2023 Cambridge IGCSE Chemistry (0620) Practice Paper with Answers

Between 20 °C and 100 °C, the temperature is below the melting point of sulfur (115 °C), so the sulfur remains in the solid state. In a solid, particles are held in a fixed lattice and vibrate about fixed positions. As temperature increases, the particles gain kinetic energy, causing them to vibrate with greater frequency.

1 mark for the correct option B.

Both isotopes have the atomic number 17, which means a neutral atom of either isotope has 17 protons and 17 electrons. When they form a Z^{-} ion, they gain one electron, resulting in 18 electrons. Option A is incorrect because isotopes of the same element have the same electronic configuration. Option B is incorrect because chemical properties depend on the number of outer shell electrons, which is identical for isotopes. Option C is incorrect because physical properties depend on mass, which differs due to the different number of neutrons.

1 mark for the correct option D.

Both silicon(IV) oxide and diamond have giant covalent structures with strong covalent bonds extending throughout the macromolecular lattice, giving them high melting points and making them hard. Option A is incorrect because each silicon atom is bonded tetrahedrally to four oxygen atoms. Option C is incorrect because they do not have free-moving ions or delocalised electrons to conduct electricity. Option D is incorrect because silicon(IV) oxide is a giant covalent structure and does not consist of simple molecules.

1 mark for the correct option B.

In the electrolysis of dilute aqueous sodium chloride: At the anode (positive electrode), OH^{-} ions are preferentially discharged over Cl^{-} ions because the solution is dilute, producing oxygen gas (O2). At the cathode (negative electrode), H^{+} ions are preferentially discharged over Na^{+} ions because hydrogen is lower in the reactivity series, producing hydrogen gas (H2).

1 mark for the correct option B.

At a higher temperature, particles gain kinetic energy and move faster. This leads to a greater frequency of collisions, but more importantly, a much larger proportion of the colliding particles possess energy equal to or greater than the activation energy (E_a), leading to a higher frequency of successful collisions. Option A is incorrect because only a catalyst can lower the activation energy. Option B is incorrect because the concentration (particle spacing) is unchanged.

1 mark for the correct option C.

Reactivity decreases down Group VII. Therefore, bromine is more reactive than iodine and can displace it from its salt (forming iodine and potassium bromide), but bromine is less reactive than chlorine and cannot displace it. Option A is incorrect because state changes from gas to liquid to solid down the group; astatine is a solid. Option C is incorrect because reactivity decreases down the group. Option D is incorrect because fluorine is a gas at room temperature.

1 mark for the correct option B.

The polymer contains ester linkages (-COO-), making it a polyester. It is a condensation polymer formed by the reaction between a dicarboxylic acid (providing the -CO- groups) and a diol (providing the -O-CH2-CH2-O- groups), eliminating water molecules. Option A is incorrect because it is a condensation polymer. Option B is incorrect because it is a polyester, not a polyamide. Option D is incorrect because proteins contain amide (peptide) linkages.

1 mark for the correct option C.

Test 1 shows that sulfate ions (SO4^{2-}) are absent because no white precipitate of barium sulfate is formed. Test 2 shows the presence of iron(II) (Fe^{2+}) ions because Fe^{2+} reacts with NaOH to form a green precipitate of iron(II) hydroxide, which is insoluble in excess. Test 3 shows the presence of carbonate (CO3^{2-}) ions because the acid reacts with the carbonate to produce carbon dioxide gas, which turns limewater cloudy. Combining these results, the salt is iron(II) carbonate.

1 mark for the correct option B.

Lighter gas molecules diffuse faster than heavier gas molecules. Methylamine has a lower relative molecular mass (\(M_r = 31\)) than hydrogen chloride (\(M_r = 36.5\)). Therefore, methylamine molecules diffuse faster and travel further along the tube than hydrogen chloride molecules in the same time. The white cloud of methylammonium chloride therefore forms closer to the hydrochloric acid end.

1 mark for the correct option B.

The atomic number of oxygen is 8, which means an oxygen atom or ion has 8 protons. The mass number is 18, so the number of neutrons is \(18 - 8 = 10\). The \(2-\)-charged ion has gained 2 electrons compared to the neutral atom, so it has \(8 + 2 = 10\) electrons.

