2025 Cambridge IGCSE Chemistry (0620) Practice Paper with Answers

\(4.2\text{ g}\) of \(\text{MgCO}_3\) corresponds to \(4.2 / 84 = 0.05\text{ mol}\). According to the stoichiometric ratio, \(1\text{ mol}\) of \(\text{MgCO}_3\) produces \(1\text{ mol}\) of \(\text{CO}_2\). Therefore, \(0.05\text{ mol}\) of \(\text{CO}_2\) is produced. Volume = \(0.05 \times 24\text{ dm}^3 = 1.2\text{ dm}^3\), which is equivalent to \(1200\text{ cm}^3\).

1 mark for correct answer A.

Because the forward reaction is exothermic, a lower temperature shifts the equilibrium in the exothermic direction (to the right) to increase the yield of \(\text{SO}_3\). Since there are fewer moles of gas on the product side (2 moles compared to 3 moles on the reactant side), a higher pressure shifts the equilibrium towards the side with fewer gas moles (to the right), further increasing the yield.

1 mark for correct answer B.

In the electrolysis of concentrated aqueous sodium chloride, chloride ions (\(\text{Cl}^-\)) are selectively discharged at the anode to form chlorine gas. Hydrogen ions (\(\text{H}^+\)) from water are discharged at the cathode to form hydrogen gas, because hydrogen is lower in the reactivity series than sodium.

1 mark for correct answer A.

Increasing the concentration increases the number of acid particles per unit volume. This directly increases the frequency of collisions between the reacting particles, leading to a faster rate of reaction.

1 mark for correct answer C.

In this reaction, the oxidation number of iron decreases from \(+3\) to \(0\) (it is reduced), while the oxidation number of carbon increases from \(+2\) in \(\text{CO}\) to \(+4\) in \(\text{CO}_2\) (it is oxidised). Therefore, carbon monoxide acts as the reducing agent because it reduces the iron(III) oxide.

1 mark for correct answer C.

The \(R_f\) value is calculated as: \(R_f = \frac{\text{distance travelled by the substance}}{\text{distance travelled by the solvent front}} = \frac{3.2\text{ cm}}{8.0\text{ cm}} = 0.40\).

1 mark for correct answer A.

Nylon is formed by condensation polymerisation to produce a polyamide containing amide linkages (\(-\text{CONH}-\)). Terylene is formed by condensation polymerisation to produce a polyester containing ester linkages (\(-\text{COO}-\)).

1 mark for correct answer C.

Evaporating the filtrate to dryness (removing all of the water) would yield an anhydrous powder rather than hydrated crystals. To obtain hydrated crystals, the solution should only be heated until it is saturated (the crystallisation point), then allowed to cool.

1 mark for correct answer B.

According to the kinetic theory of gases, decreasing the temperature reduces the kinetic energy of the gas particles, causing them to move closer together and decrease the volume. Increasing the external pressure compresses the gas particles closer together, also decreasing the volume. Therefore, combining a decrease in temperature with an increase in pressure yields the largest decrease in volume.

Award 1 mark for the correct option B.

First, write the balanced equation: \(\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}\). Find the moles of Mg reacting: \(\text{moles of Mg} = \frac{1.2\text{ g}}{24\text{ g/mol}} = 0.05\text{ mol}\). From the stoichiometry, 1 mole of Mg produces 1 mole of \(\text{H}_2\). Therefore, \(0.05\text{ mol}\) of \(\text{H}_2\) is produced. Calculate the volume at r.t.p.: \(\text{Volume} = 0.05\text{ mol} \times 24\text{ dm}^3\text{/mol} = 1.20\text{ dm}^3\).

Award 1 mark for the correct option B.

During the electrolysis of concentrated aqueous sodium chloride, chloride ions (\(\text{Cl}^-\)) are discharged at the positive anode to form chlorine gas (\(\text{Cl}_2\)) because of their high concentration. Hydrogen ions (\(\text{H}^+\)) from water are discharged at the negative cathode to form hydrogen gas (\(\text{H}_2\)) because hydrogen is less reactive than sodium.

Award 1 mark for the correct option A.

Increasing the temperature increases the kinetic energy of the reactant particles, causing more frequent and more energetic collisions. Using smaller pieces of calcium carbonate (higher surface area) increases the number of exposed particles, which increases the frequency of collisions. Both changes increase the reaction rate.

Award 1 mark for the correct option D.

An oxidising agent is a substance that oxidises another substance by accepting electrons (gaining electrons itself). In this reaction, each chlorine atom in the \(\text{Cl}_2\) molecule gains an electron to form a chloride ion (\(\text{Cl}^-\)), so chlorine is reduced and acts as the oxidising agent.

