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

B: amount of Mg = 0.12 ÷ 24 = 0.0050 mol. The equation shows that 1 mol of Mg gives 1 mol of H₂, so 0.0050 mol of hydrogen is made, a volume of 0.0050 × 24 000 = 120 cm³. Way 2 can hold and measure 120 cm³: the insoluble hydrogen pushes water out of the 250 cm³ measuring cylinder, and its scale is read. Way 1: a gas syringe does measure gas volumes, but this one holds only 100 cm³, so the plunger would be pushed out before all the hydrogen was collected. Ways 3 and 4: a gas jar has no scale, so it can collect the gas (way 4 can, since hydrogen is less dense than air) but cannot measure its volume. A: way 2 can be used as well. C: way 1 is too small. D: ways 3 and 4 have no scale.

C: R melts over a range just below the melting point of Z (37–40 °C compared with 41 °C) and boils over a range just above the boiling point of Z (292–298 °C compared with 290 °C). An impurity lowers the melting point, raises the boiling point and makes both spread over a range, so R is Z with an impurity. A: P melts and boils sharply at 41 °C and 236 °C, so it is pure X (Z has the same melting point but a much higher boiling point). B: Q has a boiling point above that of Z, but its melting point (78–81 °C) is just below that of Y (82 °C) and its boiling point is also just above that of Y (286 °C), so it is impure Y. D: S melts and boils sharply at 41 °C and 290 °C, so it is pure Z, not impure Z. Both the melting point and the boiling point are needed to identify a compound.

D: a charge of 2+ means the atom has lost 2 electrons, so the ion has 18 + 2 = 20 protons (the element is calcium) and the nucleon number is 20 protons + 22 neutrons = 42, giving ⁴²₂₀Ca²⁺. A: sulfur has 16 protons and its ion S²⁻ has 18 electrons, but the charge here must be 2+ (electrons lost), not 2−. B: argon has 18 protons; this treats the number of protons as the number of electrons and ignores the charge. C: 20 protons is right, but the nucleon number is 40 = 18 + 22, which adds the neutrons to the electrons instead of to the protons.

D: in the dot-and-cross diagram of CO₂ (O=C=O) the carbon atom shares all 4 of its outer electrons and each oxygen atom keeps two non-bonding pairs, so there are 2 × 4 = 8 non-bonding electrons. In O₂, each oxygen atom also keeps two non-bonding pairs (the atoms share two pairs), so there are 8 non-bonding electrons as well. A, B: CH₄ and H₂ have no non-bonding electrons: every outer electron is in a shared pair. C: in H₂O the oxygen atom has two non-bonding pairs, which is only 4 non-bonding electrons.

D: the atoms have 1 + 4 + 5 = 10 outer electrons altogether, which is 5 pairs. Hydrogen shares one pair with carbon. Carbon needs 8 electrons, so it shares three more pairs with nitrogen (a triple bond). Nitrogen then has 6 electrons from these three shared pairs, so its last pair (5 − 3 = 2 electrons) is one pair that is not shared (a lone pair), which completes its 8. A: with only 2 shared pairs carbon would have just 6 electrons around it. B: nitrogen would have 8 electrons around it, but carbon would have only 6, so carbon would not have a noble gas configuration. C: with 3 shared pairs and no pair that is not shared, nitrogen would have only 6 electrons around it and one pair of electrons would be missing.

B: in (NH₄)₃PO₄ the 3 outside the bracket multiplies everything inside it: 3 N + 12 H + 1 P + 4 O = 20 atoms. In CO(NH₂)₂ the 2 multiplies the NH₂: 1 C + 1 O + 2 N + 4 H = 8 atoms. The difference is 20 − 8 = 12. A: 8 is the difference between the numbers of hydrogen atoms only (12 − 4), or the total number of atoms in urea. C: 15 comes from counting urea as only 5 atoms, ignoring the 2 outside its bracket. D: 28 is the two totals added together (20 + 8), not subtracted.

B: amount of HCl = 0.0200 dm³ × 0.050 mol / dm³ = 0.00100 mol. The equation shows 2 mol HCl react with 1 mol Ca(OH)₂, so the amount of Ca(OH)₂ = 0.000500 mol in 25.0 cm³. Concentration = 0.000500 mol ÷ 0.0250 dm³ = 0.0200 mol / dm³; mass concentration = 0.0200 × 74 = 1.48 g / dm³. A: 0.74 comes from halving twice (the 1 : 2 ratio applied a second time). C: 1.85 comes from dividing by 20.0 cm³ (the acid volume) instead of 25.0 cm³. D: 2.96 comes from assuming that 1 mol of HCl reacts with 1 mol of Ca(OH)₂.

D: neutralisation is H⁺ + OH⁻ → H₂O, so H⁺ and OH⁻ ions react in a 1 : 1 ratio. The mixture has 0.025 × 0.10 = 0.0025 mol of H⁺ and 0.025 × 0.20 = 0.0050 mol of OH⁻, so all the H⁺ is used up and 0.0025 mol of OH⁻ is left over. The mixture now has a volume of 25 + 25 = 50 cm³, so the concentration of OH⁻ is 0.0025 ÷ 0.050 = 0.050 mol / dm³. A, B: the acid is the smaller amount, so all the H⁺ ions are used up and none is left over. C: 0.10 mol / dm³ comes from dividing 0.0025 mol by 0.025 dm³, the volume of the sodium hydroxide only; after mixing the volume is 0.050 dm³.

B: sulfur dioxide would turn acidified potassium manganate(VII) from purple to colourless, so it is absent. A white precipitate with limewater shows carbon dioxide (and, because sulfur dioxide is absent, it cannot be sulfur dioxide). Hydrogen ‘pops’ with a lighted splint. So the sample contains carbon dioxide and hydrogen only. A, C, D: each includes sulfur dioxide (3), which the first test has ruled out.

