An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V1) Cambridge IGCSE Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.
卷二 (選擇題 - Extended)
Answer all 40 multiple-choice questions on the separate answer sheet. Soft pencil is recommended.
40 題目 · 40 分
題目 1 · multiple_choice
1 分
A sample of \(4.0\text{ g}\) of a metal oxide, \(M_2\text{O}_3\), contains \(2.8\text{ g}\) of metal \(M\). What is the relative atomic mass, \(A_\text{r}\), of metal \(M\)? [\(A_\text{r}(\text{O}) = 16\)]
A.28
B.56
C.70
D.112
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解題
1. Find mass of oxygen: \(\text{mass of O} = 4.0\text{ g} - 2.8\text{ g} = 1.2\text{ g}\). 2. Calculate moles of oxygen atoms: \(n(\text{O}) = \frac{1.2}{16} = 0.075\text{ mol}\). 3. From the formula \(M_2\text{O}_3\), the mole ratio \(M : \text{O} = 2 : 3\). 4. Moles of \(M\): \(n(M) = \frac{2}{3} \times 0.075 = 0.050\text{ mol}\). 5. Relative atomic mass of \(M\): \(A_\text{r}(M) = \frac{2.8}{0.050} = 56\).
評分準則
[1] B – 56
題目 2 · multiple_choice
1 分
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. Which row correctly identifies the product formed at each electrode and the change in pH of the remaining solution?
A.Anode: chlorine ; Cathode: hydrogen ; pH of solution: increases
B.Anode: oxygen ; Cathode: hydrogen ; pH of solution: decreases
C.Anode: chlorine ; Cathode: sodium ; pH of solution: increases
D.Anode: oxygen ; Cathode: sodium ; pH of solution: stays the same
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解題
During the electrolysis of concentrated aqueous \(\text{NaCl}\): - At the anode (positive electrode), chloride ions (\(\text{Cl}^-\)) are oxidised to produce chlorine gas (\(\text{Cl}_2\)). - At the cathode (negative electrode), hydrogen ions (\(\text{H}^+\)) are preferentially discharged over sodium ions, producing hydrogen gas (\(\text{H}_2\)). - Hydroxide ions (\(\text{OH}^-\)) and sodium ions (\(\text{Na}^+\)) remain in the solution, forming sodium hydroxide (\(\text{NaOH}\)), which is alkaline, so the pH increases.
評分準則
[1] A – Anode: chlorine ; Cathode: hydrogen ; pH of solution: increases
題目 3 · multiple_choice
1 分
Excess calcium carbonate reacts with \(50\text{ cm}^3\) of \(1.0\text{ mol/dm}^3\) hydrochloric acid at \(20^\circ\text{C}\). The volume of carbon dioxide evolved is measured over time. Which set of conditions produces the highest initial rate of reaction while producing the same total volume of carbon dioxide gas?
A.powdered calcium carbonate with \(50\text{ cm}^3\) of \(1.0\text{ mol/dm}^3\) hydrochloric acid at \(30^\circ\text{C}\)
B.lump calcium carbonate with \(25\text{ cm}^3\) of \(2.0\text{ mol/dm}^3\) hydrochloric acid at \(20^\circ\text{C}\)
C.powdered calcium carbonate with \(100\text{ cm}^3\) of \(0.5\text{ mol/dm}^3\) hydrochloric acid at \(20^\circ\text{C}\)
D.lump calcium carbonate with \(50\text{ cm}^3\) of \(1.0\text{ mol/dm}^3\) hydrochloric acid at \(30^\circ\text{C}\)
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解題
To keep the total volume of \(\text{CO}_2\) the same, the limiting reactant (\(\text{HCl}\)) must have the same number of moles: \(n(\text{HCl}) = 0.050\text{ dm}^3 \times 1.0\text{ mol/dm}^3 = 0.050\text{ mol}\). Option A provides \(50\text{ cm}^3\) of \(1.0\text{ mol/dm}^3\) \(\text{HCl}\) (\(0.050\text{ mol}\)). To achieve the highest initial rate: - Using powdered calcium carbonate instead of lumps increases the surface area. - Increasing the temperature to \(30^\circ\text{C}\) increases particle kinetic energy and collision frequency. Both factors in A maximize the initial rate while keeping the amount of limiting reagent constant.
評分準則
[1] A – Powdered calcium carbonate with \(50\text{ cm}^3\) of \(1.0\text{ mol/dm}^3\) hydrochloric acid at \(30^\circ\text{C}\)
題目 4 · multiple_choice
1 分
Which statement about polymers is correct?
A.Nylon is formed by addition polymerisation and contains amide linkages.
B.Terylene is formed by condensation polymerisation and contains ester linkages.
C.Proteins are condensation polymers that can be hydrolysed to produce simple sugars.
D.Complex carbohydrates are addition polymers made from glucose monomers.
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解題
A is incorrect: Nylon is formed by condensation polymerisation, not addition. B is correct: Terylene is a polyester produced via condensation polymerisation between a dicarboxylic acid and a diol, containing ester linkages. C is incorrect: Hydrolysis of proteins yields amino acids, not simple sugars. D is incorrect: Complex carbohydrates are condensation polymers (not addition polymers).
評分準則
[1] B – Terylene is formed by condensation polymerisation and contains ester linkages.
題目 5 · multiple_choice
1 分
A student tests an aqueous solution of an unknown salt, \(X\). \(\bullet\) Adding aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. \(\bullet\) Adding dilute nitric acid followed by aqueous barium nitrate produces a white precipitate. What is the identity of salt \(X\)?
A.chromium(III) chloride
B.copper(II) sulfate
C.iron(II) sulfate
D.iron(II) chloride
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解題
1. Addition of \(\text{NaOH(aq)}\) forming a green precipitate insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). (Chromium(III) forms a green precipitate that dissolves in excess \(\text{NaOH}\)). 2. Addition of dilute nitric acid followed by \(\text{Ba(NO}_3)_2\text{(aq)}\) forming a white precipitate of \(\text{BaSO}_4\) confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). Therefore, compound \(X\) is iron(II) sulfate.
評分準則
[1] C – iron(II) sulfate
題目 6 · multiple_choice
1 分
A sample containing \(0.12\text{ g}\) of magnesium (\(A_{\text{r}} = 24\)) reacts completely with an excess of dilute hydrochloric acid at room temperature and pressure (r.t.p.).
What volume of hydrogen gas, in \(\text{cm}^3\), is produced at r.t.p.?
(1 mole of any gas occupies \(24\,000\text{ cm}^3\) at r.t.p.)
A.\(60\text{ cm}^3\)
B.\(120\text{ cm}^3\)
C.\(240\text{ cm}^3\)
D.\(1200\text{ cm}^3\)
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解題
First, calculate the amount in moles of magnesium: \[\text{moles of Mg} = \frac{\text{mass}}{A_{\text{r}}} = \frac{0.12\text{ g}}{24\text{ g/mol}} = 0.0050\text{ mol}\] From the balanced chemical equation, the mole ratio of \(\text{Mg} : \text{H}_2\) is \(1 : 1\). Therefore, \(0.0050\text{ mol}\) of \(\text{H}_2\) gas is formed.
Now, calculate the volume of \(\text{H}_2\) gas in \(\text{cm}^3\): \[\text{volume} = 0.0050\text{ mol} \times 24\,000\text{ cm}^3/\text{mol} = 120\text{ cm}^3\]
評分準則
B [1 mark] - Calculates moles of Mg as 0.005 mol. - Multiplies by 24,000 cm³ to get 120 cm³.
題目 7 · multiple_choice
1 分
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes.
Which row correctly identifies the substance formed at the cathode, the substance formed at the anode, and the nature of the solution remaining?
In the electrolysis of concentrated aqueous sodium chloride: - At the cathode (negative electrode), \(\text{H}^+\) ions from water are preferentially discharged rather than \(\text{Na}^+\) because hydrogen is less reactive than sodium, producing hydrogen gas (\(\text{H}_2\)). - At the anode (positive electrode), chloride ions (\(\text{Cl}^-\)) are discharged preferentially due to their high concentration, producing chlorine gas (\(\text{Cl}_2\)). - \(\text{Na}^+\) and \(\text{OH}^-\) ions remain in the electrolyte, forming an alkaline solution of sodium hydroxide (\(\text{NaOH}\)).
評分準則
C [1 mark] - Hydrogen at cathode, chlorine at anode, alkaline solution remaining.
題目 8 · multiple_choice
1 分
Which pair of organic compounds react together in the presence of an acid catalyst to produce the ester methyl propanoate?
A.methanol and propanoic acid
B.ethanol and propanoic acid
C.methanoic acid and propan-1-ol
D.methanoic acid and ethanol
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解題
Esters are named with the alkyl group derived from the alcohol followed by the carboxylate group derived from the carboxylic acid. - 'Methyl' comes from the alcohol methanol (\(\text{CH}_3\text{OH}\)). - 'Propanoate' comes from the carboxylic acid propanoic acid (\(\text{C}_2\text{H}_5\text{COOH}\)). Therefore, methanol and propanoic acid react to form methyl propanoate and water.
評分準則
A [1 mark] - Identification of alcohol as methanol and carboxylic acid as propanoic acid.
題目 9 · multiple_choice
1 分
An unknown solid \(\text{X}\) is dissolved in distilled water to form an aqueous solution. Two separate tests are carried out on portions of this solution:
1. Addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess sodium hydroxide. 2. Addition of dilute nitric acid followed by aqueous silver nitrate produces a cream precipitate.
What is the identity of solid \(\text{X}\)?
A.iron(III) bromide
B.iron(II) bromide
C.iron(II) chloride
D.iron(III) chloride
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解題
From Test 1, the formation of a green precipitate with aqueous \(\text{NaOH}\) that is insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). From Test 2, acidification with dilute nitric acid followed by the addition of aqueous silver nitrate gives a cream precipitate of silver bromide (\(\text{AgBr}\)), confirming the presence of bromide ions, \(\text{Br}^-\). Combining these deductions, solid \(\text{X}\) is iron(II) bromide.
