2024 Cambridge IGCSE Chemistry (0620) 模擬試題連答案詳解

Copper(II) carbonate is insoluble, so it is obtained as a residue when filtered. Washing with distilled water removes impurities, and drying produces a pure solid. Copper(II) sulfate is soluble, passing into the filtrate. To obtain crystals, the filtrate is evaporated to a hot saturated solution, then cooled to crystallize. The crystals are filtered and dried. Evaporating to complete dryness (Option B) would produce anhydrous copper(II) sulfate powder instead of hydrated crystals.

Award 1 mark for selecting option A. Reject all other options.

Ethanol has a lower boiling point (78 °C) than water (100 °C). During fractional distillation, the substance with the lower boiling point vaporizes more easily, passes up the fractionating column, and condenses first in the condenser, where it is collected as the first fraction.

Award 1 mark for selecting option A. Reject all other options.

Percentage of O = 100% - 70.0% = 30.0%. Moles of Fe = 70.0 / 56 = 1.25. Moles of O = 30.0 / 16 = 1.875. Dividing both values by the smallest number of moles (1.25) gives a ratio of Fe : O of 1 : 1.5. Multiplying by 2 to obtain whole numbers gives \(\text{Fe}_2\text{O}_3\).

Award 1 mark for selecting option B. Reject all other options.

Moles of \(\text{CaCO}_3\) = 25.0 g / 100 g/mol = 0.25 mol. The stoichiometry of the reaction is 1:1, so 0.25 mol of \(\text{CO}_2\) gas is produced. Volume of \(\text{CO}_2\) = 0.25 mol x 24 \(\text{dm}^3/\text{mol}\) = 6.0 \(\text{dm}^3\).

Award 1 mark for selecting option B. Reject all other options.

A weak acid is only partially dissociated in aqueous solution, which leads to a lower concentration of hydrogen ions than in a strong acid of the same concentration. This lower concentration of hydrogen ions results in a higher pH (less acidic) and a slower rate of reaction with metals such as magnesium.

Award 1 mark for selecting option A. Reject all other options.

Carbon dioxide (CO2) is a non-metal oxide that dissolves in water to form an acidic solution. Calcium oxide (CaO) is a metal oxide with basic properties. Zinc oxide (ZnO) is an amphoteric oxide that reacts with both acids and bases. Carbon monoxide (CO) is a neutral oxide that reacts with neither acids nor bases.

Award 1 mark for selecting option A. Reject all other options.

The formation of a green precipitate with aqueous sodium hydroxide which is insoluble in excess confirms the presence of iron(II) (Fe2+) ions. The formation of a white precipitate with acidified barium nitrate confirms the presence of sulfate (SO4^2-) ions. Therefore, the substance X is iron(II) sulfate (FeSO4).

Award 1 mark for selecting option A. Reject all other options.

Sulfur dioxide is a colorless, pungent, choking gas. It is a reducing agent that reduces the purple manganate(VII) ions to colorless manganese(II) ions.

Award 1 mark for selecting option D. Reject all other options.

Since the liquids are miscible and have boiling points that are relatively close to each other, fractional distillation is the correct method. Fractional distillation requires a fractionating column to separate miscible liquids with close boiling points, along with a condenser to cool and collect the vapor. Simple distillation is only effective when boiling points are very far apart. Filtration is for insoluble solids, and chromatography is for separation of small amounts of mixtures.

1 mark for the correct option (C).

Test 1: Dilute hydrochloric acid is added to solid W, producing a gas that turns limewater cloudy. This gas is carbon dioxide, confirming that solid W contains the carbonate ion. This rules out calcium chloride. Test 2: Adding aqueous sodium hydroxide to a solution of W until in excess produces a white precipitate that is insoluble in excess sodium hydroxide. Calcium ions form a white precipitate with aqueous NaOH that is insoluble in excess. Zinc and aluminium ions both form white precipitates with aqueous NaOH that dissolve in excess to give a colorless solution. Therefore, the cation present is calcium. Combining the cation and anion gives calcium carbonate.

1 mark for the correct option (A).

Calculate the ratio of moles of Fe to O: Moles of Fe = 70.0 / 56 = 1.25 mol. Moles of O = 30.0 / 16 = 1.875 mol. Divide by the smaller value (1.25): Ratio of Fe = 1.25 / 1.25 = 1. Ratio of O = 1.875 / 1.25 = 1.5. Multiply by 2 to obtain whole numbers: Fe = 2, O = 3. The empirical formula is \(\text{Fe}_2\text{O}_3\).

1 mark for the correct option (B).

Dilute nitric acid reacts with a metal carbonate (calcium carbonate) to form a salt (calcium nitrate), water, and carbon dioxide gas. Copper is below hydrogen in the reactivity series and does not react with dilute acids to release hydrogen. The reaction of an acid and an alkali (neutralisation) is exothermic, not endothermic. Acids turn blue litmus paper red, not red litmus paper blue.

1 mark for the correct option (C).

To prepare a soluble salt (copper(II) sulfate) from an insoluble base (copper(II) oxide) and an acid: React copper(II) oxide with dilute sulfuric acid, heating to speed up the reaction, until the copper(II) oxide is in excess. Filter the hot mixture to remove the unreacted, excess copper(II) oxide (this leaves copper(II) sulfate solution as the filtrate). Evaporate the filtrate to the crystallisation point (heating until a saturated solution is formed; do not evaporate to dryness, otherwise anhydrous powder is formed, not hydrated crystals). Allow to cool so that crystals form. Filter the crystals from the remaining liquid and dry them gently between sheets of filter paper (heating strongly would dehydrate the crystals).

