2024 CCEA GCSE Chemistry 1110 Practice Paper with Answers

(a) Relative atomic mass is the weighted mean of isotope masses by abundance: \( A_r = (0.79 \times 24) + (0.10 \times 25) + (0.11 \times 26) = 18.96 + 2.50 + 2.86 = 24.32 \). (b)(i) Magnesium, atomic number 12, has electrons arranged 2,8,2. (ii) Every element in Group 2 has 2 electrons in its outer shell (this is what defines it as Group 2); since chemical reactivity is governed largely by the number and arrangement of outer-shell electrons, all Group 2 elements react in a similar way — by losing their 2 outer electrons to form a 2+ ion with a stable, noble-gas-like configuration. (c)(i) Reactivity of the Group 2 metals increases going down the group. (ii) As you go down Group 2, each successive element has an additional electron shell; this means the 2 outer electrons are progressively further from the positively charged nucleus and are shielded by more inner, filled electron shells, making them easier for the atom to lose. Since Group 2 metals react by losing these 2 outer electrons, easier loss of electrons means greater reactivity, so reactivity increases down the group. (d) Boiling point increases going down Group 0, from helium (lowest) to radon (highest), as the atoms get larger and the intermolecular (van der Waals) forces between atoms get stronger. The noble gases are very unreactive because their atoms already have a full outer shell of electrons (a stable, low-energy arrangement); since they have no tendency to lose, gain, or share electrons to become more stable, they rarely take part in chemical reactions. (e) Copper is a transition metal (transition element); transition metals characteristically form ions with more than one possible charge (such as copper(I) and copper(II)) and typically form coloured compounds and solutions, unlike most Group 1/Group 2 metal compounds. (f) Compared with the Group 1 metals, transition metals generally have much higher melting points (Group 1 metals have relatively low melting points) and are much denser (Group 1 metals have unusually low densities, with the first three even less dense than water).
Final answer: (a) 24.32; (b) 2,8,2; same number (2) of outer electrons gives similar properties; (c) reactivity increases down Group 2 as the outer electrons become easier to lose (further from nucleus, more shielding); (d) boiling point increases down Group 0; noble gases are unreactive because they already have a full/stable outer shell; (e) transition metal; (f) transition metals have higher melting points and higher densities than Group 1 metals.

(a) 1 mark for correct method (weighted sum); 1 mark for correct final answer 24.32 — max 2. (b)(i) 1 mark for '2,8,2'. (ii) 1 mark for 'same number of outer electrons (2)'; 1 mark for correctly linking this to similar reactions/ion formation — max 2. (c)(i) 1 mark for 'increases'. (ii) 1 mark for 'more shells/outer electrons further from nucleus' down the group; 1 mark for 'more shielding by inner shells'; 1 mark for correctly concluding electrons are lost more easily, increasing reactivity — max 3. (d) 1 mark for 'boiling point increases down the group'; 1 mark for 'full/stable outer shell'; 1 mark for correctly linking this to no tendency to lose/gain/share electrons, hence unreactive — max 3. (e) 1 mark for 'transition metal/element'. (f) 1 mark for 'higher melting point'; 1 mark for 'higher density' (both relative to Group 1) — max 2.

(a)(i) Magnesium bromide, MgBr₂, requires one magnesium atom to react with two bromine atoms. Magnesium (electronic structure 2,8,2) loses its 2 outer electrons to achieve the stable configuration 2,8, forming a Mg²⁺ ion; each of two separate bromine atoms (electronic structure 2,8,18,7) gains one of these electrons to complete its outer shell to 2,8,18,8, each forming a Br⁻ ion. In a dot-and-cross diagram, the 2 electrons lost by magnesium (crosses) are redrawn into the outer shells of two separate bromine atoms, with square brackets and charges shown around each resulting ion. The resulting Mg²⁺ ion and two Br⁻ ions are held together by strong electrostatic attraction between the oppositely charged ions. (ii) The ions formed are Mg²⁺ and Br⁻. (iii) Because ionic bonding extends in all directions between alternating positive and negative ions, magnesium bromide forms a giant ionic lattice. In this lattice, every ion is strongly attracted electrostatically to several oppositely charged neighbouring ions; overcoming all of these many strong attractions throughout the lattice, in order to melt the compound, requires a large amount of energy, which is why ionic compounds such as magnesium bromide typically have high melting points. (b)(i) Carbon (electronic structure 2,4) has 4 outer electrons and needs 4 more to complete its outer shell; each oxygen atom (electronic structure 2,6) has 6 outer electrons and needs 2 more. To satisfy both atoms, carbon shares 2 pairs of electrons (a double covalent bond) with each of the two oxygen atoms — using all 4 of carbon's outer electrons (2 with each oxygen) and completing each oxygen's outer shell to 8. In the dot-and-cross diagram, each double bond is shown as two shared pairs (2 dots from oxygen and 2 crosses from carbon, or vice versa) between C and each O, and each oxygen atom also retains two non-bonding lone pairs of electrons (shown as its own dots not involved in bonding). (ii) Solid carbon dioxide exists as a molecular covalent (simple molecular) structure, made up of individual CO₂ molecules held together only by weak intermolecular forces (van der Waals' forces) between separate molecules — the strong covalent bonds within each molecule are not what needs to be overcome on melting. Because these intermolecular forces between molecules are much weaker than the strong electrostatic forces holding together every ion throughout a giant ionic lattice like magnesium bromide, only a small amount of energy is needed to overcome them and allow the molecules to move past each other, so carbon dioxide has a much lower melting point than an ionic compound such as magnesium bromide.
Final answer: (a) Mg loses 2e⁻ to form Mg²⁺, each of 2 Br atoms gains 1e⁻ to form Br⁻, held by electrostatic attraction; Mg²⁺, Br⁻; giant ionic lattice — strong attraction throughout needs much energy to overcome, giving a high melting point; (b) C shares 2 pairs with each O (2 double bonds), each O keeps 2 lone pairs; molecular covalent structure — only weak intermolecular forces between molecules need to be overcome, giving a much lower melting point than an ionic lattice.

(a)(i) 1 mark for Mg losing 2 electrons; 1 mark for one Mg atom transferring one electron to each of two separate Br atoms; 1 mark for both ions reaching a stable/noble-gas configuration; 1 mark for describing the ionic bond as electrostatic attraction between oppositely charged ions — max 4. (ii) 1 mark for 'Mg²⁺'; 1 mark for 'Br⁻' — max 2. (iii) 1 mark for 'giant ionic lattice'; 1 mark for reference to strong electrostatic attraction between (many) oppositely charged ions; 1 mark for correctly linking this to requiring a large amount of energy to overcome, hence a high melting point — max 3. (b)(i) 1 mark for identifying 2 double covalent bonds (2 shared pairs with each O); 1 mark for correctly using all 4 of carbon's outer electrons; 1 mark for each oxygen atom's outer shell being completed to 8; 1 mark for showing/describing lone pairs remaining on each oxygen — max 4. (ii) 1 mark for 'molecular covalent/simple molecular structure'; 1 mark for 'weak intermolecular/van der Waals forces' between molecules; 1 mark for correctly contrasting with strong electrostatic forces throughout an ionic lattice; 1 mark for correctly concluding this gives CO₂ a much lower melting point — max 4.

