted mean/average taking isotope abundance into account. (b) 2 marks: full correct definition (simplest whole-number ratio of atoms of each element); 1 mark for a partial definition (e.g. omits 'whole-number' or 'simplest'). (c) 1 mark: correct empirical formula mass (30) and correct multiplying factor (6) shown; 1 mark: correct molecular formula C6H12O6.

(a) Mr(CaCO3) = 40 + 12 + (3 × 16) = 100. Moles = mass ÷ Mr = 10 ÷ 100 = 0.1 mol.
(b) The equation CaCO3 → CaO + CO2 shows a 1:1 mole ratio between CaCO3 and CO2, so moles of CO2 produced = 0.1 mol.
(c) Mr(CO2) = 12 + (2 × 16) = 44. Mass = moles × Mr = 0.1 × 44 = 4.4 g.

(a) 1 mark: correct Mr(CaCO3) = 100 shown; 1 mark: correct moles = 0.1 mol. (b) 1 mark: correct 1:1 ratio identified from the equation; 1 mark: correct moles CO2 = 0.1 mol (ECF from (a)). (c) 1 mark: correct Mr(CO2) = 44; 1 mark: correct method (moles × Mr); 1 mark: correct final answer 4.4 g (ECF from (b)).

(a) Mr(Al) = 27. Moles of Al = mass ÷ Mr = 5.4 ÷ 27 = 0.2 mol.
(b) From the equation, the mole ratio of Al : H2 is 2 : 3. Moles of H2 = 0.2 × (3/2) = 0.3 mol. Mr(H2) = 2, so theoretical mass of H2 = 0.3 × 2 = 0.6 g. Percentage yield = (actual yield ÷ theoretical yield) × 100 = (0.45 ÷ 0.6) × 100 = 75%.
(c) The percentage yield may be less than 100% because some of the gas produced escapes or is lost during collection before it can be measured, or because side reactions occur that use up some of the reactant to form other, unwanted products.

(a) 1 mark: correct Mr(Al) = 27; 1 mark: correct moles = 0.2 mol. (b) 1 mark: correct theoretical mass 0.6 g shown using correct mole ratio; 1 mark: correct percentage yield 75% (ECF from theoretical mass calculated). (c) 1 mark each for any two valid distinct reasons (maximum 3 marks: 2 reasons at up to 1–2 marks each, or one well-developed reason for 2 and a second basic reason for 1) e.g. loss of gas/product during collection, reaction reversible/does not go to completion, side reactions occurring, apparatus leaks. All other valid answers will be credited.

(a) Mr(CuSO4) = Ar(Cu) + Ar(S) + 4 × Ar(O) = 64 + 32 + (4 × 16) = 64 + 32 + 64 = 160.
(b) The mass of water of crystallisation in the hydrated compound = total Mr − Mr(CuSO4) = 250 − 160 = 90. Since Mr(H2O) = 18, the number of moles of water, x = 90 ÷ 18 = 5.
(c) Percentage by mass of water of crystallisation = (mass of water ÷ total Mr) × 100 = (90 ÷ 250) × 100 = 36%.

(a) 1 mark: correct individual Ar values substituted; 1 mark: correct total 160 shown as working (not just quoted). (b) 1 mark: correct mass of water (90) found by subtraction; 1 mark: correct x = 5 (ECF). (c) 1 mark: correct method (90/250 × 100); 1 mark: correct answer 36% (ECF from (a) and (b)).

(a) Rate = change in volume ÷ change in time = (32 − 0) ÷ 40 = 32/40 = 0.8 cm³/s.
(b) The volume of gas remains constant at 48 cm³ from 80 s to 100 s, showing that no further gas is being produced, so the reaction has finished (stopped) at 80 s.
(c) A higher temperature increases the rate of reaction (more frequent, more energetic collisions), which alone would make the graph rise more steeply at the start. However, larger marble chips have a smaller surface area to volume ratio than the original chips, which decreases the rate of reaction, partially offsetting the effect of the higher temperature. The final total volume of gas produced would be unchanged, because the same mass (moles) of calcium carbonate and the same concentration/volume of excess acid are used, and the amount of reactant, not the rate, determines the final volume.

(a) 1 mark: correct reading of volume at 40 s (32 cm³); 1 mark: correct rate 0.8 cm³/s. (b) 1 mark: 80 s stated; 1 mark: correct reason (volume constant / no further gas produced after this time). (c) 1 mark: valid comment on rate (steeper/less steep, with correct reasoning referencing temperature and/or surface area); 1 mark: correct statement that the final volume of gas is unchanged. All other valid answers will be credited.

(a) One molecule of CH4 contains 4 C–H bonds and 2 molecules of O2 contain 2 O=O bonds. Energy to break bonds in reactants = (4 × 413) + (2 × 498) = 1652 + 996 = 2648 kJ/mol.
(b) One molecule of CO2 contains 2 C=O bonds and 2 molecules of H2O contain a total of 4 O–H bonds. Energy released forming bonds in products = (2 × 805) + (4 × 464) = 1610 + 1856 = 3466 kJ/mol.
(c) Overall energy change = energy to break bonds − energy released forming bonds = 2648 − 3466 = −818 kJ/mol. Since the energy released when new bonds form is greater than the energy needed to break the old bonds, more energy is given out than taken in, so the reaction is exothermic (negative energy change confirms this).

(a) 1 mark: correct identification of bonds broken (4 × C–H, 2 × O=O); 1 mark: correct total 2648 kJ/mol. (b) 1 mark: correct identification of bonds formed (2 × C=O, 4 × O–H); 1 mark: correct total 3466 kJ/mol. (c) 1 mark: correct energy change −818 kJ/mol (ECF from (a) and (b)); 1 mark: correctly identified as exothermic with valid reasoning (energy released > energy absorbed / negative value).

