CHEMISTRY – Paper 1 Multiple Choice – mark scheme

B: mixture 1 is a solid-solid mixture in which only one solid dissolves, so water is added, the silicon(IV) oxide is filtered off and the copper(II) sulfate is crystallised from the filtrate; mixture 2 is two miscible liquids with different boiling points, so they are separated by fractional distillation; mixture 3 is two immiscible liquids, which form two layers that are run off one at a time with a separating funnel. A: sublimation only separates a solid that changes straight into a gas on heating; neither copper(II) sulfate nor silicon(IV) oxide does this. C: filtration separates an insoluble solid from a liquid; both oil and water are liquids and both pass through the filter paper. D: the methods for mixtures 2 and 3 are swapped: liquids that mix completely cannot be separated in a separating funnel because they do not form layers, and immiscible liquids do not need distillation.

B: in solvent 1 the R_f value is 4.8 ÷ 12.0 = 0.40, which matches P and Q; in solvent 2 it is 5.2 ÷ 8.0 = 0.65, which matches Q and R. Only Q matches in both solvents. A: P matches in solvent 1 only; in solvent 2 its R_f value is 0.30, not 0.65. C: R matches in solvent 2 only; in solvent 1 its R_f value is 0.72, not 0.40. D: S matches in neither solvent.

A: statements 1 and 2 are correct. The bromine vapour is denser than air, yet the brown colour spreads up into the upper jar, so its particles must move on their own (statement 1). The colour becomes even in a shorter time as the temperature rises (280 s, 170 s, 100 s), which shows that the particles move faster at higher temperatures (statement 2). Statement 3 is wrong: a gas with a smaller relative molecular mass has lighter particles, which move faster at the same temperature, so it spreads faster, not more slowly (hydrogen, M_r = 2, would spread much faster than bromine, M_r = 160). Statement 4 is wrong: the particles keep moving in all directions after the colour has become even; there is simply no more change in the overall colour. B, C and D each include statement 3 or statement 4.

C: isotopes are atoms of the same element (the same proton number, 14) with different numbers of neutrons (different nucleon numbers): silicon-28 has 14 neutrons and silicon-30 has 16. A: the atoms have different proton numbers (13 and 14), so they are atoms of different elements. B: the nucleon numbers are the same (28) but the proton numbers differ (13 and 14), so these are atoms of different elements, not isotopes. D: the nucleon numbers are the same (30) but the proton numbers differ (14 and 15), so these are atoms of different elements.

D: solid magnesium oxide is a giant lattice of Mg²⁺ and O²⁻ ions in which each ion is held by electrostatic attraction to the oppositely charged ions around it, as the diagram shows. There are no separate MgO molecules, so the mass is the relative formula mass of the simplest ratio of ions, MgO. A: the state of a substance does not decide which term is used; the structure does. B: a relative molecular mass is used for substances made of molecules, including molecular compounds such as water; being a compound is not the reason. C: magnesium oxide is ionic, not covalent: it has ions held by electrostatic attraction, not atoms sharing electrons.

A: ethanol is made of neutral molecules; it has no ions and no free electrons, so there are no mobile charged particles to carry a current. In molten sodium chloride the Na⁺ and Cl⁻ ions are free to move, so it conducts. B: the strength of the bonds does not decide whether a substance conducts; ionic bonds (electrostatic attraction between ions) are also strong. C: sodium chloride has no delocalised electrons; its current is carried by ions. D: ethanol contains no ions at all, so it cannot have ions fixed in place.

C: copper is a lattice of positive ions in a sea of delocalised electrons. The electrons are free to move and carry heat energy through the metal, so copper is a good conductor of heat (statement 1); the attraction between the positive ions and the delocalised electrons is strong, so a lot of energy is needed to overcome it and the melting point is high (statement 3). Statement 2 is wrong: the positive ions are held in fixed positions in the lattice and only vibrate; they do not move through it. A, B and D each include statement 2.