(a)(i) Phenolphthalein is colourless in acidic solution and turns pink (magenta) in alkaline solution. (ii) Methyl orange is red in acidic solution and yellow in alkaline solution. (b) Applying the given classification directly: pH 2.0 falls in the 0–2 range, so Solution A is a strong acid; pH 4.5 falls in the 3–6 range, so Solution B is a weak acid; pH 10.5 falls in the 8–11 range, so Solution C is a weak alkali; pH 14.0 falls in the 12–14 range, so Solution D is a strong alkali. Since a lower pH corresponds to a higher concentration of hydrogen ions, and Solution A has the lowest pH (2.0) of the four, Solution A has the highest concentration of hydrogen ions. (c) Zinc carbonate is a carbonate reacting with a strong acid, nitric acid, so the general pattern acid + carbonate → salt + water + carbon dioxide applies, giving zinc nitrate as the salt: \( \text{ZnCO}_3(s) + 2\text{HNO}_3(aq) \rightarrow \text{Zn(NO}_3)_2(aq) + \text{H}_2\text{O(l)} + \text{CO}_2(g) \); checking atoms confirms this is balanced (Zn 1=1, C 1=1, O 3+6=2+1+2, H 2=2, N 2=2). As the reaction proceeds, the solid zinc carbonate is seen dissolving/reacting, with fizzing/effervescence as bubbles of carbon dioxide gas are released; bubbling this gas through limewater and observing it turn from colourless to milky confirms the gas is carbon dioxide. (d) Since sodium hydroxide (an alkali) and the salt sodium chloride are both soluble, the excess-solid method cannot be used, so a titration method with an indicator is used instead: the student first titrates a known volume of sodium hydroxide solution against the hydrochloric acid, using an indicator such as phenolphthalein or methyl orange, noting the exact volume of acid needed to just neutralise the alkali (shown by the indicator's colour change). Since the indicator itself would contaminate the final salt if left in the mixture, the titration is then repeated exactly (using the same measured volumes of acid and alkali that were found to react completely) but with no indicator added, giving a pure solution of sodium chloride. This pure solution is then evaporated/crystallised (heated to concentrate it, then cooled to allow crystals to form) to obtain dry sodium chloride crystals.
Final answer: (a) phenolphthalein: colourless (acid), pink (alkali); methyl orange: red (acid), yellow (alkali); (b) A=strong acid, B=weak acid, C=weak alkali, D=strong alkali; A has the highest H⁺ concentration; (c) ZnCO₃(s)+2HNO₃(aq)→Zn(NO₃)₂(aq)+H₂O(l)+CO₂(g); fizzing, gas turns limewater milky; (d) titrate with indicator to find the volumes needed, repeat without indicator, then evaporate/crystallise.

(a)(i) 1 mark for 'colourless' in acid; 1 mark for 'pink/magenta' in alkali — max 2. (ii) 1 mark for 'red' in acid; 1 mark for 'yellow' in alkali — max 2. (b) 1 mark each for correctly classifying A, B, C and D (max 4); 1 mark for correctly identifying A as having the highest H⁺ concentration — max 5. (c) 1 mark for correct formulae; 1 mark for correct balancing and state symbols; 1 mark for 'fizzing/effervescence'/gas produced; 1 mark for correct gas test (limewater turns milky) — max 4. (d) 1 mark for titrating with an indicator to find the volume needed; 1 mark for identifying a suitable indicator; 1 mark for repeating without indicator (or removing it, e.g. with charcoal) to obtain a pure solution; 1 mark for evaporation/crystallisation to obtain dry crystals — max 4.

(a) At 70 °C, a saturated solution in 40 g of water contains \( \dfrac{40}{100} \times 40 = 16 \text{ g} \) of dissolved copper(II) sulfate. At 10 °C, the maximum mass that can remain dissolved in 40 g of water is \( \dfrac{17}{100} \times 40 = 6.8 \text{ g} \). Since the solution originally held 16 g dissolved, but only 6.8 g can remain dissolved at the lower temperature, the mass that crystallises out is \( 16 - 6.8 = 9.2 \text{ g} \). (b) On a solubility curve, solubility (in g per 100 g water) is plotted on the y-axis against temperature (°C) on the x-axis; a curve that rises (slopes upward) from left to right shows that as temperature increases, solubility also increases, which is the typical pattern for most solids dissolving in water. (c) Sodium hydroxide solution is added dropwise to each unknown solution in turn, and the colour of any precipitate that forms is observed: a blue precipitate confirms Cu²⁺; a green precipitate confirms Fe²⁺; a reddish-brown (orange-brown) precipitate confirms Fe³⁺. However, Al³⁺, Zn²⁺ and Mg²⁺ all initially give a similar white precipitate with sodium hydroxide, so cannot yet be told apart by colour alone. To distinguish between these three white precipitates, more sodium hydroxide solution is added, in excess: the precipitates formed by Al³⁺ and Zn²⁺ both redissolve in excess sodium hydroxide, forming a colourless solution again, whereas the precipitate formed by Mg²⁺ does not redissolve even in excess, allowing Mg²⁺ to be identified at this stage. Finally, to distinguish Al³⁺ from Zn²⁺ (since both redissolve in excess sodium hydroxide), a fresh sample of the solution is instead tested with ammonia solution: both Al³⁺ and Zn²⁺ again give a white precipitate with a small amount of ammonia solution, but on adding excess ammonia solution, the zinc hydroxide precipitate redissolves (forming a soluble complex), while the aluminium hydroxide precipitate remains insoluble even in excess ammonia — allowing Zn²⁺ and Al³⁺ to finally be told apart. (d) Iron(III) ions react with hydroxide ions in a precipitation reaction to form insoluble iron(III) hydroxide: \( \text{Fe}^{3+}(aq) + 3\text{OH}^-(aq) \rightarrow \text{Fe(OH)}_3(s) \). (e) Copper(II) ions give a blue precipitate (copper(II) hydroxide) with sodium hydroxide solution.
Final answer: (a) 9.2 g; (b) an upward-sloping curve shows solubility increasing with temperature; (c) NaOH added dropwise — Cu²⁺ blue, Fe²⁺ green, Fe³⁺ reddish-brown precipitates identified directly; Al³⁺/Zn²⁺/Mg²⁺ all give white precipitates, distinguished by excess NaOH (Mg²⁺ precipitate insoluble, Al³⁺/Zn²⁺ dissolve) then excess ammonia (Zn²⁺ precipitate dissolves, Al³⁺ does not); (d) Fe³⁺(aq)+3OH⁻(aq)→Fe(OH)₃(s); (e) blue.

(a) 1 mark for correct mass dissolved at 70 °C (16 g); 1 mark for correct maximum mass soluble at 10 °C (6.8 g); 1 mark for correct method (subtraction); 1 mark for correct final answer (9.2 g) — max 4 (ECF applied). (b) 1 mark for 'solubility on y-axis, temperature on x-axis'; 1 mark for 'curve slopes upward/increases left to right' — max 2. (c) Banded mark scheme (6 marks, QWC assessed). Indicative content: add NaOH solution — blue ppt = Cu²⁺; green ppt = Fe²⁺; reddish-brown ppt = Fe³⁺; white ppt (Al³⁺/Zn²⁺/Mg²⁺) needs further tests; add excess NaOH — Mg²⁺ ppt insoluble, Al³⁺/Zn²⁺ ppt dissolves; use ammonia solution to distinguish Al³⁺ (insoluble in excess) from Zn²⁺ (dissolves in excess). Band A (5–6 marks): at least 5 of the indicative points, in a clear, systematic 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. (d) 1 mark for correct species/formulae; 1 mark for correct balancing (3OH⁻) and state symbols — max 2. (e) 1 mark for 'blue' — max 2 (1 mark for stating a plausible colour, 1 mark for correctly stating 'blue' specifically).

(a) Mass of water lost = mass of hydrated sample − mass of anhydrous salt = \( 2.46 - 1.20 = 1.26 \text{ g} \). (b) Moles of water lost \( = \dfrac{1.26}{18} = 0.07 \text{ mol} \); moles of anhydrous MgSO₄ remaining \( = \dfrac{1.20}{120} = 0.01 \text{ mol} \). (c) The value of x is the ratio of moles of water to moles of MgSO₄: \( x = \dfrac{0.07}{0.01} = 7 \) (consistent with the well-known formula of hydrated magnesium sulfate, MgSO₄·7H₂O, Epsom salt). (d) The percentage by mass of water of crystallisation is the mass of water lost divided by the original hydrated mass, as a percentage: \( \dfrac{1.26}{2.46} \times 100 = 51.2\% \) (to 1 decimal place). (e) Using the equation Mg + H₂SO₄ → MgSO₄ + H₂, the mole ratio of Mg to MgSO₄ is 1:1. Moles of Mg \( = \dfrac{3.6}{24} = 0.15 \text{ mol} \), so moles of MgSO₄ formed = 0.15 mol. Mass of MgSO₄ \( = 0.15 \times 120 = 18.0 \text{ g} \). (f) In practice, some of the product is often lost during the experiment — for example, some magnesium sulfate solution may be lost when filtering off any unreacted magnesium, or some solid may be lost when transferring the crystallised product between pieces of apparatus, meaning the actual mass obtained is usually somewhat less than the maximum (theoretical) mass calculated.
Final answer: (a) 1.26 g; (b) 0.07 mol water, 0.01 mol MgSO₄; (c) x = 7; (d) 51.2%; (e) 18.0 g; (f) e.g. loss of product during filtration/transfer, or incomplete reaction.