C: the elements in the Periodic Table are arranged in order of increasing proton number, and each element has one more proton than the one before it, so the element after uranium has 92 + 1 = 93 protons. A: 91 is the proton number of the element before uranium. B: 92 is the proton number of uranium itself. D: 94 is two places after uranium.

C: the first two results show that chlorine is more reactive than bromine, and bromine is more reactive than iodine, so reactivity decreases down Group 17. Astatine, below iodine, is the least reactive, so iodine is more reactive than astatine and displaces astatine from potassium astatide. A: astatine is less reactive than iodine, so it cannot displace iodine. B: bromine is less reactive than chlorine, so it cannot displace chlorine from potassium chloride. D: iodine is less reactive than bromine, so it cannot displace bromine from potassium bromide.

A: the reaction of the Group 1 metals with water becomes more vigorous down the group, so rubidium, below potassium, is predicted to react more vigorously. B: the density of potassium (0.86) is lower than that of sodium (0.97), so the densities do not increase steadily down the group. C: sodium (0.97) and potassium (0.86) are also less dense than water (1.0), so they float as well; only rubidium sinks. D: lithium has the highest melting point and is the least reactive of these metals with water.

A: powdered limestone has a larger surface area, so more acid particles collide with the limestone each second and the reaction is faster: the curve is steeper than curve X. The acid is in excess, so the amount of carbon dioxide made depends on the limestone, and 1.0 g is half of 2.0 g: the curve levels off at half the volume of curve X. Only W is both steeper and levels off lower. B: X is the original curve, the result if nothing were changed. C: Y is steeper (faster) but levels off at the same volume as X, which would need the same mass of limestone as before (2.0 g of powder). D: Z levels off at half the volume, but it is less steep, so the reaction is slower, which is not the effect of powdering.

C: bioethanol is made from plants such as sugarcane, which can be grown again, so it is a renewable fuel and an alternative to the non-renewable fuels natural gas and crude oil. Like any fuel that contains carbon, it forms carbon dioxide when it burns. A: bioethanol is made from plants, not from crude oil, and a fuel made from crude oil is non-renewable. B: bioethanol does form carbon dioxide when it burns (the sugarcane takes in the same gas as it grows, which is why it is sustainable). D: sugarcane regrows, so the fuel made from it is renewable, not non-renewable.

A: fuel oil has large molecules and more is obtained (40%) than is needed (25%), while the demand for petrol (40%) is greater than the share obtained (25%). Cracking breaks large hydrocarbon molecules into smaller, more useful ones (petrol and alkenes), so fuel oil is cracked to match the supply to the demand. B: petrol already has small molecules; cracking breaks molecules down and cannot be used to make more of the fraction that is cracked. C: 20% of diesel is obtained and 20% is needed, so there is no surplus to crack. D: more fuel oil is obtained (40%) than is needed (25%), not less.

C: all the alcohols contain the –OH group, and each member of the series differs from the one before by one CH₂ group, so the formula of pentanol is C₅H₁₁OH (C₄H₉OH + CH₂). Its M_r = 5 × 12 + 12 × 1 + 16 = 88 (or 74 for butanol + 14 for one CH₂ group). A, B: these come from formulae with too few hydrogen atoms, C₅H₉OH (86) and C₅H₁₀OH (87), which do not fit a series that adds CH₂ each time. D: 89 comes from C₅H₁₂OH, which has one hydrogen atom too many: five carbon atoms with one –OH group have only 11 hydrogen atoms on the carbon atoms.

B: a polymer is a large molecule built up from many small units (monomers). The relative molecular mass of one ethene molecule, C₂H₄, is 2 × 12 + 4 × 1 = 28, so the number of ethene molecules joined is 28 000 ÷ 28 = 1000. Each ethene molecule contributes 2 carbon atoms, so there are 1000 × 2 = 2000 carbon atoms. A: 1000 is the number of ethene molecules (monomers), not the number of carbon atoms. C: 24 000 is the total relative mass of the carbon atoms (2000 × 12), not their number. D: 28 000 is the relative molecular mass of the whole polymer molecule.

A: in an addition polymer the C=C bond of each monomer has opened, so the chain has only single bonds, and each pair of neighbouring carbon atoms in the chain came from one monomer molecule. In the chain the carbon atoms alternate: one has two H atoms (CH₂), the next has one H and one F (CHF). So each monomer is CH₂=CHF, with the C=C bond restored. B: CHF=CHF would give a chain in which every carbon atom has one H and one F. C: CH₂=CF₂ would give a chain of alternating CH₂ and CF₂ carbon atoms. D: ethene would give a chain with no fluorine atoms at all.

A: dry air is approximately 78% nitrogen and 21% oxygen, and the remainder (about 1%) is the noble gases, mainly argon, with a trace of carbon dioxide. Neither stage removes these gases. After both stages about 1% of 2.00 dm³ = 0.020 dm³ is left. B: 0.42 dm³ (21%) is the volume of oxygen that stage 1 removes. C: 1.56 dm³ (78%) is the volume of nitrogen that stage 2 removes. D: 1.58 dm³ (79%) is the volume left after stage 1 only; stage 2 has been ignored.

B: respiration releases carbon dioxide into the atmosphere, so it increases the amount; photosynthesis takes carbon dioxide in to make glucose, so it decreases the amount. Together with combustion, these processes in the carbon cycle keep the amount of carbon dioxide in the atmosphere roughly constant. A, C: respiration does not remove carbon dioxide from the atmosphere; it releases it. C, D: photosynthesis does not add carbon dioxide to the atmosphere; it takes it in.