An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V1) Cambridge International A Level Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.
PastPaper.section Theory (Extended)
Answer all questions. Show your working where appropriate. Use a calculator if needed. Refer to the Periodic Table provided.
A mixture contains ethanol (boiling point \(78\ ^\circ\text{C}\)) and water (boiling point \(100\ ^\circ\text{C}\)). Name the separation technique used to separate these two miscible liquids.
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PastPaper.workedSolution
Fractional distillation is used to separate miscible liquids that have different boiling points. Since ethanol and water are miscible and have different boiling points, they can be separated using a fractionating column.
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Award 1 mark for 'fractional distillation'. Reject: 'simple distillation' or 'distillation'.
PastPaper.question 2 · Short structural matching
1 PastPaper.marks
An atom of an element has an atomic number of 16. State the electronic configuration of this atom.
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PastPaper.workedSolution
An atomic number of 16 means the neutral atom has 16 electrons. The first electron shell can hold up to 2 electrons, the second shell can hold up to 8, and the remaining 6 electrons reside in the third shell. Thus, the electronic configuration is 2,8,6.
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Award 1 mark for '2,8,6' (accept '2.8.6' or '2 8 6').
PastPaper.question 3 · Short structural matching
1 PastPaper.marks
State the chemical formula of the product formed at the anode during the electrolysis of concentrated aqueous sodium chloride.
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PastPaper.workedSolution
During the electrolysis of concentrated aqueous sodium chloride (brine), both hydroxide ions (\(\text{OH}^-\)) and chloride ions (\(\text{Cl}^-\)) migrate to the anode (positive electrode). Because the solution is concentrated, chloride ions are selectively discharged to form chlorine gas, \(\text{Cl}_2\).
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Award 1 mark for 'Cl2' (accept 'Cl2(g)' or '\(\text{Cl}_2\)'). Do not accept 'Cl' or 'chlorine' as a formula.
PastPaper.question 4 · Short structural matching
1 PastPaper.marks
State the chemical formula of the ionic compound formed between calcium ions and nitride ions.
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PastPaper.workedSolution
Calcium is in Group II and forms a \(\text{Ca}^{2+}\) ion. Nitrogen is in Group V and forms a nitride ion, \(\text{N}^{3-}\). To form a neutral compound, three calcium ions (total charge of \(+6\)) are needed to balance two nitride ions (total charge of \(-6\)). The resulting chemical formula is \(\text{Ca}_3\text{N}_2\).
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Award 1 mark for 'Ca3N2' (accept '\(\text{Ca}_3\text{N}_2\)').
PastPaper.question 5 · Short structural matching
1 PastPaper.marks
Give the IUPAC name of the unbranched carboxylic acid that contains exactly four carbon atoms.
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PastPaper.workedSolution
A carboxylic acid with four carbon atoms has the stem 'butan-' representing the four-carbon chain and the suffix '-oic acid' representing the carboxyl functional group (\(-\text{COOH}\)). Therefore, the IUPAC name is butanoic acid.
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Award 1 mark for 'butanoic acid'. Reject: 'butyric acid'.
PastPaper.question 6 · Short structural matching
1 PastPaper.marks
Identify the type of polymerisation reaction that occurs when monomers containing carbon-to-carbon double bonds (\(\text{C}=\text{C}\)) join together to form a polymer as the only product.
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PastPaper.workedSolution
When unsaturated monomers containing double bonds (such as alkenes) join together to form a macromolecule with no other product formed, the process is known as addition polymerisation.
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Award 1 mark for 'addition polymerisation' (accept 'addition'). Reject: 'condensation'.
PastPaper.question 7 · Theory (Extended)
12.33 PastPaper.marks
Electrolysis is the breakdown of an ionic compound, molten or in aqueous solution, by the passage of electricity.
(a) During the electrolysis of molten lead(II) bromide, \(\text{PbBr}_2\), using inert carbon electrodes: (i) State the observations at the anode and cathode. [2] (ii) Write ionic half-equations, including state symbols, for the reactions occurring at each electrode. [2]
(b) Concentrated aqueous sodium chloride is electrolysed using inert electrodes (the Chlor-alkali process). (i) Name the product formed at the anode and the cathode. [2] (ii) Explain why the electrolyte becomes alkaline as the electrolysis proceeds. [2]
(c) Describe how electrolysis is used to purify a sample of impure copper, specifying the anode, the cathode, and the electrolyte used. [4]
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PastPaper.workedSolution
(a) (i) Anode: brown fumes / brown gas of bromine is evolved. Cathode: silvery/grey liquid bead of lead is formed. (ii) Anode: \(2\text{Br}^-(\text{l}) \rightarrow \text{Br}_2(\text{g}) + 2\text{e}^-\), Cathode: \(\text{Pb}^{2+}(\text{l}) + 2\text{e}^- \rightarrow \text{Pb}(\text{l})\). (b) (i) Anode: chlorine gas. Cathode: hydrogen gas. (ii) Hydrogen ions and chloride ions are discharged at the electrodes, leaving sodium ions and hydroxide ions in solution. These form sodium hydroxide, which is alkaline. (c) Anode: impure copper. Cathode: pure copper. Electrolyte: aqueous copper(II) sulfate. Copper atoms at the anode lose electrons to form copper(II) ions, which go into solution. At the cathode, copper(II) ions gain electrons and deposit as pure copper.
