An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V1) Cambridge IGCSE Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.
Paper 2 (Multiple Choice Extended)
Answer all 40 multiple choice questions. Choose the single correct option from A, B, C, or D and record on the answer sheet.
40 Question · 40 marks
Question 1 · Multiple Choice
1 marks
Aqueous copper(II) sulfate is electrolysed using inert graphite electrodes.
Which row correctly identifies the product formed at the positive electrode (anode) and the change observed in the electrolyte?
A.Anode product: oxygen; Electrolyte observation: blue colour fades
B.Anode product: copper; Electrolyte observation: blue colour remains unchanged
C.Anode product: oxygen; Electrolyte observation: blue colour remains unchanged
D.Anode product: hydrogen; Electrolyte observation: blue colour fades
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Worked solution
At the positive electrode (anode), hydroxide ions (\(\text{OH}^-\)) from water are discharged in preference to sulfate ions (\(\text{SO}_4^{2-}\)), producing oxygen gas: \(4\text{OH}^-(\text{aq}) \rightarrow \text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + 4\text{e}^-\).
At the negative electrode (cathode), copper(II) ions (\(\text{Cu}^{2+}\)) are discharged to form copper metal: \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})\).
Because \(\text{Cu}^{2+}\) ions are removed from the solution and not replaced by the inert anode, the blue colour of the solution gradually fades.
Marking scheme
A is correct [1]. B is incorrect because copper is deposited at the cathode, not formed at the anode. C is incorrect because the blue colour fades as copper(II) ions are discharged. D is incorrect because oxygen, not hydrogen, is formed at the anode.
Question 2 · Multiple Choice
1 marks
A \(4.8\text{ g}\) sample of magnesium ribbon reacts completely with an excess of dilute hydrochloric acid at r.t.p.
What volume of hydrogen gas, measured at r.t.p., is produced?
[\(A_r\): \(\text{Mg} = 24\); molar gas volume at r.t.p. = \(24\text{ dm}^3\)]
A.\(2.4\text{ dm}^3\)
B.\(4.8\text{ dm}^3\)
C.\(12\text{ dm}^3\)
D.\(48\text{ dm}^3\)
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Worked solution
1. Calculate amount of magnesium in moles: \(\text{amount of Mg} = \frac{\text{mass}}{A_r} = \frac{4.8}{24} = 0.20\text{ mol}\).
2. Use the molar ratio from the balanced equation (\(1\text{ mol Mg} : 1\text{ mol H}_2\)): \(\text{amount of H}_2 = 0.20\text{ mol}\).
3. Calculate the volume of gas at r.t.p.: \(\text{volume} = 0.20\text{ mol} \times 24\text{ dm}^3/\text{mol} = 4.8\text{ dm}^3\).
Marking scheme
B is correct [1]. A is incorrect because it corresponds to 0.10 mol. C is incorrect because it corresponds to 0.50 mol. D is incorrect because it incorrectly multiplies by 10 or inverts molar mass.
Question 3 · Multiple Choice
1 marks
An excess of marble chips (calcium carbonate) is reacted with a fixed volume of dilute hydrochloric acid.
Which alteration increases the initial rate of the reaction AND increases the total volume of carbon dioxide gas collected?
A.increasing the temperature of the acid while keeping its volume and concentration constant
B.adding water to the acid while keeping the total amount of hydrochloric acid constant
C.increasing the concentration of the acid while keeping its volume constant
D.grinding the marble chips into a fine powder while keeping their mass constant
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Worked solution
Since marble chips are in excess, hydrochloric acid is the limiting reactant.
- Increasing the concentration of hydrochloric acid while keeping the volume constant increases the collision frequency of \(\text{H}^+\) ions per unit volume, which increases the initial rate. - It also increases the total moles of \(\text{HCl}\) (\(\text{moles} = \text{concentration} \times \text{volume}\)), thereby increasing the total volume of \(\text{CO}_2\) gas produced.
Marking scheme
C is correct [1]. A is incorrect because increasing temperature increases rate but does not change the total volume of gas. B is incorrect because adding water dilutes the acid, decreasing the initial rate. D is incorrect because grinding chips into powder increases surface area and initial rate, but the total volume of gas remains unchanged.
Question 4 · Multiple Choice
1 marks
An unknown solid compound \(Q\) is dissolved in distilled water to give a colourless solution.
- Addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess sodium hydroxide. - Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate.
What is the identity of compound \(Q\)?
A.iron(II) sulfate
B.iron(III) sulfate
C.chromium(III) sulfate
D.iron(II) chloride
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Worked solution
- The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). (Chromium(III) forms a green precipitate that dissolves in excess to give a green solution). - The formation of a white precipitate upon adding dilute nitric acid and aqueous barium nitrate confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\).
Therefore, compound \(Q\) is iron(II) sulfate.
Marking scheme
A is correct [1]. B is incorrect because iron(III) forms a red-brown precipitate with aqueous sodium hydroxide. C is incorrect because chromium(III) precipitate is soluble in excess sodium hydroxide. D is incorrect because aqueous barium nitrate tests for sulfate ions, not chloride ions.
Which row describes the effect of increasing the pressure on the equilibrium yield of ammonia and on the rate of the forward reaction?
A.Equilibrium yield of ammonia: increases; Rate of forward reaction: increases
B.Equilibrium yield of ammonia: increases; Rate of forward reaction: decreases
C.Equilibrium yield of ammonia: decreases; Rate of forward reaction: increases
D.Equilibrium yield of ammonia: decreases; Rate of forward reaction: remains unchanged
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Worked solution
- The left side has 4 moles of gas molecules (\(1 + 3\)) and the right side has 2 moles of gas molecules. Increasing the pressure favours the forward reaction (the side with fewer gas molecules), shifting equilibrium to the right and increasing the equilibrium yield of ammonia. - Increasing the pressure increases the concentration of gas particles, causing them to collide more frequently, which increases the rate of reaction.
Marking scheme
A is correct [1]. B is incorrect because increasing pressure increases reaction rate. C and D are incorrect because equilibrium shifts to the right, increasing the yield.
Question 6 · Multiple Choice
1 marks
Which statement about structural isomers is correct?
A.Butane and 2-methylpropane have different molecular formulae.
B.But-1-ene and but-2-ene have the same molecular formula and the same structural formula.
C.They are compounds that have the same molecular formula but different structural formulae.
D.Propane has two structural isomers.
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Worked solution
Structural isomers are defined as compounds having the same molecular formula but different structural formulae (different spatial arrangements of atoms).
- Butane and 2-methylpropane both have the molecular formula \(\text{C}_4\text{H}_{10}\). - But-1-ene and but-2-ene have different structural formulae. - Propane (\(\text{C}_3\text{H}_8\)) has only one possible structural arrangement.
Marking scheme
C is correct [1]. A is incorrect because butane and 2-methylpropane have the same molecular formula (\(\text{C}_4\text{H}_{10}\)). B is incorrect because but-1-ene and but-2-ene have different structural formulae. D is incorrect because propane has no structural isomers.
Question 7 · Multiple Choice
1 marks
Methane burns completely in oxygen according to the equation:
Selected average bond energies are provided: - \(\text{C–H} = 410\text{ kJ/mol}\) - \(\text{O=O} = 495\text{ kJ/mol}\) - \(\text{C=O} = 800\text{ kJ/mol}\) - \(\text{O–H} = 460\text{ kJ/mol}\)
What is the enthalpy change, \(\Delta H\), for this reaction?
A.\(-810\text{ kJ/mol}\)
B.\(+810\text{ kJ/mol}\)
C.\(-410\text{ kJ/mol}\)
D.\(-1620\text{ kJ/mol}\)
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Worked solution
1. Calculate energy required to break bonds (reactants): - \(4 \times (\text{C–H}) = 4 \times 410 = 1640\text{ kJ/mol}\) - \(2 \times (\text{O=O}) = 2 \times 495 = 990\text{ kJ/mol}\) - Total bond breaking energy = \(1640 + 990 = +2630\text{ kJ/mol}\)
2. Calculate energy released when new bonds form (products): - \(2 \times (\text{C=O}) = 2 \times 800 = 1600\text{ kJ/mol}\) - \(4 \times (\text{O–H}) = 4 \times 460 = 1840\text{ kJ/mol}\) - Total bond making energy = \(1600 + 1840 = 3440\text{ kJ/mol}\)
A is correct [1]. B is incorrect because it has an incorrect positive sign (endothermic). C is incorrect due to an arithmetic error in counting \(\text{O–H}\) or \(\text{C=O}\) bonds. D is incorrect because it multiplies the reaction enthalpy by two.
Question 8 · Multiple Choice
1 marks
Which method is the most suitable for preparing a pure, dry sample of barium sulfate?
A.adding excess solid barium carbonate to dilute sulfuric acid, filtering, and crystallising the filtrate
B.mixing aqueous barium chloride with aqueous sodium sulfate, filtering the precipitate, washing with distilled water, and drying
C.titrating aqueous barium hydroxide with dilute sulfuric acid using an indicator, then evaporating the solution to crystallise
D.reacting barium metal with dilute sulfuric acid, filtering off unreacted metal, and evaporating all water
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Worked solution
Barium sulfate is an insoluble salt. Insoluble salts are prepared by precipitation: 1. Mix two aqueous solutions containing the required ions (e.g. aqueous barium chloride and aqueous sodium sulfate). 2. Filter the mixture to collect the insoluble barium sulfate precipitate. 3. Wash the residue with distilled water to remove spectator ions. 4. Dry the precipitate in a warm oven or between filter papers.
Option A fails because barium sulfate forms an insoluble impermeable coating over the solid barium carbonate, preventing complete reaction.
Marking scheme
B is correct [1]. A is incorrect because an insoluble coating of barium sulfate forms and stops the reaction. C is incorrect because barium sulfate is insoluble and cannot be obtained by crystallisation. D is incorrect because evaporating to dryness does not remove excess unreacted reagents.
Question 9 · multiple_choice
1 marks
Concentrated aqueous potassium bromide is electrolysed using inert carbon electrodes.
Which row correctly identifies the product formed at each electrode and the change in the pH of the solution?
A.Anode: bromine | Cathode: hydrogen | Change in pH: increases
B.Anode: bromine | Cathode: potassium | Change in pH: decreases
C.Anode: oxygen | Cathode: hydrogen | Change in pH: stays the same
D.Anode: oxygen | Cathode: potassium | Change in pH: increases
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Worked solution
At the positive anode, bromide ions are oxidised to produce bromine gas/liquid: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\). At the negative cathode, hydrogen ions from water are discharged in preference to potassium ions: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\). As \(\text{H}^+\) ions are removed from the solution, hydroxide ions (\(\text{OH}^-\)) remain in excess with \(\text{K}^+\), forming aqueous potassium hydroxide. Therefore, the solution becomes alkaline and the pH increases.