D: concrete is made by mixing cement, sand and gravel with water; these substances are not chemically combined in a fixed ratio and the proportions can vary, so concrete is a mixture. Silicon(IV) oxide, SiO₂, contains silicon and oxygen chemically combined in a fixed ratio, so it is a compound. Copper contains only one kind of atom, so it is an element. A: concrete is not a compound; its parts are not chemically combined in a fixed ratio. B: silicon(IV) oxide is not a mixture; its silicon and oxygen are chemically combined in a fixed ratio of 1 : 2. C: copper is an element, not a compound, because it contains only one kind of atom.

B: X is soft and conducts electricity, so it is graphite; Y is very hard and does not conduct, so it is diamond. Statement 1 is correct: in graphite each carbon atom is joined to three others in a layer, leaving one outer electron free to move, which is why graphite conducts. Statement 2 is correct: in diamond each carbon atom is joined to four others by strong covalent bonds in all directions, so diamond is very hard. Statement 4 is correct: both have a giant structure, and a lot of energy is needed to break the many strong covalent bonds (in graphite, those within each layer), so both have very high melting points. Statement 3 is wrong: the covalent bonds within each layer of graphite are strong; graphite is soft because the forces between the layers are weak, so the layers slide over each other. A leaves out the correct statement 4; C and D include statement 3.

C: the state symbol (g) after CO₂ shows that carbon dioxide is formed as a gas. A: the state symbol (s) after CaCO₃ shows that calcium carbonate is a solid; (aq) would show a solution. B: the state symbol (s) after CaO shows that calcium oxide is a solid. D: there is no number in front of CO₂, so one mole of carbon dioxide is formed from one mole of calcium carbonate.

B: the relative formula mass is the sum of the relative atomic masses of all the atoms in the formula: 2 × A_r(M) + 3 × 16 = 160, so 2 × A_r(M) = 160 − 48 = 112 and A_r(M) = 56. A: this divides by 2 twice (112 ÷ 2 ÷ 2). C: this is the mass of two atoms of M (160 − 48); it has not been divided by 2. D: this subtracts the mass of only one oxygen atom (160 − 16); there are three oxygen atoms.

C: amount of Si = 8.4 ÷ 28 = 0.30 mol; amount of N = 5.6 ÷ 14 = 0.40 mol. The ratio Si : N = 0.30 : 0.40 = 3 : 4, so the empirical formula is Si₃N₄. A: this comes from rounding 1 : 1.33 down to 1 : 1; the simplest whole-number ratio is 3 : 4. B: this uses the ratio of the masses (8.4 : 5.6 = 3 : 2) instead of the ratio of the amounts in moles. D: this has the ratio the wrong way round; it has more silicon than nitrogen, but the sample contains less silicon (0.30 mol) than nitrogen (0.40 mol).

C: HCl + NaOH → NaCl + H₂O, so the acid and the alkali react in a 1 : 1 ratio. Moles of NaOH = 0.80 × 25.0 ÷ 1000 = 0.020 mol, so there is 0.020 mol of HCl in 20.0 cm³. Concentration of HCl = 0.020 ÷ 0.0200 = 1.0 mol / dm³; in g / dm³ this is 1.0 × 36.5 = 36.5. A: this uses the volumes the wrong way round (0.80 × 20.0 ÷ 25.0 = 0.64 mol / dm³) before multiplying by 36.5. B: this is the concentration of the sodium hydroxide (0.80 mol / dm³) multiplied by 36.5; the volumes were not used. D: this uses the M_r of sodium hydroxide (40) instead of the M_r of hydrochloric acid.

D: the strength of an acid depends on how far it ionises in water, not on how much acid is dissolved. Hydrochloric acid is a strong acid: its molecules are fully ionised, even in a dilute solution. Ethanoic acid is a weak acid: only a small fraction of its molecules ionise, even in a concentrated solution. A and B: strong and weak describe the extent of ionisation; dilute and concentrated describe the amount of acid in a given volume of solution. C: ethanoic acid is only partly ionised however concentrated it is; a higher concentration does not make an acid ionise more fully.