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(a)(i) 1 mark for brown fumes at anode; 1 mark for grey/silvery liquid at cathode. [2] (ii) 1 mark for anode equation with state symbols; 1 mark for cathode equation with state symbols. [2] (b)(i) 1 mark for chlorine at anode; 1 mark for hydrogen at cathode. [2] (ii) 1 mark for identifying that hydrogen and chloride ions are discharged; 1 mark for stating that sodium and hydroxide ions remain, forming sodium hydroxide (alkaline). [2] (c) 1 mark for anode: impure copper; 1 mark for cathode: pure copper; 1 mark for electrolyte: aqueous copper(II) sulfate; 1 mark for describing the dissolution of copper from anode and deposition on cathode. [4]
PastPaper.question 8 · Theory (Extended)
12.33 PastPaper.marks
A student investigates the rate of reaction between calcium carbonate, \(\text{CaCO}_3\), and dilute hydrochloric acid, \(\text{HCl}\). \(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\)
(a) The student repeats the experiment using the same mass of calcium carbonate but as a fine powder instead of large lumps. (i) State how the rate of reaction changes when the powder is used. [1] (ii) Explain this change in rate in terms of collision theory. [3]
(b) The reaction is carried out at a higher temperature, keeping all other variables constant. (i) State the effect of increasing temperature on the rate of reaction. [1] (ii) Explain this effect in terms of collision theory, referring to activation energy. [3]
(c) The total volume of carbon dioxide gas evolved is measured over time. (i) Describe the shape of the graph of volume of gas (y-axis) against time (x-axis) for this reaction, and explain why the gradient of the curve decreases over time. [3] (ii) State one other experimental method, besides measuring gas volume, to follow the progress of this reaction. [1]
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PastPaper.workedSolution
(a) (i) The rate of reaction increases. (ii) Powdered calcium carbonate has a larger surface area than large lumps. This means there are more exposed particles, leading to more frequent collisions per unit time between the reactant particles. (b) (i) The rate of reaction increases. (ii) Particles gain kinetic energy and move faster, so there are more frequent collisions. Also, a greater proportion of colliding particles have energy greater than or equal to the activation energy, leading to a higher rate of successful collisions. (c) (i) The graph starts steep from the origin, becomes less steep (curves downwards), and eventually levels off (becomes horizontal). The gradient decreases because the concentration of hydrochloric acid decreases as it is used up, leading to fewer successful collisions per unit time. (ii) Measuring the loss in mass of the reaction flask and its contents over time.
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(a)(i) 1 mark for stating that the rate increases. [1] (ii) 1 mark for larger surface area; 1 mark for more exposed particles; 1 mark for more frequent collisions / more collisions per unit time. [3] (b)(i) 1 mark for stating that the rate increases. [1] (ii) 1 mark for particles having more kinetic energy; 1 mark for more frequent collisions; 1 mark for more particles having energy greater than or equal to the activation energy. [3] (c)(i) 1 mark for describing the curve starting steep and leveling off; 1 mark for explaining that reactants are being used up / concentration of acid decreases; 1 mark for fewer collisions per unit time. [3] (ii) 1 mark for measuring change in mass / loss in mass of the flask. [1]
PastPaper.question 9 · Theory (Extended)
12.33 PastPaper.marks
Carboxylic acids are an important homologous series of organic compounds.
(a) Ethanoic acid can be prepared from ethanol by chemical oxidation. (i) Name a suitable oxidizing agent and the condition required for this preparation. [2] (ii) State the color change observed during this oxidation. [1]
(b) Propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\), reacts with methanol, \(\text{CH}_3\text{OH}\), in the presence of an acid catalyst to form an ester and water. (i) State the name of the catalyst used. [1] (ii) Name the ester formed in this reaction. [1] (iii) Draw the structural formula of this ester, showing all atoms and bonds. [2]
(c) Ethanoic acid is a weak acid, whereas hydrochloric acid is a strong acid. (i) Explain what is meant by a 'weak acid'. [2] (ii) Describe an experiment (other than using a pH meter) to show that ethanoic acid is a weaker acid than hydrochloric acid of the same concentration. Include the expected results for both. [3]
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PastPaper.workedSolution
(a) (i) Acidified potassium manganate(VII) (or acidified potassium dichromate(VI)) and heating under reflux. (ii) Purple to colorless (or orange to green if dichromate is used). (b) (i) Concentrated sulfuric acid. (ii) Methyl propanoate. (iii) Structural formula: \(\text{CH}_3-\text{CH}_2-\text{C}(=\text{O})-\text{O}-\text{CH}_3\) showing all atoms and bonds. (c) (i) A weak acid only partially dissociates / ionizes in aqueous solution to produce hydrogen ions. (ii) Add a piece of magnesium ribbon (or a carbonate) of the same size to equal volumes of both acids of the same concentration. Hydrochloric acid will show rapid effervescence/fizzing, while ethanoic acid will show slow/gentle effervescence.
