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From the first experiment, metal X displaces metal Y, so X is more reactive than Y (X > Y). From the second experiment, X does not displace Z, so Z is more reactive than X (Z > X). From the third experiment, Zinc displaces Z, so Zinc is more reactive than Z (Zn > Z). Combining these gives: Zn > Z > X > Y.

1 mark for the correct option.

First, calculate the moles of \(\text{NaOH}\): \(n(\text{NaOH}) = 0.0250\text{ dm}^3 \times 0.100\text{ mol/dm}^3 = 0.00250\text{ mol}\). According to the equation, the mole ratio of \(\text{NaOH}\) to \(\text{H}_2\text{SO}_4\) is 2:1. Therefore, \(n(\text{H}_2\text{SO}_4) = 0.00250 / 2 = 0.00125\text{ mol}\). Finally, calculate the concentration of \(\text{H}_2\text{SO}_4\): \(c = 0.00125\text{ mol} / 0.0200\text{ dm}^3 = 0.0625\text{ mol/dm}^3\).

1 mark for the correct option.

Convert mass percentages to moles: \(n(\text{C}) = 52.2 / 12 = 4.35\), \(n(\text{H}) = 13.0 / 1 = 13.0\), \(n(\text{O}) = 34.8 / 16 = 2.175\). Divide each by the smallest value (2.175): \(\text{C} = 4.35 / 2.175 = 2\), \(\text{H} = 13.0 / 2.175 = 5.98 \approx 6\), \(\text{O} = 2.175 / 2.175 = 1\). Thus, the empirical formula is \(\text{C}_2\text{H}_6\text{O}\).

1 mark for the correct option.

The green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\) (chromium(III) ions also form a green precipitate but it is soluble in excess NaOH to form a green solution). The white precipitate with aqueous barium nitrate in acidic conditions confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). Therefore, the salt is iron(II) sulfate.

1 mark for the correct option.

Fermentation is a batch process that produces a dilute aqueous mixture of ethanol (low purity) because yeast is killed at high alcohol concentrations, requiring further fractional distillation. Hydration of ethene is a continuous process that produces highly pure ethanol directly. Fermentation uses renewable resources (glucose from crops) and occurs at low temperatures (around 30 to 40 degrees Celsius), while hydration uses non-renewable resources (crude oil) and occurs at high temperatures (300 degrees Celsius).

1 mark for the correct option.

The thermal stability of metal carbonates increases with the reactivity of the metal. Since \(P\text{CO}_3\) does not decompose, P is the most reactive metal (e.g. sodium or potassium). Since \(Q\text{CO}_3\) decomposes very easily, Q is the least reactive metal (e.g. copper, which forms black copper(II) oxide). \(R\text{CO}_3\) decomposes moderately (forming zinc oxide, which is yellow when hot and white when cold), indicating R (zinc) is of intermediate reactivity. Therefore, the order of reactivity is P > R > Q.

1 mark for the correct option.

In a titration, a volumetric pipette is used to accurately measure a fixed volume (e.g., 25.0 cm³) of the alkali (NaOH) into a conical flask. A burette is used to deliver the acid (HCl) because it allows for variable, accurate measurements. Methyl orange is a suitable indicator for a strong acid-strong base titration, showing a sharp colour change. Measuring cylinders are not sufficiently accurate for titrations.

1 mark for the correct option.

The carboxylic acid produced has the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\), represented structurally as \(\text{C}_3\text{H}_7\text{COOH}\) (butanoic acid). Since the carbon chain length remains the same during oxidation, the starting alcohol must contain 4 carbon atoms. Primary alcohols are oxidised to carboxylic acids, so the alcohol must be the primary alcohol with 4 carbons, which is butan-1-ol. Butan-2-ol is a secondary alcohol and would oxidise to a ketone.

1 mark for the correct option.

1. Metal \(X\) reduces the oxide of metal \(W\), which means \(X\) is more reactive than \(W\) (\(X > W\)).
2. Metal \(W\) displaces metal \(Y\) from its nitrate solution, which means \(W\) is more reactive than \(Y\) (\(W > Y\)).
3. Metal \(Y\) reduces the oxide of metal \(Z\), which means \(Y\) is more reactive than \(Z\) (\(Y > Z\)).
Combining these findings gives the reactivity series order: \(X > W > Y > Z\).