(a) A catalyst is a substance that increases the rate of a chemical reaction without itself being permanently chemically changed or used up by the reaction.
(b) Decrease in activation energy = 75 − 45 = 30 kJ/mol. Percentage decrease = (decrease ÷ original) × 100 = (30 ÷ 75) × 100 = 40%.
(c) A catalyst works by providing an alternative reaction pathway that has a lower activation energy than the uncatalysed reaction. Since activation energy is the minimum energy that colliding particles need in order to react, lowering it means that a greater proportion of the particle collisions occurring at a given temperature now have enough energy to react successfully. This increases the frequency of successful collisions, so the rate of reaction increases.

(a) 1 mark: increases rate of reaction; 1 mark: not used up / chemically unchanged. (b) 1 mark: correct decrease (30 kJ/mol) shown; 1 mark: correct percentage 40%. (c) 1 mark: catalyst provides alternative pathway with lower activation energy; 1 mark: more particles/collisions have sufficient energy to react, increasing rate.

(a) Doubling the concentration of the hydrochloric acid doubles the number of acid particles present in a given volume of solution. This increases the frequency of collisions between the acid particles and the particles of the other reactant. Since the proportion of collisions that are successful (have enough energy, i.e. exceed the activation energy) stays the same, the increased collision frequency means more successful collisions occur per second, so the initial rate of reaction increases.
(b) If the rate of gas production doubles throughout the reaction, the same total volume of gas (24 cm³) is produced in half the time: new time = 120 ÷ 2 = 60 s.

(a) 1 mark: rate increases; 1 mark: more particles in same volume/increased frequency of collisions; 1 mark: correctly links to more successful collisions per unit time. (b) 1 mark: correct method (halving the time); 1 mark: correct answer 60 s.

Le Châtelier's Principle states that if a system at dynamic equilibrium is subjected to a change in conditions, the position of equilibrium shifts to counteract that change. In the Haber process, there are 4 moles of gas on the reactant side (1 N2 + 3 H2) and only 2 moles of gas on the product side (2 NH3). Increasing the pressure shifts the equilibrium position towards the side with fewer gas molecules, i.e. towards the products, increasing the yield of ammonia; therefore a high pressure favours ammonia production. The forward reaction (formation of ammonia) is exothermic. Decreasing the temperature shifts the equilibrium towards the exothermic (forward) direction, which also increases the equilibrium yield of ammonia. However, industrially the process does not simply use the lowest possible temperature and highest possible pressure. A very low temperature, although it favours a higher equilibrium yield, makes the rate of reaction far too slow to be economically viable, since dynamic equilibrium would take a very long time to be reached. A very high pressure gives a higher yield but requires extremely strong, thick-walled reaction vessels and pipework, which are very expensive to build and maintain, and pressurising the gases uses large amounts of energy, adding to running costs. For these reasons, industry uses a compromise: a moderately high temperature (around 450°C) that gives an acceptable reaction rate even though it slightly reduces the equilibrium yield compared with a lower temperature, together with a high but manageable pressure (around 200 atmospheres) that increases the yield without making the plant prohibitively expensive to build and run. An iron catalyst is also used to speed up the rate of reaching equilibrium without affecting its position.

Marked using three levels of response. Level 3 (5–6 marks): a full, coherent explanation using Le Châtelier's Principle correctly for both pressure (fewer gas moles on product side, so high pressure favours products) and temperature (exothermic forward reaction, so low temperature favours products), with a clear discussion of the industrial trade-off between yield, rate and cost, using accurate specialist terminology throughout with few or no errors of spelling, punctuation or grammar. Level 2 (3–4 marks): correct use of Le Châtelier's Principle for at least one of pressure or temperature, with some discussion of the compromise conditions, but less complete or with minor inaccuracies; reasonably clear use of specialist terms. Level 1 (1–2 marks): basic statements about increasing pressure or decreasing temperature increasing yield, with little or no explanation of the compromise; limited use of specialist terms. 0 marks: no relevant content.

(a) At the cathode (negative electrode), positive lead ions are attracted and gain electrons (reduction): Pb²⁺(l) + 2e⁻ → Pb(l).
(b) At the anode (positive electrode), negative bromide ions are attracted and lose electrons (oxidation) to form bromine: 2Br⁻(l) → Br2(g) + 2e⁻. Bromine is observed as an orange/red-brown vapour forming at the anode.

(a) 1 mark: correct species (Pb²⁺, Pb); 1 mark: correctly balanced half equation showing gain of 2 electrons. (b) 1 mark: correctly balanced half equation showing loss of 2 electrons from 2Br⁻; 1 mark: correct observation (orange/red-brown bromine vapour). Reject equations with incorrect electron balancing.

(a) At the cathode, positive hydrogen ions (from the water/acid) are attracted and gain electrons (reduction) to form hydrogen gas: 2H⁺(aq) + 2e⁻ → H2(g).
(b) At the anode, hydroxide ions (from the water) are attracted and lose electrons (oxidation) to form oxygen gas: 4OH⁻(aq) → O2(g) + 2H2O(l) + 4e⁻.

(a) 1 mark: hydrogen named; 1 mark: correctly balanced half equation. (b) 1 mark: oxygen named; 1 mark: correctly balanced half equation (either accepted form). Accept ionic half equations expressed either from OH⁻ or H2O as starting species provided correctly balanced.

(a) At the cathode, aluminium ions gain electrons (reduction) to form molten aluminium: Al³⁺(l) + 3e⁻ → Al(l).
(b) At the anode, oxide ions lose electrons (oxidation) to form oxygen gas: 2O²⁻(l) → O2(g) + 4e⁻. Because the electrolysis is carried out at a high temperature, the oxygen gas produced at the anode reacts with the hot carbon (graphite) anode itself: C(s) + O2(g) → CO2(g). This gradually oxidises away the carbon of the anode, so the anodes must be replaced periodically to maintain efficient electrolysis.

