Cambridge IGCSE · thinka-original Practice Paper

2025 Cambridge IGCSE Chemistry (0620) Practice Paper with Answers

Thinka Nov 2025 (V2) Cambridge IGCSE-Style Mock — Chemistry (0620)

160 marks180 mins2025
An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V2) Cambridge IGCSE Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.

Paper 2 - Multiple Choice (Extended)

Answer all forty multiple-choice questions. For each question, choose the single correct option from A, B, C or D.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
A pure substance is cooled from a gas to a solid. At which stages does the temperature of the substance remain constant as thermal energy is released to the surroundings?
  1. A.during condensation only
  2. B.during freezing only
  3. C.during condensation and freezing
  4. D.during the cooling of the liquid phase only
Show answer & marking scheme

Worked solution

During phase changes of a pure substance (such as condensation from gas to liquid, and freezing from liquid to solid), the temperature remains constant as forces of attraction are formed between particles, releasing energy.

Marking scheme

Award 1 mark for the correct answer C.
Question 2 · multiple-choice
1 marks
When \( 0.20\text{ mol} \) of a hydrocarbon is completely combusted in an excess of oxygen, \( 17.6\text{ g} \) of carbon dioxide and \( 10.8\text{ g} \) of water are produced.

What is the formula of the hydrocarbon?
  1. A.\( \text{CH}_4 \)
  2. B.\( \text{C}_2\text{H}_4 \)
  3. C.\( \text{C}_2\text{H}_6 \)
  4. D.\( \text{C}_3\text{H}_8 \)
Show answer & marking scheme

Worked solution

First, calculate the number of moles of \( \text{CO}_2 \) and \( \text{H}_2\text{O} \) produced:
- Moles of \( \text{CO}_2 = \frac{17.6\text{ g}}{44\text{ g/mol}} = 0.40\text{ mol} \)
- Moles of \( \text{H}_2\text{O} = \frac{10.8\text{ g}}{18\text{ g/mol}} = 0.60\text{ mol} \)

Next, find the mole ratio of the elements in the hydrocarbon relative to \( 0.20\text{ mol} \) of the hydrocarbon:
- Carbon atoms: \( \frac{0.40\text{ mol}}{0.20\text{ mol}} = 2 \) carbon atoms per molecule.
- Hydrogen atoms: Since each water molecule contains two hydrogen atoms, moles of \( \text{H} = 2 \times 0.60\text{ mol} = 1.20\text{ mol} \). Thus, \( \frac{1.20\text{ mol}}{0.20\text{ mol}} = 6 \) hydrogen atoms per molecule.

Therefore, the formula is \( \text{C}_2\text{H}_6 \).

Marking scheme

1 mark for correct calculation of moles and identifying option C.
Question 3 · multiple-choice
1 marks
An aqueous solution of a metal sulfate, \( \text{MSO}_4 \), is electrolysed using inert electrodes.

During the electrolysis, a pink-brown metal is deposited at the cathode, and a colourless gas which relights a glowing splint is evolved at the anode.

Which row identifies the metal \( \text{M} \) and the gas evolved?
  1. A.metal M: copper; gas: hydrogen
  2. B.metal M: copper; gas: oxygen
  3. C.metal M: iron; gas: oxygen
  4. D.metal M: zinc; gas: hydrogen
Show answer & marking scheme

Worked solution

The pink-brown solid deposited at the cathode indicates copper metal is formed (\( \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \)). The colourless gas that relights a glowing splint is oxygen (\( \text{O}_2 \)), which is formed at the anode by the oxidation of hydroxide ions from water: \( 4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \).

Marking scheme

1 mark for identifying M as copper and the gas as oxygen (Option B).
Question 4 · multiple-choice
1 marks
The equations for four chemical reactions are shown. In which equation is the bold reactant acting as an oxidising agent?
  1. A.**\( \text{CO}\text{(g)} \)** \( + \text{Fe}_2\text{O}_3\text{(s)} \rightarrow 2\text{FeO}\text{(s)} + \text{CO}_2\text{(g)} \)
  2. B.**\( \text{Mg}\text{(s)} \)** \( + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Mg}^{2+}\text{(aq)} + \text{Cu}\text{(s)} \)
  3. C.**\( \text{Cl}_2\text{(g)} \)** \( + 2\text{Br}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{Br}_2\text{(aq)} \)
  4. D.**\( \text{H}_2\text{(g)} \)** \( + \text{CuO}\text{(s)} \rightarrow \text{Cu}\text{(s)} + \text{H}_2\text{O}\text{(g)} \)
Show answer & marking scheme

Worked solution

An oxidising agent oxidises another substance and is itself reduced (gains electrons or decreases in oxidation number).

In option C, chlorine (\( \text{Cl}_2 \)) has an oxidation state of 0, which decreases to -1 in chloride (\( \text{Cl}^- \)). Since chlorine is reduced, it acts as the oxidising agent. In all other options, the bold reactant is oxidised (its oxidation state increases), so it acts as a reducing agent.

Marking scheme

1 mark for selecting C.
Question 5 · multiple-choice
1 marks
Which method is most suitable for preparing a pure, dry sample of the insoluble salt lead(II) sulfate?
  1. A.Add lead metal to dilute sulfuric acid, filter off the excess lead, and evaporate the filtrate to dryness.
  2. B.Mix aqueous lead(II) nitrate and dilute sulfuric acid, filter the mixture, wash the residue with distilled water, and dry it.
  3. C.Add lead(II) oxide to dilute sulfuric acid, filter off the excess oxide, and crystallise the filtrate.
  4. D.Heat lead(II) carbonate with dilute sulfuric acid until no more carbon dioxide is evolved, then evaporate the solution.
Show answer & marking scheme

Worked solution

Lead(II) sulfate is an insoluble salt. Insoluble salts are prepared by precipitation, which involves mixing two soluble reactants (aqueous lead(II) nitrate and dilute sulfuric acid). The precipitate is filtered to obtain the residue of lead(II) sulfate, washed with distilled water to remove soluble impurities, and dried. Methods involving insoluble solid reactants (like lead metal, lead(II) oxide, or lead(II) carbonate) with sulfuric acid will fail because the insoluble lead(II) sulfate forms a coating on the solid, preventing further reaction.

Marking scheme

1 mark for selecting B.
Question 6 · multiple-choice
1 marks
Purified aluminium oxide is electrolysed to extract aluminium metal.

Which statement about this industrial process is correct?
  1. A.Molten cryolite is added to chemically reduce the aluminium ions to aluminium metal.
  2. B.The anode is made of carbon and must be replaced regularly because it reacts with oxygen to form carbon dioxide.
  3. C.Aluminium ions are oxidised to aluminium metal at the negative electrode.
  4. D.Oxygen gas is released at the cathode.
Show answer & marking scheme

Worked solution

During the electrolysis of aluminium oxide dissolved in molten cryolite, aluminium ions are reduced to aluminium metal at the cathode (negative electrode). At the anode (positive electrode), oxide ions are oxidised to oxygen gas. The carbon (graphite) anodes react with this oxygen at high temperatures to produce carbon dioxide gas, meaning they gradually burn away and must be replaced regularly. Cryolite is used to lower the melting point of the electrolyte, not to reduce the aluminium ions.

Marking scheme

1 mark for selecting B.
Question 7 · multiple-choice
1 marks
The structure of an ester is shown below.

\( \text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_2\text{CH}_3 \)

Which row gives the correct name of the ester and the reactants used to prepare it?
  1. A.name: propyl propanoate | alcohol reactant: propan-1-ol | carboxylic acid reactant: propanoic acid
  2. B.name: propyl propanoate | alcohol reactant: propanoic acid | carboxylic acid reactant: propan-1-ol
  3. C.name: ethyl propanoate | alcohol reactant: ethanol | carboxylic acid reactant: propanoic acid
  4. D.name: propyl ethanoate | alcohol reactant: propan-1-ol | carboxylic acid reactant: ethanoic acid
Show answer & marking scheme

Worked solution

An ester is named as 'alkyl alkanoate'. The 'alkyl' part comes from the alcohol, and the 'alkanoate' part comes from the carboxylic acid.
- The alkyl group attached to the oxygen atom is \( -\text{CH}_2\text{CH}_2\text{CH}_3 \) (3 carbons), which comes from the alcohol propan-1-ol, giving the first part of the name: 'propyl'.
- The alkanoate group contains the carbonyl group, \( \text{CH}_3\text{CH}_2\text{COO}- \) (3 carbons in total, including the carbonyl carbon), which comes from propanoic acid, giving the second part of the name: 'propanoate'.
Therefore, the ester is propyl propanoate, made from propan-1-ol and propanoic acid.

Marking scheme

1 mark for selecting A.
Question 8 · multiple-choice
1 marks
Catalytic converters are fitted to car exhaust systems to reduce the emission of toxic gases into the atmosphere.