Marking scheme
A is correct [1]. Bromine is produced at the anode, hydrogen gas at the cathode, and potassium hydroxide remains, causing the pH to increase.
Question 10 · multiple_choice
1 marks
The reaction between magnesium ribbon and excess dilute hydrochloric acid is investigated.
Which statement explains why an increase in temperature increases the rate of this reaction?
A.The activation energy of the reaction is decreased.
B.A greater proportion of colliding particles possess energy equal to or greater than the activation energy.
C.The reactant particles collide less frequently but with significantly higher energy.
D.The reactant particles move more slowly, allowing them to remain in contact longer.
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Worked solution
Increasing the temperature increases the average kinetic energy of the reacting particles. As a result, a significantly larger fraction of colliding particles have kinetic energy equal to or greater than the activation energy (\(E_a\)), leading to a higher frequency of successful collisions.
Marking scheme
B is correct [1]. Increasing temperature increases the proportion of particles with energy \(\ge E_a\). Activation energy is only altered by catalysts.
Question 11 · multiple_choice
1 marks
A sample of \(4.8\text{ g}\) of magnesium is added to \(75\text{ cm}^3\) of \(2.0\text{ mol/dm}^3\) hydrochloric acid.
What is the volume of hydrogen gas evolved, measured at room temperature and pressure (r.t.p.)? [Molar gas volume at r.t.p. = \(24\text{ dm}^3\text{/mol}\); \(A_r\text{: Mg} = 24\)]
A.\(1.8\text{ dm}^3\)
B.\(2.4\text{ dm}^3\)
C.\(3.6\text{ dm}^3\)
D.\(4.8\text{ dm}^3\)
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Worked solution
1. Calculate moles of reactants: \(\text{Moles of Mg} = \frac{4.8\text{ g}}{24\text{ g/mol}} = 0.20\text{ mol}\) \(\text{Moles of HCl} = 2.0\text{ mol/dm}^3 \times \frac{75}{1000}\text{ dm}^3 = 0.15\text{ mol}\)
2. Determine the limiting reactant: According to the equation, \(1\text{ mol Mg}\) reacts with \(2\text{ mol HCl}\). \(0.20\text{ mol Mg}\) requires \(0.40\text{ mol HCl}\). Since only \(0.15\text{ mol HCl}\) is available, \(\text{HCl}\) is the limiting reactant.
3. Calculate moles and volume of \(\text{H}_2\): \(\text{Moles of } \text{H}_2 = \frac{0.15}{2} = 0.075\text{ mol}\) \(\text{Volume of } \text{H}_2 = 0.075\text{ mol} \times 24\text{ dm}^3\text{/mol} = 1.8\text{ dm}^3\).
Marking scheme
A is correct [1]. Correct identification of HCl as the limiting reactant (\(0.15\text{ mol}\)), giving \(0.075\text{ mol of } \text{H}_2\) and \(1.8\text{ dm}^3\).
Question 12 · multiple_choice
1 marks
Two tests are carried out on an aqueous solution of compound \(X\):
1. Aqueous sodium hydroxide is added dropwise until in excess: a green precipitate forms which is insoluble in excess. 2. Dilute nitric acid is added followed by aqueous barium nitrate: a white precipitate forms.
What is the identity of compound \(X\)?
A.chromium(III) sulfate
B.iron(II) sulfate
C.iron(II) chloride
D.iron(III) sulfate
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Worked solution
- The formation of a green precipitate with aqueous sodium hydroxide that does not dissolve in excess confirms the presence of iron(II) ions (\(\text{Fe}^{2+}\)). Note that chromium(III) also gives a green/grey-green precipitate, but it dissolves in excess sodium hydroxide to give a green solution. - The formation of a white precipitate upon adding dilute nitric acid and aqueous barium nitrate confirms the presence of sulfate ions (\(\text{SO}_4^{2-}\)). Therefore, compound \(X\) is iron(II) sulfate.
Marking scheme
B is correct [1]. \(\text{Fe}^{2+}\) gives an insoluble green precipitate with NaOH, and \(\text{SO}_4^{2-}\) gives a white precipitate with acidified \(\text{Ba(NO}_3)_2\).
Question 13 · multiple_choice
1 marks
How many structural isomers with the molecular formula \(\text{C}_4\text{H}_8\) are alkenes?
A.2
B.3
C.4
D.5
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Worked solution
Alkenes with the formula \(\text{C}_4\text{H}_8\) contain one carbon-carbon double bond: 1. \(\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3\) (but-1-ene) 2. \(\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3\) (but-2-ene) 3. \(\text{CH}_2=\text{C(CH}_3)_2\) (2-methylpropene / methylpropene) Cyclobutane and methylcyclopropane are cyclic alkanes (not alkenes). Thus, there are exactly 3 alkene structural isomers.
Marking scheme
B is correct [1]. The 3 alkene isomers of \(\text{C}_4\text{H}_8\) are but-1-ene, but-2-ene, and 2-methylpropene.
Question 14 · multiple_choice
1 marks
The reversible reaction used in the Contact process is shown.
Which single change increases both the rate of the reaction and the equilibrium yield of sulfur trioxide?
A.adding a vanadium(V) oxide catalyst
B.decreasing the temperature
C.increasing the pressure
D.increasing the volume of the reaction vessel
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Worked solution
- Increasing the pressure increases the concentration of gas molecules, resulting in more frequent collisions per unit time and thus an increased reaction rate. - In terms of equilibrium, the forward reaction produces 2 moles of gas from 3 moles of gaseous reactants. An increase in pressure favours the side with fewer gas moles (the right-hand side), increasing the equilibrium yield of \(\text{SO}_3\). - Temperature decrease increases yield but decreases rate. Adding a catalyst increases rate but does not affect yield.
Marking scheme
C is correct [1]. Increasing pressure increases collision frequency (rate) and shifts the position of equilibrium to the side with fewer moles of gas (increasing yield).
Question 15 · multiple_choice
1 marks
Which pair of substances is most suitable for preparing a pure, dry sample of barium sulfate?
A.aqueous barium chloride and aqueous sodium sulfate
B.solid barium carbonate and dilute sulfuric acid
C.solid barium metal and dilute sulfuric acid
D.solid barium oxide and dilute sulfuric acid
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Worked solution
Barium sulfate is an insoluble salt. Insoluble salts are prepared by precipitation mixing two soluble salt solutions (e.g. aqueous barium chloride and aqueous sodium sulfate). The insoluble barium sulfate precipitates out, is filtered off, washed with distilled water, and dried. Reactions between solid barium compounds (carbonate, oxide, or metal) and dilute sulfuric acid are unsuitable because an insoluble layer of barium sulfate quickly forms on the surface of the solid, preventing the reaction from going to completion.
Marking scheme
A is correct [1]. Preparation of an insoluble salt requires precipitation between two soluble aqueous solutions.
Question 16 · multiple_choice
1 marks
Which statement correctly describes synthetic polymers?
A.Nylon is a polyester formed by the reaction of a dicarboxylic acid with a diol.
B.PET is a polyamide formed by addition polymerisation.
C.Condensation polymerisation forms a polymer along with a small molecule such as water.
D.Poly(ethene) is formed by condensation polymerisation of ethane molecules.
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Worked solution
- A is incorrect: Nylon is a polyamide formed from a dicarboxylic acid and a diamine. - B is incorrect: PET is a polyester formed by condensation polymerisation. - C is correct: Condensation polymerisation involves monomer molecules joining together with the loss of a small molecule, such as water. - D is incorrect: Poly(ethene) is formed by addition polymerisation of ethene monomers.
Marking scheme
C is correct [1]. Condensation polymerisation yields the polymer and eliminates a small molecule such as water.
Question 17 · multiple_choice
1 marks
Aqueous copper(II) sulfate is electrolysed using inert platinum electrodes.
Which row correctly identifies the product formed at each electrode and the change in the electrolyte?
A.cathode: hydrogen; anode: oxygen; change in electrolyte: colour fades
B.cathode: copper; anode: oxygen; change in electrolyte: solution becomes acidic
D.cathode: oxygen; anode: copper; change in electrolyte: colour remains unchanged
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Worked solution
At the cathode (negative electrode), \(\text{Cu}^{2+}\) ions are discharged in preference to \(\text{H}^+\) ions because copper is less reactive than hydrogen: \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\). At the anode (positive electrode), \(\text{OH}^-\) ions are discharged producing oxygen gas: \(4\text{OH}^-\text{(aq)} \rightarrow \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^-\). As \(\text{Cu}^{2+}\) ions are removed from the solution, the blue colour fades and the remaining solution contains \(\text{H}^+\) and \(\text{SO}_4^{2-}\) ions, becoming acidic (forming dilute sulfuric acid).
Marking scheme
B is correct [1]. A is incorrect: hydrogen is not formed at the cathode when aqueous copper(II) ions are present. C is incorrect: sulfur dioxide is not the product at the anode. D is incorrect: oxygen is produced at the anode, not copper, and copper is discharged at the cathode.
Question 18 · multiple_choice
1 marks
A \(4.80\text{ g}\) sample of magnesium ribbon is reacted with \(100\text{ cm}^3\) of \(2.00\text{ mol/dm}^3\) hydrochloric acid.
What is the maximum volume of hydrogen gas, measured at room temperature and pressure (r.t.p.), produced in this reaction?
A.\(2.40\text{ dm}^3\)
B.\(4.80\text{ dm}^3\)
C.\(7.20\text{ dm}^3\)
D.\(9.60\text{ dm}^3\)
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Worked solution
First calculate the amounts in moles: Amount of \(\text{Mg} = \frac{4.80\text{ g}}{24.0\text{ g/mol}} = 0.200\text{ mol}\). Amount of \(\text{HCl} = 2.00\text{ mol/dm}^3 \times \frac{100}{1000}\text{ dm}^3 = 0.200\text{ mol}\). According to the stoichiometry, \(1\text{ mol}\) of \(\text{Mg}\) requires \(2\text{ mol}\) of \(\text{HCl}\). Therefore, \(0.200\text{ mol}\) of \(\text{HCl}\) reacts with only \(0.100\text{ mol}\) of \(\text{Mg}\), making \(\text{HCl}\) the limiting reactant. Amount of \(\text{H}_2\) produced = \(\frac{0.200\text{ mol}}{2} = 0.100\text{ mol}\). Volume of \(\text{H}_2\text{ at r.t.p.} = 0.100\text{ mol} \times 24.0\text{ dm}^3\text{/mol} = 2.40\text{ dm}^3\).
