SCIENCE (CHEMISTRY) – Paper 1 Multiple Choice (Chemistry section) – mark scheme

C: propanone dissolves Q and R, so the first filtrate holds Q and R together; evaporating it leaves a mixture of the two. P and S stay on the filter paper. Water then dissolves P (R has already gone), so evaporating the second filtrate leaves pure P. S dissolves in neither liquid, so it is the final residue, which is pure once it is washed and dried. Only P and S are pure. A: Q is not pure, because R also dissolves in propanone and the two are mixed in the first filtrate. B: R is not pure, for the same reason: it is in the first filtrate with Q. D: Q and R are the two solids that end up together in the first filtrate, so neither is obtained on its own.

B: in the nuclide notation the upper number is the nucleon number and the lower number is the proton number. Protons: 7 in nitrogen + 4 × 1 in the four hydrogen atoms = 11. Neutrons = nucleon number − proton number: nitrogen 14 − 7 = 7, and each hydrogen atom 1 − 1 = 0, so the ion has 7 neutrons. Electrons: the ion has a charge of 1+, so it has one electron fewer than protons, 11 − 1 = 10. A: the charge has been taken off the number of protons; the protons do not change, only the electrons do. C: 11 is the number of electrons in the neutral group of atoms; the 1+ charge means one electron has been lost. D: this gives each hydrogen atom 1 neutron; a hydrogen atom of nucleon number 1 and proton number 1 has no neutron.

B: oxygen has 6 outer electrons and each hydrogen atom has 1. Each hydrogen atom shares its electron with oxygen in a covalent bond, so there are two shared pairs (a dot and a cross in each overlap). Oxygen keeps its other four electrons as two lone pairs, giving it 8 electrons in its outer shell, and each hydrogen atom has 2. A: the lone pairs on oxygen are missing, so oxygen has only 4 electrons around it. C: only one lone pair is shown, so oxygen has 6 electrons around it instead of 8. D: this shows the transfer of electrons from hydrogen to oxygen to make ions, which is ionic bonding; water is a covalent molecule.

C: nitrogen has 5 and each fluorine has 7 outer electrons, 5 + 3 × 7 = 26 in all. Each atom ends with 8 electrons. The three N–F bonds are three shared pairs, which use 6 electrons. The other 26 − 6 = 20 electrons are non-bonding: one lone pair on nitrogen (2 electrons) and three lone pairs on each fluorine atom (3 × 6 = 18). A: 6 is the number of electrons that are shared in the three bonds. B: 18 forgets the lone pair on the nitrogen atom. D: 26 is the total number of outer shell electrons, including the ones that are shared.

C: amount of X = 5.6 ÷ 56 = 0.10 mol and amount of Y = 4.8 ÷ 32 = 0.15 mol. X and Y react one atom to one atom, so 0.10 mol of X uses 0.10 mol of Y and X is used up. 0.05 mol of Y (0.05 × 32 = 1.6 g) is left over. So the solid is a mixture of the compound Z and the element Y. Statement 2 is correct: some Y is left. Statement 4 is correct: nothing escapes from the sealed tube, so the mass is still 5.6 + 4.8 = 10.4 g. 1: the solid contains left-over Y as well as the compound Z, so it is a mixture, not a pure compound. 3: all of X has reacted, so none is left to be attracted to a magnet, and Z is not magnetic.

A: in each formula the positive and negative charges must cancel. Aluminium nitrate: one Al³⁺ (+3) needs three NO₃⁻ (−3), written Al(NO₃)₃ with a bracket round the nitrate. Ammonium sulfate: two NH₄⁺ (+2) balance one SO₄²⁻ (−2), written (NH₄)₂SO₄. Magnesium nitride: three Mg²⁺ (+6) balance two N³⁻ (−6), written Mg₃N₂. B: Mg₂N₃ has the numbers the wrong way round: +4 and −9 do not cancel. C: NH₄SO₄ has +1 and −2, so the charges do not cancel; two ammonium ions are needed. D: AlNO₃ has +3 and −1, so the charges do not cancel; three nitrate ions are needed.

