CHEMISTRY – Paper 1 Multiple Choice – mark scheme

A: time is measured with a stopwatch, temperature with a thermometer, and a volume of liquid such as 100 cm³ of water with a measuring cylinder. B: the thermometer and the measuring cylinder are swapped; a measuring cylinder does not measure temperature. C: the stopwatch and the thermometer are swapped; a thermometer does not measure time. D: a gas syringe measures the volume of a gas, not the volume of a liquid.

B: add water and stir (2) so that the sodium chloride dissolves; filter (1) to remove the chalk, which is insoluble and stays in the filter paper; heat the filtrate (4) to evaporate some of the water and make the solution more concentrated; then leave it to cool (3) so that crystals of sodium chloride form. A: the mixture must be dissolved before it is filtered; filtering the dry mixture separates nothing. C: there is no filtrate to heat until the mixture has been filtered, so step 4 cannot come before step 1; heating the undissolved mixture instead would leave the sodium chloride mixed with the chalk. D: the solution must be made more concentrated by heating before it is cooled; a dilute solution does not form crystals when it cools.

C: ammonia molecules (relative molecular mass 17) are lighter than hydrogen chloride molecules (relative molecular mass 36.5), so ammonia diffuses faster and travels further along the tube in the same time. The gases meet in the half of the tube nearer the hydrochloric acid, at about 60% of the way from the ammonia end, which is position Y. W, X: these are nearer the ammonia end; the white solid would form there only if the heavier hydrogen chloride diffused faster than ammonia. Z: this is too close to the hydrochloric acid end; the white solid would form there only if ammonia diffused about four times faster than hydrogen chloride.

D: an electron has a relative charge of −1 and a very small relative mass, about 1 / 1840 of the mass of a proton. A: a proton has a relative charge of +1, not −1. B: a neutron has a relative mass of 1, the same as a proton, not 1 / 1840. C: an electron is negatively charged (−1), not +1.

B: the proton number is the number of protons in the nucleus, 26. The nucleon number is the total number of protons and neutrons in the nucleus, 26 + 30 = 56. A: 30 is the number of neutrons, not the nucleon number. C: 30 is the number of neutrons, not the proton number. D: the two numbers are the wrong way round.

A: a magnesium atom has the electronic configuration 2,8,2. It loses its 2 outer electrons to form Mg²⁺, which has 10 electrons arranged 2,8: the same as the noble gas neon. B: the atom has only 2 outer electrons; a metal atom loses electrons to form a positive ion and does not gain them, and 2,8,4 would not be a noble gas configuration. C: the loss of 2 electrons and the 2+ charge are right, but the ion has the configuration 2,8 of neon, not 2,8,8 of argon. D: gaining 6 electrons would give the configuration of argon, but metal atoms do not form negative ions in this way; magnesium reaches a full outer shell by losing 2 electrons.

C: a chlorine atom has 7 outer electrons. Two chlorine atoms share one pair of electrons (a single covalent bond), so each atom has 7 + 1 = 8 outer electrons, the configuration of argon. A: electrons are transferred when ions form between a metal and a non-metal; two non-metal atoms of the same element share electrons. B: sharing two pairs would give each atom 9 outer electrons; one shared pair is enough to give each atom 8. D: a sea of delocalised electrons is found in metals, not in a molecule of a non-metal.

B: each oxygen atom has 6 outer electrons. In an oxygen molecule the atoms share two pairs of electrons (a double bond) and each atom keeps two non-bonding pairs, so each atom has 4 + 4 = 8 electrons around it and has contributed 2 + 4 = 6 electrons. A: one shared pair with 5 other electrons on each atom gives each atom only 7 electrons around it, and so it is not a full outer shell. C: one shared pair and 6 other electrons on each atom has too many electrons: each atom would have provided 7 electrons, but an oxygen atom has only 6 outer electrons. D: two shared pairs and 6 other electrons on each atom has far too many electrons: each atom would have provided 8 electrons, and would have 10 around it.

