Cambridge IGCSE · thinka-original Practice Paper

2025 Cambridge IGCSE Chemistry (0620) Practice Paper with Answers

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

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

Paper 22

Answer all 40 multiple-choice questions. For each question, choose the single correct alternative from A, B, C or D.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
A pure solid substance, Y, is heated at a constant rate until it has completely boiled. The temperature of Y is recorded at regular intervals. Which statement describes the behavior of Y during the period when it is melting?
  1. A.The temperature increases as the particles gain kinetic energy.
  2. B.The temperature remains constant while energy is used to break forces between particles.
  3. C.The temperature remains constant because the particles have stopped moving.
  4. D.The temperature decreases as the particles lose potential energy.
Show answer & marking scheme

Worked solution

During melting, the temperature of Y remains constant because the heat energy supplied is used to overcome the intermolecular forces holding the solid lattice together, rather than to increase the kinetic energy of the particles.

Marking scheme

Award 1 mark for the correct answer B.
Question 2 · multiple-choice
1 marks
Which row correctly identifies the number of protons, neutrons, and electrons in the phosphide ion, \(^{31}\text{P}^{3-}\)?
  1. A.Protons = 15, Neutrons = 16, Electrons = 15
  2. B.Protons = 15, Neutrons = 16, Electrons = 18
  3. C.Protons = 16, Neutrons = 15, Electrons = 18
  4. D.Protons = 15, Neutrons = 31, Electrons = 18
Show answer & marking scheme

Worked solution

Phosphorus has an atomic number of 15, so a phosphorus atom has 15 protons. The mass number is 31, so the number of neutrons is \(31 - 15 = 16\). The \(\text{P}^{3-}\) ion has a 3- charge, meaning it has gained 3 electrons, so the number of electrons is \(15 + 3 = 18\).

Marking scheme

Award 1 mark for the correct answer B.
Question 3 · multiple-choice
1 marks
The formula of the nitride ion is \(\text{N}^{3-}\) and the formula of the calcium ion is \(\text{Ca}^{2+}\). What is the correct formula of calcium nitride?
  1. A.\(\text{CaN}\)
  2. B.\(\text{Ca}_2\text{N}_3\)
  3. C.\(\text{Ca}_3\text{N}_2\)
  4. D.\(\text{Ca}_3\text{N}\)
Show answer & marking scheme

Worked solution

To form a neutral ionic compound, the total positive charge must equal the total negative charge. Three calcium ions (\(3 \times 2+ = +6\)) combine with two nitride ions (\(2 \times 3- = -6\)) to give a neutral compound with the formula \(\text{Ca}_3\text{N}_2\).

Marking scheme

Award 1 mark for the correct answer C.
Question 4 · multiple-choice
1 marks
An aqueous solution of concentrated copper(II) chloride is electrolysed using inert carbon electrodes. Which products are formed at each electrode?
  1. A.Anode = oxygen, Cathode = copper
  2. B.Anode = chlorine, Cathode = copper
  3. C.Anode = chlorine, Cathode = hydrogen
  4. D.Anode = oxygen, Cathode = hydrogen
Show answer & marking scheme

Worked solution

At the cathode, copper ions are discharged in preference to hydrogen ions because copper is lower in the reactivity series, forming copper metal. At the anode, chloride ions are discharged in preference to hydroxide ions because of their high concentration, forming chlorine gas.

Marking scheme

Award 1 mark for the correct answer B.
Question 5 · multiple-choice
1 marks
The equation for the combustion of hydrazine is shown.
\(\text{N}_2\text{H}_4(\text{g}) + \text{O}_2(\text{g}) \rightarrow \text{N}_2(\text{g}) + 2\text{H}_2\text{O}(\text{g})\)
Some bond energies are given in the list:
- \(\text{N}-\text{H}\): \(391\text{ kJ/mol}\)
- \(\text{N}-\text{N}\): \(158\text{ kJ/mol}\)
- \(\text{O}=\text{O}\): \(498\text{ kJ/mol}\)
- \(\text{N}\equiv\text{N}\): \(945\text{ kJ/mol}\)
- \(\text{O}-\text{H}\): \(463\text{ kJ/mol}\)
What is the enthalpy change, \(\Delta H\), for the reaction?
  1. A.\(-577\text{ kJ/mol}\)
  2. B.\(-116\text{ kJ/mol}\)
  3. C.\(-1002\text{ kJ/mol}\)
  4. D.\(+577\text{ kJ/mol}\)
Show answer & marking scheme

Worked solution

Energy to break bonds:
- \(4 \times (\text{N}-\text{H}) = 4 \times 391 = 1564\text{ kJ}\)
- \(1 \times (\text{N}-\text{N}) = 158\text{ kJ}\)
- \(1 \times (\text{O}=\text{O}) = 498\text{ kJ}\)
Total input = \(1564 + 158 + 498 = 2220\text{ kJ}\).

Energy released by forming bonds:
- \(1 \times (\text{N}\equiv\text{N}) = 945\text{ kJ}\)
- \(4 \times (\text{O}-\text{H}) = 4 \times 463 = 1852\text{ kJ}\)
Total output = \(945 + 1852 = 2797\text{ kJ}\).

\(\Delta H = 2220 - 2797 = -577\text{ kJ/mol}\).

Marking scheme

Award 1 mark for the correct answer A.
Question 6 · multiple-choice
1 marks
The following reversible reaction reaches dynamic equilibrium in a closed container:
\(\text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g}) \leftrightharpoons \text{PCl}_5(\text{g})\) \(\Delta H = -88\text{ kJ/mol}\)
Which change would increase the yield of \(\text{PCl}_5(\text{g})\) at equilibrium?
  1. A.Increasing the temperature at constant pressure.
  2. B.Decreasing the pressure at constant temperature.
  3. C.Increasing the pressure at constant temperature.
  4. D.Adding a catalyst at constant temperature.
Show answer & marking scheme

Worked solution

According to Le Chatelier's principle, increasing the pressure shifts the equilibrium position toward the side with fewer gas molecules. The reactant side has 2 moles of gas and the product side has 1 mole of gas, so increasing the pressure increases the yield of \(\text{PCl}_5(\text{g})\).

Marking scheme

Award 1 mark for the correct answer C.
Question 7 · multiple-choice
1 marks
Three metals, X, Y, and Z, are tested with the following results:
- X reacts vigorously with cold water.
- Y does not react with dilute hydrochloric acid.
- Z reacts slowly with dilute hydrochloric acid but does not react with cold water.
What is the order of reactivity of the metals, from most reactive to least reactive?
  1. A.X \(\rightarrow\) Y \(\rightarrow\) Z
  2. B.X \(\rightarrow\) Z \(\rightarrow\) Y
  3. C.Z \(\rightarrow\) X \(\rightarrow\) Y
  4. D.Y \(\rightarrow\) Z \(\rightarrow\) X
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Worked solution

Metal X is the most reactive since it reacts with cold water. Metal Z is moderately reactive because it reacts with dilute acid but not cold water. Metal Y is the least reactive as it does not react with dilute acid. Thus, the correct order is X \(\rightarrow\) Z \(\rightarrow\) Y.

Marking scheme

Award 1 mark for the correct answer B.
Question 8 · multiple-choice
1 marks
Which statement about alkenes is correct?
  1. A.They are saturated hydrocarbons that contain carbon-carbon double bonds.
  2. B.They react with aqueous bromine to turn the solution from colourless to orange-brown.
  3. C.They undergo addition reactions with steam in the presence of a catalyst to produce alcohols.
  4. D.They have the general formula \(\text{C}_n\text{H}_{2n+2}\).
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Worked solution

Alkenes undergo catalytic hydration (addition with steam) to produce alcohols. They are unsaturated, decolorise bromine water from orange-brown to colourless, and have the general formula \(\text{C}_n\text{H}_{2n}\).

Marking scheme

Award 1 mark for the correct answer C.
Question 9 · multiple-choice
1 marks
An element, E, has two isotopes: \(^{63}\text{E}\) and \(^{65}\text{E}\).

The relative atomic mass of E is 63.6.

What is the percentage abundance of the heavier isotope, \(^{65}\text{E}\)?
  1. A.20%
  2. B.30%
  3. C.70%
  4. D.80%
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Worked solution

Let the fractional abundance of \(^{65}\text{E}\) be \(x\).
Therefore, the fractional abundance of \(^{63}\text{E}\) is \(1 - x\).

