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

First, the mixture is filtered to remove the insoluble sand. This leaves a liquid mixture of ethanol and water. Since ethanol (boiling point \(78\ ^\circ\text{C}\)) and water (boiling point \(100\ ^\circ\text{C}\)) are miscible liquids with relatively close boiling points, they are separated by fractional distillation using a fractionating column.

1 mark for the correct option A.
- Reject other options because simple distillation does not effectively separate miscible liquids with close boiling points, and filtration must be performed first to remove the solid sand.

To find the empirical formula, divide the percentage of each element by its relative atomic mass:
- \(\text{Cu} = \frac{40.0}{64.0} = 0.625\)
- \(\text{S} = \frac{20.0}{32.0} = 0.625\)
- \(\text{O} = \frac{40.0}{16.0} = 2.50\)

Next, divide each value by the smallest value (\(0.625\)) to find the simplest whole-number ratio:
- \(\text{Cu} = \frac{0.625}{0.625} = 1\)
- \(\text{S} = \frac{0.625}{0.625} = 1\)
- \(\text{O} = \frac{2.50}{0.625} = 4\)

The empirical formula is \(\text{CuSO}_4\).

1 mark for the correct option B.
- 1 mark for calculating the correct mole ratios and obtaining the ratio \(1 : 1 : 4\).

1. The reaction with aqueous sodium hydroxide produces a green precipitate. Both \(\text{Fe}^{2+}\) and \(\text{Cr}^{3+}\) form green precipitates. However, chromium(III) hydroxide is soluble in excess sodium hydroxide (forming a green solution), while iron(II) hydroxide is insoluble in excess. This confirms the presence of \(\text{Fe}^{2+}\) ions.
2. The reaction with dilute nitric acid and aqueous barium nitrate produces a white precipitate of barium sulfate (\(\text{BaSO}_4\)), which confirms the presence of sulfate ions (\(\text{SO}_4^{2-}\)).

Therefore, salt \(X\) is iron(II) sulfate, \(\text{FeSO}_4\).

1 mark for the correct option B.
- Identify \(\text{Fe}^{2+}\) from the green precipitate insoluble in excess sodium hydroxide.
- Identify \(\text{SO}_4^{2-}\) from the white precipitate with acidified barium nitrate.

Barium sulfate is insoluble in water, while sodium chloride is highly soluble. When water is added to the mixture, the sodium chloride dissolves to form an aqueous solution, whereas the barium sulfate remains as an insoluble solid. Filtration separates the solid barium sulfate (as the residue) from the sodium chloride solution (as the filtrate). Washing the residue with distilled water is necessary to remove any remaining traces of the sodium chloride solution. Finally, drying the residue in an oven or leaving it to evaporate yields a pure, dry sample of barium sulfate.

1 mark for the correct option A.
- Reject option B because it isolates sodium chloride instead of barium sulfate.
- Reject option C because crystallisation would not separate an insoluble solid from a soluble one efficiently.
- Reject option D because neither substance sublimes upon heating.

To find the empirical formula:
1. Find the number of moles of each element in \(100\text{ g}\) of the compound:
- Moles of \(\text{Fe} = \frac{70.0}{56} = 1.25\text{ mol}\)
- Moles of \(\text{O} = \frac{30.0}{16} = 1.875\text{ mol}\)

2. Find the simplest ratio by dividing both values by the smaller number of moles (\(1.25\)):
- Ratio of \(\text{Fe} = \frac{1.25}{1.25} = 1\)
- Ratio of \(\text{O} = \frac{1.875}{1.25} = 1.5\)

3. Convert the ratio to whole numbers by multiplying both parts by 2:
- \(\text{Fe} = 1 \times 2 = 2\)
- \(\text{O} = 1.5 \times 2 = 3\)

Thus, the empirical formula is \(\text{Fe}_2\text{O}_3\).

1 mark for selecting B.
- Deduct/reject other options due to incorrect calculation of mole ratios.

In Test 1, the formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). (Note: Chromium(III) also produces a green precipitate, but it dissolves in excess sodium hydroxide to give a green solution). In Test 2, the formation of a white precipitate upon adding acidified barium nitrate confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). Combining these two results, solution \(X\) is iron(II) sulfate.

1 mark for selecting C.
- Reject A because iron(III) forms a red-brown precipitate.
- Reject B because chloride ions would give a white precipitate with silver nitrate, not barium nitrate.
- Reject D because chromium(III) precipitate is soluble in excess aqueous sodium hydroxide.

The test with dilute hydrochloric acid produces carbon dioxide gas which turns limewater cloudy, confirming the presence of the carbonate ion, \(CO_3^{2-}\). The reaction of the aqueous salt with aqueous sodium hydroxide produces a green precipitate that is insoluble in excess, which is characteristic of the iron(II) ion, \(Fe^{2+}\). Chromium(III) also forms a green precipitate with sodium hydroxide, but it is soluble in excess to give a green solution. Therefore, the salt is iron(II) carbonate.

1 mark for selecting the correct option B. Reject option A (copper(II) carbonate forms a light blue precipitate), C (iron(III) carbonate forms a red-brown precipitate), and D (chromium(III) carbonate forms a green precipitate that is soluble in excess NaOH).

Step 1: Calculate the mass of oxygen in the sample. Mass of \(O\) = 4.80 g - 2.88 g = 1.92 g. Step 2: Calculate the number of moles of each element. Moles of \(Ti\) = 2.88 / 48 = 0.060 mol. Moles of \(O\) = 1.92 / 16 = 0.120 mol. Step 3: Find the simplest whole-number ratio of the atoms. Ratio of \(Ti : O\) = 0.060 : 0.120 = 1 : 2. Therefore, the empirical formula is \(TiO_2\).

1 mark for the correct option B. Method: Award marks for calculating the mass of oxygen (1.92 g), converting both elements to their mole values (0.06 mol of Ti and 0.12 mol of O), and finding the simplest whole-number ratio of 1:2.