D: sulfur dioxide is the oxide of a non-metal, so it is acidic and reacts with alkalis; zinc oxide is an amphoteric oxide, so it reacts with both acids and alkalis. A: calcium oxide and iron(III) oxide are oxides of metals and are basic; they react with acids but not with alkalis. B: carbon monoxide is a neutral oxide, and copper(II) oxide is basic; neither reacts with alkalis. C: nitrogen dioxide is the oxide of a non-metal and forms acid rain, so it is acidic and reacts with alkalis, but magnesium oxide is basic and does not.

C: lead(II) iodide is insoluble (all iodides are soluble except those of silver and lead), so it is made by precipitation: mix two soluble salts such as lead(II) nitrate and potassium iodide, filter, wash and dry. Potassium nitrate is soluble and every potassium compound is soluble, so there is no insoluble solid that can be filtered off; acid and alkali (potassium hydroxide) are neutralised exactly by titration with an indicator, then mixed again without the indicator and crystallised. Calcium nitrate is soluble, but calcium carbonate is an insoluble carbonate, so excess calcium carbonate can be added to dilute nitric acid until no more reacts, filtered off, and the filtrate crystallised. A: lead(II) iodide is insoluble, so it is not made by titration (which gives a solution of a soluble salt), and potassium nitrate is soluble, so no solid precipitates when its solutions are mixed. B: lead(II) iodide is not made by adding an excess of a solid to an acid, and calcium nitrate is soluble, so no solid precipitates when its solutions are mixed. D: there is no insoluble potassium compound to add in excess to an acid, so potassium nitrate cannot be made in this way.

D: in cracking, a large hydrocarbon molecule is broken into smaller molecules, and hydrogen can be one of the products; here propane breaks down into propene and hydrogen. A: this is the manufacture of ammonia, which uses up hydrogen. B: this is the combustion of propane in oxygen; it does not produce hydrogen. C: this adds hydrogen to propene to make propane; it uses up hydrogen.

B: in alkaline solution aluminium reduces nitrate ions to ammonia; ammonia is an alkaline gas, so it turns damp red litmus paper blue. A: ammonia is alkaline, so it turns damp red litmus paper blue; damp blue litmus paper would stay blue. C: aluminium also reacts with aqueous sodium hydroxide to give hydrogen, but hydrogen is formed whether or not nitrate ions are present, so it does not show nitrate ions. D: nitrogen dioxide is not formed in this test.

A: oxidation is loss of electrons, which is also an increase in oxidation state: magnesium atoms lose two electrons each and the oxidation state of magnesium changes from 0 to +2. B: this is the reverse change, with magnesium ions gaining electrons; gain of electrons is reduction. C: iron(II) ions gain two electrons to become iron atoms (oxidation state +2 to 0), which is reduction. D: iron(III) ions gain an electron to become iron(II) ions, which is also reduction (and iron(III) ions are not involved in this reaction).

A: the solid does not conduct because its ions are held in fixed positions in a lattice; when it is molten the ions are free to move and carry the current. Lead forms at the negative electrode, so the lead particles are positive ions that move there and gain electrons; bromine forms at the positive electrode, so the bromine particles are negative bromide ions that move there and lose electrons. B: the charges are the wrong way round; opposite charges attract, so negative ions cannot move to the negative electrode. C: lead(II) bromide is not made of atoms and molecules; neutral particles would not move to the electrodes, and the lead and bromine are formed at the electrodes from ions. D: delocalised electrons are found in metals; lead(II) bromide does not conduct when solid, and the lead and bromine form from ions, not from electrons moving through the liquid.