Award 1 mark for the correct option C.

A Brønsted-Lowry base is defined as a proton (\(\text{H}^+\)) acceptor. In the forward reaction, the water molecule (\(\text{H}_2\text{O}\)) accepts a proton from the hydrogen chloride molecule (\(\text{HCl}\)) to form the hydronium ion (\(\text{H}_3\text{O}^+\)). Thus, water acts as a Brønsted-Lowry base.

Award 1 mark for the correct option B.

Reactivity decreases down Group VII: \(\text{chlorine} > \text{bromine} > \text{iodine}\). Bromine is less reactive than chlorine, so it cannot displace chlorine from potassium chloride (no reaction / no change). Bromine is more reactive than iodine, so it displaces iodine from potassium iodide: \(\text{Br}_2\text{(aq)} + 2\text{KI(aq)} \rightarrow 2\text{KBr(aq)} + \text{I}_2\text{(aq)}\). The displaced iodine turns the solution brown.

Award 1 mark for the correct option A.

Structural isomers are defined as compounds that share the exact same molecular formula (the same number and type of atoms) but have different structural formulae (a different arrangement of covalent bonds).

Award 1 mark for the correct option B.

The forward reaction is endothermic (\(\Delta H > 0\)), which means the reverse reaction is exothermic. Placing the mixture in ice-water lowers the temperature. According to Le Chatelier's principle, decreasing the temperature shifts the equilibrium position in the exothermic direction (to the left) to release heat. This increases the concentration of colourless \(\text{N}_2\text{O}_4\) and decreases the concentration of brown \(\text{NO}_2\), making the mixture paler brown.

1 mark for selecting B.

1. The remaining gas after passing through NaOH is unreacted oxygen, which is \(\text{15.0 cm}^3\). Thus, the oxygen that reacted is \(\text{50.0} - \text{15.0} = \text{35.0 cm}^3\).
2. The volume reduction of \(\text{35.0} - \text{15.0} = \text{20.0 cm}^3\) represents the volume of carbon dioxide absorbed by the NaOH. Therefore, \(\text{20.0 cm}^3\) of \(\text{CO}_2\) was produced.
3. Since \(\text{10.0 cm}^3\) of \(\text{C}_x\text{H}_y\) produced \(\text{20.0 cm}^3\) of \(\text{CO}_2\), each molecule of hydrocarbon contains 2 carbon atoms (\(x = 2\)).
4. The combustion equation is: \(\text{C}_2\text{H}_y + (2 + \frac{y}{4})\text{O}_2 \rightarrow 2\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}\).
5. The ratio of reacted oxygen to hydrocarbon is \(\frac{35.0}{10.0} = 3.5\). Thus, \(2 + \frac{y}{4} = 3.5 \Rightarrow \frac{y}{4} = 1.5 \Rightarrow y = 6\).
6. The formula is \(\text{C}_2\text{H}_6\).

1 mark for selecting C.

During the electrolysis of dilute sulfuric acid, \(\text{H}^+\) ions are discharged at the cathode (negative electrode) to form hydrogen gas: \(2\text{H}^+\text{(aq)} + 2\text{e}^- \rightarrow \text{H}_2\text{(g)}\). At the anode (positive electrode), hydroxide ions, \(\text{OH}^-\), from water are discharged in preference to sulfate ions to form oxygen gas and water: \(4\text{OH}^-\text{(aq)} \rightarrow 2\text{H}_2\text{O(l)} + \text{O}_2\text{(g)} + 4\text{e}^-\).

1 mark for selecting C.

\(\Delta H = \text{Energy absorbed in bond breaking} - \text{Energy released in bond making}\)
\(\Delta H = [E(\text{H–H}) + E(\text{F–F})] - [2 \times E(\text{H–F})]\)
\(-542 = [436 + 158] - 2 \times E(\text{H–F})\)
\(-542 = 594 - 2 \times E(\text{H–F})\)
\(2 \times E(\text{H–F}) = 594 + 542 = 1136\)
\(E(\text{H–F}) = 568\text{ kJ/mol}\).

1 mark for selecting B.

Element X (Group II, Period 3) is magnesium (\(\text{Mg}\)), which forms \(\text{Mg}^{2+}\) ions. Element Y (Group VI, Period 2) is oxygen (\(\text{O}\)), which forms \(\text{O}^{2-}\) ions. The resulting ionic compound is magnesium oxide, which has the formula \(XY\) (\(\text{MgO}\)). Since it is an ionic compound, it forms a giant ionic lattice with strong electrostatic forces of attraction between oppositely charged ions, giving it a very high melting point.