(a) 1 mark for 1.26 g. (b) 1 mark for correct moles of water (0.07); 1 mark for correct moles of MgSO₄ (0.01); 1 mark for both correct — max 3 (award 2 marks total if only method shown with one arithmetic slip, ECF applied). (c) 1 mark for correct method (dividing moles of water by moles of MgSO₄); 1 mark for correct final answer x = 7 — max 2 (ECF from (b)). (d) 1 mark for correct method (mass of water ÷ original mass × 100); 1 mark for correct substitution; 1 mark for correct final answer 51.2% — max 3. (e) 1 mark for correct moles of Mg (0.15); 1 mark for correctly applying the 1:1 mole ratio; 1 mark for correct method (moles × Mr); 1 mark for correct final answer 18.0 g — max 4 (ECF applied). (f) 1 mark for any valid, chemically sound reason for a lower-than-theoretical yield — max 2 (1 mark for a valid idea, 1 mark for a clear, correctly linked explanation).

(a) The reactivity series for these metals, from most to least reactive, is: potassium (K), sodium (Na), calcium (Ca), magnesium (Mg), aluminium (Al), zinc (Zn), iron (Fe), copper (Cu). (b)(i) A small piece of magnesium ribbon is placed in a hard glass (combustion) tube, and steam is passed over it — commonly generated by heating water in a separate flask and directing the steam along the tube using a delivery tube, with the magnesium itself gently heated using a Bunsen burner. As the reaction occurs, hydrogen gas (mixed with any excess steam) passes out of the tube through a further delivery tube and can be collected over water in an inverted test tube or gas jar. The gas collected is tested by removing the tube (keeping it inverted) and quickly applying a lighted splint to its mouth; a squeaky 'pop' confirms the gas is hydrogen. Because hot magnesium reacting with steam can produce hydrogen gas which is flammable/potentially explosive when mixed with air, the experiment should be carried out behind a safety screen, with eye protection worn throughout. (ii) Using the equation Mg + H₂O → MgO + H₂, the mole ratio of Mg to MgO is 1:1. Moles of Mg \( = \dfrac{3.6}{24} = 0.15 \text{ mol} \), so moles of MgO formed = 0.15 mol. Mass of MgO \( = 0.15 \times 40 = 6.0 \text{ g} \). (c) Oxidation is defined, in terms of electrons, as the loss of electrons by a species; reduction is defined as the gain of electrons by a species. In the reaction of magnesium with steam, magnesium atoms lose 2 electrons each to form Mg²⁺ ions (within MgO), so magnesium is oxidised; the oxygen atoms (originally in H₂O) gain these electrons to form O²⁻ ions (within MgO), so oxygen is reduced. (d) The rate and readiness with which a metal reacts with water/steam is a direct indicator of its reactivity: magnesium reacting reasonably well with steam (and to some extent with cold water) shows it is fairly reactive; iron reacting only slowly with cold water, but reacting properly with the more energetic conditions of steam, shows it is less reactive than magnesium but still more reactive than a metal that does not react with water/steam at all; copper's complete lack of reaction with either cold water or steam shows it is the least reactive of the three. Together, these observations place the three metals in the order magnesium (most reactive) > iron > copper (least reactive), consistent with their known positions in the reactivity series. (e) A metal's position in the reactivity series reflects how strongly it holds onto the electrons in its compounds — the more reactive the metal, the more strongly its ions are bonded within their ore, and the more energy is needed to extract the pure metal. Aluminium, being a reactive metal high in the series, forms very stable compounds that cannot be reduced using a chemical reducing agent like carbon; instead, a large amount of electrical energy is used to force electrolysis to occur, breaking the compound down into aluminium and oxygen. Iron, being a less reactive metal lower in the series, forms compounds that are less stable and can be more easily reduced; carbon (or carbon monoxide), a cheaper reducing agent, is reactive enough to remove the oxygen from iron ore (haematite) directly, avoiding the greater cost and energy requirement of electrolysis.
Final answer: (a) K>Na>Ca>Mg>Al>Zn>Fe>Cu; (b) heat Mg in a tube with steam passed over, collect gas over water, test with a lit splint (pop confirms H₂), use a safety screen; 6.0 g MgO; (c) oxidation=loss of electrons, reduction=gain of electrons; Mg oxidised, O reduced; (d) confirms Mg>Fe>Cu in reactivity; (e) more reactive metals (Al) need electrolysis since their compounds are too stable for carbon to reduce; less reactive metals (Fe) can be reduced more cheaply using carbon.

(a) 1 mark for correct order (all 8 metals correctly placed); 1 mark awarded within this if at least 6 are correctly ordered relative to each other, even with 1–2 out of place — max 2 (award full marks for the fully correct order K>Na>Ca>Mg>Al>Zn>Fe>Cu). (b)(i) 1 mark for heating Mg with steam passed over it (e.g. in a combustion tube); 1 mark for collecting the gas (e.g. over water); 1 mark for correct gas test (lit splint, 'pop'); 1 mark for a valid safety precaution (safety screen/eye protection) — max 4. (ii) 1 mark for correct moles of Mg (0.15); 1 mark for correctly applying the 1:1 mole ratio; 1 mark for correct method (moles × Mr); 1 mark for correct final answer 6.0 g — max 4 (ECF applied). (c) 1 mark for 'oxidation = loss of electrons'; 1 mark for 'reduction = gain of electrons'; 1 mark for 'magnesium is oxidised'; 1 mark for 'oxygen is reduced' — max 4. (d) 1 mark for correctly concluding Mg is more reactive than Fe; 1 mark for correctly concluding Fe is more reactive than Cu; 1 mark for a valid supporting reason drawn from the given observations (e.g. Mg reacts with steam readily/some reaction with cold water; Fe only reacts slowly with cold water but readily with steam; Cu does not react with either); 1 mark for stating the overall order Mg>Fe>Cu — max 4. (e) 1 mark for 'more reactive metals form more stable compounds, needing more energy/a stronger method to extract'; 1 mark for correctly linking aluminium's high reactivity to needing electrolysis; 1 mark for correctly linking iron's lower reactivity to being extractable using (cheaper) chemical reduction with carbon; 1 mark for a further valid, well-explained point — max 4.

(a) To carry out a fair test, the student should first weigh a conical flask containing a fixed, known mass of calcium carbonate (starting with large marble chips) on a mass balance and record this starting mass. A fixed volume and concentration of dilute hydrochloric acid (kept the same in every repeat) is then added to the flask; the flask is quickly plugged with cotton wool, which allows carbon dioxide gas to escape (so the mass loss reflects gas released) while preventing acid spray from being lost from the flask (which would also cause a loss in mass unrelated to the reaction). A stopwatch is started at the same moment the acid is added, and the mass shown on the balance is recorded at regular time intervals (for example every 20 seconds) until the mass stops decreasing, showing the reaction has finished. This entire procedure is then repeated using the same mass of calcium carbonate and the same volume and concentration of acid, changing only the surface area of the calcium carbonate — first using smaller marble chips, then using powdered calcium carbonate — keeping every other variable (mass of carbonate, acid volume/concentration, temperature) the same so that surface area is the only variable being tested. The results for each surface area are then plotted as a graph of mass loss against time, and the rate of reaction for each (for example, from the initial gradient of each graph) can be compared. (b) The rate over the first 20 seconds is the change in mass divided by the change in time: \( \text{rate} = \dfrac{150.00 - 149.52}{20 - 0} = \dfrac{0.48}{20} = 0.024 \text{ g/s} \). (c) Comparing the masses recorded, the mass falls steadily from 150.00 g at t = 0 s down to 149.04 g at t = 80 s, but then stays exactly the same (149.04 g) at t = 100 s; since the mass is no longer decreasing between these two readings, this shows that carbon dioxide gas is no longer being produced, so the reaction is complete by approximately 80–100 s. (d) For the same total mass of calcium carbonate, powdered calcium carbonate has a much larger total surface area exposed to the acid than the same mass in the form of large chips, since breaking a solid into smaller pieces greatly increases the surface area to volume ratio. A larger exposed surface area means more calcium carbonate particles are available at the surface to be struck by acid particles at any moment, so the frequency of successful collisions between acid particles and calcium carbonate particles is higher, giving a faster initial rate of reaction for the powder compared with the large chips.
Final answer: (a) fixed mass of CaCO₃ + fixed volume/conc. acid in a cotton-wool-plugged flask on a balance, recording mass loss over time, repeated with different surface areas (chips vs powder) as the only changed variable; (b) 0.024 g/s; (c) complete by about 80–100 s, since mass stops changing; (d) powder has a greater surface area, giving more frequent collisions and a faster rate.