Marking scheme
A is correct [1]. B is incorrect: calculated assuming magnesium is the limiting reactant (\(0.200 \times 24.0 = 4.80\text{ dm}^3\)). C is incorrect: arithmetic error involving incomplete molar volume conversion. D is incorrect: calculated assuming excess acid and full conversion of both reagents without limiting stoichiometry.
Question 19 · multiple_choice
1 marks
Which statement correctly explains why increasing the temperature increases the rate of a chemical reaction?
A.The activation energy of the reaction is lowered.
B.The activation energy is increased, leading to more energetic collisions.
C.The particles collide less frequently but with much higher kinetic energy.
D.A greater proportion of colliding particles have energy greater than or equal to the activation energy.
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Worked solution
Increasing the temperature increases the kinetic energy of the particles, so particles move faster and collide more frequently. Crucially, a significantly greater proportion of colliding particles possess energy equal to or greater than the activation energy (\(E_a\)), resulting in a higher frequency of successful collisions.
Marking scheme
D is correct [1]. A is incorrect: increasing temperature does not lower the activation energy; only a catalyst lowers activation energy. B is incorrect: the activation energy of the reaction remains unchanged. C is incorrect: the frequency of collisions increases, not decreases.
Question 20 · multiple_choice
1 marks
Solid \(\text{W}\) is dissolved in water to make an aqueous solution. Separate portions of the solution are tested:
1. Addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. 2. Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate.
What is the identity of solid \(\text{W}\)?
A.iron(II) chloride
B.iron(III) sulfate
C.iron(II) sulfate
D.chromium(III) sulfate
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Worked solution
Test 1: Aqueous sodium hydroxide giving a green precipitate insoluble in excess confirms the presence of iron(II) ions (\(\text{Fe}^{2+}\)). Chromium(III) also gives a green precipitate but it dissolves in excess sodium hydroxide. Test 2: Dilute nitric acid followed by aqueous barium nitrate giving a white precipitate confirms the presence of sulfate ions (\(\text{SO}_4^{2-}\)). Therefore, solid \(\text{W}\) is iron(II) sulfate.
Marking scheme
C is correct [1]. A is incorrect: iron(II) chloride gives a white precipitate with aqueous silver nitrate, not barium nitrate. B is incorrect: iron(III) sulfate gives a red-brown precipitate with aqueous sodium hydroxide. D is incorrect: chromium(III) sulfate forms a green precipitate that is soluble in excess sodium hydroxide.
Question 21 · multiple_choice
1 marks
Which statement regarding the structural isomers of \(\text{C}_4\text{H}_{10}\) and \(\text{C}_4\text{H}_8\) is correct?
A.There are three structural isomers of \(\text{C}_4\text{H}_{10}\).
B.All structural isomers of \(\text{C}_4\text{H}_{10}\) are saturated hydrocarbons.
C.All structural isomers of \(\text{C}_4\text{H}_8\) do not decolourise aqueous bromine.
D.The structural isomers of \(\text{C}_4\text{H}_{10}\) have identical boiling points.
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Worked solution
Butane (\(\text{C}_4\text{H}_{10}\)) has two structural isomers: butane and 2-methylpropane. Both are alkanes, meaning they are saturated hydrocarbons (they contain only single covalent bonds). \(\text{C}_4\text{H}_8\) has alkenes as isomers (e.g. but-1-ene, but-2-ene, 2-methylpropene) which are unsaturated, as well as cyclobutane (cycloalkane).
Marking scheme
B is correct [1]. A is incorrect: \(\text{C}_4\text{H}_{10}\) has 2 structural isomers, not 3. C is incorrect: \(\text{C}_4\text{H}_8\) isomers such as but-1-ene decolourise bromine water; they are not saturated. D is incorrect: structural isomers have different physical properties such as boiling points.
Question 22 · multiple_choice
1 marks
Which pair of monomers can undergo condensation polymerisation to form a polyester?
A.a dicarboxylic acid and a diol
B.a dicarboxylic acid and a diamine
C.a dialkene and a diol
D.a diamine and a diol
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Worked solution
A polyester is formed by a condensation reaction between a dicarboxylic acid (containing two \(-\text{COOH}\) groups) and a diol (containing two \(-\text{OH}\) groups), eliminating small molecules of water at each ester linkage.
Marking scheme
A is correct [1]. B is incorrect: a dicarboxylic acid and a diamine form a polyamide (such as nylon). C is incorrect: dialkenes and diols do not undergo condensation polymerisation together. D is incorrect: diamines and diols do not react to form polyesters.
Question 23 · multiple_choice
1 marks
The Contact process involves the following reversible reaction:
Which combination of conditions produces the highest equilibrium yield of sulfur trioxide, \(\text{SO}_3\)?
A.high temperature and high pressure
B.high temperature and low pressure
C.low temperature and high pressure
D.low temperature and low pressure
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Worked solution
The forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's principle, lowering the temperature shifts the equilibrium position to the right in the exothermic direction, increasing the yield of \(\text{SO}_3\). The forward reaction involves a decrease in the number of moles of gas (3 moles of gaseous reactants form 2 moles of gaseous products). Increasing the pressure shifts the equilibrium position to the side with fewer gas molecules (the right), thereby increasing the yield of \(\text{SO}_3\).
Marking scheme
C is correct [1]. A is incorrect: high temperature shifts an exothermic reaction towards the left, lowering yield. B is incorrect: high temperature and low pressure both decrease the yield. D is incorrect: low pressure shifts equilibrium towards the left (more moles of gas).
Question 24 · multiple_choice
1 marks
Hydrogen reacts with chlorine according to the equation:
D is correct [1]. A is incorrect: calculated as \(+184\text{ kJ/mol}\) (energy out - energy in, incorrect sign convention). B is incorrect: failing to multiply the \(\text{H}-\text{Cl}\) bond energy by 2 (\(678 - 431 = +247\)). C is incorrect: \(-247\text{ kJ/mol}\) from only forming one mole of \(\text{HCl}\).
Question 25 · multiple_choice
1 marks
Concentrated aqueous copper(II) sulfate is electrolysed using inert graphite electrodes.
What is observed at the anode and what is the change in the electrolyte?
A.pink-brown solid deposited; electrolyte remains blue
B.colourless gas bubbles formed; electrolyte fades from blue to colourless
C.brown liquid formed; electrolyte turns pale green
D.colourless gas bubbles formed; electrolyte remains blue
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Worked solution
At the anode (positive electrode), hydroxide ions (\(\text{OH}^-\)) from water are preferentially discharged over sulfate ions (\(\text{SO}_4^{2-}\)), producing oxygen gas: \(4\text{OH}^-\text{(aq)} \rightarrow \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^-\). Therefore, colourless gas bubbles are observed at the anode.
At the cathode, \(\text{Cu}^{2+}\) ions are discharged to form copper metal: \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\). As \(\text{Cu}^{2+}\) ions are removed from the solution, the blue colour of the electrolyte gradually fades until it becomes colourless.
Marking scheme
B [1 mark] - Deduce that oxygen gas (colourless bubbles) is evolved at the anode and \(\text{Cu}^{2+}\) ions are removed from solution causing the blue colour to fade.
Question 26 · multiple_choice
1 marks
Which statement correctly explains why increasing the temperature increases the rate of a chemical reaction?
A.The activation energy of the reaction is decreased.
B.A greater proportion of colliding particles have energy equal to or greater than the activation energy.
C.The particles collide with lower frequency but have greater mass.
D.The reactant bonds are weakened without any change in particle speeds.
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Worked solution
Increasing the temperature increases the average kinetic energy of the particles. Consequently, a significantly greater proportion of colliding particles possess kinetic energy equal to or greater than the activation energy (\(E_a\)), leading to a higher frequency of successful collisions. Catalysts lower activation energy, not temperature.
Marking scheme
B [1 mark] - Correctly links temperature increase to the proportion of particles with energy \(\ge E_a\).
Question 27 · multiple_choice
1 marks
A volume of \(20\text{ cm}^3\) of a gaseous hydrocarbon \(\text{C}_x\text{H}_y\) reacts completely with \(100\text{ cm}^3\) of oxygen gas, \(\text{O}_2\).
\(60\text{ cm}^3\) of carbon dioxide gas, \(\text{CO}_2\), is formed.
All gas volumes are measured at room temperature and pressure (r.t.p.).
What is the formula of the hydrocarbon?
A.\(\text{C}_2\text{H}_4\)
B.\(\text{C}_3\text{H}_6\)
C.\(\text{C}_3\text{H}_8\)
D.\(\text{C}_4\text{H}_{10}\)
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Worked solution
By Avogadro's law, the reacting volume ratios equal the molar ratios:
C [1 mark] - Determine mole ratio 1 : 5 : 3 and deduce formula \(\text{C}_3\text{H}_8\).
Question 28 · multiple_choice
1 marks
A student carries out tests on an aqueous solution of compound \(\text{X}\).
- Addition of aqueous sodium hydroxide gives a white precipitate that dissolves in excess to give a colourless solution. - Addition of aqueous ammonia gives a white precipitate that is insoluble in excess. - Addition of dilute nitric acid followed by aqueous silver nitrate gives a cream precipitate.
What is compound \(\text{X}\)?
A.aluminium bromide
B.aluminium chloride
C.zinc bromide
D.zinc chloride
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Worked solution
1. Cation tests: - \(\text{Al}^{3+}\) gives a white precipitate with \(\text{NaOH(aq)}\) soluble in excess, and a white precipitate with \(\text{NH}_3\text{(aq)}\) insoluble in excess. - (\(\text{Zn}^{2+}\) would dissolve in excess \(\text{NH}_3\text{(aq)}\); \(\text{Ca}^{2+}\) does not form a precipitate with \(\text{NH}_3\text{(aq)}\)). Thus, the cation is \(\text{Al}^{3+}\).
2. Anion tests: - Dilute nitric acid followed by aqueous silver nitrate gives a cream precipitate, identifying the bromide ion (\(\text{Br}^-\)).
Therefore, compound \(\text{X}\) is aluminium bromide.
Marking scheme
A [1 mark] - Identify \(\text{Al}^{3+}\) from precipitation reactions with \(\text{NaOH}\) and \(\text{NH}_3\). - Identify \(\text{Br}^-\)
Question 29 · multiple_choice
1 marks
The production of sulfur trioxide in the Contact process is represented by the equation:
Which combination of changes increases both the equilibrium yield of \(\text{SO}_3\text{(g)}\) and the rate of the forward reaction?