Award 1 mark for the correct option (A).
- Reject other options as they represent incorrect orders of reactivity based on displacement and reduction rules.

1. Calculate the moles of \(\text{KOH}\) used:
\(n(\text{KOH}) = \text{volume} \times \text{concentration} = 0.0250\text{ dm}^3 \times 0.150\text{ mol/dm}^3 = 0.00375\text{ mol}\).
2. Use the stoichiometric ratio from the balanced chemical equation (\(1\text{ mol H}_2\text{SO}_4 : 2\text{ mol KOH}\)) to find the moles of \(\text{H}_2\text{SO}_4\):
\(n(\text{H}_2\text{SO}_4) = \frac{0.00375}{2} = 0.001875\text{ mol}\).
3. Calculate the concentration of \(\text{H}_2\text{SO}_4\):
\(C = \frac{\text{moles}}{\text{volume}} = \frac{0.001875\text{ mol}}{0.01875\text{ dm}^3} = 0.100\text{ mol/dm}^3\).

Award 1 mark for the correct option (B).
- If option C is selected, the 1:2 mole ratio was inverted.
- If option D is selected, the mole ratio was ignored.

1. Express mass percent as masses in grams (out of 100g): \(70.0\text{ g of Fe}\) and \(30.0\text{ g of O}\).
2. Convert mass to moles:
\(n(\text{Fe}) = \frac{70.0}{56} = 1.25\text{ mol}\)
\(n(\text{O}) = \frac{30.0}{16} = 1.875\text{ mol}\).
3. Divide by the smallest mole value (1.25):
\(\text{Fe} = \frac{1.25}{1.25} = 1\)
\(\text{O} = \frac{1.875}{1.25} = 1.5\).
4. Multiply both values by 2 to obtain whole numbers: \(\text{Fe} = 2\), \(\text{O} = 3\).
The empirical formula is \(\text{Fe}_2\text{O}_3\).

Award 1 mark for the correct option (B).
- Option A is incorrect (represents 1:1 ratio if atomic mass is ignored).
- Option C represents magnetite, which has a different composition by mass.

1. The reaction with silver nitrate forming a cream precipitate indicates the presence of bromide ions (\(\text{Br}^-\)). (White = chloride, cream = bromide, yellow = iodide).
2. The reaction with sodium hydroxide forming a green precipitate that is insoluble in excess indicates the presence of iron(II) ions (\(\text{Fe}^{2+}\)). Chromium(III) also forms a green precipitate, but it is soluble in excess sodium hydroxide to form a dark green solution.
Therefore, the salt \(X\) is iron(II) bromide.

Award 1 mark for the correct option (C).
- Option A: incorrect precipitate color for chloride (would be white).
- Option B: iron(III) would form a red-brown precipitate with sodium hydroxide.
- Option D: chromium(III) hydroxide is soluble in excess sodium hydroxide.

The catalytic hydration of ethene is carried out industrially by reacting ethene and steam in the presence of a phosphoric(V) acid catalyst at a temperature of \(300\ ^\circ\text{C}\) and a pressure of \(60\text{ atm}\) (or \(6000\text{ kPa}\)).

Award 1 mark for the correct option (A).
- Option B and C describe conditions closer to fermentation of glucose using yeast.
- Option D describes the conditions for the Haber process.

The thermal stability of a metal carbonate is directly related to the reactivity of the metal. More reactive metals form more stable carbonates that require higher temperatures to decompose.
- \(Q\text{CO}_3\) is the most stable (no decomposition), so metal \(Q\) is the most reactive.
- \(P\text{CO}_3\) is the least stable (decomposes very easily), so metal \(P\) is the least reactive.
- \(R\) is of intermediate reactivity.
Therefore, the order of reactivity is \(Q > R > P\).

Award 1 mark for the correct option (B).
- Reject option A as it represents least to most reactive.
- Reject options C and D as they position the intermediate or least reactive metals incorrectly.

- The pipette and burette must be rinsed with water to clean them, but then must be rinsed with the solution they will contain (alkali and acid respectively) so that any remaining water drops do not dilute the solution and alter the measured concentrations.
- The conical flask where the reaction takes place should only be rinsed with distilled water. If it is rinsed with acid or alkali, additional unmeasured amounts of acid/alkali would be present, leading to systematic titration errors.

