CHEMISTRY – Paper 1 Multiple Choice – mark scheme

C: a gas syringe measures the volume of a gas directly and a stopwatch measures the time. A: a balance measures mass, not the volume of a gas. B: a burette measures volumes of liquid and a thermometer measures temperature; neither measures time. D: a pipette delivers a fixed volume of liquid and a thermometer measures temperature; neither measures the volume of a gas or the time.

B: in fractional distillation a mixture of liquids that mix completely is heated; the vapours repeatedly condense and re-evaporate as they rise up the column, and the liquid with the lowest boiling point reaches the top first. A: this describes paper chromatography. C: this describes sublimation. D: this describes the use of a separating funnel for immiscible liquids.

B: the spot in the ink that is level with the spot of dye R is at 40 mm, and the solvent front has moved 80 mm. R_f = distance moved by the dye ÷ distance moved by the solvent = 40 ÷ 80 = 0.50. A: uses the lowest spot in the ink (20 ÷ 80), which is not level with R. C: uses the highest spot in the ink (60 ÷ 80), which is level with dye S, not R. D: inverts the ratio (80 ÷ 40); an R_f value is never greater than 1.

D: a pure substance melts and boils sharply at fixed temperatures, and sample Z melts at exactly 112 °C and boils at exactly 305 °C, the values for pure K. A and C: these samples melt and boil over a range of temperatures, which shows that they are impure (impurities lower the melting point, raise the boiling point and make the sample melt and boil over a range). B: sample X melts at 112 °C, like K, but its boiling point is 249 °C, so it is a different substance.

A: 1 and 2 are diffusion: the copper(II) sulfate particles spread through the water and the ammonia particles spread through the air because the particles are moving randomly all the time, without any stirring or draught. 3: squeezing a gas into a smaller volume is evidence that there are large spaces between the particles of a gas, not that they move. 4: a solid expanding shows that its particles vibrate more and take up more room; it is not evidence about particles in a liquid or a gas. B, C, D each include 3 or 4.

C: methane has the smaller relative molecular mass (16 compared with 71), so its molecules move faster at the same temperature and it diffuses faster; raising the temperature gives the molecules more kinetic energy, so they move faster and the rate of diffusion increases. Methane at 40 °C has both advantages. A and B: chlorine molecules are heavier and move more slowly, so chlorine diffuses more slowly than methane at the same temperature. D: methane diffuses faster than chlorine, but it diffuses more slowly at 20 °C than at 40 °C.

A: the sample contains two kinds of molecule: one kind contains two atoms of the same element (an element) and the other contains atoms of two different elements joined together (a compound); two substances that are not joined together make a mixture. B: all of the molecules contain atoms of one element only, so this is an element. C: all of the molecules contain two different atoms joined together, so this is a compound. D: the two kinds of molecule are both elements (one for each kind of atom), so this is a mixture of two elements, with no compound.

C: one molecule contains 1 carbon atom, 1 oxygen atom (joined by a double bond, but still one atom), 2 nitrogen atoms and 4 hydrogen atoms (2 on each nitrogen atom): CH₄N₂O, which is also written CO(NH₂)₂. A: counts the hydrogen atoms on one nitrogen atom only. B: leaves out the oxygen atom. D: gives each nitrogen atom three hydrogen atoms, as in ammonia, NH₃.

D: M_r of Fe₂O₃ = 2 × 56 + 3 × 16 = 160, so n = 16.0 ÷ 160 = 0.100 mol. The ratio Fe₂O₃ : Fe is 1 : 2, so the maximum amount of iron is 0.200 mol = 0.200 × 56 = 11.2 g. Percentage yield = 8.96 ÷ 11.2 × 100 = 80%. A: this is the percentage that was not obtained (100 − 80). B: the 1 : 2 ratio has been applied twice (maximum mass 22.4 g). C: compares the mass of iron obtained with the mass of iron(III) oxide used (8.96 ÷ 16.0); masses of different substances must be converted using moles.

B: potassium hydroxide and nitric acid are both fully ionised in solution, and the only change is that H⁺ ions and OH⁻ ions combine to form water; the K⁺ and NO₃⁻ ions stay in solution. A: barium ions and sulfate ions also react, to form a precipitate of barium sulfate, so there is a second reaction. C: the carbonate ions react with H⁺ ions to form water and carbon dioxide; there are no OH⁻ ions. D: the hydroxide ions are in a solid, not free in solution, and the magnesium ions are formed too: Mg(OH)₂(s) + 2H⁺(aq) → Mg²⁺(aq) + 2H₂O(l).

