2023 Cambridge IGCSE Chemistry (0620) Practice Paper with Answers

An ion \( X^{2-} \) has a charge of \( 2- \), which means it has gained 2 electrons compared to its neutral atom. Number of protons = number of electrons in the neutral atom = \( 18 - 2 = 16 \). Nucleon number = protons + neutrons = 34. Number of neutrons = \( 34 - 16 = 18 \). Thus, there are 16 protons and 18 neutrons in the nucleus.

Award 1 mark for selecting the correct option A.

During fractional distillation, the liquid with the lower boiling point (heptane) evaporates more easily and is distilled first. The glass beads in the fractionating column provide a large surface area for repeated condensation and evaporation, allowing the higher boiling point component (octane) to condense and run back down into the flask. Therefore, option C is correct.

Award 1 mark for selecting the correct option C.

The balanced chemical equation is: \( 2\text{C}_2\text{H}_6\text{(g)} + 7\text{O}_2\text{(g)} \rightarrow 4\text{CO}_2\text{(g)} + 6\text{H}_2\text{O(l)} \). The molar mass of ethane, \( \text{C}_2\text{H}_6 \), is \( (2 \times 12) + (6 \times 1) = 30\text{ g/mol} \). Moles of ethane = \( 3.0 / 30 = 0.10\text{ mol} \). The mole ratio of ethane to oxygen is \( 2 : 7 \). Thus, moles of oxygen required = \( 0.10 \times 3.5 = 0.35\text{ mol} \). Volume of oxygen at r.t.p. = \( 0.35\text{ mol} \times 24\text{ dm}^3/\text{mol} = 8.4\text{ dm}^3 \).

Award 1 mark for selecting the correct option C.

Energy needed to break reactant bonds = \( 436\text{ (H-H)} + 242\text{ (Cl-Cl)} = 678\text{ kJ/mol} \). Energy released in forming product bonds = \( 2 \times 431\text{ (H-Cl)} = 862\text{ kJ/mol} \). Enthalpy change, \( \Delta H = \text{energy absorbed} - \text{energy released} = 678 - 862 = -184\text{ kJ/mol} \).

Award 1 mark for selecting the correct option A.

Decreasing the temperature reduces the average kinetic energy of the particles, resulting in fewer successful collisions per second and a lower rate of reaction. Since the quantities of limiting reactant (hydrochloric acid) are identical, the total volume of carbon dioxide gas produced remains the same.

Award 1 mark for selecting the correct option C.

Aluminium oxide is amphoteric because it reacts with both acids and bases. Sulfur dioxide is a non-metal oxide and is acidic. Calcium oxide is a metal oxide and is basic.

Award 1 mark for selecting the correct option B.

\( X \) reacts with cold water, so it is highly reactive and ranks highest. \( Y \) is above carbon in reactivity since its oxide cannot be reduced by carbon, but it is below \( X \). \( Z \) is below carbon but above hydrogen since it reacts with dilute acid. \( W \) is below hydrogen because it does not react with dilute acid. The correct order is \( X \rightarrow Y \rightarrow Z \rightarrow W \).

Award 1 mark for selecting the correct option A.

A homologous series is defined as a family of compounds with the same general formula, identical functional groups, similar chemical properties, and a graduation in physical properties. Successive members differ by a \( -\text{CH}_2- \) group.

Award 1 mark for selecting the correct option B.

At the anode, chloride ions (\(\text{Cl}^-\)) are preferentially discharged over hydroxide ions (\(\text{OH}^-\)) because the solution is concentrated, producing chlorine gas. At the cathode, hydrogen ions (\(\text{H}^+\)) are preferentially discharged over sodium ions (\(\text{Na}^+\)) because hydrogen is less reactive than sodium, producing hydrogen gas. As hydrogen ions are discharged, the remaining solution contains a relative excess of hydroxide ions, causing the electrolyte to become alkaline and the pH to increase.

1 mark for the correct option A. [Reject incorrect products or incorrect pH change]

Energy absorbed to break bonds in reactants = \(1 \times 945 + 3 \times 436 = 2253\text{ kJ/mol}\). Energy released in forming bonds in products = \(2 \times 3 \times 391 = 2346\text{ kJ/mol}\). Enthalpy change \(\Delta H = \text{Energy absorbed} - \text{Energy released} = 2253 - 2346 = -93\text{ kJ/mol}\). Since \(\Delta H\) is negative, the reaction is exothermic.

1 mark for option A. [0 marks for any incorrect calculation or reaction classification]

Experiment 2 uses powder (greater surface area) and a higher temperature (30 °C vs 20 °C), both of which increase the rate of reaction, so the initial rate is faster. To find the limiting reactant: moles of \(\text{CaCO}_3 = 5.0 / 100 = 0.05\text{ mol}\); moles of \(\text{HCl} = (50/1000) \times 1.0 = 0.05\text{ mol}\). According to the equation \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\), 1 mole of \(\text{CaCO}_3\) reacts with 2 moles of \(\text{HCl}\). Therefore, \(\text{HCl}\) is the limiting reactant in both experiments. Since the amount of \(\text{HCl}\) is identical in both, the total volume of carbon dioxide produced remains the same.

1 mark for option B. [Reject options predicting different gas volumes or slower rates]

Statement 1 is correct because hydrochloric acid is a strong acid that fully dissociates in water, yielding a higher concentration of \(\text{H}^+\) ions and therefore a lower pH. Statement 2 is correct because the fully dissociated strong acid contains a higher concentration of mobile ions, conducting electricity better. Statement 3 is incorrect because the lower concentration of hydrogen ions in the weak ethanoic acid means the initial rate of reaction with magnesium is slower. Statement 4 is correct because both are monoprotic acids of the same concentration and volume, meaning they contain the same total moles of acidic protons available for neutralisation.