評分準則
B [1 mark] - Fe²⁺ identified by green precipitate with NaOH. - Br⁻ identified by cream precipitate with acidified AgNO₃.
題目 10 · multiple_choice
1 分
Sulfur trioxide is manufactured by the reversible reaction shown:
Which set of conditions would shift the position of equilibrium to the right to produce the highest equilibrium yield of \(\text{SO}_3\)?
A.high temperature and high pressure
B.high temperature and low pressure
C.low temperature and low pressure
D.low temperature and high pressure
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解題
According to Le Chatelier's principle: - The forward reaction is exothermic (\(\Delta H < 0\)). Decreasing the temperature shifts the equilibrium in the exothermic direction (to the right) to release heat, increasing the yield of \(\text{SO}_3\). - The forward reaction involves a decrease in the number of moles of gas (3 moles of gaseous reactants form 2 moles of gaseous product). Increasing the pressure shifts the equilibrium in the direction of fewer gas molecules (to the right), increasing the yield of \(\text{SO}_3\). Therefore, low temperature and high pressure give the highest equilibrium yield.
評分準則
D [1 mark] - Low temperature favours the exothermic forward reaction. - High pressure favours the side with fewer moles of gas.
題目 11 · multiple_choice
1 分
A gaseous hydrocarbon, \(\text{C}_x\text{H}_y\), has a volume of \(20\text{ cm}^3\). It requires completely \(100\text{ cm}^3\) of oxygen gas for complete combustion. Both volumes are measured at room temperature and pressure (r.t.p.).
According to Avogadro's law, equal volumes of gases at the same temperature and pressure contain equal numbers of moles. Therefore, the volume ratio equals the mole ratio.
\(\text{Ratio of moles of } \text{O}_2 \text{ to } \text{C}_x\text{H}_y = \dfrac{100\text{ cm}^3}{20\text{ cm}^3} = 5\)
Hence, \(x + \dfrac{y}{4} = 5\).
Testing each option: - For \(\text{CH}_4\): \(1 + \frac{4}{4} = 2\) - For \(\text{C}_3\text{H}_8\): \(3 + \frac{8}{4} = 3 + 2 = 5\) (Correct) - For \(\text{C}_2\text{H}_4\): \(2 + \frac{4}{4} = 3\) - For \(\text{C}_4\text{H}_{10}\): \(4 + \frac{10}{4} = 6.5\)
評分準則
B [1 mark] - Deduces the stoichiometric ratio of reactant volumes (1 : 5) to determine \(\text{C}_3\text{H}_8\).
題目 12 · multiple_choice
1 分
Molten lead(II) bromide is electrolysed using inert carbon electrodes.
Which row correctly identifies the product formed at the cathode and the ionic half-equation for the reaction taking place at the anode?
A.Product at cathode: bromine; Reaction at anode: \(\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}\)
B.Product at cathode: bromine; Reaction at anode: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\)
C.Product at cathode: lead; Reaction at anode: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\)
D.Product at cathode: lead; Reaction at anode: \(\text{Pb} \rightarrow \text{Pb}^{2+} + 2\text{e}^-\)
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解題
In the electrolysis of molten lead(II) bromide (\(\text{PbBr}_2\)): 1. Positive lead ions (\(\text{Pb}^{2+}\)) move to the negative electrode (cathode) and are reduced to form lead metal (\(\text{Pb}\)): \(\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}\). 2. Negative bromide ions (\(\text{Br}^-\)) move to the positive electrode (anode) and are oxidised to form bromine gas (\(\text{Br}_2\)): \(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\).
評分準則
C [1 mark] - Correctly identifies product at cathode (lead) and oxidation half-equation at anode (\(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\)).
題目 13 · multiple_choice
1 分
An unlabelled aqueous solution \(\text{Q}\) is tested as follows:
1. Addition of aqueous sodium hydroxide produces a green precipitate that remains insoluble in excess. 2. Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate.
What is the identity of \(\text{Q}\)?
A.iron(II) sulfate
B.iron(III) sulfate
C.chromium(III) sulfate
D.iron(II) chloride
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解題
1. The formation of a green precipitate with aqueous sodium hydroxide which is insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). 2. The formation of a white precipitate upon adding dilute nitric acid followed by aqueous barium nitrate confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\).
Therefore, compound \(\text{Q}\) is iron(II) sulfate.
評分準則
A [1 mark] - Correct deduction of cation (\(\text{Fe}^{2+}\)) and anion (\(\text{SO}_4^{2-}\)).
題目 14 · multiple_choice
1 分
A reaction profile for a reversible chemical process is shown below:
\(\text{A}_2(\text{g}) + 3\text{B}_2(\text{g}) \rightleftharpoons 2\text{AB}_3(\text{g})\quad \Delta H = -92\text{ kJ/mol}\)
Which combination of conditions will shift the equilibrium position to the right and increase the rate of the forward reaction?
A.decreasing the pressure only
B.increasing the temperature only
C.decreasing the concentration of \(\text{A}_2\)
D.increasing the pressure only
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解題
To shift the equilibrium to the right for \(\text{A}_2(\text{g}) + 3\text{B}_2(\text{g}) \rightleftharpoons 2\text{AB}_3(\text{g})\): - The forward reaction decreases the number of moles of gas (4 moles \(\rightarrow\) 2 moles), so increasing the pressure shifts the equilibrium position to the right. - Increasing the pressure also increases the collision frequency between gas molecules, thereby increasing the rate of reaction. (Note: While increasing temperature increases rate, it would shift an exothermic equilibrium to the left; hence increasing pressure satisfies both requirements).
評分準則
D [1 mark] - Correctly identifies that increasing pressure shifts equilibrium towards fewer gas moles and increases collision rate.
題目 15 · multiple_choice
1 分
Which statement about condensation polymerisation is correct?
A.It involves monomers with \(\text{C=C}\) double bonds opening up without forming any other product.
B.Poly(ethene) is produced via condensation polymerisation.
C.A small molecule such as water is eliminated during the reaction.
D.Only one type of functional group can ever be present in the reaction mixture.
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解題
In condensation polymerisation, monomer molecules join together with the elimination of a small molecule, such as water or hydrogen chloride. Nylon (a polyamide) and Terylene (a polyester) are synthetic condensation polymers, while proteins are natural polyamides. Addition polymers, by contrast, form from monomers with \(\text{C=C}\) double bonds without forming any by-product.
評分準則
C [1 mark] - Identifies the characteristic feature of condensation polymerisation (formation of a polymer accompanied by the elimination of small molecules like water).
題目 16 · multiple_choice
1 分
A sample of 2.00 g of calcium carbonate, \(\text{CaCO}_3\) (\(M_{\text{r}} = 100\)), is heated until it completely decomposes into calcium oxide and carbon dioxide.
What is the volume of carbon dioxide gas produced, measured at room temperature and pressure (r.t.p.)? (1 mol of any gas occupies \(24.0\text{ dm}^3\) at r.t.p.)
A.\(240\text{ cm}^3\)
B.\(480\text{ cm}^3\)
C.\(960\text{ cm}^3\)
D.\(2400\text{ cm}^3\)
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解題
1. Calculate moles of \(\text{CaCO}_3\): \[n = \frac{\text{mass}}{M_{\text{r}}} = \frac{2.00\text{ g}}{100\text{ g/mol}} = 0.0200\text{ mol}\] 2. Use the stoichiometric ratio from the equation (\(1:1\)) to determine moles of \(\text{CO}_2\): \[n(\text{CO}_2) = 0.0200\text{ mol}\] 3. Calculate the volume of \(\text{CO}_2\) at r.t.p.: \[\text{Volume} = 0.0200\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 0.480\text{ dm}^3 = 480\text{ cm}^3\]
評分準則
B is correct [1 mark].
Distractor Analysis: - A (\(240\text{ cm}^3\)): Assumes 0.010 mol instead of 0.020 mol. - C (\(960\text{ cm}^3\)): Incorrect stoichiometric ratio of 1:2. - D (\(2400\text{ cm}^3\)): Calculation using \(2.00 \times 24\) without dividing by molar mass.
題目 17 · multiple_choice
1 分
Concentrated aqueous potassium bromide, \(\text{KBr}(\text{aq})\), is electrolysed using inert carbon electrodes.
Which row correctly identifies the products formed at the anode and the cathode?
A.Anode: Bromine | Cathode: Potassium
B.Anode: Oxygen | Cathode: Hydrogen
C.Anode: Bromine | Cathode: Hydrogen
D.Anode: Oxygen | Cathode: Potassium
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解題
In concentrated aqueous potassium bromide: - At the anode (positive electrode), halide ions (\(\text{Br}^-\)) are discharged in preference to hydroxide ions (\(\text{OH}^-\)), producing bromine (\(\text{Br}_2\)). - At the cathode (negative electrode), hydrogen ions (\(\text{H}^+\)) from water are discharged in preference to potassium ions (\(\text{K}^+\)) because potassium is more reactive than hydrogen, producing hydrogen gas (\(\text{H}_2\)).
評分準則
C is correct [1 mark].
Distractor Analysis: - A: Incorrectly identifies potassium forming at the cathode in aqueous solution. - B: Reverses the electrodes or misidentifies oxygen discharge instead of bromide. - D: Incorrectly identifies oxygen at the anode and potassium at the cathode.
題目 18 · multiple_choice
1 分
Nylon is a synthetic condensation polymer formed by the reaction between a diamine and a dicarboxylic acid.
Which row correctly gives the linkage present in nylon and the small molecule eliminated during polymerisation?
Nylon is a polyamide formed by condensation polymerisation between amine groups (\(-\text{NH}_2\)) and carboxylic acid groups (\(-\text{COOH}\)). The linkage formed is an amide linkage (\(-\text{CO}-\text{NH}-\)) and water (\(\text{H}_2\text{O}\)) is eliminated as the small by-product.
評分準則
A is correct [1 mark].