C: statement 1 is correct: in a concentrated solution of sodium chloride the chloride ions are discharged in preference to the hydroxide ions from the water, so chlorine forms at the anode. Statement 4 is correct: copper is below hydrogen in the reactivity series, so copper(II) ions are discharged at the cathode in preference to hydrogen ions from the water. Statement 2 is incorrect: in a dilute solution the hydroxide ions are discharged in preference to the chloride ions, so oxygen (4OH⁻ → O₂ + 2H₂O + 4e⁻), not chlorine, forms at the anode. Statement 3 is incorrect: sodium is above hydrogen in the reactivity series, so hydrogen ions from the water are discharged at the cathode and hydrogen forms, not sodium. A and D each include statement 2, B and D each include statement 3.

A: the electrolyte contains Mg²⁺, SO₄²⁻, H⁺ and OH⁻ ions. At the cathode (negative) magnesium is above hydrogen in the reactivity series, so the hydrogen ions from the water are discharged instead: 2H⁺ + 2e⁻ → H₂. At the anode (positive) sulfate ions are not discharged, so OH⁻ ions are: 4OH⁻ → O₂ + 2H₂O + 4e⁻, a loss of electrons (oxidation). B: magnesium ions are not discharged because magnesium is more reactive than hydrogen; magnesium metal is never formed from an aqueous solution. C: the anode equation is not balanced for charge: four OH⁻ ions carry a charge of 4−, so four electrons must be lost, not two. D: the two equations are at the wrong electrodes; electrons are gained (reduction) at the cathode, not the anode.

B: a metal atom forms a positive ion by losing its outer electrons, so R, whose ion has a charge of 2+, loses two electrons and has 2 outer electrons. A non-metal atom forms a negative ion by gaining electrons to fill its outer shell with eight, so S, whose ion has a charge of 3−, gains three electrons and has 8 − 3 = 5 outer electrons (it is in Group 15). A: the number of outer electrons equals the size of the charge only for a positive ion; the three electrons of S are gained, not the number it started with. C: 6 outer electrons would make R gain two electrons and form a 2− ion, not a 2+ ion; and 3 outer electrons for S takes the size of the charge as the number of outer electrons. D: this applies the rule "8 minus the charge" to R as well, but R forms a positive ion: it loses its 2 outer electrons.

D: elements in the same group have the same number of outer electrons, which is why their chemical properties are similar. X has six outer electrons. An atom with 16 protons has the configuration 2,8,6, so it also has six outer electrons and is in the same group as X. A: 6 protons gives 2,4: the atom has two electron shells like X, but four outer electrons, and having the same number of shells does not make the properties similar. B: 10 protons gives 2,8, a full outer shell, so this is an unreactive noble gas. C: 15 protons gives 2,8,5, with five outer electrons.

D: reactivity decreases down Group 17 (chlorine, bromine, iodine), and a halogen displaces a less reactive halogen from a solution of its halide. Bromine is less reactive than chlorine, so it cannot displace chloride ions: there is no reaction with potassium chloride. Bromine is more reactive than iodine, so it displaces iodide ions, forming iodine: Br₂ + 2I⁻ → 2Br⁻ + I₂. A: this says bromine cannot displace iodine, but iodine is below bromine in the group. B: this reverses the order: it has bromine displacing chlorine but not iodine. C: bromine does displace iodine, but it cannot displace the more reactive chlorine.

B: hydrogen can only remove the oxygen from the oxide of a metal that is below hydrogen in the reactivity series. Copper is below hydrogen, so copper(II) oxide is reduced to copper and steam forms. A, C, D: calcium, magnesium and sodium are all above hydrogen in the reactivity series, so they hold on to their oxygen more strongly than hydrogen does and their oxides are not reduced.

A: the more reactive a metal, the higher it is in the reactivity series and the more strongly it is held in its compounds, so the harder it is to obtain from its ore. Of the four metals, calcium is the most reactive (calcium, iron, copper, silver in order of decreasing reactivity), so it is the most difficult to obtain; it needs electrolysis of a molten compound. B, C, D: copper, iron and silver are all lower in the series than calcium, so their compounds are easier to break down; silver, the least reactive, is even found uncombined in nature.