D: a base reacts with an acid to form a salt and water (sodium nitrate and water); a base also reacts with an ammonium salt, releasing ammonia gas: NaOH + NH₄NO₃ → NaNO₃ + NH₃ + H₂O. A, B: hydrogen is formed when an acid reacts with a metal, not with a base. A, C: students think that a base cannot react with a salt, but it reacts with an ammonium salt to release ammonia.

A: step 1 is correct: copper(II) oxide is insoluble, so it is added until it is in excess; then all of the acid has reacted and the solution contains no acid. Step 2 is correct: the unreacted solid is removed by filtration. Step 3 is not correct: heating until all of the water has evaporated gives a powder, not crystals (and drives off the water of crystallisation); the filtrate should be heated only until it is saturated, then left to cool so that crystals form. B, C, D each include step 3.

B: the nitrogen comes from the air (78% of the air is nitrogen), and the hydrogen is made by cracking hydrocarbons from crude oil; cracking also gives alkenes. 2: the air contains no hydrogen gas (it is not a source of hydrogen). 3: cracking hydrocarbons gives alkenes and hydrogen; hydrocarbons contain no nitrogen. A, C, D each include 2 or 3.

C: at 300 °C the mixture takes 600 minutes to reach equilibrium and contains 67% ammonia; at 450 °C it takes only 20 minutes but contains only 28% ammonia. This is the compromise made in industry: a temperature high enough for the reaction to be fast, but not so high that the percentage of ammonia becomes too low. A: the experiments at 450 °C and 200 atmospheres with and without the catalyst both give 28% ammonia; the catalyst only shortens the time (20 minutes instead of 800). B: at 200 atmospheres the time is 20 minutes, shorter than the 30 minutes at 100 atmospheres. D: the percentage of ammonia falls from 67% to 28% when the temperature is raised from 300 °C to 450 °C.

A: both ions give a white precipitate with a few drops of aqueous ammonia, but with excess aqueous ammonia the precipitate of zinc hydroxide dissolves to give a colourless solution, while the precipitate of aluminium hydroxide does not dissolve. B, D: a few drops of either alkali gives a white precipitate with both solutions. C: zinc hydroxide and aluminium hydroxide both dissolve in excess aqueous sodium hydroxide to give a colourless solution, so there is no difference.

D: sulfur dioxide is a reducing agent, so it turns acidified potassium manganate(VII) from purple to colourless. A: ammonia is alkaline and turns damp red litmus paper blue. B: a glowing splint relights in oxygen; hydrogen burns with a squeaky pop when a lighted splint is held in it. C: oxygen relights a glowing splint; the squeaky pop with a lighted splint is the test for hydrogen.

C: an oxidising agent oxidises colourless iodide ions to brown iodine, so aqueous potassium iodide turns from colourless to brown; a reducing agent reduces purple manganate(VII) ions to a colourless solution of manganese(II) ions, so acidified potassium manganate(VII) turns from purple to colourless. A, B: aqueous potassium iodide is colourless at the start; it cannot turn from brown to colourless when an oxidising agent is added. B, D: acidified potassium manganate(VII) is purple at the start; it does not turn from colourless to purple when a reducing agent is added.

B: X is an ionic compound: its ions are held in a lattice when it is solid, so they cannot carry the current, but they are free to move when it is molten or dissolved, so it conducts (and is decomposed at the electrodes). A: delocalised electrons are present in metals; a compound that is decomposed when it conducts contains ions, not delocalised electrons. C: the ions are already present in the solid, in the lattice; melting or dissolving only frees them to move. D: an ionic compound does not contain molecules; the charged particles are ions.

C: P is connected to the negative terminal, so the positive magnesium ions are attracted to P and gain electrons to form magnesium; Q is connected to the positive terminal, so the negative chloride ions are attracted to Q and lose electrons to form chlorine gas. A: this has the products at the wrong electrodes (positive ions go to the negative electrode). B, D: hydrogen and oxygen come from water; molten magnesium chloride contains only magnesium ions and chloride ions.

A: chemical properties depend on the electrons in the outer shell, so atoms with the same number of outer shell electrons (the same electronic configuration in the outer shell) react in similar ways; for example, they gain or lose the same number of electrons to form ions. B: the number of shells is the same for elements in the same period, not the same group; it increases down a group. C: the number of protons increases down a group, so atoms in a group have different numbers of protons. D: only the atoms of noble gases have full outer shells.