1 mark for the correct option (B).

Iron(II) ions form a green precipitate with aqueous sodium hydroxide which is insoluble in excess. They also form a green precipitate with aqueous ammonia which is insoluble in excess. Iron(III) forms a red-brown precipitate with both. Chromium(III) forms a green precipitate with sodium hydroxide which is soluble in excess to form a green solution. Copper(II) forms a blue precipitate.

1 mark for the correct option (A).

According to Avogadro's Law, the volume ratio of gases is equal to their mole ratio. The reaction can be written as: \(\text{C}_x\text{H}_y\text{(g)} + \text{O}_2\text{(g)} \rightarrow x\text{CO}_2\text{(g)} + \frac{y}{2}\text{H}_2\text{O(g)}\). Given volumes: \(\text{C}_x\text{H}_y = 10\ \text{cm}^3\) (1 volume), \(\text{CO}_2 = 40\ \text{cm}^3\) (4 volumes), \(\text{H}_2\text{O} = 50\ \text{cm}^3\) (5 volumes). Therefore, \(x = 4\) and \(\frac{y}{2} = 5 \implies y = 10\). The molecular formula of the hydrocarbon is \(\text{C}_4\text{H}_{10}\).

1 mark for the correct option (C).

An oxide that reacts with both acids and bases is classified as an amphoteric oxide. Aluminium oxide is amphoteric and reacts with both acids and bases. Calcium oxide is a basic oxide and only reacts with acids. Carbon dioxide and sulfur dioxide are acidic oxides and only react with bases. Therefore, only oxide 1 (aluminium oxide) reacts with both.

1 mark for the correct option (A).

Iodine undergoes sublimation when heated, transforming directly from a solid to a gas and then condensing on a cool surface, separating it from the sand and sodium chloride. The remaining mixture of sand and sodium chloride can be separated by adding water to dissolve the soluble sodium chloride. Filtration then removes the insoluble sand as a residue. Finally, evaporating the water from the filtrate yields pure, dry sodium chloride.

1 mark for the correct option B. Reject option A because dissolving first wet-capsulates the iodine making sublimation inefficient. Reject option C because filtration cannot separate dry solids without a liquid medium. Reject option D because dissolving first and then evaporating is redundant and does not separate sand.

Simple distillation is used to separate a volatile solvent (water, boiling point 100 degrees Celsius) from a non-volatile dissolved solute (copper(II) sulfate). Fractional distillation is used to separate miscible liquids with close boiling points. Filtration is used to separate insoluble solids from liquids. Crystallisation is used to obtain the solute (copper(II) sulfate crystals), not the solvent.

1 mark for the correct option A. Reject B because fractional distillation is for mixtures of volatile liquids. Reject C because copper(II) sulfate is soluble, so filtration will not separate it. Reject D because crystallisation collects the salt, not water.

To find the empirical formula: 1. Determine the mass of each element in 100 g of the compound: Mass of Fe = 70.0 g and Mass of O = 30.0 g. 2. Calculate the number of moles of each element: Moles of Fe = 70.0 / 56 = 1.25 mol and Moles of O = 30.0 / 16 = 1.875 mol. 3. Find the simplest whole-number ratio by dividing by the smallest value (1.25): Ratio of Fe = 1.25 / 1.25 = 1 and Ratio of O = 1.875 / 1.25 = 1.5. Multiply by 2 to get whole numbers: Fe2O3.

1 mark for the correct option B. Award 1 mark for correct calculation steps leading to Fe2O3.

According to Avogadro's Law, the volume ratio of reacting gases is equal to their mole ratio in the balanced equation. Ratio of volumes: CxHy : O2 : CO2 = 20 : 100 : 60 = 1 : 5 : 3. Thus, the reaction equation is CxHy + 5O2 -> 3CO2 + zH2O. From the carbon atoms on both sides: x = 3. From the oxygen atoms on both sides: 2 * 5 = (2 * 3) + z, which gives 10 = 6 + z, so z = 4. Therefore, there are 4 molecules of H2O. From the hydrogen atoms on both sides: y = 2 * z = 2 * 4 = 8. Thus, the formula of the hydrocarbon is C3H8.

1 mark for the correct option C. Award 1 mark for correct calculation of mole ratio 1 : 5 : 3 and solving for x and y.

All acids, whether strong or weak, react with metal carbonates (such as sodium carbonate) to produce a salt, carbon dioxide gas, and water. Weak acids only partially dissociate in water, so option A is incorrect. Copper is low in the reactivity series and does not react with dilute acids to produce hydrogen gas, so option B is incorrect. Acids turn blue litmus paper red, not red litmus paper blue, so option D is incorrect.

1 mark for the correct option C. Reject A because weak acids only partially dissociate. Reject B because copper does not react with dilute acids. Reject D because acids turn blue litmus red.

Amphoteric oxides are oxides of some metalloids or metals that can react as both acids and bases (e.g., Al2O3 and ZnO). Carbon monoxide (CO) is a neutral oxide. Calcium oxide (CaO) is a basic metallic oxide. Sulfur dioxide (SO2) is an acidic non-metallic oxide.

1 mark for the correct option D. Reject A (neutral oxide). Reject B (basic oxide). Reject C (acidic oxide).

1. The reaction with aqueous sodium hydroxide produces a green precipitate insoluble in excess, which indicates the presence of iron(II) ions. Chromium(III) also produces a green precipitate, but it dissolves in excess sodium hydroxide to form a green solution. 2. The reaction with acidified barium nitrate produces a white precipitate of barium sulfate, which indicates the presence of sulfate ions. Therefore, salt X is iron(II) sulfate.