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(a)(i) 1 mark for acidified potassium manganate(VII) / acidified potassium dichromate(VI); 1 mark for heat / reflux. [2] (ii) 1 mark for purple to colorless / orange to green. [1] (b)(i) 1 mark for concentrated sulfuric acid. [1] (ii) 1 mark for methyl propanoate. [1] (iii) 2 marks for drawing methyl propanoate showing all bonds. Deduct 1 mark for any missing bond. [2] (c)(i) 1 mark for 'partial ionization / dissociation'; 1 mark for 'in aqueous solution to yield hydrogen ions'. [2] (ii) 1 mark for adding magnesium / zinc / carbonate; 1 mark for stating that HCl gives fast effervescence / gas evolved rapidly; 1 mark for stating ethanoic acid gives slower effervescence / gas evolved slowly. [3]
PastPaper.question 10 · Theory (Extended)
12.33 PastPaper.marks
Soluble and insoluble salts can be prepared using different experimental methods.
(a) Describe how to prepare a pure, dry sample of hydrated copper(II) sulfate crystals, \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\), starting from solid copper(II) oxide, \(\text{CuO}\), and dilute sulfuric acid, \(\text{H}_2\text{SO}_4\). [6]
(b) Barium sulfate, \(\text{BaSO}_4\), is an insoluble salt. (i) Name two soluble salts that could be reacted together to prepare a precipitate of barium sulfate. [2] (ii) Write the ionic equation, including state symbols, for the precipitation of barium sulfate. [2]
(c) Sodium chloride can be prepared by reacting dilute hydrochloric acid with aqueous sodium hydroxide. Explain why an indicator is needed to prepare a pure sample of sodium chloride by titration, and how the pure salt is obtained after titration. [2]
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PastPaper.workedSolution
(a) Add excess copper(II) oxide to warm dilute sulfuric acid and stir. Filter the mixture to remove the unreacted copper(II) oxide. Heat the filtrate (copper(II) sulfate solution) until the crystallization point is reached. Leave the solution to cool and crystallize slowly. Filter off the crystals and dry them between pieces of filter paper. (b) (i) Barium chloride (or barium nitrate) and sodium sulfate (or potassium sulfate). (ii) \(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\). (c) An indicator is needed to determine the exact volume of acid needed to neutralize the alkali. Once the volume is known, the titration is repeated without the indicator using the same volumes, and the resulting solution is evaporated to obtain pure sodium chloride crystals without indicator contamination.
PastPaper.markingScheme
(a) 1 mark for heating/warming the acid; 1 mark for adding copper(II) oxide in excess; 1 mark for filtering to remove excess reactant; 1 mark for heating filtrate to saturation point / crystallization point; 1 mark for allowing to cool to form crystals; 1 mark for filtering and drying crystals between filter papers. [6] (b)(i) 1 mark for a soluble barium salt (e.g., barium chloride / barium nitrate); 1 mark for a soluble sulfate salt (e.g., sodium sulfate / potassium sulfate). [2] (ii) 1 mark for formulas: \(\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4\); 1 mark for correct state symbols: \((\text{aq})\), \((\text{aq})\), and \((\text{s})\). [2] (c) 1 mark for explaining that indicator determines the end-point/exact neutralization volume; 1 mark for explaining that repeating without indicator prevents contamination of the product. [2]
PastPaper.question 11 · Theory (Extended)
12.33 PastPaper.marks
A gaseous hydrocarbon, **X**, has a percentage composition by mass of 85.7% carbon and 14.3% hydrogen.
