Cambridge IGCSE · Thinka 原創模擬試題

2023 Cambridge IGCSE Chemistry (0620) 模擬試題連答案詳解

Thinka Jun 2023 (V2) Cambridge International A Level-Style Mock — Chemistry (0620)

160 180 分鐘2023
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V2) Cambridge International A Level Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.

Paper 22 選擇題 (Extended)

There are forty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D.
40 題目 · 40
題目 1 · 選擇題
1
What is the maximum volume of carbon dioxide, measured at room temperature and pressure (rtp), produced when \(10.0\text{ g}\) of calcium carbonate (\(M_r = 100\)) reacts completely with excess dilute hydrochloric acid? (1 mol of gas occupies \(24.0\text{ dm}^3\) at rtp)
  1. A.\(0.24\text{ dm}^3\)
  2. B.\(1.20\text{ dm}^3\)
  3. C.\(2.40\text{ dm}^3\)
  4. D.\(24.0\text{ dm}^3\)
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解題

First, calculate the moles of calcium carbonate: \(\text{moles} = 10.0 / 100 = 0.100\text{ mol}\). According to the reaction equation: \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\), 1 mole of calcium carbonate produces 1 mole of carbon dioxide. Therefore, 0.100 moles of \(\text{CO}_2\) are produced. Volume of gas = \(0.100\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 2.40\text{ dm}^3\).

評分準則

Award 1 mark for the correct option C.
題目 2 · 選擇題
1
Which statement explains why increasing the temperature increases the rate of a chemical reaction?
  1. A.The activation energy of the reaction is lowered.
  2. B.A greater proportion of colliding particles possess energy equal to or greater than the activation energy.
  3. C.The reactant particles become closer together, increasing collision frequency.
  4. D.The enthalpy change of the reaction becomes more exothermic, releasing more energy.
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解題

Increasing the temperature increases the average kinetic energy of the particles. Consequently, a much larger proportion of the colliding particles have kinetic energy equal to or exceeding the activation energy, leading to more successful collisions per unit time. While the overall collision frequency also increases slightly, the significant change in the proportion of successful collisions is the primary factor driving the rate increase.

評分準則

Award 1 mark for the correct option B.
題目 3 · 選擇題
1
During the electrolysis of concentrated aqueous sodium chloride using inert electrodes, which products are formed at the electrodes?
  1. A.Anode: hydrogen gas; Cathode: chlorine gas
  2. B.Anode: chlorine gas; Cathode: hydrogen gas
  3. C.Anode: oxygen gas; Cathode: sodium metal
  4. D.Anode: chlorine gas; Cathode: sodium metal
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解題

At the anode (positive electrode), chloride ions (\(\text{Cl}^-\)) are selectively discharged over hydroxide ions (\(\text{OH}^-\)) because they are in high concentration, forming chlorine gas. At the cathode (negative electrode), hydrogen ions (\(\text{H}^+\)) are discharged rather than sodium ions (\(\text{Na}^+\)) because hydrogen is lower in the reactivity series, forming hydrogen gas.

評分準則

Award 1 mark for the correct option B.
題目 4 · 選擇題
1
Which statement about the halogens and their halide ions is correct?
  1. A.Fluorine is a darker green gas than chlorine at room temperature.
  2. B.Aqueous bromine can displace iodide ions from an aqueous solution of potassium iodide.
  3. C.The reactivity of the halogens increases down Group VII.
  4. D.Silver chloride is a cream-coloured precipitate that is soluble in dilute nitric acid.
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解題

Bromine is more reactive than iodine because reactivity decreases down Group VII. Therefore, bromine can displace iodide ions from potassium iodide solution to form iodine and potassium bromide. Fluorine is a pale yellow gas. Silver chloride is a white precipitate (silver bromide is cream) and is insoluble in dilute nitric acid.

評分準則

Award 1 mark for the correct option B.
題目 5 · 選擇題
1
Which linkage is present in nylon?
  1. A.ester linkage
  2. B.amide linkage
  3. C.ether linkage
  4. D.glycosidic linkage
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解題

Nylon is a polyamide made by condensation polymerization of a dicarboxylic acid and a diamine. The linkage formed between the monomers is an amide linkage (\(-\text{CO}-\text{NH}-\)).

評分準則

Award 1 mark for the correct option B.
題目 6 · 選擇題
1
Two separate experiments are carried out. In Experiment 1, \(10\text{ cm}^3\) of \(1.0\text{ mol/dm}^3\) hydrochloric acid is reacted with excess zinc. In Experiment 2, \(10\text{ cm}^3\) of \(1.0\text{ mol/dm}^3\) ethanoic acid is reacted with excess zinc. Which statement is correct?
  1. A.Experiment 1 produces a larger total volume of hydrogen gas than Experiment 2.
  2. B.Experiment 2 has a faster initial rate of reaction than Experiment 1.
  3. C.Both experiments produce the same total volume of hydrogen gas, but Experiment 1 has a faster initial rate.
  4. D.Both experiments have the same initial rate, but Experiment 1 produces a larger total volume of hydrogen gas.
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解題

Both hydrochloric acid and ethanoic acid are monobasic (monoprotic) acids. Since both solutions have the same concentration and volume, they contain the same number of moles of replaceable hydrogen ions and will produce the same total volume of hydrogen gas. However, hydrochloric acid is a strong acid (fully ionised) and ethanoic acid is a weak acid (partially ionised). The higher concentration of hydrogen ions in Experiment 1 gives it a faster initial rate.

評分準則

Award 1 mark for the correct option C.
題目 7 · 選擇題
1
The equation for the Haber process is: \(\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})\) with a forward enthalpy change of \(-92\text{ kJ/mol}\). Which set of changes will both shift the position of equilibrium to the right?
  1. A.increasing the temperature and increasing the pressure
  2. B.decreasing the temperature and increasing the pressure
  3. C.increasing the temperature and decreasing the pressure
  4. D.decreasing the temperature and decreasing the pressure
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解題

The forward reaction is exothermic (\(\Delta H < 0\)). Decreasing the temperature shifts the equilibrium in the exothermic direction (to the right) to release heat. There are 4 moles of gas on the reactant side and 2 moles of gas on the product side. Increasing the pressure shifts the equilibrium towards the side with fewer gas moles (to the right).

評分準則

Award 1 mark for the correct option B.
題目 8 · 選擇題
1
A solid compound, X, is dissolved in water. When aqueous sodium hydroxide is added to the solution of X, a green precipitate is formed which is insoluble in excess. When dilute nitric acid and aqueous barium nitrate are added to another portion of the solution, a white precipitate is formed. What is compound X?
  1. A.iron(II) chloride
  2. B.iron(II) sulfate
  3. C.iron(III) sulfate
  4. D.chromium(III) sulfate
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解題

A green precipitate with sodium hydroxide that is insoluble in excess NaOH indicates the presence of iron(II) ions (\(\text{Fe}^{2+}\)). Chromium(III) also gives a green precipitate but is soluble in excess NaOH to form a green solution. A white precipitate with barium nitrate in acidic conditions indicates the presence of sulfate ions (\(\text{SO}_4^{2-}\)). Therefore, compound X is iron(II) sulfate.

評分準則

Award 1 mark for the correct option B.
題目 9 · 選擇題
1
A 1.50 g sample of an organic compound, X, contains only carbon, hydrogen and oxygen. Complete combustion of X in excess oxygen produces 2.20 g of carbon dioxide, \(CO_2\), and 0.90 g of water, \(H_2O\). What is the empirical formula of X?
  1. A.\(CHO\)
  2. B.\(CH_2O\)
  3. C.\(C_2H_4O\)
  4. D.\(CH_3O\)
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解題

1. Find the mass of C in 2.20 g \(CO_2\): \((12 / 44) \times 2.20 = 0.60\) g. 2. Find the mass of H in 0.90 g \(H_2O\): \((2 / 18) \times 0.90 = 0.10\) g. 3. Find the mass of O: \(1.50\text{ g} - 0.60\text{ g} - 0.10\text{ g} = 0.80\) g. 4. Calculate the moles of each element: Moles of C = \(0.60 / 12 = 0.05\) mol; Moles of H = \(0.10 / 1 = 0.10\) mol; Moles of O = \(0.80 / 16 = 0.05\) mol. 5. Determine the simplest ratio: C : H : O = 1 : 2 : 1. Therefore, the empirical formula is \(CH_2O\).

評分準則

1 mark for selecting the correct option B. Reject all other options.
題目 10 · 選擇題
1
Excess calcium carbonate (marble chips) reacts with 50 \(cm^3\) of 1.0 \(mol/dm^3\) hydrochloric acid to produce carbon dioxide gas. The volume of gas released is plotted against time to give Curve X. The experiment is repeated under different conditions, keeping the mass of calcium carbonate the same and in excess. Which set of conditions produces Curve Y, where the reaction rate is slower but the final volume of gas collected is the same?
  1. A.50 \(cm^3\) of 1.0 \(mol/dm^3\) hydrochloric acid with powdered marble chips
  2. B.50 \(cm^3\) of 1.0 \(mol/dm^3\) hydrochloric acid with the same sized marble chips at a higher temperature
  3. C.100 \(cm^3\) of 0.5 \(mol/dm^3\) hydrochloric acid with the same sized marble chips
  4. D.25 \(cm^3\) of 2.0 \(mol/dm^3\) hydrochloric acid with the same sized marble chips
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解題

To obtain the same final volume of carbon dioxide gas, the number of moles of the limiting reactant (hydrochloric acid) must remain the same: 50 \(cm^3\) of 1.0 \(mol/dm^3\) HCl contains 0.05 mol of HCl. Option C (100 \(cm^3\) of 0.5 \(mol/dm^3\) HCl) also contains 0.05 mol of HCl, so the final volume is unchanged. Because the concentration of hydrochloric acid is lower, the rate of reaction is slower, which produces Curve Y.