Which equation represents a reaction that occurs in a catalytic converter to convert toxic gases into non-toxic gases?
  1. A.\( 2\text{NO}\text{(g)} + 2\text{CO}\text{(g)} \rightarrow \text{N}_2\text{(g)} + 2\text{CO}_2\text{(g)} \)
  2. B.\( 2\text{CO}\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} \)
  3. C.\( \text{N}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{NO}\text{(g)} \)
  4. D.\( \text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + \text{H}_2\text{O}\text{(g)} \)
Show answer & marking scheme

Worked solution

In a catalytic converter, nitrogen monoxide (\( \text{NO} \)) and carbon monoxide (\( \text{CO} \)), which are both toxic gases, react on the surface of the catalyst to form nitrogen (\( \text{N}_2 \)) and carbon dioxide (\( \text{CO}_2 \)), which are non-toxic. The balanced chemical equation for this reaction is \( 2\text{NO} + 2\text{CO} \rightarrow \text{N}_2 + 2\text{CO}_2 \).

Marking scheme

1 mark for selecting A.
Question 9 · multiple-choice
1 marks
Two gas syringes are filled with equal volumes of different gases at room temperature and pressure. Syringe 1 contains carbon monoxide, CO. Syringe 2 contains sulfur dioxide, SO2. The gases are allowed to diffuse through identical small apertures. Which statement about the rate of diffusion of the two gases is correct?
  1. A.CO diffuses faster than SO2 because it has a lower relative molecular mass.
  2. B.CO diffuses slower than SO2 because it has a lower relative molecular mass.
  3. C.SO2 diffuses faster than CO because it has a higher relative molecular mass.
  4. D.Both gases diffuse at the same rate because they are at the same temperature.
Show answer & marking scheme

Worked solution

The rate of diffusion of a gas is inversely proportional to the square root of its relative molecular mass. Carbon monoxide has a relative molecular mass of 28 (12 + 16), which is lower than that of sulfur dioxide, which is 64 (32 + 2 * 16). Therefore, carbon monoxide molecules move faster and diffuse more rapidly than sulfur dioxide molecules at the same temperature.

Marking scheme

Award 1 mark for selecting option A, which correctly identifies that CO diffuses faster due to its lower relative molecular mass.
Question 10 · multiple-choice
1 marks
The equation for the reaction between hydrogen sulfide gas and chlorine gas is shown. H2S(g) + Cl2(g) -> S(s) + 2HCl(g). Which statement correctly describes the redox process?
  1. A.Chlorine is oxidised because it loses hydrogen.
  2. B.Chlorine is reduced because it gains hydrogen.
  3. C.Hydrogen sulfide is reduced because it gains chlorine.
  4. D.Hydrogen sulfide is oxidised because it gains oxygen.
Show answer & marking scheme

Worked solution

In this reaction, chlorine gas (Cl2) gains hydrogen to form hydrogen chloride (HCl). In terms of hydrogen transfer, reduction is the gain of hydrogen. Therefore, chlorine is reduced. Alternatively, the oxidation state of chlorine decreases from 0 in Cl2 to -1 in HCl, confirming it is reduced.

Marking scheme

Award 1 mark for selecting option B, recognizing that chlorine gains hydrogen and is therefore reduced.
Question 11 · multiple-choice
1 marks
Iron is extracted from its ore, hematite, in a blast furnace. Calcium carbonate is added to the furnace to remove acidic impurities. Which row correctly describes the role of calcium carbonate?
  1. A.decomposes thermally to form an acidic oxide which reacts with basic impurities
  2. B.decomposes thermally to form a basic oxide which reacts with acidic impurities
  3. C.reacts directly with iron(III) oxide to reduce it to iron metal
  4. D.reacts with carbon monoxide to produce carbon dioxide
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Worked solution

In the blast furnace, calcium carbonate undergoes thermal decomposition to form calcium oxide (a basic metal oxide) and carbon dioxide. The basic calcium oxide then reacts with the main acidic impurity, silicon(IV) oxide (silica), to form molten slag (calcium silicate) in a neutralization reaction.

Marking scheme

Award 1 mark for selecting option B, which correctly states that calcium carbonate decomposes to a basic oxide which reacts with acidic impurities.
Question 12 · multiple-choice
1 marks
Ethane reacts with chlorine in a substitution reaction. What are the correct conditions and the formula of an organic product formed in this reaction?
  1. A.darkness, C2H4Cl2
  2. B.darkness, C2H5Cl
  3. C.ultraviolet light, C2H5Cl
  4. D.ultraviolet light, C2H4
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Worked solution

Alkanes such as ethane are generally unreactive but react with halogens like chlorine in substitution reactions. This reaction requires ultraviolet (UV) light to provide the activation energy to break the chlorine-chlorine bond. A hydrogen atom in ethane is replaced by a chlorine atom to form chloroethane, C2H5Cl.

Marking scheme

Award 1 mark for selecting option C, identifying ultraviolet light as the correct condition and C2H5Cl as the correct substitution product.
Question 13 · multiple-choice
1 marks
A concentrated aqueous solution of sodium chloride is electrolysed using inert platinum electrodes. Which row correctly identifies the products formed at each electrode?
  1. A.cathode: sodium, anode: chlorine
  2. B.cathode: hydrogen, anode: chlorine
  3. C.cathode: hydrogen, anode: oxygen
  4. D.cathode: oxygen, anode: hydrogen
Show answer & marking scheme

Worked solution

During the electrolysis of concentrated aqueous sodium chloride, H+ and Na+ ions migrate to the cathode. Since hydrogen is lower in the reactivity series than sodium, H+ ions are preferentially discharged, forming hydrogen gas. Cl- and OH- ions migrate to the anode. Because the solution is concentrated, Cl- ions are preferentially discharged over OH- ions, forming chlorine gas.

Marking scheme

Award 1 mark for selecting option B, which correctly identifies hydrogen at the cathode and chlorine at the anode.
Question 14 · multiple-choice
1 marks
Gaseous pollutants from car engines are passed through a catalytic converter. Which reaction takes place in the catalytic converter to remove these pollutants?
  1. A.2CO + 2NO -> 2CO2 + N2
  2. B.CH4 + 2O2 -> CO2 + 2H2O
  3. C.2NO2 -> N2 + 2O2
  4. D.CO + H2O -> CO2 + H2
Show answer & marking scheme

Worked solution

A catalytic converter facilitates the reaction between carbon monoxide (CO) and nitrogen monoxide (NO), both of which are toxic gases. They react to form non-toxic carbon dioxide (CO2) and nitrogen (N2) gases: 2CO + 2NO -> 2CO2 + N2.

Marking scheme

Award 1 mark for selecting option A, which shows the balanced equation representing the reduction of nitrogen oxides and oxidation of carbon monoxide in a catalytic converter.
Question 15 · multiple-choice
1 marks
Which statement about the chemical properties of aqueous ethanoic acid is correct?
  1. A.It reacts with copper metal to produce hydrogen gas.
  2. B.It turns methyl orange indicator yellow.
  3. C.It reacts with sodium carbonate to produce carbon dioxide gas.
  4. D.It decomposes on heating to form ethene and water.
Show answer & marking scheme

Worked solution

Ethanoic acid is a weak acid. Acids react with metal carbonates, such as sodium carbonate, to produce a salt (sodium ethanoate), water, and carbon dioxide gas. Ethanoic acid does not react with unreactive metals like copper, and it turns methyl orange red (not yellow, which is the colour of methyl orange in alkaline solutions).

Marking scheme

Award 1 mark for selecting option C, identifying that ethanoic acid reacts with carbonates to produce carbon dioxide.
Question 16 · multiple-choice
1 marks
An excess of dilute hydrochloric acid is added to 4.8 g of magnesium ribbon. Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g). What volume of hydrogen gas, measured at room temperature and pressure, is produced? (The relative atomic mass of Mg is 24. One mole of any gas occupies 24 dm3 at r.t.p.)
  1. A.2.4 dm3
  2. B.4.8 dm3
  3. C.12.0 dm3
  4. D.24.0 dm3
Show answer & marking scheme

Worked solution

First, calculate the number of moles of magnesium: moles = mass / Ar = 4.8 g / 24 g/mol = 0.2 mol. From the stoichiometry of the balanced equation, 1 mole of Mg produces 1 mole of H2. Therefore, 0.2 mol of Mg produces 0.2 mol of H2. Finally, calculate the volume of gas: volume = moles * molar volume = 0.2 mol * 24 dm3/mol = 4.8 dm3.