1 mark for option B (statements 1, 2, and 4 are correct). [Reject other combinations]

The molar mass of \(\text{CH}_4\) is \(12 + 4(1) = 16\text{ g/mol}\). Moles of \(\text{CH}_4\) molecules = \(1.6\text{ g} / 16\text{ g/mol} = 0.1\text{ mol}\). Each \(\text{CH}_4\) molecule contains 5 atoms (1 carbon atom + 4 hydrogen atoms). Therefore, total moles of atoms = \(0.1\text{ mol} \times 5 = 0.5\text{ mol}\). Total number of atoms = \(0.5 \times 6.0 \times 10^{23} = 3.0 \times 10^{23}\).

1 mark for option C. [Reject calculations that only determine the number of molecules (A) or ignore the stoichiometry of the molecule]

The ion has a \(2-\) charge and 18 electrons, meaning the neutral atom has \(18 - 2 = 16\) electrons. Thus, the proton number of \(X\) is 16. The nucleon number (mass number) is the sum of protons and neutrons: \(16 + 18 = 34\). An element with proton number 16 has the electron configuration 2, 8, 6. Because it has 6 valence electrons, it is in Group VI.

1 mark for option B. [Reject options predicting proton number 18 (argon) or 20 (calcium)]

Metal \(Y\) is the most reactive because it reacts violently with cold water. Metal \(X\) is also highly reactive because its oxide cannot be reduced by carbon, but it is less reactive than \(Y\). Metal \(W\) is moderately unreactive because it does not react with steam but its carbonate decomposes. Metal \(Z\) is very unreactive because its nitrate decomposes all the way to the metal itself on heating. Thus, the order of decreasing reactivity is \(Y \rightarrow X \rightarrow W \rightarrow Z\).

1 mark for option A. [Reject any other sequence of reactivities]

There are exactly three structural isomers for \(\text{C}_5\text{H}_{12}\):
1. Pentane (straight chain): \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3\)
2. 2-methylbutane (branched chain): \(\text{CH}_3\text{CH(CH}_3)\text{CH}_2\text{CH}_3\)
3. 2,2-dimethylpropane (highly branched chain): \(\text{CH}_3\text{C(CH}_3)_2\text{CH}_3\)

1 mark for option B. [Reject 2, 4, or 5 isomers]

To obtain hydrated crystals of a soluble salt, the solution must first be heated to saturate it (concentration). Heating to dryness would remove all water of crystallization, forming anhydrous powder rather than hydrated crystals. The saturated solution is then allowed to cool, during which crystals form because solubility decreases at lower temperatures. The crystals are then separated from the remaining solution by filtration and dried gently using filter paper (or in a warm oven at low temperature, but not heated strongly as that would dehydrate them).

Award 1 mark for the correct option B. Reject options A, C, and D because they involve heating to dryness or washing with inappropriate reagents which would destroy or fail to form the hydrated crystals.

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

\[R_f = \frac{5.2\text{ cm}}{8.0\text{ cm}} = 0.65\]

A larger \(R_f\) value indicates that the component travels further up the chromatogram. This is because it is more soluble in the mobile phase (the moving solvent) relative to the stationary phase (the paper).

Award 1 mark for option A. Option B has the correct calculation but incorrect solubility interpretation. Option C has an incorrect calculation (> 1, which is impossible for \(R_f\) values). Option D has an incorrect calculation.

To calculate the energy change of the reaction:
1. Energy required to break bonds (reactants):
We break 1 \(\text{C–H}\) bond and 1 \(\text{Cl–Cl}\) bond.
Energy input = \(413 + 242 = 655\text{ kJ/mol}\).

2. Energy released when bonds are formed (products):
We form 1 \(\text{C–Cl}\) bond and 1 \(\text{H–Cl}\) bond.
Energy output = \(339 + 431 = 770\text{ kJ/mol}\).

3. Enthalpy change (\(\Delta H\)):
\(\Delta H = \text{Energy input} - \text{Energy output} = 655 - 770 = -115\text{ kJ/mol}\).

Since \(\Delta H\) is negative, the reaction is exothermic.

Award 1 mark for option A. Option B represents the positive value of the change (endothermic). Option C is a distractor with an incorrect value. Option D is a distractor with an incorrect calculation.

The conical flask contains sodium hydroxide, which is alkaline. Therefore, the starting solution has a high pH (alkaline) and methyl orange is yellow in alkaline solution. As sulfuric acid (acid) is added from the burette, the pH decreases because the alkali is being neutralised. At the end-point (neutralization), methyl orange changes from yellow to orange. (If excess acid is added, it turns red).

Award 1 mark for option A. Option B represents the reverse titration (adding alkali to acid). Options C and D contain incorrect combinations of colour change or pH trend.

First, write the balanced equation for the reaction:
\[\text{CaCO}_3\text{(s)} + 2\text{HCl}\text{(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O}\text{(l)} + \text{CO}_2\text{(g)}\]
From the equation, 1 mole of \(\text{CaCO}_3\) produces 1 mole of \(\text{CO}_2\).

Calculate the number of moles of \(\text{CaCO}_3\):
\[\text{moles} = \frac{\text{mass}}{M_r} = \frac{10.0\text{ g}}{100} = 0.10\text{ mol}\]

Therefore, \(0.10\text{ mol}\) of \(\text{CO}_2\) is produced.

Calculate the volume of \(\text{CO}_2\) at rtp:
\[\text{Volume} = 0.10\text{ mol} \times 24.0\text{ dm}^3\text{/mol} = 2.40\text{ dm}^3\]

Award 1 mark for option A. Option B is incorrect (uses a 1:2 ratio). Option C is off by a factor of 10. Option D does not divide by the molar mass.