(a) 1 mark: correct species (Al³⁺, Al); 1 mark: correctly balanced half equation showing gain of 3 electrons. (b) 1 mark: correctly balanced anode half equation (2O²⁻ → O2 + 4e⁻); 1 mark: oxygen produced reacts with the hot carbon anode; 1 mark: carbon anode is oxidised to CO2/burns away, requiring periodic replacement.

(a) Iron is extracted from its ore (haematite) by chemical reduction using carbon (in the form of carbon monoxide) in the blast furnace, as iron is less reactive than carbon.
(b) Aluminium is extracted by electrolysis of molten aluminium oxide (purified from bauxite) dissolved in molten cryolite. This method is necessary because aluminium is more reactive than carbon, so carbon is not a strong enough reducing agent to displace aluminium from aluminium oxide; a more powerful method (electrolysis) is therefore required.
(c) A metal's position in the reactivity series relative to carbon determines the extraction method used: metals below carbon in the reactivity series (less reactive than carbon, e.g. iron, zinc, copper) can be economically extracted by reduction with carbon, since carbon can displace them from their oxides. Metals above carbon (more reactive than carbon, e.g. aluminium, potassium, sodium, calcium, magnesium) cannot be displaced by carbon, so they must be extracted by electrolysis of their molten compounds, which is a far more expensive process because of the large amounts of electricity needed.

(a) 1 mark: method (reduction with carbon/blast furnace); 1 mark: correct reducing agent (carbon/carbon monoxide). (b) 1 mark: method (electrolysis); 1 mark: correct reason (aluminium more reactive than carbon, cannot be displaced by it). (c) 1 mark: metals below carbon extracted by reduction with carbon; 1 mark: metals above carbon extracted by electrolysis (more expensive/energy-intensive). All other valid answers will be credited.

(a) In phytomining, plants that can absorb metal compounds (such as copper compounds) are grown on land containing low-grade copper ore. The plants take up the copper compounds through their roots as they grow. Once grown, the plants are harvested and then burned, producing an ash that contains the absorbed copper compounds. An acid is added to this ash, which reacts with the copper compounds to produce a solution (called a leachate) containing dissolved copper compounds. Copper metal can then be obtained from this leachate solution by displacement, using a more reactive metal such as scrap iron (iron displaces the less reactive copper from solution), or alternatively by electrolysis of the solution.
(b) Phytomining avoids the environmental damage caused by traditional mining methods, such as digging, moving and disposing of large quantities of rock, and it allows copper to be extracted economically from low-grade ores that would not be worth mining by conventional methods.

(a) 1 mark each for: plants absorb copper compounds from low-grade ore; plants harvested and burned to produce ash; acid added to ash to produce leachate solution; copper obtained from leachate by displacement (e.g. with scrap iron) or electrolysis. (b) 2 marks for a fully explained advantage (e.g. avoids digging/moving/disposing of large amounts of rock, reducing environmental impact); 1 mark for a basic, undeveloped advantage. All other valid answers will be credited.

(a) Carbon monoxide gas, formed in the blast furnace from the reaction of carbon (coke) with oxygen (and further reaction of carbon dioxide with more carbon), acts as the reducing agent that reduces the iron(III) oxide in haematite to iron. Balanced equation: Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g).
(b) Limestone (calcium carbonate, CaCO3) is added to the blast furnace along with the iron ore and coke. On heating, the limestone thermally decomposes to form calcium oxide (a basic oxide) and carbon dioxide: CaCO3 → CaO + CO2. The calcium oxide then reacts with the acidic silica (sand, SiO2) impurity present in the ore, in an acid–base reaction, to form calcium silicate (slag): CaO + SiO2 → CaSiO3. This molten slag is less dense than the molten iron, floats on top of it, and can be run off/tapped separately from the furnace, removing the impurities from the iron.

(a) 1 mark: carbon monoxide named as reducing agent; 1 mark: correct formulae; 1 mark: correctly balanced equation Fe2O3 + 3CO → 2Fe + 3CO2. (b) 1 mark: limestone decomposes to form calcium oxide (basic oxide); 1 mark: calcium oxide reacts with acidic silica to form calcium silicate (slag); 1 mark: slag is molten/less dense, floats on iron and is run off separately. All other valid answers will be credited.

(a) Zinc is more reactive than copper, so a displacement reaction occurs. Zinc atoms on the surface of the metal lose two electrons each (oxidation) to form Zn²⁺ ions, which pass into solution. These electrons are gained by Cu²⁺ ions from the copper(II) sulfate solution (reduction), forming copper atoms that are deposited as a reddish-brown/orange coating on the surface of the zinc. As the reaction proceeds, the blue colour of the copper(II) sulfate solution fades (as Cu²⁺ ions are used up and replaced by colourless Zn²⁺ ions) and the zinc metal is gradually eaten away.
(b) Combining the two half equations (Zn → Zn²⁺ + 2e⁻ and Cu²⁺ + 2e⁻ → Cu) gives the overall ionic equation: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).
(c) Iron is more reactive than copper (higher in the reactivity series), so it readily displaces/reduces copper ions from solution to form copper metal. Iron is also a cheap and abundant metal, making it a cost-effective choice for recovering copper from dilute leachate solutions on an industrial scale.

(a) 1 mark: blue colour fades; 1 mark: reddish-brown copper coating deposited on zinc; 1 mark: correct explanation in terms of electron transfer (Zn loses electrons/oxidised, Cu²⁺ gains electrons/reduced). (b) 1 mark: correctly balanced ionic equation. (c) 1 mark: iron more reactive than copper, displaces it from solution; 1 mark: iron is cheap/readily available.