During fractional distillation, the liquid with the lowest boiling point vaporises most easily and passes up the fractionating column first. Therefore, hydrocarbon P (boiling point 36 °C) is distilled and collected first. The temperature on the thermometer will remain at around 36 °C while hydrocarbon P is distilling. In the fractionating column, higher-boiling liquids (like R and S) condense at lower, hotter parts of the column and drip back down, while the lowest-boiling liquid (P) reaches the cooler top of the column and passes into the condenser.

1 mark for the correct option B. A is incorrect because the substance with the lowest boiling point is collected first. C is incorrect because P will condense and distill at 36 °C. D is incorrect because the lowest boiling point component (P) reaches the top of the column first while higher boiling point components condense lower down.

First, the mixture must be filtered to remove the insoluble copper(II) oxide, leaving a filtrate of copper(II) sulfate solution. Next, the filtrate is heated to evaporate some of the water until a saturated solution is formed (crystallisation point). The solution is then left to cool, allowing hydrated copper(II) sulfate crystals to form. Finally, the crystals are separated from the remaining liquid by filtration and dried gently with filter paper.

Award 1 mark for the correct option (B). [1 mark]

To find the empirical formula, first calculate the percentage of oxygen: 100% - 70.0% = 30.0%. Next, find the mole ratio by dividing the percentage of each element by its relative atomic mass: Fe = 70.0 / 56 = 1.25 moles, O = 30.0 / 16 = 1.875 moles. Divide both values by the smallest number of moles (1.25): Fe = 1.25 / 1.25 = 1, O = 1.875 / 1.25 = 1.5. To obtain whole numbers, multiply the ratio by 2, which gives Fe = 2 and O = 3. Therefore, the empirical formula is Fe2O3.

Award 1 mark for the correct option (B). [1 mark]

A green precipitate with aqueous sodium hydroxide that is insoluble in excess indicates the presence of iron(II) ions, Fe2+. (Note that chromium(III) also forms a green precipitate, but it is soluble in excess NaOH). The formation of a white precipitate with aqueous barium nitrate after acidification with dilute nitric acid indicates the presence of sulfate ions, SO42-. Therefore, salt X is iron(II) sulfate.

Award 1 mark for the correct option (C). [1 mark]

To obtain a pure solvent (water) from a solution, simple distillation is used because water evaporates and is condensed back into a liquid, leaving the non-volatile solute behind. To obtain pure crystals of a soluble salt (copper(II) sulfate), crystallisation is used, where the solution is heated to concentration and then cooled to allow crystals to form. Filtration is not suitable as the solute is fully dissolved, and evaporation to dryness would not yield large, pure crystals but rather anhydrous powder.

1 mark for selecting option A, indicating simple distillation for pure water and crystallisation for copper(II) sulfate crystals.

First, find the percentage of oxygen by mass: \(100.0\% - 70.0\% = 30.0\%\). Next, calculate the moles of each element in a 100 g sample: Moles of Fe = \(70.0 / 56 = 1.25\), Moles of O = \(30.0 / 16 = 1.875\). Find the simplest ratio of moles by dividing both values by the smaller value (1.25): Fe = \(1.25 / 1.25 = 1\), O = \(1.875 / 1.25 = 1.5\). To get whole numbers, multiply the ratio by 2: Fe = 2, O = 3. Therefore, the empirical formula is \(\text{Fe}_2\text{O}_3\).

1 mark for selecting option B, which represents the correct empirical formula of iron(III) oxide.

In Test 1, the formation of a green precipitate that is soluble in excess sodium hydroxide to form a green solution is the characteristic test for the chromium(III) ion, \(\text{Cr}^{3+}\). (Iron(II) ions would form a green precipitate that is insoluble in excess sodium hydroxide). In Test 2, acidification with dilute nitric acid followed by the addition of aqueous barium nitrate forms a white precipitate of barium sulfate, which confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). Combining these results, the salt X is chromium(III) sulfate.

1 mark for selecting option A, which correctly identifies both the cation as chromium(III) and the anion as sulfate.

First, filtration is used to separate the insoluble sand (the residue) from the aqueous sodium chloride solution (the filtrate). The sand can then be washed and dried. Second, simple distillation is performed on the filtrate (sodium chloride solution). The water boils and evaporates, then condenses in the condenser to be collected as pure liquid water, leaving the solid sodium chloride behind in the distillation flask.

1 mark for selecting the correct sequence (filtration followed by simple distillation) which separates insoluble solids and recovers pure solvent from a solution.

To find the empirical formula:
1. Determine the number of moles of each element in 100 g of the compound:
- Moles of \(\text{Cu} = \frac{47.4}{64.0} \approx 0.741\text{ mol}\)
- Moles of \(\text{Cl} = \frac{52.6}{35.5} \approx 1.482\text{ mol}\)

2. Divide by the smallest number of moles to find the simplest ratio:
- Ratio of \(\text{Cu} = \frac{0.741}{0.741} = 1\)
- Ratio of \(\text{Cl} = \frac{1.482}{0.741} \approx 2\)

Therefore, the empirical formula is \(\text{CuCl}_2\).

1 mark for the correct calculation of moles and identifying the 1:2 ratio to give \(\text{CuCl}_2\).

1. The reaction with aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. This confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). (Chromium(III) ions also form a green precipitate, but it dissolves in excess sodium hydroxide to form a green solution).
2. The reaction with dilute nitric acid and aqueous barium nitrate produces a white precipitate of barium sulfate, which confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\).

Combining these ions gives iron(II) sulfate, \(\text{FeSO}_4\).

1 mark for identifying \(\text{Fe}^{2+}\) from the sodium hydroxide test and \(\text{SO}_4^{2-}\) from the barium nitrate test, leading to iron(II) sulfate.

The test with acidified silver nitrate produces a cream precipitate, which indicates the presence of bromide ions (\(Br^-\)). A white precipitate would indicate chloride, and a yellow precipitate would indicate iodide.

The test with aqueous sodium hydroxide produces a white precipitate that dissolves in excess, which points to either zinc (\(Zn^{2+}\)) or aluminium (\(Al^{3+}\)) ions.

The test with aqueous ammonia produces a white precipitate that is insoluble in excess, which is characteristic of aluminium (\(Al^{3+}\)) ions (zinc ions would dissolve in excess ammonia to form a colorless solution).