A.increasing temperature and increasing pressure
B.increasing pressure and adding more \(\text{O}_2\text{(g)}\)
C.decreasing temperature and adding a catalyst
D.decreasing pressure and removing \(\text{SO}_3\text{(g)}\)
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Worked solution
- Increasing pressure shifts the equilibrium position to the right because there are fewer moles of gas on the right (2 moles) than on the left (3 moles), thereby increasing the yield of \(\text{SO}_3\text{(g)}\). It also increases the concentration of gas molecules, which increases the collision frequency and thus the rate. - Increasing the concentration of \(\text{O}_2\text{(g)}\) drives the equilibrium to the right (increasing yield) and also increases the rate of collision between \(\text{SO}_2\) and \(\text{O}_2\). - Increasing temperature would increase the rate but decrease the equilibrium yield because the forward reaction is exothermic.
Marking scheme
B [1 mark] - Correctly apply Le Chatelier's principle and collision theory.
Question 30 · multiple_choice
1 marks
Barium sulfate, \(\text{BaSO}_4\), is an insoluble salt.
Which procedure is most suitable for preparing a pure, dry sample of barium sulfate in the laboratory?
A.Add excess barium carbonate to dilute sulfuric acid, filter, evaporate the filtrate to crystallisation point, and dry the crystals.
B.Mix aqueous barium chloride and dilute sulfuric acid, filter the mixture, wash the residue with distilled water, and dry on filter paper.
C.Titrate aqueous barium hydroxide with dilute sulfuric acid, and evaporate the neutral solution to dryness.
D.Add barium metal to dilute sulfuric acid, filter off excess metal, and heat the filtrate until all water has evaporated.
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Worked solution
Barium sulfate is an insoluble salt, which must be prepared by a precipitation reaction between two soluble salts (or a soluble salt and an acid):
1. Mix solutions of aqueous barium chloride, \(\text{BaCl}_2\text{(aq)}\), and dilute sulfuric acid, \(\text{H}_2\text{SO}_4\text{(aq)}\). 2. Filter off the insoluble barium sulfate residue. 3. Wash the residue with distilled water to remove soluble impurities. 4. Dry the solid residue between filter papers or in a warm oven.
Methods involving crystallisation from the filtrate are used for soluble salts, not insoluble salts.
Marking scheme
B [1 mark] - Identify the correct precipitation, filtration, washing and drying procedure for an insoluble salt.
Question 31 · multiple_choice
1 marks
Which statement regarding synthetic polymers is correct?
A.Nylon is a polyester formed from a dicarboxylic acid and a diol.
B.Poly(ethene) is formed by condensation polymerisation with the release of water.
C.PET is a polyester formed from a dicarboxylic acid and a diol.
D.Proteins are synthetic polyamides formed from monomers that contain only carboxylic acid groups.
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Worked solution
- Poly(ethylene terephthalate) (PET) is a polyester formed by condensation polymerisation between a dicarboxylic acid monomer and a diol monomer with the elimination of water molecules. - Nylon is a polyamide, not a polyester. - Poly(ethene) is formed by addition polymerisation without eliminating small molecules. - Proteins are natural polyamides formed from amino acid monomers containing both amino and carboxylic acid functional groups.
Marking scheme
C [1 mark] - Recognize that PET is a polyester formed from a dicarboxylic acid and a diol.
Question 32 · multiple_choice
1 marks
Which structural formula represents a branched isomer of but-1-ene?
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Worked solution
But-1-ene has the molecular formula \(\text{C}_4\text{H}_8\) and is an unbranched alkene. - \(\text{CH}_3\text{C(CH}_3\text{)}=\text{CH}_2\) (2-methylpropene) has the molecular formula \(\text{C}_4\text{H}_8\), contains a carbon-carbon double bond, and is branched, making it a branched structural isomer of but-1-ene. - \(\text{CH}_3\text{CH}=\text{CHCH}_3\) is unbranched (but-2-ene). - \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3\) is butane (\(\text{C}_4\text{H}_{10}\)), an alkane.
Marking scheme
C [1 mark] - Identify the branched structural isomer with molecular formula \(\text{C}_4\text{H}_8\).
Question 33 · multiple_choice
1 marks
A concentrated aqueous solution of sodium chloride is electrolysed using inert carbon electrodes.
Which row correctly identifies the substance produced at each electrode and the change in pH of the electrolyte during the electrolysis?
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Worked solution
In the electrolysis of concentrated aqueous sodium chloride (brine): - At the cathode (negative electrode), $\text{H}^+$ ions are discharged preferentially to $\text{Na}^+$ ions, forming hydrogen gas, $\text{H}_2(\text{g})$. - At the anode (positive electrode), chloride ions ($\text{Cl}^-$) are discharged preferentially due to their high concentration, forming chlorine gas, $\text{Cl}_2(\text{g})$. - As $\text{H}^+$ and $\text{Cl}^-$ ions are removed from the solution, $\text{Na}^+$ and $\text{OH}^-$ ions remain in solution, forming aqueous sodium hydroxide ($\text{NaOH}$). The presence of excess hydroxide ions causes the pH of the solution to increase (become more alkaline).
Therefore, row $\mathbf{A}$ is correct.
Marking scheme
A is correct [1]. - B is incorrect because chlorine is discharged at the anode in concentrated solution, and the pH increases. - C is incorrect because sodium is too reactive to be discharged from an aqueous solution. - D is incorrect for both electrode products and pH change.
Question 34 · multiple_choice
1 marks
Excess calcium carbonate chips are added to $50.0\text{ cm}^3$ of $1.00\text{ mol/dm}^3$ dilute hydrochloric acid at $25\,^\circ\text{C}$. The volume of carbon dioxide gas evolved is recorded at regular time intervals.
The experiment is repeated under four different single changes of conditions.
Which change will produce a graph with a steeper initial gradient but will reach the same final volume of carbon dioxide gas?
A.Using $100.0\text{ cm}^3$ of $1.00\text{ mol/dm}^3$ dilute hydrochloric acid instead of $50.0\text{ cm}^3$
B.Using the same mass of calcium carbonate as a fine powder instead of chips
C.Carrying out the reaction at $15\,^\circ\text{C}$ instead of $25\,^\circ\text{C}$
D.Using larger chips of calcium carbonate of the same total mass
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Worked solution
A steeper initial gradient means the initial rate of reaction is increased. The same final volume of gas means the amount (in moles) of the limiting reactant is unchanged. In this reaction, calcium carbonate is in excess, so hydrochloric acid is the limiting reactant.
- Using powdered calcium carbonate increases surface area, which increases the rate (steeper initial gradient). Since the amount of $\text{HCl}$ is unchanged ($50.0\text{ cm}^3$ of $1.00\text{ mol/dm}^3$, $n = 0.050\text{ mol}$), the total volume of $\text{CO}_2$ produced remains identical. - Using $25.0\text{ cm}^3$ of $2.00\text{ mol/dm}^3$ $\text{HCl}$ also produces the same total volume and increases rate, but powdered carbonate is a direct physical modification of the excess reagent without altering solution stoichiometry. - Increasing the volume of $1.00\text{ mol/dm}^3$ $\text{HCl}$ would increase the final volume of gas. - Decreasing the temperature would decrease the initial rate (shallower gradient).
Marking scheme
B is correct [1]. - A is incorrect as using $100.0\text{ cm}^3$ doubles the moles of limiting reactant, producing twice the volume of gas. - C is incorrect as lower temperature decreases the rate of reaction (shallower gradient). - D is incorrect as using larger chips decreases the surface area and lowers the initial rate.
Question 35 · multiple_choice
1 marks
A sample of $4.80\text{ g}$ of magnesium ribbon is reacted completely with $100\text{ cm}^3$ of $2.50\text{ mol/dm}^3$ sulfuric acid.
[$A_\text{r}: \text{Mg}, 24$; molar gas volume at r.t.p. = $24.0\text{ dm}^3$]
What is the maximum volume of hydrogen gas, measured at room temperature and pressure (r.t.p.), produced in this reaction?
A.$2.40\text{ dm}^3$
B.$3.60\text{ dm}^3$
C.$4.80\text{ dm}^3$
D.$6.00\text{ dm}^3$
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Worked solution
1. Calculate moles of each reactant: $$\text{Moles of } \text{Mg} = \frac{4.80\text{ g}}{24\text{ g/mol}} = 0.20\text{ mol}$$ $$\text{Moles of } \text{H}_2\text{SO}_4 = \text{concentration} \times \text{volume} = 2.50\text{ mol/dm}^3 \times \frac{100}{1000}\text{ dm}^3 = 0.25\text{ mol}$$
2. Determine the limiting reactant: The molar ratio of $\text{Mg} : \text{H}_2\text{SO}_4$ is $1:1$. Since $0.20\text{ mol} < 0.25\text{ mol}$, $\text{Mg}$ is the limiting reactant.
3. Calculate the volume of $\text{H}_2$ gas produced: $$\text{Moles of } \text{H}_2 = 0.20\text{ mol}$$ $$\text{Volume of } \text{H}_2 = 0.20\text{ mol} \times 24.0\text{ dm}^3\text{/mol} = 4.80\text{ dm}^3$$
Therefore, the volume produced is $4.80\text{ dm}^3$.
Marking scheme
C is correct [1]. - A ($2.40\text{ dm}^3$) is incorrect, corresponding to half the required mole quantity. - B ($3.60\text{ dm}^3$) is incorrect. - D ($6.00\text{ dm}^3$) is incorrect as it assumes $\text{H}_2\text{SO}_4$ ($0.25\text{ mol} \times 24.0 = 6.00\text{ dm}^3$) is the limiting reactant instead of magnesium.
Question 36 · multiple_choice
1 marks
Compound $\mathbf{X}$ is a solid. Aqueous solution of $\mathbf{X}$ is tested with various reagents and the observations are recorded:
- Addition of aqueous sodium hydroxide gives a green precipitate that is insoluble in excess sodium hydroxide. - Addition of dilute nitric acid followed by aqueous barium nitrate gives a white precipitate.
What is compound $\mathbf{X}$?
A.chromium(III) sulfate
B.iron(II) sulfate
C.iron(II) chloride
D.iron(III) sulfate
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Worked solution
Analysis of the qualitative test observations: 1. With aqueous sodium hydroxide, a green precipitate that remains insoluble in excess confirms the presence of iron(II) ions ($\text{Fe}^{2+}$). - $\text{Cr}^{3+}$ produces a grey-green precipitate that is soluble in excess sodium hydroxide to give a green solution. - $\text{Fe}^{3+}$ gives a red-brown precipitate. - $\text{Cu}^{2+}$ gives a light blue precipitate.
2. With dilute nitric acid followed by aqueous barium nitrate, a white precipitate confirms the presence of sulfate ions ($\text{SO}_4^{2-}$).