Using the formula for relative atomic mass:
\(65x + 63(1 - x) = 63.6\)
\(65x + 63 - 63x = 63.6\)
\(2x + 63 = 63.6\)
\(2x = 0.6\)
\(x = 0.3\)

Converting the fractional abundance to a percentage:
\(0.3 \times 100\% = 30\%\).

Marking scheme

1 mark for the correct answer B (30%).
Reject other options: A (20%), C (70%), D (80%).
Question 10 · multiple-choice
1 marks
Concentrated aqueous potassium bromide is electrolysed using inert electrodes.

Which row shows the products formed at each electrode and the change in pH of the electrolyte during the electrolysis?

| | Product at anode (+) | Product at cathode (-) | Change in pH of electrolyte |
|---|---|---|---|
| **A** | bromine | hydrogen | increases |
| **B** | bromine | potassium | decreases |
| **| C** | oxygen | hydrogen | no change |
| **D** | oxygen | potassium | increases |
  1. A.Row A
  2. B.Row B
  3. C.Row C
  4. D.Row D
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Worked solution

During the electrolysis of concentrated aqueous potassium bromide:
- At the anode (+), bromide ions (\(\text{Br}^-\)) are discharged in preference to hydroxide ions because it is a concentrated solution of a halide, producing bromine gas.
- At the cathode (-), hydrogen ions (\(\text{H}^+\)) are discharged in preference to potassium ions (\(\text{K}^+\)) because hydrogen is lower in the reactivity series, producing hydrogen gas.
- As hydrogen ions are discharged and bromide ions are discharged, \(\text{OH}^-\)(aq) and \(\text{K}^+\)(aq) ions remain in the solution. This increases the concentration of hydroxide ions, making the solution alkaline, so the pH increases.

Marking scheme

1 mark for the correct answer A.
Reject B, C and D because they identify incorrect electrode products or an incorrect pH trend.
Question 11 · multiple-choice
1 marks
A molecule of methanol has the formula \(\text{CH}_3\text{OH}\).

How many bonding pairs of electrons and how many non-bonding outer-shell electrons are there in one molecule of methanol?

| | Bonding pairs of electrons | Non-bonding outer-shell electrons |
|---|---|---|
| **A** | 5 | 4 |
| **B** | 5 | 8 |
| **C** | 6 | 4 |
| **D** | 6 | 8 |
  1. A.Row A
  2. B.Row B
  3. C.Row C
  4. D.Row D
Show answer & marking scheme

Worked solution

In a molecule of methanol (\(\text{CH}_3\text{OH}\)):
- There are three single \(\text{C-H}\) covalent bonds, one single \(\text{C-O}\) covalent bond, and one single \(\text{O-H}\) covalent bond. This gives a total of 5 single covalent bonds, which correspond to 5 bonding pairs.
- Carbon (Group IV) uses all 4 of its outer-shell electrons in bonding.
- Hydrogen (Group I) uses its only electron in bonding.
- Oxygen (Group VI) has 6 outer-shell electrons. It uses 2 in bonding (one with carbon, one with hydrogen), which leaves 4 non-bonding outer-shell electrons (forming 2 lone pairs).
- Thus, there are 5 bonding pairs and 4 non-bonding electrons.

Marking scheme

1 mark for the correct option A.
Reject options B, C, D which incorrectly count the bonding pairs or non-bonding electrons.
Question 12 · multiple-choice
1 marks
An oxide of nitrogen contains 30.4% nitrogen by mass.

What is the empirical formula of this oxide?

(Relative atomic masses: \(A_r(\text{N}) = 14\), \(A_r(\text{O}) = 16\))
  1. A.\(\text{NO}\)
  2. B.\(\text{NO}_2\)
  3. C.\(\text{N}_2\text{O}\)
  4. D.\(\text{N}_2\text{O}_3\)
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Worked solution

1. Find the percentage of oxygen: \(100\% - 30.4\% = 69.6\%\).
2. Calculate the moles of each element in 100 g:
- Moles of N = \(30.4 / 14 = 2.17\text{ mol}\)
- Moles of O = \(69.6 / 16 = 4.35\text{ mol}\)
3. Divide by the smallest number of moles (2.17):
- Ratio of N = \(2.17 / 2.17 = 1\)
- Ratio of O = \(4.35 / 2.17 \approx 2\)
4. The empirical formula is \(\text{NO}_2\).

Marking scheme

1 mark for the correct answer B.
Reject A, C, and D.
Question 13 · multiple-choice
1 marks
The reaction between hydrogen gas and chlorine gas to form hydrogen chloride gas is represented by the following chemical equation:

$$\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{HCl(g)}$$

The bond energies are shown in the table.

| Bond | Bond energy / kJ/mol |
|---|---|
| \(\text{H-H}\) | 436 |
| \(\text{Cl-Cl}\) | 242 |
| \(\text{H-Cl}\) | 431 |

What is the energy change for this reaction?
  1. A.\(-184\text{ kJ/mol}\)
  2. B.\(-153\text{ kJ/mol}\)
  3. C.\(+153\text{ kJ/mol}\)
  4. D.\(+184\text{ kJ/mol}\)
Show answer & marking scheme

Worked solution

1. Energy required to break bonds (reactants side):
- \(1 \times \text{H-H} = 436\text{ kJ/mol}\)
- \(1 \times \text{Cl-Cl} = 242\text{ kJ/mol}\)
- Total energy input = \(436 + 242 = 678\text{ kJ/mol}\)

2. Energy released when forming bonds (products side):
- \(2 \times \text{H-Cl} = 2 \times 431 = 862\text{ kJ/mol}\)

3. Energy change (\(\Delta H\)):
- \(\Delta H = \text{Energy input} - \text{Energy output} = 678 - 862 = -184\text{ kJ/mol}\).

Marking scheme

1 mark for the correct calculation leading to option A (-184 kJ/mol).
Reject other values: B (-153), C (+153), D (+184).
Question 14 · multiple-choice
1 marks
Four metals, W, X, Y and Z, are added separately to aqueous solutions of their metal nitrates. The observations are recorded in the table below.

| Metal | Solution containing \(\text{W}^{2+}\) | Solution containing \(\text{X}^{2+}\) | Solution containing \(\text{Y}^{2+}\) | Solution containing \(\text{Z}^{2+}\) |
|---|---|---|---|---|
| **W** | - | no reaction | no reaction | Z metal formed |
| **X** | W metal formed | - | Y metal formed | Z metal formed |
| **Y** | W metal formed | no reaction | - | Z metal formed |
| **Z** | no reaction | no reaction | no reaction | - |

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

Worked solution

- Metal X displaces W, Y and Z. Therefore, X is the most reactive metal.
- Metal Y displaces W and Z, but does not react with X. Therefore, Y is less reactive than X but more reactive than W and Z.
- Metal W only displaces Z, meaning it is more reactive than Z but less reactive than X and Y.
- Metal Z does not displace any of the other metals, so it is the least reactive.

Therefore, the order of reactivity from most to least reactive is: \(\text{X} \rightarrow \text{Y} \rightarrow \text{W} \rightarrow \text{Z}\).

Marking scheme

1 mark for the correct option A.
Reject options B, C, and D, which represent incorrect reactivity orders.
Question 15 · multiple-choice
1 marks
Ethene gas reacts with steam to produce ethanol.

Which set of conditions is required for this industrial addition reaction to take place?
  1. A.\(300^\circ\text{C}\), \(60\text{ atm}\), phosphoric(V) acid catalyst
  2. B.\(150^\circ\text{C}\), \(5\text{ atm}\), nickel catalyst
  3. C.\(450^\circ\text{C}\), \(200\text{ atm}\), iron catalyst
  4. D.\(35^\circ\text{C}\), \(1\text{ atm}\), yeast catalyst
Show answer & marking scheme

Worked solution

The industrial hydration of ethene to produce ethanol requires the following conditions:
- A temperature of \(300^\circ\text{C}\)
- A pressure of \(60\text{ atm}\)
- A phosphoric(V) acid (\(\text{H}_3\text{PO}_4\)) catalyst.

These correspond to option A.

Marking scheme

1 mark for the correct answer A.
Reject options B (which refers to hydrogenation-like conditions), C (Haber process conditions), and D (yeast fermentation conditions).
Question 16 · multiple-choice
1 marks
An unknown salt, Y, is dissolved in distilled water. The following tests are performed:

- The addition of aqueous sodium hydroxide to the solution produces a green precipitate that is insoluble in excess.
- The addition of dilute nitric acid followed by aqueous barium nitrate to the solution produces a white precipitate.