(a) Producing new aluminium from bauxite ore requires electrolysis of molten aluminium oxide, a process that consumes very large quantities of electrical energy both to keep the electrolyte molten at high temperature and to drive the electrolysis reaction itself. Recycling aluminium simply involves collecting, melting down and re-casting existing aluminium metal, which requires far less energy since no electrolysis or ore-purification steps are needed.
(b) Besides saving energy, recycling aluminium conserves the Earth's limited supplies of bauxite ore, reduces the carbon dioxide emissions associated with generating the large amounts of electricity needed for electrolysis, and reduces the volume of mining waste and other extraction by-products that would otherwise need to be disposed of.
(c) Aluminium has a low density, making structures built from it lightweight, which is important for fuel efficiency in aircraft. Although aluminium is a reactive metal, it reacts rapidly with oxygen in the air to form a thin, strong, continuous layer of aluminium oxide on its surface; this oxide layer sticks firmly to the metal beneath and prevents further oxygen or moisture from reaching it, so the aluminium is protected from further corrosion.

(a) 1 mark: electrolysis to extract new aluminium requires very large amounts of electrical energy; 1 mark: recycling only needs melting, not electrolysis, so uses much less energy. (b) 1 mark each for any two valid distinct benefits (conserves ore/finite resource; reduces CO2 emissions; reduces mining waste/environmental damage; cheaper). (c) 1 mark: low density (lightweight) and/or forms a protective oxide layer preventing further corrosion. All other valid answers will be credited.

Mr(NaOH) = 23 + 16 + 1 = 40. Moles of NaOH = mass ÷ Mr = 4.0 ÷ 40 = 0.1 mol. Volume of solution = 250 cm³ = 0.25 dm³. Concentration = moles ÷ volume = 0.1 ÷ 0.25 = 0.4 mol/dm³.

1 mark: correct moles of NaOH (0.1 mol); 1 mark: correct conversion of volume to dm³ (0.25 dm³); 1 mark: correct method (moles ÷ volume); 1 mark: correct final answer 0.4 mol/dm³. (Full 4 marks require the correct final answer with working shown; award up to 3 marks for correct method/working leading to an arithmetic slip.)

Moles of NaOH used = concentration × volume (in dm³) = 0.20 × (25.0/1000) = 0.20 × 0.025 = 0.005 mol. The equation shows a 1:1 mole ratio between NaOH and HCl, so moles of HCl = 0.005 mol. Concentration of HCl = moles ÷ volume (in dm³) = 0.005 ÷ (20.0/1000) = 0.005 ÷ 0.020 = 0.25 mol/dm³.

1 mark: correct moles of NaOH (0.005 mol); 1 mark: correct 1:1 mole ratio applied to find moles HCl; 1 mark: correct conversion of 20.0 cm³ to dm³; 1 mark: correct final answer 0.25 mol/dm³.

The equation shows a 1:1 mole ratio between Mg and H2, so 0.5 mol of magnesium produces 0.5 mol of hydrogen gas. Volume of gas = moles × 24 dm³ = 0.5 × 24 = 12 dm³.

1 mark: correct 1:1 mole ratio identified; 1 mark: correct moles of H2 (0.5 mol); 1 mark: correct method (moles × 24 dm³); 1 mark: correct final answer 12 dm³.

(a) Mr(KOH) = 39 + 16 + 1 = 56. Concentration in g/dm³ = concentration in mol/dm³ × Mr = 0.25 × 56 = 14 g/dm³.
(b) Volume = 50 cm³ = 0.050 dm³. Moles = concentration × volume = 0.25 × 0.050 = 0.0125 mol.

(a) 1 mark: correct Mr(KOH) = 56; 1 mark: correct answer 14 g/dm³. (b) 1 mark: correct conversion of 50 cm³ to dm³ (0.050 dm³); 1 mark: correct answer 0.0125 mol.

(a) Using the classification pH 0–2 strong acid, pH 3–6 weak acid, pH 7 neutral, pH 8–11 weak alkali, pH 12–14 strong alkali: Solution A (pH 1) is a strong acid; Solution B (pH 7) is neutral; Solution C (pH 13) is a strong alkali; Solution D (pH 4) is a weak acid.
(b) Solution C has pH 13, making it a strong (corrosive) alkali, so the student should wear eye protection (and gloves) and avoid skin/eye contact when handling it.
(c) The pH meter should first be calibrated using one or more buffer solutions of known, accurately fixed pH. The clean probe is then rinsed with distilled water and dipped into the test solution, and the reading is allowed to stabilise before the pH is recorded, typically to at least one decimal place – giving a more precise numerical value than the approximate colour match used with universal indicator.

(a) 1 mark for two or three correct classifications; 2 marks for all four correct. (b) 1 mark: valid safety precaution (eye protection/gloves/avoid contact, since C is corrosive). (c) 1 mark: calibrate meter with buffer solution(s) of known pH; 1 mark: reading given to at least one decimal place / more precise than colour-matching. All other valid answers will be credited.

(a) A lit splint producing a squeaky pop is the standard test for hydrogen, so Gas X is hydrogen. Limewater (calcium hydroxide solution) turning milky is the standard test for carbon dioxide, so Gas Y is carbon dioxide. Damp litmus paper turning red and then being bleached white is the standard test for chlorine, so Gas Z is chlorine.
(b) When carbon dioxide first reacts with limewater, it forms an insoluble white precipitate of calcium carbonate, making the solution milky: Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l). If excess carbon dioxide continues to be bubbled through, it reacts further with the calcium carbonate precipitate and water to form soluble calcium hydrogencarbonate: CaCO3(s) + H2O(l) + CO2(g) → Ca(HCO3)2(aq). Since this product is soluble, the precipitate redissolves and the milkiness disappears, leaving a colourless solution.

(a) 1 mark each for correct identification of X (hydrogen), Y (carbon dioxide), Z (chlorine). (b) 1 mark: precipitate/milkiness disappears/redissolves; 1 mark: correct reason (excess CO2 forms soluble calcium hydrogencarbonate).