D: a metal is a regular lattice of positive ions surrounded by a sea of delocalised electrons that are free to move between the ions. A: positive and negative ions alternating in a lattice is an ionic compound such as sodium chloride, not a metal; a metal has no negative ions. B: separate small molecules are a simple molecular substance such as iodine or water. C: atoms joined by covalent bonds in a continuous network is a giant covalent structure such as diamond or silicon dioxide.

A: the symbols are Na (sodium, from its Latin name natrium), K (potassium, from kalium) and Mg (magnesium). B: So is not the symbol of any element; sodium is Na. C: P is the symbol of phosphorus; potassium is K. D: Mn is the symbol of manganese; magnesium is Mg.

B: silver ions and chloride ions are in aqueous solution, and they join to form insoluble silver chloride, a solid (s). The equation is balanced for atoms and for charge. A: silver chloride is a precipitate, so its state symbol is (s), not (aq). C: the reactants are silver ions and chloride ions in solution, not solid silver and chlorine atoms (chlorine gas is Cl₂). D: the formula of silver chloride is AgCl, because silver forms Ag⁺ ions; AgCl₂ is not balanced for charge.

A: the relative molecular mass of water is 18, so 18 g of water is 1 mole. One mole of any substance contains the Avogadro constant of particles, 6.0 × 10²³; here the particles are molecules. B: each molecule of water has 3 atoms, so 1 mole of water contains 3 × 6.0 × 10²³ = 1.8 × 10²⁴ atoms. C: each molecule has 2 hydrogen atoms, so there are 2 × 6.0 × 10²³ = 1.2 × 10²⁴ hydrogen atoms. D: this multiplies the Avogadro constant by 18, the mass in grams; the number of moles is 18 / 18 = 1, not 18.

C: the equation shows that 1 mol of ZnCO₃ forms 1 mol of ZnO, so 0.50 mol of ZnO forms. The relative formula mass of ZnO is 65 + 16 = 81, so the mass is 0.50 × 81 = 40.5 g. A: 0.50 × 44 = 22 g is the mass of carbon dioxide formed, not of zinc oxide. B: 0.50 × 65 = 32.5 g is the mass of the zinc alone; the oxygen in the zinc oxide has been left out. D: 81 g is the mass of 1 mol of zinc oxide; only 0.50 mol is formed here.

C: statement 1 is correct: the lower the pH, the higher the concentration of hydrogen ions, and the sports drink (pH 3.5) is less acidic than the stomach juice (pH 1.5). Statement 3 is correct: a pH above 7 means that hydroxide ions outnumber hydrogen ions, and the higher the pH, the higher the concentration of hydroxide ions; the soap solution (pH 10.0) is alkaline and the water is neutral. Statement 2 is wrong: water at pH 7.0 is neutral, so its concentrations of hydrogen ions and hydroxide ions are equal. A and D include statement 2; B includes statement 2 and leaves out statement 3.

D: calcium hydroxide is a base, and a base reacts with an ammonium salt to release ammonia gas (together with water and calcium chloride). Ammonia dissolves in the water on the litmus paper to form an alkaline solution, so damp red litmus paper turns blue. A: hydrogen chloride is not given off; the chloride ions stay in the calcium chloride that forms, and the gas released is ammonia. A gas that left red litmus paper red would be acidic or neutral, but ammonia is alkaline. B: the gas is ammonia, not hydrogen chloride; hydrogen chloride is acidic and would not turn red litmus paper blue. C: the gas is ammonia, but ammonia is alkaline and turns damp red litmus paper blue; it does not leave it red.

D: silver chloride is one of the two insoluble chlorides (the other is lead(II) chloride); all other chlorides are soluble. A: ammonium salts, and all carbonates of ammonium and Group 1 metals, are soluble. B: most sulfates are soluble; only the sulfates of barium, calcium and lead are insoluble, and magnesium sulfate is soluble. C: all nitrates are soluble, and so are all salts of potassium.