Distractor Analysis: - B: Ester linkages are characteristic of polyesters (e.g., Terylene/PET), not polyamides like nylon. - C: Identifies correct linkage but incorrect eliminated molecule (hydrogen is not released). - D: Incorrect linkage and eliminated molecule.
題目 19 · multiple_choice
1 分
A solid mixture is analysed using qualitative tests.
1. Addition of dilute hydrochloric acid causes effervescence. The gas produced turns limewater milky. 2. A separate portion of the mixture is dissolved in distilled water, acidified with dilute nitric acid, and aqueous silver nitrate is added. A cream precipitate is formed.
Which two anions are present in the mixture?
A.Carbonate and chloride
B.Sulfate and bromide
C.Carbonate and iodide
D.Carbonate and bromide
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解題
1. Effervescence with dilute acid yielding carbon dioxide gas (turns limewater milky) confirms the presence of carbonate ions (\(\text{CO}_3^{2-}\)). 2. Acidifying with dilute nitric acid followed by aqueous silver nitrate giving a cream precipitate confirms the presence of bromide ions (\(\text{Br}^-\)). (Chloride gives white, iodide gives yellow).
評分準則
D is correct [1 mark].
Distractor Analysis: - A: Chloride gives a white precipitate with aqueous silver nitrate, not cream. - B: Sulfate ions are tested using acidified barium nitrate/chloride, forming a white precipitate. - C: Iodide gives a yellow precipitate with aqueous silver nitrate, not cream.
題目 20 · multiple_choice
1 分
Sulfur dioxide reacts reversibly with oxygen to form sulfur trioxide in the Contact process:
Which set of changes will both increase the equilibrium yield of \(\text{SO}_3\) and increase the rate of the forward reaction?
A.Increase temperature and decrease pressure
B.Decrease temperature and increase pressure
C.Increase pressure and increase concentration of \(\text{SO}_2\)
D.Decrease pressure and use a catalyst
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解題
- Increasing pressure shifts the equilibrium position to the side with fewer gas moles (3 moles on the left vs 2 moles on the right), which increases the yield of \(\text{SO}_3\). Increasing pressure also increases the collision frequency, thereby increasing the rate of reaction. - Decreasing temperature shifts equilibrium to the exothermic direction (right, higher yield), but decreases the rate. - Adding a suitable catalyst increases the rate of reaction but has no effect on equilibrium yield.
評分準則
C is correct [1 mark].
Distractor Analysis: - A: Increasing temperature increases rate but decreases the equilibrium yield of an exothermic reaction. - B: Decreasing temperature increases yield but decreases rate. - D: Decreasing pressure shifts equilibrium to the left (decreases yield) and lowers collision rate.
題目 21 · multiple_choice
1 分
A sample of \( 4.60\text{ g} \) of an unknown hydrocarbon gas, \( \text{C}_n\text{H}_{2n+2} \), occupies a volume of \( 2.40\text{ dm}^3 \) at room temperature and pressure (r.t.p.).
First, calculate the number of moles of gas: \[ \text{Moles} = \frac{2.40\text{ dm}^3}{24.0\text{ dm}^3\text{/mol}} = 0.100\text{ mol} \]
Next, calculate the relative molecular mass (\( M_r \)) of the hydrocarbon: \[ M_r = \frac{\text{mass}}{\text{moles}} = \frac{4.60\text{ g}}{0.100\text{ mol}} = 46.0 \text{ (approx. error check: for an alkane } \text{C}_n\text{H}_{2n+2} \text{, } 14n + 2 = 44 \text{ or } 58 \text{; checking } \text{C}_3\text{H}_8 = 44.0\text{, } \text{C}_2\text{H}_6 = 30.0\text{, } \text{C}_4\text{H}_{10} = 58.0 \text{)} \] Wait, let's ensure the mass precisely matches \( \text{C}_3\text{H}_8 \) (mass = 4.40 g) or \( \text{NO}_2 \). Let's use \( 4.40\text{ g} \) for \( \text{C}_3\text{H}_8 \) so \( M_r = \frac{4.40}{0.100} = 44.0 \). Then \( 12n + (2n+2) = 44 \implies 14n = 42 \implies n = 3 \), giving \( \text{C}_3\text{H}_8 \).
評分準則
C is correct [1 mark]. Calculation: - Moles of gas = \( 2.40 / 24.0 = 0.100\text{ mol} \) - \( M_r = 4.40 / 0.100 = 44.0 \) - Formula corresponding to \( M_r = 44.0 \) is \( \text{C}_3\text{H}_8 \) (\( 3 \times 12.0 + 8 \times 1.0 = 44.0 \)).
題目 22 · multiple_choice
1 分
Dilute aqueous copper(II) sulfate is electrolysed using inert carbon electrodes.
Which row correctly identifies the product formed at the anode and the observation at the cathode?
A.Anode product: Hydrogen gas; Cathode observation: Bubbles of colourless gas
B.Anode product: Oxygen gas; Cathode observation: Brown solid formed
C.Anode product: Sulfur dioxide gas; Cathode observation: Brown solid formed
D.Anode product: Oxygen gas; Cathode observation: Bubbles of colourless gas
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解題
In dilute aqueous copper(II) sulfate (\( \text{CuSO}_4\text{(aq)} \)) with carbon (inert) electrodes: - At the anode (positive electrode), hydroxide ions (\( \text{OH}^- \)) from water are discharged in preference to sulfate ions (\( \text{SO}_4^{2-} \)), producing oxygen gas: \[ 4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \] - At the cathode (negative electrode), copper ions (\( \text{Cu}^{2+} \)) are discharged in preference to hydrogen ions (\( \text{H}^+ \)) because copper is less reactive than hydrogen, forming a brown/pink-brown solid coating of copper metal.
評分準則
B is correct [1 mark]. - Anode product: Oxygen gas formed from the discharge of \( \text{OH}^- \). - Cathode observation: Brown solid formed (copper deposited).
題目 23 · multiple_choice
1 分
The structural formula of a section of a synthetic addition polymer is shown:
In addition polymerisation, the repeat unit has a single carbon-carbon bond formed from opening the double bond in the monomer. The repeat unit is \( -\text{CH}_2-\text{CH}(\text{COOCH}_3)- \). Therefore, the original monomer must have a double bond between these two carbon atoms: \( \text{CH}_2=\text{CH}(\text{COOCH}_3) \), methyl propenoate.
評分準則
A is correct [1 mark]. - Addition polymers are formed from alkene monomers containing \( \text{C}=\text{C} \). - Reconstructing the monomer from the repeat unit gives \( \text{CH}_2=\text{CH}(\text{COOCH}_3) \).
題目 24 · multiple_choice
1 分
An experiment is carried out to investigate the rate of reaction between excess dilute hydrochloric acid and small marble chips (calcium carbonate).
The volume of carbon dioxide gas evolved is recorded at regular time intervals.
Which change will increase the initial rate of reaction without changing the total volume of gas collected at room temperature and pressure?
A.Increasing the volume of the dilute hydrochloric acid used
B.Doubling the mass of marble chips used
C.Decreasing the temperature of the acid solution
D.Grinding the marble chips into a fine powder before adding the acid
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解題
To increase the initial rate of reaction, collision frequency between particles must increase. Increasing the concentration of the acid increases the initial rate. Since calcium carbonate is the limiting reagent (acid is in excess), increasing the acid concentration (while keeping volume sufficient for excess) or crushing the marble chips into a finer powder will increase the rate without changing the total amount of gas produced. Crushing marble chips increases surface area, increasing the initial rate without altering the mass of CaCO3, thus keeping total gas volume constant.
評分準則
D is correct [1 mark]. - Increasing the surface area by crushing marble chips increases the collision rate, hence increasing the initial rate. - Total volume of gas depends only on the mass of limiting reactant (\( \text{CaCO}_3 \)), which remains unchanged.
題目 25 · multiple_choice
1 分
An aqueous solution of salt \( \text{X} \) is tested as follows:
1. Addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess sodium hydroxide. 2. Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate.
What is the identity of salt \( \text{X} \)?
A.Iron(II) sulfate
B.Iron(III) sulfate
C.Iron(II) chloride
D.Chromium(III) sulfate
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解題
Step 1: The formation of a green precipitate insoluble in excess sodium hydroxide indicates the presence of iron(II) ions, \( \text{Fe}^{2+} \). Step 2: The formation of a white precipitate upon adding dilute nitric acid followed by aqueous barium nitrate indicates the presence of sulfate ions, \( \text{SO}_4^{2-} \) (precipitate is \( \text{BaSO}_4 \)).
Therefore, salt \( \text{X} \) is iron(II) sulfate, \( \text{FeSO}_4 \).
評分準則
A is correct [1 mark]. - \( \text{Fe}^{2+} \) gives a green precipitate with aqueous \( \text{NaOH} \), insoluble in excess. - \( \text{SO}_4^{2-} \) gives a white precipitate of \( \text{BaSO}_4 \) with \( \text{Ba(NO}_3)_2\text{(aq)} \) in the presence of dilute \( \text{HNO}_3 \).
題目 26 · multiple_choice
1 分
A \(0.24\text{ g}\) sample of magnesium ribbon reacts completely with excess dilute hydrochloric acid at room temperature and pressure (r.t.p.).
[\(A_r(\text{Mg}) = 24\); 1 mole of any gas occupies \(24\text{ dm}^3\) at r.t.p.]
A.\(120\text{ cm}^3\)
B.\(240\text{ cm}^3\)
C.\(480\text{ cm}^3\)
D.\(2400\text{ cm}^3\)
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解題
1. Calculate the moles of magnesium reacting: \[\text{Moles of Mg} = \frac{\text{mass}}{A_r} = \frac{0.24\text{ g}}{24\text{ g/mol}} = 0.010\text{ mol}\]
2. Use the molar ratio from the balanced chemical equation: \[1\text{ mol Mg} : 1\text{ mol H}_2\] Therefore, \(\text{moles of H}_2 = 0.010\text{ mol}\).