(a) Banded mark scheme (6 marks, QWC assessed). Indicative content: fixed mass of calcium carbonate and fixed volume/concentration of acid used each time; flask plugged with cotton wool (to release gas but limit spray loss); mass recorded at regular time intervals on a balance; timed with a stopwatch; only surface area (chip size vs powder) changed between repeats, all other variables controlled; results compared via a graph of mass loss against time / initial gradient. Band A (5–6 marks): at least 5 of the indicative points, in a clear, logical, fair-test 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. (b) 1 mark for correct method (change in mass ÷ change in time); 1 mark for correct substitution (0.48/20); 1 mark for correct final answer 0.024 g/s — max 3. (c) 1 mark for a time in the range 80–100 s; 1 mark for correct reasoning (mass no longer changes/decreasing) — max 2. (d) 1 mark for 'greater surface area (for the powder)'; 1 mark for correctly linking this to more frequent collisions and a faster rate — max 2.

(a)(i) A homologous series is a family of organic compounds sharing the same general formula, showing similar chemical properties (often due to a shared functional group, or lack of one), showing a gradual change (gradation) in physical properties such as boiling point as chain length increases, and differing from one member to the next by a CH₂ unit. (ii) The alkanes follow the general formula CₙH₂ₙ₊₂; hexane (C₆H₁₄) has one fewer carbon atom in pentane, whose formula is found by \( C_5H_{2(5)+2} = C_5H_{12} \). (b)(i) Cracking breaks down larger, saturated hydrocarbon molecules (alkanes), for which there is less demand as fuels, into smaller, more useful molecules; because the products contain more hydrogen relative to carbon than would fit a fully saturated chain of that length, some products are unsaturated alkenes. It requires a high temperature, often together with a catalyst to allow the process to occur at a lower temperature than would otherwise be needed. (ii) Balancing carbon and hydrogen: C₆H₁₄ has 6 C and 14 H; C₄H₁₀ has 4 C and 10 H; the remaining product needs \( 6-4=2 \) carbon atoms and \( 14-10=4 \) hydrogen atoms, giving C₂H₄ (ethene). (c)(i) Bromine water, an orange/brown solution, is added to a sample of each hydrocarbon and shaken; the alkene reacts with bromine water across its C=C double bond in an addition reaction, decolourising it (turning it from orange/brown to colourless); the alkane, having no C=C double bond, does not react, so the bromine water remains orange/brown. (ii) It is the C=C double covalent bond (in but-2-ene) that is broken (opened) during the addition reaction with bromine. (d)(i) To produce ethanol by fermentation, a sugar solution is mixed with yeast (which provides the enzymes needed to catalyse the reaction), and the mixture is kept at a warm temperature, typically around 25–35 °C (warm enough for the yeast's enzymes to work efficiently without being denatured by excessive heat), and crucially in anaerobic conditions (the absence of oxygen), since oxygen would otherwise allow unwanted further oxidation/microbial spoilage of the ethanol produced. (ii) Ethanoic acid reacting with the metal carbonate magnesium carbonate follows the general pattern for an acid + carbonate reaction, producing a salt (magnesium ethanoate), water and carbon dioxide gas; as the reaction occurs, the solid magnesium carbonate is seen to dissolve/react, with fizzing/effervescence as bubbles of carbon dioxide gas are given off. Since ethanoic acid, CH₃COOH, provides one replaceable hydrogen ion per molecule and magnesium carbonate provides a 2+ magnesium ion, two molecules of ethanoic acid are needed per magnesium ion: \( 2\text{CH}_3\text{COOH(aq)} + \text{MgCO}_3(s) \rightarrow (\text{CH}_3\text{COO})_2\text{Mg(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2(g) \). (e) Carboxylic acids such as ethanoic acid are classed as weak acids because, unlike strong acids (which are completely ionised in water), only a small proportion of the carboxylic acid molecules actually ionise (split up into ions) when dissolved in water — most of the acid remains as complete, un-ionised molecules in solution. (f) The presence of soot (carbon particles) and carbon monoxide, rather than only carbon dioxide and water, shows that the alcohol did not burn completely — there was insufficient oxygen available for complete combustion, so incomplete combustion occurred instead. This is dangerous primarily because carbon monoxide is a toxic gas that binds to haemoglobin in red blood cells more readily than oxygen does, reducing the blood's capacity to carry oxygen around the body, which can cause illness or, in serious cases, death — and is especially hazardous because carbon monoxide is colourless and odourless, making it hard to detect without a specific alarm.
Final answer: (a) homologous series definition; CₙH₂ₙ₊₂; pentane C₅H₁₂; (b) cracking breaks large alkanes into smaller/more useful molecules incl. alkenes, high temperature (+catalyst); C₂H₄; (c) bromine water decolourised by an alkene, not by an alkane; C=C double bond broken; (d) sugar+yeast, warm, anaerobic; fizzing/CO₂ given off, MgCO₃ dissolves; 2CH₃COOH(aq)+MgCO₃(s)→(CH₃COO)₂Mg(aq)+H₂O(l)+CO₂(g); (e) only partially ionised in solution; (f) soot/CO shows incomplete combustion (insufficient oxygen); CO is toxic, reduces blood's oxygen-carrying capacity.

(a)(i) 1 mark for 'same general formula/similar chemical properties'; 1 mark for 'gradation in physical properties'/'differ by CH₂' — max 2. (ii) 1 mark for CₙH₂ₙ₊₂; 1 mark for C₅H₁₂ — max 2. (b)(i) 1 mark for 'breakdown of larger/saturated alkanes into smaller, more useful molecules, some unsaturated/alkenes'; 1 mark for 'high temperature'; 1 mark for 'catalyst' as an additional valid condition — max 3. (ii) 1 mark for C₂H₄. (c)(i) 1 mark for 'add bromine water'; 1 mark for 'decolourised' by the alkene; 1 mark for 'stays orange/brown' with the alkane — max 3. (ii) 1 mark for 'C=C double (covalent) bond'. (d)(i) 1 mark for 'sugar + yeast'; 1 mark for 'warm (approx. 25–35 °C)'; 1 mark for 'anaerobic/no oxygen' — max 3. (ii) 1 mark for 'fizzing/effervescence'/gas given off; 1 mark for the solid dissolving/reacting; 1 mark for correct formulae; 1 mark for correct balancing (2CH₃COOH) — max 4. (e) 1 mark for 'only partially ionised in solution'. (f) 1 mark for correctly identifying incomplete combustion/insufficient oxygen; 1 mark for a valid danger of carbon monoxide (toxic, reduces oxygen-carrying capacity of blood) — max 2.