B: in a reversible reaction the ammonia breaks down to nitrogen and hydrogen as well as forming from them, so the reaction does not go to completion and all three gases are present. If 1.0 mol of ammonia has formed, then 0.5 mol of nitrogen and 1.5 mol of hydrogen have been used (1 : 3 : 2), leaving 1.0 − 0.5 = 0.5 mol of nitrogen and 4.0 − 1.5 = 2.5 mol of hydrogen. A: this would be a complete reaction: all of the nitrogen has gone, but a reversible reaction cannot use up a reactant completely. C: 1.0 mol of ammonia uses 1.5 mol of hydrogen, not 1.0 mol, so 2.5 mol of hydrogen would be left, not 3.0 mol. D: 1.0 mol of ammonia uses only 1.5 mol of hydrogen, leaving 2.5 mol, not 1.5 mol.

A: copper(II) ions form a light blue precipitate of copper(II) hydroxide with a little aqueous ammonia, and this precipitate dissolves in excess ammonia to give a dark blue solution. B: the precipitate is the right colour, but it does dissolve in excess ammonia; it is insoluble in excess aqueous sodium hydroxide, not in excess ammonia. C: a white precipitate that dissolves in excess ammonia to give a colourless solution is what zinc ions give. D: a white precipitate that does not dissolve in excess ammonia is what aluminium ions give; copper(II) ions do not give a white precipitate.

B: reduction is loss of oxygen, so copper(II) oxide, which loses oxygen to become copper, is reduced (statement 1); oxidation is gain of oxygen, so hydrogen, which gains oxygen to become water, is oxidised (statement 2). Statement 3 is wrong: loss of oxygen is reduction, not oxidation. A, C and D each include statement 3.

D: in changing from Fe²⁺ to Fe³⁺, the iron loses an electron and its oxidation state increases from +2 to +3; loss of electrons and an increase in oxidation state both mean the iron is oxidised. A: a decrease in oxidation state and a gain of electrons describe reduction, the opposite change (from Fe³⁺ to Fe²⁺). B: the iron cannot lose electrons and have a decrease in oxidation state; losing electrons raises the oxidation state. C: a gain of electrons would lower the oxidation state, not raise it.

D: the impure copper is the positive electrode. Its copper atoms lose electrons and dissolve as ions, Cu → Cu²⁺ + 2e⁻, so this electrode gets smaller. The copper(II) ions move to the negative electrode, a thin sheet of pure copper, where they gain electrons, Cu²⁺ + 2e⁻ → Cu, so the pure copper gets bigger. The copper(II) sulfate solution stays the same colour because copper dissolves from one electrode as fast as it is deposited on the other. A, B: the impure copper must be the positive electrode, so that it can lose electrons and dissolve; the negative electrode is where copper is deposited. C: the positive electrode loses copper, so it gets smaller, not bigger.

C: in a simple cell the further apart the two metals are in the reactivity series, the greater the voltage. Every metal X is more reactive than copper, so it is the negative electrode: it loses electrons, which flow through the wires to the copper strip, and the voltmeter reads the electrical energy produced by this redox reaction. Metal R gives the largest voltage (2.71 V), so it is furthest above copper in the reactivity series and is the most reactive. A, D: P (1.10 V) and S (0.78 V) give smaller voltages, so each is closer to copper in reactivity than R is. B: Q gives the smallest voltage (0.46 V), so it is the least reactive of the four.

B: write the electronic configuration of each atom. W (17) is 2,8,7; X (8) is 2,6; Y (19) is 2,8,8,1; Z (16) is 2,8,6. The group is given by the number of outer electrons: X and Z both have 6, so they are in the same group (Group 16). The period is given by the number of occupied shells: W and Z both have three shells, so they are in the same period (Period 3). A: W has 7 outer electrons and Y has 1, so they are not in the same group; X (two shells) and Z (three shells) are not in the same period. C: X and Z are in the same group, but W (three shells) and Y (four shells) are in different periods. D: W (7 outer electrons) and Z (6) are in the same period, not the same group, and X and Z are in the same group, not the same period; the two columns are swapped.

A: the Group 1 metals float on water because they have low densities, and their reaction with water becomes more vigorous down the group: lithium reacts gently, sodium reacts rapidly and potassium reacts so vigorously that the hydrogen ignites. P reacts most slowly and stays solid, so it is lithium. Q is the most vigorous (the hydrogen catches fire), so it is potassium. R is in between, melting into a ball without a flame, so it is sodium. B: this puts sodium as the most vigorous (Q) and potassium in the middle, the reverse of the trend. C: this gives the gentlest reaction (P) to sodium instead of lithium. D: this gives the gentlest reaction (P) to potassium, the most reactive of the three. The order of vigour is lithium, then sodium, then potassium, so P must be lithium and Q must be potassium.