C: concentration = amount of solute ÷ volume in dm³. 250 cm³ = 0.250 dm³, so concentration = 0.050 mol ÷ 0.250 dm³ = 0.20 mol / dm³. A: 0.00020 divides by 250 without changing cm³ to dm³. B: 0.0125 multiplies by 0.250 instead of dividing. D: 12.5 multiplies by 250 instead of dividing by 0.250.

B: calcium hydroxide is a base, so it neutralises the acid in the soil and raises the pH. The mass is read per square metre: 450 g over 10 m² is 45 g per m², and the graph gives a pH of about 7.1, inside the range 6.5 to 7.5. A: 200 g is only 20 g per m², giving a pH of about 5.6, which is still too acidic. C: 600 g is 60 g per m², giving a pH of about 7.9, too alkaline. D: 800 g is 80 g per m², giving a pH of about 8.9, much too alkaline.

C: damp blue litmus paper that turns red and is then bleached white shows chlorine, so X is chlorine. Damp red litmus paper that turns blue shows an alkaline gas, ammonia, so Z is ammonia. The gas left over is carbon dioxide, so Y is carbon dioxide (it would turn limewater cloudy). A: X cannot be ammonia; ammonia turns damp red litmus paper blue and does not bleach damp blue litmus paper. B: carbon dioxide turns damp blue litmus paper red but does not bleach it, so X is not carbon dioxide. D: Y and Z are swapped: ammonia is the gas that turns damp red litmus paper blue, so it is in cylinder Z.

A: the electronic configurations are W 2; X 2.1; Y 2.8; Z 2.8.1. X (lithium) and Z (sodium) each have 1 outer shell electron, so they are in the same group and react in a similar way. B: W (helium) and Y (neon) are both unreactive noble gases, but not because their outer shells hold the same number of electrons: W has 2 and Y has 8. They are alike because each has a full outer shell. C: X and Y both have 2 shells, but the number of shells does not decide how an element reacts: X has 1 outer electron and is a reactive metal, while Y has a full outer shell and is unreactive. D: Y has 2 shells and Z has 3, and Y (a full outer shell) and Z (1 outer electron) do not behave alike.

D: the Periodic Table puts the element with proton number 85 in Group 17, directly below iodine. Down Group 17 the elements get darker, so Q is dark in colour. Group 17 atoms each gain one electron to form 1− ions, and sodium forms Na⁺, so the compound is NaQ. Reactivity decreases down the group (chlorine displaces bromine, bromine displaces iodine), so Q is less reactive than iodine and does not displace iodine from aqueous potassium iodide. A: Na₂Q needs a 2− ion; Group 17 elements form 1− ions, so the formula is NaQ. B: Q is below iodine, so it is less reactive than iodine and cannot displace it. C: the colour gets darker down the group, not paler.

A: an argon atom (2.8.8) already has a full outer shell of 8 electrons, which is a very stable arrangement, so it does not need to gain, lose or share electrons and exists as single atoms. A chlorine atom (2.8.7) has one electron fewer than a full shell, so two chlorine atoms each share one pair of electrons and form a Cl₂ molecule. B: the outer shell of an argon atom is not empty; it holds 8 electrons. C: an argon atom has 8 outer shell electrons, not 1. D: argon does have one more proton than chlorine, but this is not the reason; argon atoms do not join because their outer shell is already full.

D: each change speeds up the reaction because the particles collide more often, but for a different reason. Experiment 1: crushing a solid exposes more of its particles to the other reactant, so there are more collisions each second. Experiment 2: a higher pressure squeezes the gas particles closer together (more particles in the same volume), so they collide more often. Experiment 3: heating makes the particles move faster, so they collide more often. A: crushing a solid does not make the particles move faster; it exposes more of them. B: raising the pressure does not make the particles move faster; it pushes them closer together. C: heating does not bring the particles closer together; it makes them move faster.