In the electrolysis of concentrated aqueous sodium chloride (brine):
- At the anode (+), chloride ions (\(\text{Cl}^-\)) are discharged in preference to hydroxide ions because of their high concentration, producing chlorine gas (\(\text{Cl}_2\)).
- At the cathode (-), hydrogen ions (\(\text{H}^+\)) from water are discharged in preference to sodium ions (\(\text{Na}^+\)) because hydrogen is lower in the reactivity series, producing hydrogen gas (\(\text{H}_2\)).
- As \(\text{H}^+\in\) water are discharged, \(\text{OH}^-\in\) water are left behind in the solution around the cathode, paired with \(\text{Na}^+\) ions. This forms sodium hydroxide (\(\text{NaOH}\)), which is alkaline, so the pH around the cathode increases.

Award 1 mark for option A. Option B is for dilute sodium chloride or acid. Option C has incorrect cathode product and pH trend. Option D is for dilute sulfuric acid.

Metal Z is the most reactive because it reacts with cold water. Metal X is moderately reactive because it does not react with cold water but does react with steam. Metal Y is the least reactive of the three because it does not react with steam, and its oxide is easily reduced by carbon (which is placed higher than Y in the reactivity series). Thus, the order of decreasing reactivity is Z, X, Y.

Award 1 mark for option A. Option B incorrect order between X and Y. Options C and D place Z as less reactive, which is incorrect since it reacts with cold water.

In test 1, the green precipitate that is insoluble in excess sodium hydroxide indicates the presence of iron(II) ions (\(\text{Fe}^{2+}\)). Note that chromium(III) also gives a green precipitate, but it is soluble in excess sodium hydroxide to give a green solution.
In test 2, the formation of a white precipitate with aqueous barium nitrate in the presence of nitric acid confirms the presence of sulfate ions (\(\text{SO}_4^{2-}\)).
Therefore, salt W is iron(II) sulfate.

Award 1 mark for option A. Option B is incorrect because iron(III) gives a red-brown precipitate. Option C is incorrect because chromium(III) hydroxide is soluble in excess NaOH. Option D is incorrect because chloride ions would not give a precipitate with barium nitrate (they give a white precipitate with silver nitrate).

During the electrolysis of concentrated aqueous copper(II) chloride: 1. At the anode (positive electrode), chloride ions are discharged in preference to hydroxide ions because they are in high concentration. This produces chlorine gas. 2. At the cathode (negative electrode), copper(II) ions are discharged in preference to hydrogen ions because copper is lower in the reactivity series. This produces copper metal. 3. As blue copper(II) ions are discharged and removed from the solution, the blue colour of the electrolyte fades.

1 mark for identifying the correct products at both electrodes and the correct change in the appearance of the electrolyte (Option B).

The energy change of a reaction is calculated as the energy absorbed in bond breaking minus the energy released in bond making. 1. Energy absorbed to break reactant bonds: 1 mole of \(H-H\) bonds (436 kJ) + 1 mole of \(Cl-Cl\) bonds (243 kJ) = +679 kJ/mol. 2. Energy released when product bonds are formed: 2 moles of \(H-Cl\) bonds = 2 x 432 = 864 kJ/mol. 3. Overall energy change: +679 - 864 = -185 kJ/mol.

1 mark for calculating total bond-breaking energy (+679 kJ/mol) and total bond-making energy (-864 kJ/mol) to arrive at the correct overall energy change of -185 kJ/mol (Option A).

A weak acid is only partially ionised (dissociated) in aqueous solution compared to a strong acid of the same concentration which is fully ionised. Because it is partially ionised, the concentration of hydrogen ions in the weak acid is lower. A lower concentration of hydrogen ions results in a higher pH (closer to neutral 7) and decreases the rate of reaction with magnesium, meaning it reacts more slowly.

1 mark for correctly identifying that a weak acid has a higher pH and reacts more slowly than a strong acid of the same concentration (Option B).

An ion with a +3 charge has 3 fewer electrons than protons. Since the ion has 10 electrons, the neutral atom has 10 + 3 = 13 electrons. Therefore, the number of protons (proton number) is 13. The nucleon number (mass number) is the sum of protons and neutrons: 13 + 14 = 27.

1 mark for determining the correct proton number (13) and nucleon number (27) (Option C).

Metal \(W\) reacts with the oxide of metal \(Y\), which means \(W\) is more reactive than \(Y\) (\(W > Y\)). Metal \(W\) does not react with the oxide of metal \(Z\), which means \(Z\) is more reactive than \(W\) (\(Z > W\)). Metal \(X\) reacts with the oxide of metal \(Z\), which means \(X\) is more reactive than \(Z\) (\(X > Z\)). Combining these relationships: \(X > Z > W > Y\).

1 mark for correctly deducing the order of reactivity as X > Z > W > Y (Option A).

Analyzing the formula of the organic compound: The group \(-OH\) attached to a carbon atom is an alcohol functional group. The group \(-COOH\) at the end of the chain is a carboxylic acid functional group. Therefore, the compound contains both alcohol and carboxylic acid functional groups.

1 mark for identifying the functional groups as alcohol and carboxylic acid (Option A).

1. Find the moles of magnesium reacting: moles of Mg = 2.4 g / 24 g/mol = 0.1 mol. 2. Use the stoichiometric ratio from the balanced equation: 1 mole of Mg produces 1 mole of H2. Therefore, 0.1 mol of Mg produces 0.1 mol of H2. 3. Calculate the volume of H2 gas at r.t.p.: Volume = 0.1 mol x 24 dm3/mol = 2.4 dm3.

1 mark for calculating the correct volume of hydrogen gas as 2.4 dm3 (Option B).