(a) The crucible containing the hydrated crystals is weighed, then heated strongly with a Bunsen burner for a set period of time. The crucible is then allowed to cool (ideally in a desiccator, to avoid reabsorbing moisture from the air) before being reweighed. This process of heating, cooling and reweighing is repeated until two successive readings are the same (constant mass), which shows that all the water of crystallisation has been driven off and no further loss of mass is occurring – if the mass were still decreasing, heating would need to continue.
(b) Since the crucible becomes very hot, the student should use tongs to move and handle it rather than touching it directly, and should place it on a heatproof mat to cool. Eye protection should be worn throughout, and the experiment carried out in a well-ventilated area away from flammable materials.

(a) 1 mark: heat, cool and reweigh; 1 mark: repeat process; 1 mark: constant mass reached when two consecutive readings are the same. (b) 1 mark: use tongs to handle the hot crucible; 1 mark: further valid precaution (eye protection / heatproof mat / well-ventilated area). All other valid answers will be credited.

(a) A pipette filler is fitted to the top of the pipette and used to draw up the alkali solution, taking care that the bottom of the meniscus lines up exactly with the graduation (calibration) mark on the pipette (viewed at eye level). The measured alkali is then allowed to run out of the pipette into the conical flask, and the tip of the pipette is touched against the inside wall of the flask to transfer the final drop.
(b) Reading the burette scale at eye level, from the bottom of the meniscus, ensures the volume is read consistently and avoids parallax error (an apparent difference in the reading caused by viewing the scale from above or below eye level), giving an accurate and repeatable titre value.
(c) Repeating the titration and obtaining two concordant results (titres agreeing within 0.10 cm³ of each other) demonstrates that the results are reliable and reproducible, allowing an accurate mean titre to be calculated from the concordant (accurate) runs only, while any inconsistent 'rough' titre is excluded from the average.

(a) 1 mark: use pipette filler to fill to the calibration mark with bottom of meniscus on the line; 1 mark: run into flask, touching tip to inside of flask for final drop. (b) 1 mark: avoids parallax error, giving an accurate reading. (c) 1 mark: checks reliability/reproducibility of results; 1 mark: allows accurate mean titre to be calculated from concordant results, excluding anomalies.

(a) Temperature change = final temperature − starting temperature = 12.5 − 20.0 = −7.5°C, i.e. a fall of 7.5°C.
(b) As the mass of ammonium nitrate dissolved increases, the final temperature of the water decreases below the starting temperature of 20.0°C. Since the temperature of the surroundings (the water) falls, this shows that dissolving ammonium nitrate absorbs heat energy from the surroundings rather than releasing it, so the process is endothermic.
(c) The data shows the temperature falls by 2.5°C for every 2.0 g of ammonium nitrate dissolved (2.0 g → 2.5°C fall; 4.0 g → 5.0°C fall; 6.0 g → 7.5°C fall), a constant fall of 1.25°C per gram. For 8.0 g, the predicted fall = 8.0 × 1.25 = 10.0°C. Predicted final temperature = 20.0 − 10.0 = 10.0°C.

(a) 1 mark: correct answer, a fall of 7.5°C (accept −7.5°C). (b) 1 mark: endothermic; 1 mark: correct reasoning (temperature/heat energy of surroundings decreases, so heat is absorbed by the dissolving process). (c) 1 mark: correct identification of the pattern (2.5°C fall per 2.0 g, i.e. 1.25°C per gram); 1 mark: correct predicted fall for 8.0 g (10.0°C); 1 mark: correct final temperature 10.0°C.

(a) To judge the end-point consistently, the student should view the cross from directly above the flask (looking straight down through the solution) each time and stop the clock at the exact moment the cross can no longer be seen through the cloudy precipitate that has formed. The same person should ideally make this judgement each time, under the same lighting conditions, to minimise differences in judgement between repeats.
(b) Rate = 1 ÷ time = 1 ÷ 75 = 0.01333... s⁻¹ ≈ 0.0133 s⁻¹ (3 significant figures).
(c) As the concentration of sodium thiosulfate solution increases, the time taken for the cross to disappear decreases, meaning the rate of reaction increases. Comparing the first two rows, doubling the concentration from 0.05 to 0.10 mol/dm³ roughly halves the time (150 s to 75 s), which very nearly doubles the rate, indicating that the rate of reaction is approximately directly proportional to the concentration of sodium thiosulfate.

(a) 1 mark: view from directly above/consistent viewing position; 1 mark: judge the moment the cross can no longer be seen, using consistent conditions/same observer. (b) 1 mark: correct method (1 ÷ 75); 1 mark: correct answer 0.0133 s⁻¹ (accept 0.013 or equivalent to at least 2 s.f.). (c) 1 mark: increasing concentration increases rate/decreases time; 1 mark: valid comment on approximate direct proportionality, supported by data from the table.

(a) Temperature change = final temperature − starting temperature = 26.0 − 19.5 = 6.5°C, i.e. a rise of 6.5°C.
(b) Because the temperature of the reacting mixture rises during the reaction, heat energy must be released from the reacting chemicals to the surroundings (the solution itself). A reaction that releases heat energy to its surroundings is exothermic, so neutralisation of hydrochloric acid with sodium hydroxide is an exothermic reaction.
(c) The accuracy of the recorded temperature change could be improved by using an insulated cup with a lid (with a small hole for the thermometer/stirrer), which reduces heat loss to the surroundings during the experiment and gives a more accurate maximum temperature reading. Other valid improvements include stirring continuously to ensure the temperature is uniform throughout the mixture, or using a more precise thermometer or temperature sensor/data logger capable of reading to at least one decimal place, which reduces the uncertainty in each individual temperature reading.