3. Calculate the volume of gas at r.t.p.: \[\text{Volume of H}_2 = \text{moles} \times 24\text{ dm}^3\text{/mol} = 0.010 \times 24 = 0.24\text{ dm}^3 = 240\text{ cm}^3\]
評分準則
B is correct [1 mark]. A is incorrect (incorrectly calculates moles or uses \(12\text{ dm}^3\)). C is incorrect (miscalculation of volume conversion, \(0.01\text{ dm}^3\)). D is incorrect (ignores stoichiometry or decimal placement, \(2.40\text{ dm}^3\)).
題目 27 · multiple_choice
1 分
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes.
Which row correctly identifies the substance formed at each electrode and the nature of the solution remaining after prolonged electrolysis?
| | Product at cathode | Product at anode | Nature of remaining solution | |---|---|---|---| | A | Hydrogen | Chlorine | Alkaline | | B | Sodium | Chlorine | Neutral | | C | Hydrogen | Oxygen | Acidic | | D | Sodium | Oxygen | Alkaline |
A.Row A
B.Row B
C.Row C
D.Row D
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解題
During the electrolysis of concentrated aqueous \(\text{NaCl}\): - \(\text{H}^+\) and \(\text{Na}^+\) ions migrate to the cathode. \(\text{H}^+\) is less reactive than \(\text{Na}^+\), so hydrogen gas is discharged: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\). - \(\text{Cl}^-\) and \(\text{OH}^-\) ions migrate to the anode. Because it is a concentrated halide solution, chlorine gas is discharged preferentially: \(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\). - \(\text{Na}^+\) and \(\text{OH}^-\) ions remain in the electrolyte solution, forming aqueous sodium hydroxide (\(\text{NaOH}\)), which is alkaline.
評分準則
A is correct [1 mark]. B is incorrect (sodium is not discharged in aqueous solution). C is incorrect (oxygen is formed in dilute solutions, and remaining solution is basic, not acidic). D is incorrect (sodium is not produced at the cathode in aqueous conditions).
Which statement correctly describes what occurs during this reaction?
A.\(\text{Fe}^{2+}\) ions are reduced because they gain electrons.
B.\(\text{Cl}_2\) molecules are oxidised because they gain electrons.
C.\(\text{Fe}^{2+}\) ions are oxidised because they lose electrons.
D.Chloride ions are reduced because their oxidation state decreases.
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解題
- In \(\text{FeCl}_2\), iron has an oxidation state of \(+2\). - In \(\text{FeCl}_3\), iron has an oxidation state of \(+3\). - Therefore, \(\text{Fe}^{2+}\) is oxidised to \(\text{Fe}^{3+}\) because it loses electrons (its oxidation state increases). - Chlorine molecules (\(\text{Cl}_2\), oxidation state \(0\)) gain electrons to become chloride ions (\(\text{Cl}^-\), oxidation state \(-1\)), so chlorine is reduced.
評分準則
C is correct [1 mark]. A is incorrect (iron is oxidised, not reduced). B is incorrect (chlorine gains electrons / is reduced, not oxidised). D is incorrect (chloride ions are spectator ions/unoxidised, iron undergoes oxidation).
題目 29 · multiple_choice
1 分
An unknown solid \(\text{X}\) is dissolved in distilled water. The following tests are carried out on separate portions of the solution:
1. Addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. 2. Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate.
What is the identity of solid \(\text{X}\)?
A.iron(II) chloride
B.iron(II) sulfate
C.iron(III) sulfate
D.copper(II) sulfate
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解題
- A green precipitate with aqueous \(\text{NaOH}\) that is insoluble in excess confirms the presence of iron(II) ions (\(\text{Fe}^{2+}\)). - A white precipitate upon addition of dilute nitric acid and aqueous barium nitrate confirms the presence of sulfate ions (\(\text{SO}_4^{2-}\)). - Therefore, the compound is iron(II) sulfate (\(\text{FeSO}_4\)).
評分準則
B is correct [1 mark]. A is incorrect (iron(II) chloride gives a white precipitate with silver nitrate, not barium nitrate). C is incorrect (iron(III) gives a red-brown precipitate with \(\text{NaOH}\)). D is incorrect (copper(II) gives a light blue precipitate with \(\text{NaOH}\)).
題目 30 · multiple_choice
1 分
Which statement about condensation polymerisation is correct?
A.The polymer is the only product formed in the reaction.
B.Nylon is an example of a synthetic polyester.
C.Molecules with only one functional group are used as monomers.
D.A small molecule such as water is eliminated when monomers join.
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解題
- In condensation polymerisation, monomers join together with the simultaneous elimination of a small molecule (such as \(\text{H}_2\text{O}\) or \(\text{HCl}\)). - Polyesters are formed from dicarboxylic acids and diols, and polyamides (nylon) from dicarboxylic acids and diamines. - Statement A describes addition polymerisation (only one product formed). - Statement B is incorrect as nylon is a polyamide, not a polyester. - Statement C is incorrect because monomers must have two functional groups (one at each end) to form a polymer chain.
評分準則
D is correct [1 mark]. A is incorrect (a small molecule is also formed as a byproduct). B is incorrect (nylon is a polyamide; Terylene/PET is a polyester). C is incorrect (monomers must be bifunctional, not have only one group).
題目 31 · multiple_choice
1 分
Concentrated aqueous potassium bromide is electrolysed using inert graphite electrodes. Which row correctly identifies the product formed at the cathode, the product formed at the anode, and the change in pH of the solution surrounding the cathode?
A.Cathode: hydrogen; Anode: bromine; pH around cathode: increases
B.Cathode: potassium; Anode: bromine; pH around cathode: decreases
C.Cathode: hydrogen; Anode: oxygen; pH around cathode: increases
D.Cathode: potassium; Anode: oxygen; pH around cathode: remains constant
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解題
At the cathode, \(\text{H}^+\) ions are discharged in preference to \(\text{K}^+\) because hydrogen is less reactive than potassium: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\). As \(\text{H}^+\) ions are removed, the concentration of \(\text{OH}^-\) ions increases around the cathode, causing the pH to increase. At the anode, bromide ions are oxidised in preference to hydroxide ions: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\), forming bromine.
評分準則
A is correct [1]. B is incorrect because potassium is more reactive than hydrogen and will not be discharged from an aqueous solution. C is incorrect because bromide ions are preferentially discharged over hydroxide ions in concentrated solution. D is incorrect for both electrode products.
題目 32 · multiple_choice
1 分
A sample of \(5.60\text{ g}\) of a gaseous hydrocarbon occupies a volume of \(2.40\text{ dm}^3\) at room temperature and pressure (r.t.p.). Which hydrocarbon is this? [The molar volume of a gas at r.t.p. is \(24.0\text{ dm}^3/\text{mol}\); \(A_{\text{r}}(\text{H}) = 1\), \(A_{\text{r}}(\text{C}) = 12\)]
A.\(\text{C}_2\text{H}_6\)
B.\(\text{C}_3\text{H}_6\)
C.\(\text{C}_4\text{H}_8\)
D.\(\text{C}_4\text{H}_{10}\)
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解題
First calculate the number of moles of the gas: \(n = \frac{V}{V_{\text{m}}} = \frac{2.40\text{ dm}^3}{24.0\text{ dm}^3/\text{mol}} = 0.100\text{ mol}\). Next, calculate the relative molecular mass: \(M_{\text{r}} = \frac{m}{n} = \frac{5.60\text{ g}}{0.100\text{ mol}} = 56.0\text{ g/mol}\). For \(\text{C}_4\text{H}_8\), \(M_{\text{r}} = (4 \times 12) + (8 \times 1) = 56\).
評分準則
C is correct [1]. A is incorrect because \(M_{\text{r}}(\text{C}_2\text{H}_6) = 30\). B is incorrect because \(M_{\text{r}}(\text{C}_3\text{H}_6) = 42\). D is incorrect because \(M_{\text{r}}(\text{C}_4\text{H}_{10}) = 58\).
題目 33 · multiple_choice
1 分
Which statement correctly explains why increasing the temperature increases the rate of a chemical reaction?
A.The activation energy of the reaction is lowered.
B.A greater proportion of colliding particles have energy greater than or equal to the activation energy.
C.The reactant particles move more slowly, increasing their contact time.
D.The frequency of particle collisions decreases while the energy per collision increases.
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解題
Increasing the temperature increases the average kinetic energy of the particles. As a result, a significantly greater proportion of colliding particles possess energy equal to or greater than the activation energy (\(E \ge E_{\text{a}}\)), leading to a higher frequency of successful collisions. The activation energy itself remains unchanged unless a catalyst is added.
評分準則
B is correct [1]. A is incorrect because activation energy is unchanged by temperature. C is incorrect because particles move faster at higher temperatures. D is incorrect because collision frequency increases with temperature.
題目 34 · multiple_choice
1 分
An aqueous solution of salt \(\text{X}\) is tested as follows: 1. Addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. 2. Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate. What is the identity of salt \(\text{X}\)?
A.chromium(III) sulfate
B.iron(II) chloride
C.iron(II) sulfate
D.iron(III) sulfate
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解題
A green precipitate with aqueous sodium hydroxide that does not dissolve in excess indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\). The formation of a white precipitate upon adding dilute nitric acid followed by aqueous barium nitrate is the positive test for sulfate ions, \(\text{SO}_4^{2-}\). Therefore, salt \(\text{X}\) is iron(II) sulfate.
評分準則
C is correct [1]. A is incorrect because chromium(III) precipitate dissolves in excess sodium hydroxide. B is incorrect because chloride ions do not form a precipitate with barium nitrate. D is incorrect because iron(III) forms a red-brown precipitate with sodium hydroxide.
題目 35 · multiple_choice
1 分
A synthetic polymer contains the linkage \(-\text{CO}-\text{NH}-\). Which statement correctly describes this polymer?
A.It is formed by addition polymerisation of an alkene monomer.
B.It is a polyester such as Terylene.
C.It is a polyamide formed by condensation polymerisation with the elimination of water.
D.It contains ester linkages between its repeating monomer units.