Award 1 mark for the correct option (A).
- Option B is incorrect because rinsing the conical flask with alkali would add unmeasured moles of base.
- Option C is incorrect because un-rinsed pipettes/burettes would have diluted solutions due to remaining water droplets.
- Option D swaps the chemicals used for rinsing the pipette and burette.

According to the qualitative analysis tests, sulfite ions (\(\text{SO}_3^{2-}\)) are identified by adding dilute acid (such as hydrochloric acid) and warming gently. The sulfur dioxide (\(\text{SO}_2\)) gas produced turns acidified potassium manganate(VII) solution from purple to colourless.

Award 1 mark for the correct option (A).
- Option B describes the test for sulfate ions (\(\text{SO}_4^{2-}\)).
- Option C describes the test for carbonate ions (\(\text{CO}_3^{2-}\)).
- Option D describes the test for ammonium ions (\(\text{NH}_4^+\)).

Metal X displaces both iron and copper from their salts, meaning X is more reactive than both Fe and Cu. However, X does not displace zinc, meaning zinc is more reactive than X. Since the standard order of reactivity of the other metals is Zn > Fe > Cu, the complete order of reactivity from most to least reactive is Zn > X > Fe > Cu.

1 mark for the correct option A.

Methyl orange is yellow in alkaline solutions (such as potassium hydroxide). The acid is added from a burette to allow variable and precise volume measurements, while the pipette is used to deliver a fixed volume of the alkali.

1 mark for the correct option A.

First, calculate the percentage of oxygen: 100% - 70.0% = 30.0%. Next, find the moles of each element in 100 g of the compound: Moles of Fe = 70.0 / 56.0 = 1.25 mol. Moles of O = 30.0 / 16.0 = 1.875 mol. Divide both values by the smaller 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: Fe = 2, O = 3. Therefore, the empirical formula is Fe2O3.

1 mark for the correct option B.

A green precipitate with aqueous sodium hydroxide that is insoluble in excess indicates the presence of Fe2+ ions (chromium(III) forms a green precipitate that is soluble in excess). A white precipitate with barium nitrate in acidic conditions indicates the presence of sulfate ions (SO42-). Therefore, salt Y is iron(II) sulfate.

1 mark for the correct option A.

The hydration of ethene uses phosphoric acid as a catalyst and produces pure ethanol in a continuous process. Fermentation is a batch process (not continuous) using a renewable resource (glucose) under mild conditions because yeast enzymes denature at high temperatures.

1 mark for the correct option B.

Carbonate stability increases up the reactivity series. Sodium is the most reactive of the three metals, so sodium carbonate does not decompose at standard laboratory (Bunsen) temperatures. Copper is the least reactive, so copper(II) carbonate decomposes most easily.

1 mark for the correct option A.

The balanced chemical equation is: 2NaOH + H2SO4 -> Na2SO4 + 2H2O. First, calculate the moles of NaOH: moles = 0.0250 dm3 * 0.100 mol/dm3 = 0.00250 mol. According to the stoichiometry of the equation, 2 moles of NaOH react with 1 mole of H2SO4. Thus, moles of H2SO4 = 0.00250 mol / 2 = 0.00125 mol. Finally, calculate the concentration of H2SO4: concentration = moles / volume = 0.00125 mol / 0.0200 dm3 = 0.0625 mol/dm3.

1 mark for the correct option B.

First, calculate the relative molecular mass (Mr) of nitrogen gas, N2: Mr(N2) = 2 * 14.0 = 28.0 g/mol. Next, calculate the number of moles of N2 in 14.0 g: moles = mass / Mr = 14.0 / 28.0 = 0.500 mol. Finally, calculate the volume of gas: volume = moles * molar volume = 0.500 mol * 24.0 dm3/mol = 12.0 dm3.

1 mark for the correct option A.

(a) From the table, X displaces all other metals (most reactive). Z displaces W and Y. W only displaces Y. Y displaces none (least reactive). Order of reactivity: X > Z > W > Y.

(b) Copper(II) carbonate, a green powder, decomposes on heating to form copper(II) oxide (a black solid) and carbon dioxide gas (which turns limewater cloudy/milky):
\(CuCO_3(s) \rightarrow CuO(s) + CO_2(g)\).