1 mark for selecting C.

Experiment 2 uses powder instead of chips, meaning it has a much larger surface area. This increases the frequency of collisions, so the initial rate of reaction is faster. Since calcium carbonate is in excess in both cases, the limiting reactant is the hydrochloric acid. The volume and concentration of hydrochloric acid are identical in both experiments, so the total yield (total volume of carbon dioxide gas produced) remains the same.

1 mark for selecting B.

Melting points increase down Group VII. Therefore, the melting point of astatine (\(x\)) must be significantly higher than that of iodine (\(114^\circ\text{C}\)). Thus, \(x = 300\) is a reasonable value. The colours of the halogens become progressively darker down the group (yellow-green to red-brown to grey-black to black). Therefore, astatine (\(y\)) must be black.

1 mark for selecting C.

Structural isomers have the same molecular formula but different structural arrangements. But-1-ene has the molecular formula \(\text{C}_4\text{H}_8\) and contains one double bond. But-2-ene (\(\text{CH}_3\text{CH}=\text{CHCH}_3\)) also has the molecular formula \(\text{C}_4\text{H}_8\), but the double bond is in a different position, making them structural isomers.

1 mark for selecting B.

Isotopes of the same element have the same number of protons but different numbers of neutrons. Since the mass numbers of the two isotopes are 65 and 63, the difference in the number of neutrons is \(65 - 63 = 2\). Option A is incorrect because the relative atomic mass (63.5) is closer to 63, meaning \(^{63}X\) is more abundant. Option B is incorrect because physical properties depend on mass and differ slightly. Option D is incorrect because chemical properties depend on the electronic configuration, which is identical for isotopes of the same element.

1 mark for the correct option C.

First, calculate the number of moles of \(\text{CaCO}_3\): \(\text{moles} = \frac{\text{mass}}{M_{\text{r}}} = \frac{10.0}{100} = 0.1\text{ mol}\). From the balanced equation: \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\), the mole ratio of \(\text{CaCO}_3\) to \(\text{CO}_2\) is 1:1. Therefore, 0.1 mol of \(\text{CO}_2\) is produced. Finally, calculate the volume of \(\text{CO}_2\): \(\text{volume} = 0.1\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 2.4\text{ dm}^3\).

1 mark for the correct option B.

A catalyst provides an alternative reaction pathway with a lower activation energy. This does not change the kinetic energy of the particles, but it increases the proportion of particles that have enough energy to react (energy greater than or equal to the activation energy), leading to a higher frequency of successful collisions.

1 mark for the correct option B.

During the electrolysis of concentrated aqueous \(\text{NaCl}\), chloride ions (\(\text{Cl}^-\)) are selectively discharged at the anode to form chlorine gas. Hydrogen ions (\(\text{H}^+\)) from water are discharged at the cathode to form hydrogen gas. The remaining sodium ions (\(\text{Na}^+\)) and hydroxide ions (\(\text{OH}^-\)) stay in the solution, making it alkaline (sodium hydroxide).

1 mark for the correct option C.

An exothermic reaction releases thermal energy to the surroundings, causing the temperature of the surroundings to increase. Energetically, the process of bond breaking is endothermic (absorbs energy) and bond making is exothermic (releases energy). For a reaction to be overall exothermic, the energy absorbed to break bonds must be less than the energy released when new bonds are formed.

1 mark for the correct option D.

Since the forward reaction is exothermic, decreasing the temperature shifts the equilibrium in the forward direction to produce more heat, increasing the yield of ammonia. Increasing the pressure shifts the equilibrium to the side with fewer gas molecules (from 4 moles of gas on the left to 2 moles of gas on the right), which also increases the yield of ammonia. Therefore, decreasing the temperature and increasing the pressure both increase the yield.

1 mark for the correct option C.

As you go down Group VII: 1) Melting points and boiling points increase because the molecules get larger, resulting in stronger intermolecular forces. 2) The colors become darker (fluorine is pale yellow, chlorine is pale green, bromine is red-brown, and iodine is grey-black). 3) Reactivity decreases because the outer shell is further from the nucleus, making it harder to attract an incoming electron.

1 mark for the correct option B.

Alkanes have the general formula \(\text{C}_n\text{H}_{2n+2}\). Therefore, \(\text{C}_3\text{H}_8\), \(\text{C}_4\text{H}_{10}\), and \(\text{C}_6\text{H}_{14}\) are all alkanes. \(\text{C}_5\text{H}_{10}\) fits the general formula \(\text{C}_n\text{H}_{2n}\), which represents an alkene, belonging to a different homologous series.

1 mark for the correct option C.