(a) 1 mark: correct answer, a rise of 6.5°C. (b) 1 mark: exothermic; 1 mark: correct reasoning (temperature rises, so heat energy is released to the surroundings). (c) up to 3 marks for one clearly explained improvement (e.g. 1 mark for stating a valid improvement such as using a lid on the insulated container; 1 mark for explaining it reduces heat loss to the surroundings; 1 mark for linking this to a more accurate maximum temperature/temperature change being recorded). All other valid answers will be credited.

(a) Lead(II) nitrate solution and potassium iodide solution are both colourless. When mixed, the lead ions and iodide ions combine to form insoluble lead(II) iodide, which is seen as a bright yellow precipitate forming immediately in the mixture.
(b) This is a precipitation reaction (a type of double decomposition reaction), in which two soluble ionic compounds in solution react to form an insoluble solid (precipitate). The spectator ions (K+ and NO3–) do not appear in the ionic equation, which shows only the ions that combine to form the precipitate: Pb²⁺(aq) + 2I⁻(aq) → PbI2(s).

(a) 1 mark: yellow (or bright yellow) precipitate; 1 mark: forms immediately/on mixing (from colourless solutions). (b) 1 mark: precipitation (or double decomposition) reaction named; 1 mark: correct ionic species (Pb²⁺, I⁻, PbI2); 1 mark: correctly balanced ionic equation.

(a) A small sample of each solution is acidified with a few drops of dilute nitric acid (to remove any carbonate or hydroxide ions that would otherwise also give a precipitate with silver ions and interfere with the result), and then silver nitrate solution is added.
(b) The silver halide formed depends on which halide ion is present: chloride ions give a white precipitate of silver chloride (Ag+ + Cl– → AgCl); bromide ions give a cream precipitate of silver bromide (Ag+ + Br– → AgBr); iodide ions give a yellow precipitate of silver iodide (Ag+ + I– → AgI).

(a) 1 mark: dilute nitric acid added first; 1 mark: silver nitrate solution added. (b) 1 mark each for correct precipitate colour: chloride (white), bromide (cream), iodide (yellow).

(a) Chlorine is above iodine in Group 7, and reactivity of the halogens decreases going down the group, so chlorine is more reactive than iodine. A more reactive halogen can displace a less reactive halogen from a solution of its halide ions. Here, chlorine displaces iodide ions, oxidising them to iodine, which is observed as the colourless potassium iodide solution turning a brown/orange colour (the colour of aqueous iodine).
(b) Chlorine molecules gain electrons (are reduced) while iodide ions lose electrons (are oxidised): Cl2(aq) + 2I⁻(aq) → 2Cl⁻(aq) + I2(aq).

(a) 1 mark: solution turns brown/orange; 1 mark: chlorine displaces iodine/iodide; 1 mark: correct reasoning (chlorine more reactive, reactivity decreases down group). (b) 1 mark: correct species; 1 mark: correctly balanced ionic equation.

(a) When dilute acid reacts with a carbonate, carbon dioxide gas is produced (acid + carbonate → salt + water + carbon dioxide). Since the gas produced in Test 1 turns limewater milky – the standard positive test for carbon dioxide – this confirms that the anion present in the solid is the carbonate ion, CO3²⁻.
(b) The standard flame test colours show that a lilac flame is produced specifically by potassium ions, so Test 2 confirms the cation present is K+.
(c) Combining the identified cation (K+) and anion (CO3²⁻) gives the compound potassium carbonate, with formula K2CO3 (two K+ ions balance the 2– charge of the carbonate ion).

(a) 1 mark: carbonate ion identified; 1 mark: correct reasoning (gas turns limewater milky = CO2 = confirms carbonate). (b) 1 mark: potassium ion identified from lilac flame. (c) 1 mark: correct name potassium carbonate; 1 mark: correct formula K2CO3.

(a) From the table, the only metal ion listed whose white precipitate dissolves in excess sodium hydroxide to give a colourless solution is Al³⁺, so the unknown solution contains aluminium ions.
(b) Aluminium ions react with hydroxide ions to form insoluble aluminium hydroxide: Al³⁺(aq) + 3OH⁻(aq) → Al(OH)3(s). (In excess sodium hydroxide, this precipitate then redissolves as the amphoteric aluminium hydroxide reacts further with hydroxide ions, but this second equation is not required to describe the initial precipitate.)
(c) Both Fe²⁺ and Fe³⁺ ions form a precipitate with sodium hydroxide that does not dissolve when excess sodium hydroxide is added, unlike Al³⁺. However, the two iron precipitates have distinctly different colours: Fe²⁺(aq) forms a green precipitate of iron(II) hydroxide, while Fe³⁺(aq) forms an orange-brown precipitate of iron(III) hydroxide. By observing the colour of the insoluble precipitate formed, the student can therefore distinguish between solutions containing Fe²⁺ and those containing Fe³⁺.

(a) 1 mark: Al³⁺ correctly identified. (b) 1 mark: correct species (Al³⁺, OH⁻, Al(OH)3); 1 mark: correctly balanced equation. (c) 1 mark: both form a precipitate that does not dissolve in excess NaOH; 1 mark: Fe²⁺ gives green precipitate; 1 mark: Fe³⁺ gives orange-brown precipitate.