1 mark for the correct option A.

During the electrolysis of concentrated aqueous copper(II) chloride: At the anode (positive electrode), chloride ions (\(\text{Cl}^-\)) are discharged in preference to hydroxide ions because it is a concentrated halide solution. This produces chlorine gas, which is observed as bubbles of a pale green gas. At the cathode (negative electrode), copper ions (\(\text{Cu}^{2+}\)) are discharged in preference to hydrogen ions because copper is lower in the reactivity series. This produces copper metal, which is observed as a pink-brown solid deposit.

1 mark for the correct option C.

Energy needed to break reactant bonds: Breaking 1 \(\text{C}=\text{C}\) bond: \(1 \times 614 = 614\) kJ/mol. Breaking 4 \(\text{C}-\text{H}\) bonds: \(4 \times 413 = 1652\) kJ/mol. Breaking 1 \(\text{H}-\text{H}\) bond: \(1 \times 436 = 436\) kJ/mol. Total energy input = \(614 + 1652 + 436 = 2702\) kJ/mol. Energy released forming product bonds: Forming 1 \(\text{C}-\text{C}\) bond: \(1 \times 347 = 347\) kJ/mol. Forming 6 \(\text{C}-\text{H}\) bonds: \(6 \times 413 = 2478\) kJ/mol. Total energy output = \(347 + 2478 = 2825\) kJ/mol. Energy change = \(\text{Total energy input} - \text{Total energy output} = 2702 - 2825 = -123\) kJ/mol.

1 mark for the correct option A.

To keep the final volume of hydrogen gas the same, the total number of moles of sulfuric acid (the limiting reactant) must remain constant. Initial moles of sulfuric acid = \(0.025 \text{ dm}^3 \times 0.5 \text{ mol/dm}^3 = 0.0125\) mol. Option A: Zinc lumps have a smaller surface area than powder, which decreases the rate of reaction. Option B: Raising the temperature to 35 °C increases the kinetic energy of the particles and the frequency of successful collisions, increasing the initial rate. Since the quantity of acid is unchanged (0.0125 mol), the final volume of hydrogen is the same. Option C: \(0.050 \text{ dm}^3 \times 0.25 \text{ mol/dm}^3 = 0.0125\) mol, so the final volume of hydrogen is the same, but the concentration of acid is lower, which decreases the initial rate. Option D: \(0.025 \text{ dm}^3 \times 0.25 \text{ mol/dm}^3 = 0.00625\) mol, which would reduce the final volume of hydrogen gas produced.

1 mark for the correct option B.

1. Temperature: The forward reaction is exothermic (\(\Delta H = -197\text{ kJ/mol}\)). According to Le Chatelier's principle, decreasing the temperature shifts the equilibrium in the exothermic direction (to the right) to release heat, thereby increasing the yield of \(\text{SO}_3\). 2. Pressure: There are 3 moles of gas on the reactant side (\(2\text{SO}_2 + \text{O}_2\)) and 2 moles of gas on the product side (\(2\text{SO}_3\)). Increasing the pressure shifts the equilibrium to the side with fewer moles of gas (to the right) to reduce pressure, thereby increasing the yield of \(\text{SO}_3\).

1 mark for the correct option C.

Chlorine is more reactive than bromine because reactivity decreases down Group VII. Therefore, chlorine can displace bromine from an aqueous solution of a bromide salt: \(\text{Cl}_2(aq) + 2\text{KBr}(aq) \rightarrow 2\text{KCl}(aq) + \text{Br}_2(aq)\). Other options are incorrect: Bromine is a red-brown liquid at room temperature and pressure. Chlorine is more reactive than iodine. Melting points of the halogens increase down the group.

1 mark for the correct option D.

1. The green precipitate that is insoluble in excess sodium hydroxide indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\). (Note: Chromium(III) also forms a green precipitate, but it dissolves in excess sodium hydroxide to form a green solution). 2. The cream precipitate formed with acidified silver nitrate indicates the presence of bromide ions, \(\text{Br}^-\). Therefore, salt X is iron(II) bromide.

1 mark for the correct option B.