During a phase change, such as condensation, the temperature of a substance remains constant because the energy released by the forming of intermolecular attractions compensates for the cooling. Since temperature is a direct measure of the average kinetic energy of the particles, the average kinetic energy also remains constant during this transition.

1 mark for the correct option B.

First, find the moles of sulfate ions: 1 mole of \(\text{Al}_2(\text{SO}_4)_3\) produces 3 moles of \(\text{SO}_4^{2-}\). Therefore, \(0.05\text{ mol} \times 3 = 0.15\text{ mol}\) of \(\text{SO}_4^{2-}\). Next, calculate the concentration of \(\text{SO}_4^{2-}\) in \(\text{mol/dm}^3\): Volume is \(250\text{ cm}^3 = 0.25\text{ dm}^3\). Concentration = \(0.15\text{ mol} / 0.25\text{ dm}^3 = 0.6\text{ mol/dm}^3\). Calculate the relative formula mass of \(\text{SO}_4^{2-}\): \(M_r = 32 + (4 \times 16) = 96\text{ g/mol}\) . Finally, convert the concentration to \(\text{g/dm}^3\): \(0.6\text{ mol/dm}^3 \times 96\text{ g/mol} = 57.6\text{ g/dm}^3\).

1 mark for the correct option C.

When aqueous copper(II) sulfate is electrolysed using active copper electrodes, the anode (positive electrode) dissolves as copper atoms are oxidised: \(\text{Cu(s)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2\text{e}^-\). At the cathode (negative electrode), copper ions are reduced to copper metal: \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\).

1 mark for the correct option C.

Adding more catalyst (manganese(IV) oxide) provides a greater surface area/amount of catalyst, which increases the rate of reaction. Since the catalyst does not affect the yield, and the volume and concentration of the limiting reactant (hydrogen peroxide) remain unchanged, the final volume of oxygen produced is the same.

1 mark for the correct option D.

Because the forward reaction is exothermic, decreasing the temperature shifts the position of equilibrium to the right (exothermic direction), increasing the yield of Z. There are 3 moles of gaseous reactants and 1 mole of gaseous product; therefore, increasing the pressure shifts the equilibrium to the side with fewer gas moles (the right), also increasing the yield of Z.

1 mark for the correct option C.

A reducing agent reduces another substance and is itself oxidised (its oxidation state increases). In option A, carbon monoxide (CO) reduces iron(III) oxide (\(\text{Fe}_2\text{O}_3\)) to iron (Fe), while the carbon in CO is oxidised from +2 to +4. Therefore, CO is acting as the reducing agent.

1 mark for the correct option A.

Insoluble salts are prepared by precipitation. Mixing soluble salts (barium chloride and sodium sulfate) produces insoluble barium sulfate. Filtration collects the insoluble salt as the residue. Washing with distilled water removes any remaining soluble impurities, and drying yields the pure dry salt.

1 mark for the correct option B.

First, adding ethanol dissolves only naphthalene, leaving sand and sodium chloride as the solid residue. Filtering separates this liquid ethanol solution from the solids. Evaporating the ethanol leaves pure naphthalene. Next, adding water to the solid residue dissolves only sodium chloride, leaving sand insoluble. Filtering collects sand as the residue, and evaporating the water from the filtrate recovers pure sodium chloride.

1 mark for the correct option B.

(a) Isotopes are atoms of the same element that have the same proton number (number of protons) but different nucleon numbers (number of neutrons).

(b)
- For \(^{35}_{17}\text{Cl}\):
- Protons = 17
- Neutrons = 35 - 17 = 18
- Electrons = 17 (neutral atom)
- For \(^{37}_{17}\text{Cl}^-\):
- Protons = 17
- Neutrons = 37 - 17 = 20
- Electrons = 17 + 1 = 18 (has a 1- charge, meaning it gained 1 electron)

(c) \(\text{Relative atomic mass} = \frac{(35 \times 75.0) + (37 \times 25.0)}{100} = \frac{2625 + 925}{100} = 35.5\)

(d) Phosphorus is in Group V and has 5 outer-shell electrons. Chlorine is in Group VII and has 7 outer-shell electrons. In \(\text{PCl}_3\):
- There are three single covalent bonds (each consisting of one shared pair of electrons, one from P and one from Cl).
- The phosphorus atom has one remaining lone pair (2 non-bonding electrons).
- Each chlorine atom has three remaining lone pairs (6 non-bonding electrons).

(e) Phosphorus trichloride has a simple molecular structure. There are weak intermolecular forces (forces of attraction between molecules). Very little thermal energy is needed to overcome these weak forces.