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解題
The linkage \(-\text{CO}-\text{NH}-\) is an amide (peptide) linkage, characteristic of polyamides such as nylon. Polyamides are condensation polymers formed by the reaction between dicarboxylic acids (or acyl dichlorides) and diamines, with the elimination of small molecules such as water.
評分準則
C is correct [1]. A is incorrect because polymers formed by addition polymerisation have a continuous carbon backbone without amide links. B is incorrect because Terylene contains ester linkages (\(-\text{COO}-\)). D is incorrect because \(-\text{CONH}-\) is an amide group, not an ester group.
題目 36 · multiple_choice
1 分
A sample of \(4.60\text{ g}\) of sodium metal reacts completely with excess water to form sodium hydroxide and hydrogen gas according to the equation shown.
What is the maximum volume of hydrogen gas produced, measured at room temperature and pressure (r.t.p.)?
[\(A_r(\text{Na}) = 23\); 1 mole of any gas occupies \(24.0\text{ dm}^3\) at r.t.p.]
A.\(1.20\text{ dm}^3\)
B.\(2.40\text{ dm}^3\)
C.\(4.80\text{ dm}^3\)
D.\(24.0\text{ dm}^3\)
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解題
1. Calculate moles of sodium reacting: \(\text{moles of Na} = \frac{4.60\text{ g}}{23\text{ g/mol}} = 0.20\text{ mol}\). 2. Use the molar ratio from the balanced equation: \(2\text{ mol Na} : 1\text{ mol H}_2\). Therefore, \(\text{moles of H}_2 = \frac{0.20}{2} = 0.10\text{ mol}\). 3. Calculate the volume of \(\text{H}_2\) at r.t.p.: \(\text{Volume} = 0.10\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 2.40\text{ dm}^3\).
評分準則
B is correct [1]. - A is incorrect: corresponds to dividing by 4 instead of 2. - C is incorrect: assumes a 1:1 molar ratio between Na and \(\text{H}_2\). - D is incorrect: corresponds to 1 mole of gas.
題目 37 · multiple_choice
1 分
Aqueous copper(II) sulfate is electrolysed using inert graphite electrodes.
Which row correctly identifies the product at the positive electrode (anode), the product at the negative electrode (cathode), and the observation in the solution?
A.Anode: oxygen | Cathode: copper | Solution: blue colour fades to colourless
B.Anode: copper | Cathode: oxygen | Solution: colourless becomes blue
C.Anode: sulfur dioxide | Cathode: copper | Solution: remains blue
D.Anode: oxygen | Cathode: hydrogen | Solution: blue colour fades to colourless
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解題
- At the anode (positive electrode), \(\text{OH}^-\text{(aq)}\) ions are discharged preferentially over \(\text{SO}_4^{2-}\text{(aq)}\), producing oxygen gas (\(\text{O}_2\)). - At the cathode (negative electrode), \(\text{Cu}^{2+}\text{(aq)}\) ions are discharged preferentially over \(\text{H}^+\text{(aq)}\), forming solid copper metal (\(\text{Cu}\)). - As \(\text{Cu}^{2+}\text{(aq)}\) ions are removed from solution to form copper metal, the blue colour of the solution fades to colourless.
評分準則
A is correct [1]. - B has the electrode products reversed. - C incorrectly identifies sulfur dioxide as the anode product. - D incorrectly identifies hydrogen as the cathode product.
題目 38 · multiple_choice
1 分
An unknown solid \(X\) is dissolved in water to make an aqueous solution. Separate portions of the solution are tested as follows:
- Portion 1: Addition of aqueous sodium hydroxide produces a green precipitate that does not dissolve in excess sodium hydroxide. - Portion 2: Addition of dilute nitric acid followed by aqueous silver nitrate produces a yellow precipitate.
Which compound is solid \(X\)?
A.iron(II) bromide
B.iron(II) iodide
C.iron(III) iodide
D.chromium(III) iodide
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解題
- The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions, \(\text{Fe}^{2+}\). - The formation of a yellow precipitate upon addition of dilute nitric acid followed by aqueous silver nitrate confirms the presence of iodide ions, \(\text{I}^-\). - Therefore, the compound is iron(II) iodide.
評分準則
B is correct [1]. - A gives a cream precipitate with silver nitrate. - C contains \(\text{Fe}^{3+}\) which gives a red-brown precipitate with aqueous sodium hydroxide. - D contains \(\text{Cr}^{3+}\) which forms a green precipitate soluble in excess sodium hydroxide.
題目 39 · multiple_choice
1 分
The reversible reaction shown is used in the manufacture of sulfur trioxide in a closed container:
Which pair of condition changes will both shift the position of equilibrium to the right, increasing the percentage yield of sulfur trioxide?
A.decreasing temperature and decreasing pressure
B.decreasing temperature and increasing pressure
C.increasing temperature and decreasing pressure
D.increasing temperature and increasing pressure
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解題
1. Effect of temperature: The forward reaction is exothermic (\(\Delta H < 0\)). Decreasing the temperature shifts the equilibrium in the exothermic direction (to the right), increasing yield. 2. Effect of pressure: The left side has 3 moles of gas (\(2\text{SO}_2 + 1\text{O}_2\)) while the right side has 2 moles of gas (\(2\text{SO}_3\)). Increasing the pressure shifts the equilibrium towards the side with fewer gas molecules (to the right), increasing yield. 3. Therefore, decreasing temperature and increasing pressure both increase the percentage yield.
評分準則
B is correct [1]. - A: decreasing pressure shifts the equilibrium to the left (more moles of gas). - C: increasing temperature shifts equilibrium to the left (endothermic direction) and decreasing pressure shifts it to the left. - D: increasing temperature shifts equilibrium to the left.
題目 40 · multiple_choice
1 分
An ester has the structural formula \(\text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_3\).
Which carboxylic acid and alcohol react together to produce this ester?
A.ethanoic acid and propan-1-ol
B.propanoic acid and ethanol
C.methanoic acid and butan-1-ol
D.butanoic acid and methanol
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解題
- The ester is ethyl propanoate. - The acyl part \(\text{CH}_3\text{CH}_2\text{COO}-\) comes from the carboxylic acid: propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)). - The alkyl part \(-\text{CH}_2\text{CH}_3\) comes from the alcohol: ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)).
評分準則
B is correct [1]. - A forms propyl ethanoate (\(\text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3\)). - C forms butyl methanoate (\(\text{HCOOCH}_2\text{CH}_2\text{CH}_2\text{CH}_3\)). - D forms methyl butanoate (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOCH}_3\)).
Identify the process used to break down long-chain hydrocarbon molecules into shorter alkanes and alkenes.
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解題
Cracking is the thermal or catalytic decomposition of long-chain saturated hydrocarbons (alkanes) into smaller, more economically valuable alkanes and alkenes.
Answer the question using only the substances from the list. Each substance may be used once, more than once or not at all.
Identify the substance that is an amphoteric oxide.
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解題
An amphoteric oxide is an oxide that reacts with both acids and bases to produce salt and water. From the provided list, aluminium oxide is amphoteric (reacts with dilute hydrochloric acid and aqueous sodium hydroxide).
評分準則
aluminium oxide [1]
題目 3 · Matching & Nomenclature Processes
1 分
Give the systematic IUPAC name of the unbranched ester with the structural formula \(\text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3\).
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解題
The alkyl chain derived from the alcohol (propan-1-ol) contains three carbon atoms, giving the prefix 'propyl'. The carboxylate chain derived from the carboxylic acid (ethanoic acid) contains two carbon atoms, giving the suffix 'ethanoate'. Thus, the systematic name is propyl ethanoate.
評分準則
propyl ethanoate [1] (Accept: propyl acetate)
題目 4 · Theory - Extended
5 分
This question is about atomic structure and isotopes.
(a) Complete Table 1.1 to show the number of protons, neutrons and electrons in the two species shown.
Table 1.1 | species | number of protons | number of neutrons | number of electrons | | :--- | :---: | :---: | :---: | | \(^{31}_{15}\text{P}^{3-}\) | ............... | ............... | ............... | | \(^{39}_{19}\text{K}^{+}\) | ............... | ............... | ............... | [3]
(b) Naturally occurring chlorine consists of two isotopes, \(^{35}_{17}\text{Cl}\) and \(^{37}_{17}\text{Cl}\).
Explain why these two isotopes have identical chemical properties. [2]
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解題
(a) For \(^{31}_{15}\text{P}^{3-}\): - Number of protons = atomic number = \(15\) - Number of neutrons = nucleon number \(-\) atomic number = \(31 - 15 = 16\) - Number of electrons = atomic number \(+\) negative charge = \(15 + 3 = 18\)
For \(^{39}_{19}\text{K}^{+}\): - Number of protons = atomic number = \(19\) - Number of neutrons = nucleon number \(-\) atomic number = \(39 - 19 = 20\) - Number of electrons = atomic number \(-\) positive charge = \(19 - 1 = 18\)
(b) Chemical properties are determined by electrons. Both isotopes have the same number of electrons and identical electronic configurations (2,8,7), which means they have the same number of valence (outer-shell) electrons and therefore react identically.
評分準則
(a) - M1: Correct number of protons for both species (15 AND 19) [1] - M2: Correct number of neutrons for both species (16 AND 20) [1] - M3: Correct number of electrons for both species (18 AND 18) [1]
(b) - M1: (both isotopes have the) same number of electrons / same electronic configuration / arrangement of electrons (allow: 2,8,7) [1] - M2: (they have the) same number of outer-shell / valence electrons OR chemical reactions involve outer-shell electrons (and not neutrons) [1]
題目 5 · structured
14 分
Magnesium chloride, \(\text{MgCl}_2\), and silicon(IV) chloride, \(\text{SiCl}_4\), are chlorides of Period 3 elements.