Marking scheme

Award 1 mark for selecting option B, which is the correct calculated volume of hydrogen gas (4.8 dm3).
Question 17 · multiple-choice
1 marks
During a paper chromatography experiment to separate a mixture of amino acids, the solvent front travels 8.0 cm from the baseline. A spot of amino acid X travels 3.2 cm from the baseline. Why is a locating agent sprayed onto the chromatogram, and what is the Rf value of amino acid X?
  1. A.The locating agent makes the colourless spots visible; Rf = 0.40
  2. B.The locating agent makes the colourless spots visible; Rf = 2.50
  3. C.The locating agent prevents the spots from dissolving; Rf = 0.40
  4. D.The locating agent prevents the spots from dissolving; Rf = 2.503
Show answer & marking scheme

Worked solution

The Rf value is calculated by dividing the distance travelled by the substance by the distance travelled by the solvent front: Rf = 3.2 cm / 8.0 cm = 0.40. Amino acids are colourless substances, so a locating agent must be sprayed on the chromatogram to react with them and make the spots visible.

Marking scheme

1 mark for selecting option A, which correctly calculates Rf as 0.40 and identifies that the locating agent is used because amino acids are colourless.
Question 18 · multiple-choice
1 marks
The ionic equation for a redox reaction is shown: 2Fe3+(aq) + 2I-(aq) -> 2Fe2+(aq) + I2(aq). Which statement correctly describes this reaction in terms of electron transfer?
  1. A.Fe3+ ions act as the reducing agent because they gain electrons.
  2. B.Fe3+ ions act as the oxidising agent because they lose electrons.
  3. C.I- ions act as the reducing agent because they lose electrons.
  4. D.I- ions act as the oxidising agent because they gain electrons.
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Worked solution

In this reaction, iodide ions (I-) lose electrons to form iodine (I2). Since loss of electrons is oxidation, iodide ions are oxidised. Therefore, I- acts as the reducing agent. Iron(III) ions (Fe3+) gain electrons to form iron(II) ions (Fe2+), so Fe3+ is reduced and acts as the oxidising agent.

Marking scheme

1 mark for selecting option C, which correctly identifies that iodide ions act as the reducing agent by losing electrons.
Question 19 · multiple-choice
1 marks
What is the maximum volume of carbon dioxide gas, measured at r.t.p., produced when 10.0 g of calcium carbonate, CaCO3, reacts completely with an excess of dilute hydrochloric acid? (Mr of CaCO3 = 100; the volume of one mole of any gas is 24 dm3 at r.t.p.)
  1. A.1.2 dm3
  2. B.2.4 dm3
  3. C.12.0 dm3
  4. D.24.0 dm3
Show answer & marking scheme

Worked solution

First, calculate the moles of CaCO3 used: moles = mass / Mr = 10.0 g / 100 g/mol = 0.10 mol. From the equation: CaCO3 + 2HCl -> CaCl2 + H2O + CO2, the mole ratio of CaCO3 to CO2 is 1:1. Therefore, 0.10 mol of CO2 is produced. Finally, calculate the volume of CO2 gas: volume = moles * 24 dm3/mol = 0.10 mol * 24 dm3/mol = 2.4 dm3.

Marking scheme

1 mark for selecting option B, which represents the correct calculation of 2.4 dm3.
Question 20 · multiple-choice
1 marks
Three metals, W, X and Y, are tested. Only W reacts with cold water. Both W and X react with dilute hydrochloric acid, but Y does not. Heating the oxide of Y with carbon produces metal Y, but heating the oxide of X with carbon does not produce metal X. What is the order of reactivity of the three metals, starting with the least reactive?
  1. A.Y -> X -> W
  2. B.W -> X -> Y
  3. C.Y -> W -> X
  4. D.X -> Y -> W
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Worked solution

W is the most reactive because it reacts with cold water. Y is the least reactive because it does not react with dilute acid, and its oxide can be reduced by carbon (meaning Y is less reactive than carbon). X is intermediate because it reacts with acid but its oxide cannot be reduced by carbon (meaning X is more reactive than carbon). Therefore, the order of reactivity starting from the least reactive is Y -> X -> W.

Marking scheme

1 mark for selecting option A, which correctly orders the metals by reactivity from least to most reactive.
Question 21 · multiple-choice
1 marks
Concentrated aqueous sodium chloride is electrolysed using inert electrodes. Which row correctly identifies the products formed at each electrode and the change in pH of the remaining solution?
  1. A.Anode product: chlorine; Cathode product: hydrogen; pH of solution: decreases
  2. B.Anode product: chlorine; Cathode product: hydrogen; pH of solution: increases
  3. C.Anode product: oxygen; Cathode product: sodium; pH of solution: stays the same
  4. D.Anode product: oxygen; Cathode product: hydrogen; pH of solution: increases
Show answer & marking scheme

Worked solution

During the electrolysis of concentrated aqueous NaCl, chloride ions (Cl-) are discharged at the anode to form chlorine gas. Hydrogen ions (H+) from water are discharged at the cathode to form hydrogen gas. The remaining ions in solution are Na+ and OH-, forming sodium hydroxide, which is alkaline. Therefore, the pH of the remaining solution increases.

Marking scheme

1 mark for selecting option B, which correctly states that chlorine forms at the anode, hydrogen forms at the cathode, and the pH increases (becomes alkaline).
Question 22 · multiple-choice
1 marks
Methane reacts with chlorine in a substitution reaction. What is the essential condition for this reaction to occur, and what is a gaseous inorganic product of this reaction?
  1. A.Condition: high pressure; Product: hydrogen
  2. B.Condition: phosphoric acid catalyst; Product: chloromethane
  3. C.Condition: high temperature only; Product: water
  4. D.Condition: ultraviolet light; Product: hydrogen chloride
Show answer & marking scheme

Worked solution

The substitution reaction between an alkane (such as methane) and a halogen (such as chlorine) requires ultraviolet (UV) light to provide the activation energy to split the chlorine molecules into radicals. The reaction produces chloromethane (an organic gas) and hydrogen chloride (an inorganic gas).

Marking scheme

1 mark for selecting option D, which correctly identifies ultraviolet light as the condition and hydrogen chloride as the gaseous inorganic product.
Question 23 · multiple-choice
1 marks
A student wants to prepare a pure, dry sample of the insoluble salt lead(II) sulfate, PbSO4. Which pair of aqueous solutions should the student mix together to achieve this?
  1. A.lead(II) nitrate and sodium sulfate
  2. B.lead(II) carbonate and dilute sulfuric acid
  3. C.lead(II) oxide and dilute sulfuric acid
  4. D.lead(II) chloride and copper(II) sulfate
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Worked solution

To prepare an insoluble salt, a precipitation reaction is used. This requires mixing two soluble salt solutions. Lead(II) nitrate is a soluble nitrate salt, and sodium sulfate is a soluble sodium salt. Mixing them forms a precipitate of insoluble lead(II) sulfate. The other options involve insoluble reactants (lead(II) carbonate, lead(II) oxide, or lead(II) chloride) which are not suitable for a straightforward precipitation preparation.

Marking scheme

1 mark for selecting option A, which specifies the correct pair of soluble solutions required for precipitation.
Question 24 · multiple-choice
1 marks
Which statement about the chemical properties of ethanoic acid is correct?
  1. A.It is a strong acid that completely dissociates in aqueous solution.
  2. B.It reacts with sodium carbonate to produce carbon dioxide gas.
  3. C.It is oxidized by acidified potassium manganate(VII) to form ethanol.
  4. D.It turns thymolphthalein indicator blue.
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Worked solution

Ethanoic acid is a weak carboxylic acid. It reacts with metal carbonates, such as sodium carbonate, to produce a salt (sodium ethanoate), water, and carbon dioxide gas. It is a weak acid, so it only partially dissociates in water. It turns thymolphthalein indicator colourless, not blue (blue indicates an alkali).

Marking scheme

1 mark for selecting option B, which correctly states that ethanoic acid reacts with sodium carbonate to produce carbon dioxide.
Question 25 · multiple-choice
1 marks
What is the oxidation number of manganese in potassium manganate(VI), \( \text{K}_2\text{MnO}_4 \), and potassium manganate(VII), \( \text{KMnO}_4 \)?
  1. A.+4 and +7
  2. B.+6 and +7
  3. C.+6 and +5
  4. D.+2 and +7
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Worked solution

In potassium manganate(VI), \( \text{K}_2\text{MnO}_4 \):
\( 2(+1) + x + 4(-2) = 0 \)
\( +2 + x - 8 = 0 \)
\( x = +6 \)

In potassium manganate(VII), \( \text{KMnO}_4 \):
\( 1(+1) + x + 4(-2) = 0 \)
\( +1 + x - 8 = 0 \)
\( x = +7 \)

Therefore, the oxidation numbers are +6 and +7 respectively.