Therefore, salt \(Y\) is aluminium bromide.

Award 1 mark for the correct option (A).
- Deduce the anion is bromide from the cream precipitate with acidified silver nitrate.
- Deduce the cation is aluminium by distinguishing between zinc and aluminium using aqueous ammonia.

To find the empirical formula of the compound:

1. State the mass of each element in \(100\text{ g}\) of the compound:
- Mass of \(S = 25.2\text{ g}\)
- Mass of \(F = 74.8\text{ g}\)

2. Calculate the number of moles of each element:
- \(\text{Moles of S} = \frac{25.2}{32} = 0.7875\text{ mol}\)
- \(\text{Moles of F} = \frac{74.8}{19} = 3.937\text{ mol}\)

3. Determine the simplest ratio by dividing both values by the smaller number of moles (\(0.7875\)):
- For \(S\): \(\frac{0.7875}{0.7875} = 1\)
- For \(F\): \(\frac{3.937}{0.7875} \approx 5\)

Thus, the empirical formula of the compound is \(SF_5\).

Award 1 mark for the correct option (C).
- Correct working showing moles of S as 0.7875 and moles of F as 3.937, leading to a 1:5 ratio.

Let us evaluate the products of each reaction:

- Option A: Heating solid ammonium chloride with aqueous sodium hydroxide produces ammonia gas (\(NH_3\)):
\(NH_4Cl(s) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l) + NH_3(g)\)

- Option B: Adding dilute nitric acid to copper(II) oxide (a basic oxide) is a neutralisation reaction that yields copper(II) nitrate solution and water; no gas is released:
\(CuO(s) + 2HNO_3(aq) \rightarrow Cu(NO_3)_2(aq) + H_2O(l)\)

- Option C: Adding hydrochloric acid to magnesium ribbon produces hydrogen gas (\(H_2\)):
\(Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)\)

- Option D: Adding sulfuric acid to sodium carbonate produces carbon dioxide gas (\(CO_2\)):
\(Na_2CO_3(s) + H_2SO_4(aq) \rightarrow Na_2SO_4(aq) + H_2O(l) + CO_2(g)\)

Award 1 mark for the correct option (B).
- Correct identification that a neutralisation reaction between a metal oxide and an acid produces only salt and water, without generating any gaseous byproduct.

Filtration is used to separate an insoluble solid (sand) from a liquid mixture (water and ethanol). Once the sand is removed as the residue, fractional distillation is used to separate the two miscible liquids (ethanol and water) because they have different boiling points (78 °C and 100 °C).

1 mark for the correct option A.

Assume a 100 g sample of the oxide. This contains 73.4 g of cobalt (Co) and 26.6 g of oxygen (O). Calculate the number of moles of each element: Moles of \(\text{Co} = 73.4 / 59 \approx 1.24\). Moles of \(\text{O} = 26.6 / 16 \approx 1.66\). Find the simplest whole number ratio by dividing both values by the smaller number: \(\text{Co} = 1.24 / 1.24 = 1\), \(\text{O} = 1.66 / 1.24 \approx 1.33\). Multiply both by 3 to get whole numbers: \(\text{Co} = 3\), \(\text{O} = 4\). Therefore, the empirical formula is \(\text{Co}_3\text{O}_4\).

1 mark for the correct option C.

The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess is characteristic of iron(II) ions, \(\text{Fe}^{2+}\). Chromium(III) ions also form a green precipitate, but it dissolves in excess sodium hydroxide to form a green solution. The formation of a white precipitate with aqueous barium nitrate in the presence of dilute nitric acid confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). Therefore, salt \(X\) is iron(II) sulfate.

1 mark for the correct option C.

First, heat the mixture. Only ammonium chloride sublimes, and the vapor can be condensed and collected as a pure solid. Next, add water to the residue containing copper(II) oxide and sodium chloride. Sodium chloride dissolves, while copper(II) oxide remains insoluble. Filter the mixture; the insoluble copper(II) oxide is collected as the residue on the filter paper, which is washed and dried. Finally, evaporate the water from the filtrate to obtain dry sodium chloride crystals.

1 mark for option A. Reject B, C, and D because they do not separate the mixtures based on the correct solubilities and sublimation properties.

Step 1: Calculate the percentage of hydrogen by mass. \( \% \text{ H} = 100\% - 85.7\% = 14.3\% \). Step 2: Calculate the mole ratio of the elements. Moles of \( \text{C} = 85.7 / 12 = 7.14 \), moles of \( \text{H} = 14.3 / 1 = 14.3 \). Step 3: Simplify the mole ratio to find the empirical formula. \( \text{C} : \text{H} = 7.14 / 7.14 : 14.3 / 7.14 \approx 1 : 2 \), so the empirical formula is \( \text{CH}_2 \). Step 4: Find the molecular formula. The empirical formula mass of \( \text{CH}_2 = 12 + (2 \times 1) = 14 \). The number of empirical units \( n = M_r / 14 = 84 / 14 = 6 \). Therefore, the molecular formula is \( 6 \times (\text{CH}_2) = \text{C}_6\text{H}_{12} \).

1 mark for option C. Options A, B, and D do not match both the percentage composition and the relative molecular mass of 84.

The first test produces carbon dioxide gas (which turns limewater cloudy), indicating the presence of carbonate ions (\( \text{CO}_3^{2-} \)). The second test with aqueous ammonia produces a light blue precipitate that dissolves in excess to give a deep blue solution, which is the characteristic test for copper(II) ions (\( \text{Cu}^{2+} \)). Therefore, solid \( Y \) is copper(II) carbonate.

1 mark for option A. Reject B and D because iron(II) compounds form a green precipitate with ammonia that is insoluble in excess. Reject C because chlorides do not release carbon dioxide gas upon reaction with dilute hydrochloric acid.

To obtain hydrated crystals of copper(II) sulfate, the soluble copper(II) sulfate must first be dissolved in water. Filtration removes the insoluble copper(II) oxide as residue. The filtrate containing copper(II) sulfate solution is then heated to the crystallisation point (saturated solution). Upon cooling, crystals of the hydrated salt form. These crystals are separated by filtration and dried.