(a) In its solid form, the Al³⁺ and O²⁻ ions in aluminium oxide are held rigidly in a fixed lattice and are not free to move, so solid aluminium oxide cannot conduct electricity or undergo electrolysis; the ions must be made free to move, either by melting the compound or dissolving it, before electrolysis is possible. However, pure aluminium oxide has an extremely high melting point, so melting it on its own would require an impractically large amount of energy (and cost) to maintain it in the molten state throughout the process. Dissolving the aluminium oxide in molten cryolite instead significantly lowers the temperature at which the mixture becomes and remains molten, while still freeing the aluminium and oxide ions to move and carry charge, allowing electrolysis to proceed at a lower, more economical temperature. (b) At the cathode, aluminium ions gain electrons (are reduced) to form aluminium metal: \( \text{Al}^{3+}(l) + 3e^- \rightarrow \text{Al}(l) \); at the anode, oxide ions lose electrons (are oxidised) to form oxygen gas: \( 2\text{O}^{2-}(l) \rightarrow \text{O}_2(g) + 4e^- \). (c) During electrolysis, oxygen gas is produced at the carbon anodes; because the process is carried out at a high operating temperature, this oxygen reacts directly with the carbon of the anodes, converting them into carbon dioxide gas and gradually burning them away; consequently, the anodes are steadily worn down and need to be periodically replaced to keep the process running efficiently. (d) Extracting new aluminium from bauxite ore by electrolysis is an extremely energy-intensive process, since it requires a continuous, large electrical current to be passed through the molten mixture over a long period. Recycling aluminium, in contrast, simply requires melting down aluminium metal that has already been extracted, which uses only a small fraction of the energy needed for full extraction from ore. Recycling therefore saves a significant amount of energy (and the associated cost/environmental impact of generating that electricity), and also reduces the amount of bauxite ore that needs to be mined (reducing the environmental damage of mining) and the amount of aluminium waste that would otherwise go to landfill. (e) Carbon can only be used to extract a metal by chemical reduction if the metal is less reactive than carbon itself, since a less reactive element cannot displace/reduce a more reactive one from its compound. Aluminium is more reactive than carbon, meaning aluminium oxide is too chemically stable to be reduced by carbon under normal conditions, so a more powerful method — supplying electrical energy directly via electrolysis — is required to extract aluminium instead.
Final answer: (a) solid ions can't move/conduct; dissolving in cryolite lowers the melting point needed, saving energy, while freeing the ions to move; (b) Al³⁺(l)+3e⁻→Al(l); 2O²⁻(l)→O₂(g)+4e⁻; (c) anodes react with the oxygen produced, burning away as CO₂; (d) recycling uses far less energy than extraction from ore, and reduces mining/waste; (e) aluminium is more reactive than carbon, so carbon cannot reduce its oxide — electrolysis is needed instead.

(a) 1 mark for 'solid ions are fixed/cannot move to conduct'; 1 mark for 'pure Al₂O₃ has a very high melting point'; 1 mark for correctly explaining cryolite lowers the (energy-costly) melting/operating temperature while still allowing ion movement — max 3. (b) 1 mark for correct cathode half-equation (species, charge, balancing); 1 mark for correct state symbol(s); 1 mark for correct anode half-equation (species, charge, balancing); 1 mark for correct state symbol(s) — max 4. (c) 1 mark for 'oxygen produced at the anode reacts with the carbon'; 1 mark for 'forming carbon dioxide'; 1 mark for correctly concluding the anode is worn away/needs replacing — max 3. (d) 1 mark for 'recycling uses much less energy than extraction from ore'; 1 mark for 'reduces the amount of ore that needs to be mined'; 1 mark for 'reduces waste (e.g. to landfill)' — max 3. (e) 1 mark for 'aluminium is more reactive than carbon'; 1 mark for correctly concluding carbon therefore cannot reduce/displace aluminium from its oxide — max 2.

(a) The student places the limestone sample in a conical flask and adds the excess dilute hydrochloric acid; the flask is quickly sealed with a bung carrying a delivery tube, connected to a gas syringe, so that no gas is lost before it can be measured. As the reaction proceeds, the carbon dioxide gas produced flows along the delivery tube into the gas syringe, pushing its plunger outward along the syringe's graduated scale; once the reaction is complete (the plunger stops moving), the total volume of gas collected can be read directly from the syringe. (b) First, the moles of calcium carbonate are calculated, assuming the whole 5.00 g sample is pure: \( \text{moles CaCO}_3 = \dfrac{5.00}{100} = 0.05 \text{ mol} \). From the balanced equation, the mole ratio of CaCO₃ to CO₂ is 1:1, so moles of CO₂ = 0.05 mol. Using the fact that 1 mole of any gas occupies 24 dm³ at room temperature and pressure: \( \text{volume of CO}_2 = 0.05 \times 24 = 1.2 \text{ dm}^3 \) (1200 cm³). (c) Percentage purity compares the actual volume of gas collected with the theoretical (maximum) volume calculated in (b), assuming any gas produced comes only from the pure calcium carbonate present (the insoluble impurity does not react): \( \text{percentage purity} = \dfrac{\text{actual volume}}{\text{theoretical volume}} \times 100 = \dfrac{1080}{1200} \times 100 = 90.0\% \). (d) The moles of sodium hydroxide are found first: \( \text{moles NaOH} = \text{concentration} \times \text{volume (dm}^3) = 0.150 \times \dfrac{24.0}{1000} = 0.150 \times 0.0240 = 0.0036 \text{ mol} \). Since the neutralisation reaction NaOH + HCl → NaCl + H₂O shows a 1:1 mole ratio, moles of HCl = moles of NaOH = 0.0036 mol. The concentration of the hydrochloric acid is then found using \( \text{concentration} = \dfrac{\text{moles}}{\text{volume (dm}^3)} = \dfrac{0.0036}{20.0/1000} = \dfrac{0.0036}{0.0200} = 0.180 \text{ mol/dm}^3 \).
Final answer: (a) limestone + acid in a sealed conical flask, gas collected and measured in a gas syringe via a delivery tube; (b) 1.2 dm³ (1200 cm³); (c) 90.0% pure; (d) 0.180 mol/dm³.

(a) 1 mark for 'conical flask' with limestone and acid; 1 mark for 'bung and delivery tube' sealing the flask; 1 mark for 'gas syringe' correctly connected to collect the gas; 1 mark for reading the volume from the syringe's scale (once the plunger stops moving/reaction is complete) — max 4. (b) 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 (or 1200 cm³) — max 4. (c) 1 mark for correct method (actual ÷ theoretical × 100), with theoretical converted to the same units as actual (cm³); 1 mark for correct substitution (1080/1200); 1 mark for correct final answer 90.0%; 1 mark for a numerical answer with the correct unit/form (percentage, to an appropriate number of significant figures) — max 4 (ECF applied from (b)). (d) 1 mark for correct moles of NaOH (0.0036); 1 mark for correctly applying the 1:1 mole ratio; 1 mark for correct method (moles ÷ volume in dm³); 1 mark for correct final answer 0.180 mol/dm³ — max 4.

(a) Le Châtelier's Principle states that if a system at equilibrium experiences a change in conditions, the equilibrium will shift in the direction that acts to oppose/counteract that change. Continuously removing ammonia gas as it forms reduces the concentration of ammonia present in the equilibrium mixture; the system responds by shifting the position of equilibrium towards the side that produces more ammonia (the forward, exothermic reaction), since this is the direction that opposes/replaces the loss of ammonia. Because ammonia is continually being removed, the equilibrium never actually re-establishes itself with a constant ammonia concentration but is instead constantly being 'pulled' towards producing more product; overall, this results in a greater total yield of ammonia being produced than if the ammonia were simply allowed to remain in the mixture. (b) Atom economy for the formation of the desired product, calcium oxide, is calculated using \( \text{atom economy} = \dfrac{\text{mass of desired product}}{\text{total mass of products}} \times 100 \). The total mass of products is the mass of CaO plus the mass of CO₂: \( 56 + 44 = 100 \) (which, for this particular reaction, is conveniently equal to the relative formula mass of the single reactant, CaCO₃, since no atoms are lost overall). Atom economy \( = \dfrac{56}{100} \times 100 = 56.0\% \). (c) Using \( \Delta H = \text{(energy to break bonds in reactants)} - \text{(energy released forming bonds in products)} \): bonds broken (reactants) = 2 mol H–H + 1 mol O=O \( = (2 \times 436) + 498 = 872 + 498 = 1370 \text{ kJ} \); bonds formed (products) = 2 mol H₂O, each containing 2 O–H bonds, so 4 mol O–H total \( = 4 \times 463 = 1852 \text{ kJ} \); overall energy change \( = 1370 - 1852 = -482 \text{ kJ/mol} \) (for the reaction as written, involving 2 mol of H₂). Since the value is negative (more energy is released forming the new O–H bonds than is needed to break the H–H and O=O bonds), the reaction is exothermic.
Final answer: (a) removing ammonia shifts equilibrium further towards the product side (Le Châtelier), increasing the overall yield; (b) atom economy = 56.0%; (c) ΔH = −482 kJ/mol, exothermic.