(c) Aluminium reacts rapidly with atmospheric oxygen to form a very thin, tough, and non-porous layer of aluminium oxide (\(Al_2O_3\)) on its surface. This oxide layer is unreactive and prevents the underlying metal from coming into contact with dilute acids.

(d) (i) Sodium nitrate (Group I nitrate) decomposes to form sodium nitrite and oxygen:
\(2NaNO_3(s) \rightarrow 2NaNO_2(s) + O_2(g)\).
(ii) Copper(II) nitrate (transition metal nitrate) decomposes to form copper(II) oxide, nitrogen dioxide, and oxygen:
\(2Cu(NO_3)_2(s) \rightarrow 2CuO(s) + 4NO_2(g) + O_2(g)\).

(a) [3 marks]
- 1 mark for identifying X as the most reactive.
- 1 mark for identifying Y as the least reactive.
- 1 mark for the correct full sequence: X > Z > W > Y.

(b) [3 marks]
- 1 mark for observation: green solid/powder turns black.
- 1 mark for observation: gas evolved turns limewater cloudy.
- 1 mark for correct chemical equation with state symbols: \(CuCO_3(s) \rightarrow CuO(s) + CO_2(g)\).

(c) [2 marks]
- 1 mark for mentioning the protective oxide layer (aluminium oxide).
- 1 mark for stating that this layer prevents the acid from reaching the metal underneath.

(d) [5 marks]
- (i) 1 mark for correct products (\(NaNO_2\) and \(O_2\)) and reactants; 1 mark for correct balancing and state symbols.
- (ii) 1 mark for correct products (\(CuO\), \(NO_2\), \(O_2\)); 1 mark for correct balancing; 1 mark for correct state symbols on all species.

(a) Titration procedure:
1. Use a volumetric pipette to measure exactly 25.0 \(\text{cm}^3\) of potassium hydroxide solution into a clean conical flask.
2. Add a few drops of a suitable indicator, such as methyl orange, to the flask.
3. Fill a clean, rinsed burette with the standard sulfuric acid solution, ensuring the jet space is filled.
4. Record the initial burette reading.
5. Run the acid from the burette slowly into the conical flask while swirling continuously until the indicator permanently changes color (e.g., from yellow to orange for methyl orange).
6. Record the final burette reading to find the volume of acid added.

(b)
(i) \(\text{Moles of } H_2SO_4 = \text{concentration} \times \text{volume in } \text{dm}^3 = 0.150 \times \frac{18.40}{1000} = 0.00276 \text{ mol}\).

(ii) From the chemical equation, 1 mole of \(H_2SO_4\) reacts with 2 moles of \(KOH\).
\(\text{Moles of } KOH = 2 \times 0.00276 = 0.00552 \text{ mol}\).

(iii) \(\text{Concentration of } KOH = \frac{\text{moles}}{\text{volume in } \text{dm}^3} = \frac{0.00552}{0.0250} = 0.221 \text{ mol/dm}^3\).

(iv) \(\text{Concentration in } \text{g/dm}^3 = \text{concentration in } \text{mol/dm}^3 \times M_r = 0.221 \times 56 = 12.376 \approx 12.4 \text{ g/dm}^3\).

(a) [5 marks]
- 1 mark: Mention pipette used to measure 25.0 \(\text{cm}^3\) of KOH solution into a conical flask.
- 1 mark: Add a named indicator (e.g., methyl orange / phenolphthalein).
- 1 mark: Mention burette filled with sulfuric acid.
- 1 mark: Add acid dropwise near the end-point with swirling until a sharp color change occurs (e.g., yellow to orange/red for methyl orange or pink to colorless for phenolphthalein).
- 1 mark: Record initial and final readings to calculate titre, repeating to obtain concordant results.

(b) [6 marks]
- (i) 1 mark for correct calculation: \(2.76 \times 10^{-3}\) mol (accept 0.00276 mol).
- (ii) 1 mark for multiplying moles of acid by 2: \(5.52 \times 10^{-3}\) mol (accept 0.00552 mol).
- (iii) 1 mark for dividing by volume in \(\text{dm}^3\) (0.0250); 1 mark for correct concentration: 0.221 \(\text{mol/dm}^3\) (allow error carried forward from (ii)).
- (iv) 1 mark for multiplying the molar concentration by 56; 1 mark for final value: 12.4 \(\text{g/dm}^3\) (accept range 12.37 to 12.40; allow error carried forward).