1 mark for the correct option B. Option A is incorrect as evaporation to dryness produces anhydrous copper(II) sulfate powder. Option C is incorrect as filtration is done before dissolving. Option D is incorrect as copper(II) oxide does not sublime.

First, calculate the percentage of sulfur: 100% - 60.0% = 40.0%. Next, find the mole ratio of the elements: Moles of S = 40.0 / 32.0 = 1.25. Moles of O = 60.0 / 16.0 = 3.75. Divide by the smallest number of moles: S = 1.25 / 1.25 = 1, O = 3.75 / 1.25 = 3. Therefore, the empirical formula is \(\text{SO}_3\).

1 mark for the correct option C. Incorrect options are based on arithmetic errors or inverted ratios: A (1:1), B (1:2), D (2:3).

Test 1: The formation of a white precipitate with acidified barium nitrate confirms the presence of sulfate, \(\text{SO}_4^{2-}\), ions. Test 2: The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II), \(\text{Fe}^{2+}\), ions (chromium(III) ions form a green precipitate that dissolves in excess to form a green solution). Therefore, salt Y is iron(II) sulfate.

1 mark for the correct option C. Option A is incorrect as chromium(III) precipitate dissolves in excess NaOH. Option B is incorrect as chloride ions do not form a precipitate with barium nitrate. Option D is incorrect as iron(III) forms a red-brown precipitate.

Fractional distillation separates miscible liquids based on their boiling points. The liquid with the lowest boiling point evaporates most easily and passes out of the top of the fractionating column first. Since Liquid X has the lowest boiling point (56 °C), it is collected first. Liquid Y (78 °C) is collected next, and Liquid Z (100 °C) is collected last.

1 mark for selecting the correct order based on increasing boiling points.

To find the empirical formula:
1. Divide the percentage mass of each element by its relative atomic mass:
- For Carbon: \(\frac{85.7}{12} = 7.14\)
- For Hydrogen: \(\frac{14.3}{1} = 14.3\)

2. Divide each value by the smallest calculated value (7.14):
- Carbon: \(\frac{7.14}{7.14} = 1\)
- Hydrogen: \(\frac{14.3}{7.14} \approx 2\)

Therefore, the simplest ratio of \(\text{C} : \text{H}\) is \(1 : 2\), which gives the empirical formula \(\text{CH}_2\).

1 mark for the correct empirical formula derived from stoichiometric ratios.

1. The green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). (Note: Chromium(III) also produces a green precipitate but it dissolves in excess sodium hydroxide to form a green solution).
2. The white precipitate formed upon adding dilute nitric acid followed by barium nitrate confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\).
Combining these gives iron(II) sulfate.

1 mark for identifying the cation as iron(II) and the anion as sulfate to determine the correct salt name.

1. Determine the mass of iron and oxygen in the compound:
- Mass of \(\text{Fe} = 2.24\text{ g}\)
- Mass of \(\text{O} = 3.20\text{ g} - 2.24\text{ g} = 0.96\text{ g}\)

2. Calculate the number of moles of each element:
- Moles of \(\text{Fe} = \frac{2.24\text{ g}}{56\text{ g/mol}} = 0.04\text{ mol}\)
- Moles of \(\text{O} = \frac{0.96\text{ g}}{16\text{ g/mol}} = 0.06\text{ mol}\)

3. Determine the simplest whole-number ratio:
- Ratio of \(\text{Fe} : \text{O} = 0.04 : 0.06 = 1 : 1.5\)
- Multiplying by 2 to get whole numbers gives \(2 : 3\).

Therefore, the empirical formula is \(\text{Fe}_2\text{O}_3\).

Award 1 mark for the correct option B.
- Correctly calculating the mass of oxygen (0.96 g) and finding moles of Fe (0.04 mol) and O (0.06 mol).
- Correctly simplifying the mole ratio to 2:3 to obtain the formula \(\text{Fe}_2\text{O}_3\).

1. Analyze Test 1: Both \(\text{Fe}^{2+}\) and \(\text{Cr}^{3+}\) form green precipitates with sodium hydroxide. However, chromium(III) hydroxide is soluble in excess sodium hydroxide to form a green solution, whereas iron(II) hydroxide is insoluble in excess. This confirms the cation is \(\text{Fe}^{2+}\).
2. Analyze Test 2: The addition of acidified barium nitrate tests for sulfate ions. The formation of a white precipitate (barium sulfate) confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\).
3. Combining these findings, compound \(X\) is iron(II) sulfate.

Award 1 mark for the correct option D.
- Reject option A and B because chromium(III) hydroxide precipitate dissolves in excess aqueous sodium hydroxide.
- Reject option C because chloride ions react with silver nitrate, not barium nitrate, to give a precipitate.

Since the mixture consists of miscible liquids with different boiling points, they are best separated by fractional distillation. Liquid P has the lowest boiling point (56 °C), meaning it will evaporate first, rise up the fractionating column, condense in the condenser, and be collected first in its purest form.

Award 1 mark for the correct option C.
- Crystallization is for separating soluble solids from solutions.
- Simple distillation is used to separate a solvent from a solution or liquids with extremely large differences in boiling point, but is less effective than fractional distillation for separating multiple miscible liquids cleanly.
- Chromatography is used to separate and identify components of a mixture, not to obtain bulk liquid samples efficiently.

The glass beads inside a fractionating column provide a large surface area for repeated condensation and evaporation, which allows for highly efficient separation of liquids with different boiling points. Liquid P has the lowest boiling point (56 \u00b0C) and will distil first, meaning the temperature at the top of the column will remain at 56 \u00b0C while P is being collected. Distillation separates substances based on differences in boiling points, not solubility. Therefore, option C is correct.

[1 mark] C is correct. Reject A, B, and D.

1. Identify the cation: Aqueous sodium hydroxide reacts with transition metal ions to form insoluble hydroxides. Both \(\text{Fe}^{2+}\) and \(\text{Cr}^{3+}\) form green precipitates. However, chromium(III) hydroxide dissolves in excess sodium hydroxide to give a green solution, while iron(II) hydroxide is insoluble in excess. Thus, the cation is \(\text{Fe}^{2+}\). 2. Identify the anion: The test using dilute nitric acid followed by aqueous barium nitrate is specific for sulfate ions, \(\text{SO}_4^{2-}\). The white precipitate formed is barium sulfate, \(\text{BaSO}_4\). 3. Combine the ions: \(\text{Fe}^{2+}\) and \(\text{SO}_4^{2-}\) combine to form iron(II) sulfate, \(\text{FeSO}_4\).

