CHEMISTRY – Paper 1 Multiple Choice – mark scheme

C: a drying agent must not react with the gas. Statement 2 is correct: a gas that turns limewater milky (carbon dioxide or sulfur dioxide) is acidic, and concentrated sulfuric acid, an acid, does not react with an acidic gas. Statement 4 is correct: a gas that burns with a squeaky pop is hydrogen, which is neutral and does not react with calcium oxide. Statement 1 is wrong: a gas that turns damp red litmus paper blue is ammonia, a basic gas, which reacts with concentrated sulfuric acid. Statement 3 is wrong: a gas that turns damp blue litmus paper red is acidic, and it reacts with calcium oxide, a basic oxide. A includes statement 1; B includes statements 1 and 3, both wrong; D includes statement 3.

C: before spraying, only the yellow pigment can be seen, because the amino acids and the sugar are colourless: 1 spot. A locating agent is needed to see colourless substances. After spraying, the two amino acids react with it and show as purple spots, and the yellow pigment is still visible: 3 spots. The sugar does not react with this locating agent, so it stays colourless and cannot be seen. A: after spraying, the yellow pigment is still on the paper and still visible, so there are 3 spots and not 2. B: the yellow pigment is coloured, so it can be seen before spraying; it is not 0 spots. D: this assumes that the locating agent shows up every colourless substance, but it reacts only with amino acids, so the sugar is not shown.

A: the top number is the nucleon number (protons + neutrons) and the bottom number is the proton number. Statement 1 is correct: all three have a nucleon number of 40. Statement 2 is correct: argon-40 has 40 − 18 = 22 neutrons and calcium-40 has 40 − 20 = 20 neutrons, a difference of 2. Statement 3 is wrong: isotopes are atoms of the same element with the same proton number but different nucleon numbers; potassium (19 protons) and calcium (20 protons) are different elements. Statement 4 is wrong: potassium-40 has 19 protons and 40 − 19 = 21 neutrons. B, C and D each contain statement 3 or statement 4.

B: an ion with a charge of 2+ has lost two electrons, so it has two more protons than electrons: 36 + 2 = 38 protons. The nucleon number is protons + neutrons, so neutrons = 90 − 38 = 52. The ion has 36 electrons, so it has 52 − 36 = 16 more neutrons than electrons. A: 14 = 52 − 38 compares the neutrons with the protons, not with the electrons. C: 18 comes from ignoring the charge: 36 protons would give 54 neutrons, and 54 − 36 = 18. D: 20 comes from applying the charge the wrong way round: 34 protons would give 56 neutrons, and 56 − 36 = 20.

C: solid sodium chloride is a giant lattice of ions held in fixed positions, so no charged particles are free to move and it does not conduct. When it is molten or dissolved in water the lattice breaks up and the ions are free to move, so it conducts. A: molten sodium chloride also conducts. B: aqueous sodium chloride also conducts. D: solid sodium chloride does not conduct, because its ions cannot move.

B: hydrogen has 1 outer electron, carbon 4 and oxygen 6, so there are 4 + 6 + 2 × 1 = 12 outer electrons in the molecule. Carbon needs 8 electrons: it shares a pair with each of the 2 hydrogen atoms, so it must share two pairs with oxygen (a double bond). Oxygen then has 4 electrons from the shared pairs and needs 4 more, which are two pairs that are not shared (two lone pairs). So 4 pairs (8 electrons) are used in bonding and 12 − 8 = 4 electrons are not involved in bonding. A: this counts only one lone pair on the oxygen atom, which would then have only 6 electrons around it. C: three lone pairs would come from a single C–O bond, which would leave carbon with only 6 electrons around it. D: 8 is the number of electrons that are involved in bonding (4 shared pairs), not the number that are not.