(a) To calculate empirical formula:
- Moles of Carbon: \(\frac{40.0}{12} = 3.33\)
- Moles of Hydrogen: \(\frac{6.7}{1} = 6.7\)
- Moles of Oxygen: \(\frac{53.3}{16} = 3.33\)

Divide each by the smallest value (3.33):
- C = 1, H = 2, O = 1
Empirical formula = \(CH_2O\).

(b) Empirical formula mass of \(CH_2O = 12 + 2(1) + 16 = 30\).
Multiplier \(n = \frac{M_r}{\text{empirical mass}} = \frac{180}{30} = 6\).
Molecular formula = \(6 \times CH_2O = C_6H_{12}O_6\).

(c) Moles of glucose reacting = \(\frac{45.0}{180} = 0.250 \text{ mol}\).
According to the equation, 1 mole of glucose yields 2 moles of \(CO_2\).
Moles of \(CO_2\) produced = \(2 \times 0.250 = 0.500 \text{ mol}\).
Volume of \(CO_2\) at r.t.p. = \(0.500 \times 24 = 12.0 \text{ dm}^3\).

(d) From the equation, 1 mole of glucose yields 2 moles of \(C_2H_5OH\).
Moles of ethanol produced = \(2 \times 0.250 = 0.500 \text{ mol}\).
Mass of ethanol = \(0.500 \times 46 = 23.0 \text{ g}\).

(a) [3 marks]
- 1 mark for dividing percentages by correct relative atomic masses (\(A_r\)).
- 1 mark for correct mole ratio (1 : 2 : 1).
- 1 mark for correct empirical formula: \(CH_2O\).

(b) [2 marks]
- 1 mark for empirical formula unit mass of 30.
- 1 mark for molecular formula: \(C_6H_{12}O_6\).

(c) [5 marks]
- 1 mark for \(M_r\) of glucose = 180.
- 1 mark for moles of glucose = 0.250 mol.
- 1 mark for moles of \(CO_2\) = 0.500 mol (multiplied moles of glucose by 2).
- 1 mark for multiplying moles of gas by 24.
- 1 mark for correct volume with unit: 12.0 \(\text{dm}^3\) (accept 12 \(\text{dm}^3\)).

(d) [2 marks]
- 1 mark for correct moles of ethanol = 0.500 mol.
- 1 mark for correct mass of ethanol: 23.0 g (accept 23 g).

(a) The green color of the solution suggests the presence of iron(II) ions, \(Fe^{2+}\). When aqueous sodium hydroxide is added to a solution containing \(Fe^{2+}\), a green precipitate of iron(II) hydroxide, \(Fe(OH)_2\), is formed. This precipitate is insoluble in excess sodium hydroxide.

(b) Similarly, adding aqueous ammonia to a solution containing \(Fe^{2+}\) produces a green precipitate of iron(II) hydroxide, which is insoluble in excess ammonia.

(c) (i) The test using dilute nitric acid followed by aqueous barium nitrate is the diagnostic test for sulfate ions, \(SO_4^{2-}\). The white precipitate is barium sulfate.
(ii) The ionic equation showing state symbols is:
\(Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)\).

(d) (i) The gas that turns damp red litmus paper blue is alkaline, which is ammonia gas, \(NH_3\).
(ii) Heating a compound with aqueous sodium hydroxide and aluminium foil reduces nitrate ions (\(NO_3^-\)) to ammonia gas. Therefore, the other anion present in Compound X is nitrate.

(a) [3 marks]
- 1 mark for "precipitate".
- 1 mark for "green" color.
- 1 mark for "insoluble in excess".

(b) [3 marks]
- 1 mark for "precipitate".
- 1 mark for "green" color.
- 1 mark for "insoluble in excess".

(c) [4 marks]
- (i) 1 mark for sulfate (accept formula: \(SO_4^{2-}\)).
- (ii) 1 mark for reactant ions: \(Ba^{2+}(aq) + SO_4^{2-}(aq)\); 1 mark for correct product formula: \(BaSO_4(s)\); 1 mark for correct state symbols on both sides.