[1 mark] C is the correct answer. - Award 1 mark for the correct option. - Reject other options: A contains \(\text{Cr}^{3+}\) (soluble in excess NaOH); B contains \(\text{Cl}^-\); D contains \(\text{Fe}^{3+}\).

(a) Sublimation.
(b) Sodium chloride is soluble in water, whereas sand (silicon(IV) oxide) is insoluble in water.
(c) Filter the mixture using a filter funnel and filter paper. The sand remains on the paper as residue. Wash the residue with a small amount of distilled water to remove any remaining salt solution. Dry the sand in a warm oven or leave it to dry between filter papers.
(d) Heat the sodium chloride solution in an evaporating basin to evaporate some water until a saturated solution is formed (crystallization point). Leave the solution to cool and allow crystals to form. Filter the crystals from the remaining liquid and dry them using filter paper or in a desiccator.
(e) Measure the melting point of the crystals. Pure sodium chloride has a sharp melting point (at 801 °C), whereas impurities will lower the melting point and make it melt over a range of temperatures.

(a) 1 mark: Sublimation. [Reject: evaporation, distillation]
(b) 1 mark: Sodium chloride is soluble (in water).
1 mark: Sand/silicon(IV) oxide is insoluble (in water).
(c) 1 mark: Filter the mixture (to obtain sand as residue).
1 mark: Wash the residue/sand with distilled water.
1 mark: Dry the sand in a warm oven / between filter papers.
(d) 1 mark: Heat/evaporate the solution to the point of crystallization / to obtain a saturated solution. [Reject: evaporate to dryness]
1 mark: Leave to cool (and crystallize).
1 mark: Filter/separate crystals and dry them with filter paper / in a warm oven / desiccator.
(e) 1 mark: Measure the melting point (and compare to the literature value / look for a sharp melting point).

(a) Moles of C = \(85.7 / 12 = 7.14\). Moles of H = \(14.3 / 1 = 14.3\). Divide by the smallest value: C = \(7.14 / 7.14 = 1\), H = \(14.3 / 7.14 = 2\). The empirical formula is \(CH_2\).
(b) (i) Empirical formula mass of \(CH_2 = 12 + 2 = 14\). Ratio = \(84 / 14 = 6\). Molecular formula is \(C_6H_{12}\).
(ii) \(C_nH_{2n}\)
(c) (i) From orange/brown/yellow to colourless.
(ii) Moles of X = \(4.2 / 84 = 0.05\text{ mol}\). Moles of \(Br_2\) required = \(0.05\text{ mol}\). Mass of \(Br_2 = 0.05 \times 160 = 8.0\text{ g}\).

(a) 1 mark: Divide percentages by relative atomic masses (\(85.7/12\) and \(14.3/1\)).
1 mark: Obtain simplest mole ratio of \(1 : 2\).
1 mark: Correct empirical formula \(CH_2\).
(b) (i) 1 mark: Calculation of empirical formula mass as 14.
1 mark: Correct molecular formula \(C_6H_{12}\).
(ii) 1 mark: \(C_nH_{2n}\).
(c) (i) 1 mark: Orange / brown / yellow.
1 mark: Colourless / decolourised. [Reject: clear]
(ii) 1 mark: Calculate moles of X as \(0.05\text{ mol}\) OR state 1:1 reacting ratio.
1 mark: Correct mass of \(8.0\text{ g}\) (or \(8\text{ g}\)).

(a) Iron(II) ion / \(Fe^{2+}\)
(b) (i) Bromide ion / \(Br^-\)
(ii) \(Ag^+(aq) + Br^-(aq) \rightarrow AgBr(s)\)
(c) (i) Carbon dioxide / \(CO_2\)
(ii) Carbonate ion / \(CO_3^{2-}\)
(iii) Add warm aqueous sodium hydroxide to the solid Y. Test the gas produced with damp red litmus paper, which should turn blue if ammonium ions are present.
(d) Nitric acid reacts with and removes any carbonate ions (or other interfering ions like hydroxide) present in the solution. This prevents the formation of other silver precipitates (like silver carbonate) which would give a false positive result.

(a) 1 mark: Iron(II) / \(Fe^{2+}\) [Reject: iron / Iron(III)]
(b) (i) 1 mark: Bromide / \(Br^-\)
(ii) 1 mark: Correct reactants and product formula (\(Ag^+(aq) + Br^-(aq) \rightarrow AgBr(s)\)).
1 mark: Correct state symbols (\((aq)\) for reactants, \((s)\) for product).
(c) (i) 1 mark: Carbon dioxide / \(CO_2\)
(ii) 1 mark: Carbonate / \(CO_3^{2-}\)
(iii) 1 mark: Add warm aqueous sodium hydroxide.
1 mark: Test gas with damp red litmus paper turning blue.
(d) 1 mark: To react with/remove carbonate ions / sulfite ions.
1 mark: To prevent a false positive / to prevent the precipitation of silver carbonate.

(a) Filter the mixture using a filter funnel and filter paper. The silicon(IV) oxide remains as the residue on the filter paper. Wash the residue with a small volume of distilled water to remove any remaining salt solution, and then dry it in a warm oven or between dry filter papers.

(b) Set up a simple distillation apparatus. Place the filtrate in a distillation flask and heat it until it boils. The water evaporates to form water vapour, which passes into a condenser cooled by flowing water. The water vapour condenses back into liquid water and is collected as the distillate in a receiving flask.

(c) The process is crystallisation. Heat the sodium chloride solution in an evaporating basin to evaporate some of the water until a saturated solution is formed (the crystallisation point). Allow the hot saturated solution to cool slowly so that crystals form. Filter the crystals to separate them from any remaining solution, and dry them using filter paper.