1 mark for the correct option B. Reject A because chloride ions would not give a precipitate with barium nitrate. Reject C because iron(III) gives a red-brown precipitate. Reject D because chromium(III) precipitate is soluble in excess sodium hydroxide.

Chlorine gas is a strong bleaching agent and bleaches damp litmus paper. Ammonia is alkaline and turns damp red litmus paper blue. Carbon dioxide is weakly acidic and turns damp blue litmus paper red but does not bleach it. Oxygen sustains combustion but does not bleach litmus paper.

1 mark for the correct option C. Reject A, B, and D because none of these gases has a bleaching action on damp litmus paper.

In fractional distillation, the thermometer bulb is positioned at the top of the fractionating column where vapor enters the condenser to measure the boiling point of the vapor being distilled. The component with the lower boiling point (propanone, 56 °C) vaporizes more readily and distills first.

Award 1 mark for the correct answer A.

To obtain pure, dry crystals, the solution is heated to saturate it, then allowed to cool slowly to form crystals. Filtration separates the crystals from the remaining liquid. Washing with a small amount of cold distilled water removes impurities without dissolving too much product, and drying with filter paper removes surface water without decomposing the salt.

Award 1 mark for the correct answer B.

Step 1: Divide the percentage of each element by its relative atomic mass. For C: \(40.0 / 12 = 3.33\). For H: \(6.7 / 1 = 6.7\). For O: \(53.3 / 16 = 3.33\). Step 2: Divide each number by the smallest value (3.33). C: \(3.33 / 3.33 = 1\), H: \(6.7 / 3.33 \approx 2\), O: \(3.33 / 3.33 = 1\). The empirical formula is therefore \(\text{CH}_2\text{O}\).

Award 1 mark for the correct answer B.

On the right side of the balanced equation, there are \(8 \times 2 = 16\) oxygen atoms in carbon dioxide, and \(10 \times 1 = 10\) oxygen atoms in water. This gives a total of \(16 + 10 = 26\) oxygen atoms. Since oxygen exists as diatomic molecules (\(\text{O}_2\)), we need \(26 / 2 = 13\) molecules of oxygen. Therefore, \(x = 13\).

Award 1 mark for the correct answer C.

Copper(II) carbonate is a green solid that reacts with dilute sulfuric acid to form copper(II) sulfate (a blue solution), carbon dioxide (which causes bubbling/effervescence), and water. Since the copper(II) carbonate is in excess, some unreacted green solid will remain at the bottom of the beaker.

Award 1 mark for the correct answer C.

Weak acids only partially dissociate (ionize) in aqueous solutions. Consequently, they release fewer hydrogen ions (\(\text{H}^+\)) into the solution, resulting in a lower concentration of hydrogen ions and a higher pH value than a strong acid of the equal concentration.

Award 1 mark for the correct answer B.

Aluminium ions (\(\text{Al}^{3+}\)) form a white precipitate of aluminium hydroxide, \(\text{Al(OH)}_3\), with both sodium hydroxide and ammonia. This precipitate is soluble in excess sodium hydroxide but insoluble in excess ammonia. Zinc ions also form a white precipitate with both reagents, but it is soluble in excess of BOTH reagents.

Award 1 mark for the correct answer A.

Acidified silver nitrate forms a yellow precipitate of silver iodide when iodide ions (\(\text{I}^-\)) are present. Acidified barium chloride forms a white precipitate of barium sulfate when sulfate ions (\(\text{SO}_4^{2-}\)) are present; because no precipitate was formed here, sulfate ions are absent.

Award 1 mark for the correct answer B.

Methanol and ethanol are miscible liquids with close boiling points (boiling point difference of 13 °C). Fractional distillation is the most effective method to separate miscible liquids with close boiling points because the fractionating column allows repeated condensation and vaporization.

1 mark for selecting B (Fractional distillation).

First, calculate the moles of each element in 100 g: Moles of \(C = \frac{85.7}{12.0} \approx 7.14\) mol. Moles of \(H = \frac{14.3}{1.0} = 14.3\) mol. Next, divide by the smallest number of moles (7.14): \(C = \frac{7.14}{7.14} = 1\), \(H = \frac{14.3}{7.14} \approx 2\). Therefore, the empirical formula is \(\text{CH}_2\).

1 mark for selecting B (CH2).

A red colour with universal indicator and a rapid reaction with magnesium indicates a strong acid (X). A blue colour and no reaction with magnesium indicates an alkali (Y). An orange-yellow colour and a slow reaction with magnesium indicates a weak acid (Z).

1 mark for selecting C (X is a strong acid, Y is a strong alkali, Z is a weak acid).

The green precipitate insoluble in excess sodium hydroxide confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). The white precipitate formed with barium nitrate after acidification confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). Therefore, the salt is iron(II) sulfate.

1 mark for selecting C (Iron(II) sulfate).

Pencil lead contains graphite, which is insoluble in water and organic chromatography solvents. It does not dissolve or move up the paper, so it will not interfere with the chromatogram. Ink is soluble and would separate into its own dye components.

1 mark for selecting C (Pencil graphite is insoluble in the solvent and will not run or contaminate the chromatogram).

