CHEMISTRY – Paper 1 Multiple Choice – mark scheme

D: calcium oxide is a basic oxide, so it dries gases that are neutral or basic. Propane is neutral, so it does not react with calcium oxide. The gas jar is upright, so the gas is collected by downward delivery, which needs a gas denser than air; the M_r of propane, C₃H₈, is 44, which is more than 29. A: ammonia is basic, so calcium oxide can dry it, but its M_r of 17 is less than 29, so it is lighter than air and would rise out of an upright jar; it is collected by upward delivery. B: carbon dioxide is denser than air (M_r 44) but it is an acidic gas and reacts with calcium oxide to form calcium carbonate, so it cannot be dried by calcium oxide. C: hydrogen is neutral, but its M_r of 2 is far below 29, so it is lighter than air and would escape from an upright jar.

B: at 10 minutes the temperature is about 125 °C. The sample began to melt at about 60 °C and had finished melting by 72 °C, and it does not start to boil until about 210 °C, so at 10 minutes it is a liquid. The temperature keeps rising while the sample melts (60 to 72 °C) and while it boils (210 to 222 °C), instead of staying constant. A pure substance melts and boils at one fixed temperature each, so this sample is impure: its melting range is below the melting point of pure Y (80 °C) and its boiling range is above the boiling point of pure Y (200 °C). A, D: a solid has melted completely by 72 °C, so at 10 minutes (about 125 °C) the sample is not a solid. C: the sample is a liquid, but a pure sample would melt and boil at fixed temperatures (80 °C and 200 °C) with the temperature constant during each change of state; the graph shows neither, so it is not pure.

A: electrons = proton number (lower number) minus the positive charge; neutrons = nucleon number (upper number) minus proton number. Particle 1: 28 − 2 = 26 electrons, 58 − 28 = 30 neutrons. Particle 2 is an atom: 26 electrons, 54 − 26 = 28 neutrons. Particle 3: 26 − 3 = 23 electrons, 56 − 26 = 30 neutrons. Particle 4: 25 − 2 = 23 electrons, 55 − 25 = 30 neutrons. Particles 1 and 2 both have 26 electrons but have 30 and 28 neutrons. B: particles 1 and 3 have the same number of neutrons (30) but different numbers of electrons (26 and 23). C: particles 2 and 4 have different numbers of electrons (26 and 23). D: particles 3 and 4 have the same number of electrons (23) but also the same number of neutrons (30).

A: the electron arrangement 2,8,8 means that the ion has 18 electrons. A 2− ion has gained two electrons, so it has two more electrons than protons: 18 − 2 = 16 protons. Every atom of Z has 16 protons, so the isotope with nucleon number 32 has 32 − 16 = 16 neutrons, and a neutral atom has 16 electrons. B: the protons and neutrons are right but the atom is given the 18 electrons of the ion; an atom is neutral, so its electrons equal its protons. C: the protons have been taken as the 18 electrons of the ion; the ion has gained two electrons, so the atom has two fewer protons than 18. D: the charge has been applied the wrong way round (a 2− ion has two more electrons than protons, not two fewer).

C: in the lattice each ion is held by electrostatic attraction to ions of opposite charge. When a layer slides along by one ion, as in the diagram, ions with the same charge come next to each other and repel; the lattice splits, so the crystal is brittle. A: ionic compounds have ionic bonds (electrostatic attraction between oppositely charged ions), not covalent bonds. B: an ionic compound has no delocalised electrons; these are found in metals. D: ions become free to move only when the compound is molten or dissolved; the force on the crystal does not melt it.

B: count all the electrons that each element supplies, including those in the shared pairs. Each atom of X supplies 6 outer electrons: 4 dots in the two double bonds (two shared pairs with each Y atom) and 2 dots in the lone pair, so X is in Group 16. Each atom of Y supplies 6 crosses: 2 in the shared pairs of its double bond and 4 in its two lone pairs, so Y is also in Group 16. A: this counts only the 4 dots in the shared pairs for X and leaves out the lone pair, which gives Group 14. C: this counts only the 4 crosses in the lone pairs of Y and leaves out the 2 in the shared pairs, which gives Group 14. D: both slips together: only the shared dots for X and only the lone-pair crosses for Y.