For Curve Y to have a faster initial rate, the rate of reaction must be increased. This can be achieved by raising the temperature to \(35^\circ\text{C}\). Since the total volume of gas produced is determined by the number of moles of the limiting reactant (hydrochloric acid), using the same volume and concentration of acid (\(50\text{ cm}^3\) of \(1.0\text{ mol/dm}^3\)) yields the same final amount of gas. Therefore, option A is correct.

[1] award for correct option A.

According to Le Chatelier's principle: 1. The forward reaction is exothermic (\(\Delta H < 0\)), so decreasing the temperature shifts the equilibrium to the right to produce more heat. 2. There are 4 moles of gaseous reactants on the left and 2 moles of gaseous product on the right. Increasing the pressure shifts the equilibrium to the side with fewer gas moles (to the right). Thus, decreasing temperature and increasing pressure will shift the position of equilibrium to the right.

[1] award for correct option B.

1. Moles of \(\text{Al} = \frac{5.4\text{ g}}{27\text{ g/mol}} = 0.2\text{ mol}\). 2. From the equation, \(2\text{ moles of Al}\) produce \(3\text{ moles of H}_2\). Therefore, \(0.2\text{ moles of Al}\) produce \(0.2 \times \frac{3}{2} = 0.3\text{ mol of H}_2\). 3. Volume of gas at r.t.p. = \(0.3\text{ mol} \times 24\text{ dm}^3\text{/mol} = 7.2\text{ dm}^3\).

[1] award for correct option C.

During the electrolysis of dilute \(\text{Na}_2\text{SO}_4\text{(aq)}\): 1. At the anode (positive electrode), \(\text{OH}^-\text{(aq)}\) ions are preferentially discharged over \(\text{SO}_4^{2-}\text{(aq)}\) ions to form oxygen gas (\(\text{O}_2\)). 2. At the cathode (negative electrode), \(\text{H}^+\text{(aq)}\) ions are preferentially discharged over \(\text{Na}^+\text{(aq)}\) ions to form hydrogen gas (\(\text{H}_2\)).

[1] award for correct option A.

Polyamides are formed by condensation polymerisation when a dicarboxylic acid monomer (providing \(\text{-CO-}\) groups) reacts with a diamine monomer (providing \(\text{-NH-}\) groups), eliminating water molecules.

[1] award for correct option B.

The reaction of an alkane (such as ethane) with a halogen is a photochemical substitution reaction. The ultraviolet (UV) light provides the activation energy required to break the covalent bond in chlorine molecules (\(\text{Cl-Cl}\)), producing reactive chlorine radicals that initiate the substitution.

[1] award for correct option D.

1. \(\text{SO}_2\) is a non-metal oxide that reacts with bases to form salts, making it an acidic oxide. 2. \(\text{CaO}\) is a metal oxide that reacts with acids, making it a basic oxide. 3. \(\text{Al}_2\text{O}_3\) is amphoteric because it reacts with both acids and bases. 4. \(\text{CO}\) is a neutral oxide because it does not react with acids or bases. This matches the classification in option A.

[1] award for correct option A.

As Group VII is descended: 1. The density of the elements increases. 2. The elements become less reactive (reactivity decreases). 3. The color of the elements becomes darker (not lighter). 4. The melting and boiling points increase, meaning they become less volatile (not more volatile).

[1] award for correct option B.

The rate of diffusion of a gas is inversely proportional to the square root of its relative molecular mass (\(M_r\)). This means that gases with a higher \(M_r\) diffuse slower than gases with a lower \(M_r\).

Let's calculate the relative molecular mass of each gas:
- Methane, \(CH_4\): \(12 + (4 \times 1) = 16\)
- Neon, \(Ne\): \(20\)
- Carbon monoxide, \(CO\): \(12 + 16 = 28\)
- Carbon dioxide, \(CO_2\): \(12 + (2 \times 16) = 44\)

Carbon dioxide has the highest relative molecular mass (44) and therefore diffuses the slowest.

1 mark for the correct option D.

The atomic number (bottom number) of cobalt is 27, which represents the number of protons. Since it is cobalt, it always has 27 protons.

The nucleon number (top number) is 60, which is the total number of protons and neutrons. Therefore, the number of neutrons is \(60 - 27 = 33\).

The ion has a \(2+\) charge, meaning it has lost 2 electrons compared to its neutral atomic state. A neutral cobalt atom has 27 electrons, so this ion has \(27 - 2 = 25\) electrons.