Marking scheme

1 mark for the correct option B.
Question 26 · multiple-choice
1 marks
Which statement about the structure and properties of silicon(IV) oxide, \( \text{SiO}_2 \), is correct?
  1. A.It has a simple molecular structure with weak intermolecular forces.
  2. B.Each silicon atom is covalently bonded to two oxygen atoms.
  3. C.It has a giant structure with strong covalent bonds, giving it a high melting point.
  4. D.It conducts electricity when molten because it has free-moving ions.
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Worked solution

Silicon(IV) oxide has a giant macromolecular structure with strong covalent bonds between silicon and oxygen atoms, which requires a large amount of energy to break, resulting in a high melting point. It does not conduct electricity as it lacks free ions or delocalised electrons.

Marking scheme

1 mark for the correct option C.
Question 27 · multiple-choice
1 marks
An increase in temperature increases the rate of a chemical reaction. Which statement explains this effect?
  1. A.The activation energy of the reaction is lowered.
  2. B.The reactant particles are closer together, increasing collision frequency.
  3. C.A higher proportion of colliding particles have energy equal to or greater than the activation energy.
  4. D.The average distance between the reactant particles increases, leading to more energetic collisions.
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Worked solution

Increasing the temperature increases the kinetic energy of the particles, meaning they move faster and collide more frequently. More importantly, a much higher proportion of the colliding particles now possess kinetic energy equal to or greater than the activation energy, leading to a significant increase in the frequency of successful collisions.

Marking scheme

1 mark for the correct option C.
Question 28 · multiple-choice
1 marks
The equation represents a reversible reaction.

\( \text{X(g)} + 2\text{Y(g)} \rightleftharpoons 2\text{Z(g)} \) (forward reaction is exothermic)

Which conditions of temperature and pressure will produce the highest yield of \( \text{Z(g)} \) at equilibrium?
  1. A.high temperature and high pressure
  2. B.high temperature and low pressure
  3. C.low temperature and high pressure
  4. D.low temperature and low pressure
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Worked solution

1. Pressure: There are 3 moles of gas on the left and 2 moles of gas on the right. Increasing the pressure shifts the equilibrium to the side with fewer gas moles (the right), increasing the yield of Z.
2. Temperature: The forward reaction is exothermic. Decreasing the temperature shifts the equilibrium in the exothermic direction (to the right) to release heat, increasing the yield of Z.

Thus, low temperature and high pressure yield the most Z.

Marking scheme

1 mark for the correct option C.
Question 29 · multiple-choice
1 marks
Concentrated aqueous sodium chloride (brine) is electrolysed using inert electrodes. Which row correctly identifies the products formed at the electrodes and the nature of the remaining solution?
  1. A.Cathode: hydrogen; Anode: chlorine; Solution: becomes acidic.
  2. B.Cathode: sodium; Anode: chlorine; Solution: remains neutral.
  3. C.Cathode: hydrogen; Anode: chlorine; Solution: becomes alkaline.
  4. D.Cathode: oxygen; Anode: chlorine; Solution: becomes alkaline.
Show answer & marking scheme

Worked solution

During the electrolysis of concentrated aqueous sodium chloride:
- At the cathode, hydrogen ions (\( \text{H}^+ \)) are discharged in preference to sodium ions (\( \text{Na}^+ \)) to form hydrogen gas.
- At the anode, chloride ions (\( \text{Cl}^- \)) are discharged in preference to hydroxide ions (\( \text{OH}^- \)) because they are in high concentration, forming chlorine gas.
- The remaining ions in solution are sodium (\( \text{Na}^+ \)) and hydroxide (\( \text{OH}^- \)), which form sodium hydroxide, making the solution alkaline.

Marking scheme

1 mark for the correct option C.
Question 30 · multiple-choice
1 marks
An atom of element X is represented as \( ^{37}_{17}\text{X} \). Which statement about a neutral isotope of X is correct?
  1. A.It has 17 protons, 17 electrons, and a nucleon number of 35.
  2. B.It has 18 protons and a nucleon number of 37.
  3. C.It has different chemical properties because it has a different mass.
  4. D.It has 17 protons, 18 electrons, and a nucleon number of 37.
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Worked solution

An isotope of element X must have the same proton number (atomic number) of 17, and therefore the same number of electrons (17) in a neutral atom. However, it must have a different number of neutrons, which changes its nucleon (mass) number (e.g., to 35). Thus, option A is correct.

Marking scheme

1 mark for the correct option A.
Question 31 · multiple-choice
1 marks
An ester has the structural formula \( \text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3 \). What is the name of this ester and which carboxylic acid is used to prepare it?
  1. A.ethyl propanoate, propanoic acid
  2. B.propyl ethanoate, ethanoic acid
  3. C.propyl ethanoate, propanol
  4. D.ethyl propanoate, ethanoic acid
Show answer & marking scheme

Worked solution

The ester part derived from the alcohol has 3 carbons (propyl group: \( -\text{CH}_2\text{CH}_2\text{CH}_3 \)), and the part derived from the carboxylic acid has 2 carbons (acetate/ethanoate group: \( \text{CH}_3\text{COO}- \)). Therefore, the ester is propyl ethanoate and the carboxylic acid used to prepare it is ethanoic acid.

Marking scheme

1 mark for the correct option B.
Question 32 · multiple-choice
1 marks
What volume of carbon dioxide, measured at room temperature and pressure (r.t.p.), is produced when \( 10.0\text{ g} \) of calcium carbonate, \( \text{CaCO}_3 \), reacts completely with excess dilute hydrochloric acid?

[Relative formula mass: \( \text{CaCO}_3 = 100 \); the volume of one mole of gas at r.t.p. is \( 24\text{ dm}^3 \)]
  1. A.0.24 dm³
  2. B.2.4 dm³
  3. C.24.0 dm³
  4. D.4.8 dm³
Show answer & marking scheme

Worked solution

The equation for the reaction is:
\( \text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)} \)

1. Calculate the moles of \( \text{CaCO}_3 \):
\( \text{Moles} = \frac{10.0\text{ g}}{100\text{ g/mol}} = 0.1\text{ mol} \)

2. From the stoichiometry of the equation, 1 mole of \( \text{CaCO}_3 \) produces 1 mole of \( \text{CO}_2 \). So, \( 0.1\text{ mol} \) of \( \text{CO}_2 \) is produced.

3. Calculate the volume of \( \text{CO}_2 \) gas:
\( \text{Volume} = 0.1\text{ mol} \times 24\text{ dm}^3\text{/mol} = 2.4\text{ dm}^3 \)

Marking scheme

1 mark for the correct option B.
Question 33 · multiple-choice
1 marks
A sample of pure naphthalene solid is heated steadily from \(20\ ^\circ\text{C}\) until it melts at \(80\ ^\circ\text{C}\) and eventually boils at \(218\ ^\circ\text{C}\).

Which statement describes the naphthalene at \(80\ ^\circ\text{C}\) while it is melting?
  1. A.The temperature increases while the thermal energy is used to break intermolecular forces.
  2. B.The temperature remains constant and the particles gain kinetic energy.
  3. C.The temperature remains constant and the particles change from a regular arrangement to a random arrangement.
  4. D.The temperature increases and the particles move further apart.
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Worked solution

During melting, the temperature remains constant at the melting point (\(80\ ^\circ\text{C}\)) because the thermal energy absorbed is used to overcome the intermolecular forces holding the particles in a regular lattice. As a result, the arrangement of naphthalene particles changes from a regular solid lattice to a random liquid structure, while their average kinetic energy (and thus temperature) remains unchanged.

Marking scheme

Award 1 mark for the correct option C.
Question 34 · multiple-choice
1 marks
Under the same conditions of temperature and pressure, which gas diffuses the fastest?
  1. A.carbon dioxide, \(\text{CO}_2\)
  2. B.sulfur dioxide, \(\text{SO}_2\)
  3. C.nitrogen, \(\text{N}_2\)
  4. D.argon, \(\text{Ar}\)
Show answer & marking scheme

Worked solution

The rate of diffusion of a gas is inversely proportional to the square root of its relative molecular mass (\(M_{\text{r}}\)). Therefore, the gas with the lowest \(M_{\text{r}}\) will diffuse the fastest.

Let us calculate the \(M_{\text{r}}\) for each gas:
- Carbon dioxide (\(\text{CO}_2\)): \(12 + (2 \times 16) = 44\)
- Sulfur dioxide (\(\text{SO}_2\)): \(32 + (2 \times 16) = 64\)
- Nitrogen (\(\text{N}_2\)): \(2 \times 14 = 28\)
- Argon (\(\text{Ar}\)): \(40\)

Nitrogen has the lowest relative molecular mass (28) and therefore diffuses the fastest.

Marking scheme

Award 1 mark for the correct option C.
Question 35 · multiple-choice
1 marks
An ion of element \(X\) contains 16 protons, 18 neutrons and 18 electrons.