(a)
- Same number of protons / same atomic number [1]
- Different number of neutrons / different mass number / different nucleon number [1]

(b)
- Protons: 17 and 17 (both correct for 1 mark) [1]
- Neutrons: 18 and 20 (both correct for 1 mark) [1]
- Electrons: 17 and 18 (both correct for 1 mark) [1]

(c)
- Correct formula / working shown: \(\frac{(35 \times 75) + (37 \times 25)}{100}\) [1]
- Correct answer: 35.5 [1]

(d)
- One shared pair of electrons between the P atom and each of the three Cl atoms [1]
- One lone pair of electrons (2 non-bonding electrons) on the P atom [1]
- Three lone pairs of electrons (6 non-bonding electrons) on each Cl atom [1]

(e)
- Mention of simple molecular structure / simple covalent molecules [1]
- Weak forces between molecules / weak intermolecular forces [1]
- Little energy required to break these forces [1] (reject: weak covalent bonds)

(a)(i) When sodium is added to water, it floats on the surface, melts into a shiny spherical ball due to the heat of reaction, moves rapidly across the surface, and fizzes (effervesces) as hydrogen gas is produced.
(a)(ii) The chemical equation is: \(2\text{Na(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)} + \text{H}_2\text{(g)}\)
(a)(iii) The resulting solution is sodium hydroxide, which is highly alkaline. Methyl orange turns yellow in alkaline solutions.

(b)(i) Transition metals have much higher densities, higher melting points, and are much harder and stronger than the soft, low-density alkali metals of Group I.
(b)(ii) Iron (Fe) is the catalyst used in the Haber process.

(c)(i) Pure copper has a regular lattice arrangement of metal atoms of the same size. These layers can easily slide over one another when a force is applied. In brass, the zinc atoms are larger than the copper atoms. This disrupts the regular rows and layers, preventing them from sliding over each other easily, making the alloy harder.
(c)(ii) Copper is an excellent conductor of electricity and is highly ductile (can be drawn into wires).

(a)(i)
- Any two from: floats / melts (into a ball) / moves on the surface / effervescence / bubbles / disappears [2]

(a)(ii)
- Correct formulae of reactants and products: \(\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2\) [1]
- Correctly balanced equation: \(2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\) [1]
- Correct state symbols: \(\text{s}\), \(\text{l}\), \(\text{aq}\), \(\text{g}\) [1]

(a)(iii)
- Yellow [1]

(b)(i)
- Any two from: higher density / higher melting point / harder / stronger [2] (do not accept chemical properties)

(b)(ii)
- Iron / \(\text{Fe}\) [1]

(c)(i)
- Pure copper has layers of atoms of the same size that slide over each other [1]
- Zinc atoms are a different size [1]
- Disrupts the regular arrangement of layers / prevents sliding of layers [1]

(c)(ii)
- Good conductor of electricity / ductile [1]

(a)(i) When calcium carbonate is heated, it decomposes to form calcium oxide and carbon dioxide:
\(\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})\)
(a)(ii) This is a thermal decomposition reaction.

(b)(i) \(M_{\text{r}}(\text{CaCO}_3) = 40.0 + 12.0 + (3 \times 16.0) = 100.0\)
\(\text{Moles of } \text{CaCO}_3 = \frac{\text{mass}}{\text{M}_{\text{r}}} = \frac{5.00}{100.0} = 0.050\text{ mol}\)
(b)(ii) From the equation, 1 mole of \(\text{CaCO}_3\) produces 1 mole of \(\text{CO}_2\).
\(\text{Moles of } \text{CO}_2 = 0.050\text{ mol}\)
\(\text{Volume of } \text{CO}_2 = 0.050\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 1.20\text{ dm}^3\)

(c)(i) \(\text{Moles of } \text{HCl} = \text{Concentration} \times \text{Volume (in dm}^3) = 2.00\text{ mol/dm}^3 \times \frac{25.0}{1000}\text{ dm}^3 = 0.0500\text{ mol}\)
(c)(ii) From the equation: 2 moles of \(\text{HCl}\) produce 1 mole of \(\text{CO}_2\).
\(\text{Moles of } \text{CO}_2 = \frac{0.0500}{2} = 0.0250\text{ mol}\)
(c)(iii) \(\text{Volume of } \text{CO}_2 = 0.0250\text{ mol} \times 24000\text{ cm}^3/\text{mol} = 600\text{ cm}^3\)
(c)(iv) The test is to bubble the gas through limewater (aqueous calcium hydroxide). The positive result is that the limewater turns cloudy / milky.