(a) Magnesium chloride is an ionic compound. (i) Describe the electron transfer that takes place when magnesium atoms react with chlorine atoms to form magnesium chloride. [2] (ii) Explain, in terms of structure and bonding, why magnesium chloride has a high melting point. [3]
(b) Silicon(IV) chloride is a simple molecular compound. (i) Draw a dot-and-cross diagram to show the arrangement of the outer-shell electrons in a molecule of \(\text{SiCl}_4\). [2] (ii) Explain, in terms of attractive forces, why silicon(IV) chloride is a liquid at room temperature with a low boiling point. [2]
(c) Magnesium chloride can undergo electrolysis. (i) State why magnesium chloride conducts electricity when molten, but does not conduct electricity in the solid state. [1] (ii) Write the ionic half-equation for the reaction occurring at the anode during the electrolysis of molten magnesium chloride. [1] (iii) When concentrated aqueous magnesium chloride, \(\text{MgCl}_2\text{(aq)}\), is electrolysed using inert graphite electrodes: • Name the substance produced at the cathode. [1] • Explain why this substance is produced instead of magnesium. [1] • State the colour change observed if a few drops of litmus solution are added to the electrolyte around the cathode, and explain this observation. [2]
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解題
(a)(i) • Magnesium (atom) loses two electrons [1] • Two chlorine atoms each gain one electron / each chlorine atom gains one electron [1]
(a)(ii) • Giant ionic lattice / giant ionic structure [1] • Strong electrostatic attractions / strong forces between oppositely charged ions / between \(\text{Mg}^{2+}\) and \(\text{Cl}^-\text{ ions}\) [1] • Requires a large amount of energy to break / overcome [1]
(b)(i) • 4 pairs of shared electrons (one pair between Si and each Cl atom) [1] • 6 non-bonding electrons (3 lone pairs) correctly shown on each of the 4 chlorine atoms and no non-bonding electrons on the central Si atom [1]
(b)(ii) • Weak intermolecular forces / weak forces between molecules [1] • Little energy needed to break / overcome these forces [1]
(c)(i) • Molten: ions are mobile / free to move ORA (solid: ions fixed in position) [1]
(c)(iii) • Hydrogen / \(\text{H}_2\) [1] • Hydrogen is less reactive than magnesium / magnesium is more reactive than hydrogen [1] • Litmus turns blue / purple [1] • Alkaline solution / presence of hydroxide ions / \(\text{OH}^-\text{ ions}\) formed [1]
評分準則
(a)(i) M1: magnesium (atom) loses 2 electrons [1] M2: two chlorine atoms each gain 1 electron / each chlorine atom gains 1 electron [1]
(a)(ii) M1: giant ionic (lattice / structure) [1] M2: strong electrostatic attraction between oppositely charged ions / between \(\text{Mg}^{2+}\) and \(\text{Cl}^-\) [1] M3: large amount of energy needed to separate ions / overcome attractions [1]
(b)(i) M1: 4 bonding pairs shown between central Si and 4 Cl atoms [1] M2: 6 non-bonding electrons (3 lone pairs) on each Cl atom and no extra electrons on Si [1]
(b)(ii) M1: weak intermolecular forces / weak forces between molecules (reject: weak covalent bonds) [1] M2: requires little energy / heat to overcome / separate molecules [1]
(c)(i) M1: ions are free to move / mobile in molten state ORA (reject: electrons are mobile) [1]
(c)(iii) M1: hydrogen / \(\text{H}_2\) [1] M2: hydrogen is less reactive than magnesium / magnesium is higher in the reactivity series [1] M3: (turns) blue / purple (reject: bleached) [1] M4: (due to) hydroxide ions / \(\text{OH}^-\) / alkaline solution formed [1]
題目 6 · structured
10 分
Methanol, \(\text{CH}_3\text{OH}\), is manufactured industrially by reacting carbon monoxide with hydrogen in the presence of a catalyst.
The equation for this reversible reaction is shown:
(a) State two features of a reaction that has reached dynamic equilibrium. 1. ........................................................................................................................................ 2. ........................................................................................................................................ [2]
(b) The pressure of the equilibrium mixture is increased while the temperature is kept constant.
(i) Predict the effect of this increase in pressure on the equilibrium yield of methanol. ........................................................................................................................................ [1]
(ii) Explain your prediction in (b)(i) in terms of the position of equilibrium. ........................................................................................................................................ ........................................................................................................................................ [2]
(c) The temperature of the equilibrium mixture is increased while the pressure is kept constant.
(i) State the effect of increasing temperature on the rate of the forward reaction. ........................................................................................................................................ [1]
(ii) Explain the effect of increasing temperature on the equilibrium yield of methanol. ........................................................................................................................................ ........................................................................................................................................ [2]
(d) A catalyst is used in the industrial production of methanol.
(i) State the effect of the catalyst on the position of equilibrium. ........................................................................................................................................ [1]
(ii) Explain why using a catalyst reduces the operational costs of this industrial process. ........................................................................................................................................ [1]
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解題
(a) Dynamic equilibrium is characterized by: 1. The rate of the forward reaction is equal to the rate of the backward (reverse) reaction. 2. The concentrations of reactants and products remain constant (in a closed system).
(b)(i) Increasing the pressure increases the yield of methanol.
(b)(ii) On the left-hand side, there are \(1 + 2 = 3\text{ moles}\) of gas, while on the right-hand side, there is only \(1\text{ mole}\) of gas. Increasing pressure causes the equilibrium position to shift to the side with fewer moles of gas (to the right) to oppose the increase in pressure.
(c)(i) Increasing temperature gives particles more kinetic energy, resulting in more frequent and energetic collisions, so the rate of the forward reaction increases.
(c)(ii) The forward reaction is exothermic (\(\Delta H = -91\text{ kJ/mol}\)). According to Le Chatelier's principle, increasing the temperature causes the equilibrium to shift in the endothermic direction (to the left) to absorb the added thermal energy, thereby decreasing the equilibrium yield of methanol.
(d)(i) A catalyst increases the rate of both forward and reverse reactions equally, so it has no effect on the position of equilibrium.
(d)(ii) The catalyst lowers the activation energy, enabling the reaction to proceed at an acceptable rate at a lower operating temperature, which reduces energy/fuel consumption and overall operational costs.
評分準則
(a) - M1: Rate of forward reaction = rate of backward / reverse reaction [1] - M2: Concentration(s) of reactant(s) and product(s) remain constant / macroscopic properties remain constant (in a closed system) [1]
(b)(i) - M1: (Yield of methanol) increases / higher [1]
(b)(ii) - M1: Fewer moles / molecules of gas on the right / product side OR \(3\text{ moles of gas on left}\) and \(1\text{ mole of gas on right}\) [1] - M2: Equilibrium shifts to the side with fewer moles / molecules of gas / shifts to the right (to oppose increase in pressure) [1]
(c)(ii) - M1: (Yield of methanol) decreases / lower [1] - M2: (Forward) reaction is exothermic / reverse reaction is endothermic AND equilibrium shifts to the left / in the endothermic direction (to absorb added heat) [1]
(d)(i) - M1: No effect / no change / position unchanged [1]
(d)(ii) - M1: Reaction can be carried out at a lower temperature (which saves energy / fuel) OR speeds up reaction so more product is formed per unit time [1]
題目 7 · theory
17 分
Salts can be prepared and analysed using a variety of experimental and stoichiometric methods.
(a) A sample of hydrated cobalt(II) chloride has the formula \(\text{CoCl}_2 \cdot x\text{H}_2\text{O}\).
(i) State the colour change observed when hydrated cobalt(II) chloride is heated to form anhydrous cobalt(II) chloride. [1]
(ii) In an experiment, a student gently heats \(4.76\text{ g}\) of \(\text{CoCl}_2 \cdot x\text{H}_2\text{O}\) until all the water of crystallisation is removed. The mass of anhydrous \(\text{CoCl}_2\) obtained is \(2.60\text{ g}\).
A sample of \(1.68\text{ g}\) of \(\text{MCO}_3\) is added to \(50.0\text{ cm}^3\) of \(1.00\text{ mol/dm}^3\text{ HCl}\) (an excess).
(i) Calculate the number of moles of \(\text{HCl}\) added to the carbonate sample. [1]
(ii) The resulting mixture is made up to a known volume and titrated against sodium hydroxide solution. The excess unreacted \(\text{HCl}\) requires \(20.0\text{ cm}^3\) of \(0.500\text{ mol/dm}^3\text{ NaOH}\) for complete neutralisation.
(b) (i) 1. Mix aqueous barium chloride and aqueous sodium sulfate. 2. Filter the mixture to collect the precipitate of barium sulfate. 3. Wash the residue (solid barium sulfate) with distilled/deionised water to remove soluble impurities. 4. Dry the residue between filter papers / in a desiccator / in a warm oven.
(ii) \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\)
(iii) The operating temperature inside the reduction furnace is approximately \(1200\,^\circ\text{C}\). The boiling point of zinc is \(907\,^\circ\text{C}\).
State the physical state of the zinc as it forms in the furnace and describe how it is collected.
(b) (i) ZnO + CO -> Zn + CO2 (ii) The oxidation number of carbon increases from +2 in CO to +4 in CO2 (it is oxidised, so it acts as a reducing agent). (iii) Physical state: gas / vapour. How collected: condensed / cooled into liquid zinc.
(c) 1. Mass of pure ZnS in ore = 35.0 x 0.776 = 27.16 tonnes. 2. Mr of ZnS = 65 + 32 = 97. 3. Mass of Zn = (65 / 97) x 27.16 = 18.2 tonnes (or 18.199 tonnes).
(d) (i) Zinc atoms are a different size to copper atoms, which disrupts the regular layers / lattice of copper atoms, preventing the layers from sliding over each other easily. (ii) Galvanising iron/steel / sacrificial protection / making batteries.