First, calculate the \(M_r\) of \(\text{CaCO}_3 = 40 + 12 + (3 \times 16) = 100\). Moles of \(\text{CaCO}_3 = \frac{10.0}{100} = 0.10\) mol. From the equation, 1 mole of \(\text{CaCO}_3\) produces 1 mole of \(\text{CO}_2\), so 0.10 mol of \(\text{CO}_2\) is produced. The \(M_r\) of \(\text{CO}_2 = 12 + (2 \times 16) = 44\). Mass of \(\text{CO}_2 = 0.10 \times 44 = 4.4\) g.

1 mark for selecting A (4.4 g).

A Brønsted-Lowry acid is defined as a proton (\(\text{H}^+\)) donor. In the forward reaction, \(\text{H}_2\text{O}\) donates a proton to \(\text{NH}_3\) to form \(\text{OH}^-\), so \(\text{H}_2\text{O}\) is the acid. In the reverse reaction, \(\text{NH}_4^+\) donates a proton to \(\text{OH}^-\), so \(\text{NH}_4^+\) is the acid.

1 mark for selecting B (H2O (forward) and NH4+ (reverse)).

Chlorine is an acidic gas with bleaching properties; it bleaches damp litmus paper. Carbon dioxide turns limewater cloudy, not relighting a glowing splint (which is oxygen). Ammonia is alkaline and turns damp red litmus paper blue. Hydrogen gas is tested with a lighted splint and burns with a 'pop' sound.

1 mark for selecting C (Chlorine | Bleaches damp litmus paper).

(a)(i)
1. Filter the mixture. The sand remains on the filter paper as residue.
2. Wash the sand residue with distilled water to remove any remaining salt solution.
3. Dry the sand in a warm oven or leave it to dry in the air.

(a)(ii)
1. Heat the sodium chloride solution (filtrate) in an evaporating basin to evaporate water until a saturated solution is formed (the crystallization point).
2. Allow the hot saturated solution to cool slowly so that sodium chloride crystals form.
3. Filter the mixture to separate the crystals from the remaining liquid.
4. Dry the crystals by pressing them gently between sheets of dry filter paper or in a warm oven.

(b)(i)
Fractional distillation

(b)(ii)
1. When the mixture is heated, the liquid with the lower boiling point (ethanol) vaporizes more readily and passes up the fractionating column.
2. The ethanol vapour is condensed in the condenser and collected as a pure liquid first, while water remains in the flask or condenses back.

(a)(i) [Total: 3 marks]
- Filter the mixture / filtration (1)
- Wash the residue with distilled water (1)
- Dry the residue / sand (1)

(a)(ii) [Total: 4 marks]
- Heat the solution / evaporate water until crystallization point / saturated solution is reached (1)
- Leave to cool / crystallize (1)
- Filter off the crystals (1)
- Dry crystals with filter paper / in a warm oven (1)

(b)(i) [Total: 1 mark]
- Fractional distillation (1)

(b)(ii) [Total: 2 marks]
- Low-boiling-point liquid / ethanol vaporizes first / goes up the column first (1)
- Vapour is condensed in the condenser / liquid is collected first (1)

(a)
1. Draw a straight line in pencil near the bottom of a piece of chromatography paper.
2. Place a small concentrated spot of the green food colouring onto the pencil line.
3. Suspend the paper in a beaker containing a suitable solvent, ensuring that the solvent level is below the pencil line.
4. Allow the solvent to travel up the paper near the top, remove the paper, and mark the position of the solvent front.

(b)(i)
Ink is soluble and contains dyes that would dissolve and separate in the solvent, interfering with the chromatogram, whereas pencil lead (graphite) is insoluble and will not run.

(b)(ii)
If the solvent is above the start line, the food colouring spots will dissolve and wash off into the solvent pool instead of travelling up the paper.

(c)(i)
\(R_f = \frac{\text{distance travelled by spot}}{\text{distance travelled by solvent front}}\)
\(R_f = \frac{6.4\text{ cm}}{8.0\text{ cm}} = 0.80\)

(c)(ii)
1. Spray the chromatogram with a locating agent (such as ninhydrin).
2. Heat the paper in an oven to develop the colour of the spots.

(a) [Total: 4 marks]
- Draw pencil line / baseline (1)
- Spot sample on the pencil line (1)
- Put paper in solvent with solvent level below the pencil line (1)
- Allow solvent to run up the paper and mark solvent front (1)

(b)(i) [Total: 1 mark]
- Pencil / graphite is insoluble / ink will dissolve and separate / run (1)

(b)(ii) [Total: 1 mark]
- To stop sample spots dissolving into the solvent / washing off (1)

(c)(i) [Total: 2 marks]
- Formula / correct working: \(6.4 / 8.0\) (1)
- Final value: \(0.8\) or \(0.80\) (1)

(c)(ii) [Total: 2 marks]
- Spray with a locating agent / ninhydrin (1)
- Heat / dry in oven (1)

(a)(i)
Mass of oxygen = mass of copper oxide - mass of copper
Mass of oxygen = \(1.59\text{ g} - 1.27\text{ g} = 0.32\text{ g}\)

(a)(ii)
1. Moles of Cu: \(\frac{1.27}{64} = 0.0198\text{ mol}\) (or approx 0.02)
2. Moles of O: \(\frac{0.32}{16} = 0.0200\text{ mol}\)
3. Ratio Cu : O = 0.0198 : 0.0200 \(\approx 1 : 1\)
Empirical formula is \(\text{CuO}\).