What is the identity of salt Y?
  1. A.iron(II) sulfate
  2. B.iron(III) sulfate
  3. C.chromium(III) chloride
  4. D.iron(II) chloride
Show answer & marking scheme

Worked solution

- The green precipitate formed with sodium hydroxide which is insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). (Note: Chromium(III) also gives a green precipitate with sodium hydroxide, but it dissolves in excess to give a green solution).
- The white precipitate formed with acidified barium nitrate confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\).

Therefore, the unknown salt Y is iron(II) sulfate.

Marking scheme

1 mark for the correct answer A.
Reject B (iron(III) gives a red-brown precipitate), C (chromium(III) is soluble in excess NaOH and chloride does not react with barium nitrate), and D (chloride does not give a precipitate with barium nitrate).
Question 17 · multiple-choice
1 marks
Some bond energies are shown in the table.

| Bond | Bond energy / kJ/mol |
| :--- | :--- |
| \(\text{H}-\text{H}\) | 436 |
| \(\text{Br}-\text{Br}\) | 193 |
| \(\text{H}-\text{Br}\) | 366 |

What is the enthalpy change, \(\Delta H\), for the reaction shown below?

\[\text{H}_2\text{(g)} + \text{Br}_2\text{(g)} \rightarrow 2\text{HBr(g)}\]
  1. A.\(-103\text{ kJ/mol}\)
  2. B.\(+103\text{ kJ/mol}\)
  3. C.\(-263\text{ kJ/mol}\)
  4. D.\(-1361\text{ kJ/mol}\)
Show answer & marking scheme

Worked solution

To calculate the enthalpy change of the reaction:
1. Find the energy required to break the bonds in the reactants:
Energy in = \(1 \times \text{bond energy(H}-\text{H)} + 1 \times \text{bond energy(Br}-\text{Br)}\)
Energy in = \(436 + 193 = 629\text{ kJ/mol}\)

2. Find the energy released when new bonds are formed in the products:
Energy out = \(2 \times \text{bond energy(H}-\text{Br)}\)
Energy out = \(2 \times 366 = 732\text{ kJ/mol}\)

3. Calculate the enthalpy change (\(\Delta H\)):
\(\Delta H = \text{Energy in} - \text{Energy out}\)
\(\Delta H = 629 - 732 = -103\text{ kJ/mol}\)

Marking scheme

Award 1 mark for the correct answer A.
Question 18 · multiple-choice
1 marks
Concentrated aqueous copper(II) chloride is electrolysed using platinum electrodes.

Which row correctly describes the products at the electrodes and the visual change in the electrolyte?
  1. A.Anode product: oxygen | Cathode product: copper | Color of electrolyte: turns darker blue
  2. B.Anode product: chlorine | Cathode product: hydrogen | Color of electrolyte: stays blue
  3. C.Anode product: chlorine | Cathode product: copper | Color of electrolyte: turns paler blue
  4. D.Anode product: oxygen | Cathode product: hydrogen | Color of electrolyte: turns paler blue
Show answer & marking scheme

Worked solution

In the electrolysis of concentrated aqueous copper(II) chloride:
- At the cathode (negative electrode), \(\text{Cu}^{2+}\) ions are selectively discharged in preference to \(\text{H}^{+}\) ions because copper is lower in the reactivity series. Copper metal forms as a pink/brown solid.
- At the anode (positive electrode), halide ions (\(\text{Cl}^{-}\)) are present in a high concentration, so chlorine gas is discharged and evolved.
- The blue color of the electrolyte is due to the presence of aqueous copper(II) ions. As \(\text{Cu}^{2+}\) ions are discharged and removed from the solution, the blue color of the electrolyte fades, turning paler blue.

Marking scheme

Award 1 mark for the correct answer C.
Question 19 · multiple-choice
1 marks
Ethene reacts with chlorine gas at room temperature and pressure.

What is the structural formula of the organic product formed?
  1. A.\(\text{CH}_3-\text{CH}_2\text{Cl}\)
  2. B.\(\text{CH}_2\text{Cl}-\text{CH}_2\text{Cl}\)
  3. C.\(\text{CHCl}=\text{CHCl}\)
  4. D.\(\text{CH}_3-\text{CHCl}_2\)
Show answer & marking scheme

Worked solution

The reaction between ethene (an alkene) and chlorine is an addition reaction. The double bond in ethene opens up, and one chlorine atom adds to each carbon atom of the double bond.

\[\text{CH}_2=\text{CH}_2 + \text{Cl}_2 \rightarrow \text{CH}_2\text{Cl}-\text{CH}_2\text{Cl}\]

The product is 1,2-dichloroethane, which has the structural formula \(\text{CH}_2\text{Cl}-\text{CH}_2\text{Cl}\).

Marking scheme

Award 1 mark for the correct answer B.
Question 20 · multiple-choice
1 marks
What is the volume of \(0.88\text{ g}\) of carbon dioxide gas, \(\text{CO}_2\), at room temperature and pressure (r.t.p.)?

(Molar volume of a gas at r.t.p. is \(24\text{ dm}^3\); \(M_r\) of \(\text{CO}_2\) is 44)
  1. A.\(480\text{ cm}^3\)
  2. B.\(240\text{ cm}^3\)
  3. C.\(4.8\text{ dm}^3\)
  4. D.\(48\text{ cm}^3\)
Show answer & marking scheme

Worked solution

1. Find the number of moles of \(\text{CO}_2\):
\[\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.88\text{ g}}{44\text{ g/mol}} = 0.02\text{ mol}\]

2. Calculate the volume of gas at r.t.p.:
\[\text{volume} = \text{moles} \times 24\text{ dm}^3\text{/mol} = 0.02 \times 24 = 0.48\text{ dm}^3\]

3. Convert \(\text{dm}^3\) to \(\text{cm}^3\):
\[0.48 \times 1000 = 480\text{ cm}^3\]

Marking scheme

Award 1 mark for the correct answer A.
Question 21 · multiple-choice
1 marks
Which statement about the giant structures of diamond, graphite, and silicon(IV) oxide is correct?
  1. A.Both diamond and silicon(IV) oxide have giant covalent structures where every atom is bonded to four other atoms of the same type.
  2. B.Graphite conducts electricity because it has delocalised electrons between its layers, whereas diamond has no delocalised electrons.
  3. C.Silicon(IV) oxide is a simple molecular compound with a low melting point.
  4. D.Diamond is soft and slippery because its atoms are arranged in hexagonal rings that can slide over each other.
Show answer & marking scheme

Worked solution

- Diamond and silicon(IV) oxide do not have the same bonding structures; in silicon(IV) oxide, each silicon atom is bonded to four oxygen atoms, and each oxygen is bonded to two silicon atoms.
- Graphite conducts electricity because it contains delocalised electrons between its layers of carbon atoms. Diamond has all its outer-shell electrons shared in localized covalent bonds, so it has no delocalised electrons and does not conduct electricity.
- Silicon(IV) oxide has a giant covalent macromolecular structure, not a simple molecular structure, and has a very high melting point.
- Graphite is soft and slippery because its carbon atoms are arranged in hexagonal layers with weak intermolecular forces between the layers, allowing them to slide over each other. Diamond is extremely hard.

Marking scheme

Award 1 mark for the correct answer B.
Question 22 · multiple-choice
1 marks
Three metals, \(W\), \(X\) and \(Y\), are tested for their reactivity.

- Metal \(W\) reacts with aqueous copper(II) sulfate, but does not react with aqueous zinc sulfate.
- Metal \(X\) reacts with aqueous zinc sulfate.
- Metal \(Y\) does not react with aqueous copper(II) sulfate.

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

Worked solution

To find the order of reactivity:
1. Metal \(W\) reacts with aqueous copper(II) sulfate, which means \(W\) is more reactive than copper (\(W > \text{Cu}\)). However, \(W\) does not react with aqueous zinc sulfate, meaning it is less reactive than zinc (\(\text{Zn} > W\)).
2. Metal \(X\) reacts with aqueous zinc sulfate, meaning \(X\) is more reactive than zinc (\(X > \text{Zn}\)).
3. Metal \(Y\) does not react with aqueous copper(II) sulfate, which means \(Y\) is less reactive than copper (\(\text{Cu} > Y\)).