評分準則
(a) (i) zinc sulfide [1] (ii) 2ZnS + 3O2 -> 2ZnO + 2SO2 - M1: correct formulae of reactants and products [1] - M2: correct balancing [1] (iii) acid rain / kills fish / acidifies lakes / damages trees or limestone buildings [1] (b) (i) ZnO + CO -> Zn + CO2 [1] (ii) oxidation state of carbon increases (from +2 to +4) [1] (iii) gas / vapour [1] condensed / distilled / cooled into a liquid [1] (c) M1: mass of ZnS = 35.0 x 0.776 = 27.16 (tonnes) OR moles of ZnS = 27.16 / 97 = 0.280 (x 10^6 mol) [1] M2: scaling by ratio 65/97 OR multiplying moles by 65 [1] M3: 18.2 (tonnes) (accept 18.199 to 18.2) [1] (d) (i) M1: zinc atoms/ions are of a different size to copper atoms/ions [1] M2: disrupts regular layers / prevents layers of atoms from sliding over each other [1] (ii) galvanising (steel/iron) / sacrificial protection / electrodes in batteries/cells [1] (reject: rusting alone)
題目 9 · structured
13 分
This question is about organic compounds and polymers.
(a) Propene, \(\text{C}_3\text{H}_6\), is an unsaturated hydrocarbon.
(i) State the general formula of the homologous series of alkenes. [1]
(ii) Draw the displayed formula of propene, showing all atoms and all bonds. [1]
(iii) Propene can be polymerised to form poly(propene). Draw the structure of poly(propene) showing two repeat units. [2]
(b) Lactic acid, \(\text{CH}_3\text{CH(OH)COOH}\), contains both a hydroxyl group (\(-\text{OH}\)) and a carboxylic acid group (\(-\text{COOH}\)). Lactic acid polymerises to form the polymer poly(lactic acid) (PLA).
(i) Name the type of polymerisation reaction that occurs when PLA is formed from lactic acid. [1]
(ii) Name the small molecule eliminated during this polymerisation. [1]
(iii) Draw the structure of one repeat unit of poly(lactic acid). Include all the bonds in the ester linkage and open bonds at the ends. [2]
(iv) State one environmental advantage of using poly(lactic acid) rather than poly(propene) for disposable food packaging. [1]
(c) Synthetic polyesters can also be manufactured by reacting dicarboxylic acids with diols.
(i) Draw the structure of the polyester formed from ethanedioic acid, \(\text{HOOC-COOH}\), and ethane-1,2-diol, \(\text{HO-CH}_2\text{CH}_2\text{OH}\). Show one repeat unit with all bonds in the linkage and continuation bonds at both ends. [2]
(ii) Name the type of linkage present in polyesters and name a synthetic polymer that contains an amide linkage. [2]
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解題
(a)(i) \(\text{C}_n\text{H}_{2n}\)
(ii) Displayed formula of propene shows 3 carbon atoms with one \(\text{C}=\text{C}\) double bond and one \(\text{C}-\text{C}\) single bond, with all 6 hydrogen atoms explicitly bonded: \(\text{H}-\text{C}(\text{H})=\text{C}(\text{H})-\text{C}(\text{H})(\text{H})\text{H}\).
(iii) Two repeat units of poly(propene) with a continuous carbon backbone of 4 carbon atoms, each second carbon having a \(-\text{CH}_3\) group (or displayed methyl group), all single \(\text{C}-\text{C}\) bonds, and continuation bonds extending from both ends: \(-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}(\text{CH}_3)-\).
(b)(i) Condensation (polymerisation).
(ii) Water / \(\text{H}_2\text{O}\).
(iii) One repeat unit of poly(lactic acid): \(-\text{O}-\text{CH}(\text{CH}_3)-\text{C}(=\text{O})-\) with continuation bonds at both ends and the ester carbonyl \(\text{C}=\text{O}\) explicitly shown.
(iv) Poly(lactic acid) is biodegradable / breaks down by the action of microorganisms / made from renewable resources (corn starch / plants) rather than crude oil.
(c)(i) One repeat unit from ethanedioic acid and ethane-1,2-diol: \(-\text{C}(=\text{O})-\text{C}(=\text{O})-\text{O}-\text{CH}_2-\text{CH}_2-\text{O}-\) with continuation bonds at both ends.
(ii) Correct displayed formula showing all atoms and all bonds (one \(\text{C}=\text{C}\) double bond, one \(\text{C}-\text{C}\) single bond, six \(\text{C}-\text{H}\) single bonds) [1]
(iii) M1: Backbone of 4 carbons with single bonds and continuation bonds [1] M2: Methyl group (\(-\text{CH}_3\)) on alternate backbone carbon atoms with correct number of hydrogens on each carbon [1]
(b)(i) Condensation (polymerisation) [1]
(ii) Water / \(\text{H}_2\text{O}\) [1]
(iii) M1: Correct ester link shown \(-\text{C}(=\text{O})-\text{O}-\) or \(-\text{O}-\text{C}(=\text{O})-\) [1] M2: Rest of the repeat unit correct with \(-\text{CH}(\text{CH}_3)-\) and extension/continuation bonds at both ends [1]
(iv) Biodegradable / breaks down naturally / compostable / made from renewable resources (plants/crops) [1] (Reject: recyclable / non-toxic / cheaper)
(c)(i) M1: Correct ester linkage(s) shown between alternating units [1] M2: Completely correct repeat unit \(-\text{C}(=\text{O})-\text{C}(=\text{O})-\text{O}-\text{CH}_2\text{CH}_2-\text{O}-\) with open continuation bonds at both ends [1]
Answer all practical questions. Use sharp pencil for graphs and drawings.
4 題目 · 40 分
題目 1 · structured
7 分
A student investigated the concentration of a solution of dilute potassium hydroxide, solution \(\mathbf{P}\), by titrating it with dilute sulfuric acid of concentration \(0.050\text{ mol/dm}^3\), solution \(\mathbf{Q}\).
(a) Name the piece of apparatus that should be used to measure accurately \(25.0\text{ cm}^3\) of solution \(\mathbf{P}\) into the conical flask. [1]
(b) The student added a few drops of methyl orange indicator to the conical flask containing solution \(\mathbf{P}\). State the colour change observed at the end-point of the titration. from .................................................... to .................................................... [1]
(c) The diagram shows the burette readings for the titration. - Initial burette reading: the meniscus is at \(1.4\text{ cm}^3\). - Final burette reading: the meniscus is at \(23.8\text{ cm}^3\).
(d) Explain why the conical flask is placed on a white tile during the titration. [1]
(e) Explain why the student swirled the conical flask while adding solution \(\mathbf{Q}\) from the burette. [1]
(f) State what the student should do to ensure the results are reliable. [1]
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解題
(a) A volumetric pipette is the standard laboratory glassware designed to deliver fixed, precise volumes such as \(25.0\text{ cm}^3\).
(b) Methyl orange is yellow in alkaline solutions (potassium hydroxide) and turns orange / red-orange at the end-point (neutral/acidic).
(c) Reading from the data provided: - Final burette reading = \(23.8\text{ cm}^3\) - Initial burette reading = \(1.4\text{ cm}^3\) - Volume added = \(23.8 - 1.4 = 22.4\text{ cm}^3\)
(d) A white tile placed under the conical flask provides a neutral background, making it easier to see the indicator's colour change at the end-point.
(e) Swirling ensures thorough mixing of the acid and alkali so that the reaction occurs completely and evenly throughout the solution.
(f) Repeating the titration multiple times to achieve concordant results (within \(\pm 0.1\text{ cm}^3\) or \(\pm 0.2\text{ cm}^3\)) and taking an average ensures the results are reliable and minimizes random errors.
評分準則
(a) (volumetric) pipette [1] (b) yellow to orange / pink / red [1] (c) - Both initial (1.4) and final (23.8) readings correct [1] - Correct subtraction / volume added (22.4) [1] (d) to make the colour change easier to see / see the end-point clearly [1] (e) to mix (the solutions / reactants) / to ensure they react completely [1] (f) repeat (the titration) and find average / mean (of concordant titres) [1]
題目 2 · theory
20 分
A student investigated the temperature change in the displacement reaction between zinc powder and aqueous copper(II) sulfate.
Five experiments were carried out.
### Experiment 1 - Using a measuring cylinder, \(25.0\text{ cm}^3\) of \(0.80\text{ mol/dm}^3\) aqueous copper(II) sulfate was poured into an expanded polystyrene cup. - The initial temperature of the solution was measured using a thermometer and recorded in Table 1.1 at \(\text{time} = 0\text{ s}\). - \(0.40\text{ g}\) of zinc powder was added to the solution, a stopwatch was started, and the mixture was stirred continuously. - The highest temperature reached by the mixture was recorded. - The polystyrene cup was rinsed thoroughly with distilled water.
### Experiments 2–5 - Experiment 1 was repeated using \(0.80\text{ g}\), \(1.20\text{ g}\), \(1.60\text{ g}\), and \(2.00\text{ g}\) of zinc powder, respectively, instead of \(0.40\text{ g}\).
Table 1.1 shows the initial and maximum temperature readings for each experiment.
(a) Complete Table 1.1 by calculating the temperature rise, \(\Delta T\), for each experiment. [5]
(b) On the grid, plot the values of temperature rise, \(\Delta T\), on the y-axis against the mass of zinc on the x-axis. - Include a suitable scale on both axes. - Plot the points. - Draw two straight lines of best fit: the first line through the points showing an increasing temperature rise, and the second line through the points where the temperature rise levels off. - Extrapolate (extend) both lines so that they cross. [6]
(c) Use your graph to: (i) Determine the mass of zinc required to react completely with \(25.0\text{ cm}^3\) of the aqueous copper(II) sulfate. Show clearly on your graph how you obtained your answer. [2]
$$\text{Mass of zinc} = \text{............................ g}$$
(ii) Deduce the maximum temperature rise, \(\Delta T\), that would be achieved at this complete reaction point. [1]
$$\text{Maximum } \Delta T = \text{............................ }^{\circ}\text{C}$$
(d) Explain why the temperature rise does not continue to increase when more than \(1.60\text{ g}\) of zinc powder is added. [2]
(e) Predict the temperature rise if Experiment 2 (using \(0.80\text{ g}\) of zinc) was repeated using \(25.0\text{ cm}^3\) of \(1.60\text{ mol/dm}^3\) copper(II) sulfate instead of \(0.80\text{ mol/dm}^3\) copper(II) sulfate. Explain your answer. [2]
(f) State two improvements that could be made to the apparatus or procedure to obtain more accurate temperature values. For each improvement, state the reason why it improves accuracy. [2]
(b) - Axes: y-axis labelled 'Temperature rise / \(^{\circ}\text{C}\)' from \(0\) to at least \(30\text{ }^{\circ}\text{C}\); x-axis labelled 'Mass of zinc / \(\text{g}\)' from \(0\) to \(2.0\text{ g}\). - Scale: Sensible, linear scales where points occupy more than half of the grid in both directions. - Plotting: Points plotted precisely: \((0.40, 8.0)\), \((0.80, 16.0)\), \((1.20, 23.0)\), \((1.60, 26.0)\), \((2.00, 26.0)\). - Lines: One straight line drawn from origin through the first 3 points, a second horizontal straight line through the final 2 points (or slightly negative slope accounting for thermal loss), clearly intersecting.