(d) [3 marks]
- (i) 1 mark for ammonia (accept formula: \(NH_3\)).
- (ii) 2 marks for nitrate (accept formula: \(NO_3^-\)). (1 mark for identifying nitrate, 1 mark for linking the reduction test to the presence of nitrate).

(a) (i) \(C_2H_4 + H_2O \rightarrow C_2H_5OH\).
(ii) Catalyst: phosphoric acid (\(H_3PO_4\)). Temperature: 300 °C.

(b) Advantages of fermentation over hydration:
1. It uses renewable raw materials (e.g., glucose from sugar cane/crops) whereas ethene is obtained from non-renewable crude oil.
2. It occurs at low temperatures (around 30-37 °C) and atmospheric pressure, requiring much less energy.

(c) (i) The laboratory oxidizing agent is acidified potassium manganate(VII). The color changes from purple to colorless when ethanol is oxidized.
(ii) Ethanoic acid structural formula:
```
H O
| //
H-C-C
| \
H O-H
```
(All C-H, C-C, C=O, C-O, and O-H bonds must be clearly shown).

(d) (i) The ester formed is ethyl ethanoate.
(ii) Ethyl ethanoate structural formula:
```
H O H H
| // | |
H-C-C-O - C - C-H
| | |
H H H
```

(a) [3 marks]
- (i) 1 mark for correct chemical equation: \(C_2H_4 + H_2O \rightarrow C_2H_5OH\).
- (ii) 1 mark for catalyst: phosphoric acid (accept \(H_3PO_4\)); 1 mark for temperature: 300 °C (accept range 250 °C to 350 °C).

(b) [2 marks]
- 1 mark for stating that it uses renewable resources.
- 1 mark for stating that it requires less energy / operates at lower temperature/pressure.

(c) [5 marks]
- (i) 1 mark for acidified potassium manganate(VII) (accept acidified potassium dichromate(VI)); 1 mark for starting color (purple / orange); 1 mark for ending color (colorless / green).
- (ii) 2 marks for drawing ethanoic acid showing all bonds: 1 mark for correct methyl group (\(CH_3-\)); 1 mark for correct carboxylic acid functional group (\(-C(=O)-O-H\)) showing the O-H bond.

(d) [3 marks]
- (i) 1 mark for naming: ethyl ethanoate.
- (ii) 2 marks for drawing: 1 mark for correct ester linkage showing all bonds (\(-C(=O)-O-\)); 1 mark for correct alkyl groups (ethyl and methyl) with all bonds shown.

(a) The other three raw materials are:
1. Coke (carbon)
2. Limestone (calcium carbonate)
3. Air (oxygen)

(b)
(i) Coke reacts with carbon dioxide in an endothermic reaction to form carbon monoxide:
\(C(s) + CO_2(g) \rightarrow 2CO(g)\).

(ii) Carbon monoxide reduces iron(III) oxide to iron liquid:
\(Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(l) + 3CO_2(g)\).

(c)
(i) Limestone undergoes thermal decomposition in the hot furnace to produce calcium oxide and carbon dioxide:
\(CaCO_3(s) \rightarrow CaO(s) + CO_2(g)\)
Calcium oxide is a basic oxide, which reacts with the acidic silicon(IV) oxide (silica) impurity to form calcium silicate (slag):
\(CaO(s) + SiO_2(s) \rightarrow CaSiO_3(l)\).

(ii) Slag is used in road construction or making cement.

(d) Carbon can reduce iron because carbon is more reactive than iron, meaning it has a higher affinity for oxygen and can displace iron from its oxide. However, aluminium is more reactive than carbon, so carbon cannot displace aluminium from aluminium oxide.

(a) [3 marks]
- 1 mark for coke / carbon.
- 1 mark for limestone / calcium carbonate.
- 1 mark for air / oxygen.

(b) [3 marks]
- (i) 1 mark for correct equation: \(C + CO_2 \rightarrow 2CO\).
- (ii) 2 marks: 1 mark for correct reactant and product formulas (\(Fe_2O_3 + CO \rightarrow Fe + CO_2\)); 1 mark for correct balancing: \(Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2\).