What is the correct symbol for this ion?
  1. A.\(^{34}_{16}\text{S}^{2+}\)
  2. B.\(^{34}_{16}\text{S}^{2-}\)
  3. C.\(^{18}_{16}\text{Ar}^{2-}\)
  4. D.\(^{34}_{18}\text{Ar}^{2+}\)
Show answer & marking scheme

Worked solution

1. The number of protons determines the identity of the element. With 16 protons, the atomic number is 16, which corresponds to sulfur (\(\text{S}\)).
2. The nucleon number (mass number) is the sum of protons and neutrons: \(16 + 18 = 34\). This goes at the top-left of the elemental symbol: \(^{34}\text{S}\).
3. The charge of the ion is determined by the difference between the number of protons and electrons: \(16\text{ protons} - 18\text{ electrons} = -2\), which is written as \(2-\).

Combining these, we get \(^{34}_{16}\text{S}^{2-}\).

Marking scheme

Award 1 mark for the correct option B.
Question 36 · multiple-choice
1 marks
An oxide of sulfur contains 40.0% sulfur and 60.0% oxygen by mass.

What is the empirical formula of this oxide?

[Relative atomic masses, \(A_{\text{r}}\): \(\text{S} = 32\), \(\text{O} = 16\)]
  1. A.\(\text{SO}_2\)
  2. B.\(\text{SO}_3\)
  3. C.\(\text{S}_2\text{O}_3\)
  4. D.\(\text{S}_2\text{O}_5\)
Show answer & marking scheme

Worked solution

To find the empirical formula:
1. Divide the mass percentage of each element by its relative atomic mass:
- For sulfur: \(40.0 \div 32 = 1.25\)
- For oxygen: \(60.0 \div 16 = 3.75\)

2. Divide each value by the smallest result (1.25) to get the simplest whole-number ratio:
- For sulfur: \(1.25 \div 1.25 = 1\)
- For oxygen: \(3.75 \div 1.25 = 3\)

Thus, the simplest ratio of \(\text{S} : \text{O}\) is \(1 : 3\), giving the empirical formula \(\text{SO}_3\).

Marking scheme

Award 1 mark for the correct option B.
Question 37 · multiple-choice
1 marks
Consider the following single-displacement reaction:

\(\text{Mg(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Mg}^{2+}\text{(aq)} + \text{Cu(s)}\)

Which statement is correct?
  1. A.Magnesium is reduced because it loses electrons.
  2. B.Copper(II) ions are oxidized because they gain electrons.
  3. C.Magnesium is oxidized because its oxidation number increases.
  4. D.Copper(II) ions act as a reducing agent because they are reduced.
Show answer & marking scheme

Worked solution

- Magnesium metal (\(\text{Mg}\)) loses electrons and its oxidation state increases from 0 to +2. Therefore, magnesium is oxidized.
- Copper(II) ions (\(\text{Cu}^{2+}\)) gain electrons and their oxidation state decreases from +2 to 0. Therefore, copper(II) ions are reduced.
- Option A is incorrect because losing electrons is oxidation (OIL), not reduction.
- Option B is incorrect because gaining electrons is reduction (RIG), not oxidation.
- Option D is incorrect because the reactant that is reduced (\(\text{Cu}^{2+}\)) acts as the oxidizing agent, not the reducing agent.
- Option C is correct because magnesium's oxidation number increases, which defines oxidation.

Marking scheme

Award 1 mark for the correct option C.
Question 38 · multiple-choice
1 marks
Concentrated aqueous sodium chloride (brine) is electrolysed using inert carbon electrodes.

Which row shows the products obtained at the cathode and the anode?
  1. A.cathode: hydrogen; anode: chlorine
  2. B.cathode: sodium; anode: chlorine
  3. C.cathode: hydrogen; anode: oxygen
  4. D.cathode: sodium; anode: oxygen
Show answer & marking scheme

Worked solution

During the electrolysis of concentrated aqueous sodium chloride:
- At the cathode (negative electrode), both \(\text{H}^+\) and \(\text{Na}^+\) ions are attracted. \(\text{H}^+\) is discharged preferentially because hydrogen is lower in the reactivity series than sodium. Thus, hydrogen gas (\(\text{H}_2\)) is produced.
- At the anode (positive electrode), both \(\text{Cl}^-\) and \(\text{OH}^-\) ions are attracted. Because it is a concentrated solution, \(\text{Cl}^-\) is discharged preferentially, producing chlorine gas (\(\text{Cl}_2\)).

Marking scheme

Award 1 mark for the correct option A.
Question 39 · multiple-choice
1 marks
Three metals, \(X\), \(Y\), and \(Z\), are reacted with aqueous solutions of their nitrates. The following observations are recorded:

- Metal \(X\) displaces metal \(Y\) from its salt solution.
- Metal \(X\) does not displace metal \(Z\) from its salt solution.

What is the order of reactivity of the metals, from most reactive to least reactive?
  1. A.\(X \rightarrow Y \rightarrow Z\)
  2. B.\(Y \rightarrow X \rightarrow Z\)
  3. C.\(Z \rightarrow X \rightarrow Y\)
  4. D.\(Z \rightarrow Y \rightarrow X\)
Show answer & marking scheme

Worked solution

1. Since metal \(X\) can displace metal \(Y\) from its salt solution, \(X\) is more reactive than \(Y\) (\(X > Y\)).
2. Since metal \(X\) cannot displace metal \(Z\) from its salt solution, \(Z\) is more reactive than \(X\) (\(Z > X\)).

Combining these two observations gives the reactivity order: \(Z > X > Y\).

Therefore, from most reactive to least reactive, the order is \(Z \rightarrow X \rightarrow Y\).

Marking scheme

Award 1 mark for the correct option C.
Question 40 · multiple-choice
1 marks
An ester has the structural formula shown below:

\(\text{CH}_3-\text{CO}-\text{O}-\text{CH}_2-\text{CH}_2-\text{CH}_3\)

What is the name of this ester?
  1. A.ethyl propanoate
  2. B.propyl ethanoate
  3. C.methyl butanoate
  4. D.butyl methanoate
Show answer & marking scheme

Worked solution

An ester name consists of two parts:
1. The alkyl part (derived from the alcohol group): The portion bonded to the single-bonded oxygen atom is \(-\text{CH}_2-\text{CH}_2-\text{CH}_3\), which contains 3 carbon atoms and is therefore 'propyl'.
2. The carboxylate part (derived from the carboxylic acid group): The portion containing the carbonyl carbon \(\text{CH}_3-\text{C}=\text{O}\) has 2 carbon atoms, which corresponds to 'ethanoate'.

Combining these two parts, the name of the ester is propyl ethanoate.

Marking scheme

Award 1 mark for the correct option B.

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Paper 4 - Theory (Extended)

Answer all six structured theory questions. Write chemical formulas clearly and state the units in stoichiometric calculations.
6 Question · 79.98 marks
Question 1 · structured
13.33 marks
A student determines the concentration of a sample of dilute sulfuric acid, \(\text{H}_2\text{SO}_4\), by titrating it with a standard solution of sodium hydroxide, \(\text{NaOH}\).

(a) State the colour change of methyl orange indicator when the end-point is reached as sodium hydroxide is added to the acid in the conical flask.
From ............................................................. to ............................................................. [2]

(b) Describe why a volumetric pipette is preferred over a measuring cylinder to measure the volume of the acid. [1]

(c) The student uses \(25.0\text{ cm}^3\) of the sulfuric acid solution. The titration requires \(18.40\text{ cm}^3\) of \(0.150\text{ mol/dm}^3\) sodium hydroxide solution for complete neutralisation.

(i) Write the balanced chemical equation for the reaction. [2]
(ii) Calculate the number of moles of \(\text{NaOH}\) used in the titration. [1]
(iii) Determine the number of moles of \(\text{H}_2\text{SO}_4\) that reacted. [1]
(iv) Calculate the concentration of the sulfuric acid in \(\text{mol/dm}^3\). [2]
(v) Calculate the concentration of the sulfuric acid in \(\text{g/dm}^3\). (Relative molecular mass, \(M_r\), of \(\text{H}_2\text{SO}_4 = 98.0\)) [2]

(d) Explain why the student must rinse the burette with the sodium hydroxide solution before filling it, rather than just with distilled water. [2]
Show answer & marking scheme

Worked solution

(a) Methyl orange is red in acidic solutions and yellow in basic solutions. The transition end-point is orange.
From red to orange.

(b) A volumetric pipette has a much smaller margin of error / is more precise/accurate than a measuring cylinder.