(a)(i)
- \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\) [1]

(a)(ii)
- Thermal decomposition [1]

(b)(i)
- Relative formula mass calculation shown as 100.0 [1]
- \(0.050\text{ mol}\) [1]

(b)(ii)
- Multiply moles of gas by 24.0 [1]
- \(1.2\text{ dm}^3\) (accept \(1.20\text{ dm}^3\)) [1]

(c)(i)
- Working: \(2.00 \times \frac{25.0}{1000}\) [1]
- \(0.05\text{ mol}\) (accept \(0.0500\text{ mol}\)) [1]

(c)(ii)
- \(0.025\text{ mol}\) (accept \(0.0250\text{ mol}\)) [1]

(c)(iii)
- Multiply moles of gas by 24000 [1]
- \(600\text{ cm}^3\) [1]

(c)(iv)
- Bubble gas through limewater [1]
- Limewater turns milky / cloudy [1]

(a)(i) A catalyst provides an alternative reaction pathway with a lower activation energy. Consequently, a greater proportion of the colliding particles have energy greater than or equal to this lower activation energy, resulting in a higher frequency of successful collisions.
(a)(ii) Increasing the temperature increases the kinetic energy of the particles, causing them to move faster. This leads to two effects:
1. The particles collide more frequently (more collisions per unit time).
2. The collisions are more energetic, so a greater proportion of collisions are successful (possess energy greater than or equal to the activation energy).

(b)(i) Typical conditions for the Haber process are:
- Temperature: \(450\ ^\circ\text{C}\)
- Pressure: \(200\text{ atm}\) (or \(20\text{ MPa}\))
- Catalyst: Iron (Fe)
(b)(ii) Increasing the pressure shifts the equilibrium position to the right (towards products). This is because the right-hand side has fewer moles of gas (2 moles of \(\text{NH}_3\)) compared to the left-hand side (4 moles of reactants: 1 mole of \(\text{N}_2\) and 3 moles of \(\text{H}_2\)).
(b)(iii) Increasing the pressure increases the rate of reaction.
(b)(iv) Although a lower temperature shifts the equilibrium to the right (increasing the equilibrium yield of ammonia because the forward reaction is exothermic), a low temperature makes the rate of reaction too slow to be industrially viable.

(a)(i)
- Alternative reaction pathway [1]
- Lower activation energy [1]
- More particles have energy \(\ge\) activation energy / more successful collisions per unit time [1]

(a)(ii)
- Particles have more kinetic energy / move faster [1]
- More frequent collisions / more collisions per unit time [1]
- More successful collisions / greater proportion of collisions have energy \(\ge\) activation energy [1]

(b)(i)
- \(450\ ^\circ\text{C}\) (accept \(350\text{--}500\ ^\circ\text{C}\)) [1]
- \(200\text{ atm}\) (accept \(150\text{--}250\text{ atm}\) or \(15\text{--}25\text{ MPa}\)) [1]
- Iron / \(\text{Fe}\) [1]

(b)(ii)
- Shifts right / towards products [1]
- Fewer (moles of) gas molecules on the right-hand side (2 vs 4) [1]

(b)(iii)
- Increases [1]

(b)(iv)
- Low temperature makes reaction rate too slow [1]
- Economically unviable / not productive [1]

(a)(i) Sulfur dioxide is an oxide of a non-metal, so it is an acidic oxide.
(a)(ii) Zinc oxide reacts with both acids and bases, so it is an amphoteric oxide.
(a)(iii) Carbon monoxide is a neutral oxide.
(a)(iv) Magnesium oxide is a metal oxide, so it is a basic oxide.

(b)(i) \(\text{copper(II) carbonate} + \text{sulfuric acid} \rightarrow \text{copper(II) sulfate} + \text{carbon dioxide} + \text{water}\)
(b)(ii)
- Addition: Add excess copper(II) carbonate powder to the acid. You will know all the acid has reacted when no more bubbles of gas are produced and some green powder remains unreacted at the bottom of the beaker.
- Separation: Filter the mixture using a filter funnel and filter paper. The unreacted copper(II) carbonate is collected as the residue on the filter paper, leaving the blue copper(II) sulfate solution as the filtrate.
- Crystallization: Heat the filtrate in an evaporating basin to evaporate water until the saturation point is reached. Leave the hot solution to cool slowly so that crystals form. Filter off the crystals.
- Drying: Gently pat the crystals dry between sheets of filter paper or leave them in a warm oven.

(c) Mix aqueous barium chloride and aqueous sodium sulfate in a beaker to form a white precipitate of barium sulfate. Filter the mixture to obtain the barium sulfate as a residue on the filter paper. Wash the residue with distilled water to remove any soluble impurities (sodium chloride). Dry the solid between sheets of filter paper or in a warm oven.