To obtain Curve Y, the initial rate must be higher because the curve is steeper, which requires a higher concentration of acid. However, the total volume of gas collected must remain the same, meaning the moles of the limiting reactant (hydrochloric acid) must remain unchanged. Calculating the moles of HCl: Original has 50 cm3 of 1.0 mol/dm3 HCl = 0.05 mol. Option C has 25 cm3 of 2.0 mol/dm3 HCl = 0.05 mol. Since the concentration is doubled (2.0 mol/dm3 vs 1.0 mol/dm3), the reaction rate is higher (steeper curve), but because the number of moles of acid is the same, the final volume of gas is identical.

1 mark for identifying that higher concentration with the same number of moles of acid yields a faster initial rate but the same final volume of gas (Option C).

During the electrolysis of concentrated aqueous copper(II) chloride:
- At the cathode (negative electrode), copper ions (\(Cu^{2+}\)) are preferentially discharged over hydrogen ions (\(H^+\)) because copper is lower in the reactivity series. This forms copper metal (a pink-brown solid coating).
- At the anode (positive electrode), chloride ions (\(Cl^-\)) are preferentially discharged over hydroxide ions (\(OH^-\)) because they are in a high concentration. This forms chlorine gas.
- As blue copper(II) ions are discharged and removed from the solution, the intensity of the blue color decreases, meaning the blue color fades (eventually becoming colorless).

1 mark for correct option A.
- Reject other options that show incorrect electrode products or incorrect color changes.

1. Calculate the number of moles of magnesium used:
\(\text{moles of Mg} = \frac{\text{mass}}{\text{relative atomic mass}} = \frac{0.36\text{ g}}{24\text{ g/mol}} = 0.015\text{ mol}\)

2. Use the molar ratio from the balanced chemical equation:
\(1\text{ mole of Mg}\) produces \(1\text{ mole of H}_2\).
Therefore, \(0.015\text{ mol of Mg}\) produces \(0.015\text{ mol of H}_2\).

3. Calculate the volume of hydrogen gas at r.t.p.:
\(\text{volume of gas} = \text{moles} \times 24\text{ dm}^3\text{/mol} = 0.015\text{ mol} \times 24\text{ dm}^3\text{/mol} = 0.36\text{ dm}^3\).

1 mark for correct option B.
- Award 1 mark for correct calculation showing 0.36 dm³.

1. Calculate the energy required to break reactants' bonds:
- \(4 \times \text{C–H}\) bonds = \(4 \times 410 = 1640\text{ kJ/mol}\)
- \(2 \times \text{O=O}\) bonds = \(2 \times 496 = 992\text{ kJ/mol}\)
Total energy absorbed to break bonds = \(1640 + 992 = 2632\text{ kJ/mol}\)

2. Calculate the energy released when products' bonds are formed:
- \(2 \times \text{C=O}\) bonds = \(2 \times 805 = 1610\text{ kJ/mol}\)
- \(4 \times \text{O–H}\) bonds = \(4 \times 460 = 1840\text{ kJ/mol}\)
Total energy released from forming bonds = \(1610 + 1840 = 3450\text{ kJ/mol}\)

3. Calculate \(\Delta H\):
\(\Delta H = \text{Energy absorbed} - \text{Energy released} = 2632 - 3450 = -818\text{ kJ/mol}\).

1 mark for correct option A.
- Confirm the negative sign for exothermic combustion reaction.

- Statement 1 is incorrect: Increasing temperature does not lower the activation energy. Activation energy can only be lowered by adding a catalyst.
- Statement 2 is correct: At a higher temperature, particles have more kinetic energy and move faster, leading to more frequent collisions per unit time.
- Statement 3 is correct: Because particles have higher kinetic energies, a much larger fraction of colliding molecules possess energy equal to or greater than the activation energy (\(E \ge E_a\)), resulting in a higher proportion of successful collisions. Therefore, only statements 2 and 3 are correct.

1 mark for correct option C.
- Statement 1 must be excluded, as temperature has no effect on activation energy.

Oxides that react with both acids (such as hydrochloric acid) and alkalis (such as sodium hydroxide) to form salt and water are called amphoteric oxides. Elements that form amphoteric oxides (such as aluminium in \(\text{Al}_2\text{O}_3\) and zinc in \(\text{ZnO}\)) are metals. Therefore, the oxide is amphoteric, and \(X\) is a metal.

1 mark for correct option A.
- Reject B because amphoteric oxides are formed by metals.
- Reject C and D because basic and acidic oxides only react with one of the classes (acids or bases respectively).

- Metal \(Z\) must be highly reactive because it reacts vigorously with cold water, placing it at the top of the list.
- Metal \(X\) is very reactive since its oxide is too stable to be reduced by either carbon or hydrogen.
- Metal \(W\) is moderately reactive because carbon can reduce its oxide, but hydrogen cannot. This placing is below carbon but above hydrogen.
- Metal \(Y\) is the least reactive because hydrogen is reactive enough to reduce its oxide, indicating \(Y\) lies below hydrogen in the reactivity series.
- Therefore, the correct order of reactivity is \(Z \rightarrow X \rightarrow W \rightarrow Y\).

1 mark for correct option A.
- Reject options where Z is not the most reactive, or where Y is not the least reactive.

Structural isomers are compounds that have the same molecular formula but different structural arrangements.
- Propan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)) and propan-2-ol (\(\text{CH}_3\text{CH(OH)CH}_3\)) both share the molecular formula \(\text{C}_3\text{H}_8\text{O}\) but have different structures (the position of the hydroxyl functional group differs). Therefore, they are structural isomers.
- Methane and ethane (option B) are different alkanes with different molecular formulas (\(\text{CH}_4\) and \(\text{C}_2\text{H}_6\)).
- Ethene and propene (option C) are different alkenes with different molecular formulas (\(\text{C}_2\text{H}_4\) and \(\text{C}_3\text{H}_6\)).
- Ethanoic acid and ethyl methanoate (option D) have different molecular formulas: \(\text{C}_2\text{H}_4\text{O}_2\) vs \(\text{C}_3\text{H}_6\text{O}_2\).