A: bronze is an alloy, a mixture of a metal (copper) with another element (tin). The diagram shows atoms of two different elements mixed together in the layer, not joined in a fixed ratio. B: the atoms in an alloy are mixed, not chemically combined, so an alloy is not a compound. C: bronze contains atoms of two elements, so it is not an element. D: bronze is made of atoms held by metallic bonding; it contains no negative ions.

D: X has a very high melting point and conducts when molten but not when solid, so it contains ions that can move only when the lattice breaks up: it is a giant ionic lattice (like magnesium oxide). Z has a very high melting point because strong covalent bonds must be broken, and it has no ions or free electrons, so it does not conduct even when molten: it is a giant covalent structure (like silicon dioxide). A: X conducts when molten, so it is ionic and not giant covalent; Y has a low melting point and does not conduct, so it is simple molecular and not ionic. B: W conducts when solid, so it has free electrons: it is a metal, not an ionic compound. C: Y has a low melting point, so it is simple molecular (like sulfur) and not giant covalent; Z has a very high melting point, so it is not simple molecular.

B: carbon dioxide is a simple molecular substance: only the weak forces between its molecules are overcome when it changes to a gas, so little energy is needed. Silicon dioxide is a giant covalent structure: every atom is joined to its neighbours by strong covalent bonds, and a huge amount of energy is needed to break them. A: the covalent bonds inside a carbon dioxide molecule are not broken when it changes to a gas, so their strength does not explain the difference. C: silicon dioxide is made of non-metal atoms joined by covalent bonds; it does not contain ions. D: silicon dioxide has no delocalised electrons (it does not conduct electricity); its atoms are held by covalent bonds.

D: (NH₄)₂SO₄ has 2 nitrogen atoms and 2 × 4 = 8 hydrogen atoms, a ratio of 2 : 8 = 1 : 4. A: CO(NH₂)₂ has 2 nitrogen atoms and 2 × 2 = 4 hydrogen atoms, a ratio of 1 : 2. B: NH₄HCO₃ has 1 nitrogen atom and 4 + 1 = 5 hydrogen atoms (one more hydrogen atom in HCO₃), a ratio of 1 : 5. C: NH₄NO₃ has 2 nitrogen atoms (one in NH₄ and one in NO₃) and 4 hydrogen atoms, a ratio of 1 : 2; counting only the NH₄ part gives 1 : 4, but the nitrogen atom in NO₃ must be counted too.

A: in M₃(PO₄)₂ the positive and negative charges balance: 3 ions of M balance 2 phosphate ions, so 3 × (charge on M) = 2 × 3 = 6 and the charge on M is 2+. The sulfate ion has a charge of 2−, so one M²⁺ ion balances one sulfate ion: MSO₄. B: M₂SO₄ would need M to form a 1+ ion. C: M₂(SO₄)₃ would need M to form a 3+ ion. D: M₃(SO₄)₂ copies the ratio 3 : 2 from the phosphate, but sulfate has a different charge from phosphate.

C: M_r of Ca(NO₃)₂ = 40 + 2 × (14 + 3 × 16) = 164. It contains 2 nitrogen atoms, 2 × 14 = 28, so the percentage of nitrogen by mass = 28 ÷ 164 × 100 = 17.1%. Mass of nitrogen in 500 g = 17.07% of 500 = 85 g. A: 17 is the percentage of nitrogen, which has not been used to find the mass in 500 g. B: 43 g counts only one nitrogen atom (14 ÷ 164 × 500). D: 171 g uses 56 as the mass of nitrogen in one formula unit, as if there were four nitrogen atoms instead of two (56 ÷ 164 × 500).

D: the mass lost is the carbon dioxide given off: 1.76 g ÷ 44 = 0.0400 mol CO₂. From the equation, 1 mol CaCO₃ gives 1 mol CO₂, so 0.0400 mol CaCO₃ decomposed: 0.0400 × 100 = 4.00 g. Percentage purity = 4.00 ÷ 5.00 × 100 = 80%. A: 35% is the mass lost as a percentage of the sample (1.76 ÷ 5.00), without using the equation. B: 45% is the mass of calcium oxide formed (0.0400 × 56 = 2.24 g) as a percentage of the sample. C: 65% treats the whole residue (5.00 − 1.76 = 3.24 g), which includes the impurities, as calcium carbonate.