Combining these deductions shows that compound $\mathbf{X}$ is iron(II) sulfate.
Marking scheme
B is correct [1]. - A is incorrect because chromium(III) sulfate produces a precipitate that dissolves in excess $\text{NaOH}$. - C is incorrect because iron(II) chloride would give no precipitate with barium nitrate (chloride gives a precipitate with silver nitrate). - D is incorrect because iron(III) sulfate gives a red-brown precipitate with $\text{NaOH}$.
Question 37 · multiple_choice
1 marks
How many structural isomers have the unbranched or branched molecular formula $\text{C}_4\text{H}_8$ that contain a carbon-to-carbon double bond?
A.2
B.3
C.4
D.5
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Worked solution
The molecular formula $\text{C}_4\text{H}_8$ contains an alkene group ($\,\text{C}=\text{C}\,$, degree of unsaturation = 1).
The possible structural isomers containing a $\text{C}=\text{C}$ double bond are: 1. But-1-ene: $\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3$ 2. But-2-ene: $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3$ 3. 2-Methylpropene: $\text{CH}_2=\text{C}(\text{CH}_3)_2$
Thus, there are exactly 3 structural isomers with a $\text{C}=\text{C}$ double bond.
Marking scheme
B is correct [1]. - A (2) is incorrect because it omits branched 2-methylpropene. - C (4) is incorrect as it incorrectly counts stereoisomers (cis/trans) or cyclobutane, but the question specifies structural isomers containing a $\text{C}=\text{C}$ bond. - D (5) is incorrect.
Question 38 · multiple_choice
1 marks
Which statement correctly describes the preparation of pure, dry crystals of barium sulfate?
A.Add excess barium carbonate to dilute sulfuric acid, filter off the excess solid, and heat the filtrate until crystals form.
B.Titrate aqueous barium hydroxide with dilute sulfuric acid using an indicator, then evaporate the neutral solution to the point of crystallisation.
C.React barium metal with dilute sulfuric acid, filter the mixture, and collect the crystals from the evaporated filtrate.
D.Mix aqueous barium chloride with aqueous sodium sulfate, filter the mixture, wash the residue with distilled water, and dry the residue.
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Worked solution
Barium sulfate is an insoluble salt. - Insoluble salts are prepared by precipitation reactions, which involve mixing two aqueous solutions containing the relevant ions (e.g., aqueous barium chloride and aqueous sodium sulfate). - The precipitate formed is separated by filtration. - The residue is washed with a small amount of distilled water to remove soluble impurities and then dried between sheets of filter paper (or in a low-temperature oven). - Soluble salts are prepared by titration or reaction of acid with an excess insoluble solid, followed by crystallisation, but insoluble salts must not be crystallised from the filtrate.
Marking scheme
D is correct [1]. - A is incorrect because barium sulfate is insoluble and cannot be obtained by evaporating the filtrate to crystallisation point. - B is incorrect because titrating an acid and alkali is used to prepare soluble salts. - C is incorrect because barium metal does not dissolve to give crystallisable barium sulfate in this manner, as $\text{BaSO}_4$ forms an insoluble coating preventing complete reaction.
Question 39 · multiple_choice
1 marks
The reversible reaction between sulfur dioxide and oxygen in the Contact process is shown.
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Worked solution
According to Le Chatelier's principle: 1. Effect of Temperature: The forward reaction is exothermic ($\Delta H$ is negative). Lowering the temperature shifts the equilibrium position in the exothermic direction (to the right), which increases the yield of $\text{SO}_3$.
2. Effect of Pressure: The left-hand side has $2 + 1 = 3$ moles of gas, while the right-hand side has $2$ moles of gas. Increasing the pressure shifts the equilibrium position to the side with fewer gas moles (to the right), which increases the yield of $\text{SO}_3$.
Therefore, a low temperature and a high pressure give the highest equilibrium yield of $\text{SO}_3$ (Row $\mathbf{C}$).
Marking scheme
C is correct [1]. - A is incorrect because high temperature favours the endothermic backward reaction, decreasing yield. - B is incorrect because high temperature and low pressure both decrease yield. - D is incorrect because low pressure shifts equilibrium to the left.
Question 40 · multiple_choice
1 marks
Which statement about condensation polymers is correct?
A.Condensation polymers are formed exclusively from monomer molecules containing carbon-to-carbon double bonds.
B.Nylon is a synthetic polyamide formed by the reaction between a dicarboxylic acid and a diamine.
C.PET is a condensation polymer containing amide linkages along its polymer backbone.
D.During condensation polymerisation, the polymer is the only product formed.
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Worked solution
- Condensation polymerisation occurs when monomers with two functional groups react together, eliminating a small molecule (such as water) for each linkage formed. - Polyamides, such as nylon, contain the amide linkage ($-\text{CONH}-$), which is formed from the reaction between a dicarboxylic acid and a diamine. - Polyesters, such as PET, contain the ester linkage ($-\text{COO}-$), formed from a dicarboxylic acid and a diol. - Addition polymerisation (not condensation) involves monomer molecules containing $\,\text{C}=\text{C}\,$, where no small molecule is eliminated.
Therefore, statement $\mathbf{B}$ is correct.
Marking scheme
B is correct [1]. - A is incorrect because the monomers in addition polymerisation have carbon-to-carbon double bonds, whereas condensation polymers are made from monomers with functional groups like $-\text{COOH}$, $-\text{NH}_2$, or $-\text{OH}$. - C is incorrect because polyesters contain ester linkages ($-\text{COO}-$), not amide linkages. - D is incorrect because a small molecule such as water or hydrogen chloride is eliminated during condensation polymerisation.
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7 Question · 80 marks
Question 1 · Equation & Reagent Matching
7 marks
Salts can be prepared in the laboratory by selecting suitable starting reagents and experimental methods.
Table 1.1 gives information about the preparation of three different salts.
Table 1.1
| Salt to prepare | Starting reagents | Method of preparation | | :--- | :--- | :--- | | barium sulfate, \(\text{BaSO}_4\) | aqueous barium nitrate and reagent A | precipitation | | magnesium chloride, \(\text{MgCl}_2\) | dilute hydrochloric acid and excess solid reagent B | reaction of an acid with an insoluble substance | | potassium sulfate, \(\text{K}_2\text{SO}_4\) | dilute sulfuric acid and aqueous potassium hydroxide | method C |
(a) Identify reagent A, reagent B, and method C.
(i) reagent A [1]
(ii) reagent B [1]
(iii) method C [1]
(b) Write a balanced chemical equation, including state symbols, for the reaction between dilute hydrochloric acid and the solid reagent B named in (a)(ii). [2]
(c) Write an ionic equation, including state symbols, for the precipitation reaction used to prepare barium sulfate. [2]
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Worked solution
(a)(i) Barium sulfate is an insoluble salt prepared by precipitation. This requires mixing two soluble solutions containing barium ions and sulfate ions. Since barium nitrate provides \(\text{Ba}^{2+}\text{(aq)}\), a suitable soluble source of sulfate ions is aqueous sodium sulfate (or potassium sulfate / dilute sulfuric acid).
(ii) Magnesium chloride is a soluble salt prepared from an acid and an insoluble substance. Suitable insoluble substances containing magnesium include magnesium oxide (\(\text{MgO}\)), magnesium carbonate (\(\text{MgCO}_3\)), or magnesium hydroxide (\(\text{Mg(OH)}_2\)). Excess magnesium metal (\(\text{Mg}\)) is also acceptable as an insoluble reactant.
(iii) Potassium sulfate is a soluble salt formed from a soluble acid (dilute sulfuric acid) and a soluble base (aqueous potassium hydroxide). The method used is titration.
(b) For magnesium oxide reacting with dilute hydrochloric acid: \[\text{MgO(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{O(l)}\] (If \(\text{MgCO}_3\) is used: \(\text{MgCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\); if \(\text{Mg}\) is used: \(\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}\)).
(c) The ionic equation for the precipitation of insoluble barium sulfate is: \[\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\]
(b) [2 marks] - M1: Correct formulae of reactants and products for the chosen reagent in (a)(ii) [1] - M2: Correct balancing and correct state symbols throughout [1] - e.g. \(\text{MgO(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{O(l)}\) - e.g. \(\text{MgCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}\) - e.g. \(\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}\)
(c) [2 marks] - M1: \(\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4\) [1] - M2: Correct state symbols: \(\text{(aq)} + \text{(aq)} \rightarrow \text{(s)}\) [1] (dependent on correct or recognisable formulae)
Question 2 · theory
10 marks
This question is about atomic structure and the Periodic Table.
(a) (i) Define the term isotopes. [2]
(ii) Complete Table 1.1 to show the number of protons, neutrons and electrons in the given particles.
Table 1.1 particlenumber of protonsnumber of neutronsnumber of electrons\(^{31}_{15}\text{P}^{3-}\) \(^{39}_{19}\text{K}^{+}\) [3]
(b) Phosphorus has an atomic number of 15.
(i) State the electronic configuration of a phosphorus atom. [1]
(ii) Explain, in terms of its electronic configuration, why phosphorus is placed in Group V and in Period 3 of the Periodic Table. [2]
(c) A naturally occurring sample of magnesium contains three isotopes: \(^{24}\text{Mg}\), \(^{25}\text{Mg}\) and \(^{26}\text{Mg}\). The percentage abundances are shown in Table 1.2.
Table 1.2 isotopepercentage abundance / %\(^{24}\text{Mg}\)79.0\(^{25}\text{Mg}\)10.0\(^{26}\text{Mg}\)11.0 Calculate the relative atomic mass, \(A_r\), of this sample of magnesium. Give your answer to two decimal places.
\(A_r\) = .............................. [2]
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Worked solution
(a) (i) Isotopes are atoms of the same element having the same number of protons (or same atomic number) but different numbers of neutrons (or different nucleon / mass numbers).
(ii) For \(^{31}_{15}\text{P}^{3-}\): Protons = 15 Neutrons = 31 − 15 = 16 Electrons = 15 + 3 = 18
(a) (i) M1: atoms of the same element with the same number of protons / same atomic number [1] M2: different number of neutrons / different nucleon number / different mass number [1] REJECT 'different elements'
(ii) Row 1 (\(^{31}_{15}\text{P}^{3-}\)): 15 protons AND 16 neutrons [1] Row 1 (\(^{31}_{15}\text{P}^{3-}\)): 18 electrons [1] Row 2 (\(^{39}_{19}\text{K}^{+}\)): 19 protons, 20 neutrons AND 18 electrons [1]
(b) (i) 2,8,5 or 2.8.5 [1]
(ii) M1: (Group V because it has) 5 outer-shell electrons / 5 valence electrons [1] M2: (Period 3 because it has) 3 occupied electron shells / 3 energy levels [1]
(c) M1: correct expression: \(\frac{(24 \times 79.0) + (25 \times 10.0) + (26 \times 11.0)}{100}\) or \(18.96 + 2.50 + 2.86\) [1] M2: 24.32 [1] (Note: 24.32 must be to 2 decimal places for M2. Allow ECF on arithmetic error in M1).