D: the reactants are at 200 kJ / mol and the products at 50 kJ / mol, so the products have 200 − 50 = 150 kJ / mol less energy than the reactants. The enthalpy change is therefore −150 kJ / mol, and the reaction is exothermic. A: 300 kJ / mol is the energy at the top of the peak; the activation energy is measured from the level of the reactants, 300 − 200 = 100 kJ / mol. B: the products are lower in energy than the reactants, so energy is given out and the reaction is exothermic, not endothermic. C: the size of the enthalpy change is right but its sign is wrong; ΔH is negative when the products are lower in energy than the reactants.

D: an enzyme is a biological catalyst. A catalyst increases the rate of a reaction and is chemically unchanged at the end, so it is not used up. A, B: a catalyst does not slow a reaction down. A, C: a catalyst is not a reactant, so none of it is used up in the reaction, even though it may be present in a small amount.

A: enzymes are biological catalysts that work best at a particular temperature (here 40 °C) and are damaged by higher temperatures, so their catalytic action is lost. Catalyst Q gives the shortest time at 40 °C, then slows down, and at 80 °C it takes more than 600 s to collect the 20 cm³ of oxygen, so Q is the enzyme. The time taken with P halves with each 20 °C rise: P speeds the reaction up all the way, as a compound such as manganese(IV) oxide, which is a catalyst, does. B: Q is indeed faster than P at 20 °C (90 s against 160 s), but that does not show that it is an enzyme. C: a shorter time means a faster reaction, so P is getting faster as the temperature rises, which is what a non-enzyme catalyst does. D: P works at all four temperatures, which is true of an ordinary catalyst; it is the enzyme that stops working at a high temperature.

A: natural gas and crude oil formed over millions of years from the remains of living things, and they are being used up much faster than new supplies are formed, so they are non-renewable (statements 1 and 2). Statement 3 is true of the fuels but it is not the reason: a renewable fuel such as bioethanol also gives out carbon dioxide when it burns. Statement 4 is not the reason either: where a fuel is found does not decide whether it is renewable. B, C and D each include statement 3 or 4.

C: the members of a homologous series show a gradation in their physical properties as the molecules get bigger. Each extra CH₂ group raises the boiling point by about 20 °C (by 13, 19 and 21 °C in the table), so pentan-1-ol, with one more CH₂ than butan-1-ol, boils at about 118 + 20 = 138 °C. A: this is lower than the boiling point of butan-1-ol; the boiling point rises as the molecules get bigger. B: this is the boiling point of butan-1-ol; pentan-1-ol has a larger molecule, so its boiling point is higher. D: this needs a rise of 60 °C, about three times the rise for any earlier step.

B: alkanes are generally unreactive, but they burn and they undergo substitution with chlorine in ultraviolet light. Statement 1 is correct: a hydrogen atom of ethane is replaced by a chlorine atom, forming chloroethane and hydrogen chloride. Statement 4 is correct: the equation is balanced (4 C, 12 H and 14 O atoms on each side) and shows complete combustion. Statement 2 is incorrect: although the equation balances, the hydrogen atoms that are replaced leave as hydrogen chloride, not as hydrogen gas. Statement 3 is incorrect: ethane is saturated and has no C=C bond, so bromine cannot add to it, and it does not react with aqueous bromine in the dark. A and C each include statement 3, C and D each include statement 2.

C: alkenes have the general formula C_nH_2n, so a molecule has 3n atoms in total. 3n = 24, so n = 8 and the molecule is C₈H₁₆, with 2 × 8 = 16 hydrogen atoms. A: 8 is the number of carbon atoms. B: 12 is half of the atoms, which assumes that there are equal numbers of carbon and hydrogen atoms. D: 18 is the number of hydrogen atoms in an alkane with 8 carbon atoms, C₈H₁₈ (C_nH_2n+2), not in an alkene.