C: stainless steel is an alloy, a mixture of a metal (iron) with other elements (such as chromium). Its elements are not chemically combined. A, D: copper(II) oxide and zinc oxide are compounds of a metal with oxygen; their elements are chemically combined in a fixed ratio, so they are not mixtures. B: sodium chloride is a compound of a metal with a non-metal, with its ions in a fixed ratio; it is not a mixture.

D: iodine is a simple molecular substance; its small molecules are held together only by weak forces between the molecules, which are easily overcome. Silicon(IV) oxide is a giant covalent structure, so melting it means breaking many strong covalent bonds. Poly(ethene) is a macromolecule: its long molecules are held together by forces between the molecules; these are overcome when the plastic softens, and the covalent bonds within each chain stay intact. A: this treats all three as giant covalent; the covalent bonds within iodine molecules and along polymer chains are not broken on melting or softening. B: iodine does not melt by breaking its covalent bonds; only the forces between its molecules are overcome. C: this gives poly(ethene) a giant covalent structure; its covalent bonds along the chains stay intact when it softens.

C: statement 1 is correct: R does not conduct and has a very high melting point, so it is a giant covalent substance in which many strong covalent bonds must be broken. Statement 3 is correct: P does not conduct when solid but conducts when molten, so it is an ionic compound whose ions can move only when the lattice is melted. Statement 4 is correct: Q conducts when solid, so it contains delocalised electrons that are free to move. Statement 2 is not correct: S has a very low melting point because the forces between its molecules are weak and easily overcome; the covalent bonds within its molecules are strong and are not broken on melting. So A, B and D each include statement 2 or leave out a correct statement.

B: the compound is neutral, so the total positive charge equals the total negative charge. Four oxide ions give 4 × 2− = 8−, so the three cobalt ions must give 8+ altogether. One Co²⁺ and two Co³⁺ give 2 + 6 = 8+, and there are 3 cobalt ions in all. A: three Co³⁺ ions give 9+, which is too much. C: two Co²⁺ and one Co³⁺ give 7+, which is too little. D: three Co²⁺ ions give 6+, which is too little.

B: percentage of iron = (mass of iron in one formula unit ÷ M_r) × 100. FeCO₃: 56 ÷ 116 = 48.3%. FeS₂: 56 ÷ 120 = 46.7%. Fe₂O₃: 112 ÷ 160 = 70.0%. Fe₃O₄: 168 ÷ 232 = 72.4%, the highest. A: Fe₂O₃ is the nearest, but 70.0% is lower than 72.4%. C, D: each of these ores has less than half of its mass as iron, because the heavier carbonate and disulfide parts make up more than half of M_r.

D: moles of NaOH = 8.0 ÷ 40 = 0.20 mol; moles of CuSO₄ = 24.0 ÷ 160 = 0.15 mol. 0.20 mol of NaOH needs only 0.10 mol of CuSO₄, so sodium hydroxide is the limiting reactant. The ratio NaOH : Cu(OH)₂ is 2 : 1, so the maximum is 0.10 mol of Cu(OH)₂ = 0.10 × 98 = 9.8 g. Percentage yield = 6.86 ÷ 9.8 × 100 = 70%. A: 21% is 6.86 g divided by the total mass of the two reactants (32.0 g). B: 35% uses a ratio of 1 mol of Cu(OH)₂ to 1 mol of NaOH (0.20 mol = 19.6 g), which ignores the 2 in 2NaOH. C: 47% takes copper(II) sulfate as the limiting reactant (0.15 mol gives 0.15 mol = 14.7 g of Cu(OH)₂), but there is more than enough of it.

D: an alkali produces hydroxide ions, OH⁻, in aqueous solution; an alkaline solution turns Universal Indicator blue or purple. A, B: hydrogen ions, H⁺, are produced by acids, not alkalis. C: Universal Indicator is red in a strongly acidic solution; an alkaline solution turns it purple.

A: magnesium oxide is a base (the oxide of a metal); a base reacts with an acid to form a salt and water, so the only other product is water: MgO + 2HCl → MgCl₂ + H₂O. B: a carbonate reacts with an acid to form a salt, carbon dioxide and water, so carbon dioxide is not the only other product. C: magnesium hydroxide is also a base, so it forms a salt and water, not hydrogen; hydrogen forms only when a metal reacts with an acid. D: magnesium is a metal, so it reacts with the acid to form magnesium chloride and hydrogen only; no water is formed.