1 mark for the correct option A.

In the electrolysis of concentrated aqueous potassium bromide:
- At the anode (positive electrode), halide ions (\(Br^-\)) are discharged in preference to hydroxide ions (\(OH^-\)) because the solution is concentrated. This produces bromine gas, \(Br_2(g)\).
- At the cathode (negative electrode), hydrogen ions (\(H^+\)) from water are discharged in preference to potassium ions (\(K^+\)) because potassium is high in the reactivity series. This produces hydrogen gas, \(H_2(g)\).

1 mark for the correct option A.

The energy change of a reaction is calculated using the formula:
\(\Delta H = \text{Energy absorbed to break bonds} - \text{Energy released when bonds are formed}\)

1. Energy absorbed to break reactant bonds:
- Break 1 mol of \(H-H\) bonds = \(436\text{ kJ}\)
- Break 1 mol of \(Cl-Cl\) bonds = \(242\text{ kJ}\)
- Total energy absorbed = \(436 + 242 = 678\text{ kJ}\)

2. Energy released when product bonds are formed:
- Form 2 mol of \(H-Cl\) bonds = \(2 \times 431 = 862\text{ kJ}\)

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

1 mark for the correct option A.

Let's analyse the tests:
1. The reaction with silver nitrate under acidic conditions producing a cream precipitate confirms the presence of bromide ions (\(Br^-\)).
2. The reaction with sodium hydroxide producing a green precipitate could indicate either \(Fe^{2+}\) or \(Cr^{3+}\) ions. However, chromium(III) hydroxide is soluble in excess sodium hydroxide to give a green solution, whereas iron(II) hydroxide is insoluble in excess. Since the precipitate is insoluble in excess, the cation must be \(Fe^{2+}\).

Therefore, the salt Y is iron(II) bromide.

1 mark for the correct option C.

The polymer contains the ester linkage, \(-\text{CO}-\text{O}-\), linking the repeating units together, so it is a polyester. Polyesters are formed by condensation polymerisation of a diol (which provides the \(-\text{O}-\text{R}-\text{O}-\) part) and a dicarboxylic acid (which provides the \(-\text{CO}-\text{R}'-\text{CO}-\) part).

1 mark for the correct option B.

First, find the number of moles of magnesium that reacted:
\(\text{Moles of Mg} = \frac{\text{mass}}{\text{relative atomic mass}} = \frac{1.2\text{ g}}{24\text{ g/mol}} = 0.05\text{ mol}\)

From the balanced equation, \(1\text{ mol}\) of \(Mg\) reacts to produce \(1\text{ mol}\) of \(H_2\). Therefore, \(0.05\text{ mol}\) of \(H_2\) is produced.

Now calculate the volume of hydrogen gas at r.t.p.:
\(\text{Volume} = \text{moles} \times 24\text{ dm}^3\text{/mol} = 0.05 \times 24 = 1.2\text{ dm}^3\).

1 mark for the correct option A.

A reducing agent is a substance that reduces another substance while itself being oxidized.

- In A, \(CuO\) loses oxygen to become \(Cu\) (it is reduced, so it acts as an oxidizing agent).
- In B, \(CO\) gains oxygen to become \(CO_2\) (it is oxidized, so it reduces \(Fe_2\text{O}_3\) to \(Fe\). Thus, it acts as a reducing agent).
- In C, \(Cl_2\) gains electrons to become \(Cl^-\), so it is reduced (it acts as an oxidizing agent).
- In D, \(H_2\text{SO}_4\) provides \(H^+\) ions which gain electrons to become \(H_2\), so it is reduced (it acts as an oxidizing agent).

1 mark for the correct option B.

The number of moles of HCl in both experiments is the same (0.050 mol). Since marble chips are in excess, the total amount of carbon dioxide gas produced depends on the limiting reactant (HCl) and will be the same for both. Experiment 2 has a higher temperature, higher concentration of acid, and smaller particle size (greater surface area), so its initial rate of reaction is faster.

1 mark for the correct option (B).

Energy change = (energy of bonds broken) - (energy of bonds formed). Bonds broken = (1 * H-H) + (1 * F-F) = 436 + 158 = 594 kJ/mol. Bonds formed = (2 * H-F) = 2 * 562 = 1124 kJ/mol. Energy change = 594 - 1124 = -530 kJ/mol.