(a) Calculate the empirical formula of hydrocarbon **X**. [3]
(b) The relative molecular mass (\(M_r\)) of **X** is 56. Deduce the molecular formula of **X**. [2]
(c) Hydrocarbon **X** undergoes complete combustion in excess oxygen. \(\text{C}_4\text{H}_8(\text{g}) + 6\text{O}_2(\text{g}) \rightarrow 4\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{g})\) Calculate the volume of carbon dioxide gas, in \(\text{dm}^3\), produced at room temperature and pressure (r.t.p.) when 5.6 g of hydrocarbon **X** is completely burned. (Assume 1 mole of gas occupies \(24\text{ dm}^3\) at r.t.p.) [4]
(d) Calculate the total number of hydrogen atoms present in 5.6 g of hydrocarbon **X**. (The Avogadro constant, \(L = 6.02 \times 10^{23}\) per mole) [3]
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PastPaper.workedSolution
(a) Let the mass of compound be 100 g. Moles of C = \(85.7 / 12 = 7.14\) mol Moles of H = \(14.3 / 1 = 14.3\) mol Dividing by the smallest value (7.14) gives a ratio of 1 C : 2 H. Empirical formula is \(\text{CH}_2\). (b) Empirical formula mass of \(\text{CH}_2 = 12 + 2 = 14\). Factor = \(56 / 14 = 4\). Molecular formula is \(\text{C}_4\text{H}_8\). (c) Moles of \(\text{C}_4\text{H}_8 = 5.6 / 56 = 0.1\) mol. From the equation, 1 mole of \(\text{C}_4\text{H}_8\) produces 4 moles of \(\text{CO}_2\). Moles of \(\text{CO}_2 = 0.1 \times 4 = 0.4\) mol. Volume of \(\text{CO}_2 = 0.4 \times 24 = 9.6\) \(\text{dm}^3\). (d) Moles of \(\text{C}_4\text{H}_8 = 0.1\) mol. Moles of H atoms = \(0.1 \times 8 = 0.8\) mol. Number of H atoms = \(0.8 \times 6.02 \times 10^{23} = 4.816 \times 10^{23}\) atoms (or \(4.82 \times 10^{23}\)).
PastPaper.markingScheme
(a) 1 mark for dividing percentages by atomic masses (85.7/12 and 14.3/1); 1 mark for obtaining mole ratio 7.14 : 14.3; 1 mark for simplifying to simplest whole-number ratio \(\text{CH}_2\). [3] (b) 1 mark for empirical mass of 14; 1 mark for correct molecular formula \(\text{C}_4\text{H}_8\). [2] (c) 1 mark for calculating moles of \(\text{C}_4\text{H}_8 = 0.1\) mol; 1 mark for using stoichiometry ratio 1 : 4 to find moles of \(\text{CO}_2 = 0.4\) mol; 1 mark for multiplying moles of gas by 24; 1 mark for final answer 9.6 \(\text{dm}^3\). [4] (d) 1 mark for identifying there are 8 H atoms per molecule (or 0.8 moles of H atoms); 1 mark for multiplying by Avogadro's constant; 1 mark for correct calculation: \(4.82 \times 10^{23}\) (accept \(4.8 \times 10^{23}\) or \(4.816 \times 10^{23}\)). [3]
PastPaper.question 12 · Theory (Extended)
12.33 PastPaper.marks
The Periodic Table organizes elements according to their proton numbers and electronic structures.
(a) Complete the sub-parts below to deduce the particles found in the given atoms or ions. [3] (i) State the number of protons, neutrons, and electrons in a neutral atom of \(^{31}\text{P}\). (ii) State the number of protons, neutrons, and electrons in the chloride ion, \(^{37}\text{Cl}^-\).
(b) Describe the trend in reactivity of the Group I alkali metals as you descend the group, and explain this trend in terms of their electronic configuration and ease of losing electrons. [5]
(c) Define the term *isotopes*. Explain why isotopes of the same element have identical chemical properties. [4]
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PastPaper.workedSolution
(a) (i) For \(^{31}\text{P}\): Protons = 15, Neutrons = 16, Electrons = 15. (ii) For \(^{37}\text{Cl}^-\): Protons = 17, Neutrons = 20, Electrons = 18. (b) Reactivity increases down Group I. As you go down the group, the number of electron shells increases, so the single outer shell electron is further from the nucleus. This results in weaker electrostatic attraction between the positive nucleus and the negative outer electron. Therefore, less energy is required to remove the outer electron, making it easier to lose. (c) Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. Isotopes have the same chemical properties because they have the same number of electrons in their outer shell (same electronic configuration), and chemical reactions depend on outer shell electrons.
PastPaper.markingScheme
(a) 1.5 marks for \(^{31}\text{P}\) row (all three correct: 15, 16, 15); 1.5 marks for \(^{37}\text{Cl}^-\) row (all three correct: 17, 20, 18). Deduct 0.5 marks per incorrect value. [3] (b) 1 mark for stating reactivity increases down Group I; 1 mark for mentioning more electron shells / atomic radius increases; 1 mark for outer electron being further from nucleus; 1 mark for weaker attraction between nucleus and outer electron; 1 mark for stating that the outer electron is lost more easily. [5] (c) 1 mark for 'atoms of same element with same number of protons'; 1 mark for 'different number of neutrons'; 1 mark for stating chemical properties depend on outer shell electrons / electronic configuration; 1 mark for stating isotopes have the same electronic configuration. [4]