(c) (i) \(\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\)
(ii) \(\text{Moles} = \text{concentration} \times \text{volume} = 0.150 \times \frac{18.40}{1000} = 0.00276\text{ mol}\)
(iii) Ratio of \(\text{H}_2\text{SO}_4 : \text{NaOH}\) is \(1:2\).
\(\text{Moles of } \text{H}_2\text{SO}_4 = \frac{0.00276}{2} = 0.00138\text{ mol}\)
(iv) \(\text{Concentration} = \frac{\text{moles}}{\text{volume}} = \frac{0.00138}{0.0250} = 0.0552\text{ mol/dm}^3\)
(v) \(\text{Concentration in g/dm}^3 = 0.0552 \times 98.0 = 5.41\text{ g/dm}^3\) (rounded to 3 significant figures).

(d) Distilled water left in the burette would dilute the sodium hydroxide solution, decreasing its concentration and increasing the volume needed for titration, leading to an inaccurate result.

Marking scheme

(a) From red [1] to orange [1]. (Reject: yellow).
(b) Volumetric pipette has a higher precision / smaller error margin [1].
(c) (i) Correct reactants and products [1], correctly balanced [1].
(ii) \(0.00276\text{ mol}\) [1].
(iii) \(0.00138\text{ mol}\) (allow ecf from c(ii)) [1].
(iv) Shows division by \(0.0250\text{ dm}^3\) [1], answer \(0.0552\text{ mol/dm}^3\) [1].
(v) Multiplication of c(iv) by 98.0 [1], answer \(5.41\text{ g/dm}^3\) (allow \(5.4096\)) [1].
(d) distils water dilutes the NaOH solution [1], making the concentration lower / requiring a larger volume of titrant [1].
Question 2 · structured
13.33 marks
Zinc is extracted from its main ore, zinc blende, which contains zinc sulfide, \(\text{ZnS}\).

(a) (i) Name the gas produced when zinc blende is heated strongly in air (roasted). [1]
(ii) Write the balanced chemical equation for this roasting reaction. [2]

(b) The zinc oxide formed is then reduced in a furnace using carbon (coke).

(i) Explain why carbon can be used to reduce zinc oxide, with reference to the reactivity series. [1]
(ii) State the two roles of carbon in this furnace. [2]
(iii) Write the balanced chemical equation for the reduction of zinc oxide by carbon monoxide. [2]

(c) Define the term redox in terms of electron transfer. [2]

(d) In the extraction furnace, zinc is produced as a gas and is then condensed.

(i) Explain why zinc is produced as a gas in a furnace operating at \(1000\text{ }^\circ\text{C}\). (Melting point of zinc is \(420\text{ }^\circ\text{C}\), boiling point is \(907\text{ }^\circ\text{C}\).) [1]
(ii) State how zinc is separated from other remaining solid materials in the furnace. [1]
(iii) Suggest why aluminum cannot be extracted from its oxide using carbon. [1]
Show answer & marking scheme

Worked solution

(a) (i) Heating zinc sulfide in oxygen produces zinc oxide and sulfur dioxide gas.
(ii) \(2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2\)

(b) (i) Carbon is higher than zinc in the reactivity series, so it can displace zinc from its oxide.
(ii) Carbon reacts with oxygen to produce heat (exothermic combustion acting as a fuel) and acts as/produces the reducing agent (carbon monoxide).
(iii) \(\text{ZnO} + \text{CO} \rightarrow \text{Zn} + \text{CO}_2\)

(c) Redox involves both oxidation and reduction occurring simultaneously. In terms of electrons, oxidation is the loss of electrons and reduction is the gain of electrons.

(d) (i) The operating temperature of the furnace (\(1000\text{ }^\circ\text{C}\)) is higher than the boiling point of zinc (\(907\text{ }^\circ\text{C}\)).
(ii) It vaporises and is separated by fractional distillation / condensation of the gas.
(iii) Aluminum is more reactive than carbon, so carbon is not strong enough to reduce aluminum oxide.

Marking scheme

(a) (i) Sulfur dioxide / \(\text{SO}_2\) [1].
(ii) Correct species [1], correctly balanced [1].
(b) (i) Carbon is more reactive than zinc / higher in the reactivity series [1].
(ii) Burning to provide heat / acting as fuel [1], acting as a reducing agent [1].
(iii) Correct species [1], correctly balanced [1].
(c) Oxidation is loss of electrons [1], reduction is gain of electrons [1].
(d) (i) Furnace temperature is higher than the boiling point of zinc / \(1000 > 907\) [1].
(ii) Distillation / condensation of vapour [1].
(iii) Aluminum is more reactive than carbon / carbon cannot displace aluminum [1].
Question 3 · structured
13.33 marks
Alkanes are a homologous series of saturated hydrocarbons.

(a) Describe two characteristics of a homologous series, other than having similar chemical properties. [2]

(b) Explain why alkanes are generally unreactive. [1]

(c) Propane, \(\text{CH}_3\text{CH}_2\text{CH}_3\), reacts with bromine in the presence of ultraviolet (UV) light.

(i) State the name of this type of organic reaction. [1]
(ii) Explain the role of ultraviolet light in this reaction. [1]
(iii) Give the structural formula and state the IUPAC name of the two different monosubstituted organic products that can form when propane reacts with bromine.

• Product 1 structural formula:
Name:
• Product 2 structural formula:
Name: [4]

(d) Define the term structural isomers. [2]

(e) Write the molecular formula of an alkane containing 8 carbon atoms, and state its physical state at room temperature and pressure. [2]
Show answer & marking scheme

Worked solution

(a) Members of a homologous series share the same general formula, exhibit a gradual trend in physical properties, and adjacent members differ by a \(-\text{CH}_2-\). (Any two)

(b) Alkanes contain only strong, single \(\text{C}-\text{C}\) and \(\text{C}-\text{H}\) covalent bonds, which require significant energy to break, and they lack a reactive functional group.

(c) (i) Substitution reaction.
(ii) Provides the energy needed to break the covalent bond in the bromine molecule (\(\text{Br}_2\)) to form bromine radicals.
(iii) Bromine can substitute a hydrogen atom on either the end carbons (C1/C3) or the middle carbon (C2):
- Product 1: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}\), Name: 1-bromopropane
- Product 2: \(\text{CH}_3\text{CHBrCH}_3\), Name: 2-bromopropane

(d) Structural isomers are compounds that have the exact same molecular formula but different structural formulas (different connectivity of atoms).

(e) The general formula for alkanes is \(\text{C}_n\text{H}_{2n+2}\). For \(n = 8\), the formula is \(\text{C}_8\text{H}_{18}\). At room temperature and pressure, octanes are liquids.

Marking scheme

(a) Any two from: Same general formula [1], Gradual trend in physical properties [1], Neighboring members differ by \(-\text{CH}_2-\)[1].
(b) Saturated / only contain strong single bonds / no functional group [1].
(c) (i) Substitution [1].
(ii) Provides energy / breaks Br–Br bond / forms radicals [1].
(iii) Product 1 formula: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}\) [1], Name: 1-bromopropane [1];
Product 2 formula: \(\text{CH}_3\text{CHBrCH}_3\) [1], Name: 2-bromopropane [1].
(d) Same molecular formula [1], different structural formula [1].
(e) \(\text{C}_8\text{H}_{18}\) [1], liquid [1].
Question 4 · structured
13.33 marks
A student prepares a pure, dry sample of the insoluble salt lead(II) sulfate, \(\text{PbSO}_4\), using aqueous lead(II) nitrate, \(\text{Pb(NO}_3)_2\), and aqueous sodium sulfate, \(\text{Na}_2\text{SO}_4\).

(a) (i) Write the ionic equation, including state symbols, for this precipitation reaction. [2]
(ii) Describe the experimental steps the student should take to obtain a pure, dry sample of lead(II) sulfate once the precipitate has formed in the mixture. [3]

(b) The student is provided with an unknown solid, salt W. The student carries out two tests on W and records the observations in the table below.

• Test 1: Add aqueous sodium hydroxide to a solution of W and warm gently. | Observation 1: A gas is evolved which turns damp red litmus paper blue.
• Test 2: Add dilute nitric acid followed by aqueous barium nitrate to a solution of W. | Observation 2: A white precipitate forms.

(i) Identify the gas evolved in Test 1. [1]
(ii) Identify the cation present in W. [1]
(iii) Identify the anion present in W. [1]
(iv) State the chemical formula of salt W. [1]

(c) A different salt, Z, contains chloride ions.

(i) Describe a chemical test to confirm the presence of chloride ions in a solution of Z. [2]
(ii) If salt Z also gives a lilac colour in a flame test, identify salt Z. [2]
Show answer & marking scheme

Worked solution

(a) (i) In precipitation, lead(II) ions and sulfate ions react to form solid lead(II) sulfate:
\(\text{Pb}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{PbSO}_4(\text{s})\)
(ii) Once the precipitate is formed, the mixture must be filtered to collect the residue. The residue is washed with distilled water to remove soluble impurities (such as sodium nitrate). Finally, the residue is dried in a warm oven or using filter paper.