1 mark for the correct option (A).

Moles of Zn = 1.30 g / 65 g/mol = 0.02 mol. From the stoichiometry of the equation, 1 mole of Zn produces 1 mole of H2 gas. Therefore, 0.02 mol of Zn produces 0.02 mol of H2 gas. Volume of H2 gas at r.t.p. = 0.02 mol * 24 dm3/mol = 0.48 dm3.

1 mark for the correct option (B).

Chlorine is more reactive than bromine, so bromine cannot displace chlorine from potassium chloride (no reaction). Bromine is more reactive than iodine, so bromine displaces iodine from potassium iodide, forming aqueous iodine which appears as a brown solution.

1 mark for the correct option (A).

According to Le Chatelier's principle, lowering the temperature shifts the equilibrium in the exothermic direction (forward reaction), thus shifting the equilibrium to the right. Increasing the pressure shifts the equilibrium in the direction with fewer moles of gas (2 moles of gas on the left vs 1 mole on the right), which is to the right. Therefore, low temperature and high pressure yield more N2O4.

1 mark for the correct option (B).

A reducing agent reduces another substance and itself gets oxidized. In reaction B, carbon monoxide (CO) is oxidized to carbon dioxide (CO2) while reducing copper(II) oxide (CuO) to copper metal (Cu). Thus, CO acts as the reducing agent.

1 mark for the correct option (B).

Nylon is a polyamide, meaning it contains amide linkages (-CO-NH-). Terylene is a polyester, meaning it contains ester linkages (-CO-O-). Therefore, the correct row is B.

1 mark for the correct option (B).

In dilute aqueous sodium chloride, the ions present are Na+, Cl-, H+, and OH-. At the positive electrode (anode), hydroxide ions (OH-) are preferentially discharged over chloride ions (Cl-) to produce oxygen gas. At the negative electrode (cathode), hydrogen ions (H+) are preferentially discharged over sodium ions (Na+) to produce hydrogen gas. Thus, oxygen is formed at the anode and hydrogen at the cathode.

1 mark for the correct option (C).

(a) A magnesium atom loses two electrons to form a magnesium ion, Mg2+. A sulfur atom gains these two electrons to form a sulfide ion, S2-. (b) Sulfur forms a double covalent bond with each of the two oxygen atoms by sharing two pairs of electrons with each. There is one lone pair (two non-bonding electrons) left on the sulfur atom's outer shell. (c)(i) Magnesium sulfide has a giant ionic lattice structure with strong electrostatic forces of attraction between oppositely charged ions, which require a large amount of thermal energy to break. (c)(ii) Sulfur dioxide has a simple molecular structure with weak intermolecular forces of attraction between molecules, which require very little energy to overcome. (c)(iii) In molten magnesium sulfide, the ionic lattice is broken, and the ions are free to move and carry charge. In solid magnesium sulfide, the ions are held in fixed positions within the lattice and cannot move.

(a) Mg loses two electrons [1]; S gains two electrons [1]; correct charges Mg2+ and S2- specified [1]. (b) Two shared pairs of electrons in each S=O bond [1]; four non-bonding electrons on each O atom [1]; two non-bonding electrons on the S atom [1]. (c)(i) Giant ionic lattice [1]; strong electrostatic attraction between oppositely charged ions [1]; requires a lot of energy to break [1]. (c)(ii) Simple covalent/molecular [1]; weak intermolecular forces [1]; requires little energy to overcome [1]. (c)(iii) Molten has mobile ions [1]; solid has ions fixed in lattice [1].

(a) Increasing the pressure increases the yield of methanol. This is because the equilibrium shifts to the side with fewer moles of gas to oppose the increase in pressure. The reactant side has 3 moles of gas while the product side has only 1 mole. (b)(i) Increasing the temperature increases the rate of the reaction. This is because the particles gain kinetic energy, move faster, and collide more frequently. Additionally, a significantly larger proportion of the colliding particles have energy equal to or greater than the activation energy, leading to more successful collisions per unit time. (b)(ii) Increasing the temperature decreases the yield of methanol. Because the forward reaction is exothermic, the equilibrium shifts to the left (the endothermic direction) to absorb the added thermal energy. (c)(i) A catalyst has no effect on the position of the equilibrium. (c)(ii) A catalyst provides an alternative reaction pathway with a lower activation energy, so a higher proportion of reacting particles have energy greater than or equal to the activation energy.