評分準則

1 mark for selecting the correct option C. Reject all other options.
題目 11 · 選擇題
1
An aqueous solution of copper(II) sulfate is electrolysed using inert carbon electrodes. Which row correctly describes the observations at each electrode and the change in pH of the solution?
  1. A.Cathode: bubbles of colourless gas; Anode: pink solid forms; pH of solution: decreases
  2. B.Cathode: pink solid forms; Anode: bubbles of colourless gas; pH of solution: decreases
  3. C.Cathode: pink solid forms; Anode: bubbles of colourless gas; pH of solution: increases
  4. D.Cathode: pink solid forms; Anode: bubbles of brown gas; pH of solution: remains constant
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解題

During the electrolysis of aqueous copper(II) sulfate using inert carbon electrodes: 1. Copper ions are discharged at the cathode because copper is lower than hydrogen in the reactivity series, forming a pink/brown solid. 2. Hydroxide ions are discharged at the anode to produce oxygen gas, which is seen as bubbles of a colourless gas. 3. Hydrogen and sulfate ions remain in the solution, forming sulfuric acid, which causes the pH to decrease.

評分準則

1 mark for selecting the correct option B. Reject all other options.
題目 12 · 選擇題
1
In which reaction does the underlined substance act as a proton acceptor?
  1. A.underlined HCl in: \(HCl(aq) + H_2O(l) \rightarrow H_3O^+(aq) + Cl^-(aq)\)
  2. B.underlined HSO4- in: \(HSO_4^-(aq) + H_2O(l) \rightleftharpoons SO_4^{2-}(aq) + H_3O^+(aq)\)
  3. C.underlined CH3COOH in: \(CH_3COOH(aq) + H_2O(l) \rightleftharpoons CH_3COO^-(aq) + H_3O^+(aq)\)
  4. D.underlined NH3 in: \(NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)\)
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解題

According to the proton theory of acids and bases, a base is a proton acceptor. In the reaction of option D, the underlined ammonia molecule accepts a proton from water to form an ammonium ion. In the other options, the underlined substances act as proton donors (acids).

評分準則

1 mark for selecting the correct option D. Reject all other options.
題目 13 · 選擇題
1
The structure of a synthetic polymer is represented as: \(-NH-\square-NH-CO-\Delta-CO-NH-\square-NH-CO-\Delta-CO-\). Which monomers are used to make this polymer?
  1. A.\(HO-\square-OH\) and \(HOOC-\Delta-COOH\)
  2. B.\(H_2N-\square-NH_2\) and \(HOOC-\Delta-COOH\)
  3. C.\(H_2N-\square-COOH\) and \(HOOC-\Delta-COOH\)
  4. D.\(H_2N-\square-NH_2\) and \(HO-\Delta-OH\)
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解題

The polymer contains amide linkages (-NH-CO-) and is a polyamide. Polyamides are synthetic condensation polymers formed from a diamine monomer containing two amine groups (\(H_2N-\square-NH_2\)) and a dicarboxylic acid monomer containing two carboxylic acid groups (\(HOOC-\Delta-COOH\)).

評分準則

1 mark for selecting the correct option B. Reject all other options.
題目 14 · 選擇題
1
The results of three displacement experiments involving three halogens, \(X_2\), \(Y_2\), and \(Z_2\), and their halide ions, \(X^-\), \(Y^-\), and \(Z^-\), are described. Experiment 1: \(X_2\) reacts with \(Y^-\) to turn a colourless solution orange-brown. Experiment 2: \(Y_2\) reacts with \(Z^-\) to turn a colourless solution dark brown. Experiment 3: \(Z_2\) is added to \(X^-\), and no reaction occurs. What are the correct identities of halogens X, Y and Z?
  1. A.X = chlorine, Y = bromine, Z = iodine
  2. B.X = iodine, Y = bromine, Z = chlorine
  3. C.X = bromine, Y = chlorine, Z = iodine
  4. D.X = chlorine, Y = iodine, Z = bromine
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解題

In Experiment 1, halogen \(X_2\) displaces \(Y^-\), so X is more reactive than Y. In Experiment 2, halogen \(Y_2\) displaces \(Z^-\), so Y is more reactive than Z. Thus, the order of reactivity is X > Y > Z. Since reactivity decreases down Group VII, X must be chlorine, Y must be bromine, and Z must be iodine.

評分準則

1 mark for selecting the correct option A. Reject all other options.
題目 15 · 選擇題
1
An unknown salt, W, is dissolved in water to make a solution. Portions of this solution are tested. Test 1: Addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess sodium hydroxide. Test 2: Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate. What is the identity of W?
  1. A.chromium(III) sulfate
  2. B.iron(III) sulfate
  3. C.iron(II) chloride
  4. D.iron(II) sulfate
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解題

The cation forms a green precipitate with aqueous sodium hydroxide that is insoluble in excess, identifying it as iron(II), \(Fe^{2+}\). (Chromium(III) also forms a green precipitate, but it dissolves in excess sodium hydroxide to form a green solution). The anion forms a white precipitate with aqueous barium nitrate, identifying it as sulfate, \(SO_4^{2-}\). Therefore, W is iron(II) sulfate.

評分準則

1 mark for selecting the correct option D. Reject all other options.
題目 16 · 選擇題
1
The equation for the reversible reaction to produce methanol is shown: \(CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)\) (where the forward reaction is exothermic). The temperature of the equilibrium mixture is increased while keeping the pressure constant. How do the yield of methanol and the rate of the forward reaction change?
  1. A.Yield of methanol decreases; Rate of forward reaction increases
  2. B.Yield of methanol decreases; Rate of forward reaction decreases
  3. C.Yield of methanol increases; Rate of forward reaction increases
  4. D.Yield of methanol increases; Rate of forward reaction decreases
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解題

Since the forward reaction is exothermic, increasing the temperature shifts the equilibrium position to the left (the endothermic direction) to absorb the added heat, decreasing the yield of methanol. Increasing the temperature increases the kinetic energy of the particles, leading to more frequent and successful collisions, thereby increasing the rate of the forward reaction.

評分準則

1 mark for selecting the correct option A. Reject all other options.
題目 17 · 選擇題
1
An electrolysis experiment is set up using inert carbon electrodes in concentrated aqueous sodium chloride. Which row correctly identifies the products formed at each electrode and the change in pH of the electrolyte during the electrolysis?
  1. A.Cathode: hydrogen; Anode: chlorine; pH: increases
  2. B.Cathode: sodium; Anode: chlorine; pH: decreases
  3. C.Cathode: hydrogen; Anode: oxygen; pH: remains constant
  4. D.Cathode: sodium; Anode: oxygen; pH: increases
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解題

During the electrolysis of concentrated aqueous sodium chloride (brine), hydrogen ions are discharged at the cathode to produce hydrogen gas because hydrogen is less reactive than sodium. Chloride ions are discharged at the anode to produce chlorine gas because they are in high concentration. This leaves sodium ions and hydroxide ions in the solution, forming sodium hydroxide, which is alkaline, causing the pH to increase.

評分準則

1 mark for the correct option. Correctly identifying that hydrogen is produced at the cathode, chlorine at the anode, and pH increases.
題目 18 · 選擇題
1
The contact process involves the reversible reaction: \(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\), where the forward reaction is exothermic. Which set of reaction conditions would produce the highest equilibrium yield of sulfur trioxide?
  1. A.high temperature and high pressure
  2. B.high temperature and low pressure
  3. C.low temperature and high pressure
  4. D.low temperature and low pressure
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解題

Since the forward reaction is exothermic, a lower temperature shifts the position of equilibrium to the right to favor the forward reaction. Since there are 3 moles of gaseous reactants and 2 moles of gaseous products, a higher pressure shifts the equilibrium to the side with fewer gas moles (the right). Thus, low temperature and high pressure yield the maximum amount of sulfur trioxide.

評分準則

1 mark for the correct option. Correctly applying Le Chatelier's principle to temperature and pressure changes.
題目 19 · 選擇題
1
A student investigates the reaction between excess calcium carbonate and dilute hydrochloric acid. Which change in conditions increases the initial rate of reaction but does not change the total volume of carbon dioxide gas collected at the end of the reaction?
  1. A.Using a more concentrated solution of hydrochloric acid with the same volume.
  2. B.Increasing the temperature of the hydrochloric acid.
  3. C.Using a larger mass of calcium carbonate with the same particle size.
  4. D.Adding a catalyst to the mixture containing a larger volume of the same acid.
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解題

Increasing the temperature increases the kinetic energy of the particles, leading to more frequent and successful collisions, which increases the initial rate of reaction. Since the quantity of the limiting reactant (hydrochloric acid) is unchanged, the total moles of product and therefore the total volume of carbon dioxide remain the same.