(a)(iii)
Reactants: \(\text{CuO}\) and \(\text{H}_2\)
Products: \(\text{Cu}\) and \(\text{H}_2\text{O}\)
Equation: \(\text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O}\)

(b)(i)
Balanced equation:
\(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\)

(b)(ii)
1. \(M_r\) of \(\text{SiO}_2 = 28 + (16 \times 2) = 60\)
Moles of \(\text{SiO}_2 = \frac{15.0}{60} = 0.25\text{ mol}\)
2. From balanced equation, 1 mole of \(\text{SiCl}_4\) produces 1 mole of \(\text{SiO}_2\).
Moles of \(\text{SiCl}_4\) needed = 0.25 mol.
\(M_r\) of \(\text{SiCl}_4 = 28 + (4 \times 35.5) = 170\)
Mass of \(\text{SiCl}_4 = 0.25 \times 170 = 42.5\text{ g}\)

(a)(i) [Total: 1 mark]
- \(0.32\text{ g}\) (1)

(a)(ii) [Total: 3 marks]
- Calculation of moles of Cu AND O (e.g. \(0.02\text{ mol}\) each) (1)
- Simplest ratio calculation (1:1) (1)
- Correct empirical formula \(\text{CuO}\) (1)

(a)(iii) [Total: 2 marks]
- Correct formulas for reactants and products (1)
- Correctly balanced (1)

(b)(i) [Total: 2 marks]
- Correct reactants and products (1)
- Correct balancing (\(2\text{H}_2\text{O}\) and \(4\text{HCl}\)) (1)

(b)(ii) [Total: 2 marks]
- Calculating moles of \(\text{SiO}_2 = 0.25\text{ mol}\) (1)
- Correct mass of \(\text{SiCl}_4 = 42.5\text{ g}\) (1)

(a)(i)
Mass of water lost = \(4.92\text{ g} - 2.40\text{ g} = 2.52\text{ g}\)

(a)(ii)
1. Moles of anhydrous \(\text{MgSO}_4 = \frac{2.40}{120} = 0.020\text{ mol}\)
2. Moles of water lost = \frac{2.52}{18} = 0.14\text{ mol}
3. Ratio of moles of \(\text{MgSO}_4\) to moles of water:
\(\frac{0.14}{0.020} = 7\)
So, \(x = 7\).

(b)(i)
1. Moles of \(\text{CaCO}_3 = \frac{5.0}{100} = 0.05\text{ mol}\)
2. According to the equation, 1 mole of \(\text{CaCO}_3\) produces 1 mole of \(\text{CO}_2\).
Moles of \(\text{CO}_2 = 0.05\text{ mol}\)
3. Volume of \(\text{CO}_2 = 0.05\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 1.20\text{ dm}^3\) (or \(1200\text{ cm}^3\)).

(b)(ii)
1. According to the equation, 1 mole of \(\text{CaCO}_3\) produces 1 mole of \(\text{CaO}\).
Moles of \(\text{CaO} = 0.05\text{ mol}\)
2. Mass of \(\text{CaO} = 0.05\text{ mol} \times 56\text{ g/mol} = 2.80\text{ g}\).

(a)(i) [Total: 1 mark]
- \(2.52\text{ g}\) (1)

(a)(ii) [Total: 3 marks]
- Moles of \(\text{MgSO}_4 = 0.020\text{ mol}\) (1)
- Moles of water = \(0.14\text{ mol}\) (1)
- Ratio = 1 : 7 / correct value of \(x = 7\) (1)

(b)(i) [Total: 3 marks]
- Moles of \(\text{CaCO}_3 = 0.05\text{ mol}\) (1)
- Moles of \(\text{CO}_2 = 0.05\text{ mol}\) (1)
- Volume = \(1.2\text{ dm}^3\) or \(1200\text{ cm}^3\) (must include correct unit) (1)

(b)(ii) [Total: 3 marks]
- Identifies 1:1 mole ratio (moles of \(\text{CaO} = 0.05\text{ mol}\)) (1)
- Multiplication of moles by 56 (1)
- Correct mass of \(2.80\text{ g}\) (1)

(a)(i)
An acid is a proton (or \(\text{H}^+\)) donor.

(a)(ii)
A strong acid is completely ionized or dissociated in aqueous solution, whereas a weak acid is only partially ionized or dissociated in aqueous solution.

(a)(iii)
- Hydrochloric acid (strong acid): pH = 1 (accept pH 0, 1, or 2)
- Ethanoic acid (weak acid): pH = 3 (accept pH 3, 4, 5, or 6)

(b)(i)
1. Rapid effervescence / bubbling / fizzing occurs.
2. The magnesium ribbon dissolves / disappears to form a colourless solution.
(Note: stating 'hydrogen gas is produced' is not an observation)

(b)(ii)
1. Effervescence / bubbling / fizzing occurs.
2. The green copper(II) carbonate powder dissolves to form a blue solution.

(c)
Sulfur dioxide is a non-metal oxide, so it is an acidic oxide.

(a)(i) [Total: 1 mark]
- Proton donor / \(\text{H}^+\) donor (1)

(a)(ii) [Total: 2 marks]
- Strong acid completely dissociated / ionized (1)
- Weak acid partially dissociated / ionized (1)

(a)(iii) [Total: 2 marks]
- Hydrochloric acid pH: 0 to 2 (1)
- Ethanoic acid pH: 3 to 6 (1)

(b)(i) [Total: 2 marks]
- Effervescence / bubbling / fizzing (1)
- Magnesium dissolves / disappears / colourless solution formed (1)

(b)(ii) [Total: 2 marks]
- Effervescence / bubbling / fizzing (1)
- Green powder dissolves / blue solution formed (1)

(c) [Total: 1 mark]
- Acidic (oxide) (1)

(a)
\(\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})\)