(a) Yield increases [1]; equilibrium shifts to side with fewer moles of gas [1]; 3 moles on left vs 1 mole on right [1]. (b)(i) Rate increases [1]; particles have more kinetic energy/move faster, leading to higher collision frequency [1]; higher proportion of collisions have energy >= activation energy [1]. (b)(ii) Yield decreases [1]; equilibrium shifts in endothermic direction to absorb heat [1]. (c)(i) No effect [1]. (c)(ii) Provides alternative pathway [1]; with lower activation energy [1]; more particles have sufficient energy to react [1].

(a) Moles of NaOH = concentration * volume = 0.150 mol/dm3 * (25.0 / 1000) dm3 = 0.00375 mol. (b) From the equation, 2 moles of NaOH react with 1 mole of H2SO4. Therefore, moles of H2SO4 = 0.00375 / 2 = 0.001875 mol. (c) Concentration of H2SO4 = moles / volume = 0.001875 mol / (18.5 / 1000) dm3 = 0.10135... mol/dm3, which rounds to 0.101 mol/dm3 (to 3 significant figures). (d) Relative formula mass (Mr) of H2SO4 = 2(1.0) + 32.1 + 4(16.0) = 98.1 g/mol. Concentration in g/dm3 = 0.10135 mol/dm3 * 98.1 g/mol = 9.94 g/dm3 (or 9.93 g/dm3 if using Mr = 98). (e)(i) A strong acid is a proton donor that completely dissociates/ionises in aqueous solution. (e)(ii) H+(aq) + OH-(aq) -> H2O(l).

(a) Moles of NaOH = 0.00375 [2] (1 mark for correct working). (b) Moles of H2SO4 = 0.001875 [1] (allow ecf from part a). (c) Concentration of H2SO4 = 0.101 (or 0.1014) mol/dm3 [2] (1 mark for correct working). (d) Mr of H2SO4 calculation (98 or 98.1) [1]; Concentration in g/dm3 = 9.93 or 9.94 [1] (allow ecf from part c). (e)(i) Proton donor [1]; completely dissociated/ionised in water [1]. (e)(ii) H+(aq) and OH-(aq) on left [1]; H2O(l) on right [1]; correct state symbols [1].

(a) The colourless solution turns orange-brown. (b) Cl2(aq) + 2Br-(aq) -> 2Cl-(aq) + Br2(aq). (c) Bromide ions, Br-, lose electrons to form bromine molecules, Br2. Loss of electrons is oxidation, so bromide ions are oxidised. Chlorine molecules, Cl2, gain electrons to form chloride ions, Cl-. Gain of electrons is reduction, so chlorine molecules are reduced. (d) Bromine is less reactive than chlorine, so it cannot displace chlorine from its compound. (e) Reactivity of halogens decreases down Group VII. As the atomic radius increases and shielding increases, the outer shell is further from the positive nucleus, making it harder for the atom to attract and gain an incoming electron.

(a) Colourless [1] to orange / brown / orange-brown [1]. (b) Cl2 and 2Br- on left [1]; 2Cl- and Br2 on right [1]; correct state symbols (all aq) [1]. (c) Bromide ions lose electrons / are oxidised [1]; chlorine gains electrons / is reduced [1]; oxidation is loss, reduction is gain [1]. (d) Bromine is less reactive than chlorine [2] (1 mark for saying bromine cannot displace chlorine). (e) Reactivity decreases down the group [1]; because atomic size/shielding increases, making it harder to attract/gain an electron [1].

(a) Condensation polymerisation. (b) Alcohol (or hydroxyl) group and carboxylic acid (or carboxyl) group. (c) Monomer 1 (ethane-1,2-diol): HO-CH2-CH2-OH, showing O-H and C-O single bonds clearly. Monomer 2 (butanedioic acid): HO-C(=O)-CH2-CH2-C(=O)-OH, showing the C=O and O-H bonds clearly. (d) Addition polymerisation involves unsaturated monomers (alkenes) joining together with no other product formed. Condensation polymerisation involves monomers with two functional groups joining with the elimination of a small molecule, such as water or hydrogen chloride. (e) Making synthetic fibres / clothing / ropes / plastic bottles.