(a) 1 mark for correctly stating/applying Le Châtelier's Principle (equilibrium shifts to oppose the change); 1 mark for 'removing ammonia reduces its concentration'; 1 mark for correctly identifying the equilibrium shifts towards the product/forward side; 1 mark for correctly concluding this increases the overall yield of ammonia — max 4. (b) 1 mark for correct total mass of products (100); 1 mark for correct method (desired product mass ÷ total product mass × 100); 1 mark for correct substitution (56/100); 1 mark for correct final answer 56.0% — max 4. (c) 1 mark for correct bonds-broken total (1370); 1 mark for correct bonds-formed total (1852, correctly identifying 4 mol O–H bonds); 1 mark for correct method (broken − formed) giving −482 kJ/mol; 1 mark for correctly stating 'exothermic' — max 4 (ECF applied).

(a) To carry out a flame test, the student first cleans a nichrome wire by dipping it into concentrated hydrochloric acid, then dips the clean wire into a sample of the solid compound, so a small amount sticks to it; the wire (with the sample) is then held in a hot, roaring/blue Bunsen flame, and the colour produced in the flame is observed and recorded. If the compound contains lithium ions, the flame would appear crimson. (b) Since pure substances (unlike mixtures) melt and boil at a single, specific temperature, the student could measure the melting point of the solid liquid (using a thermometer as it is cooled/frozen) or its boiling point (as it is heated); pure water melts at exactly 0 °C and boils at exactly 100 °C, so obtaining these exact values would support that the liquid is pure water. As a simpler chemical test, the student could instead add a small amount of the liquid to white, anhydrous copper(II) sulfate; if the liquid contains water, the white anhydrous copper(II) sulfate turns blue as it becomes hydrated. (c) A white precipitate formed with silver nitrate solution is the characteristic result for chloride ions, which react with silver ions to form insoluble silver chloride: \( \text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl(s)} \) (a cream precipitate would instead indicate bromide, and a yellow precipitate would indicate iodide).
Final answer: (a) clean nichrome wire dipped in conc. HCl then the sample, held in a Bunsen flame — lithium gives a crimson flame; (b) test melting point (0°C)/boiling point (100°C), or add to anhydrous copper(II) sulfate (turns blue) to confirm water; (c) chloride, Cl⁻; Ag⁺(aq)+Cl⁻(aq)→AgCl(s).

(a) 1 mark for 'nichrome wire' cleaned in concentrated hydrochloric acid; 1 mark for dipping the wire in the sample and holding it in a Bunsen flame; 1 mark for observing the flame colour; 1 mark for 'crimson' as the lithium flame colour — max 4. (b) 1 mark for a valid property tested (melting point or boiling point, or use of anhydrous copper(II) sulfate); 1 mark for the correct expected result for that property (0 °C / 100 °C, or 'turns blue'); 1 mark for a second valid detail/property — max 3. (c) 1 mark for 'chloride'/'Cl⁻'; 1 mark for correct ionic species; 1 mark for correct formula (AgCl); 1 mark for correct state symbols — max 4.

(a) The disappearing cross experiment is set up by placing a conical flask directly on top of a sheet of paper marked with a clear black cross underneath the flask. A fixed, measured volume of sodium thiosulfate solution — diluted with a measured volume of water to give the required concentration, while keeping the total volume of liquid the same in every repeat — is added to the flask. A fixed volume of dilute hydrochloric acid (the same volume and concentration used in every repeat) is then added to the flask, and a stopwatch is started at the same moment, with the flask given a brief swirl to mix the solutions. The student looks down vertically through the top of the open flask at the black cross beneath it; as the reaction proceeds, a cloudy yellow precipitate of sulfur gradually forms, making the solution increasingly opaque, and the stopwatch is stopped at the exact moment the cross can no longer be seen through the solution, with this time recorded. This whole procedure is then repeated using different concentrations of sodium thiosulfate solution (made by diluting with different volumes of water, while keeping the total volume of liquid, and the volume and concentration of the acid, the same each time), so that concentration of sodium thiosulfate is the only variable being deliberately changed between repeats — this ensures a fair test. To improve consistency, the same person should judge when the cross disappears in every repeat, and the flask/paper combination should be viewed against the same background/lighting conditions each time. (b) Rate is calculated as \( \dfrac{1}{\text{time}} \) for each concentration: at 10 g/dm³, rate \( = \dfrac{1}{200} = 0.005 \text{ s}^{-1} \); at 20 g/dm³, rate \( = \dfrac{1}{100} = 0.010 \text{ s}^{-1} \); at 30 g/dm³, rate \( = \dfrac{1}{67} \approx 0.015 \text{ s}^{-1} \); at 40 g/dm³, rate \( = \dfrac{1}{50} = 0.020 \text{ s}^{-1} \). Examining these values, doubling the concentration from 10 to 20 g/dm³ doubles the rate (0.005 to 0.010), and doubling again from 20 to 40 g/dm³ doubles the rate again (0.010 to 0.020); this pattern shows that the rate of reaction is directly proportional to the concentration of sodium thiosulfate — a graph of rate against concentration would give a straight line passing through the origin. (c) Manganese(IV) oxide acts as a catalyst by providing an alternative pathway for the decomposition of hydrogen peroxide 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 this energy barrier means a much greater proportion of the particle collisions occurring at any given temperature now have enough energy to react successfully; because more collisions are successful per second, the reaction proceeds at a faster rate with the catalyst present than without it. Despite taking part in providing this alternative pathway, the catalyst itself is not permanently used up or chemically changed by the reaction — the same mass and chemical identity of manganese(IV) oxide remains at the end of the reaction as at the start, so it could, in principle, be reused. (d) Judging the precise instant at which the cross becomes completely invisible through the increasingly cloudy solution is inherently subjective — different observers (or even the same observer on different occasions) might judge the cross to have 'just disappeared' at slightly different moments, introducing an element of human/random error into each recorded time, which could affect the precision and reliability of the results.
Final answer: (a) fixed volumes of diluted thiosulfate + acid in a flask over a cross, timing until the cross disappears, repeated with only concentration varied; (b) rates 0.005, 0.010, 0.015, 0.020 s⁻¹; rate is directly proportional to concentration; (c) catalyst provides a lower-activation-energy pathway so more collisions succeed, faster rate; catalyst is chemically unchanged at the end; (d) judging exactly when the cross disappears is subjective, a source of random/human error.

(a) 1 mark for fixed/measured volume of thiosulfate (diluted as needed); 1 mark for total volume of liquid kept the same; 1 mark for fixed volume/concentration of acid; 1 mark for timing from adding acid to the cross disappearing; 1 mark for viewing the cross from directly above through the flask; 1 mark for only concentration changed between repeats (fair test); 1 mark for a valid consistency measure (e.g. same observer); 1 mark for a further valid procedural detail — max 8. (b) 1 mark for at least two correct rate values; 1 mark for all four correct rate values; 1 mark for correctly identifying rate increases as concentration increases; 1 mark for correctly identifying the proportional relationship (rate doubles as concentration doubles); 1 mark for correctly describing this as direct proportionality (straight line through the origin) — max 5. (c) 1 mark for 'lower activation energy pathway'; 1 mark for correctly linking this to a greater proportion of successful collisions; 1 mark for correctly concluding the rate is faster; 1 mark for 'the catalyst is chemically unchanged/not used up' at the end — max 4. (d) 1 mark for identifying the judgement of the exact disappearance point as subjective; 1 mark for correctly linking this to random/human error — max 2.