評分準則

1 mark for the correct option. Identifying that temperature increases rate without changing the total yield when the limiting reactant is kept constant.
題目 20 · 選擇題
1
Calcium carbide, \(CaC_2\), reacts with water to form ethyne gas, \(C_2H_2\), and calcium hydroxide, \(Ca(OH)_2\). The equation for the reaction is: \(CaC_2(s) + 2H_2O(l) \rightarrow Ca(OH)_2(aq) + C_2H_2(g)\). What volume of ethyne gas, measured at room temperature and pressure (rtp), is produced when \(16.0 \text{ g}\) of calcium carbide (\(M_r = 64.0\)) reacts completely with excess water? (The molar volume of any gas at rtp is \(24.0 \text{ dm}^3\).)
  1. A.1.5 dm3
  2. B.3.0 dm3
  3. C.6.0 dm3
  4. D.12.0 dm3
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解題

First, calculate the number of moles of \(CaC_2\): moles = mass / \(M_r\) = \(16.0 / 64.0 = 0.25 \text{ mol}\). From the stoichiometry of the equation, 1 mole of \(CaC_2\) produces 1 mole of \(C_2H_2\). Therefore, \(0.25 \text{ mol}\) of \(C_2H_2\) is produced. Volume of gas = moles * molar volume = \(0.25 \times 24.0 \text{ dm}^3 = 6.0 \text{ dm}^3\).

評分準則

1 mark for the correct option. Awarded for calculating 0.25 mol of CaC2 and multiplying by 24.0 to get 6.0 dm3.
題目 21 · 選擇題
1
In the reaction shown, which substance is oxidized and what is the change in the oxidation state of sulfur? \(2FeCl_3(aq) + H_2S(g) \rightarrow 2FeCl_2(aq) + S(s) + 2HCl(aq)\)
  1. A.FeCl3 is oxidized; iron oxidation state changes from +3 to +2
  2. B.H2S is oxidized; sulfur oxidation state changes from -2 to 0
  3. C.FeCl3 is reduced; iron oxidation state changes from +2 to +3
  4. D.H2S is reduced; sulfur oxidation state changes from 0 to -2
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解題

In \(H_2S\), sulfur has an oxidation state of -2. In elemental sulfur (S), its oxidation state is 0. An increase in oxidation state indicates oxidation. Therefore, \(H_2S\) is oxidized, and the oxidation state of sulfur changes from -2 to 0.

評分準則

1 mark for the correct option. Correctly identifying that H2S is oxidized and the change in sulfur's oxidation state is from -2 to 0.
題目 22 · 選擇題
1
Terylene is a synthetic polyester. Which statement about the formation of Terylene is correct?
  1. A.It is made from a dicarboxylic acid and a diol, with the elimination of water molecules.
  2. B.It is made from a diamine and a dicarboxylic acid, with the elimination of hydrogen chloride.
  3. C.It is made by addition polymerization of a single monomer containing a carbon-carbon double bond.
  4. D.It is a polyamide formed with the elimination of oxygen molecules.
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解題

Polyesters like Terylene are condensation polymers formed by reacting monomer mixtures containing dicarboxylic acids and diols. Each ester linkage is formed with the elimination of a water molecule.

評分準則

1 mark for the correct option. Correctly identifying the monomers and the eliminated molecule for a polyester.
題目 23 · 選擇題
1
Which statement correctly compares a \(0.1 \text{ mol/dm}^3\) solution of hydrochloric acid with a \(0.1 \text{ mol/dm}^3\) solution of ethanoic acid?
  1. A.Hydrochloric acid has a higher pH than ethanoic acid.
  2. B.Hydrochloric acid has a lower electrical conductivity than ethanoic acid.
  3. C.Hydrochloric acid reacts more rapidly with magnesium ribbon than ethanoic acid.
  4. D.Hydrochloric acid requires a larger volume of 0.1 mol/dm3 sodium hydroxide for complete neutralization than ethanoic acid.
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解題

Hydrochloric acid is a strong acid and is fully dissociated, meaning it has a higher concentration of hydrogen ions than the weak, partially dissociated ethanoic acid. This higher concentration of hydrogen ions results in a lower pH, higher electrical conductivity, and a faster rate of reaction with magnesium. However, because both are monoprotic acids of the same concentration, they require the exact same volume of base for neutralization.

評分準則

1 mark for the correct option. Correctly comparing the properties of a strong acid and a weak acid of the same concentration.
題目 24 · 選擇題
1
An aqueous solution of salt X is tested. Addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate. What is the identity of salt X?
  1. A.Chromium(III) sulfate
  2. B.Iron(II) chloride
  3. C.Iron(II) sulfate
  4. D.Copper(II) sulfate
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解題

A green precipitate with aqueous sodium hydroxide that is insoluble in excess indicates the presence of iron(II) ions (\(Fe^{2+}\)). Chromium(III) also gives a green precipitate, but it dissolves in excess sodium hydroxide to form a green solution. A white precipitate with barium nitrate in the presence of nitric acid indicates the presence of sulfate ions (\(SO_4^{2-}\)). Therefore, salt X is iron(II) sulfate.

評分準則

1 mark for the correct option. Identifying iron(II) from the insoluble green precipitate and sulfate from the barium nitrate test.
題目 25 · 選擇題
1
An element X has two main isotopes, \(^{63}\text{X}\) and \(^{65}\text{X}\). The relative atomic mass of X is 63.6. Which statement about these isotopes is correct?
  1. A.The isotope \(^{63}\text{X}\) has a greater percentage abundance than the isotope \(^{65}\text{X}\).
  2. B.Atoms of \(^{65}\text{X}\) have two more protons than atoms of \(^{63}\text{X}\).
  3. C.The physical properties of \(^{63}\text{X}\) and \(^{65}\text{X}\) are identical.
  4. D.The isotope \(^{65}\text{X}\) has a percentage abundance of 60%.
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解題

To find the relative abundance of each isotope, let the fractional abundance of \(^{65}\text{X}\) be \(x\). The fractional abundance of \(^{63}\text{X}\) is \(1 - x\). We can set up the equation: \(63(1 - x) + 65x = 63.6\). Simplifying this gives: \(63 - 63x + 65x = 63.6\) which simplifies to \(63 + 2x = 63.6\). Therefore, \(2x = 0.6\), which means \(x = 0.3\). Thus, \(^{65}\text{X}\) has an abundance of 30%, and \(^{63}\text{X}\) has an abundance of 70%. Since 70% is greater than 30%, \(^{63}\text{X}\) has a greater percentage abundance. Option B is incorrect because isotopes of the same element have the same number of protons. Option C is incorrect because physical properties of isotopes (such as mass and density) differ slightly due to the difference in mass number. Option D is incorrect as the abundance of \(^{65}\text{X}\) is 30%.

評分準則

1 mark for the correct answer A. 0 marks for any other option.
題目 26 · 選擇題
1
A sample of 2.10 g of a group II metal carbonate, \(\text{XCO}_3\), is fully decomposed by heating to form a metal oxide and carbon dioxide gas. The volume of carbon dioxide gas collected at rtp is 600 cm\(^3\). What is the identity of metal X?
  1. A.Calcium
  2. B.Magnesium
  3. C.Strontium
  4. D.Barium
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解題

The equation for the thermal decomposition is: \(\text{XCO}_3(\text{s}) \rightarrow \text{XO}(\text{s}) + \text{CO}_2(\text{g})\). First, calculate the moles of \(\text{CO}_2\) collected: \(\text{moles of CO}_2 = 600\text{ cm}^3 / 24000\text{ cm}^3/\text{mol} = 0.025\text{ mol}\). Since the molar ratio of \(\text{XCO}_3\) to \(\text{CO}_2\) is 1:1, the moles of \(\text{XCO}_3\) is also 0.025 mol. Next, calculate the relative formula mass (\(M_r\)) of \(\text{XCO}_3\): \(M_r = 2.10\text{ g} / 0.025\text{ mol} = 84\). Using the formula mass of the carbonate group, we get \(M_r(\text{X}) + 12 + (3 \times 16) = 84\), which simplifies to \(M_r(\text{X}) + 60 = 84\), so \(M_r(\text{X}) = 24\). Looking at the periodic table, the group II metal with a relative atomic mass of approximately 24 is Magnesium.

評分準則

1 mark for the correct answer B. 0 marks for any other option.
題目 27 · 選擇題
1
A student investigates the rate of reaction between marble chips (calcium carbonate) and dilute hydrochloric acid: \(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\). In Experiment 1, 5.0 g of large marble chips are reacted with 50 cm\(^3\) of 1.0 mol/dm\(^3\) \(\text{HCl}\) at 20 \(^{\circ}\)C. Which change in conditions for Experiment 2 would result in a faster initial rate of reaction but produce the same final volume of carbon dioxide gas?
  1. A.Using 5.0 g of powdered calcium carbonate instead of large chips, with all other conditions kept the same.
  2. B.Heating the acid to 40 \(^{\circ}\)C and using 100 cm\(^3\) of 1.0 mol/dm\(^3\) \(\text{HCl}\).
  3. C.Using 50 cm\(^3\) of 2.0 mol/dm\(^3\) \(\text{HCl}\) instead of 1.0 mol/dm\(^3\) \(\text{HCl}\) at 20 \(^{\circ}\)C.
  4. D.Adding 50 cm\(^3\) of distilled water to the acid before starting the reaction.
查看答案詳解

解題

In Experiment 1, the moles of \(\text{CaCO}_3\) is \(5.0\text{ g} / 100\text{ g/mol} = 0.05\text{ mol}\), and the moles of \(\text{HCl}\) is \(0.050\text{ dm}^3 \times 1.0\text{ mol/dm}^3 = 0.05\text{ mol}\). Because 1 mole of \(\text{CaCO}_3\) reacts with 2 moles of \(\text{HCl}\), \(\text{HCl}\) is the limiting reactant. Using powdered calcium carbonate increases the surface area, which increases the frequency of successful collisions and thus increases the initial rate of reaction. However, because the total masses and moles of both reactants are unchanged, the final volume of carbon dioxide gas produced remains the same. Options B and C change the moles of the limiting reactant, which would increase the final volume of gas. Option D decreases the concentration of acid, which would slow down the reaction.