Combining these facts together:
\[X > \text{Zn} > W > \text{Cu} > Y\]

Therefore, the order of reactivity of the three metals from most reactive to least reactive is \(X \rightarrow W \rightarrow Y\).

Marking scheme

Award 1 mark for the correct answer A.
Question 23 · multiple-choice
1 marks
A student is given an unknown colourless solution.

- When dilute hydrochloric acid is added and the mixture is warmed, a gas is evolved that turns acidified aqueous potassium manganate(VII) from purple to colourless.
- When a fresh sample of the solution is acidified with dilute nitric acid and aqueous barium nitrate is added, no precipitate is formed.

Which ion is present in the solution?
  1. A.carbonate, \(\text{CO}_3^{2-}\)
  2. B.chloride, \(\text{Cl}^{-}\)
  3. C.sulfate, \(\text{SO}_4^{2-}\)
  4. D.sulfite, \(\text{SO}_3^{2-}\)
Show answer & marking scheme

Worked solution

- When sulfite ions (\(\text{SO}_3^{2-}\)) react with dilute acids, sulfur dioxide gas (\(\text{SO}_2\)) is produced. Sulfur dioxide is a reducing agent that turns acidified aqueous potassium manganate(VII) from purple to colourless.
- Sulfate ions (\(\text{SO}_4^{2-}\)) would give a white precipitate of barium sulfate when reacted with barium nitrate, but sulfite ions do not yield a precipitate under these acidic conditions.
- Therefore, the solution contains sulfite ions.

Marking scheme

Award 1 mark for the correct answer D.
Question 24 · multiple-choice
1 marks
Which row correctly pairs a mixture with its most suitable separation technique?
  1. A.Mixture: ethanol and water | Separation technique: simple distillation
  2. B.Mixture: glycine and alanine | Separation technique: chromatography
  3. C.Mixture: copper(II) sulfate from its aqueous solution | Separation technique: filtration
  4. D.Mixture: sand and salt | Separation technique: fractional distillation
Show answer & marking scheme

Worked solution

- Ethanol and water are miscible liquids with close boiling points, so they require **fractional distillation** (simple distillation is not sufficient).
- Glycine and alanine are amino acids. Since they are soluble but have different properties, they can be separated by **paper chromatography**.
- Copper(II) sulfate is soluble in water, so filtration cannot separate it from its aqueous solution. **Crystallisation** or evaporation is required.
- Sand and salt can be separated by dissolving in water, **filtering** to remove sand, and then **evaporating/crystallising** to recover the salt (fractional distillation is completely incorrect).

Marking scheme

Award 1 mark for the correct answer B.
Question 25 · multiple-choice
1 marks
A substance Y is heated from \(-10\text{ }^\circ\text{C}\) to \(110\text{ }^\circ\text{C}\). It melts at \(-5\text{ }^\circ\text{C}\) and boils at \(85\text{ }^\circ\text{C}\). Which statement describes the arrangement and movement of the particles of Y at \(25\text{ }^\circ\text{C}\)?
  1. A.They are regularly arranged and vibrate about fixed positions.
  2. B.They are randomly arranged and can slide past each other.
  3. C.They are randomly arranged and are far apart moving rapidly.
  4. D.They are regularly arranged and can move freely in any direction.
Show answer & marking scheme

Worked solution

Since \(25\text{ }^\circ\text{C}\) is between the melting point (\(-5\text{ }^\circ\text{C}\)) and boiling point (\(85\text{ }^\circ\text{C}\)), substance Y is a liquid at this temperature. The particles in a liquid are randomly arranged and can move or slide past each other.

Marking scheme

Award 1 mark for correct option B.
Question 26 · multiple-choice
1 marks
An atom of element Z has 16 protons, 18 neutrons and 16 electrons. What is the nucleon number of this atom and to which group of the Periodic Table does it belong?
  1. A.nucleon number 32, Group VI
  2. B.nucleon number 34, Group VI
  3. C.nucleon number 32, Group VIII
  4. D.nucleon number 34, Group VIII
Show answer & marking scheme

Worked solution

The nucleon number is the sum of protons and neutrons: \(16 + 18 = 34\). The element has 16 protons, so its electronic configuration is 2, 8, 6. Since it has 6 electrons in its outer shell, it belongs to Group VI.

Marking scheme

Award 1 mark for correct option B.
Question 27 · multiple-choice
1 marks
Which row describes the number of shared pairs of electrons and the total number of non-bonding outer shell electrons in a molecule of methanol, \(\text{CH}_3\text{OH}\)?
  1. A.shared pairs = 5, non-bonding outer electrons = 4
  2. B.shared pairs = 5, non-bonding outer electrons = 8
  3. C.shared pairs = 6, non-bonding outer electrons = 4
  4. D.shared pairs = 6, non-bonding outer electrons = 8
Show answer & marking scheme

Worked solution

In \(\text{CH}_3\text{OH}\), there are 5 single covalent bonds (3 \(\text{C–H}\) bonds, 1 \(\text{C–O}\) bond, and 1 \(\text{O–H}\) bond). Each single covalent bond consists of one shared pair of electrons, giving 5 shared pairs. Oxygen (Group VI) has 6 outer shell electrons: 2 are used in bonding, leaving 4 non-bonding outer electrons (2 lone pairs). Carbon uses all 4 outer electrons in bonding, and hydrogen uses its only electron. Thus, there are 4 non-bonding outer electrons in total.

Marking scheme

Award 1 mark for correct option A.
Question 28 · multiple-choice
1 marks
Molten lead(II) bromide is electrolysed using inert carbon electrodes. What is observed at each electrode?
  1. A.anode: brown gas; cathode: grey liquid
  2. B.anode: grey liquid; cathode: brown gas
  3. C.anode: silver liquid; cathode: orange gas
  4. D.anode: orange gas; cathode: bubbles of colourless gas
Show answer & marking scheme

Worked solution

During the electrolysis of molten lead(II) bromide: at the cathode (negative electrode), lead ions (\(\text{Pb}^{2+}\)) are reduced to lead metal, which appears as a grey liquid. At the anode (positive electrode), bromide ions (\(\text{Br}^-\)) are oxidised to bromine gas, which is observed as a brown gas.

Marking scheme

Award 1 mark for correct option A.
Question 29 · multiple-choice
1 marks
The reaction between nitrogen and hydrogen to form ammonia is exothermic.

$$\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) \quad \Delta H = -92\text{ kJ/mol}$$

Which statement about this reaction is correct?
  1. A.The total energy needed to break the bonds in the reactants is greater than the total energy released when new bonds are formed.
  2. B.The temperature of the surroundings decreases as the reaction proceeds.
  3. C.The energy level of the reactants is higher than the energy level of the products in a reaction pathway diagram.
  4. D.The activation energy for the forward reaction is larger than the activation energy for the reverse reaction.
Show answer & marking scheme

Worked solution

In an exothermic reaction, heat is released because the products are more stable and lower in energy than the reactants. Therefore, in a reaction pathway diagram, the energy level of the reactants is higher than the energy level of the products. Options A and B describe endothermic behavior, and D is incorrect because exothermic reactions have lower activation energy in the forward direction compared to the reverse.

Marking scheme

Award 1 mark for correct option C.
Question 30 · multiple-choice
1 marks
Equal masses of four metals, W, X, Y and Z, are separately added to equal volumes of dilute hydrochloric acid of the same concentration. The temperature of each mixture is measured before the metal is added and after the reaction is complete. The results are shown in the table.

| Metal | Initial temperature / $^\circ\text{C}$ | Final temperature / $^\circ\text{C}$ |
| :--- | :--- | :--- |
| W | 20 | 25 |
| X | 20 | 20 |
| Y | 20 | 41 |
| Z | 20 | 32 |

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

Worked solution

The temperature rise (\(\Delta T\)) indicates the rate and vigor of the reaction with acid:
- W: \(25 - 20 = 5\text{ }^\circ\text{C}\)
- X: \(20 - 20 = 0\text{ }^\circ\text{C}\) (no reaction)
- Y: \(41 - 20 = 21\text{ }^\circ\text{C}\) (highest temperature rise)
- Z: \(32 - 20 = 12\text{ }^\circ\text{C}\)

Ordering from greatest temperature change to lowest gives: Y $\rightarrow$ Z $\rightarrow$ W $\rightarrow$ X.