(a)(i)
- Acidic [1]
(a)(ii)
- Amphoteric [1]
(a)(iii)
- Neutral [1]
(a)(iv)
- Basic [1]

(b)(i)
- \(\text{copper(II) carbonate} + \text{sulfuric acid} \rightarrow \text{copper(II) sulfate} + \text{carbon dioxide} + \text{water}\) [1]

(b)(ii)
- Excess copper(II) carbonate is added until bubbling stops / solid stops dissolving [1]
- Filter the mixture to remove unreacted solid [1]
- Heat filtrate / solution until saturated / to crystallization point [1]
- Cool to allow crystals to form [1]
- Filter off the crystals [1]
- Dry crystals with filter paper / warm oven (not hot) [1]

(c)
- Mix the two solutions together [1]
- Filter (to obtain barium sulfate precipitate as residue) [1]
- Wash residue with distilled water and dry (with filter paper / in warm oven) [1]

(a) A homologous series is characterized by:
- Same general formula (e.g., \(\text{C}_n\text{H}_{2n+2}\) for alkanes)
- Similar chemical properties because they have the same functional group
- A gradual trend in physical properties (e.g., boiling point increases with molecular mass)
- Successive members differ by a \(\text{CH}_2\) group.

(b)(i) Alkenes have the general formula \(\text{C}_n\text{H}_{2n}\).
(b)(ii) Hydrocarbon means a compound containing carbon and hydrogen only. Unsaturated means it contains at least one carbon-carbon double bond (\(\text{C=C}\)).
(b)(iii) The displayed formula of but-1-ene showing all atoms and all bonds is:
```
H H H H
| | | |
H-C = C - C - C-H
| |
H H
```

(c)(i) The reaction of propene with steam is an addition reaction (specifically hydration). The product is propanol (either propan-1-ol or propan-2-ol).
(c)(ii) The industrial reaction uses a phosphoric acid (\(\text{H}_3\text{PO}_4\)) catalyst and a temperature of \(300\ ^\circ\text{C}\) (with pressure around 60 atm).

(d)(i) But-2-ene is an isomer with the double bond between carbon 2 and carbon 3:
```
H H H H
| | | |
H-C - C = C - C-H
| |
H H
```
(d)(ii) Methylpropene has a branched chain:
```
H H H
| | |
H-C - C = C-H
| |
H C-H
|
H
```
(d)(iii) The carboxylic acid with 4 carbon atoms is butanoic acid. Its molecular formula is \(\text{C}_4\text{H}_8\text{O}_2\).

(a)
- Any two from: same general formula / similar chemical properties / same functional group / consecutive members differ by \(\text{CH}_2\) / trend in physical properties [2]

(b)(i)
- \(\text{C}_n\text{H}_{2n}\) [1]

(b)(ii)
- Hydrocarbon: contains carbon and hydrogen only [1]
- Unsaturated: contains double bond(s) / \(\text{C=C}\) bonds [1]

(b)(iii)
- Displayed formula of but-1-ene showing all atoms and bonds correctly [1]

(c)(i)
- Addition [1]
- Propanol / propan-1-ol / propan-2-ol [1]

(c)(ii)
- Phosphoric acid / \(\text{H}_3\text{PO}_4\) [1]
- \(300\ ^\circ\text{C}\) (accept \(250\text{--}350\ ^\circ\text{C}\)) [1]

(d)(i)
- Correct displayed formula of but-2-ene [1]

(d)(ii)
- Correct displayed formula of methylpropene [1]

(d)(iii)
- \(\text{C}_4\text{H}_8\text{O}_2\) [1]

(a) Boiling tube or test-tube.

(b) To prevent 'suck-back' of the cold limewater into the hot boiling tube as it cools, which would cause the glass tube to crack/shatter.

(c) Turns cloudy / milky / forms a white precipitate.

(d) From green to black.

(e) Replace the second test-tube of limewater with a gas syringe (or collect the gas over water using an inverted measuring cylinder).

(f) Mass of the metal carbonate, rate of heating (or strength of Bunsen flame), and the particle size/surface area of the carbonate solid.

(a) Test-tube / boiling tube [1]

(b) To prevent 'suck-back' / limewater being drawn back into the hot tube [1]; because the contracting air/gas on cooling creates a partial vacuum, which would crack the hot glass [1]

(c) Turns cloudy / milky / white precipitate forms [1]

(d) Green [1] to black [1]

(e) Replace the second tube with a gas syringe / collect over water in a graduated container [1]; connect it to the delivery tube with a gas-tight seal [1]

(f) Any two from: mass of carbonate, particle size / surface area of carbonate, distance of flame from tube / rate of heating [2]

(a) Pencil lead (graphite) is insoluble in the solvent and will not run or contaminate the chromatogram, whereas ink is soluble and would separate.