C: lead(II) sulfate is insoluble, so it is the solid left on the filter paper (the residue). Washing the residue with distilled water removes the dissolved sodium nitrate that wets it, and drying it leaves pure, dry lead(II) sulfate. A: the filtrate is the solution of sodium nitrate; evaporating it would give sodium nitrate, not lead(II) sulfate. B: evaporating and crystallising is used for a soluble salt; the lead(II) sulfate is already a solid in the mixture, and the sodium nitrate would crystallise with it. D: without washing, the residue is left contaminated with sodium nitrate, so it is not pure.

C: conditions P and S differ only in the pressure. At 200 atm (S) the yield is 55%, higher than the 42% at 100 atm (P), and equilibrium is reached in 30 minutes instead of 45 minutes, so the time is shorter. A: conditions P and Q differ only in the catalyst. Both give a yield of 42%, so the catalyst does not change the yield; it only makes equilibrium reached sooner (45 minutes instead of 2,400 minutes). B: conditions P and R differ only in the temperature. At 450 °C the equilibrium is reached sooner (5 minutes), but the yield falls from 42% to 14%. D: the yield does increase at 200 atm, but the time to reach equilibrium becomes shorter (30 minutes), not longer.

D: sulfur dioxide is tested with acidified potassium manganate(VII): it is a reducing agent, so it turns the purple solution colourless. A: ammonia is tested with damp red litmus paper, which turns blue. B: carbon dioxide is tested with limewater, which turns milky; it does not decolourise acidified potassium manganate(VII). C: hydrogen is tested with a lighted splint, which goes out with a ‘squeaky pop’.

A: chlorine is a more powerful oxidising agent than iodine; it oxidises colourless iodide ions to brown iodine (Cl₂ + 2I⁻ → 2Cl⁻ + I₂), so the potassium iodide turns brown. A reducing agent is needed to turn purple acidified potassium manganate(VII) colourless; chlorine is not one, so there is no change. Chlorine acts as an oxidising agent only. B: chlorine does not decolourise acidified potassium manganate(VII), so it is not a reducing agent as well. C: this is the pattern for a reducing agent, such as sulfur dioxide; chlorine oxidises the iodide ions, so the potassium iodide does turn brown. D: chlorine does oxidise iodide ions, so aqueous potassium iodide does change colour.

B: molten lithium chloride contains Li⁺ and Cl⁻ ions; lithium forms at the negative electrode and chlorine gas, Cl₂, forms at the positive electrode: 2LiCl → 2Li + Cl₂. Moles of Li = 1.4 ÷ 7 = 0.20 mol. The ratio Li : Cl₂ is 2 : 1, so 0.10 mol of Cl₂ forms, and 0.10 × 24 = 2.4 dm³. C: 4.8 dm³ takes 0.20 mol of gas, which ignores that two lithium ions are needed for each Cl₂ molecule. D: 9.6 dm³ turns the ratio round (2 mol of Cl₂ for 1 mol of Li). A: 1.2 dm³ halves the amount twice (0.05 mol).

A: concentrated aqueous R chloride. At the negative electrode, aqueous R sulfate gives hydrogen, so R ions are not discharged and hydrogen forms. At the positive electrode, concentrated aqueous sodium chloride gives chlorine, so chloride ions are discharged from a concentrated solution. Aqueous Q sulfate. At the negative electrode, aqueous Q nitrate gives Q, so Q ions are discharged and Q forms. At the positive electrode, sulfate ions are not discharged (aqueous R sulfate gives oxygen), so hydroxide ions from the water give oxygen. B: chlorine, not oxygen, forms at the positive electrode from a concentrated chloride solution (as in concentrated sodium chloride). C: R forms at the negative electrode only if its ions are discharged in preference to hydrogen ions; aqueous R sulfate gives hydrogen, so they are not. D: aqueous Q nitrate gives Q at the negative electrode, so Q ions are discharged in preference to hydrogen ions and hydrogen does not form.