Marking scheme

Award 1 mark for correct option B.
Question 31 · multiple-choice
1 marks
An organic compound has the molecular formula \(\text{C}_5\text{H}_{10}\). Which statement about this compound is correct?
  1. A.It is a saturated hydrocarbon and belongs to the alkane homologous series.
  2. B.It reacts with aqueous bromine to cause a colour change from colourless to orange.
  3. C.It can undergo addition polymerisation to form a polymer.
  4. D.It has a lower boiling point than butane, \(\text{C}_4\text{H}_{10}\).
Show answer & marking scheme

Worked solution

The molecular formula \(\text{C}_5\text{H}_{10}\) fits the general formula \(\text{C}_n\text{H}_{2n}\) for alkenes. Alkenes contain a double carbon-carbon bond (unsaturation) and can undergo addition polymerisation (C is correct). Alkenes change the colour of aqueous bromine from orange to colourless (B is incorrect). They are unsaturated (A is incorrect) and have a higher boiling point than butane (\(\text{C}_4\text{H}_{10}\)) because they are larger molecules with stronger intermolecular forces (D is incorrect).

Marking scheme

Award 1 mark for correct option C.
Question 32 · multiple-choice
1 marks
A student performs a paper chromatography experiment to identify the food dyes present in a green sweet. The chromatogram shows two spots, one blue and one yellow. The solvent front travelled \(8.0\text{ cm}\) from the baseline. The blue spot travelled \(6.0\text{ cm}\) and the yellow spot travelled \(3.2\text{ cm}\). What is the \(R_f\) value of the yellow dye?
  1. A.0.40
  2. B.0.53
  3. C.0.75
  4. D.2.50
Show answer & marking scheme

Worked solution

The \(R_f\) value is calculated by dividing the distance travelled by the spot by the distance travelled by the solvent front:

$$R_f = \frac{\text{distance travelled by dye}}{\text{distance travelled by solvent}} = \frac{3.2\text{ cm}}{8.0\text{ cm}} = 0.40$$

Marking scheme

Award 1 mark for correct option A.
Question 33 · multiple-choice
1 marks
The melting points and boiling points of four substances, A, B, C and D, are shown in the table.

$$\begin{array}{|c|c|c|} \hline \text{Substance} & \text{Melting point / } ^\circ\text{C} & \text{Boiling point / } ^\circ\text{C} \\ \hline \text{A} & -95 & 36 \\ \hline \text{B} & -115 & -85 \\ \hline \text{C} & 0 & 100 \\ \hline \text{D} & -5 & 80 \\ \hline \end{array}$$

Which substance is a liquid at \(-10\ ^\circ\text{C}\) and a gas at \(50\ ^\circ\text{C}\)?
  1. A.A
  2. B.B
  3. C.C
  4. D.D
Show answer & marking scheme

Worked solution

At \(-10\ ^\circ\text{C}\), substance A is a liquid because its temperature is above its melting point (\(-95\ ^\circ\text{C}\)) but below its boiling point (\(36\ ^\circ\text{C}\)). At \(50\ ^\circ\text{C}\), substance A is a gas because its temperature is above its boiling point (\(36\ ^\circ\text{C}\)).

Marking scheme

1 mark for the correct option A.
Question 34 · multiple-choice
1 marks
An atom of element X contains 18 protons, 22 neutrons and 18 electrons. Which row correctly identifies the nucleon number of X, its Group in the Periodic Table, and the number of electron shells in X?

$$\begin{array}{|c|c|c|c|} \hline & \text{nucleon number} & \text{Group number} & \text{number of electron shells} \\ \hline \text{A} & 40 & \text{VIII} & 3 \\ \hline \text{B} & 40 & \text{VI} & 3 \\ \hline \text{C} & 22 & \text{VIII} & 4 \\ \hline \text{D} & 18 & \text{VIII} & 3 \\ \hline \end{array}$$
  1. A.A
  2. B.B
  3. C.C
  4. D.D
Show answer & marking scheme

Worked solution

The nucleon number is the sum of protons and neutrons: \(18 + 22 = 40\). The electron configuration of an atom with 18 electrons is 2,8,8. This configuration has 8 valence electrons, placing it in Group VIII (noble gases), and uses 3 electron shells.

Marking scheme

1 mark for the correct option A.
Question 35 · multiple-choice
1 marks
Which statement about the formation of ions in magnesium oxide is correct?
  1. A.Each magnesium atom loses two electrons to form a \(Mg^{2+}\) ion.
  2. B.Each oxygen atom loses two electrons to form an \(O^{2-}\) ion.
  3. C.A magnesium ion has the same electronic configuration as an argon atom.
  4. D.The ions are held together by weak covalent bonds.
Show answer & marking scheme

Worked solution

Magnesium is in Group II and has 2 valence electrons, which it loses to achieve a stable octet, forming a \(Mg^{2+}\) ion. Oxygen is in Group VI and gains 2 electrons to form an \(O^{2-}\) ion. The magnesium ion \(Mg^{2+}\) has 10 electrons (configuration 2,8), which is the same as neon, not argon. The ions are held together by strong electrostatic forces (ionic bonding), not covalent bonds.

Marking scheme

1 mark for the correct option A.
Question 36 · multiple-choice
1 marks
Concentrated aqueous copper(II) chloride is electrolysed using inert carbon electrodes. Which products are formed at each electrode?
  1. A.Cathode: copper; Anode: chlorine
  2. B.Cathode: hydrogen; Anode: chlorine
  3. C.Cathode: copper; Anode: oxygen
  4. D.Cathode: hydrogen; Anode: oxygen
Show answer & marking scheme

Worked solution

During the electrolysis of concentrated aqueous copper(II) chloride, \(Cu^{2+}\) ions are discharged at the cathode because copper is less reactive than hydrogen, depositing copper metal. At the anode, chloride ions (\(Cl^-\)) are discharged because the solution is concentrated, producing chlorine gas.

Marking scheme

1 mark for the correct option A.
Question 37 · multiple-choice
1 marks
The reaction of hydrogen with chlorine to form hydrogen chloride gas is exothermic.

\[H_2(g) + Cl_2(g) \rightarrow 2HCl(g)\]

The bond energies are shown in the table.

$$\begin{array}{|c|c|} \hline \text{Bond} & \text{Bond energy / kJ/mol} \\ \hline \text{H–H} & 436 \\ \hline \text{Cl–Cl} & 242 \\ \hline \text{H–Cl} & 431 \\ \hline \end{array}$$

What is the overall energy change for this reaction?
  1. A.-184 kJ/mol
  2. B.+184 kJ/mol
  3. C.-247 kJ/mol
  4. D.+247 kJ/mol
Show answer & marking scheme

Worked solution

Energy absorbed to break bonds: \(1 \times \text{H–H} + 1 \times \text{Cl–Cl} = 436 + 242 = 678\text{ kJ/mol}\). Energy released when bonds are formed: \(2 \times \text{H–Cl} = 2 \times 431 = 862\text{ kJ/mol}\). Overall energy change = Energy absorbed - Energy released = \(678 - 862 = -184\text{ kJ/mol}\).

Marking scheme

1 mark for the correct option A.
Question 38 · multiple-choice
1 marks
A student tests a gaseous hydrocarbon to determine whether it is an alkane or an alkene. Which test and observation would confirm that the hydrocarbon is an alkene?
  1. A.Add aqueous bromine; the orange colour remains.
  2. B.Add aqueous bromine; the colour changes from orange to colourless.
  3. C.Ignite the gas; it burns with a clean blue flame.
  4. D.Bubble the gas through water; the water turns cloudy.
Show answer & marking scheme

Worked solution

Alkenes are unsaturated hydrocarbons containing a carbon-carbon double bond (\(C=C\)). They undergo an addition reaction with aqueous bromine (bromine water), which decolourises it (turning it from orange to colourless). Alkanes do not react with bromine water in the absence of UV light, so the orange colour would remain.

Marking scheme

1 mark for the correct option B.
Question 39 · multiple-choice
1 marks
A student wants to investigate the rate of reaction between calcium carbonate and dilute hydrochloric acid by measuring the volume of gas produced over time. Which set of apparatus is most suitable for this experiment?
  1. A.electronic balance and a thermometer
  2. B.gas syringe, flask, and a stopwatch
  3. C.burette, pipette, and a pH meter
  4. D.condenser, beaker, and a Bunsen burner
Show answer & marking scheme

Worked solution

The reaction produces carbon dioxide gas. To measure the volume of gas produced over time, a flask is needed to contain the reaction, a gas syringe is used to collect and measure the gas volume, and a stopwatch is required to measure time intervals.