B: statement 1 is correct: a metal above hydrogen in the reactivity series reacts with an acid to form a salt and hydrogen. Statement 4 is correct: a carbonate reacts with an acid to form a salt, water and carbon dioxide. Statement 2 is wrong: copper(II) oxide is a base, and a base reacts with an acid to form only a salt and water; no hydrogen is formed. Statement 3 is wrong: iron reacts with dilute sulfuric acid to form iron(II) sulfate (FeSO₄), not iron(III) sulfate. A, C and D each include statement 2 or statement 3.

B: neutralisation is the reaction between hydrogen ions from the acid and hydroxide ions from the alkali to form water, H⁺ + OH⁻ → H₂O. The potassium ions and nitrate ions are unchanged: they stay in solution as potassium nitrate. A: nitrate ions do not combine with hydrogen ions (nitric acid is a strong acid, fully ionised), and potassium ions do not combine with hydroxide ions (potassium hydroxide is fully ionised). C: the two pairs are the wrong way round. D: potassium ions and hydroxide ions do not combine to form water; the water comes from H⁺ and OH⁻.

D: 1 → 3 → 4 → 2 → 5. Excess zinc is added so that all the acid reacts (the zinc is in excess when bubbling stops); the unreacted zinc is filtered off to leave a solution of zinc sulfate; some water is evaporated to make a saturated solution; the hot saturated solution is cooled, so that crystals form; the crystals are then filtered off and dried. A: the solution is cooled while the excess zinc is still in it and before it has been evaporated, so it is unsaturated and no crystals can form. B: the solution is cooled while the excess zinc is still in it, before it has been evaporated, so no crystals form; the hot solution would then be evaporated and filtered without being cooled again. C: the solution is cooled before it has been evaporated to saturation, so no crystals form; after evaporating, the hot solution would then be filtered without being cooled again.

B: chlorine is an acidic gas that bleaches: damp blue litmus paper turns red and is then bleached white. A: ammonia turns damp red litmus paper blue. C: hydrogen does not change damp litmus paper; it burns with a squeaky pop when a lighted splint is held to it. D: oxygen does not change damp litmus paper; it relights a glowing splint.

C: Z turns aqueous potassium iodide brown, so it oxidises iodide ions to iodine (it is an oxidising agent); Z also turns acidified potassium manganate(VII) from purple to colourless, so it reduces manganate(VII) ions (it is a reducing agent). Z shows both behaviours. A: a brown colour means iodine has formed from iodide ions, so X has oxidised the iodide: X is an oxidising agent. B: purple to colourless means the manganate(VII) ions have been reduced, so Y is the reducing agent. D: Y reduces manganate(VII) and has no effect on iodide ions, so Y is a reducing agent, not an oxidising agent.

A: electrode P is joined to the negative terminal of the power supply, so the positive sodium ions are attracted to it; each sodium ion gains an electron to form a sodium atom (reduction): Na⁺ + e⁻ → Na. B: sodium ions are positive, so they move to the negative electrode (P), not to Q, and they gain electrons rather than losing them. C: negative chloride ions move to the positive electrode (Q), and at Q they lose electrons to form chlorine. D: electrons flow through the wires of the circuit; the current in the molten sodium chloride is carried by ions, not by electrons.

A: from the Periodic Table, the elements with proton numbers 12 and 35 are magnesium (a metal) and bromine (a non-metal), and those with proton numbers 19 and 53 are potassium (a metal) and iodine (a non-metal). In each molten ionic compound the positive metal ions (Mg²⁺, K⁺) go to the negative electrode, where they gain electrons to form the metal; the negative non-metal ions (Br⁻, I⁻) go to the positive electrode, where they lose electrons to form bromine and iodine. B: both pairs of products are the wrong way round: metals form at the negative electrode and non-metals at the positive electrode. C: the products for compound R are correct, but those for compound S are the wrong way round. D: the products for compound S are correct, but those for compound R are the wrong way round.