(a) The crucible, with the coiled strip of magnesium ribbon placed inside it and its lid resting on top, is set on a pipeclay triangle supported by a tripod, and heated strongly using a Bunsen burner. As the magnesium begins to burn, the student uses tongs to lift the lid slightly at regular intervals, allowing a small amount of fresh air (oxygen) to enter the crucible to keep the reaction going, before quickly replacing the lid. The lid must not be left fully open throughout, because this would allow the burning magnesium and the fine, white magnesium oxide 'smoke' it produces to escape from the crucible, resulting in a loss of product and giving an inaccurately low final mass; however, the lid also cannot be left fully closed throughout, since this would starve the reaction of the oxygen it needs to react completely, leaving some magnesium unreacted. Lifting the lid only periodically strikes a balance, supplying enough oxygen for the reaction to reach completion while minimising the loss of product as smoke. (b) The mass of oxygen combined is the difference between the final mass of magnesium oxide and the original mass of magnesium: \( 3.00 - 1.80 = 1.20 \text{ g} \). (c) Using \( \text{moles} = \dfrac{\text{mass}}{A_r} \): moles Mg \( = \dfrac{1.80}{24} = 0.075 \text{ mol} \); moles O \( = \dfrac{1.20}{16} = 0.075 \text{ mol} \). (d) Dividing both values by the smaller value (0.075): Mg \( \frac{0.075}{0.075} = 1 \), O \( \frac{0.075}{0.075} = 1 \); the simplest whole-number ratio of Mg:O is 1:1, giving the empirical formula MgO. (e) Reweighing the crucible and its contents repeatedly — heating, allowing it to cool, then weighing, and repeating this cycle — and continuing until two consecutive mass readings are identical, provides direct evidence from the data itself that the reaction has gone to completion, since a further increase in mass would only occur if unreacted magnesium were still combining with more oxygen. Heating for a single, fixed period of time instead relies on an assumption about how long the reaction takes, which may not be accurate for every sample or set of heating conditions — heating for too short a time could leave the reaction incomplete (giving an inaccurately low final mass and an incorrect empirical formula), while heating for far longer than necessary would simply waste time and fuel without any additional benefit.
Final answer: (a) crucible on a pipeclay triangle/tripod, heated strongly with the lid lifted periodically (using tongs) to admit air without losing product as smoke; (b) 1.20 g; (c) 0.075 mol Mg, 0.075 mol O; (d) empirical formula MgO; (e) reweighing to constant mass directly confirms the reaction is complete, unlike an assumed fixed heating time.

(a) 1 mark for 'crucible on a pipeclay triangle/tripod'; 1 mark for heating strongly with a Bunsen burner; 1 mark for lifting the lid periodically (with tongs); 1 mark for correctly explaining fully open risks losing product as smoke; 1 mark for correctly explaining fully closed would starve the reaction of oxygen; 1 mark for correctly concluding periodic lifting balances both concerns — max 6. (b) 1 mark for 1.20 g. (c) 1 mark for correct method for Mg; 1 mark for correct method for O; 1 mark for both correct values (0.075 each) — max 3. (d) 1 mark for correctly dividing by the smaller value; 1 mark for stating 'MgO' as the empirical formula — max 2. (e) 1 mark for 'confirms the reaction is complete/no more mass is being gained'; 1 mark for correctly contrasting this with the uncertainty of a fixed heating time; 1 mark for a further valid point (e.g. risk of stopping too early, or wasting time if too long) — max 3.

(a) To collect and measure the gas, the container holding the copper(II) carbonate (a boiling tube or similar, suitable for heating) is fitted with a bung carrying a delivery tube, which is connected to a gas syringe; as the copper(II) carbonate decomposes on heating, the carbon dioxide gas produced travels along the delivery tube and collects in the gas syringe, pushing its plunger outward, so the volume of gas produced can be read directly from the syringe's scale. The reaction can be confirmed to be complete in two ways: the plunger of the gas syringe stops moving (no more gas is being produced), and/or the mass of the solid remaining in the container, when repeatedly reweighed after further heating and cooling, stops decreasing (i.e. reaches a constant mass), showing no further carbon dioxide is being lost from the solid. (b) Moles of copper(II) carbonate is found using \( \text{moles} = \dfrac{\text{mass}}{M_r} = \dfrac{12.4}{124} = 0.1 \text{ mol} \). From the balanced equation, the mole ratio of CuCO₃ to CO₂ is 1:1, so moles of CO₂ produced = 0.1 mol. Mass of CO₂ \( = 0.1 \times 44 = 4.4 \text{ g} \). (c) The percentage by mass of carbon dioxide in the original sample is the mass of CO₂ produced divided by the original mass of the sample, as a percentage: \( \dfrac{4.4}{12.4} \times 100 = 35.5\% \) (to 1 decimal place). (d) Copper(II) oxide, the solid product remaining once all the copper(II) carbonate has decomposed, is a black solid, so the crucible would contain a black powder/solid at the end of the reaction.
Final answer: (a) delivery tube from the heated container to a gas syringe; reaction complete once the plunger stops moving/mass of solid is constant; (b) 0.1 mol CuCO₃, 4.4 g CO₂; (c) 35.5%; (d) black (copper(II) oxide).

(a) 1 mark for 'delivery tube' connecting the heated container to a gas syringe; 1 mark for the gas collecting in/measured by the syringe; 1 mark for a valid indicator that gas production has stopped (plunger stops moving); 1 mark for a valid alternative indicator (mass of solid becomes constant) — max 4. (b) 1 mark for correct moles of CuCO₃ (0.1); 1 mark for correctly applying the 1:1 mole ratio; 1 mark for correct method (moles × Mr of CO₂); 1 mark for correct mass of CO₂ (4.4 g) — max 4. (c) 1 mark for correct method (mass CO₂ ÷ original mass × 100); 1 mark for correct substitution; 1 mark for correct final answer 35.5%; 1 mark for the answer given to the correct number of decimal places — max 4 (ECF applied). (d) 1 mark for 'black'; 1 mark for correctly identifying the solid as copper(II) oxide — max 2.

The oxygen gas is prepared by adding a small amount of manganese(IV) oxide, which acts as a catalyst, to hydrogen peroxide solution in a conical flask; the manganese(IV) oxide speeds up the decomposition of the hydrogen peroxide without itself being used up. The flask is fitted with a bung carrying a delivery tube, and the oxygen gas produced is collected over water: the delivery tube leads to an inverted, water-filled test tube or gas jar resting on a beehive shelf within a trough of water, so that as oxygen gas is produced, it bubbles up and displaces the water in the tube, since oxygen (unlike some other gases) is only slightly soluble in water and so is not significantly lost by this collection method. Once a test tube (or gas jar) has filled with gas, it is removed and tested using a glowing (not flaming) wooden splint inserted into the mouth of the tube; if the gas is oxygen, it relights/reignites the glowing splint, causing it to burst back into flame, confirming the gas is oxygen. (a) Oxygen is a colourless, odourless gas which is only slightly soluble in water. (b) Oxygen's ability to support combustion and respiration gives it several uses — for example, it is used in hospitals to help patients who are having difficulty breathing, since it supports respiration in the body, or it is used in welding, where its ability to support very vigorous, hot combustion allows metals to be cut or joined. (c) The decomposition of hydrogen peroxide (catalysed by manganese(IV) oxide) is summarised by the word equation: hydrogen peroxide → water + oxygen. (d) A catalyst is defined as a substance that increases the rate of a chemical reaction without itself being permanently used up or chemically changed by the reaction; because the manganese(IV) oxide is not consumed in catalysing the decomposition of hydrogen peroxide, the same mass and chemical form of it remains present at the end of the reaction as at the start, meaning it could, in principle, be recovered (for example, by filtration) and reused to catalyse further batches of the same reaction.
Final answer: manganese(IV) oxide added to hydrogen peroxide in a flask, oxygen collected over water via a delivery tube, tested with a glowing splint (relights); oxygen is colourless and odourless; used in hospitals/welding; hydrogen peroxide → water + oxygen; the catalyst is chemically unchanged, so can be reused.