(b) If the solvent is above the baseline, the dye samples will dissolve into the solvent in the beaker rather than travelling up the paper.

(c) \(R_f = \frac{\text{distance travelled by substance}}{\text{distance travelled by solvent front}} = \frac{6.0}{10.0} = 0.60\)

(d) Red 40 and Blue 1. Dye X produced two spots with the exact same distances (4.0 cm and 8.5 cm) and therefore the same \(R_f\) values as Red 40 and Blue 1.

(e) It produces only one spot on the chromatogram.

(f) Yes, the \(R_f\) values will change because the dyes will have different solubilities in a different solvent.

(g) Use a locating agent / spray (e.g., ninhydrin) or view under ultraviolet (UV) light.

(a) Pencil is insoluble / will not dissolve/run [1]

(b) To prevent the dyes from dissolving in the solvent / washing off the paper [1]

(c) Calculation: \(6.0 / 10.0\) [1]; \(R_f = 0.60\) (no units) [1]

(d) Red 40 and Blue 1 [1]; because the spots from X have the same distance travelled / same \(R_f\) values as Red 40 and Blue 1 [1]

(e) Only one spot / no separation [1]

(f) Yes, because different substances have different solubilities in different solvents / different affinities to the paper in different solvents [1]; this changes the relative distance travelled [1]

(g) Use a locating agent / spray / UV light [1]

(a) Carbon dioxide gas is produced and escapes from the flask.

(b) To allow the carbon dioxide gas to escape while preventing any liquid spray/droplets from being lost.

(c) Initial mass after adding chips = 250.00 + 5.00 = 255.00 g. Final mass = 248.40 g. Total loss in mass = 255.00 - 248.40 = 6.60 g.

(d) The mass remains constant at 248.40 g from 180 s onwards.

(e) The rate of reaction decreases. The mass loss in the first 30 s is 5.60 g, between 30 and 60 s is 0.40 g, between 60 and 90 s is 0.28 g, and between 90 and 120 s is 0.17 g. Since the mass loss per unit time is decreasing, the rate of reaction is slowing down.

(f) (i) The reaction would be faster, so the curve of mass loss would be steeper at the start, but would level off at the same final mass loss (6.60 g).
(ii) The reaction would be faster, so the curve of mass loss would be steeper at the start.

(a) Carbon dioxide gas is evolved and escapes [1]

(b) To prevent loss of liquid spray / acid droplets [1]

(c) Initial mass = 255.00 g, final mass = 248.40 g [1]; loss = 6.60 g [1]

(d) Mass becomes constant / does not change after 180 s [1]

(e) Rate decreases / slows down [1]; because the mass loss in successive 30 s intervals is decreasing (e.g. 5.60 g -> 0.40 g -> 0.28 g) [1]

(f) (i) Curve is steeper / reaction is faster [1]; levels off at the same final mass loss / 6.60 g [1]
(ii) Curve is steeper / reaction is faster [1]

1. Dissolve the solid barium chloride in distilled water in a beaker, stirring with a glass rod until it is completely dissolved to form barium chloride solution.
2. Add dilute sulfuric acid to the barium chloride solution.
3. Continue adding dilute sulfuric acid until no more precipitate forms (ensuring it is in excess and the reaction is complete).
4. Stir the mixture well to ensure thorough reaction.
5. Filter the mixture using a filter funnel and filter paper to separate the insoluble barium sulfate precipitate from the solution.
6. Keep the barium sulfate residue on the filter paper and discard the filtrate.
7. Wash the residue of barium sulfate with distilled water several times.
8. This washing removes any unreacted barium chloride or excess sulfuric acid/hydrochloric acid impurities.
9. Dry the barium sulfate precipitate by pressing it between pieces of filter paper or placing it in a warm oven.
10. Safety precaution: Wear safety goggles and gloves because dilute sulfuric acid is corrosive and barium compounds are toxic.

Marking points (max 10):
1. Dissolve solid barium chloride in distilled water [1]
2. Use a glass rod to stir to assist dissolving [1]
3. Add dilute sulfuric acid [1]
4. Add excess sulfuric acid / until no more precipitate forms [1]
5. Stir the reaction mixture [1]
6. Filter to obtain the precipitate / barium sulfate as residue [1]
7. Wash the residue with distilled water [1]
8. To remove soluble impurities / excess acid / chloride ions [1]
9. Dry the residue (e.g. in a warm oven / between filter papers) [1]
10. Safety precaution: Wear safety goggles (acid is corrosive) / gloves (barium is toxic) [1]