1 mark for correct option A.
- Reject options B, C, and D as they do not have the same molecular formula.

1. Cation test: Aqueous sodium hydroxide reacts with \(Fe^{2+}\) ions to produce a green precipitate of iron(II) hydroxide, which is insoluble in excess sodium hydroxide. (Note: Chromium(III) also forms a green precipitate, but it is soluble in excess sodium hydroxide to form a green solution).
2. Anion test: Dilute nitric acid followed by aqueous barium nitrate is the standard test for sulfate ions (\(SO_4^{2-}\)). The formation of a white precipitate of barium sulfate confirms the presence of sulfate ions.
Combining these observations, salt \(S\) is iron(II) sulfate.

1 mark for correct option B.
- Reject A: chromium(III) hydroxide is soluble in excess NaOH.
- Reject C: chloride ions form a precipitate with silver nitrate, not barium nitrate.
- Reject D: iron(III) forms a red-brown precipitate with NaOH.

(a) Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons.

(b) (i)
- An atom of \(^{26}\text{Mg}\): 12 protons, 14 neutrons, 12 electrons.
- A magnesium ion, \(^{24}\text{Mg}^{2+}\): 12 protons, 12 neutrons, 10 electrons.

(ii) They have the same number of electrons in their outer shell / same electronic configuration.

(c) (i) The magnesium ion \(\text{Mg}^{2+}\) should be drawn with a \(2+\) charge and either a full second shell (8 electrons) or empty outer shell. The oxide ion \(\text{O}^{2-}\) should be drawn with a \(2-\) charge and 8 electrons in its outer shell (6 of one symbol e.g., crosses, and 2 of another e.g., dots, representing the electrons transferred from magnesium).

(ii) Magnesium oxide has a giant ionic lattice structure with strong electrostatic forces of attraction between the oppositely charged ions (\(\text{Mg}^{2+}\) and \(\text{O}^{2-}\)), which require a large amount of thermal energy to overcome.

(a) 1 mark for: atoms of the same element / same number of protons (or atomic number); 1 mark for: different number of neutrons (or mass number).

(b)(i) 1 mark for protons of both (12 and 12); 1 mark for neutrons of \(^{26}\text{Mg}\) (14); 1 mark for neutrons of \(^{24}\text{Mg}^{2+}\) (12); 1 mark for electrons of both (12 and 10).

(b)(ii) 1 mark for: same number of outer shell electrons; 1 mark for: chemical reactions depend on outer shell electrons / electronic configuration.

(c)(i) 1 mark for: correct charge on magnesium ion (\(2+\)) and oxide ion (\(2-\)); 1 mark for: correct outer shell electronic structure of magnesium ion (empty outer shell or 8 electrons); 1 mark for: correct outer shell of oxide ion with 8 electrons showing transfer (e.g. 6 dots and 2 crosses).

(c)(ii) 1 mark for: giant ionic lattice; 1 mark for: strong electrostatic attraction between oppositely charged ions; 0.33 mark for: requires a lot of energy to break.

(a) (i) An exothermic reaction is one where more energy is released when making new bonds than is absorbed when breaking existing bonds.
(ii) The energy level diagram should show: reactants at a higher energy level than products; a curve going up to a maximum (representing the activation energy) and then down to the products; the activation energy, \(E_a\), represented by an upward arrow from the reactants to the peak; the enthalpy change, \(\Delta H\), represented by a downward arrow from reactants to products.

(b) Energy absorbed to break bonds: \(436 + 242 = 678\text{ kJ/mol}\).
Energy released to make bonds: \(2 \times 431 = 862\text{ kJ/mol}\).
\(\Delta H = 678 - 862 = -184\text{ kJ/mol}\).

(c) \(\Delta H = E_{\text{break}} - E_{\text{make}}\)
\(+180 = E_{\text{break}} - 1260\)
\(E_{\text{break}} = 180 + 1260 = 1440\text{ kJ}\).

(a)(i) 1 mark for: energy is absorbed to break bonds and released to make bonds; 1 mark for: more energy released than absorbed.

(a)(ii) 1 mark for: reactants drawn at a higher level than products; 1 mark for: a curve rising from reactants to a peak and falling to products; 1 mark for: \(E_a\) correctly labeled with an arrow from reactant level to peak; 1 mark for: \(\Delta H\) correctly labeled with an arrow pointing down from reactant level to product level.

(b) 1 mark for: calculation of energy to break bonds (678 kJ/mol); 1 mark for: calculation of energy to make bonds (862 kJ/mol); 1 mark for: subtraction to get -184; 0.33 mark for: correct negative sign and units (kJ/mol).

(c) 1 mark for: correct formula or relationship (\(\Delta H = E_{\text{break}} - E_{\text{make}}\)); 1 mark for: substitution of values (\(180 = E_{\text{break}} - 1260\)); 1 mark for: transposition (\(E_{\text{break}} = 180 + 1260\)); 1 mark for: final value of 1440 kJ.

(a) Indicator: Phenolphthalein. Color change: from pink to colorless. (Alternatively: Methyl orange, from yellow to orange/red).

(b) A pipette is more accurate/precise (or has a lower percentage error/uncertainty) than a measuring cylinder for measuring a fixed volume.

(c) (i) \(\text{Moles of NaOH} = \text{concentration} \times \text{volume} = 0.150\text{ mol/dm}^3 \times \frac{25.0}{1000}\text{ dm}^3 = 3.75 \times 10^{-3}\text{ mol}\).