評分準則

1 mark for the correct answer A. 0 marks for any other option.
題目 28 · 選擇題
1
Equal volumes of two different acids are prepared: Acid P is 0.1 mol/dm\(^3\) hydrochloric acid, and Acid Q is 0.1 mol/dm\(^3\) ethanoic acid. Which statement about these two solutions is correct?
  1. A.Acid P has a higher pH than Acid Q.
  2. B.Acid P is a better electrical conductor than Acid Q.
  3. C.When reacted with excess zinc, Acid Q produces a greater volume of hydrogen gas than Acid P.
  4. D.Acid Q reacts more vigorously with magnesium than Acid P.
查看答案詳解

解題

Hydrochloric acid (P) is a strong acid and is fully ionised in aqueous solution, resulting in a high concentration of hydrogen ions. Ethanoic acid (Q) is a weak acid and is only partially ionised, resulting in a much lower concentration of hydrogen ions. Because electrical conductivity depends on the concentration of mobile ions, Acid P is a much better conductor than Acid Q. Acid P has a lower pH (more acidic) than Acid Q, making option A incorrect. Since both acids have the same volume and concentration (same number of moles of acid molecules), they will produce the exact same total volume of hydrogen gas when reacted with excess zinc, making option C incorrect. Acid P reacts more vigorously due to its higher concentration of hydrogen ions, making option D incorrect.

評分準則

1 mark for the correct answer B. 0 marks for any other option.
題目 29 · 選擇題
1
Halogens and their halide solutions undergo displacement reactions based on their reactivity. Which mixture results in a displacement reaction and what is the final appearance of the mixture?
  1. A.Reactants: \(\text{Br}_2(\text{aq}) + \text{NaCl}(\text{aq})\), Displacement occurs: No, Final appearance: colourless
  2. B.Reactants: \(\text{Cl}_2(\text{aq}) + \text{KBr}(\text{aq})\), Displacement occurs: Yes, Final appearance: orange-brown
  3. C.Reactants: \(\text{I}_2(\text{aq}) + \text{KCl}(\text{aq})\), Displacement occurs: Yes, Final appearance: colourless
  4. D.Reactants: \(\text{Br}_2(\text{aq}) + \text{KI}(\text{aq})\), Displacement occurs: No, Final appearance: brown
查看答案詳解

解題

Chlorine is more reactive than bromine, so it displaces bromide ions from potassium bromide: \(\text{Cl}_2(\text{aq}) + 2\text{KBr}(\text{aq}) \rightarrow 2\text{KCl}(\text{aq}) + \text{Br}_2(\text{aq})\). The formation of aqueous bromine causes the solution to change from nearly colourless/pale yellow to an orange-brown colour. Option A does not undergo a displacement reaction, but the final mixture remains orange because bromine is present. Option C does not undergo displacement because iodine is less reactive than chlorine. Option D does undergo displacement (forming iodine), but the table claims no displacement occurs.

評分準則

1 mark for the correct answer B. 0 marks for any other option.
題目 30 · 選擇題
1
A polymer has the following repeating unit: \(- \text{O} - \text{CH}_2 - \text{CH}_2 - \text{O} - \text{CO} - \text{CH}_2 - \text{CH}_2 - \text{CO} -\). Which statement about this polymer is correct?
  1. A.It is an addition polymer formed from a single unsaturated monomer.
  2. B.It is a polyamide and water is formed during its synthesis.
  3. C.It is a polyester formed from a dicarboxylic acid and a diol.
  4. D.It is a synthetic carbohydrate containing glycosidic linkages.
查看答案詳解

解題

The repeating unit of the polymer contains ester linkages (\(-\text{O}-\text{CO}-\)) along the backbone. This indicates it is a polyester, which is a type of condensation polymer. It is synthesised from a diol monomer (\(\text{HO}-\text{CH}_2-\text{CH}_2-\text{OH}\)) and a dicarboxylic acid monomer (\(\text{HOOC}-\text{CH}_2-\text{CH}_2-\text{COOH}\)), with the elimination of water molecules. Therefore, option C is correct.

評分準則

1 mark for the correct answer C. 0 marks for any other option.
題目 31 · 選擇題
1
An aqueous solution of sodium sulfate, \(\text{Na}_2\text{SO}_4\), is electrolysed using inert carbon electrodes. A few drops of universal indicator are added to the solution, which is initially green (neutral pH 7). What is observed at each electrode after the current has passed for several minutes?
  1. A.Cathode (-): colourless gas produced and indicator turns red; Anode (+): colourless gas produced and indicator turns blue/purple
  2. B.Cathode (-): colourless gas produced and indicator turns blue/purple; Anode (+): colourless gas produced and indicator turns red
  3. C.Cathode (-): yellow-green gas produced and indicator remains green; Anode (+): colourless gas produced and indicator turns red
  4. D.Cathode (-): colourless gas produced and indicator remains green; Anode (+): colourless gas produced and indicator remains green
查看答案詳解

解題

At the cathode (-), hydrogen ions (\(\text{H}^+\)) from water are discharged to form colourless hydrogen gas (\(\text{H}_2\)) because hydrogen is less reactive than sodium. This leaves an excess of hydroxide ions (\(\text{OH}^-\)) around the cathode, making the solution alkaline and turning the universal indicator blue/purple. At the anode (+), hydroxide ions (\(\text{OH}^-\)) from water are discharged to form colourless oxygen gas (\(\text{O}_2\)) because sulfate ions are more stable. This leaves an excess of hydrogen ions (\(\text{H}^+\)) around the anode, making the solution acidic and turning the universal indicator red.

評分準則

1 mark for the correct answer B. 0 marks for any other option.
題目 32 · 選擇題
1
Iron(III) chloride reacts with hydrogen sulfide gas according to the equation: \(2\text{FeCl}_3(\text{aq}) + \text{H}_2\text{S}(\text{g}) \rightarrow 2\text{FeCl}_2(\text{aq}) + \text{S}(\text{s}) + 2\text{HCl}(\text{aq})\). Which statement describes the redox processes in this reaction?
  1. A.Iron(III) ions are oxidised because their oxidation state increases from +2 to +3.
  2. B.Sulfur is reduced because it gains electrons.
  3. C.Hydrogen sulfide acts as the reducing agent because the oxidation state of sulfur increases from -2 to 0.
  4. D.Iron(III) chloride acts as the reducing agent because iron gains electrons.
查看答案詳解

解題

In this reaction, the oxidation state of iron decreases from +3 (in \(\text{FeCl}_3\)) to +2 (in \(\text{FeCl}_2\)), meaning iron(III) is reduced. The oxidation state of sulfur increases from -2 (in \(\text{H}_2\text{S}\)) to 0 (in elemental \(\text{S}\)), meaning sulfur is oxidised by losing electrons. Since hydrogen sulfide contains the species being oxidised, it acts as the reducing agent. Therefore, option C is correct.

評分準則

1 mark for the correct answer C. 0 marks for any other option.
題目 33 · 選擇題
1
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. Which row correctly describes the products formed at the electrodes and the change in pH of the electrolyte during the electrolysis?
  1. A.Anode product: \(Cl_2(g)\), Cathode product: \(H_2(g)\), pH change: increases
  2. B.Anode product: \(O_2(g)\), Cathode product: \(H_2(g)\), pH change: decreases
  3. C.Anode product: \(Cl_2(g)\), Cathode product: \(Na(s)\), pH change: remains constant
  4. D.Anode product: \(O_2(g)\), Cathode product: \(Na(s)\), pH change: increases
查看答案詳解

解題

During the electrolysis of concentrated aqueous sodium chloride using inert carbon electrodes:
- At the anode (+), chloride ions (\(Cl^-\)) are discharged in preference to hydroxide ions because they are in high concentration, producing chlorine gas: \(2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-\).
- At the cathode (-), hydrogen ions (\(H^+\)) are preferentially discharged over sodium ions because hydrogen is lower in the reactivity series, producing hydrogen gas: \(2H^+(aq) + 2e^- \rightarrow H_2(g)\).
- Since \(H^+\) and \(Cl^-\) ions are removed from the solution, sodium ions (\(Na^+\)) and hydroxide ions (\(OH^-\)) remain in the solution, forming sodium hydroxide. This increases the concentration of hydroxide ions, causing the pH of the electrolyte to increase.

評分準則

1 mark for the correct option. Award 1 mark for identifying the correct products at both electrodes and the resulting increase in pH.
題目 34 · 選擇題
1
A student investigates the rate of reaction between marble chips (excess) and dilute hydrochloric acid. The volume of carbon dioxide gas collected is measured over time.

The experiment is repeated using the same mass of marble chips and the same volume of hydrochloric acid, but with one change: the concentration of the hydrochloric acid is doubled, while its volume is halved.

How does the second experiment compare to the first?
  1. A.The initial rate of reaction is faster, and a greater volume of carbon dioxide is collected.
  2. B.The initial rate of reaction is faster, and the same volume of carbon dioxide is collected.
  3. C.The initial rate of reaction is the same, but a greater volume of carbon dioxide is collected.
  4. D.The initial rate of reaction is the same, and the same volume of carbon dioxide is collected.
查看答案詳解

解題

The rate of reaction depends on the concentration of the reactants. In the second experiment:
- The concentration of hydrochloric acid is doubled, which increases the frequency of collisions between reactant particles per unit time, resulting in a faster initial rate of reaction.
- The volume of hydrochloric acid is halved. The total number of moles of acid is given by: \(\text{moles} = \text{concentration} \times \text{volume}\). Since the concentration is doubled and the volume is halved, the total number of moles of HCl remains exactly the same.
- Since marble chips are in excess, the total volume of carbon dioxide gas produced is determined by the limiting reactant (HCl). Because the moles of HCl are the same, the total volume of carbon dioxide collected is also the same.