Question 3 · theory
10 marks
This question is about bonding, structure and properties of different substances.
(a) Magnesium fluoride, \(\text{MgF}_2\), is an ionic compound.
(i) State the electronic configuration of: - the magnesium ion in \(\text{MgF}_2\) - the fluoride ion in \(\text{MgF}_2\) [2]
(ii) Explain why magnesium fluoride conducts electricity when molten, but does not conduct electricity when solid. [2]
(b) Silicon(IV) oxide, \(\text{SiO}_2\), has a giant covalent structure.
(i) Describe the structure and bonding in silicon(IV) oxide. [2]
(ii) Explain, in terms of structure and bonding, why silicon(IV) oxide has a very high melting point. [2]
(c) Methanol, \(\text{CH}_3\text{OH}\), is a simple molecular compound.
Draw a dot-and-cross diagram to show the arrangement of the outer-shell electrons in a molecule of methanol.
Show outer-shell electrons only. [2]
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Worked solution
(a) (i) Magnesium atom has atomic number 12 (configuration 2,8,2), so \(\text{Mg}^{2+}\) loses 2 electrons giving 2,8. Fluorine atom has atomic number 9 (configuration 2,7), so \(\text{F}^-\kern-0.1em\) gains 1 electron giving 2,8.
(ii) In solid magnesium fluoride, the ions are held in fixed positions within the giant ionic lattice and are not free to move. In molten magnesium fluoride, the lattice breaks down and the ions are free to move to carry the electric current.
(b) (i) Silicon(IV) oxide has a giant covalent lattice where each silicon atom is bonded to 4 oxygen atoms and each oxygen atom is bonded to 2 silicon atoms by covalent bonds.
(ii) A very large amount of thermal energy is needed to break the many strong covalent bonds throughout the giant structure.
(c) In \(\text{CH}_3\text{OH}\): - The central carbon atom shares 1 pair of electrons with each of the three hydrogen atoms (3 single covalent bonds, 6 bonding electrons total). - The carbon atom shares 1 pair of electrons with the oxygen atom (1 C–O single bond). - The oxygen atom shares 1 pair of electrons with the fourth hydrogen atom (1 O–H single bond). - The oxygen atom has 2 lone pairs (4 non-bonding electrons) to complete its octet.
(a)(ii) - Solid: ions are in fixed positions / cannot move (in lattice) (1) - Molten: ions are mobile / free to move (1) Note: REJECT 'free electrons' / 'delocalised electrons' for either mark.
(b)(i) - Each silicon atom is covalently bonded to four oxygen atoms (1) - Each oxygen atom is bonded to two silicon atoms (1) ALLOW giant covalent / macromolecular structure (1) if silicon-oxygen coordination ratio is not fully described.
(b)(ii) - Many / strong covalent bonds (1) - (Requires) a large amount of energy to break (bonds) (1) REJECT reference to intermolecular forces.
(c) - 5 shared pairs of bonding electrons correctly shown: 3 \(\text{C}-\text{H}\), 1 \(\text{C}-\text{O}\), 1 \(\text{O}-\text{H}\) (1) - 2 lone pairs (4 non-bonding electrons) on oxygen and no other unshared electrons on carbon or hydrogen (1)
Question 4 · theory
10 marks
A student investigated the rate of reaction between magnesium metal and dilute hydrochloric acid.
(a) Write a chemical equation for the reaction of magnesium with dilute hydrochloric acid. Include state symbols. [2]
(b) The student repeated the experiment using hydrochloric acid of a higher concentration, keeping all other variables constant. State and explain the effect of increasing the concentration of hydrochloric acid on the rate of reaction. Use ideas about particles and collisions in your answer. [3]
(c) In another experiment, the temperature of the reaction mixture was increased. Explain, in terms of collision theory and activation energy, why an increase in temperature increases the rate of reaction. [3]
(d) The original experiment was repeated using the same mass of magnesium powder instead of magnesium ribbon. The acid was in excess in both experiments. State the effect of using magnesium powder instead of magnesium ribbon on: (i) the initial rate of the reaction, [1] (ii) the total volume of hydrogen gas collected when the reaction is complete. [1]
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Worked solution
(a) The reaction between magnesium and hydrochloric acid forms magnesium chloride and hydrogen gas: \[\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}\]
(b) Increasing the concentration increases the rate of reaction. A higher concentration means there are more acid particles (\(\text{H}^+\) ions) per unit volume. As a result, the particles collide more frequently (higher collision frequency / more collisions per second).
(c) Increasing temperature causes particles to gain kinetic energy and move faster, leading to more frequent collisions. More importantly, a significantly greater proportion of colliding particles possess energy greater than or equal to the activation energy (\(E_{\text{a}}\)), resulting in a much higher proportion of successful (effective) collisions per unit time.
(d) (i) Powder has a larger surface area exposed to the acid, so the initial rate of reaction increases. (ii) The mass of magnesium and the amount of acid are unchanged, and the acid is in excess; therefore, the total volume of hydrogen gas collected remains the same.
Marking scheme
(a) - \(\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\) [1] - Correct state symbols: (s), (aq), (aq), (g) dependent on correct or near correct formulae [1]
(b) - (Rate) increases [1] - More particles per unit volume / particles are closer together [1] (IGNORE: more particles alone / more concentrated) - Higher frequency of collisions / more collisions per unit time / more collisions per second [1] (REJECT: "more collisions" without reference to time/rate/frequency)
(c) - Particles have more kinetic energy / move faster [1] - Higher frequency of collisions / more collisions per unit time / per second [1] - Greater proportion / fraction of particles / collisions have energy equal to or greater than activation energy / \(E \ge E_{\text{a}}\) [1] (ALLOW: more successful collisions per second / higher proportion of collisions are successful)
(d)(i) - (Rate) increases / faster [1]
(d)(ii) - Same / unchanged / no effect [1]
Question 5 · theory
15 marks
Aluminium is manufactured on a large scale by the electrolysis of a molten mixture of aluminium oxide and cryolite.
(a) (i) Name the ore from which aluminium oxide is obtained. [1]
(ii) State two reasons why cryolite is added to aluminium oxide. [2]
(iii) Write the ionic half-equation for the reaction occurring at the cathode during this extraction. [1]
(iv) Explain why the carbon anodes must be replaced at regular intervals during the extraction of aluminium. Include a balanced chemical equation in your answer. [3]
(b) A student electrolyses concentrated aqueous copper(II) chloride, \(\text{CuCl}_2(\text{aq})\), using inert carbon electrodes.
(i) State one observation made at the cathode. [1]
(ii) Write the ionic half-equation for the reaction occurring at the anode. [1]
(iii) Describe a chemical test to confirm the identity of the gas evolved at the anode and give the positive result. [2]
(c) Copper is refined industrially by electrolysis using aqueous copper(II) sulfate as the electrolyte.
(i) Identify the material used for the anode and the material used for the cathode in this refining process. [2]
(ii) Explain why the blue colour of the aqueous copper(II) sulfate electrolyte does not fade during the refining process. [2]
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Worked solution
(a) (i) The primary aluminium ore is bauxite. (ii) Cryolite is added because: 1. It lowers the operating temperature / melting point from approximately 2000 °C to around 950 °C (reducing energy costs). 2. It improves the electrical conductivity of the electrolyte / acts as a solvent for aluminium oxide. (iii) At the cathode (negative electrode), aluminium ions gain electrons (reduction): \(\text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al}\) (iv) Oxide ions are oxidised at the anode to form oxygen gas (\(\text{O}_2\)). Due to the high operating temperature (around 950 °C), the oxygen gas reacts with the graphite (carbon) anodes, oxidising them to carbon dioxide gas (\(\text{CO}_2\)). As a result, the anodes burn away and decrease in mass over time. Equation: \(\text{C} + \text{O}_2 \rightarrow \text{CO}_2\) (or \(2\text{C} + \text{O}_2 \rightarrow 2\text{CO}\)).
(b) (i) In aqueous \(\text{CuCl}_2\), \(\text{Cu}^{2+}\) ions are discharged preferentially at the cathode over \(\text{H}^+\) ions because copper is less reactive than hydrogen. Copper metal forms as a pink/brown/red-brown solid deposit on the cathode. (ii) In concentrated chloride solution, chloride ions are discharged preferentially at the anode: \(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\) or \(2\text{Cl}^- - 2\text{e}^- \rightarrow \text{Cl}_2\) (iii) Chlorine gas is identified by placing damp litmus paper (or damp indicator paper) into the gas; it bleaches (turns white).
(c) (i) For copper purification: Anode (positive electrode): Impure copper Cathode (negative electrode): Pure copper (ii) Copper atoms from the anode lose electrons to form \(\text{Cu}^{2+}\) ions (\(\text{Cu} \rightarrow \text{Cu}^{2+} + 2\text{e}^-\)) at the exact same rate as \(\text{Cu}^{2+}\) ions gain electrons to form copper metal at the cathode (\(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\)). Therefore, the concentration of \(\text{Cu}^{2+}\) ions in the electrolyte remains constant, so the intensity of the blue colour remains unchanged.
(ii) [2 marks] • M1: Rate of copper dissolving / \(\text{Cu}^{2+}\) forming at anode is equal to rate of copper depositing / \(\text{Cu}^{2+}\) discharging at cathode [1] • M2: Concentration of \(\text{Cu}^{2+}\) / copper(II) ions in solution remains constant [1]
Question 6 · theory
16 marks
Basic copper(II) carbonate, \( \text{Cu}_2\text{CO}_3(\text{OH})_2 \), reacts with dilute sulfuric acid, \( \text{H}_2\text{SO}_4 \), to form aqueous copper(II) sulfate, carbon dioxide, and water.
(a) Write a balanced chemical equation for the reaction between basic copper(II) carbonate and dilute sulfuric acid. [2]
(b) In an experiment, an excess of basic copper(II) carbonate is added to \( 50.0\text{ cm}^3 \) of \( 1.60\text{ mol/dm}^3 \) \( \text{H}_2\text{SO}_4 \).