(b) (i) The gas that turns damp red litmus paper blue is ammonia, \(\text{NH}_3\).
(ii) The presence of ammonia gas upon warming with aqueous sodium hydroxide indicates the ammonium ion, \(\text{NH}_4^+\).
(iii) The test with barium nitrate in acidic conditions produces a white precipitate, indicating the sulfate ion, \(\text{SO}_4^{2-}\).
(iv) Combining ammonium and sulfate ions: \((\text{NH}_4)_2\text{SO}_4\).

(c) (i) To test for chloride ions, add dilute nitric acid to acidify the solution, then add aqueous silver nitrate. A white precipitate of silver chloride (\(\text{AgCl}\)) forms.
(ii) A lilac flame test indicates potassium ions (\(\text{K}^+\)). Since it also contains chloride ions, the salt is potassium chloride, \(\text{KCl}\).

Marking scheme

(a) (i) Correct ionic species [1], correct state symbols [1].
(ii) Filter the mixture [1], wash the residue with distilled water [1], dry the residue (in oven / filter paper) [1].
(b) (i) Ammonia / \(\text{NH}_3\) [1].
(ii) Ammonium / \(\text{NH}_4^+\) [1].
(iii) Sulfate / \(\text{SO}_4^{2-}\) [1].
(iv) \((\text{NH}_4)_2\text{SO}_4\) [1].
(c) (i) Add dilute nitric acid and aqueous silver nitrate [1], white precipitate [1].
(ii) Lilac flame indicates potassium / \(\text{K}^+\) [1], name is potassium chloride / \(\text{KCl}\) [1].
Question 5 · structured
13.33 marks
Electroplating is used to coat a steel object with a layer of nickel. The apparatus consists of an electrolysis cell with an aqueous solution of nickel(II) sulfate, \(\text{NiSO}_4\).

(a) State the name given to the positive electrode and the negative electrode. [2]

(b) To coat the steel object with nickel:
(i) State which electrode (positive or negative) the steel object must be connected to. [1]
(ii) Name the metal that should be used as the positive electrode. [1]

(c) Write the ionic half-equation for the reaction occurring at the negative electrode during this electroplating process. [2]

(d) State and explain any changes that occur to the concentration of nickel(II) ions in the electrolyte during this electroplating process. [2]

(e) The student replaces the nickel positive electrode with an inert carbon electrode.
(i) State what gas is observed bubbling at this carbon positive electrode. [1]
(ii) Write the half-equation for the reaction occurring at this carbon positive electrode. [2]
(iii) State what change, if any, will now happen to the colour of the blue nickel(II) sulfate solution as the electrolysis continues. Explain your answer. [2]
Show answer & marking scheme

Worked solution

(a) Positive electrode: anode. Negative electrode: cathode.

(b) (i) The object to be plated (steel) must be the cathode / negative electrode.
(ii) The active anode must be made of the metal to be plated, which is nickel.

(c) Nickel ions gain electrons to form nickel metal on the cathode: \(\text{Ni}^{2+} + 2\text{e}^- \rightarrow \text{Ni}\).

(d) The concentration of nickel(II) ions remains constant. Nickel atoms from the anode lose electrons to form nickel ions (\(\text{Ni} \rightarrow \text{Ni}^{2+} + 2\text{e}^-\)) at the exact same rate that nickel ions are reduced to nickel metal at the cathode.

(e) (i) When an inert electrode is used at the anode in aqueous solution, hydroxide ions/water are oxidized to produce oxygen gas.
(ii) \(2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^-\)
(iii) The blue colour of the solution will fade. Nickel ions are discharged at the cathode but are not replenished at the anode because the carbon electrode is inert and does not dissolve.

Marking scheme

(a) Positive electrode = anode [1], Negative electrode = cathode [1].
(b) (i) Negative electrode / cathode [1].
(ii) Nickel [1].
(c) Correct species [1], correctly balanced with electrons on the left [1].
(d) Concentration remains constant [1], rate of oxidation at anode equals rate of reduction at cathode [1].
(e) (i) Oxygen [1].
(ii) Correct species [1], correctly balanced with electrons on the right [1].
(iii) Colour fades [1], nickel ions are removed at cathode but not replaced [1].
Question 6 · structured
13.33 marks
This question is about gases, their properties, and atmospheric pollutants.

(a) Two gases, methylamine (\(\text{CH}_3\text{NH}_2\), \(M_r = 31\)) and hydrogen bromide (\(\text{HBr}\), \(M_r = 81\)), are released at opposite ends of a long glass tube.

(i) Define the term diffusion. [1]
(ii) Explain, in terms of particles, why gases diffuse. [1]
(iii) State and explain at which position (closer to methylamine, in the middle, or closer to hydrogen bromide) the gases will react to form a white solid. [3]

(b) When a pure solid substance is heated, its temperature rises until it begins to melt. During melting, the temperature remains constant.

(i) Explain why the temperature remains constant during melting, even though thermal energy is still being supplied. [2]
(ii) Describe the difference in both arrangement and motion of particles between the solid state and the liquid state of this substance. [2]

(c) Carbon monoxide and oxides of nitrogen are common pollutants found in car exhaust emissions.

(i) State one adverse effect of carbon monoxide on human health. [1]
(ii) Explain how oxides of nitrogen are formed in car engines, and state one environmental problem caused by them. [3]
Show answer & marking scheme

Worked solution

(a) (i) Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration.
(ii) Gas particles are in constant, random motion, colliding with each other and spreading out.
(iii) The white solid forms closer to the hydrogen bromide end. Methylamine has a lower relative molecular mass (\(M_r = 31\)) than hydrogen bromide (\(M_r = 81\)). Lighter gas particles move/diffuse faster.

(b) (i) The thermal energy supplied is absorbed to break/overcome the intermolecular forces holding the solid lattice together, rather than increasing the average kinetic energy of the particles.
(ii) Arrangement: Solid particles have a regular, ordered arrangement, while liquid particles have a random, disordered arrangement.
Motion: Solid particles only vibrate about fixed positions, while liquid particles can slide over each other / move randomly.

(c) (i) Carbon monoxide is highly toxic because it binds strongly to hemoglobin, preventing oxygen transport in the blood.
(ii) Formation: High temperatures inside car engines provide enough energy for nitrogen and oxygen gases from the air to react together. Environmental problem: Causes acid rain / photochemical smog.

Marking scheme

(a) (i) Movement of particles from high to low concentration [1].
(ii) Particles in constant random motion [1].
(iii) Closer to hydrogen bromide [1], methylamine has a lower \(M_r\) / is lighter [1], lighter gases diffuse faster [1].
(b) (i) Energy used to overcome / break bonds or forces [1], kinetic energy (and temperature) does not increase [1].
(ii) Arrangement: regular in solid vs random in liquid [1], Motion: vibrating in solid vs sliding/moving in liquid [1].
(c) (i) Toxic / binds to hemoglobin / reduces oxygen transport [1].
(ii) Nitrogen and oxygen react [1], at high temperatures in car engines [1], Acid rain / photochemical smog [1].

Paper 6 - Alternative to Practical

Answer all four questions. Ensure diagrams of apparatus are drawn clearly with correct functional labels.
4 Question · 40 marks
Question 1 · practical_and_planning
10 marks
A student investigates the rate of reaction between magnesium ribbon and dilute sulfuric acid.

(a) Draw a labelled diagram of the apparatus used to react magnesium ribbon with dilute sulfuric acid, collect the hydrogen gas produced, and measure its volume over time. [3]

(b) State the observation, other than the cessation of bubbling, that shows the reaction is complete if the magnesium ribbon was the limiting reactant. [1]

(c) The rate of reaction decreases as the reaction progresses.
(i) Explain, in terms of collision theory, why the rate of reaction decreases. [2]
(ii) State what happens to the rate of reaction if the temperature of the acid is increased, and explain your answer in terms of particle energy. [2]

(d) The student repeats the experiment using the same mass of magnesium ribbon but cuts it into smaller pieces.
(i) State the effect of this change on the initial rate of reaction. [1]
(ii) State the effect of this change on the final volume of gas produced. [1]
Show answer & marking scheme

Worked solution

(a) The diagram should show a reaction vessel (e.g., conical flask) containing magnesium and sulfuric acid, sealed with a stopper. A delivery tube must connect the flask to a gas syringe or an inverted measuring cylinder filled with water in a trough. All components must be correctly labelled.

(b) The magnesium ribbon completely dissolves / disappears.

(c) (i) As the reaction proceeds, the concentration of hydrogen ions / acid decreases. This results in fewer acid particles per unit volume, which decreases the frequency of successful collisions between reactant particles.
(ii) The rate of reaction increases. At higher temperatures, particles have more kinetic energy and move faster, leading to a higher frequency of successful collisions, and a greater proportion of particles have energy greater than or equal to the activation energy.