(c) [5 marks]
- (i) 1 mark for explaining thermal decomposition of limestone.
- 1 mark for equation: \(CaCO_3 \rightarrow CaO + CO_2\).
- 1 mark for explaining basic calcium oxide reacting with acidic silica.
- 1 mark for equation: \(CaO + SiO_2 \rightarrow CaSiO_3\).
- (ii) 1 mark for stating use: road building / concrete / cement manufacture.

(d) [2 marks]
- 1 mark for stating carbon is more reactive than iron.
- 1 mark for stating carbon is less reactive than aluminium.

To carry out the investigation:
1. Use a measuring cylinder to measure a fixed volume (e.g., \(25\text{ cm}^3\)) of the \(1.0\text{ mol/dm}^3\) aqueous copper(II) sulfate and pour it into a polystyrene cup.
2. Place a thermometer in the solution, wait for the reading to stabilize, and record the initial temperature.
3. Weigh a fixed mass (e.g., \(1.0\text{ g}\)) of metal powder \(X\) using a balance.
4. Add metal powder \(X\) to the polystyrene cup, stir the mixture thoroughly with the thermometer, and record the maximum temperature reached.
5. Repeat steps 1 to 4 using the exact same volume of copper(II) sulfate and the same mass of metal powder for metals \(Y\) and \(Z\) separately (using fresh solutions and washing the thermometer/cup in between).
6. Calculate the temperature rise for each metal: \(\Delta T = T_{\text{maximum}} - T_{\text{initial}}\).
7. The metal that produces the highest temperature rise is the most reactive, and the metal with the lowest temperature rise is the least reactive.

1 mark: Measure a fixed volume of copper(II) sulfate using a measuring cylinder.
1 mark: Use a polystyrene cup / insulated container.
1 mark: Measure and record the initial temperature of the copper(II) sulfate solution.
1 mark: Weigh and add a fixed mass of metal powder (using a balance).
1 mark: Stir the mixture during the reaction.
1 mark: Record the maximum temperature reached.
1 mark: Repeat the procedure for the other two metals.
1 mark: State a key controlled variable (e.g., same starting volume/concentration of copper(II) sulfate solution, same mass of metal powder, or same particle size/surface area).
1 mark: Calculate the temperature change (\(\Delta T\)) for each metal.
1 mark: Explain how to deduce the order of reactivity (largest temperature rise = most reactive; smallest temperature rise = least reactive).

(a) Calculate volumes of acid added:
- Titration 1: \(24.7 - 0.5 = 24.2\text{ cm}^3\)
- Titration 2: \(36.2 - 12.1 = 24.1\text{ cm}^3\)
- Titration 3: \(48.5 - 24.2 = 24.3\text{ cm}^3\)

(b) Titrations are repeated to ensure the results are reliable and to identify and discard any anomalous runs to get concordant values.

(c) All three values (\(24.2, 24.1, 24.3\)) are within \(0.2\text{ cm}^3\) of each other, so they are all concordant.
Average volume = \(\frac{24.2 + 24.1 + 24.3}{3} = 24.2\text{ cm}^3\).

(d) Methyl orange is yellow in alkaline solutions (sodium hydroxide) and turns red/pink in acids. At the neutralization endpoint, the color changes from yellow to orange.

(e) Volumetric pipettes have a much smaller percentage error than measuring cylinders, making them highly accurate for quantitative volumetric analysis.

(f) To prevent dilution of the acid, the burette must be rinsed first with distilled water and then with a small amount of the hydrochloric acid itself.

(a) 3 marks: 1 mark for each correct volume calculation (24.2, 24.1, 24.3 cm^3).
(b) 1 mark: To obtain concordant results / improve reliability / identify anomalies.
(c) 2 marks: 1 mark for identifying concordant results (all three titres) and showing the working, 1 mark for the correct calculation (24.2 cm^3).
(d) 2 marks: 1 mark for initial color (yellow), 1 mark for final color at the endpoint (orange or pink/red).
(e) 1 mark: Pipette is more accurate / has a smaller error than a measuring cylinder.
(f) 1 mark: Rinse with the hydrochloric acid (after washing with distilled water).