(ii) From the chemical equation, \(1\text{ mol}\) of \(\text{H}_2\text{SO}_4\) reacts with \(2\text{ mol}\) of \(\text{NaOH}\).
\(\text{Moles of H}_2\text{SO}_4 = \frac{3.75 \times 10^{-3}}{2} = 1.875 \times 10^{-3}\text{ mol}\).

(iii) \(\text{Concentration of H}_2\text{SO}_4 = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{1.875 \times 10^{-3}}{0.01875} = 0.100\text{ mol/dm}^3\).

(iv) \(M_r(\text{H}_2\text{SO}_4) = 2(1.0) + 32.1 + 4(16.0) = 98.1\) (or 98).
\(\text{Concentration in g/dm}^3 = 0.100\text{ mol/dm}^3 \times 98.1\text{ g/mol} = 9.81\text{ g/dm}^3\) (or \(9.80\text{ g/dm}^3\) if using \(M_r = 98\)).

(a) 1 mark for: Phenolphthalein (accept Methyl orange); 1 mark for: pink (or yellow); 1 mark for: colorless (or orange/red).

(b) 1 mark for: more accurate / precise / lower uncertainty than a measuring cylinder.

(c)(i) 1 mark for: correct formula or division by 1000 (\(0.150 \times 25/1000\)); 1 mark for: \(3.75 \times 10^{-3}\text{ mol}\).

(c)(ii) 1 mark for: 1:2 mole ratio (halving the moles of NaOH); 1 mark for: \(1.875 \times 10^{-3}\text{ mol}\).

(c)(iii) 1 mark for: dividing moles of acid by \(18.75/1000\) (or 0.01875); 1 mark for: \(0.100\text{ mol/dm}^3\).

(c)(iv) 1 mark for: calculating \(M_r(\text{H}_2\text{SO}_4) = 98\) or \(98.1\); 1 mark for: multiplying concentration in mol/dm3 by \(M_r\); 1.33 marks for: final value \(9.8\) to \(9.81\) with unit \(\text{g/dm}^3\).

(a) B, C, A, D (most reactive to least reactive).

(b) (i) \(\text{C(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{CSO}_4\text{(aq)} + \text{Cu(s)}\) (or ionic: \(\text{C} + \text{Cu}^{2+} \rightarrow \text{C}^{2+} + \text{Cu}\)).
(ii) Any two from:
- Blue solution fades/becomes colorless.
- Pink/brown/red-brown solid is deposited on the metal.
- The metal C dissolves/disappears/shrinks.
- Temperature increases (exothermic reaction).

(c) (i) \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\).
(ii) Calcium carbonate is added to remove silicon dioxide (silica), which is an acidic impurity. First, calcium carbonate undergoes thermal decomposition to form calcium oxide and carbon dioxide:
\(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\)
Then, calcium oxide (a basic oxide) reacts with silicon dioxide (an acidic oxide) to form calcium silicate (slag):
\(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\)
The molten slag floats on top of the molten iron and is tapped off.

(a) 2 marks for: B, C, A, D (1 mark if order is reversed or if one error is made).

(b)(i) 2 marks for: \(\text{C} + \text{CuSO}_4 \rightarrow \text{CSO}_4 + \text{Cu}\) (1 mark for correct formulas of reactants, 1 mark for correct products. Allow ionic equation).

(b)(ii) 2 marks for any two observations (1 mark each):
- blue solution turns lighter / fades / becomes colorless
- pink/brown/red-brown solid formed
- metal C dissolves / gets smaller
- mixture gets warm / temperature increases.

(c)(i) 2 marks for: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\) (1 mark for correct reactants and products, 1 mark for correct balancing).

(c)(ii) 1.5 marks for: thermal decomposition of calcium carbonate to form calcium oxide / equation \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\); 1.5 marks for: CaO reacts with silica / SiO2 (acidic impurity); 1.5 marks for: reaction equation \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\); 0.83 marks for: slag floats on the molten iron and is tapped off.

(a) (i) Structural isomers are molecules that have the same molecular formula but different structural arrangements (different structural formulas).
(ii)
- Isomer 1: Butanoic acid (carboxylic acid homologous series). Structure should show: \(\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-COOH}\) showing all C-H, C-C, C=O, C-O, and O-H bonds.
- Isomer 2: Ethyl ethanoate (ester homologous series). Structure should show: \(\text{CH}_3\text{-COO-CH}_2\text{-CH}_3\) showing all C-H, C-C, C=O, C-O bonds.

(b) (i) Ethyl propanoate.
(ii) Structural formula of ethyl propanoate should show all atoms and bonds: \(\text{CH}_3\text{-CH}_2\text{-C(=O)-O-CH}_2\text{-CH}_3\).
(iii) Water.

(c) Add aqueous bromine (bromine water) to the sample. If the compound is unsaturated, the orange/brown/yellow bromine water will be decolorized / turn colorless.

(a)(i) 1 mark for: same molecular formula; 1 mark for: different structural formula.

(a)(ii) 1 mark for: fully displayed structure of butanoic acid; 1 mark for: 'carboxylic acid' as homologous series; 1 mark for: fully displayed structure of ethyl ethanoate (or methyl propanoate / propyl methanoate); 1 mark for: 'ester' as homologous series.

(b)(i) 1 mark for: ethyl propanoate (reject other names).

(b)(ii) 2 marks for: correct displayed structure showing all bonds (1 mark for correct ester linkage \(\text{-C(=O)-O-}\), 1 mark for correct alkyl groups ethyl and propyl attached to the correct ends).

(b)(iii) 1 mark for: water / \(\text{H}_2\text{O}\).