(i) Calculate the number of moles of \( \text{H}_2\text{SO}_4 \) present in \( 50.0\text{ cm}^3 \) of this acid solution. [1]
(ii) Calculate the volume, in \( \text{cm}^3 \), of carbon dioxide gas formed at room temperature and pressure (r.t.p.). [2]
(iii) Calculate the maximum theoretical mass of hydrated copper(II) sulfate crystals, \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \), that can be formed from this reaction. [\( A_r: \text{H}, 1;\ \text{C}, 12;\ \text{O}, 16;\ \text{S}, 32;\ \text{Cu}, 64 \)] [3]
(iv) After crystallization and drying, the student obtained \( 15.5\text{ g} \) of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \). Calculate the percentage yield of hydrated copper(II) sulfate. [2]
(c) Describe three practical steps to obtain pure, dry crystals of hydrated copper(II) sulfate from the reaction mixture after the reaction has finished. [3]
(d) A small portion of the aqueous copper(II) sulfate solution is tested in two separate test-tubes.
(i) Aqueous sodium hydroxide is added dropwise until in excess. State the observation. [1]
(ii) Aqueous ammonia is added dropwise until in excess. State the observations. [2]
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(c) 1. Filter the mixture to remove excess unreacted basic copper(II) carbonate. 2. Heat the filtrate until saturated / to crystallization point, then allow to cool. 3. Filter off the crystals and dry them between filter papers / in a desiccator / low-temperature oven.
(d) (i) Light blue precipitate (insoluble in excess). (ii) Light blue precipitate formed initially, which dissolves in excess ammonia to give a deep blue solution.
Marking scheme
(a) • Correct formulae of all reactants and products [1] • Correct balancing: 1, 2, 2, 1, 3 [1]
(c) • Filter to remove excess / unreacted basic copper(II) carbonate [1] • Heat/evaporate filtrate to crystallisation point / until saturated, then allow to cool [1] • Filter off crystals AND dry between filter papers / in a warm oven / desiccator (REJECT heat to complete dryness / strong heating of crystals) [1]
(d)(i) • Light blue precipitate [1]
(d)(ii) • Light blue precipitate [1] • Dissolves in excess (ammonia) to give a deep blue solution [1]
Question 7 · theory
12 marks
Hydrocarbons are organic compounds that contain hydrogen and carbon only.
(a) Compound A is an alkene with the molecular formula \(\text{C}_4\text{H}_8\).
(i) Define the term structural isomers. [2]
(ii) Compound A has an unbranched carbon chain. Draw the displayed formulae of the two structural isomers of compound A and state their systematic names. [4]
(b) But-1-ene can undergo addition polymerisation to produce poly(but-1-ene).
(i) Draw a section of poly(but-1-ene) showing two repeat units. [2]
(ii) State one environmental disadvantage of disposing of addition polymers by incineration (burning). [1]
(c) Synthetic polymers can also be produced by condensation polymerisation.
(i) Name the type of polymer formed when a dicarboxylic acid reacts with a diol. [1]
(ii) State two differences between addition polymerisation and condensation polymerisation. [2]
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Worked solution
(a) (i) Structural isomers are defined as compounds that possess the same molecular formula but have different structural formulae.
(ii) The unbranched alkenes with formula \(\text{C}_4\text{H}_8\) are: • Isomer 1: but-1-ene. Displayed formula shows all single bonds and the double bond between carbon-1 and carbon-2: \(\text{H}_2\text{C}=\text{CH}-\text{CH}_2-\text{CH}_3\). • Isomer 2: but-2-ene. Displayed formula shows all single bonds and the double bond between carbon-2 and carbon-3: \(\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3\).
(b) (i) During addition polymerisation of but-1-ene (\(\text{CH}_2=\text{CH}-\text{CH}_2\text{CH}_3\)), the double bond breaks to form a continuous single-bonded carbon backbone. Two repeat units require a continuous backbone of 4 carbon atoms with single bonds, each containing alternating \(-\text{H}\) and \(-\text{CH}_2\text{CH}_3\) (or \(-\text{C}_2\text{H}_5\)) groups, with open extension/continuation bonds extending beyond the brackets or terminal carbons.
(ii) Incineration produces carbon dioxide (a greenhouse gas which contributes to climate change) and can release toxic or acidic gases (e.g. carbon monoxide).
(c) (i) When a dicarboxylic acid (containing \(-\text{COOH}\) groups) reacts with a diol (containing \(-\text{OH}\) groups), ester linkages (\(-\text{COO}-\)) are formed, creating a polyester.
(ii) Key differences: 1. Addition polymerisation yields only one product (the polymer), whereas condensation polymerisation eliminates a small molecule (such as \(\text{H}_2\text{O}\)) alongside the polymer. 2. Addition polymerisation uses unsaturated monomers (containing \(\text{C}=\text{C}\) double bonds), whereas condensation polymerisation uses monomers with two reactive functional groups (e.g. \(-\text{COOH}\), \(-\text{OH}\), or \(-\text{NH}_2\)).
Marking scheme
(a)(i) • same molecular formula [1] • different structural formulae / different structures / different arrangements of atoms [1] (REJECT: different general formula / different empirical formula)
(a)(ii) • correct displayed formula of but-1-ene showing all atoms and all C–H, C–C, and C=C bonds [1] • systematic name: but-1-ene [1] (ALLOW 1-butene) • correct displayed formula of but-2-ene showing all atoms and all C–H, C–C, and C=C bonds [1] • systematic name: but-2-ene [1] (ALLOW 2-butene, cis-but-2-ene, trans-but-2-ene)
(b)(i) • continuous 4-carbon single-bonded chain backbone with continuation bonds at both ends [1] • correct positioning of two ethyl / \(-\text{CH}_2\text{CH}_3\) / \(-\text{C}_2\text{H}_5\) groups on alternating carbons and remaining valencies filled with \(-\text{H}\) [1]
(b)(ii) • produces toxic / poisonous gases OR releases carbon dioxide / greenhouse gases / contributes to global warming [1] (IGNORE: causes air pollution unqualified)
(c)(i) • polyester [1] (ALLOW PET / poly(ethylene terephthalate))
(c)(ii) Any TWO from: • condensation forms a small molecule / water (as a by-product) whereas addition forms only one product / no other product [1] • addition requires monomers with \(\text{C}=\text{C}\) / carbon-carbon double bonds / unsaturated monomers, whereas condensation requires monomers with two / different functional groups [1] • addition polymer backbone contains only carbon atoms, whereas condensation polymer backbone contains oxygen / nitrogen atoms / contains ester / amide linkages [1]
Paper 6 (Alternative to Practical)
Answer all questions. Use the Qualitative Analysis Notes provided on the final pages to identify ions and gases.
4 Question · 40 marks
Question 1 · open_response
8 marks
A student prepares pure, dry crystals of hydrated nickel(II) sulfate, \(\text{NiSO}_4\cdot 6\text{H}_2\text{O}\), by reacting solid nickel(II) carbonate with dilute sulfuric acid.
(a) Name a piece of apparatus suitable to measure \(25.0\text{ cm}^3\) of dilute sulfuric acid accurately. [1]
(b) State two observations which indicate that the reaction is complete and nickel(II) carbonate is present in excess. [2]
(c) State why the reaction mixture is filtered before crystallisation. [1]
(d) Describe the procedure used to obtain pure, dry crystals of hydrated nickel(II) sulfate from the aqueous filtrate. [3]
(e) The student dried the crystals in an oven at \(150\text{ }^\circ\text{C}\) and observed that the green crystals turned into a yellow-white powder. Explain this observation. [1]
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Worked solution
(a) A volumetric pipette or a burette is designed to measure fixed or variable volumes accurately to \(0.1\text{ cm}^3\) (or \(0.05\text{ cm}^3\)).
(b) During the reaction of a carbonate with an acid, carbon dioxide gas is produced. When the acid is completely used up and carbonate is in excess: 1. Effervescence (bubbling/fizzing) ceases. 2. Excess solid nickel(II) carbonate no longer dissolves and settles at the bottom of the beaker.
(c) Filtration removes the unreacted solid nickel(II) carbonate, ensuring only aqueous nickel(II) sulfate remains in the filtrate.
(d) To prepare hydrated crystals: 1. Heat/boil the filtrate in an evaporating basin to evaporate water until the point of crystallisation (saturation point) is reached. 2. Allow the hot saturated solution to cool and stand so that crystals form slowly. 3. Filter off the crystals and dry them gently by pressing between sheets of filter paper (or in a desiccator / warm oven below dehydration temperature).
(e) Excessive heating removes the water of crystallisation from hydrated nickel(II) sulfate, turning it into anhydrous nickel(II) sulfate, which is yellow-white in colour.
(b) [2 marks] - Effervescence / bubbling / fizzing stops (1) - Solid / powder / green solid remains / settles at the bottom / stops dissolving (1)
(c) [1 mark] - To remove excess / unreacted nickel(II) carbonate / solid (1) IGNORE: To remove impurities
(d) [3 marks] - Heat / evaporate (the filtrate) to crystallisation point / until saturated / until crystals form on a glass rod (1) - Leave to cool / crystallise (1) - Filter (off crystals) AND dry between filter papers / in a desiccator / in a warm oven (1) REJECT: Heat to dryness for M1 / M3
(e) [1 mark] - Water of crystallisation is lost / dehydrated / anhydrous nickel(II) sulfate formed (1)
Question 2 · structured
18 marks
A student investigated the temperature change during the displacement reaction between aqueous copper(II) sulfate and zinc powder.
Seven experiments were carried out using the following procedure:
1. Measure $25.0\text{ cm}^3$ of aqueous copper(II) sulfate using a measuring cylinder and pour it into a polystyrene cup. 2. Measure and record the initial temperature of the aqueous copper(II) sulfate. 3. Add $0.20\text{ g}$ of zinc powder to the cup, stir the mixture continuously using the thermometer, and record the maximum temperature reached. 4. Empty and rinse the polystyrene cup with distilled water. 5. Repeat steps 1 to 4 using $0.40\text{ g}$, $0.60\text{ g}$, $0.80\text{ g}$, $1.00\text{ g}$, $1.20\text{ g}$, and $1.40\text{ g}$ of zinc powder.