Procedure:
1. Dissolve a spatula measure of each solid (A, B, and C) in separate test-tubes using distilled water to make aqueous solutions.
2. **Test for Sodium Carbonate (Carbonate ion test)**: Add a portion of dilute nitric acid to a sample of each solution. The solution that shows effervescence (bubbles) is sodium carbonate. (The gas can be confirmed as carbon dioxide, but the fizzing is diagnostic here).
3. **Test for Ammonium Sulfate (Sulfate ion test)**: To a fresh portion of each solution, add dilute nitric acid followed by aqueous barium nitrate. Only the solution containing ammonium sulfate will form a white precipitate (barium sulfate).
4. **Test for Zinc Chloride (Chloride ion test)**: To a fresh portion of each solution, add dilute nitric acid followed by aqueous silver nitrate. Only the solution containing zinc chloride will form a white precipitate (silver chloride).
5. **Alternative confirmation of Zinc / Ammonium**: Add aqueous sodium hydroxide dropwise and then in excess to the remaining solution. The one that forms a white precipitate which dissolves in excess to give a colorless solution is zinc chloride. The one that produces a pungent gas on heating (which turns damp red litmus paper blue) is ammonium sulfate.

1 mark: Dissolve each solid in distilled water to make separate solutions.
1 mark: Add dilute nitric acid to each solution/solid.
1 mark: State that the solution which produces bubbles/effervescence is sodium carbonate.
1 mark: Add dilute nitric acid followed by aqueous barium nitrate to the solutions.
1 mark: State that ammonium sulfate produces a white precipitate (with barium nitrate).
1 mark: Add dilute nitric acid followed by aqueous silver nitrate to the solutions.
1 mark: State that zinc chloride produces a white precipitate (with silver nitrate).
1 mark: Add aqueous sodium hydroxide to a sample of the solids/solutions and warm gently.
1 mark: State that ammonium sulfate produces a gas (ammonia) that turns damp red litmus paper blue.
1 mark: State that zinc chloride with sodium hydroxide forms a white precipitate that dissolves in excess to give a colorless solution.

Experimental Method:
1. Use a measuring cylinder to measure a fixed volume of water (e.g., \(100\text{ cm}^3\)) and pour it into the copper calorimeter.
2. Record the initial temperature of the water using a thermometer.
3. Weigh the spirit burner containing methanol on a balance and record its initial mass.
4. Set up the calorimeter on a stand with a clamp so that it is positioned just above the spirit burner.
5. Light the methanol burner using a splint and allow it to heat the water for a fixed time (e.g., 3 minutes) or until a specific temperature rise (e.g., \(30^\circ\text{C}\)) is achieved.
6. Extinguish the burner and record the maximum (final) temperature of the water.
7. Immediately reweigh the spirit burner and record its final mass.
8. Empty the calorimeter, allow it to cool, and repeat the entire procedure using the same volume of fresh cold water for ethanol and propan-1-ol.

Data Analysis & Variables:
- Control variables: same volume/mass of water, same distance between the burner wick and the calorimeter base, and same stirring rate.
- Calculation: Calculate the temperature rise (\(\Delta T = T_{\text{final}} - T_{\text{initial}}\)) and the mass of alcohol burned (\(\Delta m = m_{\text{initial}} - m_{\text{final}}\)).
- Evaluation: Compute the energy indicator value, which is \(\frac{\Delta T}{\Delta m}\) (temperature rise per gram of fuel burned). The alcohol with the highest value of \(\frac{\Delta T}{\Delta m}\) releases the most heat energy per gram.

1 mark: Measure a fixed volume/mass of water into the calorimeter.
1 mark: Record the initial temperature of the water using a thermometer.
1 mark: Measure and record the initial mass of the spirit burner.
1 mark: Clamp the calorimeter securely above the spirit burner.
1 mark: Burn the alcohol to heat the water (specify a fixed heating time or temperature rise).
1 mark: Measure and record the final temperature of the water.
1 mark: Measure and record the final mass of the spirit burner.
1 mark: Repeat the exact procedure for the other two alcohols using fresh water.
1 mark: State at least two key control variables (e.g., same volume of water, same distance between wick and calorimeter, or use of a draft shield to minimize heat loss).
1 mark: Explain how to process the results (calculate temperature change, calculate mass change, and divide temperature change by mass change to find the temperature rise per gram).