C: ethene adds hydrogen in an addition reaction: C₂H₄ + H₂ → C₂H₆, so one volume of ethene reacts with one volume of hydrogen to give one volume of ethane. The 30 cm³ of ethene reacts with 30 cm³ of hydrogen to form 30 cm³ of ethane, leaving 50 − 30 = 20 cm³ of hydrogen unreacted. The gas at the end is 30 + 20 = 50 cm³. A: 20 cm³ is the unreacted hydrogen only; the ethane formed is also a gas. B: 30 cm³ is the ethane only; the unreacted hydrogen is also a gas. D: 80 cm³ is the volume at the start; two volumes of gas (ethene and hydrogen) combine to make one volume of ethane, so the total volume decreases.

B: methanoic acid and ethanoic acid are carboxylic acids, which contain the –CO₂H group, and each member of a homologous series differs from the next by one CH₂ group. The molecular formulae are CH₂O₂, C₂H₄O₂, C₃H₆O₂, C₄H₈O₂ and C₅H₁₀O₂ (propanoic, butanoic and pentanoic acid follow ethanoic acid), so the number of hydrogen atoms is twice the number of carbon atoms, and the fifth member is C₅H₁₀O₂. A: this has two hydrogen atoms too few for a member of the series (C_nH_2nO₂); it would be C_nH_2n−2O₂. C: this has two hydrogen atoms too many, the pattern of an alkane (C_nH_2n+2), not of a carboxylic acid. D: C₆H₁₂O₂ is the sixth member, which comes from counting the first member as the zeroth, or adding one CH₂ too many.

C: hydrochloric acid is a strong acid and ethanoic acid is a weak acid, a carboxylic acid that reacts with some metals but is only partly ionised in water. In each experiment the magnesium is in excess and the two acids have the same concentration and volume, so they contain the same amount of acid (0.010 mol), and each gives the same volume of hydrogen (120 cm³). The weak acid has a lower concentration of hydrogen ions, so it reacts more slowly: curve Y is less steep than curve X, but it rises to the same final volume because more of the acid ionises as the hydrogen ions are used up. A: this shows the weak acid reacting faster than the strong acid. B: the two acids have the same concentration, but not the same concentration of hydrogen ions, so the rates are not the same. D: the volumes of hydrogen differ, but the acids contain the same amount of acid, so the same volume of gas is formed; a weak acid does not give less gas, it only reacts more slowly.

A: the polymer has amide links, –CO–NH–, so it is nylon, a polyamide made by condensation polymerisation. The –CO– end of each link comes from a dicarboxylic acid (the hatched rectangle, with –CO₂H at both ends) and the –NH– end comes from a diamine (the open rectangle, with –NH₂ at both ends). B: the right two monomers, but the rectangles are swapped: in the polymer the hatched rectangle lies between two –CO– groups, so the dicarboxylic acid must contain the hatched rectangle. C: a dicarboxylic acid and a diol form ester links, –CO–O–, so they make a polyester such as Terylene, not a polyamide. D: a diol and a diamine have no –CO₂H groups and cannot form the –CO–NH– links.

B: carbon monoxide, methane and sulfur dioxide are all common atmospheric pollutants (others are the nitrogen oxides, ozone and unburned hydrocarbons). A: argon is a noble gas that makes up about 1% of clean air; it is not a pollutant. C: nitrogen is the main gas in clean air (78%); it is not a pollutant, although its oxides are. D: oxygen is a gas of clean air (21%) and not a pollutant, even though nitrogen dioxide and ozone are.

D: calcium carbonate is a base that reacts with the acids in the water: CaCO₃ + 2H⁺ → Ca²⁺ + CO₂ + H₂O. It neutralises the acid and raises the pH of the lake. A: calcium carbonate is used to absorb sulfur dioxide from the waste gases of power stations (flue gas desulfurisation), before it can reach the air; it cannot be used to absorb it from the air above a lake. B: converting nitrogen oxides into nitrogen is the job of the catalytic converter of a car. C: an acid reacting with a carbonate gives a salt, water and carbon dioxide, not hydrogen; the reaction is neutralisation.