Marking scheme

1 mark for the correct option B.
Question 40 · multiple-choice
1 marks
An unknown aqueous solution X is tested.

- Addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess.
- Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate.

What is the identity of X?
  1. A.iron(II) sulfate
  2. B.iron(III) sulfate
  3. C.iron(II) chloride
  4. D.copper(II) sulfate
Show answer & marking scheme

Worked solution

A green precipitate with aqueous sodium hydroxide that is insoluble in excess indicates the presence of iron(II) ions, \(Fe^{2+}\). A white precipitate with barium nitrate in acidic conditions indicates the presence of sulfate ions, \(SO_4^{2-}\). Thus, the salt is iron(II) sulfate.

Marking scheme

1 mark for the correct option A.

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Paper 42

Answer all structured questions on the spaces provided. Show working where necessary and use appropriate units.
6 Question · 80 marks
Question 1 · structured
13 marks
Hydrazine, \(N_2H_4\), is used as a rocket propellant. It reacts with oxygen, \(O_2\), to form nitrogen gas, \(N_2\), and water vapor, \(H_2O\).

(a) Draw the displayed formula of a molecule of hydrazine, \(N_2H_4\). [2]

(b) Write a balanced chemical equation for the combustion of hydrazine. [2]

(c) This reaction is exothermic. Draw an energy level diagram for this reaction, labelling the reactants, products, activation energy (\(E_a\)), and enthalpy change (\(\Delta H\)). [4]

(d) Use the following bond energies to calculate the enthalpy change (\(\Delta H\)) for the reaction in kJ/mol.
- N-H: 391 kJ/mol
- N-N: 160 kJ/mol
- O=O: 498 kJ/mol
- N\equiv N: 945 kJ/mol
- O-H: 463 kJ/mol [5]
Show answer & marking scheme

Worked solution

(a) Hydrazine displayed formula: H-N(H)-N(H)-H showing N-N single bond and two N-H single bonds on each nitrogen.
(b) Balanced equation: \(N_2H_4 + O_2 \rightarrow N_2 + 2H_2O\).
(c) Reactants are higher than products. \(E_a\) is the upward arrow from reactants to the peak. \(\Delta H\) is the downward arrow from reactants to products.
(d) Bonds broken: 4 \(\times\) N-H (1564) + 1 \(\times\) N-N (160) + 1 \(\times\) O=O (498) = 2222 kJ. Bonds formed: 1 \(\times\) N\equiv N (945) + 4 \(\times\) O-H (1852) = 2797 kJ. \(\Delta H = 2222 - 2797 = -575\) kJ/mol.

Marking scheme

(a) M1: single N-N bond [1]; M2: four single N-H bonds [1].
(b) M1: correct reactants and products [1]; M2: balanced equation [1].
(c) M1: reactant line higher than product line [1]; M2: y-axis labeled 'Energy' / 'Enthalpy' and x-axis labeled 'Progress of reaction' [1]; M3: activation energy labeled with upward arrow [1]; M4: enthalpy change labeled with downward arrow [1].
(d) M1: Energy required to break bonds = 2222 kJ [1]; M2: Energy released in making bonds = 2797 kJ [1]; M3: subtraction (broken - formed) [1]; M4: -575 [1]; M5: kJ/mol [1].
Question 2 · structured
14 marks
Aqueous sodium sulfate, \(Na_2SO_4\), is electrolysed using inert platinum electrodes.

(a) Describe what is meant by the term electrolysis. [3]

(b) State the observations at both electrodes during this electrolysis.
- cathode (negative electrode) [1]
- anode (positive electrode) [1]

(c) Write the ionic half-equation for the reaction at the cathode. [2]

(d) Write the ionic half-equation for the reaction at the anode. [3]

(e) Describe how the pH of the solution changes around each electrode during electrolysis. Explain your answer.
- near the cathode [2]
- near the anode [2]
Show answer & marking scheme

Worked solution

(a) Electrolysis is the decomposition of an ionic compound when molten or in aqueous solution by the passage of an electric current.
(b) Bubbles of a colorless gas are observed at both the cathode and the anode.
(c) Cathode: \(2H^+ + 2e^- \rightarrow H_2\) or \(2H_2O + 2e^- \rightarrow H_2 + 2OH^-\).
(d) Anode: \(4OH^- \rightarrow O_2 + 2H_2O + 4e^-\).
(e) Near the cathode: pH increases because \(H^+\) ions are discharged, leaving an excess of \(OH^-\) ions. Near the anode: pH decreases because \(OH^-\) ions are discharged, leaving an excess of \(H^+\) ions.

Marking scheme

(a) M1: breakdown / decomposition [1]; M2: of an ionic compound when molten or in solution [1]; M3: by electricity / electric current [1].
(b) Cathode: bubbles of colorless gas [1]; Anode: bubbles of colorless gas [1].
(c) M1: reactant side correct (\(H^+\) and \(e^-\)) [1]; M2: product side correct (\(H_2\)) and balanced [1].
(d) M1: reactant side correct (\(OH^-\)) [1]; M2: product side correct (\(O_2\) and \(H_2O\)) [1]; M3: fully balanced with electrons [1].
(e) Cathode: pH increases [1]; because hydroxide ions (\(OH^-\)) are left in excess / hydrogen ions are removed [1]. Anode: pH decreases [1]; because hydrogen ions (\(H^+\)) are left in excess / hydroxide ions are removed [1].
Question 3 · structured
13 marks
Propene, \(C_3H_6\), is an unsaturated hydrocarbon.

(a) Define the term unsaturated. [1]

(b) Describe a test to show that propene is unsaturated. Include the observations. [2]

(c) Propene can be converted into propanol by reacting with steam.
- (i) State the name of this type of reaction. [1]
- (ii) State the catalyst and conditions required for this reaction. [3]

(d) Propene undergoes addition polymerisation to form poly(propene).
- (i) Explain what is meant by addition polymerisation. [2]
- (ii) Draw the structure of a single repeat unit of poly(propene). [2]

(e) Draw the displayed formula of propene. [2]
Show answer & marking scheme

Worked solution

(a) Unsaturated means containing at least one carbon-carbon double bond.
(b) Test: Add aqueous bromine / bromine water. Observation: changes colour from orange/brown to colourless.
(c) (i) Hydration / addition reaction. (ii) Acid catalyst (phosphoric acid), 300 °C, 60 atm.
(d) (i) Monomers joining together to form a long chain molecule (polymer) as the only product.
(ii) Single C-C bond in the chain with continuation bonds on both ends: -[CH(CH3)-CH2]-.
(e) Displayed formula of propene contains a double C=C bond, three C-H on double bonded carbons, one single C-C bond, and three C-H on the methyl carbon.

Marking scheme

(a) M1: Contains at least one carbon-carbon double bond (C=C) [1].
(b) M1: add bromine water / aqueous bromine [1]; M2: turns from orange/brown to colourless [1].
(c) (i) M1: hydration / addition [1]. (ii) M1: phosphoric acid / acid catalyst [1]; M2: 300 °C [1]; M3: 60 atm [1].
(d) (i) M1: small molecules (monomers) join to form a large molecule (polymer) [1]; M2: only product / no small molecules lost [1]. (ii) M1: single C-C bond in chain with continuation bonds [1]; M2: correct substituents (one CH3, three H) [1].
(e) M1: C=C double bond and C-C single bond correctly shown [1]; M2: all hydrogens and valencies correct [1].
Question 4 · structured
14 marks
Bromine has two naturally occurring isotopes, \(^{79}\text{Br}\) and \(^{81}\text{Br}\).