D: in electroplating the object to be plated is the cathode, and the metal ions in the electrolyte gain electrons there to form the metal, so in both set-ups the steel strip is coated with copper (statement 4). In set-up X the copper anode dissolves, Cu → Cu²⁺ + 2e⁻, replacing the Cu²⁺ ions that are removed at the cathode, so the blue colour of the electrolyte stays the same (statement 2). In set-up Y the graphite anode is inert, so it cannot supply copper ions: oxygen gas forms at the anode (statement 3) and the electrolyte slowly becomes paler blue as Cu²⁺ ions are used up. Statement 1 is incorrect because the copper anode decreases in mass as it dissolves, so A, B and C, which each include statement 1, are wrong.

A: an oxide that reacts only with an acid is a basic oxide, which is formed by a metal; an oxide that reacts only with an alkali is an acidic oxide, formed by a non-metal. The oxide of Q reacts with both, so it is amphoteric and Q lies between the metals and the non-metals. Across a period, from left to right, the elements change from metallic to non-metallic, so P (a metal, on the left) is the most metallic and R (a non-metal, on the right) is the least metallic. B: Q is not the least metallic; its amphoteric oxide places it between P and R. C: Q is not the most metallic; it reacts with an alkali as well as with an acid, which a typical metal oxide does not do. D: this has the trend the wrong way round, with the non-metal R as the most metallic element.

B: down Group 1 the melting points decrease, so lithium, at the top, has a higher melting point than sodium (statement 2). Down Group 17 the elements change from gases to a liquid to a solid, and chlorine is a gas, so fluorine, above chlorine, is also a gas (statement 3). Statement 1 is incorrect: the Group 1 metals become more reactive down the group, so lithium reacts less vigorously with water than sodium does. Statement 4 is incorrect: the Group 17 elements become less reactive down the group, so fluorine, at the top, is more reactive than chlorine. A, C and D each include statement 1 or statement 4.

A: transition elements and their compounds are often able to act as catalysts. Only iron(III) oxide and copper(II) oxide make the oxygen come off much faster (150 s and 90 s instead of 640 s), so they are acting as catalysts, and iron and copper are transition elements. Calcium oxide and magnesium oxide change the time by only a few seconds, so they are not acting as catalysts; calcium and magnesium are Group 2 metals, not transition elements. B, C and D each contain calcium oxide or magnesium oxide.

D: iron is more reactive than copper and silver (the order is zinc, iron, copper, silver), so iron displaces both from their solutions: Fe + 2Ag⁺ → Fe²⁺ + 2Ag and Fe + Cu²⁺ → Fe²⁺ + Cu. Silver and copper are solids, and because the iron is in excess some unreacted iron is left as well, so the filter paper holds silver, copper and iron. Iron is less reactive than zinc, so it cannot displace zinc ions: they stay in solution beside the iron(II) ions that have formed. A: this forgets that some of the excess iron is left over, so it must also be in the filter paper. B: this leaves out both the unreacted iron and the iron(II) ions that form when iron displaces silver and copper. C: this assumes that iron displaces zinc, but iron is less reactive than zinc; no zinc forms and the zinc ions stay in solution.

C: zinc is more reactive than iron, so where the coating is scratched the zinc corrodes in preference and protects the steel (sacrificial protection); the steel does not rust while zinc remains nearby. Tin is less reactive than iron, so tin cannot protect the steel: tin is only a barrier, and once the barrier is broken the more reactive iron is the metal that corrodes, so the steel rusts. A: this ignores the sacrificial protection of zinc. B: this has the two metals the wrong way round; it would need tin to be more reactive than iron. D: this wrongly assumes that tin protects the steel when it is scratched.

B: breaking a bond takes in exactly the energy that is released when the same bond forms, so breaking 1 mol of H–Br bonds takes in 366 kJ. Each HBr molecule has one H–Br bond, so 0.50 mol of HBr has 0.50 mol of bonds: 0.50 × 366 = 183 kJ. Bond breaking is endothermic, so the enthalpy change is positive: +183 kJ. A: the size is right but the sign is wrong, because bond breaking is endothermic, not exothermic. C: +366 kJ is for 1 mol of bonds; it does not allow for only 0.50 mol. D: +732 kJ doubles the energy for 1 mol instead of halving it.