The student should first add a few drops of bromine water to a separate sample of each of the four liquids and shake. Hex-1-ene, an alkene, contains a C=C double covalent bond, which reacts with the bromine in an addition reaction, decolourising the bromine water from orange/orange-brown to colourless. Hexane, ethanol and ethanoic acid do not contain a C=C double bond and so do not decolourise the bromine water, allowing the alkene to be identified at this stage. To distinguish between the three remaining liquids, the student should add a small amount of a carbonate or hydrogencarbonate, such as sodium hydrogencarbonate, to a sample of each. Ethanoic acid, a carboxylic acid, reacts with the carbonate to produce visible fizzing/effervescence as carbon dioxide gas is released, while hexane and ethanol, which are not acidic, show no reaction, allowing the carboxylic acid to be identified. Finally, to distinguish hexane from ethanol, the student should add a few drops of acidified potassium dichromate(VI) solution to each of the two remaining liquids and gently warm them. Ethanol is oxidised by the acidified potassium dichromate, which itself is reduced and changes colour from orange to green as this reaction occurs. Hexane, an alkane with no functional group, is not oxidised by the acidified dichromate, so the orange colour remains unchanged, allowing the alcohol and alkane to be distinguished from each other and fully identifying all four liquids.

Marked using three levels of response. Level 3 (5–6 marks): a full, logical test sequence covering all four liquids, correctly identifying the alkene using bromine water (with correct colour change), the carboxylic acid using a carbonate/hydrogencarbonate (fizzing/CO2), and distinguishing the alcohol from the alkane using acidified potassium dichromate (orange to green vs no change), using accurate specialist terminology throughout with few or no errors of spelling, punctuation or grammar. Level 2 (3–4 marks): most of the above tests and observations given correctly, but with one omission or an incomplete/less precise description of one test; reasonably clear use of specialist terms. Level 1 (1–2 marks): only one or two valid tests given, with limited detail or accuracy, e.g. only the bromine water test correctly described; limited use of specialist terms. 0 marks: no relevant content.

Mr(CaCO3) = 40 + 12 + (3 × 16) = 100. Moles of CaCO3 = 2.5 ÷ 100 = 0.025 mol. The equation shows a 1:1 mole ratio between CaCO3 and CO2, so moles of CO2 = 0.025 mol. Volume of CO2 = moles × 24 dm³ = 0.025 × 24 = 0.6 dm³.

1 mark: correct Mr(CaCO3) = 100 and moles = 0.025 mol; 1 mark: correct 1:1 mole ratio applied; 1 mark: correct method (moles × 24); 1 mark: correct final answer 0.6 dm³.

Moles of H2 collected = volume ÷ 24 000 = 120 ÷ 24 000 = 0.005 mol. The equation shows a 1:1 mole ratio between Zn and H2, so moles of Zn that reacted = 0.005 mol. Mass of Zn = moles × Ar = 0.005 × 65 = 0.325 g.

1 mark: correct conversion of 120 cm³ to moles of H2 (0.005 mol); 1 mark: correct 1:1 mole ratio applied; 1 mark: correct method (moles × Ar); 1 mark: correct final answer 0.325 g.

(a) Moles of Mg = mass ÷ Ar = 0.6 ÷ 24 = 0.025 mol. The equation shows a 1:1 mole ratio between Mg and H2, so moles of H2 produced = 0.025 mol.
(b) Volume of H2 = moles × 24 000 = 0.025 × 24 000 = 600 cm³.
(c) Using powdered magnesium instead of ribbon increases the surface area of the metal exposed to the acid, which increases the frequency of collisions and therefore the rate of reaction, so the gas is produced more quickly. However, since the same mass (0.025 mol) of magnesium reacts either way, the total volume of gas produced by the time the reaction is complete is unchanged.

(a) 1 mark: correct moles of Mg (0.025 mol); 1 mark: correct 1:1 ratio applied giving moles H2 = 0.025 mol. (b) 1 mark: correct method (moles × 24 000); 1 mark: correct answer 600 cm³ (ECF from (a)). (c) 1 mark: correct answer that total volume is unchanged (only rate changes).

(a) The magnesium ribbon is placed in a conical flask together with a measured volume of hydrochloric acid of the required concentration. The flask is immediately sealed with a rubber bung fitted with a delivery tube, so that no gas can escape except through the tube. The delivery tube is connected to a gas syringe, and as hydrogen gas is produced by the reaction, it flows along the delivery tube into the gas syringe, pushing the plunger outward. A stopclock is started at the moment the bung is inserted (or the reactants are mixed), and the volume of gas collected in the syringe is read from its scale at regular timed intervals (for example, every 10 seconds) until the reaction is complete or the syringe is full.
(b) A suitable table has a first column headed 'Time / s' listing the time points at which readings are taken (0, 10, 20, 30 s, and so on, at 10-second intervals until the reaction finishes), and then one further column for each of the three acid concentrations tested, each headed 'Volume of gas / cm³' with the specific concentration used stated in the column heading (for example, 'Volume of gas / cm³ (0.5 mol/dm³ HCl)'). Each row then records the volume of gas collected at that time for each of the three concentrations, allowing the three sets of results to be compared directly and plotted on the same graph.

Time / s | Volume of gas / cm³ (0.5 mol/dm³) | Volume of gas / cm³ (1.0 mol/dm³) | Volume of gas / cm³ (1.5 mol/dm³)
0 | | |
10 | | |
20 | | |
30 | | |

(a) 1 mark: conical flask with bung and delivery tube; 1 mark: delivery tube connects to gas syringe which measures gas volume; 1 mark: stopclock/timing used with readings taken at regular intervals. (b) 1 mark: correct 'Time / s' column with regular 10 s intervals; 1 mark: separate column for each of the three concentrations; 1 mark: correct header 'Volume of gas / cm³' with units; 1 mark: each concentration clearly labelled/distinguished in its column heading. All other valid, clearly described table designs will be credited.