(a) Max 3 marks:
- Filter the mixture / filtration [1]
- Silicon(IV) oxide is the residue / remains on the filter paper [1]
- Wash with distilled water AND dry (in oven/with filter paper) [1]

(b) Max 3 marks:
- Simple distillation [1]
- Heat the flask to boil/evaporate water [1]
- Condense the water vapour using a condenser / cool to collect liquid water [1]

(c) Max 4 marks:
- Crystallisation [1]
- Heat/evaporate to crystallisation point / saturate the solution [1]
- Leave to cool (slowly) to form crystals [1]
- Filter crystals and dry (with filter paper) [1]

(a) Carbon (C): \(85.7 / 12 = 7.14\text{ mol}\).
Hydrogen (H): \(14.3 / 1 = 14.3\text{ mol}\).
Divide by the smallest value (7.14):
C: \(7.14 / 7.14 = 1\)
H: \(14.3 / 7.14 = 2\).
Therefore, the empirical formula is \(\text{CH}_2\).

(b) Empirical formula mass of \(\text{CH}_2 = 12 + (2 \times 1) = 14\).
Ratio factor = \(56 / 14 = 4\).
So, the molecular formula is \(4 \times \text{CH}_2 = \text{C}_4\text{H}_8\).

(c) (i) From the balanced equation, 1 mole of \(\text{C}_3\text{H}_8\) produces 3 moles of \(\text{CO}_2\).
Moles of \(\text{CO}_2 = 3 \times 0.25 = 0.75\text{ mol}\).
Volume of \(\text{CO}_2\) at r.t.p. = \(0.75\text{ mol} \times 24\text{ dm}^3/\text{mol} = 18\text{ dm}^3\).

(ii) From the balanced equation, 1 mole of \(\text{C}_3\text{H}_8\) produces 4 moles of \(\text{H}_2\text{O}\).
Moles of \(\text{H}_2\text{O} = 4 \times 0.25 = 1.0\text{ mol}\).
Relative molecular mass of \(\text{H}_2\text{O} = (2 \times 1) + 16 = 18\).
Mass of \(\text{H}_2\text{O} = 1.0\text{ mol} \times 18\text{ g/mol} = 18\text{ g}\).

(a) Max 3 marks:
- Correct division of masses by relative atomic masses (\(85.7/12\) and \(14.3/1\)) [1]
- Correct molar ratio obtained (approx. \(7.14 : 14.3\), which is \(1 : 2\)) [1]
- Correct empirical formula: \(\text{CH}_2\) [1]

(b) Max 2 marks:
- Calculation of empirical formula mass as 14 [1]
- Correct molecular formula: \(\text{C}_4\text{H}_8\) [1]

(c) (i) Max 3 marks:
- Calculation of moles of \(\text{CO}_2\) as 0.75 [1]
- Multiplication of moles by 24 to find volume [1]
- Correct final answer: \(18\text{ dm}^3\) (accept 18 with correct units) [1]

(c) (ii) Max 2 marks:
- Moles of \(\text{H}_2\text{O}\) calculated as 1.0 [1]
- Correct mass calculated as 18 g [1]

(a) (i) The cation is iron(II) / \(\text{Fe}^{2+}\).
(ii) A green precipitate is formed, which remains insoluble in excess aqueous ammonia.

(b) (i) The anion is sulfate / \(\text{SO}_4^{2-}\).
(ii) \(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\)

(c) (i) Carbon dioxide: Bubble the gas through limewater. The limewater turns milky/cloudy.
(ii) Chlorine: Place damp blue litmus paper in the gas. The paper is bleached / turns white (accept turns red then bleaches).

(a) (i) Max 1 mark:
- Iron(II) / \(\text{Fe}^{2+}\) [1] (Do NOT accept iron / \(\text{Fe}^{3+}\))

(a) (ii) Max 2 marks:
- Green precipitate [1]
- Insoluble in excess [1]

(b) (i) Max 1 mark:
- Sulfate / \(\text{SO}_4^{2-}\) [1]

(b) (ii) Max 2 marks:
- Correct chemical formulae: \(\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4\) [1]
- Correct state symbols: \(\text{(aq)}\) for reactants and \(\text{(s)}\) for product [1]

(c) (i) Max 2 marks:
- Bubble through limewater [1]
- Turns cloudy / milky [1]

(c) (ii) Max 2 marks:
- Damp blue litmus paper [1]
- Bleached / turns white [1]

(a) Hydrated copper(II) sulfate is soluble in water, whereas silicon dioxide (sand) is giant covalent and insoluble. Adding water dissolves the salt, leaving sand behind. Filtration removes the sand as the residue, leaving copper(II) sulfate solution as the filtrate. Heating the filtrate to evaporation/crystallisation point allows crystals to form on cooling. Washing with a minimum of cold distilled water removes impurities without dissolving too much crystal, and drying between filter papers removes excess water.

(b)(i) The standard test for sulfate ions requires acidifying the solution with dilute nitric acid (to remove carbonate impurities) followed by adding barium nitrate solution. A white precipitate of barium sulfate confirms the sulfate ions.

(b)(ii) Adding aqueous sodium hydroxide to a solution containing \(\text{Cu}^{2+}\) ions results in a light blue precipitate of copper(II) hydroxide, which is insoluble in excess NaOH. The ionic equation must have balanced charges and correct state symbols.

(c) Anhydrous cobalt(II) chloride paper is blue and turns pink in the presence of water.

(a) Max [4] marks:
- Add water / dissolve [1]
- Filter (to remove sand) [1]
- Heat / evaporate the filtrate to point of crystallisation / saturated solution [1]
- Filter crystals AND dry with filter paper / in warm oven (reject: evaporate to dryness) [1]

(b)(i) Max [2] marks:
- Test: Add dilute hydrochloric acid / nitric acid AND aqueous barium chloride / barium nitrate [1]
- Observation: White precipitate [1]

(b)(ii) Max [3] marks:
- Observation: Light blue / blue precipitate [1]
- Ionic equation: \(\text{Cu}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Cu(OH)}_2(\text{s})\)
- Correct formulae: [1]
- Correct state symbols (aq, aq, s): [1]

(c) Max [1] mark:
- Blue to pink [1]

(a) Using the molar volume of a gas at r.t.p. (\(24\text{ dm}^3\) or \(24000\text{ cm}^3\)):
\(\text{Moles of } \text{CO}_2 = \frac{240\text{ cm}^3}{24000\text{ cm}^3\text{/mol}} = 0.010\text{ mol}\)

(b) From the stoichiometry of the equation:
\(1\text{ mol of } \text{MCO}_3\) reacts to produce \(1\text{ mol of } \text{CO}_2\).
Therefore, \(\text{moles of } \text{MCO}_3 = 0.010\text{ mol}\).