D: the decomposition is the reverse of the formation of hydrogen iodide. The bonds broken are now the bonds of the two molecules of HI, which released 598 kJ when they formed, so 598 kJ is taken in to break them. The bonds formed are those of the hydrogen and iodine molecules, which took in 587 kJ to break, so 587 kJ is released when they form. The energy taken in (598 kJ) is greater than the energy released (587 kJ), so the overall change is 598 − 587 = 11 kJ taken in (endothermic). A, B: 587 kJ is the energy for breaking the bonds in the forward reaction, not in the decomposition. A, C: the decomposition takes in more energy than it releases, so it is not exothermic; the forward reaction is the one that releases 11 kJ.

A: a catalyst provides a different pathway with a lower activation energy, so a larger proportion of the collisions has at least the energy needed to react (statement 1). Raising the temperature gives the particles more energy, so a larger proportion of the collisions has at least the activation energy (statement 2). Increasing the concentration (statement 3) puts more particles in each unit of volume, and increasing the surface area (statement 4) exposes more particles of the solid; both make collisions more frequent, but the proportion of the collisions with enough energy to react does not change. B, C and D each include statement 3 or statement 4.

C: each rise of 10 °C halves the time, so at 60 °C the magnesium takes 12 ÷ 2 = 6 s to disappear. Average rate = mass ÷ time = 0.12 g ÷ 6 s = 0.020 g / s. A: 0.0050 g / s is the rate at 40 °C (0.12 ÷ 24). B: 0.010 g / s is the rate at 50 °C (0.12 ÷ 12); it is the last measured value and the pattern has not been continued. D: 0.040 g / s is for 3 s (0.12 ÷ 3), as if the time had halved twice.

B: each carbon atom in the alkane forms one molecule of carbon dioxide, so one molecule of the alkane contains 0.40 ÷ 0.10 = 4 carbon atoms. Each molecule of water has two hydrogen atoms, so 0.50 mol of water contains 1.0 mol of hydrogen atoms, which is 1.0 ÷ 0.10 = 10 hydrogen atoms per molecule of the alkane. The formula is C₄H₁₀, which fits the general formula of the alkanes, C_nH_2n+2, with n = 4. A: C₄H₈ has the right number of carbon atoms but its formula is C_nH_2n, that of an alkene, which is unsaturated and not an alkane. C, D: five carbon atoms would give 0.50 mol of carbon dioxide, not 0.40 mol.

D: ethene is an unsaturated hydrocarbon: its C=C double bond reacts with aqueous bromine, which is decolourised, so all of the ethene is removed from the gas. Ethane is a saturated hydrocarbon with only single bonds; it does not react with aqueous bromine, so the gas that leaves is ethane. Volume of ethane = 40 cm³, so the volume of ethene = 100 − 40 = 60 cm³. B: 40 cm³ is the volume of the gas that is left, which is the ethane, not the ethene. C: 50 cm³ assumes that the sample is half ethene and half ethane, which the volumes given do not show. A: 20 cm³ would leave 80 cm³ of ethane, not the 40 cm³ that is measured; the ethene is the part of the sample that disappears.

C: a saturated chain of n carbon atoms has 2n + 1 hydrogen atoms (stearic acid: 17 carbon atoms, 35 hydrogen atoms), and each C=C bond removes two hydrogen atoms. C₁₇H₃₁COOH has 4 hydrogen atoms fewer than the saturated acid (35 − 31 = 4), so it has two C=C bonds. Polyunsaturated means more than one C=C bond in each molecule, so this acid is polyunsaturated. A: a saturated chain of 15 carbon atoms has 31 hydrogen atoms, so C₁₅H₃₁COOH has no C=C bond. B: a saturated chain of 16 carbon atoms has 33 hydrogen atoms, so C₁₆H₃₁COOH has only one C=C bond; it is unsaturated but not polyunsaturated. D: C₁₇H₃₃COOH has 2 hydrogen atoms fewer than the saturated acid, so it has one C=C bond only.

D: oxidation changes the CH₂OH group of the alcohol into the CO₂H group of a carboxylic acid and does not change the number of carbon atoms. The acid with M_r = 74 is propanoic acid, CH₃CH₂COOH (3 × 12 + 6 × 1 + 2 × 16 = 74), which has three carbon atoms, so X is propan-1-ol. A: butan-1-ol has M_r = 74 as well, but it is oxidised to butanoic acid (M_r = 88); the M_r of the alcohol is not the M_r of the acid. B: ethanol is oxidised to ethanoic acid (M_r = 60). C: methanol is oxidised to methanoic acid (M_r = 46).