(b)(i)
Sulfuric acid + Sodium hydroxide -> Sodium sulfate + Water
\(\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\)

(b)(ii)
1. Moles of \(\text{H}_2\text{SO}_4 = \text{concentration} \times \text{volume} = 0.080\text{ mol/dm}^3 \times \frac{25.0}{1000}\text{ dm}^3 = 0.0020\text{ mol}\)
2. From the equation, the mole ratio of \(\text{NaOH}\) to \(\text{H}_2\text{SO}_4\) is 2 : 1.
Moles of \(\text{NaOH}\) required = \(2 \times 0.0020 = 0.0040\text{ mol}\)
3. Volume of \(\text{NaOH}\) solution = \(\frac{\text{moles}}{\text{concentration}} = \frac{0.0040\text{ mol}}{0.100\text{ mol/dm}^3} = 0.0400\text{ dm}^3 = 40.0\text{ cm}^3\).

(c)(i)
Zinc oxide (or Aluminium oxide).

(c)(ii)
An amphoteric oxide reacts with both acids and bases (alkalis) to form a salt and water. In contrast, a basic oxide only reacts with acids and does not react with bases.

(a) [Total: 2 marks]
- Correct equation: \(\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}\) (1)
- Correct state symbols: \(\text{aq}\), \(\text{aq}\), \(\text{l}\) (1)

(b)(i) [Total: 2 marks]
- Correct reactant and product chemical formulas (1)
- Correct balancing (\(2\text{NaOH}\) and \(2\text{H}_2\text{O}\)) (1)

(b)(ii) [Total: 3 marks]
- Moles of \(\text{H}_2\text{SO}_4 = 0.0020\text{ mol}\) (1)
- Moles of \(\text{NaOH} = 0.0040\text{ mol}\) (1)
- Volume of \(\text{NaOH} = 40.0\text{ cm}^3\) (must include correct unit) (1)

(c)(i) [Total: 1 mark]
- Zinc oxide / Aluminium oxide (1)

(c)(ii) [Total: 2 marks]
- Amphoteric oxide reacts with both acids and bases / alkalis (1)
- Basic oxide only reacts with acids (1)

(a)(i)
The gas is carbon dioxide (which turns limewater cloudy), which indicates that the anion is carbonate (\(\text{CO}_3^{2-}\)).

(a)(ii)
\(\text{CO}_2 + \text{Ca(OH)}_2 \rightarrow \text{CaCO}_3 + \text{H}_2\text{O}\)

(b)(i)
The formation of a light blue precipitate with sodium hydroxide, which is insoluble in excess, confirms the presence of copper(II) ions, \(\text{Cu}^{2+}\).

(b)(ii)
Compound X contains copper(II) (\(\text{Cu}^{2+}\)) and carbonate (\(\text{CO}_3^{2-}\)). Therefore, the formula is \(\text{CuCO}_3\).

(c)(i)
The formation of a cream precipitate with silver nitrate indicates the presence of bromide ions, \(\text{Br}^-\).

(c)(ii)
A white precipitate with aqueous ammonia that is soluble in excess to form a colourless solution is characteristic of zinc ions, \(\text{Zn}^{2+}\).

(d)
1. Add aqueous sodium hydroxide to the solution and warm/heat the mixture.
2. Test the gas evolved with damp red litmus paper; it will turn blue (indicating ammonia gas).

(a)(i) [Total: 1 mark]
- Carbonate / \(\text{CO}_3^{2-}\) (1)

(a)(ii) [Total: 2 marks]
- Correct reactant and product formulas (1)
- Correctly balanced (1)

(b)(i) [Total: 1 mark]
- Copper(II) / \(\text{Cu}^{2+}\) (1)

(b)(ii) [Total: 1 mark]
- \(\text{CuCO}_3\) (1)

(c)(i) [Total: 1 mark]
- Bromide / \(\text{Br}^-\) (1)

(c)(ii) [Total: 2 marks]
- Zinc / \(\text{Zn}^{2+}\) (2) (allow 1 mark for incorrect charge, e.g., \(\text{Zn}^+\))

(d) [Total: 2 marks]
- Add aqueous sodium hydroxide and warm / heat (1)
- Gas turned damp red litmus paper blue (1)

(a)(i)
1. Test: Use damp blue litmus paper.
2. Observation: The paper turns red and then bleaches (turns white).

(a)(ii)
1. Test: Bubble the gas through acidified aqueous potassium manganate(VII).
2. Observation: The solution turns from purple to colourless (decolorizes).

(a)(iii)
1. Test and Observation: Insert a glowing splint into the gas; the splint relights.

(b)(i)
1. Add aqueous sodium hydroxide dropwise to the solution.
2. Iron(II) ions, \(\text{Fe}^{2+}\), form a green precipitate.
3. Iron(III) ions, \(\text{Fe}^{3+}\), form a red-brown precipitate.

(b)(ii)
Equation: \(\text{Fe}^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_3(\text{s})\)

(c)
Lithium ions, \(\text{Li}^+\), give a red (or crimson) colour in a flame test.