(a) State the meaning of the term isotopes. [2]

(b) Complete the table below to show the number of protons, neutrons, and electrons in a bromine-79 atom and a bromine-81 bromide ion (\(^{81}\text{Br}^-\)).

| Particle | Protons | Neutrons | Electrons |
| --- | --- | --- | --- |
| \(^{79}\text{Br}\) atom | | | |
| \(^{81}\text{Br}^-\) ion | | | | [4]

(c) The relative atomic mass of this sample of bromine is 79.90. Calculate the percentage abundance of each isotope. Show your working. [4]

(d) Bromine is a Group VII element.
- (i) State the physical state of bromine at room temperature and pressure. [1]
- (ii) Describe what is observed when aqueous bromine is added to aqueous potassium iodide. Write the ionic equation for this reaction. [3]
Show answer & marking scheme

Worked solution

(a) Atoms of the same element with the same number of protons but different numbers of neutrons.
(b) Bromine-79 atom: Protons = 35, Neutrons = 44, Electrons = 35.
Bromine-81 bromide ion: Protons = 35, Neutrons = 46, Electrons = 36.
(c) Let abundance of \(^{79}\text{Br}\) be \(x\%\). \((79x + 81(100-x))/100 = 79.90 \rightarrow 79x + 8100 - 81x = 7990 \rightarrow -2x = -110 \rightarrow x = 55\). Abundances: \(^{79}\text{Br} = 55\%\) and \(^{81}\text{Br} = 45\%\).
(d) (i) Liquid.
(ii) Solution turns from colorless to brown. Ionic equation: \(Br_2 + 2I^- \rightarrow 2Br^- + I_2\).

Marking scheme

(a) M1: same proton number / number of protons [1]; M2: different nucleon number / number of neutrons [1].
(b) M1: protons both 35 [1]; M2: neutrons in Br-79 = 44 [1]; M3: neutrons in Br-81 = 46 [1]; M4: electrons 35 and 36 respectively [1].
(c) M1: setting up equation: \(79x + 81(100-x) = 7990\) [1]; M2: expansion: \(-2x = -110\) [1]; M3: \(x = 55\%\) [1]; M4: \(^{81}\text{Br} = 45\%\) [1].
(d) (i) M1: liquid [1]. (ii) M1: turns brown / yellow-brown [1]; M2: correct reactants in ionic equation (\(Br_2 + 2I^-\)) [1]; M3: correct products in ionic equation (\(2Br^- + I_2\)) [1].
Question 5 · structured
13 marks
Zinc is extracted from its main ore, zinc blende, which contains zinc sulfide, \(ZnS\).

(a) Zinc blende is roasted in air to produce zinc oxide, \(ZnO\), and sulfur dioxide.
- (i) Write the balanced chemical equation for this reaction. [2]
- (ii) State one environmental consequence of releasing sulfur dioxide into the atmosphere. [1]

(b) Zinc oxide is then reduced in a blast furnace using carbon monoxide.
- (i) Write the chemical equation for the reduction of zinc oxide by carbon monoxide. [2]
- (ii) Explain why zinc is obtained as a vapour from the top of the blast furnace. [2]

(c) Brass is an alloy of zinc and copper.
- (i) Explain why alloys like brass are stronger than pure metals in terms of their structures. [3]
- (ii) State one use of brass. [1]

(d) Explain why zinc, rather than copper, is used for galvanising iron. [2]
Show answer & marking scheme

Worked solution

(a) (i) \(2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2\). (ii) Acid rain.
(b) (i) \(ZnO + CO \rightarrow Zn + CO_2\). (ii) The temperature in the blast furnace is higher than the boiling point of zinc, so it evaporates.
(c) (i) Different sized atoms distort the regular arrangement / lattice of atoms. Layers cannot easily slide over each other.
(ii) Musical instruments / electrical fittings / screws.
(d) Zinc is more reactive than iron, so it undergoes sacrificial protection (loses electrons preferentially to protect iron).

Marking scheme

(a) (i) M1: correct formulas [1]; M2: balanced equation [1]. (ii) M1: acid rain / respiratory problems [1].
(b) (i) M1: correct reactants and products [1]; M2: balanced equation [1]. (ii) M1: zinc has a low boiling point / furnace temperature is high [1]; M2: zinc boils / vaporises [1].
(c) (i) M1: atoms of different sizes [1]; M2: distorts the regular layer structure [1]; M3: layers cannot easily slide over each other [1]. (ii) M1: musical instruments / ornaments / coins / screws [1].
(d) M1: zinc is more reactive than iron [1]; M2: sacrificial protection / zinc reacts/corrodes instead of iron [1].
Question 6 · structured
13 marks
An experiment is carried out to find the formula of a hydrated salt of copper, \(CuSO_4 \cdot yH_2O\).

(a) A student heats 6.24 g of hydrated copper(II) sulfate. After heating to constant mass, 3.99 g of anhydrous copper(II) sulfate, \(CuSO_4\), remains.
- (i) Calculate the mass of water of crystallisation lost. [1]
- (ii) Calculate the number of moles of anhydrous copper(II) sulfate remaining. (\(M_r\) of \(CuSO_4\) = 159.5) [2]
- (iii) Calculate the number of moles of water lost. (\(M_r\) of \(H_2O\) = 18.0) [2]
- (iv) Determine the value of \(y\) in the formula \(CuSO_4 \cdot yH_2O\). [2]

(b) In a different reaction, 50.0 cm³ of 0.200 mol/dm³ hydrochloric acid is neutralized by 25.0 cm³ of aqueous sodium hydroxide.
- (i) Write the balanced chemical equation for this reaction. [2]
- (ii) Calculate the concentration, in mol/dm³, of the sodium hydroxide solution. [2]

(c) Deduce the empirical formula of a compound containing 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. [2]
Show answer & marking scheme

Worked solution

(a) (i) Mass of water lost = 6.24 g - 3.99 g = 2.25 g.
(ii) Moles of \(CuSO_4\) = 3.99 / 159.5 = 0.025 mol.
(iii) Moles of \(H_2O\) = 2.25 / 18.0 = 0.125 mol.
(iv) ratio of \(H_2O : CuSO_4\) = 0.125 / 0.025 = 5. So \(y = 5\).
(b) (i) \(HCl + NaOH \rightarrow NaCl + H_2O\).
(ii) Moles of HCl = 0.050 \(\times\) 0.200 = 0.010 mol. Since the ratio is 1:1, moles of NaOH = 0.010 mol. Concentration of NaOH = 0.010 / 0.025 = 0.400 mol/dm³.
(c) Carbon: 40.0/12 = 3.33 mol. Hydrogen: 6.7/1 = 6.7 mol. Oxygen: 53.3/16 = 3.33 mol. Ratio C:H:O = 1:2:1. Empirical formula is \(CH_2O\).

Marking scheme

(a) (i) M1: 2.25 g [1]. (ii) M1: dividing mass by Mr [1]; M2: 0.025 mol [1]. (iii) M1: dividing mass by 18 [1]; M2: 0.125 mol [1]. (iv) M1: mole ratio 0.125 / 0.025 [1]; M2: y = 5 [1].
(b) (i) M1: correct reactants and products [1]; M2: balanced [1]. (ii) M1: moles of HCl = 0.010 mol [1]; M2: concentration = 0.400 mol/dm³ [1].
(c) M1: correct moles calculation for all three elements (C=3.33, H=6.7, O=3.33) [1]; M2: empirical formula \(CH_2O\) [1].

Paper 62

Answer all questions. Read instructions carefully, interpret diagrams, construct graphical data, and design a practical plan.
4 Question · 40 marks
Question 1 · structured
10 marks
A student investigates the thermal decomposition of a green copper salt, basic copper(II) carbonate, \(\text{CuCO}_3\cdot\text{Cu(OH)}_2\), to produce copper(II) oxide, carbon dioxide, and water vapor.

(a) State the name of the apparatus used to heat the solid. [1]

(b) Describe the color change of the solid in the test-tube during heating. [2]

(c) The carbon dioxide gas is bubbled into a test-tube containing limewater. State the observation in this test-tube. [1]

(d) Explain why the delivery tube must be removed from the limewater immediately when heating is stopped. [2]

(e) Suggest one safety precaution, other than wearing safety glasses, that the student should take during the heating process. Give a reason for your answer. [2]

(f) Suggest how the apparatus could be modified to collect and measure the volume of carbon dioxide gas produced. [2]
Show answer & marking scheme

Worked solution

(a) Bunsen burner
(b) Green to black
(c) Limewater turns cloudy / milky / white precipitate
(d) To prevent suck-back of liquid into the hot tube as it cools, which would crack the hot glass tube.
(e) Use heat-resistant gloves / test-tube holder because the tube becomes very hot; OR point the mouth of the tube away from people because hot solid may spurt out.
(f) Connect the test-tube containing the carbonate to a gas syringe using a delivery tube.

Marking scheme

*(a) 1 mark:*
- Bunsen burner (accept utility burner / blowtorch).