D: the noble gases are monatomic. Argon atoms have a full outer shell, so they do not bond to each other or to any other atom, and the gas is made of separate single atoms. A: pairs of joined atoms of one element are diatomic molecules, like those of oxygen or nitrogen; argon does not form molecules. B: separate atoms of two different elements would be a mixture of two gases, not argon alone. C: joined atoms of two different elements are molecules of a compound.

C: typical transition elements are metals with a high melting point, a high density, a variable oxidation state and coloured compounds. Metal R has a high melting point (1538 °C), a high density (7.87 g per cm³) and coloured compounds, so it is the transition element (it is iron). A: P has a low melting point and a low density and its compounds are white, like the Group 1 metal sodium. B, D: Q and S have lower melting points and much lower densities than R, and their compounds are white; transition elements form coloured compounds.

A: copper(II) carbonate breaks down when it is heated strongly: CuCO₃ → CuO + CO₂. The green copper(II) carbonate changes to black copper(II) oxide, and the gas given off is carbon dioxide, which turns limewater milky. B: the solid does turn black, but the gas is carbon dioxide, not oxygen; oxygen is given off by some nitrates, not by carbonates. C: carbon dioxide is right, but copper(II) oxide is black, not white. D: neither the colour nor the gas is correct.

D: the temperature of the mixture falls, so the reaction takes in energy from its surroundings. It is endothermic, and the enthalpy change, ΔH, of an endothermic reaction is positive. A, B: an exothermic reaction gives out energy and makes the surroundings warmer, which would raise the temperature. C: an endothermic reaction does take in energy, but its ΔH is positive, not negative; a negative ΔH goes with an exothermic reaction.

B: the table shows that experiments 1 and 3 differ only in temperature. At the higher temperature the particles have more energy and move faster, so they collide more often, and a greater proportion of the collisions have at least the activation energy, so the rate of reaction is greater. A: the activation energy of a reaction does not change with temperature; only a catalyst provides a pathway with a lower activation energy. C: the surface area changes only if the metal is powdered, as in experiment 2; the magnesium in experiments 1 and 3 is ribbon in both. D: the volume and the concentration of the acid are the same in both experiments, so the concentration of the reacting particles is the same.

A: crude oil is a mixture of hydrocarbons, and the fractions are obtained by fractional distillation, which separates the hydrocarbons because they have different boiling points. A fraction has competing uses as a fuel and as a source of chemicals: the more of it that is burned as a fuel, the less is left for making chemicals such as plastics. B: the first part is right, but burning more of the fraction does leave less of it for making plastics; there is only a limited amount of each fraction. C, D: breaking down large molecules is cracking, a separate process carried out on some of the fractions; fractional distillation only separates the hydrocarbons already in the crude oil.

C: for each kilometre the carbon dioxide added to the air is the exhaust and the factory emissions, 1.2 + 0.3 = 1.5 kg, and the sugarcane took 1.0 kg out of the air as it grew. The net mass is 1.5 − 1.0 = 0.5 kg per kilometre, so 0.5 × 100 = 50 kg for the journey. A bus that burns petrol adds the whole 1.2 kg per kilometre (120 kg for 100 km), because the carbon in petrol was locked underground for millions of years, so the biofuel is the more sustainable choice. A: this subtracts the factory emissions as well as the uptake, 1.2 − 1.0 − 0.3, which ignores that the factory adds carbon dioxide to the air. B: 20 kg leaves out the 0.3 kg per kilometre released by the factory. D: 150 kg ignores the carbon dioxide taken in by the sugarcane as it grew.

B: isomers are compounds with the same molecular formula but different structural formulae, that is, the atoms are joined in a different way. For example, butane and methylpropane are both C₄H₁₀. A: this describes compounds such as ethene, C₂H₄, and propene, C₃H₆, which both have the empirical formula CH₂ but are different compounds with different molecular formulae. C: the same general formula with different numbers of carbon atoms describes members of one homologous series, such as ethane and propane, not isomers. D: ethane, C₂H₆, and ethene, C₂H₄, have the same number of carbon atoms but different numbers of hydrogen atoms, so their molecular formulae differ and they are not isomers.