(a)(i) [Total: 2 marks]
- Use damp blue litmus paper (1)
- Bleaches / turns white (1)

(a)(ii) [Total: 2 marks]
- Acidified aqueous potassium manganate(VII) (1)
- Turns colourless / decolorizes (1)

(a)(iii) [Total: 1 mark]
- Glowing splint relights (1)

(b)(i) [Total: 2 marks]
- Observation for \(\text{Fe}^{2+}\): green precipitate (1)
- Observation for \(\text{Fe}^{3+}\): red-brown precipitate (1)

(b)(ii) [Total: 2 marks]
- Correct equation: \(\text{Fe}^{3+} + 3\text{OH}^- \rightarrow \text{Fe(OH)}_3\) (1)
- Correct state symbols: \(\text{aq}\), \(\text{aq}\), \(\text{s}\) (1)

(c) [Total: 1 mark]
- Red / crimson (1)

To investigate the effect of concentration on the rate of reaction:

1. **Preparation of different concentrations:** Dilute the stock \(2.0\text{ mol/dm}^3\) hydrochloric acid with distilled water. For example, prepare five different concentrations using a measuring cylinder to mix the following volumes to a total volume of \(50\text{ cm}^3\):
- \(50\text{ cm}^3\) acid + \(0\text{ cm}^3\) water (100% stock)
- \(40\text{ cm}^3\) acid + \(10\text{ cm}^3\) water (80% stock)
- \(30\text{ cm}^3\) acid + \(20\text{ cm}^3\) water (60% stock)
- \(20\text{ cm}^3\) acid + \(30\text{ cm}^3\) water (40% stock)
- \(10\text{ cm}^3\) acid + \(40\text{ cm}^3\) water (20% stock)

2. **Control variables:**
- Keep the temperature constant (conduct all experiments at room temperature).
- Use the same total volume of acid solution (\(50\text{ cm}^3\)) in each trial.
- Use the same mass (e.g., \(5.0\text{ g}\)) and size (large chips) of marble chips in each trial.

3. **Experimental procedure and measurements:**
- Weigh \(5.0\text{ g}\) of large marble chips.
- Measure \(50\text{ cm}^3\) of the prepared acid concentration and pour it into a conical flask.
- Add the marble chips to the conical flask and immediately insert the stopper connected to a gas syringe.
- Start the stopwatch immediately.
- Record the volume of carbon dioxide gas collected in the syringe at regular intervals (e.g., every 10 or 20 seconds) for 2 minutes.
- Repeat the procedure for each prepared concentration of hydrochloric acid.

4. **Data analysis:**
- Plot a graph of volume of gas produced (y-axis) against time (x-axis) for each concentration.
- Determine the initial gradient of each curve. The concentration that yields the steepest initial gradient has the fastest rate of reaction.

Total: 10 marks

- **Acid Dilution Method (2 marks):**
- Details of how to dilute the stock acid with distilled water to obtain at least 3 different concentrations [1]
- Mention of measuring specific volumes of acid and water using a measuring cylinder/pipette [1]
- **Control Variables (2 marks):**
- Keeps the total volume of acid solution constant in each experiment [1]
- Keeps the mass and surface area/size of the marble chips constant [1]
- **Experimental Setup and Procedure (3 marks):**
- Use of a conical flask connected to a gas syringe (or a flask on a balance with a cotton wool plug) [1]
- Adds marble chips to the acid and starts the stopwatch immediately [1]
- Measures the volume of gas collected (or decrease in mass) [1]
- **Measurements over Time (1 mark):**
- Records the volume of gas (or mass) at regular, specified time intervals (e.g., every 10/20/30 seconds) [1]
- **Data Analysis (2 marks):**
- Plot a graph of volume of gas vs. time [1]
- Compare the initial gradients of the curves OR compare the time taken to collect a fixed volume of gas (steeper gradient or shorter time indicates a faster rate) [1]

To separate the components and determine the percentage by mass:

1. **Weighing:** Weigh a clean, dry beaker. Add a sample of mixture **M** and weigh again to determine the initial mass of the mixture, \(m_1\text{ g}\).
2. **Dissolving:** Add distilled water to the beaker. Stir thoroughly with a glass rod (heating gently can speed up the process) to completely dissolve the copper(II) sulfate. The silicon dioxide remains insoluble.
3. **Filtration:** Filter the mixture through a filter funnel lined with filter paper into a conical flask. The silicon dioxide will be collected as the residue on the filter paper, while the copper(II) sulfate solution passes through as the filtrate.
4. **Obtaining pure, dry silicon dioxide:** Wash the residue on the filter paper with a small amount of distilled water to remove any remaining copper(II) sulfate solution. Place the filter paper containing the residue in a warm oven or leave it in a warm place to dry completely. Weigh the dry silicon dioxide, \(m_2\text{ g}\).
5. **Obtaining pure, dry copper(II) sulfate crystals:** Pour the filtrate into an evaporating basin. Heat the solution using a Bunsen burner to evaporate the water until the crystallization point is reached (tested by dipping a cold glass rod to see if crystals form). Leave the hot, saturated solution to cool slowly at room temperature to allow large crystals to form. Filter off the crystals, wash them with a small volume of cold distilled water (or cold ethanol) to remove impurities, and dry them by patting with filter paper.
6. **Calculation:** Calculate the percentage of silicon dioxide using the formula:

\(\text{Percentage of } \text{SiO}_2 = \frac{m_2}{m_1} \times 100 \%\)

Total: 10 marks

- **Initial Measurement (1 mark):**
- Weigh the initial mass of mixture **M** before doing any separations [1]
- **Dissolving (2 marks):**
- Add distilled water to the mixture in a beaker [1]
- Stir (or heat) to ensure all copper(II) sulfate is fully dissolved [1]
- **Filtration (1 mark):**
- Filter the mixture to separate the insoluble residue (\(\text{SiO}_2\)) from the filtrate (\(\text{CuSO}_4\) solution) [1]
- **Purification of Silicon Dioxide (2 marks):**
- Wash the residue with distilled water [1]
- Dry the residue in an oven/warm place and weigh it [1]
- **Purification of Copper(II) Sulfate (3 marks):**
- Heat the filtrate in an evaporating basin to the crystallization point / to evaporate some water [1]
- Leave the solution to cool and crystallize [1]
- Filter the crystals, wash with a small amount of cold solvent, and dry with filter paper [1]
- **Percentage Calculation (1 mark):**
- Correct formula: \(\frac{\text{mass of dry silicon dioxide}}{\text{initial mass of mixture}} \times 100\%\) [1]