(a) A burette is clamped vertically in a burette stand and filled with the hydrochloric acid of unknown concentration, and the initial reading is recorded (to the nearest 0.05 cm³, reading from the bottom of the meniscus at eye level). Using a pipette and pipette filler, exactly 25.0 cm³ of the standard sodium hydroxide solution is measured into a conical flask, and a few drops of phenolphthalein indicator are added, turning the alkaline solution pink. The flask is placed on a white tile (to help observe the colour change clearly) beneath the burette, and acid is run in from the burette while continuously swirling the flask, quickly at first and then a drop at a time as the end-point is approached. The end-point is reached when the solution just changes from pink to colourless and stays colourless with one further drop; at this point the final burette reading is recorded, and the titre (final − initial reading) is calculated. A first, 'rough' titration is carried out quickly to find an approximate titre, then the titration is repeated more carefully at least twice, adding the acid dropwise as the rough titre volume is approached, until two accurate ('concordant') titres agreeing within 0.10 cm³ of each other are obtained.
(b) A suitable table has row headers 'Final burette reading / cm³', 'Initial burette reading / cm³' and 'Titre / cm³' (calculated as final minus initial), with separate columns headed 'Rough', 'Accurate 1' and 'Accurate 2' for the three titration runs, followed by a row for the mean titre, calculated as the average of only the two (or more) concordant accurate titres, excluding the rough result.

Reading | Rough | Accurate 1 | Accurate 2
Final burette reading / cm³ | | |
Initial burette reading / cm³ | | |
Titre / cm³ | | |
Mean titre (accurate runs) / cm³ =

(a) 1 mark: pipette used to measure 25.0 cm³ alkali into flask with indicator added; 1 mark: acid placed in burette with initial reading recorded; 1 mark: acid added swirling, quickly then dropwise near end-point, until permanent colour change; 1 mark: titration repeated for concordant results. (b) 1 mark: correct rows (final reading, initial reading, titre); 1 mark: separate columns for rough and at least two accurate runs; 1 mark: mean titre row/calculation using only concordant accurate results. All other valid, clearly described table designs will be credited.

(a) The data shows that as the temperature rises from 10°C to 90°C, the solubility of potassium chlorate steadily increases from 5 g to 46 g per 100 g of water, so solubility increases with increasing temperature.
(b) At 50°C, 100 g of water dissolves 19 g of potassium chlorate. For 40 g of water: mass dissolved = 19 × (40/100) = 19 × 0.4 = 7.6 g.
(c) At 90°C, 100 g of water can dissolve up to 46 g of potassium chlorate; since only 23 g is dissolved, the solution is not saturated at 90°C and all 23 g remains dissolved as the solution cools, until the solubility limit is reached. At 30°C, 100 g of water can only hold 11 g of potassium chlorate dissolved (saturated). The mass of crystals that forms = mass originally dissolved − mass that remains dissolved at 30°C = 23 − 11 = 12 g.

(a) 1 mark: correct trend (solubility increases with increasing temperature). (b) 1 mark: correct method (scaling 19 g by 40/100); 1 mark: correct answer 7.6 g. (c) 1 mark: correct solubility at 30°C identified (11 g/100 g water); 1 mark: correct method (23 − 11); 1 mark: correct answer 12 g.

(a) Copper(II) sulfate is added to warm (but not boiling) water, a small amount at a time, stirring continuously, until no more dissolves and a small excess of solid remains, forming a saturated solution at that temperature. Any remaining undissolved solid is removed by filtration, so that only the clear saturated solution remains. This solution is then left undisturbed, either to cool slowly to room temperature or in a warm place to allow the water to evaporate slowly over several days. As the solution becomes supersaturated, crystals gradually form and grow. Once large, well-formed crystals have grown, they are removed by filtration (or by carefully pouring off the remaining solution) and dried gently, for example between sheets of filter paper, without heating (to avoid losing water of crystallisation).
(b) When a saturated solution cools (or evaporates) slowly, crystals form gradually at relatively few nucleation sites, and there is enough time for dissolved particles to arrange themselves in an orderly way onto these few growing crystals, so each crystal has time to grow to a large size before the solution becomes fully depleted. When cooling happens quickly, many crystals begin to form simultaneously at a large number of different nucleation sites throughout the solution, so the available dissolved solute is shared out among far more crystals, each of which only has time to grow to a small size before the solute runs out.

(a) 1 mark: add solid to warm water and stir to form a saturated solution; 1 mark: filter off any undissolved excess solid; 1 mark: leave the solution to cool/evaporate slowly, undisturbed; 1 mark: filter and dry the crystals formed (without excess heating). (b) 1 mark: slow cooling/evaporation allows fewer nucleation sites and more time for crystals to grow larger; 1 mark: rapid cooling produces many nucleation sites/crystals forming at once, giving smaller crystals. All other valid answers will be credited.

(a) Between 0°C and 50°C, the solubility of sodium chloride rises only slightly, from 36 g to 37 g per 100 g of water – an increase of just 1 g. Over the same temperature range, the solubility of potassium nitrate rises much more steeply, from 13 g to 85 g per 100 g of water – an increase of 72 g. This shows that while both solubilities increase with temperature, potassium nitrate's solubility is far more strongly (sensitively) dependent on temperature than sodium chloride's, whose solubility barely changes.
(b) At 50°C, 100 g of water can dissolve a maximum of 37 g of sodium chloride. Since 40 g of sodium chloride was added, which is more than this maximum, 40 − 37 = 3 g of sodium chloride would remain undissolved as a solid. At 50°C, 100 g of water can dissolve up to 85 g of potassium nitrate. Since only 40 g of potassium nitrate was added, which is less than this maximum, all 40 g would dissolve completely, leaving no undissolved potassium nitrate.

(a) 1 mark: NaCl solubility increases only slightly with temperature; 1 mark: KNO3 solubility increases much more sharply with temperature; 1 mark: valid comparative statement (KNO3 solubility far more temperature-sensitive than NaCl). (b) 1 mark: NaCl – 3 g remains undissolved, with correct reasoning (37 g maximum at 50°C, 40 g added); 1 mark: KNO3 – none remains undissolved, with correct reasoning (85 g maximum exceeds 40 g added); 1 mark: correct comparison/conclusion stating only NaCl leaves a solid residue. All other valid answers will be credited.