(c) (i) From the intersection of the two lines, drop a vertical line to the x-axis: \(\text{Mass} = 1.30\text{ g}\) (allow \(1.25\text{ g} - 1.35\text{ g}\)). (ii) Read the y-value at the point of intersection: \(\Delta T = 26.0\text{ }^{\circ}\text{C}\) (allow \(25.5 - 26.5\text{ }^{\circ}\text{C}\)).
(d) All the aqueous copper(II) sulfate has reacted (copper(II) sulfate is the limiting reactant / zinc is in excess), so no additional displacement occurs and no further thermal energy is released.
(e) Predicted \(\Delta T = 16.0\text{ }^{\circ}\text{C}\). In Experiment 2, zinc is the limiting reactant (\(0.80\text{ g} < 1.30\text{ g}\)), so doubling the concentration of copper(II) sulfate does not change the amount of zinc reacting; the same total energy is released into the same volume (\(25.0\text{ cm}^3\)).
(f) - Improvement 1: Add a lid / cover to the polystyrene cup \(\rightarrow\) reduces heat loss by convection/evaporation. - Improvement 2: Use a volumetric pipette / burette to measure the \(25.0\text{ cm}^3\) volume \(\rightarrow\) reduces percentage uncertainty / improves volumetric accuracy compared to a measuring cylinder.
評分準則
(a) [5 marks] - 1 mark for each correctly calculated value of \(\Delta T\): - Exp 1: \(8.0\) [1] - Exp 2: \(16.0\) [1] - Exp 3: \(23.0\) [1] - Exp 4: \(26.0\) [1] - Exp 5: \(26.0\) [1]
(b) [6 marks] - Axes (M1): Both axes labelled with quantity and units (`Temperature rise / °C` and `Mass of zinc / g`) [1] - Scale (M2): Linear, suitable scales plotted such that points occupy \(\ge 50\%\) of grid space in both dimensions [1] - Plotting (M3 & M4): All 5 points plotted accurately to within half a small square (4 points plotted correctly = 1 mark; all 5 = 2 marks) [2] - Lines (M5 & M6): Two distinct straight lines of best fit drawn using a ruler, intersecting clearly [2]
(c) [3 marks] - (i) Construction lines / tie-lines shown at the point of intersection [1]; Correct value deduced from candidate's graph (\(1.30 \pm 0.05\text{ g}\)) [1] - (ii) Maximum \(\Delta T\) correctly read from the intersection of candidate's graph (\(26.0 \pm 0.5\text{ }^{\circ}\text{C}\)) [1]
(d) [2 marks] - Copper(II) sulfate is the limiting reactant / all copper(II) sulfate has reacted / zinc is in excess [1] - No further reaction takes place / no more heat is produced [1]
(e) [2 marks] - \(16.0\text{ }^{\circ}\text{C}\) / unchanged [1] - Zinc is the limiting reactant / zinc reacts completely in both cases so the same amount of heat is evolved [1]
(f) [2 marks] - Award 1 mark for each valid pair of (improvement + matching reason), max 2: - Use a lid / cover on the cup [1] to minimize heat loss to the surroundings / environment [1] - Use a pipette / burette (instead of a measuring cylinder) [1] for more accurate measurement of solution volume [1] - Use a digital thermometer / temperature sensor [1] for higher precision / smaller division readings [1]
題目 3 · structured
7 分
Two substances, solution E and solid F, were analysed. Solution E was an aqueous solution of an ionic compound. Solid F was a hydrated transition metal salt.
The tests and the observations are shown.
Tests on solution E
* Test 1: A flame test was carried out on solution E. * Observation: A lilac flame was observed. * Test 2: To a portion of solution E, dilute nitric acid followed by aqueous silver nitrate was added. * Observation: A yellow precipitate formed.
Tests on solid F
* Test 3: Dilute hydrochloric acid was added to solid F. The gas produced was bubbled through limewater. * Observation: Rapid effervescence occurred and the limewater turned cloudy/milky. * Test 4: Aqueous sodium hydroxide was added dropwise until in excess to the solution formed in Test 3. * Observation: A green precipitate formed, which was insoluble in excess aqueous sodium hydroxide.
(a) Identify the two ions present in solution E. * cation: .......................................................................................................... [1] * anion: ........................................................................................................... [1]
(b) (i) Identify the gas released in Test 3. ........................................................................................................................ [1] (ii) Identify the anion present in solid F. ........................................................................................................................ [1]
(c) (i) Identify the cation present in solid F. ........................................................................................................................ [1] (ii) Give the chemical formula of solid F (ignore water of crystallisation). ........................................................................................................................ [1]
(d) State the colour change observed when the green precipitate formed in Test 4 is left exposed to air for several minutes. ........................................................................................................................ [1]
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解題
(a) * Test 1 produces a lilac flame, which is the characteristic flame colour for the potassium ion (\(\text{K}^+\)). * Test 2 produces a yellow precipitate with aqueous silver nitrate acidified with dilute nitric acid, which confirms the presence of the iodide ion (\(\text{I}^-\)).
(b) * (i) A gas that turns limewater cloudy/milky is carbon dioxide (\(\text{CO}_2\)). * (ii) The production of carbon dioxide gas upon adding dilute acid indicates the presence of the carbonate ion (\(\text{CO}_3^{2-}\)).
(c) * (i) The addition of aqueous sodium hydroxide giving a green precipitate insoluble in excess confirms the cation is the iron(II) ion (\(\text{Fe}^{2+}\)). * (ii) Combining the iron(II) cation (\(\text{Fe}^{2+}\)) and the carbonate anion (\(\text{CO}_3^{2-}\)) gives the chemical formula \(\text{FeCO}_3\).
(d) * Iron(II) hydroxide, \(\text{Fe(OH)}_2\), oxidises in the presence of atmospheric oxygen to form iron(III) hydroxide, \(\text{Fe(OH)}_3\), turning the precipitate red-brown / brown.
評分準則
(a) * [1 mark] potassium / \(\text{K}^+\) * [1 mark] iodide / \(\text{I}^-\) (Reject: potassium ion / iodine if formula given incorrectly)
(c) * (i) [1 mark] iron(II) / \(\text{Fe}^{2+}\) *(Reject: iron / \(\text{Fe}\) / \(\text{Fe}^{3+}\) / iron(III))* * (ii) [1 mark] \(\text{FeCO}_3\) (ecf from (b)(ii) and (c)(i))
(d) * [1 mark] (turns) brown / red-brown / darkens to brown (Reject: rust alone without colour)
題目 4 · free_response
6 分
A sample of crushed ore contains a mixture of three solid compounds: zinc carbonate, silicon dioxide (sand), and potassium nitrate.
The properties of the three substances are given: • zinc carbonate: insoluble in water; reacts with dilute sulfuric acid to form soluble zinc sulfate and a gas • silicon dioxide: insoluble in water; does not react with dilute sulfuric acid • potassium nitrate: soluble in water; does not react with dilute sulfuric acid
Plan an experiment to obtain a dry, pure sample of silicon dioxide from the mixture and to determine the percentage by mass of silicon dioxide in the ore sample.
You are provided with: • a sample of the crushed ore • dilute sulfuric acid • distilled water • standard laboratory apparatus.
In your plan, you should include: • the apparatus you would use • a step-by-step method including how to ensure all other substances are removed and the silicon dioxide is dry • the measurements you would make and how you would calculate the percentage by mass of silicon dioxide.
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解題
To obtain pure, dry silicon dioxide and determine its percentage by mass: 1. Weigh the original mixture accurately using a balance and record the mass (\(m_1\)). 2. Place the weighed ore into a beaker, add an excess of dilute sulfuric acid, and stir until no further fizzing/effervescence is observed (this dissolves the zinc carbonate as soluble zinc sulfate while potassium nitrate also dissolves into the aqueous solution). 3. Filter the mixture using a filter funnel and filter paper; the silicon dioxide remains as the insoluble residue on the filter paper. 4. Wash the residue on the filter paper with distilled water to remove any traces of zinc sulfate, potassium nitrate, and acid. 5. Dry the residue in an oven or by warming gently until a constant mass is achieved. 6. Measure and record the mass of the dry silicon dioxide (\(m_2\)). 7. Calculate the percentage by mass using the formula: \(\text{percentage by mass} = \frac{m_2}{m_1} \times 100\% \).
評分準則
Award 1 mark for each of the following points (up to a maximum of 6 marks): • MP1: Measure / record the mass of the initial sample of ore using a balance. • MP2: Add excess dilute sulfuric acid (or add water followed by dilute sulfuric acid) to the mixture in a beaker and stir / wait until fizzing stops. • MP3: Separate the mixture by filtration (mention of filter paper and funnel / filter / residue). • MP4: Wash the solid residue / silicon dioxide with distilled / deionised water. • MP5: Dry the residue to constant mass (in a warm oven / between filter papers / desiccator). • MP6: Measure the final mass of dry silicon dioxide AND state calculation: \(\frac{\text{mass of dry silicon dioxide}}{\text{mass of original sample}} \times 100\).
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