**(a)**
- Test 1: The formation of a green precipitate with sodium hydroxide that is insoluble in excess is characteristic of iron(II) ions, \(\text{Fe}^{2+}\).
- Test 2: The formation of a light blue precipitate with ammonia that dissolves in excess to form a deep blue solution is characteristic of copper(II) ions, \(\text{Cu}^{2+}\).
Therefore, the two cations are **iron(II)** and **copper(II)**.

**(b)**
- Test 3: The formation of a thick white precipitate with barium nitrate in acidic conditions indicates the presence of sulfate ions, \(\text{SO}_4^{2-}\). (This is confirmed by Test 4, which shows no halide ions are present).
Therefore, the anion is **sulfate**.

**(c)**
- (i) The green precipitate is iron(II) hydroxide:
\(\text{Fe}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_2(\text{s})\)
- (ii) The white precipitate is barium sulfate:
\(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\)

**(d)**
- **Observation:** A blue-green flame.
- **Explanation:** Copper(II) ions, \(\text{Cu}^{2+}\), produce a distinctive blue-green color in a flame test. Iron(II) ions do not produce a distinctive flame color that would mask or interfere with the copper flame test color under standard test conditions.

Total: 10 marks

- **(a) Cations (2 marks):**
- Iron(II) / \(\text{Fe}^{2+}\) [1]
- Copper(II) / \(\text{Cu}^{2+}\) [1]
- **(b) Anion (1 mark):**
- Sulfate / \(\text{SO}_4^{2-}\) [1]
- **(c) Ionic Equations (4 marks):**
- (i) \(\text{Fe}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_2(\text{s})\)
- Correct species and balancing [1]
- Correct state symbols: (aq) for reactants, (s) for product [1]
- (ii) \(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\)
- Correct species and balancing [1]
- Correct state symbols: (aq) for reactants, (s) for product [1]
- **(d) Flame Test (3 marks):**
- Blue-green flame observed [1]
- Caused by the copper(II) ions (\(\text{Cu}^{2+}\)) present [1]
- Iron(II) does not have a characteristic flame color / does not interfere with the blue-green observation [1]

**(a)**
The diagram should clearly show:
- A horizontal or slightly tilted boiling tube (hard glass test-tube) containing a sample of the metal carbonate.
- A Bunsen burner positioned directly under the sample to apply heat.
- A delivery tube sealed with a rubber bung at the mouth of the boiling tube, leading into a second test-tube containing limewater.
- The delivery tube must have its outlet submerged beneath the surface of the limewater.
- Correct labels for: metal carbonate, Bunsen burner (or heat), delivery tube, and limewater.

**(b)**
To ensure a fair test, the following three variables must be kept constant:
1. The mass of the metal carbonate used (e.g., 1.0 g).
2. The volume (and concentration) of the limewater.
3. The heating rate (e.g., using the same Bunsen burner on a roaring blue flame at a fixed distance from the boiling tube).

**(c)**
- **Order of stability (least stable to most stable):**
\(\text{copper(II) carbonate} < \text{zinc carbonate} < \text{sodium carbonate}\)
- **Expected Observations:**
- **Copper(II) carbonate:** Decomposes very easily and quickly. The green solid turns black (forming copper(II) oxide), and the limewater turns milky very quickly.
- **Zinc carbonate:** Decomposes moderately easily. The white solid turns yellow when hot and white when cold (forming zinc oxide), and the limewater turns milky more slowly than with copper(II) carbonate.
- **Sodium carbonate:** Very stable and does not decompose at Bunsen burner temperatures. No color change is observed in the solid, and the limewater remains clear.

**(d)**
- **Precaution:** Remove the delivery tube from the limewater *before* removing the Bunsen burner flame (stopping the heating).
- **Reason:** If heating is stopped first, the air inside the boiling tube will cool and contract, creating a partial vacuum. This will suck cold limewater back up the delivery tube into the hot boiling tube ('suck-back'), causing the hot glass to crack or shatter.

Total: 10 marks

- **(a) Diagram (3 marks):**
- Correct setup showing heated boiling tube with delivery tube connected to a tube of limewater [1]
- Delivery tube outlet is correctly submerged in the limewater [1]
- Complete and correct labels (metal carbonate, Bunsen burner, delivery tube, limewater) [1]
- **(b) Variables (3 marks):**
- Constant mass of metal carbonate [1]
- Constant volume of limewater [1]
- Constant heat intensity / same Bunsen burner setting and distance [1]
- **(c) Stability and Observations (3 marks):**
- Correct order of stability: copper(II) carbonate < zinc carbonate < sodium carbonate [1]
- Correct observations for both copper(II) carbonate (green to black, rapid milkiness) and zinc carbonate (white to yellow hot/white cold, slower milkiness) [1]
- Correct observation for sodium carbonate (no decomposition/no change, limewater stays clear) [1]
- **(d) Safety Precaution (1 mark):**
- Remove delivery tube from limewater before removing heat to prevent suck-back of cold limewater, which cracks the hot test-tube [1]