(a) Complete Table 1.1 by calculating the temperature change, $\Delta T$, for each experiment. [3]
(b) Plot the temperature change, $\Delta T$, against the mass of zinc on the grid below. Draw two straight lines of best fit through the points. The two lines must cross. [5]
(c) (i) Use your graph to determine the minimum mass of zinc required to react completely with $25.0\text{ cm}^3$ of the aqueous copper(II) sulfate. Show clearly on your graph how you determined your answer. [2]
(ii) State the maximum temperature change achieved when the copper(II) sulfate has reacted completely. [1]
(d) (i) Explain why the temperature change remains constant when the mass of zinc added is increased from $1.00\text{ g}$ to $1.40\text{ g}$. [2]
(ii) Describe two observations during the reaction, other than a temperature increase. [2]
(e) (i) State one advantage of using a polystyrene cup rather than a glass beaker in this experiment. [1]
(ii) Suggest one improvement to the apparatus to reduce thermal energy loss to the surroundings. [1]
(iii) Predict the temperature change if Experiment 2 ($0.40\text{ g}$ of zinc) were repeated using $50.0\text{ cm}^3$ of the same aqueous copper(II) sulfate instead of $25.0\text{ cm}^3$. Explain your prediction. [1]
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(b) - Scale on y-axis: $0$ to $30\text{ }^\circ\text{C}$ (or $0$ to $25\text{ }^\circ\text{C}$) occupying at least half the grid. - Points plotted accurately as neat crosses or encircled dots. - First line of best fit: straight line passing through $(0, 0)$ and the points at $0.20, 0.40, 0.60, 0.80, 1.00\text{ g}$. - Second line of best fit: horizontal straight line passing through points at $1.00, 1.20, 1.40\text{ g}$ at $\Delta T = 25.0\text{ }^\circ\text{C}$.
(c) (i) The two best-fit lines intersect at a mass of $1.00\text{ g}$. Vertical tie line must be drawn down to the x-axis from the point of intersection. (ii) Maximum temperature change $= 25.0\text{ }^\circ\text{C}$.
(d) (i) All the aqueous copper(II) sulfate has reacted (copper(II) sulfate is the limiting reactant), so no further reaction occurs / no additional heat is released; the additional zinc is in excess. (ii) 1. The blue solution fades and turns colourless. 2. A reddish-brown / brown solid (copper) forms on the bottom.
(e) (i) Polystyrene is a good thermal insulator (poor conductor of heat), so it reduces thermal energy loss to the surroundings. (ii) Place a lid (or stopper with a hole for the thermometer) on the polystyrene cup. (iii) The temperature change would be $5.0\text{ }^\circ\text{C}$ (half the temperature change) because the same quantity of chemical energy/heat is released into twice the mass/volume of water.
Marking scheme
(a) [3 marks] - 5 to 7 temperature changes calculated correctly `[3]` - 3 to 4 temperature changes calculated correctly `[2]` - 1 to 2 temperature changes calculated correctly `[1]` (Values: 5.0, 10.0, 15.0, 20.0, 25.0, 25.0, 25.0)
(b) [5 marks] - M1: Suitable linear scale chosen for y-axis (temperature change) extending over at least half the grid area `[1]` - M2 & M3: Points plotted accurately to within half a small square (all 7 points plotted correctly = 2 marks, 5–6 points plotted correctly = 1 mark) `[2]` - M4: First best-fit straight line drawn with a ruler through rising points $(0.20\text{ to }1.00\text{ g})$ `[1]` - M5: Second horizontal best-fit straight line drawn with a ruler through plateau points $(1.00\text{ to }1.40\text{ g})$ showing clear intersection `[1]` (REJECT: Curving lines or freehand sketching)
(c) [3 marks] - (i) Minimum mass read correctly from intersection of candidate's lines ($1.00\text{ g}$ or $\pm 0.02\text{ g}$) `[1]` - (i) Clear construction / tie lines shown on graph `[1]` - (ii) Value of maximum temperature change read correctly from candidate's graph ($25.0\text{ }^\circ\text{C}$) `[1]`
(d) [4 marks] - (i) All aqueous copper(II) sulfate has reacted / copper(II) sulfate is the limiting reagent `[1]` - (i) Zinc is in excess / no more heat is released `[1]` - (ii) Blue solution turns colourless / fades / decolourises `[1]` (ALLOW: blue colour disappears) - (ii) Red-brown / brown / pink-brown solid forms `[1]` (ALLOW: zinc disappears / dissolves; REJECT: fizzing / bubbles)
(e) [3 marks] - (i) Polystyrene is a good thermal insulator / poor conductor of heat / reduces heat loss `[1]` (REJECT: prevents all heat loss) - (ii) Add a lid / cover / place cup in a beaker packed with cotton wool `[1]` - (iii) $5.0\text{ }^\circ\text{C}$ / halved AND because the same thermal energy is transferred to double the volume/mass of solution `[1]`
Question 3 · Qualitative Analysis Deductions
8 marks
Two substances, solid L and solution M, were analysed.
### Tests on solid L
Solid L was hydrated iron(II) sulfate.
(a) Solid L was placed in a dry boiling tube and heated gently. State the name of the liquid that condensed at the top of the boiling tube. [1]
(b) Dilute sulfuric acid was added to a sample of solid L until it dissolved completely to form solution L1. Solution L1 was divided into two portions.
(i) To the first portion of solution L1, aqueous sodium hydroxide was added dropwise until in excess. Describe the observations. * with a few drops: ......................................................................................................................... * in excess: ......................................................................................................................... [2]
(ii) To the second portion of solution L1, dilute nitric acid was added followed by aqueous barium nitrate. Describe the observation. [1]
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### Tests on solution M
Tests were carried out on solution M. The observations are recorded in Table 1.1.
Table 1.1
| Test | Observations | | :--- | :--- | | test 1 Flame test carried out on solution M. | lilac flame | | test 2 To a portion of solution M, dilute nitric acid was added, followed by aqueous silver nitrate. | yellow precipitate formed | | test 3 To another portion of solution M, aqueous sodium hydroxide and aluminium foil were added. The mixture was heated gently and the gas produced was tested with damp red litmus paper. | no effervescence; damp red litmus paper remained red |
(c) Identify the cation present in solution M. [1]
(d) Identify the anion present in solution M shown by test 2. [1]
(e) State what conclusion can be drawn about solution M from the result of test 3. [1]
(f) Name the precipitate formed in test 2. [1]
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Worked solution
(a) Hydrated iron(II) sulfate contains water of crystallisation. When heated gently, it releases water vapour which condenses as liquid water on the cooler upper parts of the tube.
(b)(i) Reaction of ext{Fe}^{2+} with aqueous sodium hydroxide: - With a few drops: a green precipitate of iron(II) hydroxide, \(\text{Fe(OH)}_2\), forms. - In excess: the green precipitate is insoluble.
(b)(ii) Test for sulfate ions (\(\text{SO}_4^{2-}\)): - Adding dilute nitric acid followed by aqueous barium nitrate forms a white precipitate of barium sulfate (\(\text{BaSO}_4\)).
(c) A lilac flame in a flame test is characteristic of potassium ions (\(\text{K}^+\)).
(d) Addition of dilute nitric acid followed by aqueous silver nitrate yielding a yellow precipitate confirms the presence of iodide ions (\(\text{I}^-\)).
(e) Heating with aqueous sodium hydroxide and aluminium foil is the test for nitrate ions (\(\text{NO}_3^-\)), which produces ammonia gas (turning damp red litmus blue). Since no ammonia was evolved and red litmus remained red, nitrate ions are absent.
(f) The yellow precipitate formed from aqueous silver ions and iodide ions is silver iodide (\(\text{AgI}\)).
Marking scheme
(a) - water / \(\text{H}_2\text{O}\) `[1]` (ALLOW steam / water vapour)
(b)(i) - green precipitate `[1]` - insoluble (in excess) / precipitate remains `[1]` (REJECT red-brown / grey-green / blue precipitate)
(b)(ii) - white precipitate `[1]` (ALLOW white ppt / white solid)
(e) - nitrate (ions) absent / no nitrate / does not contain nitrate `[1]` *(ALLOW \(\text{NO}_3^-\) absent)* (REJECT ammonium absent)
(f) - silver iodide / \(\text{AgI}\) `[1]`
Question 4 · open-ended
6 marks
A gardener has a sample of plant food which is a dry mixture of water-soluble potassium sulfate and insoluble crushed coconut husk fibres.
Plan an investigation to determine the percentage by mass of potassium sulfate in the plant food mixture.
You are provided with: - a sample of the plant food mixture - distilled water - standard laboratory apparatus.
Your plan should include: - the apparatus needed - the method to be followed, including all measurements to be taken - how the measurements will be used to calculate the percentage by mass of potassium sulfate.
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Worked solution
### Experimental Method 1. Mass measurement: Use a digital balance to weigh a sample of the plant food mixture. Record this initial mass, \(m_1\). 2. Dissolving: Place the weighed sample into a beaker, add an excess of distilled water, and stir with a glass rod (warming gently if necessary) until all soluble potassium sulfate has completely dissolved. 3. Filtration: Set up a filter funnel lined with filter paper over a conical flask or beaker. Pour the mixture through the funnel to separate the insoluble coconut husk fibres (residue) from the aqueous potassium sulfate (filtrate). 4. Washing and drying: Rinse the residue remaining on the filter paper with a small volume of distilled water to wash through any residual potassium sulfate solution into the filtrate. Leave the residue to dry in an oven or warm place until constant mass is achieved. 5. Final weighing: Weigh the dry coconut husk fibres using the balance. Record this mass as \(m_2\).
(Alternatively: The filtrate could be collected in a pre-weighed evaporating basin, heated to evaporate the water to dryness, and the mass of dry potassium sulfate weighed directly.)
### Calculations - Mass of potassium sulfate = \(m_1 - m_2\) - Percentage by mass of potassium sulfate = \(\frac{m_1 - m_2}{m_1} \times 100\%\)
Marking scheme
Award 1 mark for each of the following marking points (up to a maximum of 6 marks):
- MP1 (Weighing initial sample): Weigh a sample of the plant food mixture using a balance / balance specified. [1] - MP2 (Dissolving): Add (distilled) water to the sample in a suitable container (e.g. beaker) and stir / heat / mix to dissolve the potassium sulfate. [1] - MP3 (Filtration): Filter the mixture using a filter funnel and filter paper (to separate the insoluble residue from the filtrate). [1] - MP4 (Washing and drying residue): Wash the residue with (distilled) water AND dry the residue (e.g. in an oven / warm place / between filter papers / to constant mass). [1] (ALLOW: Evaporate the filtrate to dryness in an evaporating dish / basin / crucible) - MP5 (Weighing dry product): Weigh the dry residue / fibres (or weigh the dry potassium sulfate after evaporating filtrate to dryness). [1] - MP6 (Processing results): Clearly show how to calculate the mass of potassium sulfate (initial mass - dry residue mass OR mass of dish + salt - mass of empty dish) AND percentage by mass: $$\text{percentage} = \frac{\text{mass of potassium sulfate}}{\text{initial mass of mixture}} \times 100\%$$ [1]
Guidance notes: - `IGNORE` vague references to 'measure' without specifying a balance for mass. - `REJECT` methods that attempt to evaporate the wet residue without filtration.
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