D: a biofuel is more sustainable because the carbon dioxide given out when it burns is offset by the carbon dioxide the plants took in while they grew. The net amount added to the air = given out − taken in. For Z this is 22 − 21 = 1 unit, the smallest. W: petrol comes from crude oil, with no carbon dioxide taken in, so it adds all 31 units. X: the net is 19 − 14 = 5 units; it gives out the least carbon dioxide when it burns, but the offset is smaller. Y: the net is 24 − 22 = 2 units; it takes in the most carbon dioxide, but it gives out more than Z.

B: only unsaturated hydrocarbons, which have a C=C double bond, decolourise aqueous bromine, so the alkenes are needed. Alkanes are C_nH_2n+2, with M_r = 14n + 2; alkenes are C_nH_2n, with M_r = 14n. P: 42 = 14 × 3, so C₃H₆, an alkene. Q: 30 = 14 × 2 + 2, so C₂H₆, an alkane. R: 56 = 14 × 4, so C₄H₈, an alkene. S: 58 = 14 × 4 + 2, so C₄H₁₀, an alkane. So P and R decolourise aqueous bromine. A, C, D: each pair contains a saturated alkane (Q or S), which does not react with aqueous bromine, so the orange colour stays.

B: complete combustion of an alcohol forms carbon dioxide and water: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O. M_r of ethanol = 2 × 12 + 6 × 1 + 16 = 46, so 2.3 g is 2.3 ÷ 46 = 0.050 mol. Each mole of ethanol makes 2 mol of carbon dioxide, so 0.10 mol is formed, and 0.10 × 24 = 2.4 dm³. A: 1.2 dm³ takes 1 mol of CO₂ from each mole of ethanol and ignores the 2 in the equation. C: 3.6 dm³ is the volume of oxygen used (0.15 mol), not the carbon dioxide formed. D: 6.0 dm³ counts the water as a gas as well; at room temperature and pressure the water is a liquid, so only the carbon dioxide counts.

D: poly(ethene) is made by addition polymerisation: the C=C double bond of each ethene molecule opens and the molecules join end to end, and the polymer is the only product. The polymer has only C–C single bonds, so it is saturated and does not decolourise aqueous bromine (3 is correct). No atoms are lost, so poly(ethene), (C₂H₄)_n, has the same composition as ethene, and equal masses contain the same number of hydrogen atoms (4 is correct). 1: no water or any other small molecule is formed; the molecules simply add together. 2: the ratio of carbon to hydrogen atoms is 1 : 2 in ethene, C₂H₄, and also in poly(ethene), because every atom of the ethene ends up in the polymer.

D: in addition polymerisation the C=C double bond of the monomer opens and the other atoms stay on the carbon they were on. The chain alternates between a carbon with two H atoms and a carbon with two F atoms, so each repeat unit is –CH₂–CF₂– and the monomer is CH₂=CF₂, with both H atoms on one carbon and both F atoms on the other. A: CH₂=CHF would give a chain with only one F atom on every second carbon. B: CF₂=CF₂ would give two F atoms on every carbon of the chain. C: CHF=CHF has two F atoms but one on each carbon, so it would give one H and one F on every carbon of the chain.

A: acidified potassium manganate(VII) is decolourised by sulfur dioxide, so W is sulfur dioxide, a common atmospheric pollutant (it forms acid rain). X: a gas that relights a glowing splint is oxygen, which makes up 21% of clean air. Y: a gas that burns with a ‘pop’ is hydrogen, which is not an atmospheric pollutant. Z: a gas that puts out a lighted splint and does not affect limewater is nitrogen (or a noble gas), which is a normal constituent of clean air.

A: carbon dioxide and methane are greenhouse gases; more of them in the atmosphere may contribute to global warming, which melts polar ice and raises sea levels. B: global warming is expected to cause more extreme weather events. C: warming would make polar ice thinner, not thicker. D: acid rain is caused by sulfur dioxide and nitrogen dioxide, not by carbon dioxide and methane.