B: in each solution the product at an electrode depends on which ion is discharged. Concentrated sodium chloride (1): sodium is above hydrogen in the reactivity series, so hydrogen forms at the cathode, and chloride ions are discharged at the anode to give chlorine: two gases. Dilute sodium sulfate (4): this is the electrolysis of water, giving hydrogen at the cathode and oxygen at the anode: two gases. Copper(II) sulfate (2): copper is below hydrogen in the reactivity series, so a solid metal, copper, forms at the cathode (the other product, oxygen, is a gas). Potassium iodide (3): hydrogen forms at the cathode, but iodide ions are discharged at the anode to give iodine, which is a solid (brown in solution), not a gas. A, C and D each include solution 2 or solution 3.

A: across a period the elements change from metals on the left to non-metals on the right. J conducts electricity when solid, so it is a metal, and K, which does not, is a non-metal. J is therefore to the left of K, so J has the smaller proton number and the fewer outer shell electrons (the number of outer electrons rises by one from each element to the next). B: this is the wrong way round; J, the more metallic element, has the lower proton number. C: all the elements in a period have the same number of electron shells, so J and K have the same number. D: the more metallic element forms positive ions more easily and is less likely to form negative ions, so it is K that forms negative ions more easily.

C: three predictions from the position of rubidium in Group 1. Down Group 1 the melting points decrease, so rubidium melts at a lower temperature than potassium. A potassium atom has the electronic configuration 2,8,8,1, so four occupied shells (potassium is in Period 4); rubidium is in the next period down, Period 5, so its atoms have five occupied shells. All Group 1 metals form ions with a charge of 1+, so rubidium forms Rb⁺; with the sulfate ion, SO₄²⁻, two Rb⁺ ions are needed, giving Rb₂SO₄. A: the melting point is predicted wrongly; it is lower, not higher. B: four is the number of shells in a potassium atom; going down a group adds one shell for each period, so rubidium has five. D: RbSO₄ would need a Rb²⁺ ion; the charge of the ion is 1+, as in all Group 1 compounds.

D: a helium atom has 2 protons and so 2 electrons: the electronic configuration is 2. Its only electron shell can hold no more than two electrons, so the outer shell is full, and the atom has no tendency to gain, lose or share electrons. A: a helium atom has only two electrons, so its outer shell cannot contain eight; eight is the full outer shell of the other noble gases such as neon and argon. B: hydrogen also has only one electron shell, but its atoms do form bonds; it is the full outer shell, not the number of shells, that makes helium unreactive. C: helium atoms do not share their electrons; with a full outer shell they exist as separate, unreactive atoms.

D: a more reactive metal takes the oxygen from the oxide of a less reactive metal. Zinc is above lead in the reactivity series, so zinc displaces lead: zinc + lead(II) oxide → zinc oxide + lead. A: copper is less reactive than iron. B: iron is less reactive than magnesium. C: silver is less reactive than lead. In each of these the metal is below the metal of the oxide, so there is no reaction.

A: a metal reacts with the nitrate solution of a less reactive metal, and does not react with the nitrate of a more reactive metal. P reacts with Q nitrate, so P is more reactive than Q. Q reacts with S nitrate, so Q is more reactive than S. S reacts with R nitrate, so S is more reactive than R. Together: P, then Q, then S, then R. The experiment with R and P nitrate (no reaction) agrees: R is less reactive than P. B: S and R are the wrong way round; S displaces R, so S is above R. C: P and Q are the wrong way round; P displaces Q, so P is above Q. D: this is the order reversed; it takes a reaction to mean that the metal added is less reactive.