評分準則

1 mark for the correct option. Award 1 mark for recognizing that a higher concentration increases the initial rate, while the same number of moles of limiting reactant yields the same volume of product.
題目 35 · 選擇題
1
The following reversible reaction reaches dynamic equilibrium in a sealed container:

\[2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \quad \Delta H = -197\text{ kJ/mol}\]

Which set of conditions will produce the greatest yield of sulfur trioxide, \(SO_3(g)\), at equilibrium?
  1. A.High temperature and high pressure
  2. B.High temperature and low pressure
  3. C.Low temperature and high pressure
  4. D.Low temperature and low pressure
查看答案詳解

解題

For the reaction: \(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \quad \Delta H = -197\text{ kJ/mol}\):
1. Temperature: The forward reaction is exothermic (\(\Delta H\) is negative). According to Le Chatelier's principle, decreasing the temperature shifts the equilibrium position in the exothermic direction (to the right) to release heat, thereby increasing the yield of \(SO_3\).
2. Pressure: There are 3 moles of gaseous reactants on the left side and 2 moles of gaseous products on the right side. Increasing the pressure shifts the equilibrium position to the side with fewer gas molecules (the right side) to reduce the pressure, increasing the yield of \(SO_3\).

Therefore, the greatest yield of sulfur trioxide is achieved at low temperature and high pressure.

評分準則

1 mark for the correct option. Award 1 mark for analyzing temperature and pressure effects correctly using Le Chatelier's principle.
題目 36 · 選擇題
1
What is the minimum mass of pure carbon, \(C\), required to completely reduce 80.0 g of iron(III) oxide, \(Fe_2O_3\), according to the equation below?

\[2Fe_2O_3(s) + 3C(s) \rightarrow 4Fe(s) + 3CO_2(g)\]

[Relative atomic masses, \(A_r\): \(C = 12\), \(O = 16\), \(Fe = 56\)]
  1. A.4.5 g
  2. B.6.0 g
  3. C.9.0 g
  4. D.18.0 g
查看答案詳解

解題

1. Calculate the relative formula mass (\(M_r\)) of iron(III) oxide, \(Fe_2O_3\):
\(M_r(Fe_2O_3) = (2 \times 56) + (3 \times 16) = 112 + 48 = 160\).
2. Calculate the number of moles of \(Fe_2O_3\):
\(\text{moles} = \frac{80.0\text{ g}}{160\text{ g/mol}} = 0.50\text{ mol}\).
3. Use the stoichiometric ratio from the balanced equation:
\(2\text{ mol } Fe_2O_3 : 3\text{ mol } C\).
Therefore, \(0.50\text{ mol}\) of \(Fe_2O_3\) requires:
\(\frac{3}{2} \times 0.50 = 0.75\text{ mol}\) of carbon.
4. Calculate the mass of carbon required:
\(\text{mass} = 0.75\text{ mol} \times 12\text{ g/mol} = 9.0\text{ g}\).

評分準則

1 mark for the correct option. Award 1 mark for showing correct calculation steps: calculating moles of \(Fe_2O_3\), using the stoichiometric ratio, and converting back to mass of carbon.
題目 37 · 選擇題
1
A section of a synthetic polymer is shown:

\[-CH_2-CH(Cl)-CH_2-CH(Cl)-CH_2-CH(Cl)-\]

Which statement about this polymer is correct?
  1. A.It is a polyester formed by condensation polymerization.
  2. B.It is an addition polymer made from the monomer chloroethene.
  3. C.The empirical formula of this polymer is \(CH_2Cl\).
  4. D.The monomer contains a carbon-carbon single bond.
查看答案詳解

解題

The polymer shown is poly(chloroethene), commonly known as PVC.
- It is formed through addition polymerization of the monomer chloroethene, \(CH_2=CHCl\). In addition polymerization, double bonds in the monomers open up to form single bonds in a long chain polymer with no other products.
- A is incorrect because it is an addition polymer, not a polyester (no ester linkages).
- C is incorrect because the empirical formula of the polymer is the same as the monomer, which is \(C_2H_3Cl\).
- D is incorrect because the monomer (chloroethene) must contain a carbon-carbon double bond to undergo addition polymerization.

評分準則

1 mark for the correct option. Award 1 mark for identifying the polymer as an addition polymer formed from the monomer chloroethene.
題目 38 · 選擇題
1
Which statement correctly describes a property of a weak acid, such as ethanoic acid, compared to a strong acid of the same concentration?
  1. A.It has a lower pH and reacts more slowly with magnesium ribbon.
  2. B.It has a higher pH and reacts more slowly with magnesium ribbon.
  3. C.It has a lower pH and reacts more rapidly with magnesium ribbon.
  4. D.It has a higher pH and reacts more rapidly with magnesium ribbon.
查看答案詳解

解題

A weak acid is one that only partially dissociates into ions in aqueous solution. For a weak acid compared to a strong acid of the same concentration:
- The concentration of hydrogen ions, \([H^+]\), is lower in the weak acid.
- A lower hydrogen ion concentration results in a higher pH (closer to neutral 7).
- The lower concentration of hydrogen ions also means that the rate of reaction with magnesium ribbon (which depends on the concentration of \(H^+\) ions) is slower.
Therefore, the weak acid has a higher pH and reacts more slowly.

評分準則

1 mark for the correct option. Award 1 mark for relating weak acid partial dissociation to both higher pH and slower reaction rate.
題目 39 · 選擇題
1
Aqueous chlorine, bromine water, and iodine water are added separately to solutions of potassium halides.

Which row correctly describes the observations?
  1. A.Chlorine displaces bromine from potassium bromide, forming a brown solution.
  2. B.Bromine displaces chlorine from potassium chloride, forming a pale green solution.
  3. C.Iodine displaces bromine from potassium bromide, forming a brown solution.
  4. D.Chlorine does not react with potassium iodide.
查看答案詳解

解題

In Group VII, reactivity decreases down the group (chlorine > bromine > iodine).
- Chlorine is more reactive than bromine. When chlorine is added to a solution containing bromide ions (from potassium bromide), a displacement reaction occurs: \(Cl_2(aq) + 2KBr(aq) \rightarrow 2KCl(aq) + Br_2(aq)\). The formation of bromine turns the solution brown/orange. This matches option A.
- Option B is incorrect because bromine is less reactive than chlorine and cannot displace it.
- Option C is incorrect because iodine is less reactive than bromine and cannot displace it.
- Option D is incorrect because chlorine is more reactive than iodine and will readily displace iodide ions.

評分準則

1 mark for the correct option. Award 1 mark for applying halogen reactivity trends to predict correct displacement reactions and their observations.
題目 40 · 選擇題
1
A solid mixture containing two salts is dissolved in water to make a solution.

- Addition of dilute nitric acid followed by aqueous silver nitrate produces a yellow precipitate.
- Addition of excess aqueous sodium hydroxide produces a green precipitate that is insoluble in excess.

Which two ions are present in the mixture?
  1. A.\(Fe^{3+}\) and \(Cl^-\)
  2. B.\(Fe^{3+}\) and \(I^-\)
  3. C.\(Fe^{2+}\) and \(I^-\)
  4. D.\(Cr^{3+}\) and \(Br^-\)
查看答案詳解

解題

Let us identify each ion from the observations:
1. The addition of dilute nitric acid followed by silver nitrate solution produces a yellow precipitate. This is the characteristic test for iodide ions, \(I^-\), which form a yellow precipitate of silver iodide (\(AgI\)).
2. The addition of excess sodium hydroxide produces a green precipitate that is insoluble in excess. This is the characteristic test for iron(II) ions, \(Fe^{2+}\), which form a green precipitate of iron(II) hydroxide, \(Fe(OH)_2\), that is insoluble in excess sodium hydroxide.
Note: \(Cr^{3+}\) also forms a green precipitate with sodium hydroxide, but it is soluble in excess to form a green solution. Therefore, the ions present in the mixture must be \(Fe^{2+}\) and \(I^-\).

評分準則

1 mark for the correct option. Award 1 mark for identifying the halide as iodide (yellow precipitate) and the cation as iron(II) (insoluble green precipitate).

Paper 42 Theory (Extended)

Answer all questions. Write your answers in the spaces provided on the question paper.
6 題目 · 79.98
題目 1 · structuredTheory
13.33
A student investigates the rate of reaction between calcium carbonate (marble chips) and dilute hydrochloric acid:
\( \text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) \)

(a) Describe how the volume of gas evolved could be measured over time, naming the apparatus used. [3 marks]
(b) Explain, in terms of collision theory, how increasing the concentration of hydrochloric acid increases the rate of this reaction. [3 marks]
(c) The student repeats the experiment using the same mass of calcium carbonate but as a fine powder instead of large chips.
(i) Describe and explain the effect of this change on the rate of reaction. [3 marks]
(ii) State and explain what happens to the total volume of carbon dioxide gas produced. [2 marks]
(d) Describe the role of a catalyst in terms of activation energy and collision theory. [2.33 marks]
查看答案詳解

解題

(a) The reaction mixture is placed in a conical flask, which is connected via a delivery tube to a gas syringe (or an inverted measuring cylinder filled with water). The volume of gas in the syringe/cylinder is recorded at regular time intervals.
(b) Increasing concentration increases the number of acid particles per unit volume. This results in more frequent collisions between reacting particles, leading to an increased rate of successful collisions.
(c)(i) The rate of reaction increases because a fine powder has a larger surface area than large chips. This means there are more exposed particles, leading to more frequent collisions.
(ii) The total volume of gas remains the same because the mass (and therefore the number of moles) of the limiting reactant (calcium carbonate) is unchanged.
(d) A catalyst provides an alternative pathway with lower activation energy. This means a greater fraction of colliding particles have energy greater than or equal to the activation energy, increasing the rate of successful collisions.