*(b) 2 marks:*
- Green [1]
- to black [1]

*(c) 1 mark:*
- Turns cloudy / milky / white precipitate / chalky (reject: bubbles / effervescence as this is a physical test result, not the indicator of identity).

*(d) 2 marks:*
- To prevent suck-back / water moving back into the hot tube [1]
- which would crack/shatter the hot test-tube [1]

*(e) 2 marks:*
- Point the mouth of the test-tube away from self/others [1]
- to prevent injury from hot solids spurting out [1]
- OR use heat-resistant gloves/holders [1] because the apparatus becomes extremely hot [1].

*(f) 2 marks:*
- Connect to a gas syringe [1]
- via a delivery tube / sealed bung [1]
- OR collect over water [1] into an inverted measuring cylinder [1].
Question 2 · structured
10 marks
A student investigates the temperature change during the displacement reaction between zinc powder and aqueous copper(II) sulfate.

\(50.0\text{ cm}^3\) of \(0.20\text{ mol/dm}^3\) copper(II) sulfate solution is placed in a beaker. The temperature is recorded every 30 seconds. Excess zinc powder is added at 60 seconds.

The results are shown in the table.

$$\begin{array}{|c|c|}
\hline
\text{Time / s} & \text{Temperature / }^\circ\text{C} \\
\hline
0 & 20.0 \\
\hline
30 & 20.0 \\
\hline
60 & 20.0 \\
\hline
90 & 28.5 \\
\hline
120 & 33.0 \\
\hline
150 & 35.5 \\
\hline
180 & 34.5 \\
\hline
210 & 33.5 \\
\hline
240 & 32.5 \\
\hline
270 & 31.5 \\
\hline
\end{array}$$

(a) Plot the results on a grid of temperature (y-axis) against time (x-axis). [4]

(b) Draw two straight lines of best fit:
- one through the points from 0 to 60 seconds
- one through the cooling points from 150 to 270 seconds.
Extrapolate both lines to 60 seconds. [2]

(c) Use your graph to determine the maximum temperature reached at 60 seconds and calculate the maximum temperature change, \(\Delta T\). Show clearly on your graph how you obtained your answer. [2]

(d) Explain why zinc powder was used instead of zinc granules in this experiment. [1]

(e) Suggest why the experiment would be more accurate if carried out in a polystyrene cup rather than a glass beaker. [1]
Show answer & marking scheme

Worked solution

(a) Plot points correctly.
(b) Extrapolate the flat line at 20.0 °C and the cooling line to 60 s where they meet at 38.5 °C.
(c) Maximum temperature at 60 s = 38.5 °C. Maximum temperature change = 38.5 - 20.0 = 18.5 °C.
(d) Zinc powder has a larger surface area, increasing the rate of reaction.
(e) Polystyrene is a better thermal insulator and reduces heat loss to the surroundings.

Marking scheme

*(a) 4 marks:*
- y-axis labeled with temperature/°C and x-axis with time/s [1]
- Linear scales starting at appropriate values, where plotted points occupy more than half the grid [1]
- All points plotted correctly to within half a small square [2] (1 mark if 1-2 points are incorrect).

*(b) 2 marks:*
- One horizontal line through the first three points and one straight line through the cooling points (150 s to 270 s) [1]
- Both lines extrapolated to 60 s [1]

*(c) 2 marks:*
- Temperature read at 60 s on the extrapolated cooling line = 38.5 °C (or from candidate's graph) [1]
- Temperature change calculated correctly: \(\Delta T = 18.5\text{ }^\circ\text{C}\) [1]

*(d) 1 mark:*
- Larger surface area / reaction is faster / reaches maximum temperature quicker (less heat loss during reaction) [1]

*(e) 1 mark:*
- Polystyrene is a better insulator / reduces heat loss [1]
Question 3 · structured
10 marks
A student is provided with a green crystalline solid, salt F, which contains one cation and two anions.

The student dissolves Solid F in distilled water to make Solution F, and carries out the following tests. Complete the expected observations and conclusions.

(a) To the first portion of Solution F, aqueous sodium hydroxide is added dropwise, and then in excess. [2]

(b) (i) The mixture from (a) is warmed gently in a test-tube. A pungent gas is evolved which turns damp red litmus paper blue. Identify the gas. [1]
(ii) State the conclusion that can be made about the cation present in Solid F. [1]

(c) To the second portion of Solution F, dilute nitric acid is added followed by aqueous silver nitrate. No precipitate is formed. State the conclusion about halide ions in Solid F. [1]

(d) To the third portion of Solution F, dilute hydrochloric acid is added followed by aqueous barium chloride. A white precipitate is formed. State the name of the anion identified. [1]

(e) Solid F is known to contain iron(II) ions as its other cation. Identify the three ions present in Solid F. [3]

(f) Write the chemical formula of the gas identified in (b)(i). [1]
Show answer & marking scheme

Worked solution

(a) Dropwise: green precipitate. In excess: insoluble / precipitate remains.
(b) (i) Ammonia
(ii) Ammonium ion is present
(c) Halide ions (chloride, bromide, iodide) are absent
(d) Sulfate
(e) Iron(II) ion, ammonium ion, sulfate ion
(f) NH3

Marking scheme

*(a) 2 marks:*
- dropwise: green precipitate [1]
- excess: insoluble / no change / precipitate remains [1]

*(b)(i) 1 mark:*
- Ammonia / \(\text{NH}_3\) [1]

*(b)(ii) 1 mark:*
- Ammonium (ion) / \(\text{NH}_4^+\) present [1]

*(c) 1 mark:*
- Halide ions / chloride / bromide / iodide absent [1]

*(d) 1 mark:*
- Sulfate / \(\text{SO}_4^{2-}\) [1]

*(e) 3 marks:*
- Iron(II) / \(\text{Fe}^{2+}\) [1]
- Ammonium / \(\text{NH}_4^+\) [1]
- Sulfate / \(\text{SO}_4^{2-}\) [1]

*(f) 1 mark:*
- \(\text{NH}_3\) [1]
Question 4 · structured
10 marks
Vinegar is a solution containing ethanoic acid, which is a weak carboxylic acid.

Plan an investigation to determine which of three commercial brands of vinegar, X, Y, and Z, is the most concentrated.

You are provided with:
- samples of vinegar brands X, Y, and Z
- aqueous sodium hydroxide of concentration \(0.10\text{ mol/dm}^3\)
- phenolphthalein indicator
- common laboratory apparatus.

Your plan should include:
- the apparatus you would use
- the step-by-step practical procedure
- the measurements you would make
- how you would use your results to determine which vinegar is the most concentrated. [10]
Show answer & marking scheme

Worked solution

1. Measure 25.0 cm3 of vinegar X using a volumetric pipette.
2. Transfer the measured volume of vinegar X into a clean conical flask.
3. Add 3-4 drops of phenolphthalein indicator to the conical flask.
4. Fill a clean burette with 0.10 mol/dm3 sodium hydroxide solution.
5. Record the initial volume reading of the burette.
6. Add the sodium hydroxide solution from the burette into the conical flask slowly while swirling the flask.
7. Stop adding sodium hydroxide when the indicator changes color from colorless to a permanent pale pink.
8. Record the final burette reading and calculate the volume (titer) of sodium hydroxide used.
9. Repeat the exact same steps with vinegar Y and vinegar Z using the same volume of vinegar.
10. Compare the volumes of sodium hydroxide used. The brand of vinegar that requires the largest volume of sodium hydroxide is the most concentrated in ethanoic acid.

Marking scheme

*(10 marks: 1 mark for each of the following points up to a maximum of 10):*
- Measure a fixed/known volume of vinegar [1]
- Use a pipette / measuring cylinder (pipette preferred for accuracy) [1]
- Transfer to a conical flask / beaker [1]
- Add phenolphthalein indicator (accept any suitable acid-base indicator) [1]
- Use a burette [1]
- Fill the burette with aqueous sodium hydroxide [1]
- Record initial and final readings / measure volume of NaOH added [1]
- Add NaOH slowly / dropwise near end-point while swirling [1]
- Stop at the first permanent color change / colorless to pink [1]
- Repeat the entire procedure for the other two brands (Y and Z) [1]
- Control variable: use the same volume of vinegar for each test [1]
- Conclusion: the vinegar requiring the largest volume of sodium hydroxide is the most concentrated [1]
*(Max 10 marks)*

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