(c) 1 mark for: add aqueous bromine / bromine water (reject liquid bromine/bromine gas alone); 1 mark for: initial color yellow/orange/brown; 1.33 marks for: becomes colorless / decolorized (reject 'clear').

(a) (i) Iron(II) ion / \(\text{Fe}^{2+}\).
(ii) Copper(II) ion / \(\text{Cu}^{2+}\).

(b) (i) Anion: Bromide / \(\text{Br}^-\).
Ionic equation: \(\text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)}\).

(ii) Anion: Sulfate / \(\text{SO}_4^{2-}\).
Ionic equation: \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\).

(c) To test for carbonate ions: Add dilute hydrochloric acid (or any other dilute strong acid) to the solid sample of M. Effervescence (bubbling) will be observed as a gas is produced. Bubble this gas through limewater (aqueous calcium hydroxide). The limewater will turn cloudy/milky, which confirms that the gas is carbon dioxide (\(\text{CO}_2\)), indicating the presence of carbonate ions.

(a)(i) 1 mark for: iron(II) / \(\text{Fe}^{2+}\) (reject iron / \(\text{Fe}^{3+}\)).

(a)(ii) 1 mark for: copper(II) / \(\text{Cu}^{2+}\) (reject copper / \(\text{Cu}^+\)).

(b)(i) 1 mark for: bromide / \(\text{Br}^-\); 1 mark for: correct formulas \(\text{Ag}^+ + \text{Br}^- \rightarrow \text{AgBr}\); 1 mark for: correct state symbols: (aq) for reactants and (s) for product.

(b)(ii) 1 mark for: sulfate / \(\text{SO}_4^{2-}\); 1 mark for: correct formulas \(\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4\); 1 mark for: correct state symbols: (aq) for reactants and (s) for product.

(c) 1.5 marks for: add dilute hydrochloric acid / nitric acid / any strong acid (reject concentrated acid); 1 mark for: mention of effervescence / gas evolved; 1 mark for: identifying the gas as carbon dioxide / \(\text{CO}_2\); 1 mark for: bubble the gas into limewater; 0.83 marks for: limewater turns cloudy / milky / white precipitate.

(a) The metal carbonate sample is placed in a hard-glass test-tube (or boiling tube) which is fitted with a rubber bung and a delivery tube. The other end of the delivery tube is connected to a gas syringe (or an inverted measuring cylinder filled with water). The test-tube is heated strongly using a Bunsen burner, and the volume of carbon dioxide gas evolved is recorded at regular intervals using a stopwatch. (b) The independent variable is the parameter being changed, which is the identity of the metal carbonate. The dependent variable is the parameter being measured, which is the volume of gas collected per unit of time (the rate of thermal decomposition). (c) Variables that must be kept constant include the mass of the metal carbonate used (e.g., 1.0 g), the particle size (surface area) of the carbonate powder, and the heating intensity (distance and strength of the Bunsen burner flame). (d) Sodium carbonate does not decompose under these conditions, showing that it is far more thermally stable than copper(II) carbonate. This is because sodium is a highly reactive Group 1 alkali metal which forms extremely stable ionic bonds in its carbonate lattice, whereas copper is a less reactive transition metal whose carbonate decomposes easily. (e) The gas is bubbled into limewater (aqueous calcium hydroxide). If the gas is carbon dioxide, the limewater turns cloudy or milky due to the formation of a white precipitate of calcium carbonate.

(a) [3 marks]: 1 mark for placing the carbonate in a test-tube and heating with a Bunsen burner; 1 mark for connecting it to a gas syringe (or inverted measuring cylinder over water); 1 mark for recording the volume of gas over time. (b) [2 marks]: 1 mark for identifying the independent variable as the type of metal carbonate; 1 mark for identifying the dependent variable as the volume of gas collected / rate of gas production. (c) [2 marks]: 1 mark each for any two controlled variables: mass of metal carbonate, particle size (surface area) of carbonate, or intensity/distance of Bunsen flame. (d) [2 marks]: 1 mark for stating sodium carbonate is more thermally stable than copper(II) carbonate; 1 mark for relating this to sodium being a highly reactive metal (or Group 1 metal). (e) [1 mark]: 1 mark for stating that bubbling the gas through limewater turns it cloudy.

(a) Dilution decreases the concentration of ethanoic acid in the commercial vinegar. If undiluted vinegar were used, it would require a very large and impractical volume of 0.100 mol/dm3 sodium hydroxide to reach the end-point, which would exceed the capacity of a standard 50 cm3 burette. (b) The burette must first be rinsed with distilled water to clean it of any contaminants. It must then be rinsed with a small amount of the 0.100 mol/dm3 sodium hydroxide solution to prevent dilution by the residual water. Finally, fill the burette with the sodium hydroxide solution above the zero mark and open the tap to fill the jet space below the tap completely, ensuring there are no air bubbles. (c) Phenolphthalein is colorless in acidic solutions (such as the diluted vinegar in the conical flask). At the end-point, when the acid is neutralized by the added sodium hydroxide, the solution becomes slightly alkaline, turning the indicator to a permanent pale pink. (d) The titre volumes are: Titration 1 = 24.10 - 0.60 = 23.50 cm3; Titration 2 = 25.30 - 1.20 = 24.10 cm3; Titration 3 = 23.60 - 0.20 = 23.40 cm3. Titration 2 (24.10 cm3) is the anomalous titre because it is not within 0.20 cm3 of the other two values. The concordant titres are 23.50 cm3 and 23.40 cm3. Mean titre = (23.50 + 23.40) / 2 = 23.45 cm3. (e) A volumetric pipette is designed to measure a single fixed volume extremely accurately with high precision and very low percentage error, whereas a measuring cylinder is less precise. Precaution: Read the liquid level at eye level with the bottom of the meniscus touching the graduation line, and do not blow out the final small drop remaining in the tip of the pipette.