評分準則

(a) Conical flask connected to delivery tube [1]; Gas syringe / inverted measuring cylinder over water [1]; Measurement of volume at timed intervals [1].
(b) More particles per unit volume [1]; More frequent collisions / more collisions per unit time [1]; Increased rate of successful collisions [1].
(c)(i) Rate increases [1]; Larger surface area of powder [1]; More frequent collisions [1].
(c)(ii) Volume of gas is unchanged [1]; Mass/moles of reactants is the same / same amount of limiting reactant [1].
(d) Lowers the activation energy / provides alternative route [1]; Greater proportion of particles have energy \(\ge E_a\) [1]; (0.33 marks awarded for comprehensive explanation of successful collision frequency increase [0.33]).
題目 2 · structuredTheory
13.33
Halogens are Group VII elements. They show trends in physical and chemical properties.

(a) State the trend down Group VII for:
(i) the color intensity of the halogens. [1 mark]
(ii) the boiling point of the halogens. [1 mark]
(iii) the reactivity of the halogens. [1 mark]
(b) Chlorine gas is bubbled into a solution of aqueous potassium bromide.
(i) State the observation for this reaction. [1 mark]
(ii) Write an ionic equation, including state symbols, for this reaction. [3 marks]
(iii) Explain, in terms of electron transfer, which species is oxidised. [2 marks]
(c) Draw a dot-and-cross diagram to show the arrangement of outer-shell electrons in a molecule of hydrogen fluoride, HF. [2 marks]
(d) Explain why aqueous hydrogen fluoride behaves as a weak acid. [1.33 marks]
查看答案詳解

解題

(a)(i) The color becomes darker down the group.
(ii) The boiling point increases down the group.
(iii) The reactivity decreases down the group.
(b)(i) The colorless solution turns orange/yellow-brown.
(ii) \( \text{Cl}_2(\text{g}) + 2\text{Br}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{Br}_2(\text{aq}) \)
(iii) Bromide ions (\( \text{Br}^- \)) are oxidised because they lose electrons to form bromine molecules.
(c) A single covalent bond is shown between H and F (one pair of shared electrons, with one electron from H (dot) and one from F (cross)). The F atom has 6 other non-bonding outer electrons (crosses), and H has no other electrons.
(d) Hydrogen fluoride is a weak acid because it only partially dissociates/ionises in aqueous solution to release hydrogen ions.

評分準則

(a)(i) Color gets darker [1].
(ii) Boiling point increases [1].
(iii) Reactivity decreases [1].
(b)(i) Turns orange / yellow / brown [1].
(ii) Correct formula of reactants and products [1]; Balanced [1]; State symbols: \( \text{Cl}_2(\text{g}) \), \( \text{Br}^-(\text{aq}) \), \( \text{Cl}^-(\text{aq}) \), \( \text{Br}_2(\text{aq}) \) [1].
(iii) Bromide / \( \text{Br}^- \) [1]; Loses electrons [1].
(c) One shared pair of electrons between H and F [1]; 6 non-bonding electrons on F outer shell and none on H [1].
(d) Partially ionises/dissociates [1]; to produce \( \text{H}^+ \) ions [0.33].
題目 3 · structuredTheory
13.33
A solid mixture, \( \mathbf{Y} \), contains one cation and two different anions. A student performs a series of qualitative tests on \( \mathbf{Y} \).

(a) Aqueous sodium hydroxide is added to an aqueous solution of \( \mathbf{Y} \). A green precipitate is formed, which does not dissolve in excess sodium hydroxide.
(i) Identify the cation present. [1 mark]
(ii) Write the ionic equation, including state symbols, for the formation of this green precipitate. [2 marks]
(b) Dilute nitric acid is added to solid \( \mathbf{Y} \). Effervescence is observed, producing a colorless gas that forms a white precipitate when bubbled through limewater.
(i) Identify one of the anions in \( \mathbf{Y} \). [1 mark]
(ii) Name the gas produced. [1 mark]
(c) To another sample of the solution of \( \mathbf{Y} \), dilute nitric acid is added, followed by aqueous barium nitrate. A white precipitate is formed.
(i) Identify the second anion in \( \mathbf{Y} \). [1 mark]
(ii) Write the ionic equation for the formation of this white precipitate. [2 marks]
(d) Describe a chemical test, including reagents and observations, to distinguish between:
(i) Carbon dioxide and sulfur dioxide gas. [3.33 marks]
(ii) Aqueous iron(III) ions and aqueous aluminium ions. [2 marks]
查看答案詳解

解題

(a)(i) Iron(II) ion / \( \text{Fe}^{2+} \)
(ii) \( \text{Fe}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_2(\text{s}) \)
(b)(i) Carbonate / \( \text{CO}_3^{2-} \)
(ii) Carbon dioxide / \( \text{CO}_2 \)
(c)(i) Sulfate / \( \text{SO}_4^{2-} \)
(ii) \( \text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s}) \)
(d)(i) Bubble the gases into acidified aqueous potassium manganate(VII). Carbon dioxide does not decolorize the purple solution, whereas sulfur dioxide decolorizes the acidified potassium manganate(VII) from purple to colorless.
(ii) Add aqueous sodium hydroxide to both. Iron(III) forms a red-brown precipitate that is insoluble in excess, whereas aluminium forms a white precipitate that dissolves in excess to form a colorless solution.

評分準則

(a)(i) Iron(II) / \( \text{Fe}^{2+} \) [1].
(ii) Formulae of reactants and products [1]; State symbols [1].
(b)(i) Carbonate / \( \text{CO}_3^{2-} \) [1].
(ii) Carbon dioxide [1].
(c)(i) Sulfate / \( \text{SO}_4^{2-} \) [1].
(ii) Correct equation [2] (formulae [1], balancing [1]).
(d)(i) Use acidified potassium manganate(VII) [1]; Carbon dioxide: remains purple [1]; Sulfur dioxide: turns colorless [1]; details of correct acidified reagent [0.33].
(d)(ii) Add aqueous sodium hydroxide [1]; Iron(III) gives red-brown precipitate and aluminium gives white precipitate which dissolves in excess [1].
題目 4 · structuredTheory
13.33
Methanol is produced industrially by the reversible reaction of carbon monoxide and hydrogen:
\( \text{CO}(\text{g}) + 2\text{H}_2(\text{g}) \rightleftharpoons \text{CH}_3\text{OH}(\text{g}) \) \(\Delta H = -91\text{ kJ/mol} \)

(a) State two characteristics of a reaction at dynamic equilibrium. [2 marks]
(b) Explain the effect of the following changes on the position of equilibrium:
(i) Increasing the pressure. [3 marks]
(ii) Increasing the temperature. [3 marks]
(c) The industrial process is carried out at about \( 250\ ^\circ\text{C} \) and \( 50-100\text{ atm} \). Explain why a temperature of \( 250\ ^\circ\text{C} \) is used rather than a much lower temperature, even though a lower temperature would give a higher equilibrium yield. [2 marks]
(d) Define the term catalyst and name the catalyst used in this reaction. [3.33 marks]
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解題

(a) 1. The rate of the forward reaction equals the rate of the reverse reaction. 2. The concentrations of reactants and products remain constant (in a closed system).
(b)(i) The equilibrium position shifts to the right (towards the products). This is because there are fewer moles of gas on the right-hand side (1 mole of gas) compared to the left-hand side (3 moles of gas). Increasing pressure favors the side with fewer gas moles.
(ii) The equilibrium position shifts to the left (towards the reactants). This is because the forward reaction is exothermic. Increasing temperature favors the endothermic direction to absorb the added heat.
(c) At a lower temperature, the rate of reaction is too slow. \( 250\ ^\circ\text{C} \) is a compromise temperature that provides a sufficiently fast rate of reaction while still obtaining an acceptable yield of methanol.
(d) A catalyst is a substance that increases the rate of a chemical reaction without being chemically changed or used up at the end of the reaction. The catalyst used in the manufacture of methanol is copper (or copper-zinc oxide).