(a) Condensation [1]. (b) Alcohol / hydroxyl [1]; carboxylic acid / carboxyl [1]. (c) Structure of the diol showing all bonds in functional groups (HO-CH2-CH2-OH) [2]; Structure of the dicarboxylic acid showing all bonds in functional groups (HOOC-CH2-CH2-COOH) [2]. (d) Addition forms only one product/polymer [1]; condensation forms a polymer and a small molecule (water) [1] (or addition uses C=C / condensation uses functional groups like -OH/-COOH [1]). (e) Clothing / fabrics / plastic bottles / sleeping bag fillings [1].

(a) Effervescence/fizzing/bubbles of gas are seen, and the solid calcium carbonate dissolves/disappears/shrinks. (b) The rate of reaction is fastest at the start, decreases over time, and eventually becomes zero when the reaction stops. This is because, as the reaction proceeds, the acid reactant is used up, so the concentration of hydrochloric acid decreases. This results in a lower frequency of successful collisions between reacting particles. (c) Increasing the temperature to 35 °C increases the kinetic energy of the particles, so they move faster and collide more frequently. More importantly, a much higher proportion of the colliding particles have energy equal to or greater than the activation energy, leading to a much higher frequency of successful collisions. (d) The total volume of carbon dioxide gas collected is doubled.

(a) Fizzing / bubbling / effervescence [1]; solid disappears / dissolves [1] (reject: gas given off, as it is not directly observed without bubble description). (b) Rate is fastest at start and decreases over time / becomes zero [1]; concentration of acid decreases [1]; leading to a lower frequency of successful collisions [1]. (c) Particles have more kinetic energy / move faster, leading to a higher collision frequency [1]; higher proportion of particles have energy >= activation energy [1]; higher frequency of successful collisions [1]. (d) The volume of carbon dioxide gas is doubled [2] (1 mark for 'increases').

(a) (i) Boiling tube / test-tube
(ii) Delivery tube

(b) The green solid turns into a black solid.

(c) The limewater turns cloudy / milky / chalky.

(d) To prevent suck-back of cold limewater into the hot boiling tube; which would cause the hot glass tube to crack or shatter.

(e) Use a safety screen / perform in a well-ventilated room (or fume cupboard) / point the mouth of the boiling tube away from people.

(f) \(\text{CuCO}_3(\text{s}) \rightarrow \text{CuO}(\text{s}) + \text{CO}_2(\text{g})\)

(a) (i) Boiling tube / test-tube [1]
(ii) Delivery tube [1]

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

(c) Turns cloudy / milky [1]

(d) Prevent suck-back of liquid [1]; prevent cracking/breaking of hot glass tube [1]

(e) Any one valid safety precaution, e.g., point tube away from self/others, use a fume cupboard/well-ventilated area [1]

(f) Correct formulae for reactants and products: \(\text{CuCO}_3 \rightarrow \text{CuO} + \text{CO}_2\) [1]; Correct state symbols: \((\text{s})\), \((\text{s})\), \((\text{g})\) [1]

(a) A graph is plotted with Time / s on the x-axis and Volume of gas / \(\text{cm}^3\) on the y-axis. Points are plotted at (0,0), (10,16), (20,28), (30,37), (50,48), (60,51), (70,53), (80,53). A smooth curve of best fit is drawn, leaving the point (40,18) well below the curve.

(b) Anomalous point: 40 seconds, 18 \(\text{cm}^3\). Possible error: A gas leak occurred in the apparatus / the plunger of the gas syringe got stuck temporarily / the reading was taken too early or misread.

(c) Volume at 35 s is approximately 43 \(\text{cm}^3\) (allow 42–44 \(\text{cm}^3\) depending on curve). Working is shown by drawing a vertical dashed line from 35 s on the x-axis up to the curve, and then a horizontal line to the y-axis.

(d) The rate of reaction decreases over time because the concentration of the reactant hydrochloric acid decreases as it is used up, leading to fewer successful collisions per unit time.