A: ethanol is oxidised by oxygen from the air to ethanoic acid and water: C₂H₅OH + O₂ → CH₃COOH + H₂O. B: this is the complete combustion of ethanol, which forms carbon dioxide and water, not ethanoic acid. C: this is the fermentation of glucose, which forms ethanol. D: this is the reaction of ethanoic acid with ethanol, which forms an ester; it starts from ethanoic acid instead of forming it.

C: a carboxylic acid reacts with an alcohol to form an ester and water, one molecule of each reactant giving one molecule of the ester: CH₃CH₂COOH + CH₃OH → CH₃CH₂COOCH₃ + H₂O. The M_r of the ester = 74 + 32 − 18 = 88. Amount of propanoic acid = 7.4 ÷ 74 = 0.10 mol, which can form at most 0.10 mol of ester, a mass of 0.10 × 88 = 8.8 g. Percentage yield = 5.28 ÷ 8.8 × 100 = 60%. A: 40% is 100% − 60%. B: 50% results from taking the M_r of the ester as 74 + 32 = 106, forgetting that a molecule of water is also formed. D: 71% is 5.28 ÷ 7.4: the mass of ester divided by the mass of acid, instead of by the maximum mass of ester.

A: in addition polymerisation the C=C bond of each monomer opens and the molecules join end to end, with every other atom or group staying on the carbon atom it was on, and nothing is lost. The repeating unit is a CH₂ group followed by a carbon atom that carries two CH₃ groups, so the monomer is CH₂=C(CH₃)₂, C₄H₈, with M_r = 4 × 12 + 8 × 1 = 56. Because no atoms are lost, the M_r of the polymer molecule = number of monomer molecules × 56, so the number = 8400 ÷ 56 = 150. B: the monomer is right, but 75 comes from dividing by 112, the mass of the two repeating units that are drawn, instead of by the mass of one monomer molecule. C: CH₃CH=CHCH₃ has the same M_r (56), so 150 would be right for it, but it would give a CH₃ group on every carbon atom of the chain, not two on the same carbon atom. D: propene would give a single CH₃ group on every second carbon atom; its M_r is 42, which would give 200 molecules, and it cannot give the repeating unit shown, which contains four carbon atoms.

B: when plastics burn, incomplete combustion and other products can give toxic gases such as carbon monoxide (statement 1). Non-biodegradable plastics are not broken down by bacteria, so they stay in landfill sites for many years, using up land (statement 2). Statement 3 is incorrect: carbon dioxide is a greenhouse gas, and the large amount formed when plastics burn adds to the carbon dioxide in the air and contributes to global warming, even though carbon dioxide is already present in air. A, C and D each include statement 3.

B: P takes carbon dioxide out of the air into plants, which is photosynthesis. Q puts carbon dioxide back into the air from plants: plants respire as well as photosynthesise, so Q is respiration. R puts carbon dioxide into the air from animals, which is respiration. S puts carbon dioxide into the air from fossil fuels, which is combustion. The carbon cycle keeps the amount of carbon dioxide in the air roughly steady because the carbon dioxide removed by photosynthesis is balanced by the carbon dioxide returned by respiration and combustion. A: animals do not burn fuels, so R cannot be combustion, and fossil fuels do not respire, so S cannot be respiration. C: photosynthesis takes carbon dioxide out of the air, but Q shows carbon dioxide going into the air from plants, so Q cannot be photosynthesis. D: P and Q are the wrong way round: respiration puts carbon dioxide into the air, but P shows carbon dioxide taken into plants, and photosynthesis takes it out of the air, but Q shows it entering the air.

C: the concentration of carbon dioxide has increased by (410 − 280) ÷ 280 × 100 = 46%, while that of methane has increased by (1.92 − 0.80) ÷ 0.80 × 100 = 140%, so methane has increased by the greater percentage. Carbon dioxide and methane are both greenhouse gases and may contribute to global warming. A, B: carbon dioxide has increased by more ppm (130 compared with 1.12), which makes its rise look bigger, but its percentage increase, 46%, is smaller than that of methane. B, D: methane is a greenhouse gas as well as carbon dioxide, so it is wrong to say that only one of the two contributes to global warming.