(c) Relative formula mass is mass divided by moles:
\(M_r(\text{MCO}_3) = \frac{1.97\text{ g}}{0.010\text{ mol}} = 197\)

(d) The formula mass of \(\text{MCO}_3\) is given by:
\(M_r(\text{MCO}_3) = A_r(\text{M}) + A_r(\text{C}) + 3 \times A_r(\text{O})\)
\(197 = A_r(\text{M}) + 12.0 + 3(16.0)\)
\(197 = A_r(\text{M}) + 60.0\)
\(A_r(\text{M}) = 137\)
Looking at the Periodic Table, the element with an atomic mass of approximately 137 is Barium (\(\text{Ba}\)).

(e) Step 1: Calculate moles of \(\text{HCl}\) used:
\(\text{Moles of } \text{HCl} = \frac{25.0}{1000} \times 0.150 = 0.00375\text{ mol}\)
Step 2: Use reaction ratio (\(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\)) which is 1:1.
\(\text{Moles of } \text{NaOH} = 0.00375\text{ mol}\)
Step 3: Calculate the volume of \(\text{NaOH}\) solution:
\(\text{Volume} = \frac{\text{moles}}{\text{concentration}} = \frac{0.00375\text{ mol}}{0.100\text{ mol/dm}^3} = 0.0375\text{ dm}^3 = 37.5\text{ cm}^3\)

(a) Max [2] marks:
- Use of 24000 cm3 (or conversion of volume to dm3) [1]
- Answer = 0.010 / 0.01 (mol) [1]

(b) Max [1] mark:
- 0.010 (mol) (Allow ecf from (a)) [1]

(c) Max [2] marks:
- Correct formula usage: Mass / Moles [1]
- Answer = 197 [1] (Allow ecf from (b))

(d) Max [2] marks:
- \(A_r(\text{M}) = 137\) [1]
- Identity = Barium / \(\text{Ba}\) [1]

(e) Max [3] marks:
- Moles of HCl = \(0.00375\) (mol) [1]
- Moles of NaOH = \(0.00375\) (mol) due to 1:1 ratio [1]
- Volume of NaOH = \(37.5\text{ cm}^3\) [1]

To determine the percentage by mass of insoluble calcium carbonate in a mixture of calcium carbonate and soluble sodium chloride, follow these steps: 1. Weigh the initial dry mixture of CaCO3 and NaCl using an analytical balance and record the mass. 2. Transfer the mixture to a beaker, add distilled water, and stir thoroughly with a glass rod until all the NaCl has completely dissolved. 3. Set up a filtration apparatus with a conical flask, filter funnel, and filter paper. 4. Pour the mixture into the filter funnel. The soluble NaCl solution (filtrate) will pass through, while the insoluble CaCO3 (residue) will remain on the filter paper. 5. Rinse the residue on the filter paper with a small amount of distilled water to remove any remaining dissolved NaCl. 6. Carefully remove the filter paper containing the CaCO3 residue and place it in a warm oven or a desiccator to dry. 7. Weigh the dried residue. Repeat the drying and weighing process until a constant mass is obtained to ensure all water has evaporated. 8. Calculate the percentage by mass of CaCO3 using the formula: \(\text{Percentage of } \text{CaCO}_3 = \frac{\text{mass of dry CaCO}_3}{\text{initial mass of mixture}} \times 100\).

Award 1 mark for each of the following points, up to a maximum of 10 marks: [1] Weigh the starting mixture using a balance. [2] Add distilled water to the mixture in a beaker. [3] Stir/agitate the mixture to ensure the sodium chloride dissolves. [4] Filter the mixture using filter paper and a funnel. [5] Wash the residue on the filter paper with distilled water. [6] Dry the residue (calcium carbonate) in a warm oven/dry place. [7] Re-dry and re-weigh the residue to constant mass. [8] Weigh the final dry residue of calcium carbonate. [9] State that the residue is calcium carbonate and the filtrate contains sodium chloride. [10] Provide the correct calculation formula: \(\frac{\text{mass of residue}}{\text{initial mass of mixture}} \times 100\).

To investigate how the concentration of dilute hydrochloric acid affects the rate of reaction with calcium carbonate: 1. Set up a conical flask connected via a delivery tube to a gas syringe. 2. Weigh a specific mass (e.g., 5.0 g) of marble chips of uniform size/surface area and place them into the conical flask. 3. Measure a specific volume (e.g., 50 cm³) of the first concentration of hydrochloric acid using a measuring cylinder. 4. Pour the acid into the conical flask, quickly insert the stopper connected to the gas syringe, and start a stopwatch immediately. 5. Record the volume of carbon dioxide gas collected in the syringe at regular intervals (e.g., every 10 or 20 seconds) for 3 minutes or until the reaction stops. 6. Clean the flask and repeat the experiment using the same mass and size of marble chips, the same volume of acid, and at the same temperature, but using a different concentration of hydrochloric acid. 7. Repeat for at least three other concentrations. 8. Plot a graph of volume of gas (y-axis) against time (x-axis) for each concentration. 9. Determine the rate of reaction by calculating the initial gradient of each curve. The steeper the gradient, the faster the rate of reaction.

Award 1 mark for each of the following points, up to a maximum of 10 marks: [1] Identify the use of a gas syringe (or inverted measuring cylinder over water) to collect and measure gas. [2] Use a stopwatch/timer to measure time. [3] State that a fixed volume of acid is measured. [4] State that a fixed mass of marble chips is used. [5] Describe starting the timer immediately upon mixing reactants. [6] Record the volume of gas at regular time intervals (or measure the time to collect a fixed volume). [7] Repeat the experiment with different concentrations of hydrochloric acid. [8] State that temperature of the acid/mixture must be kept constant. [9] State that the surface area/size of the marble chips must be kept constant. [10] Explain how to compare the rates (e.g., compare the gradient of the graphs of volume vs time, or compare the volume of gas produced in a fixed time).