Banded mark scheme for the main QWC description (6 marks). Indicative content: manganese(IV) oxide added as a catalyst to hydrogen peroxide solution in a conical flask; bung and delivery tube fitted; gas collected over water (inverted tube/gas jar on a beehive shelf in a trough); reasoning that oxygen is only slightly soluble in water, allowing this collection method; test with a glowing splint; positive result is the splint relighting/bursting into flame. 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. (a) 1 mark each for any two correct physical properties of oxygen (colourless, odourless, slightly soluble in water, slightly denser than air) — max 2. (b) 1 mark for a valid use; 1 mark for correctly linking it to a relevant property (supports respiration/combustion) — max 2. (c) 1 mark for 'hydrogen peroxide'; 1 mark for '→ water + oxygen' — max 2. (d) 1 mark for 'a catalyst is not used up/chemically unchanged'; 1 mark for correctly concluding it can therefore be reused — max 2.

(a) The student first measures exactly 25 cm³ of sodium hydroxide solution into an insulated cup (such as a polystyrene cup), and records its starting temperature using a thermometer; a separate, exact 25 cm³ volume of the acid being tested is also measured out, and its starting temperature recorded (both should ideally be at the same starting temperature). The acid is then added to the sodium hydroxide in the insulated cup, the mixture is gently stirred or swirled to ensure even mixing, and the temperature is monitored closely, with the highest temperature reached during the reaction recorded as the maximum temperature. Using an insulated cup (for example a polystyrene cup, ideally with a lid with a hole for the thermometer) reduces the loss of heat energy to the surroundings during the reaction, so the maximum temperature recorded more accurately reflects the true amount of heat released by the reaction itself. To improve reliability, this procedure should be repeated for each acid, and an average maximum temperature rise calculated from the repeats. (b) This reaction is exothermic; since the temperature of the reaction mixture is observed to rise during the reaction, this shows that the reaction is releasing heat energy into its surroundings (the solution itself, which is being measured by the thermometer) rather than absorbing it, which is the defining characteristic of an exothermic reaction. (c) Hydrochloric acid is a strong acid, meaning it is completely ionised in aqueous solution, so essentially all of the acid molecules originally present are available as free H⁺ ions to react with the sodium hydroxide straight away. Ethanoic acid, in contrast, is a weak acid, meaning only a small proportion of its molecules ionise in solution at any one time, so a much lower concentration of free H⁺ ions is available to react immediately, even though the same total (nominal) concentration of acid was used; this results in a somewhat smaller amount of heat being released in the same time, giving a lower maximum temperature rise for the ethanoic acid compared with the hydrochloric acid. (d) For the comparison between the two acids to be a fair, valid test of the effect of acid type alone, other variables that could also affect the temperature change must be kept the same for both acids — for example, the concentration of the sodium hydroxide solution and of each acid, the volumes of acid and alkali used, and the starting temperature of the solutions before mixing should all be kept constant, so that any difference observed in temperature rise can be confidently attributed to the type of acid used, rather than to some other varying factor.
Final answer: (a) measure equal volumes of alkali and acid, record starting and maximum temperature in an insulated cup, repeat for reliability; (b) exothermic, since temperature rises (heat released); (c) HCl is a strong acid (fully ionised, more H⁺ available immediately) while ethanoic acid is weak (only partially ionised), giving a smaller temperature rise; (d) e.g. concentration of the solutions, volumes used, or starting temperature must be kept the same.

(a) 1 mark for measuring 25 cm³ of each solution and recording starting temperature(s); 1 mark for adding acid to alkali and stirring/swirling; 1 mark for recording the maximum/highest temperature reached; 1 mark for using an insulated container to reduce heat loss; 1 mark for repeating for reliability/calculating an average — max 5. (b) 1 mark for 'exothermic'; 1 mark for correctly explaining a temperature rise shows heat is released — max 2. (c) 1 mark for 'HCl is a strong acid, fully ionised (more H⁺ ions available)'; 1 mark for correctly contrasting this with ethanoic acid being a weak/partially ionised acid, giving a smaller temperature rise — max 2. (d) 1 mark for any one valid controlled variable (concentration, volume, or starting temperature); 1 mark for a second valid controlled variable, or a clear explanation of why it must be controlled — max 2.

(a) Magnesium's reaction with cold water, while it does technically occur, is extremely slow at room temperature — magnesium's reactivity with water becomes much more apparent and rapid only under the more energetic conditions of steam, which is why the reaction with cold water is barely noticeable over the timescale of a simple classroom observation. Aluminium's apparent lack of reaction is explained differently: although aluminium is genuinely a reactive metal, high in the reactivity series, it reacts almost instantly with oxygen in the air to form a very thin, tough, unreactive layer of aluminium oxide on its surface; this protective layer effectively shields the more reactive aluminium metal underneath from coming into contact with water, so no reaction with water is observed even though the underlying metal itself would, in principle, be expected to react. (b) The thin oxide layer that naturally covers aluminium's surface is chemically unreactive/very stable and acts as a physical barrier between the acid and the aluminium metal beneath it; since the hydrochloric acid cannot pass through this layer to reach the reactive aluminium metal directly, the acid must first slowly dissolve or otherwise break through the oxide coating before any significant reaction with the aluminium metal itself can begin. This delay, caused entirely by the presence of the protective oxide layer rather than by any lack of underlying reactivity, means aluminium appears to react considerably more slowly (or less readily) with dilute hydrochloric acid than its high position in the reactivity series (above zinc and iron) would otherwise predict. (c) Using aluminium powder instead of a single solid piece greatly increases the total surface area of aluminium exposed for the same overall mass of metal; because the acid needs to break through the thin oxide layer before reaching the aluminium metal beneath, the much larger total surface area of many small powder particles means a correspondingly larger total area of the protective oxide layer is exposed to and broken down by the acid at any given time, allowing the acid to reach and react with a much greater amount of the underlying aluminium metal more quickly than with a single solid piece, where the oxide layer only needs to be broken through over the much smaller exposed surface area of the solid. (d) Since aluminium is more reactive than copper, it displaces copper from copper(II) sulfate solution: the grey aluminium powder is gradually seen to react and disappear, while a reddish-brown/pink solid deposit of copper metal forms in its place; at the same time, the blue colour of the copper(II) sulfate solution gradually fades and becomes paler (eventually becoming colourless, or very pale), as the blue Cu²⁺ ions in solution are progressively replaced by colourless Al³⁺ ions. The balanced symbol equation, using the fact that aluminium forms 3+ ions and copper forms 2+ ions (requiring 2 Al and 3 Cu for the charges to balance across the compounds involved), is: \( 2\text{Al(s)} + 3\text{CuSO}_4(aq) \rightarrow \text{Al}_2(\text{SO}_4)_3(aq) + 3\text{Cu(s)} \).
Final answer: (a) Mg reacts with cold water but only very slowly (much faster with steam); Al's protective oxide layer shields the reactive metal beneath from water; (b) the oxide layer physically blocks the acid from reaching the aluminium metal until it is broken down, so reaction is slower than its reactivity series position predicts; (c) powder has a much larger surface area, so more of the oxide layer/underlying metal is exposed to the acid at once; (d) grey Al reacts/disappears, pink/reddish-brown Cu deposit forms, blue colour fades; 2Al(s)+3CuSO₄(aq)→Al₂(SO₄)₃(aq)+3Cu(s).

(a) 1 mark for correctly explaining magnesium reacts with cold water only very slowly (reacts much faster with steam); 1 mark for identifying aluminium's protective oxide layer as the reason for its apparent lack of reaction; 1 mark for correctly explaining this layer forms rapidly in air; 1 mark for correctly explaining it shields/protects the reactive metal beneath — max 4. (b) 1 mark for 'the oxide layer is a physical barrier'; 1 mark for 'acid cannot reach the aluminium metal directly'; 1 mark for the acid needing to break through/dissolve the layer first; 1 mark for correctly concluding this makes the reaction slower than the reactivity series alone would predict — max 4. (c) 1 mark for 'powder has a greater surface area (for the same mass)'; 1 mark for correctly linking this to more of the oxide layer/underlying metal being exposed to the acid at once — max 2. (d) 1 mark for 'aluminium reacts/disappears'; 1 mark for 'reddish-brown/pink copper deposit forms'; 1 mark for 'blue colour fades/becomes paler'; 1 mark for correct formulae in the equation; 1 mark for correct balancing (2Al, 3CuSO₄, 3Cu) — max 5.