C: atoms with only 1, 2 or 3 outer shell electrons tend to lose them, so these elements are metals: X (2,8,1 has 1 outer shell electron) and Z (2,8,8,2 has 2 outer shell electrons). W (2,5 has 5 outer shell electrons) and Y (2,8,6 has 6 outer shell electrons) tend to gain or share electrons, and are non-metals. A: W has 5 outer shell electrons and is a non-metal; only X is a metal. B: these are the two non-metals, so metals and non-metals are the wrong way round. D: Y has 6 outer shell electrons and is a non-metal; only Z is a metal.

A: argon is a Group 18 noble gas. Noble gases are monoatomic (they exist as single atoms) and are chemically unreactive, so argon provides an inert environment and does not react with the hot filament, which would otherwise burn away in air. B: the particles are right, but argon does not react with anything, including oxygen; it keeps the filament safe by being unreactive. C: argon is not diatomic like oxygen, nitrogen or chlorine; its atoms do not bond to each other. D: wrong on both counts: single atoms, and no reaction with oxygen.

B: platinum and vanadium are transition elements, and transition elements and their compounds are often able to act as catalysts. Manganese is also a transition element, and manganese(IV) oxide is a catalyst (for example for the decomposition of hydrogen peroxide). A, C, D: calcium, potassium and sodium are not transition elements; their compounds are not typical catalysts.

C: zinc is more reactive than copper, so zinc atoms form positive ions more easily: Zn → Zn²⁺ + 2e⁻. The electrons are gained by the copper(II) ions, which form copper atoms (the copper that is deposited): Cu²⁺ + 2e⁻ → Cu. A: copper has the smaller tendency to form positive ions, so it cannot displace zinc, and the ions that gain electrons are the copper(II) ions. B: both columns are the wrong way round. D: the metal is right, but Zn²⁺ ions are formed in the reaction and stay in solution; they do not gain electrons.

B: the more reactive the metal, the more stable its carbonate is to heat. The order of reactivity is Na > Ca > Zn > Cu, so thermal stability increases in the order CuCO₃ < ZnCO₃ < CaCO₃ < Na₂CO₃. Copper(II) carbonate decomposes easily to copper(II) oxide and carbon dioxide, zinc carbonate and calcium carbonate need more heating, and sodium carbonate is not decomposed by heat. A: calcium and zinc are swapped (calcium is more reactive than zinc). C: the order is reversed (decreasing stability). D: zinc and copper are swapped (zinc is more reactive than copper).

C: iron rusts only when both oxygen and water are present, and a barrier that keeps them away prevents rusting. Tube 1 has water and air, so the nail rusts. Tube 2: boiled water contains no dissolved oxygen and the layer of oil keeps air out, so there is no rusting. Tube 3: anhydrous calcium chloride absorbs water, so the air is dry and there is no rusting. Tube 4: the paint is a barrier that keeps oxygen and water away from the iron, so there is no rusting. A: forgets that painting prevents rusting. B: assumes that a nail in dry air with oxygen still rusts. D: assumes that a nail in boiled water still rusts because water is present; it needs oxygen as well.

D: breaking bonds absorbs energy and forming bonds releases energy. The reaction releases energy overall (it is exothermic), so the energy released in forming the H–Cl bonds must be greater than the energy absorbed in breaking the H–H and Cl–Cl bonds. A: this would make the reaction endothermic (more energy absorbed than released). B: breaking bonds does not release energy and forming bonds does not absorb it; the two processes are the wrong way round. C: if the two energies were the same there would be no overall energy change.

B: the plants take in carbon dioxide by photosynthesis as they grow (statement 1), so the carbon dioxide released when the biofuel burns is offset by the carbon dioxide the plants took in (statement 4). Fossil fuels such as diesel release carbon dioxide that was locked away underground for millions of years, with no such offset. Statement 2 is not correct: biodiesel contains carbon, so burning it releases carbon dioxide. Statement 3 is not correct: burning a fuel releases carbon dioxide; it does not take it in. A, C and D each contain statement 2 or 3.

D: the members of a homologous series have the same general formula (for example C_nH_2n+2 for the alkanes), similar chemical properties because they have the same functional group, and physical properties such as boiling point that change gradually as the molecules get larger. A: the members share one general formula. B: the same functional group gives similar, not different, chemical reactions. C: the physical properties are not identical; boiling points, for example, increase as the molecules get bigger.

A: alkenes are unsaturated hydrocarbons with the general formula C_nH_2n, so M_r = 12n + 2n = 14n. 14n = 70, so n = 5 and the formula is C₅H₁₀. B: C₅H₁₂ is pentane, an alkane (C_nH_2n+2), with M_r = 72; it is saturated, not an alkene. C: C₆H₁₂ has the alkene formula but M_r = 84. D: C₆H₁₄ is an alkane with M_r = 86.