(d) (i) The initial rate of reaction increases (due to a larger surface area).
(ii) The final volume of gas produced remains the same.

Marking scheme

M1: Closed reaction vessel (conical flask/boiling tube) with stopper and delivery tube [1]
M2: Gas syringe or inverted cylinder in water trough [1]
M3: Correct labels for reactants (magnesium and dilute sulfuric acid) and gas collection apparatus [1]
M4: Magnesium ribbon disappears / dissolves [1]
M5: Concentration of acid / reactant particles decreases [1]
M6: Frequency of collisions / successful collisions decreases [1]
M7: Rate increases AND particles have more kinetic energy / move faster [1]
M8: More successful collisions per unit time / more collisions have energy >= activation energy [1]
M9: Initial rate increases [1]
M10: Final volume remains unchanged / same [1]
Question 2 · practical_and_planning
10 marks
A student determines the concentration of a solution of hydrochloric acid by titrating it against standard sodium hydroxide solution.

(a) The student records the burette readings for two titration experiments:
- In Experiment 1, the initial burette reading is \( 0.5\text{ cm}^3 \) and the final burette reading is \( 23.1\text{ cm}^3 \).
- In Experiment 2, the initial burette reading is \( 1.2\text{ cm}^3 \) and the final burette reading is \( 24.0\text{ cm}^3 \).

Complete the table below to record the burette readings and calculate the volume of sodium hydroxide added in each experiment. [3]

| | Experiment 1 | Experiment 2 |
|---|---|---|
| Final burette reading / \(\text{cm}^3\) | | |
| Initial burette reading / \(\text{cm}^3\) | | |
| Volume of sodium hydroxide added / \(\text{cm}^3\) | | |

(b) Calculate the average volume of sodium hydroxide added from these results. [1]

(c) The titration is performed using methyl orange indicator. State the color change of the indicator at the end-point when adding sodium hydroxide from the burette into the hydrochloric acid in the conical flask. [2]

(d) (i) State why the conical flask is placed on a white tile during the titration. [1]
(ii) State why the conical flask is swirled continuously during the addition of the sodium hydroxide. [1]

(e) The student used a measuring cylinder to measure the volume of hydrochloric acid.
(i) Name a more precise piece of apparatus to measure this volume. [1]
(ii) Explain how using this apparatus improves the accuracy of the experiment. [1]
Show answer & marking scheme

Worked solution

(a) Completed table:
- Experiment 1: Final = 23.1, Initial = 0.5, Volume = 22.6
- Experiment 2: Final = 24.0, Initial = 1.2, Volume = 22.8

(b) Average volume = (22.6 + 22.8) / 2 = 22.7 cm3

(c) The color changes from red to orange (or yellow).

(d) (i) To make the color change at the end-point easier to see clearly.
(ii) To ensure thorough mixing of the reactants so that the reaction is complete immediately upon addition.

(e) (i) Volumetric pipette / bulb pipette.
(ii) It has a smaller percentage error / measures the exact volume of acid more accurately than a measuring cylinder.

Marking scheme

M1: All four burette readings correctly entered in the table (23.1, 0.5, 24.0, 1.2) [1]
M2: Both volumes calculated correctly (22.6 and 22.8) [1]
M3: All values given to 1 decimal place consistently [1]
M4: Average volume calculated correctly as 22.7 cm3 (allow ECF from incorrect volumes) [1]
M5: Red [1]
M6: to orange (or yellow) [1]
M7: To see the color change more clearly / against a neutral background [1]
M8: To ensure complete mixing / uniform reaction [1]
M9: Volumetric pipette / pipette (Reject: dropping pipette) [1]
M10: More precise / smaller uncertainty / smaller percentage error [1]
Question 3 · practical_and_planning
10 marks
A student carries out tests on a green hydrated solid salt, **W**.

(a) **Test 1**: A sample of solid **W** is heated in a dry test-tube.
- Observation: Condensation of a colorless liquid forms at the top of the test-tube, and the green solid turns into a black powder.
(i) Name the colorless liquid that condenses. [1]
(ii) Describe a chemical test to confirm the identity of this liquid. [2]

(b) **Test 2**: The remaining black powder from Test 1 is dissolved in dilute sulfuric acid to form a blue-green solution. This solution is divided into two portions.
(i) To the first portion, aqueous sodium hydroxide is added dropwise, then in excess.
- Observation: A green precipitate forms.
State the identity of the cation shown to be present by this test. [1]
(ii) To the second portion, aqueous ammonia is added dropwise, then in excess.
Describe the observations expected. [2]

(c) **Test 3**: A fresh sample of solid **W** is dissolved in distilled water. Dilute nitric acid is added, followed by aqueous silver nitrate.
- Observation: A cream precipitate forms.
(i) State the identity of the anion present in **W**. [1]
(ii) Deduce the chemical formula of anhydrous salt **W**. [1]

(d) Write an ionic equation, including state symbols, for the reaction occurring in Test 3. [2]
Show answer & marking scheme

Worked solution

(a) (i) Water.
(ii) Add the liquid to anhydrous copper(II) sulfate. It will turn from white to blue. (Alternatively, add to blue anhydrous cobalt(II) chloride paper, which turns pink).

(b) (i) Iron(II) / Fe2+.
(ii) A green precipitate forms, which is insoluble in excess ammonia.

(c) (i) Bromide / Br-.
(ii) FeBr2.

(d) Ag+(aq) + Br-(aq) -> AgBr(s)

Marking scheme

M1: Water [1]
M2: Add to anhydrous copper(II) sulfate / anhydrous cobalt(II) chloride [1]
M3: Turns blue (for copper sulfate) / turns pink (for cobalt chloride) [1]
M4: Iron(II) / Fe2+ [1]
M5: Green precipitate [1]
M6: Insoluble in excess ammonia [1]
M7: Bromide / Br- [1]
M8: FeBr2 [1]
M9: Correct reactants and products: Ag+ + Br- -> AgBr [1]
M10: State symbols correct: (aq) for reactants, (s) for product [1]
Question 4 · practical_and_planning
10 marks
Chlorophyll is a green pigment found in plants, which is soluble in organic solvents but insoluble in water.

Design an investigation to extract the pigments from fresh grass leaves and separate them using paper chromatography to show that grass contains more than one colored pigment.

You are provided with: fresh grass leaves, sand, a mortar and pestle, propanone (a liquid organic solvent), chromatography paper, and standard laboratory apparatus.

Your plan should include:
- how to extract a concentrated solution of the pigments from the grass
- how to set up and run the chromatography experiment
- the observations that would show that grass contains more than one pigment
- two safety precautions when handling propanone. [10]
Show answer & marking scheme

Worked solution

To extract the pigment, place the fresh grass leaves into a mortar with a small amount of sand and some propanone. Grind the mixture thoroughly using a pestle to break down the cell walls and dissolve the pigments. Filter or decant the green liquid to separate it from the solid grass residue, obtaining a concentrated extract.

To perform chromatography, draw a horizontal line in pencil (the baseline) about 2 cm from the bottom of a strip of chromatography paper. Use a capillary tube or dropper to apply a spot of the concentrated extract onto the baseline. Allow the spot to dry, and repeat several times to concentrate the pigments.

Suspend the paper inside a beaker containing a small volume of propanone, ensuring the solvent level is below the baseline. Place a lid on the beaker to prevent evaporation. Allow the solvent to travel up the paper until it is near the top, then remove the paper and mark the solvent front with a pencil.

If grass contains more than one pigment, multiple spots of different colors (such as green, yellow, or orange) will be seen at different heights on the paper.

Safety precautions: Propanone is highly flammable, so keep it away from open flames / use a hot water bath instead of a Bunsen burner if heating is needed. It is also a mild skin/respiratory irritant, so wear safety goggles / gloves / work in a well-ventilated area.

Marking scheme

Award 1 mark for each of the following points up to a maximum of 10:

Extraction of pigment:
1. Grind leaves with propanone in a mortar using a pestle. [1]
2. Add sand to assist in grinding / breaking cell walls. [1]
3. Filter or decant to obtain a clear, concentrated liquid extract. [1]

Chromatography setup:
4. Draw a baseline on the chromatography paper in pencil (Reject: pen). [1]
5. Spot the extract onto the baseline and allow it to dry / repeat to concentrate. [1]
6. Place the paper in a beaker with propanone so that the solvent level is below the baseline. [1]
7. Cover the beaker with a lid to prevent evaporation of the solvent. [1]

Analysis/Observations:
8. Run the chromatogram until the solvent front is near the top, and remove the paper. [1]
9. Multiple spots of different colors at different heights indicate more than one pigment. [1]

Safety:
10. Any two correct safety precautions: e.g., wear safety goggles, keep propanone away from naked flames/heat sources (use water bath), use gloves, or work in a fume cupboard. [1]

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