An experimental investigation to find which metal oxide releases the most heat energy per mole can be designed as follows: 1. Use an insulated reaction vessel, such as a polystyrene cup supported in a beaker, to minimise heat loss. 2. Use a measuring cylinder or pipette to transfer a fixed volume (e.g., 50 cm³) of dilute hydrochloric acid into the polystyrene cup. 3. Insert a thermometer into the acid and record its initial temperature (T1). 4. Weigh a specific mass (e.g., 1.0 g) of magnesium oxide using a balance. 5. Add the magnesium oxide to the acid in the polystyrene cup, quickly place a lid on the cup, and stir the mixture gently with the thermometer. 6. Monitor the temperature and record the maximum temperature reached (T2). 7. Calculate the temperature change (\(\Delta T = T_2 - T_1\)). 8. Repeat the exact procedure using the same mass (or calculated molar equivalent mass) of calcium oxide and zinc oxide, ensuring the volume and concentration of the acid and the initial temperature of the acid are kept constant. 9. Calculate the moles of each metal oxide used using \(\text{moles} = \frac{\text{mass}}{M_r}\). 10. Calculate the heat energy released per mole (\(\frac{\Delta T}{\text{moles}}\), or use \(q = m c \Delta T\) and divide by moles). The metal oxide that shows the highest temperature rise per mole of reactant releases the most heat energy.

Award 1 mark for each of the following points, up to a maximum of 10 marks: [1] Use a polystyrene cup (or insulated container) with a lid. [2] Measure a fixed volume of dilute hydrochloric acid. [3] Measure and record the initial temperature of the acid. [4] Weigh a known mass of each metal oxide. [5] Mix the oxide and acid, stir, and record the maximum temperature reached. [6] Calculate the temperature rise (\(\Delta T = T_{\text{final}} - T_{\text{initial}}\)). [7] Repeat the experiment for all three metal oxides (MgO, CaO, ZnO). [8] Identify at least one controlled variable (e.g., constant volume of acid, constant concentration of acid, or constant initial temperature). [9] State the calculation to find the number of moles of each oxide (\(\text{moles} = \frac{\text{mass}}{M_r}\)). [10] Compare the heat released per mole (by dividing the temperature change or calculated energy change \(q\) by the number of moles used).

### 1. Apparatus Setup
- A conical flask or boiling tube is used as the reaction vessel, fitted with a rubber bung and a delivery tube.
- The delivery tube is connected to a gas syringe OR led into an inverted measuring cylinder filled with water in a water trough to collect and measure the volume of oxygen gas produced.
- A stopwatch or timer is used to record time.
- Measuring cylinders or pipettes are used to measure the volumes of liquids precisely.

### 2. Experimental Procedure
1. Measure a fixed volume (e.g., \(25\text{ cm}^3\)) of hydrogen peroxide solution using a measuring cylinder and pour it into the conical flask.
2. Measure a fixed volume (e.g., \(5\text{ cm}^3\)) of the cobalt(II) nitrate catalyst solution.
3. Add the catalyst solution to the flask, quickly insert the stopper/bung with the delivery tube, and start the stopwatch immediately.
4. Record the volume of oxygen gas collected in the gas syringe (or inverted measuring cylinder) at a specific time (e.g., after 2 minutes) OR record the time taken to collect a fixed volume of gas (e.g., \(50\text{ cm}^3\)).
5. Rinse the flask thoroughly with distilled water.
6. Repeat steps 1–5 under identical conditions using the second catalyst, nickel(II) nitrate.
7. Repeat steps 1–5 under identical conditions using the third catalyst, copper(II) nitrate.

### 3. Variables to Keep Constant (Controls)
- **Volume and concentration of hydrogen peroxide:** Changing these would change the total quantity and initial rate of oxygen production.
- **Volume and concentration of the catalyst solutions:** A higher concentration or volume of catalyst would increase the rate of reaction.
- **Temperature:** The temperature must be kept constant (e.g., using a water bath or conducting all trials at room temperature) because higher temperatures increase reaction rates.

### 4. Analysis and Conclusion
- Compare the rates of reaction for the three catalysts.
- The catalyst that produces the largest volume of gas in the specified time (or the catalyst that takes the shortest time to produce the specified volume of gas) has the highest reaction rate and is therefore the most effective catalyst.

Award up to 10 marks based on the following criteria:

**Apparatus (Max 2 marks):**
- **[1 mark]** Suitable reaction vessel (conical flask / boiling tube / pear-shaped flask) with a tight-fitting bung and delivery tube.
- **[1 mark]** Suitable gas collection apparatus (gas syringe OR inverted measuring cylinder over water in a trough).

**Method / Procedure (Max 4 marks):**
- **[1 mark]** Measure a specified/fixed volume of hydrogen peroxide into the reaction vessel.
- **[1 mark]** Add a specified/fixed volume of the first catalyst solution (cobalt(II) nitrate) and immediately seal the vessel.
- **[1 mark]** Start the timer/stopwatch immediately and record either: the volume of gas collected in a set time interval OR the time taken to collect a set volume of gas.
- **[1 mark]** Repeat the entire procedure with the other two catalysts (nickel(II) nitrate and copper(II) nitrate) using the same volumes.

**Control Variables / Fair Test (Max 2 marks):**
- **[1 mark]** Keep the volume and concentration of hydrogen peroxide solution constant.
- **[1 mark]** Keep the volume/concentration of the catalyst solutions constant OR keep the temperature constant (e.g., using a water bath / working at room temperature).

**Analysis and Conclusion (Max 2 marks):**
- **[1 mark]** Compare the volumes of gas collected in the set time (or compare the times taken to collect the set volume).
- **[1 mark]** Conclude that the most effective catalyst is the one that gives the fastest rate of reaction (the largest volume of gas in the set time, or the shortest time to collect the set volume).