B: this is sacrificial protection. Magnesium is more reactive than iron, so it corrodes in preference to the iron; the steel does not rust while some magnesium is still connected to it. Once all the magnesium has been used up, nothing protects the steel from oxygen and water in the damp soil, so it starts to rust; this is why the blocks must be replaced from time to time. A: the magnesium is the metal that reacts; it does not stay unchanged, and it does not act as a barrier like paint. C: the steel does not rust at all while the more reactive magnesium is attached; the magnesium corrodes first. D: the protection lasts only as long as there is magnesium left; it does not continue once the magnesium has all corroded.

C: the bonds that break are the bonds in the reactants, the H–H bonds in hydrogen and the O=O bond in oxygen; energy has to be supplied to break them (endothermic). The new bonds, the O–H bonds in steam, form as the atoms join, and energy is given out when bonds form (exothermic). A and B: the O–H bonds are in the product, steam, so they are formed in this reaction, not broken. A also gives the wrong energy change for bond making. D: the bonds broken are correct, but forming bonds gives out energy, so the energy change is exothermic, not endothermic.

B: the reaction is endothermic (ΔH is positive), so more energy is taken in to break the bonds in the reactants than is given out when the bonds in the products form. Statements 1 and 2 are correct: bond breaking takes in energy and bond making gives out energy. Statement 4 is correct: the energy taken in from the surroundings ends up in the products, which therefore have more energy than the reactants. Statement 3 is incorrect: it describes an exothermic reaction, the opposite of what the positive ΔH shows. A leaves out statement 4; C and D include statement 3.

A: a catalyst provides a different reaction pathway with a lower activation energy, so a greater proportion of the colliding particles have enough energy to react and the rate increases. B: a catalyst does not change the energy that the molecules have; the temperature stays the same. C: the enthalpy change depends only on the energy of the reactants and products, and a catalyst does not alter it. D: a catalyst does not change the concentration of the reactant.

A: the rate is the volume of oxygen formed divided by the time taken. Statement 1 is correct: 24 cm³ forms in the first 30 s and less in each later interval (18, 12, 6, then none), so the reaction is fastest at the start as the reactant is used up. Statement 3 is correct: between 30 s and 90 s, (54 − 24) ÷ 60 = 0.50 cm³ / s. Statement 2 is incorrect: oxygen is still being formed between 90 s and 120 s (54 to 60 cm³); the volume stops changing after 120 s. Statement 4 is incorrect: between 60 s and 120 s, (60 − 42) ÷ 60 = 0.30 cm³ / s, not 0.20. B, C and D each include statement 2 or statement 4.

C: alkanes are saturated hydrocarbons with the general formula C_nH_2n+2, so their relative molecular mass is 12n + (2n + 2) = 14n + 2. For 114: 14n = 112, so n = 8 and the formula is C₈H₁₈ (12 × 8 + 18 = 114). A: C₇H₁₆ is an alkane, but its relative molecular mass is 100. B: C₈H₁₆ has the right number of carbon atoms but only 2n hydrogen atoms, the formula of an alkene; its relative molecular mass is 112. D: C₉H₁₈ also fits C_nH_2n (an alkene), with a relative molecular mass of 126.

D: aqueous bromine is decolourised by an unsaturated hydrocarbon, which has a C=C double bond that bromine adds across. Compounds 3 (but-1-ene) and 4 (but-2-ene) each have a C=C double bond. Compound 2 (methylpropane) is saturated: it has only single bonds, so it does not decolourise aqueous bromine. Compound 1 (ethanoic acid) has a C=O double bond, but this is not a C=C bond and does not react with aqueous bromine, so it is not decolourised. A, B and C each include compound 1 or compound 2.

C: margarine is made by the addition of hydrogen to the C=C double bonds of an unsaturated vegetable oil. One molecule of hydrogen adds across each C=C double bond, so each molecule of the oil needs three molecules of hydrogen. Amount of hydrogen = 0.50 × 3 = 1.5 mol. A: 0.17 mol comes from dividing by 3 instead of multiplying. B: 0.50 mol assumes that only one molecule of hydrogen reacts with each molecule of the oil. D: 3.0 mol is the amount of hydrogen atoms (1.5 × 2), not of hydrogen molecules.