C: propene is the alkene with three carbon atoms, so it has one C=C double bond and the formula C₃H₆. A: this is ethene, C₂H₄, which has only two carbon atoms. B: this has three carbon atoms, but all its bonds are single bonds, so it is the alkane propane, C₃H₈. D: this is but-1-ene, C₄H₈, which has four carbon atoms.

B: statement 1 is correct, because the alcohols are a homologous series whose members all contain the –OH group. Statement 2 is correct, because successive members differ by one CH₂ group (CH₃OH, C₂H₅OH, C₃H₇OH). Statement 3 is incorrect: the boiling points of a homologous series increase as the molecules get larger (methanol about 65 °C, ethanol about 78 °C, propan-1-ol about 97 °C), because there are stronger attractions between larger molecules. A: statement 2 is also correct. C, D: both include statement 3.

D: ethanol is made from ethene by the catalysed addition of steam: C₂H₄ + H₂O → C₂H₅OH. A: this is fermentation of glucose by yeast, the other method of making ethanol; it does not use ethene. B: ethene and hydrogen react to form ethane, C₂H₆, not ethanol. C: ethanol is a reactant here; it is oxidised by oxygen to ethanoic acid, so ethanol is not being made.

B: a polymer is a large molecule built up from many small units (monomers) joined end to end in a long chain. A: these are separate monomer molecules, not joined together. C: only two units are joined in each molecule, so these are small molecules, not polymers. D: these are separate molecules of two different monomers, not joined together.

C: ethene is an unsaturated hydrocarbon with a C=C double bond, so it decolourises aqueous bromine. In addition polymerisation the C=C bond of every ethene molecule opens up so that the molecules can join into a long chain, and nothing else is formed. Poly(ethene) has only single bonds between its carbon atoms, so it is saturated and aqueous bromine stays orange. A: poly(ethene) has no C=C double bonds left, so it does not decolourise bromine. B: ethene has a C=C double bond and does decolourise bromine. D: this is the wrong way round; the double bonds are in the monomer, not in the polymer.

A: the units in the figure are joined by –CO–O– links, so PET is a polyester and the links broken are ester links. Heating with water and an acid catalyst (acid hydrolysis) breaks each ester link, the reverse of the reaction in which the polymer formed. The hatched part is joined to the links through its oxygen atoms, so it becomes an alcohol (with –OH groups at both ends). The open part is joined to the links through carbon atoms of C=O groups, so it becomes a carboxylic acid (with –CO₂H groups at both ends). These are the monomers, which can be used to make new polyester. B: the links are ester links, but the two products are the wrong way round. C, D: an amide link, –CO–NH–, is found in nylon (a polyamide), not in a polyester; hydrolysis of an amide link gives an amine and a carboxylic acid.

B: dry air is about 78% nitrogen by volume, which makes nitrogen the most abundant gas; about 21% is oxygen and the remaining 1% or so is argon and other noble gases and carbon dioxide. A: nitrogen is the most abundant gas, but 21% is the percentage of oxygen, not nitrogen. C: 21% is right for oxygen, but oxygen is the second most abundant gas, not the most abundant. D: oxygen makes up about 21%, not 78%, of dry air; the 78% belongs to nitrogen.

D: sulfur dioxide (from burning fossil fuels in the power station) dissolves in rainwater to form acid rain, which damages buildings and harms the respiratory system. Carbon monoxide (from the incomplete combustion of fuel in traffic) is toxic because it combines with haemoglobin in the blood, so the blood carries less oxygen. A, B: carbon monoxide does not form acid rain (it is toxic, but it is not an acidic gas that dissolves in rain), and neither nitrogen dioxide nor sulfur dioxide is the gas that reduces the oxygen carried by the blood. C: nitrogen dioxide does help to form acid rain, but sulfur dioxide is not the gas that reduces the oxygen carried by the blood; that gas is carbon monoxide.