評分準則

(a) Rate of forward reaction = rate of reverse reaction [1]; Concentrations of reactants and products remain constant [1].
(b)(i) Shift to the right [1]; Fewer gas moles on the right (1 mole vs 3 moles on left) [1]; System opposes the pressure increase by shifting to side with fewer molecules [1].
(b)(ii) Shift to the left [1]; Forward reaction is exothermic [1]; High temperature favors endothermic reaction [1].
(c) Lower temperature decreases reaction rate [1]; \( 250\ ^\circ\text{C} \) is a compromise temperature giving reasonable rate and yield [1].
(d) Increases rate of reaction [1]; remains chemically unchanged [1]; copper / copper-zinc catalyst [1]; (0.33 marks for detailed explanation of alternative pathway with lower \(E_a\) [0.33]).
題目 5 · structuredTheory
13.33
A 5.30 g sample of impure anhydrous sodium carbonate, \( \text{Na}_2\text{CO}_3 \), was dissolved in water to make a 250 \( \text{cm}^3 \) solution. A 25.0 \( \text{cm}^3 \) portion of this solution was titrated against 0.200 \( \text{mol/dm}^3 \) hydrochloric acid. It required 22.5 \( \text{cm}^3 \) of the acid for complete neutralisation:
\( \text{Na}_2\text{CO}_3(\text{aq}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) \)

(a) Calculate the number of moles of hydrochloric acid used in the titration. [2 marks]
(b) Determine the number of moles of sodium carbonate in the 25.0 \( \text{cm}^3 \) portion. [2 marks]
(c) Calculate the number of moles of sodium carbonate in the 250 \( \text{cm}^3 \) solution. [2 marks]
(d) Calculate the mass of sodium carbonate in the 250 \( \text{cm}^3 \) solution. [Relative formula mass: \( \text{Na}_2\text{CO}_3 = 106 \)] [2 marks]
(e) Calculate the percentage purity of the original sample of sodium carbonate. [2.33 marks]
(f) A separate 4.29 g sample of hydrated sodium carbonate, \( \text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} \), was heated strongly to constant mass. The mass of the anhydrous residue was 1.59 g. Calculate the value of \( x \). [3 marks]
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解題

(a) Moles of \( \text{HCl} = 0.200 \times \frac{22.5}{1000} = 0.00450\text{ mol} \).
(b) Moles of \( \text{Na}_2\text{CO}_3 = \frac{0.00450}{2} = 0.00225\text{ mol} \).
(c) Moles in 250 \( \text{cm}^3 = 0.00225 \times 10 = 0.0225\text{ mol} \).
(d) Mass of \( \text{Na}_2\text{CO}_3 = 0.0225 \times 106 = 2.385\text{ g} \).
(e) Percentage purity = \( \frac{2.385}{5.30} \times 100\% = 45.0\% \).
(f) Mass of \( \text{H}_2\text{O} \) lost = \( 4.29\text{ g} - 1.59\text{ g} = 2.70\text{ g} \).
Moles of \( \text{Na}_2\text{CO}_3 = \frac{1.59}{106} = 0.0150\text{ mol} \).
Moles of \( \text{H}_2\text{O} = \frac{2.70}{18.0} = 0.150\text{ mol} \).
Ratio of \( \text{H}_2\text{O} : \text{Na}_2\text{CO}_3 = \frac{0.150}{0.0150} = 10 \).
Therefore, \( x = 10 \).

評分準則

(a) Uses formula \( n = c \times V \) [1]; Moles of \( \text{HCl} = 0.00450\text{ mol} \) [1].
(b) Uses 1:2 ratio [1]; Moles of \( \text{Na}_2\text{CO}_3 = 0.00225\text{ mol} \) [1].
(c) Multiplies by 10 [1]; Moles in 250 \( \text{cm}^3 = 0.0225\text{ mol} \) [1].
(d) Multiplies by 106 [1]; Mass = 2.385 g [1].
(e) Uses ratio of mass of pure to impure mass [1]; 45.0% [1]; (0.33 marks for correct rounding/significant figures [0.33]).
(f) Mass of water lost = 2.70 g [1]; Moles of anhydrous salt = 0.0150 mol AND moles of water = 0.150 mol [1]; Ratio \( x = 10 \) [1].
題目 6 · structuredTheory
13.33
Synthetic and natural polymers have many important applications.

(a) Nylon-6,6 is a synthetic polyamide made by condensation polymerisation.
(i) State the meaning of the term condensation polymerisation. [2 marks]
(ii) Nylon-6,6 is made from 1,6-diaminohexane and hexanedioic acid. Draw the block structures of these two monomers, showing all the atoms and bonds in their functional groups. [2 marks]
(iii) Draw a section of the polymer chain of nylon-6,6 showing one amide linkage. Show all bonds in the linkage. [2 marks]
(b) Terylene is a polyester.
(i) State the name of the two functional groups that react together to form Terylene. [2 marks]
(ii) State one major use of Terylene. [1 mark]
(c) Proteins are natural polymers.
(i) Name the type of monomer that forms proteins. [1 mark]
(ii) State the name of the linkage in proteins and draw its structure. [2.33 marks]
(iii) Name the process by which proteins are broken down into their monomers in the laboratory. [1 mark]
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解題

(a)(i) Condensation polymerisation is the joining of monomers together to form a polymer with the elimination of a small molecule, such as water or hydrogen chloride.
(ii) Diamine monomer: \( \text{H}_2\text{N}-\text{[block]}-\text{NH}_2 \); Dicarboxylic acid monomer: \( \text{HOOC}-\text{[block]}-\text{COOH} \) (where HOOC and COOH are drawn out fully as \( -\text{C}(=\text{O})-\text{O}-\text{H} \)).
(iii) Polyamide structure: \( -[\text{block}]-\text{NH}-\text{C}(=\text{O})-[\text{block}]- \).
(b)(i) Carboxylic acid (or carboxyl) and alcohol (or hydroxyl).
(ii) Making synthetic fibers for clothing / sleeping bags / sails.
(c)(i) Amino acids.
(ii) Amide linkage (or peptide linkage). Structure is \( -\text{C}(=\text{O})-\text{NH}- \).
(iii) Acid hydrolysis / hydrolysis.

評分準則

(a)(i) Joining of monomer molecules to make a long chain [1]; loss of small molecule/water [1].
(ii) Correct amine group structure \( \text{H}-\text{N}(\text{-H})- \) on both ends of a block [1]; correct carboxylic acid structure \( -\text{C}(=\text{O})-\text{O}-\text{H} \) on both ends of a block [1].
(iii) Correct amide link \( -\text{C}(=\text{O})-\text{N}(\text{-H})- \) [1]; continuation bonds shown at both ends [1].
(b)(i) Carboxylic acid (or carboxyl) [1]; Alcohol (or hydroxyl) [1].
(ii) Clothing / fabrics / ropes [1].
(c)(i) Amino acids [1].
(ii) Amide / peptide linkage [1]; correct structural drawing \( -\text{C}(=\text{O})-\text{N}(\text{-H})- \) [1]; (0.33 marks for showing continuation bonds [0.33]).
(c)(iii) Hydrolysis [1].

Paper 62 Alternative to Practical

Answer all questions. Write your answers in the spaces provided on the question paper.
4 題目 · 40
題目 1 · alternativeToPractical
10
A student wants to plan an investigation to determine the solubility of potassium nitrate, \(\text{KNO}_3\), in water at different temperatures. Solubility is defined as the maximum mass of solute that can dissolve in \(100\text{ g}\) of water at a specified temperature. The student is provided with: solid potassium nitrate, distilled water, and standard laboratory apparatus (beakers, test-tubes, balance, thermometer, Bunsen burner, tripod, gauze, measuring cylinders, glass rod). (a) Name a suitable piece of apparatus to measure exactly \(20.0\text{ cm}^3\) of distilled water. [1] (b) Describe how the student could prepare a saturated solution of potassium nitrate at a specific temperature (e.g. \(40\text{ }^\circ\text{C}\)) and ensure that no solid remains suspended in the liquid before further steps. [3] (c) Once a clear, saturated solution at \(40\text{ }^\circ\text{C}\) is obtained, describe the experimental steps required to determine the mass of dissolved potassium nitrate in a known volume of this solution. [4] (d) Suggest two safety precautions that should be taken during this investigation. [2]
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解題

(a) A volumetric pipette or a burette is the most suitable apparatus for measuring an exact volume such as \(20.0\text{ cm}^3\). (b) To ensure saturation, excess potassium nitrate must be added to the water. The mixture should be heated and stirred to ensure dissolution of the maximum possible amount, and then allowed to cool down to the target temperature of \(40\text{ }^\circ\text{C}\) with constant stirring. Once at \(40\text{ }^\circ\text{C}\), the mixture is filtered or decanted to remove any remaining suspended solid, leaving a clear saturated solution. (c) First, weigh a clean, empty evaporating basin and record its mass. Transfer a precise volume (such as \(10.0\text{ cm}^3\)) of the saturated solution into the basin using a pipette. Heat the basin gently to evaporate all the water until only dry solid remains. Allow the basin to cool and weigh it. Subtract the initial empty basin mass from the final mass to find the mass of the dry potassium nitrate. (d) Essential safety precautions include wearing safety goggles to protect the eyes from hot spitting salt crystals during evaporation, and using crucible tongs or heat-resistant gloves when handling hot evaporating basins.

評分準則

(a) Volumetric pipette / burette [1] (Accept: measuring cylinder [1]). (b) Add excess solid potassium nitrate to water [1]; heat the mixture and then cool to the target temperature of \(40\text{ }^\circ\text{C}\) with stirring [1]; filter or decant to remove undissolved solid [1]. (c) Weigh an empty evaporating basin [1]; pipette/transfer a known volume of saturated solution into the basin and record the combined mass [1]; heat the solution to evaporate the water to dryness [1]; cool and reweigh the dry basin containing the solid, and subtract the empty basin mass [1]. (d) Wear safety goggles to prevent injury from spitting liquid [1]; use tongs / heat-resistant gloves to handle hot glassware [1].
題目 2 · alternativeToPractical
10
A student investigates the rate of reaction between marble chips (calcium carbonate) and dilute hydrochloric acid: \(\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}\). They measure the volume of carbon dioxide gas collected over time using a gas syringe. (a) Draw a labelled diagram of the assembled apparatus that could be used to carry out this experiment and collect the gas. [3] (b) The gas syringe readings at different times during the experiment are recorded: At \(0\text{ s}\): \(0\text{ cm}^3\); At \(30\text{ s}\): \(24\text{ cm}^3\); At \(60\text{ s}\): \(40\text{ cm}^3\); At \(90\text{ s}\): \(51\text{ cm}^3\); At \(120\text{ s}\): \(58\text{ cm}^3\); At \(150\text{ s}\): \(60\text{ cm}^3\); At \(180\text{ s}\): \(60\text{ cm}^3\). (i) State the time at which the reaction stopped. [1] (ii) Explain your answer to (b)(i). [1] (c) Calculate the average rate of reaction between \(0\text{ s}\) and \(60\text{ s}\). Show your working and state the units. [2] (d) The student repeats the experiment using the same mass of calcium carbonate but as a fine powder instead of large chips. (i) Describe how the rate of reaction changes. [1] (ii) Explain this change in terms of particle collision theory. [2]
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解題

(a) A complete diagram must show a reaction vessel (like a conical flask) sealed with a rubber bung. A delivery tube must connect the flask to a gas syringe (or an inverted measuring cylinder over a trough of water). The reactants inside the flask and the gas syringe must be clearly labelled. (b) (i) The reaction stopped at \(150\text{ s}\). (ii) Because after \(150\text{ s}\), the volume of gas remained constant at \(60\text{ cm}^3\), indicating no more carbon dioxide was being produced. (c) Average rate = \(\frac{\text{Volume of gas collected}}{\text{Time elapsed}} = \frac{40\text{ cm}^3}{60\text{ s}} = 0.67\text{ cm}^3\text{/s}\). (d) (i) The rate of reaction increases when fine powder is used. (ii) This is because the powder has a much larger surface area than the large marble chips, which results in a greater number of exposed reactant particles and therefore a higher frequency of successful collisions per unit time.