D: unsaturated molecules contain carbon–carbon double bonds; polyunsaturated means that there is more than one (poly means many) in the molecule. A: this describes a saturated fat or oil. B: this describes a monounsaturated one, with a single double bond. C: confuses the prefix poly with polymerisation; a polyunsaturated oil is not a polymer made from many small molecules.

D: CH₃CH₂CH₂OH is propan-1-ol: its molecules contain the –OH group joined to a carbon chain, which is the group in the alcohol homologous series. A: CH₃CH₂CO₂H is propanoic acid; its –CO₂H group contains an O–H bond, but the compound is a carboxylic acid, not an alcohol. B: CH₃CO₂CH₃ is methyl ethanoate, an ester (an isomer of propanoic acid). C: CH₃CH₂CH₃ is propane, an alkane; it contains no oxygen.

C: ethanoic acid contains the –CO₂H group (a carbon atom double-bonded to one oxygen atom and single-bonded to an –OH group). A: ethene is a hydrocarbon (an alkene) with a C=C double bond and no oxygen. B: methyl methanoate has the same molecular formula as ethanoic acid (C₂H₄O₂) but its oxygen atoms are in an ester group, –CO₂–; it has no O–H bond. D: ethanol is an alcohol; its –OH group is joined to a carbon atom that has no C=O.

B: ethanol is oxidised to ethanoic acid by oxygen in the air: C₂H₅OH + O₂ → CH₃COOH + H₂O. The same change happens when ethanol is warmed with acidified potassium manganate(VII), which is the oxidising agent. A, C: nitrogen is unreactive and takes no part in the change. C, D: ethanol gains oxygen, so it is oxidised, not reduced; the oxygen (the oxidising agent) is the substance that is reduced.

A: the name of an ester has two parts: the first part comes from the alcohol (ethanol, ethyl) and the second from the acid (butanoic acid, butanoate), so ethyl butanoate is made from butanoic acid and ethanol. B: this pair forms methyl butanoate. C: the two parts are swapped: ethanoic acid and butan-1-ol form butyl ethanoate. D: this pair has the same total number of carbon atoms (6) but forms propyl propanoate, an isomer of ethyl butanoate.

B: in addition polymerisation the ethene molecules, C₂H₄, join together without losing any atoms, so M_r of the polymer = n × M_r of ethene. M_r of C₂H₄ = 28, so n = 56 000 ÷ 28 = 2000. A: divides by 56 instead of 28 (a monomer of M_r 56). C: divides by 14, the M_r of a CH₂ group, so this counts CH₂ units, not ethene molecules. D: takes the relative molecular mass of the polymer as the number of molecules.

C: the links between the units are amide links, –CO–NH–, so this is a polyamide, nylon. Nylon is made by condensation polymerisation: two different monomers join together and a small molecule (water) is lost each time a link forms. A, B: an addition polymer is made from monomers that contain C=C double bonds, has a backbone of carbon atoms and loses no small molecule; this polymer has amide links in its chain and forms with the loss of water. D: Terylene is also made by condensation polymerisation, but it is a polyester with ester links, –CO–O–, and no nitrogen atoms.

A: breaking a polymer down into its monomers is depolymerisation (statement 1), and the hydrolysis of a polyester uses acid as a catalyst (statement 2). Statement 3 is not correct: water is a reactant, not a product. Each ester link is split by adding water; water is lost, not formed, when the polyester is made by condensation polymerisation. B and D each leave out one of the two correct statements. C includes statement 3.

B: carbon monoxide comes from the incomplete combustion of carbon-containing fuels, which happens in a limited supply of air (statement 1); nitrogen oxides are formed in vehicle engines and by lightning (statement 2); and sulfur dioxide comes from burning fossil fuels that contain sulfur and from volcanoes (statement 4). Statement 3 is not correct: lightning is a source of nitrogen oxides, not of carbon monoxide. A, C and D each include statement 3.

D: the ozone layer absorbs harmful ultraviolet radiation from the Sun, which protects living things (for example from skin cancer). It is broken down by chlorine atoms released from chlorofluorocarbons (CFCs), which were used as refrigerants and aerosol propellants. A, B: absorbing infrared radiation from the Earth is the role of greenhouse gases in the greenhouse effect, not of the ozone layer. A, C: carbon dioxide is a greenhouse gas linked with global warming; it is not the main cause of ozone depletion.