(a) [1 mark]: 1 mark for explaining that concentrated acid would require too large a volume of sodium hydroxide (or would exceed the burette capacity). (b) [3 marks]: 1 mark for rinsing the burette with distilled water; 1 mark for rinsing with the sodium hydroxide solution; 1 mark for filling the burette and ensuring that the jet space below the tap is completely filled and free of air bubbles. (c) [1 mark]: 1 mark for colorless to pink (accept pale pink). (d) [3 marks]: 1 mark for calculating all three titre volumes correctly (23.50, 24.10, and 23.40 cm3); 1 mark for identifying Titration 2 as anomalous; 1 mark for calculating the correct mean of 23.45 cm3 using Titration 1 and Titration 3. (e) [2 marks]: 1 mark for stating that a volumetric pipette has a higher accuracy / lower percentage error than a measuring cylinder; 1 mark for a valid precaution (e.g., read the bottom of the meniscus at eye level, use a pipette filler safely, or do not force out the liquid left in the tip).

(a) Distribute a measured volume of dilute hydrochloric acid into a conical flask. Weigh a 0.10 g sample of magnesium ribbon. Add the magnesium to the flask, immediately fit a rubber bung containing a delivery tube, and start the stopwatch. Connect the delivery tube to a gas syringe to measure the volume of hydrogen gas collected. Read the volume on the syringe at regular intervals (e.g., every 10 seconds). (b) (i) Experiment 1 uses a higher concentration of hydrochloric acid, so its initial rate of reaction is faster, which means its curve has a steeper initial gradient. (ii) Both experiments use the same mass (0.10 g) of magnesium ribbon with excess acid, meaning the limiting reactant is identical. Thus, the final volume of hydrogen gas collected is exactly the same. (iii) Because Experiment 2 is slower, its curve lies below the curve of Experiment 1 at all times before both curves level off at the same plateau. (c) The reactions stop when the magnesium ribbon (the limiting reactant) is completely used up. Since the same mass (0.10 g) of magnesium is used in both trials and the acid is in excess, the same chemical quantity (moles) of magnesium reacts, producing the same total volume of hydrogen gas. (d) Increasing the temperature increases the initial rate of reaction. According to collision theory, raising the temperature gives the reactant particles more kinetic energy. As a result, they move faster and collide more frequently, and a greater fraction of the colliding particles possess energy equal to or greater than the activation energy, leading to a higher frequency of successful collisions.

(a) [3 marks]: 1 mark for adding magnesium to acid in a conical flask and quickly sealing it with a bung; 1 mark for collecting the gas using a gas syringe or an inverted measuring cylinder over water; 1 mark for starting the stopwatch immediately and taking readings at regular intervals. (b) [3 marks]: 1 mark for (i) stating that Experiment 1 has a steeper initial gradient (or Experiment 2 has a shallower gradient); 1 mark for (ii) stating that the final volumes of gas collected are equal; 1 mark for (iii) stating that the curve for Experiment 2 lies below the curve for Experiment 1. (c) [2 marks]: 1 mark for explaining that the reaction stops because magnesium is the limiting reactant and is completely consumed; 1 mark for explaining that the same mass of magnesium yields the same moles of gas. (d) [2 marks]: 1 mark for stating that a higher temperature increases the initial rate of reaction; 1 mark for explaining that particles have more kinetic energy resulting in more frequent collisions (or a higher fraction of particles exceeding the activation energy).

(a) (i) The cation is iron(II), Fe2+. The formation of a green precipitate with sodium hydroxide that is insoluble in excess is characteristic of Fe2+. (ii) To confirm the presence of Fe2+, add aqueous ammonia dropwise to a sample of solution Y. A green precipitate of iron(II) hydroxide will form. When excess aqueous ammonia is added, the green precipitate remains insoluble. (b) The reaction of a salt with dilute acid that produces a gas which turns limewater cloudy (carbon dioxide) confirms that the anion is carbonate, CO3^2-. (c) To test for ammonium ions, add aqueous sodium hydroxide to solid Z or its solution, and warm the mixture gently. Test the gas evolved with damp red litmus paper. The presence of ammonium ions is confirmed if the gas evolved (ammonia) turns damp red litmus paper blue. (d) To prepare pure, dry crystals of magnesium sulfate: 1. Add excess magnesium carbonate powder to a measured volume of dilute sulfuric acid in a beaker and stir until no more effervescence is seen and some unreacted solid remains at the bottom. 2. Filter the mixture using a funnel and filter paper into an evaporating basin to remove the excess unreacted magnesium carbonate. 3. Heat the filtrate in the evaporating basin until the crystallization point is reached (tested by dipping a cold glass rod to see if crystals form). 4. Leave the hot saturated solution to cool and crystallize slowly. Filter the crystals, rinse them with a small volume of cold distilled water to remove impurities, and dry them between sheets of filter paper.

(a) [3 marks total]: (i) 1 mark for identifying the cation as iron(II) / Fe2+ (do not accept iron or Fe3+). (ii) 2 marks: 1 mark for stating that adding aqueous ammonia dropwise forms a green precipitate; 1 mark for stating that the precipitate is insoluble in excess. (b) [1 mark]: 1 mark for identifying the anion as carbonate / CO3^2-. (c) [2 marks]: 1 mark for adding aqueous sodium hydroxide and warming/heating; 1 mark for stating that the gas produced turns damp red litmus paper blue. (d) [4 marks]: 1 mark for adding excess magnesium carbonate to dilute sulfuric acid; 1 mark for filtering the mixture to remove the unreacted solid; 1 mark for heating the filtrate to the crystallization point and leaving it to cool; 1 mark for filtering the crystals, washing with cold distilled water, and drying between filter papers.