評分準則

(a) Award 1 mark for drawing a sealed reaction vessel (conical flask with bung and delivery tube); 1 mark for drawing a gas syringe connected to the delivery tube (or inverted cylinder over water); 1 mark for correct labels of reactants (calcium carbonate and hydrochloric acid) and gas syringe. (b) (i) \(150\text{ s}\) [1]. (ii) The volume of gas remains constant at \(60\text{ cm}^3\) / stops increasing [1]. (c) Award 1 mark for correct calculation: \(40 / 60 = 0.67\) (accept \(0.667\) or \(0.67\)); 1 mark for correct unit: \(\text{cm}^3\text{/s}\) (or \(\text{cm}^3\text{ s}^{-1}\)). (d) (i) Rate increases / reaction is faster [1]. (ii) Fine powder has a larger surface area [1]; resulting in more frequent collisions between reactant particles per unit time [1].
題目 3 · alternativeToPractical
10
Two substances, Solid E and Liquid F, were analysed. Solid E was a green transition metal salt. Liquid F was an aqueous solution of an acid. (a) Solid E was dissolved in distilled water to make Solution E. To a \(2\text{ cm}^3\) portion of Solution E, aqueous sodium hydroxide was added dropwise until in excess. Observation: A green precipitate was formed, which was insoluble in excess. Identify the cation present in Solid E. [1] (b) To another \(2\text{ cm}^3\) portion of Solution E, dilute nitric acid and then aqueous barium nitrate were added. Observation: A white precipitate was formed. Identify the anion present in Solid E. [1] (c) Deduce the chemical formula of Solid E. [1] (d) A sample of Solid E was heated strongly in a dry test-tube. A gas was evolved which turned a piece of damp blue litmus paper red, and then bleached it. (i) Identify this gas. [1] (ii) Describe a chemical test for this gas, other than using indicator paper, and state the expected observation. [2] (e) Liquid F was tested. (i) When a piece of magnesium ribbon was added to Liquid F, a gas was evolved. Describe a test to identify this gas and state the expected positive observation. [2] (ii) When aqueous silver nitrate was added to Liquid F, a white precipitate was formed. Identify Liquid F. [2]
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解題

(a) Aqueous sodium hydroxide reacts with \(\text{Fe}^{2+}\) ions to form a green precipitate of iron(II) hydroxide, which is insoluble in excess. This confirms the presence of \(\text{Fe}^{2+}\) (chromium(III) also forms a green precipitate, but it dissolves in excess to give a green solution). (b) Barium nitrate in acidic conditions is the standard test for sulfate ions, \(\text{SO}_4^{2-}\), which yields a white precipitate of barium sulfate. (c) Combining \(\text{Fe}^{2+}\) and \(\text{SO}_4^{2-}\) gives the formula \(\text{FeSO}_4\). (d) (i) Heating hydrated iron(II) sulfate decomposes the salt, producing sulfur dioxide gas, \(\text{SO}_2\), which is acidic (turns damp blue litmus red) and has bleaching action. (ii) Sulfur dioxide is tested by passing it into acidified aqueous potassium manganate(VII), which is reduced from purple to colourless. (e) (i) The reaction of a metal like magnesium with an acid produces hydrogen gas. The test for hydrogen is to expose it to a lighted splint, which produces a 'squeaky pop'. (ii) The white precipitate with silver nitrate indicates chloride ions. Since Liquid F is an acid, the solution is hydrochloric acid, \(\text{HCl(aq)}\).

評分準則

(a) Iron(II) / \(\text{Fe}^{2+}\) [1] (Reject: iron / \(\text{Fe}^{3+}\)). (b) Sulfate / \(\text{SO}_4^{2-}\) [1]. (c) \(\text{FeSO}_4\) [1] (Accept: \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\)). (d) (i) Sulfur dioxide / \(\text{SO}_2\) (accept Chlorine / \(\text{Cl}_2\)) [1]. (ii) Pass gas through acidified aqueous potassium manganate(VII) [1]; observation: colour change from purple to colourless / decolourised [1]. (e) (i) Test: lighted splint [1]; observation: squeaky pop / pops [1]. (ii) Chloride ion identified [1]; Liquid F is hydrochloric acid / \(\text{HCl}\) [1].
題目 4 · alternativeToPractical
10
A student determined the concentration of a solution of sodium hydroxide, \(\text{NaOH}\), by titrating it against a standard solution of sulfuric acid, \(\text{H}_2\text{SO}_4\), of concentration \(0.0500\text{ mol/dm}^3\). The student filled a burette with the sulfuric acid. They used a volumetric pipette to transfer \(25.0\text{ cm}^3\) of the sodium hydroxide solution into a conical flask, and added a few drops of methyl orange indicator. (a) Describe the colour change of the methyl orange indicator at the end-point of this titration (acid added to alkali). [2] From: ......................... To: ......................... (b) The student carried out three titrations: Titration 1: Initial reading was \(0.40\text{ cm}^3\), final reading was \(24.60\text{ cm}^3\); Titration 2: Initial reading was \(1.20\text{ cm}^3\), final reading was \(25.50\text{ cm}^3\); Titration 3: Initial reading was \(2.10\text{ cm}^3\), final reading was \(26.20\text{ cm}^3\). (i) Complete the table by filling in the values for the volume of acid added (titre) in each titration. [3] | Titration number | Final reading / \(\text{cm}^3\) | Initial reading / \(\text{cm}^3\) | Volume of acid added / \(\text{cm}^3\) | |---|---|---|---| | 1 | 24.60 | 0.40 | | | 2 | 25.50 | 1.20 | | | 3 | 26.20 | 2.10 | | (ii) Calculate the mean titre of sulfuric acid to be used in the calculation. Show your working. [2] (c) Calculate the concentration of the sodium hydroxide solution in \(\text{mol/dm}^3\). The equation for the reaction is: \(2\text{NaOH(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}\) [3]
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解題

(a) Methyl orange is yellow in alkaline solution and turns red in acidic solutions. The transition end-point is orange. Thus, when acid is titrated into alkali, the color changes from yellow to orange. (b) (i) The volume of acid added is found by subtracting the initial reading from the final reading: Titration 1: \(24.60 - 0.40 = 24.20\text{ cm}^3\); Titration 2: \(25.50 - 1.20 = 24.30\text{ cm}^3\); Titration 3: \(26.20 - 2.10 = 24.10\text{ cm}^3\). (ii) The mean titre is calculated as: \(\frac{24.20 + 24.30 + 24.10}{3} = 24.20\text{ cm}^3\). (c) First, determine the amount of substance in moles of \(\text{H}_2\text{SO}_4\): \(\text{moles} = \text{concentration} \times \text{volume in dm}^3 = 0.0500\text{ mol/dm}^3 \times \frac{24.20}{1000}\text{ dm}^3 = 0.00121\text{ mol}\). According to the chemical equation, \(1\text{ mol}\) of \(\text{H}_2\text{SO}_4\) reacts with \(2\text{ mol}\) of \(\text{NaOH}\). Therefore, \(\text{moles of NaOH} = 2 \times 0.00121 = 0.00242\text{ mol}\). Finally, calculate the concentration of \(\text{NaOH}\): \(\text{concentration} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00242\text{ mol}}{0.0250\text{ dm}^3} = 0.0968\text{ mol/dm}^3\).

評分準則

(a) From: yellow [1]; To: orange (accept: peach / pink) [1] (Reject: red). (b) (i) Award 1 mark for each correct value: Titration 1: \(24.20\text{ cm}^3\) [1]; Titration 2: \(24.30\text{ cm}^3\) [1]; Titration 3: \(24.10\text{ cm}^3\) [1]. (ii) Working showing sum divided by 3 [1]; mean titre = \(24.20\text{ cm}^3\) [1]. (c) Award 1 mark for calculating moles of acid: \(0.0500 \times (24.20 / 1000) = 0.00121\text{ mol}\) [1] (or consequential on their mean titre); 1 mark for using 1:2 mole ratio to get moles of \(\text{NaOH}\): \(0.00121 \times 2 = 0.00242\text{ mol}\) [1]; 1 mark for calculating concentration: \(0.00242 / 0.0250 = 0.0968\text{ mol/dm}^3\) [1].

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