(a) Axes labeled with units and appropriate linear scale [1]; All 8 non-anomalous points plotted correctly to within half a small square [2]; Smooth curve of best fit drawn ignoring the point at 40 s [1]

(b) Anomalous point identified as (40 s, 18 \(\text{cm}^3\)) [1]; Valid reason, e.g., gas leak / syringe stuck / misreading the scale [1]

(c) Correct reading from their graph (approx. 42–44 \(\text{cm}^3\)) [1]; Indication of working shown on the graph (lines to/from curve) [1]

(d) Rate decreases [1]; Because acid is used up / concentration of acid decreases [1]

(a) (i) Iron(II) ion / \(\text{Fe}^{2+}\)
(ii) The iron(II) hydroxide precipitate is oxidised to iron(III) hydroxide by oxygen in the air.

(b) (i) Ammonia / \(\text{NH}_3\)
(ii) Ammonium ion / \(\text{NH}_4^+\)

(c) (i) Sulfate / \(\text{SO}_4^{2-}\)
(ii) \(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\)

(d) Dissolve solid Z in distilled water, add dilute nitric acid, and then add aqueous silver nitrate. If chloride is present, a white precipitate is formed. If iodide is present, a yellow precipitate is formed.

(a) (i) Iron(II) / \(\text{Fe}^{2+}\) [1]
(ii) Oxidation / reaction with oxygen in the air (to form iron(III) hydroxide) [1]

(b) (i) Ammonia [1]
(ii) Ammonium / \(\text{NH}_4^+\)[1]

(c) (i) Sulfate / \(\text{SO}_4^{2-}\) [1]
(ii) Correct species: \(\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4\) [1]; Correct state symbols: \((\text{aq})\), \((\text{aq})\) and \((\text{s})\) [1]

(d) Add dilute nitric acid and aqueous silver nitrate [1]; White precipitate with chloride [1]; Yellow precipitate with iodide [1]

1. **Apparatus**: Balance, beaker, stirring rod, filter funnel, filter paper, evaporating basin, Bunsen burner, tripod, gauze.

2. **Weighing 1**: Weigh a known mass of the dry mixture using a balance and record this initial mass as \(m_1\).

3. **Separating Sodium Chloride**:
- Add distilled water to the beaker containing the mixture and stir thoroughly with a glass rod to dissolve all the sodium chloride.
- Filter the mixture. The sodium chloride solution passes through as the filtrate.
- Wash the residue (which contains sand and calcium carbonate) with distilled water to remove any remaining salt solution.
- Dry the residue in a warm oven.
- **Weighing 2**: Weigh the dry residue of sand and calcium carbonate and record this mass as \(m_2\).

4. **Reacting and Separating Calcium Carbonate**:
- Add excess dilute hydrochloric acid to the dry residue of sand and calcium carbonate in a beaker. Stir until effervescence stops to ensure all calcium carbonate has fully reacted and dissolved as soluble calcium chloride.
- Filter the mixture. The sand remains as the residue on the filter paper, while the calcium chloride solution passes through as filtrate.
- Wash the sand residue with distilled water.
- Dry the sand residue thoroughly.
- **Weighing 3**: Weigh the dry sand residue and record this mass as \(m_3\).

5. **Calculation**:
- The mass of calcium carbonate is calculated as: \(m_{\text{carbonate}} = m_2 - m_3\).
- The percentage by mass of calcium carbonate in the original mixture is: \(\frac{m_2 - m_3}{m_1} \times 100\%\).

Maximum [10] marks from:
- Weigh initial mixture (mass \(m_1\)) [1]
- Add distilled water to the mixture and stir to dissolve sodium chloride [1]
- Filter the mixture to separate the solution of sodium chloride from insoluble sand and calcium carbonate [1]
- Wash the residue (sand + calcium carbonate) with distilled water and dry it [1]
- Weigh the dry residue (mass \(m_2\)) [1]
- Add excess dilute hydrochloric acid to the residue to react and dissolve calcium carbonate [1]
- Filter to separate the unreacted sand from the solution of calcium chloride [1]
- Wash the sand with distilled water, dry it, and weigh it (mass \(m_3\)) [1]
- Mass of calcium carbonate calculated as \(m_2 - m_3\) [1]
- Percentage by mass calculated as \(\frac{m_2 - m_3}{m_1} \times 100\%\) [1]