B: alcohols burn completely in plentiful air to form carbon dioxide and water, and are oxidised to carboxylic acids by acidified potassium manganate(VII); ethanol gives ethanoic acid. A: an ester forms from a carboxylic acid and an alcohol with an acid catalyst, not when an alcohol is oxidised. C: carbon monoxide forms in a limited supply of air (incomplete combustion), not in plentiful air. D: the two columns are mixed up; burning gives carbon dioxide and water, and oxidation by the manganate(VII) gives the acid.

A: a carboxylic acid and an alcohol react in a 1 : 1 ratio to form an ester and water (the concentrated sulfuric acid is only a catalyst): CH₃COOH + C₄H₉OH → CH₃CO₂C₄H₉ + H₂O. Amount of ethanoic acid = 12.0 ÷ 60 = 0.200 mol; amount of butan-1-ol = 7.4 ÷ 74 = 0.100 mol. Butan-1-ol is the smaller amount, so it limits the amount of ester: 0.100 mol. M_r of the ester = 60 + 74 − 18 = 116, so the maximum mass = 0.100 × 116 = 11.6 g. B: 13.4 g uses 134 (60 + 74) as the M_r of the ester and forgets that a molecule of water is lost. C: 23.2 g is 0.200 × 116; it takes the ethanoic acid as the limiting reactant, but the acid is in excess. D: 26.8 g makes both slips: the acid as the limiting reactant and the water not subtracted.

C: an ester contains the group –CO–O–, and it can be drawn either way round. The part with the C=O (–C(=O)CH₂CH₃) comes from the carboxylic acid: counting the carbon of the C=O, it has three carbon atoms, so the acid is propanoic acid. The part joined to the single-bonded oxygen (CH₃CH₂–O–) comes from the alcohol: it has two carbon atoms, so the alcohol is ethanol. The ester is ethyl propanoate. Every pair has five carbon atoms in all, as the ester does, but only C divides them 3 + 2 in the right order. B: the acid and alcohol parts are swapped, which happens if the part on the left is taken to come from the acid; ethanoic acid and propan-1-ol would form propyl ethanoate. A would form butyl methanoate and D would form methyl butanoate.

D: in addition polymerisation the C=C double bond of each but-2-ene molecule opens, and the molecules join end to end. Each molecule becomes a repeat unit of two carbon atoms in the chain, and each of these two carbon atoms keeps its hydrogen atom and its CH₃ group, so there is a CH₃ side group on each chain carbon atom. No C=C double bond is left. A: this puts the two CH₃ groups into the chain (four carbon atoms in the chain) when they are side groups of the chain. B: this repeat unit has only three carbon atoms; it is the repeat unit of poly(propene), made from propene, not from but-2-ene. C: this repeat unit has two CH₃ groups on the same carbon atom; it would be made from methylpropene.

B: a chemical method of recycling changes the plastic into other substances. Cracking waste plastic into smaller molecules for fuel (2) and depolymerisation, the breaking down of a polyester into its monomers by hydrolysis with an acid catalyst (3), are the two chemical methods. Process 1 is a physical method: the poly(ethene) is only melted and re-shaped, and no new substance forms, so A is wrong. Process 4 is not recycling: the plastic is burned to release energy, not made into new materials, so C and D are wrong.

D: burning the trees is combustion of carbon compounds, which adds carbon dioxide to the air. With the trees gone, there is less photosynthesis in the area, so less carbon dioxide is taken in from the air. The carbon cycle can then no longer keep the carbon dioxide level in the atmosphere steady, and the level rises. A: fewer trees means less photosynthesis, not more. B, C: burning adds carbon dioxide to the air; it does not remove it.