An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V4) Cambridge International AS Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.
卷一: 選擇題
Answer all 40 four-option multiple choice questions on the separate answer sheet. Calculators and standard data booklet values may be used.
40 题目 · 40 分
题目 1 · multiple_choice
1 分
The first five successive ionisation energies of an element, \(Z\), are shown.
A.Element \(Z\) is located in Group 15 of the Periodic Table.
B.Element \(Z\) forms an oxide with the formula \(Z\text{O}_2\).
C.The fifth ionisation involves removing an electron from an outer p subshell.
D.Element \(Z\) forms a stable ion with a charge of \(+5\).
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解题
There is a very large jump in ionisation energy between the 4th and 5th ionisation energies (from \(6223\text{ kJ}\text{ mol}^{-1}\) to \(37831\text{ kJ}\text{ mol}^{-1}\)). This indicates that the 5th electron is removed from an inner quantum shell. Therefore, \(Z\) has 4 valence electrons and belongs to Group 14 of the Periodic Table. Elements in Group 14 form oxides with the formula \(Z\text{O}_2\).
评分标准
B is correct [1].
Identification of Group 14 from the large jump between the 4th and 5th successive ionisation energies.
题目 2 · multiple_choice
1 分
A \(15.0\text{ cm}^3\) sample of a gaseous hydrocarbon, \(\text{C}_x\text{H}_y\), is mixed with \(90.0\text{ cm}^3\) of oxygen gas (an excess) and ignited. After complete combustion, the total volume of gas remaining at room temperature and pressure is \(60.0\text{ cm}^3\).
When this resulting gas mixture is shaken with aqueous sodium hydroxide, the volume decreases to \(15.0\text{ cm}^3\).
Method: - Volume of \(\text{CO}_2 = 45.0\text{ cm}^3\) \(\implies x = 3\). - Volume of \(\text{O}_2\) consumed \(= 75.0\text{ cm}^3\) \(\implies 3 + y/4 = 5 \implies y = 8\).
题目 3 · multiple_choice
1 分
Which pair of species have the same shape and the same approximate bond angle?
A.\(\text{H}_3\text{O}^+\) and \(\text{NH}_3\)
B.\(\text{NH}_4^+\) and \(\text{BF}_3\)
C.\(\text{CO}_2\) and \(\text{SO}_2\)
D.\(\text{CH}_4\) and \(\text{SF}_4\)
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解题
Both \(\text{H}_3\text{O}^+\) and \(\text{NH}_3\) have 3 bonding pairs and 1 lone pair of electrons around the central atom, giving them a trigonal pyramidal shape with a bond angle of approximately \(107^\circ\).
- In B: \(\text{NH}_4^+\) is tetrahedral (\(109.5^\circ\)) while \(\text{BF}_3\) is trigonal planar (\(120^\circ\)). - In C: \(\text{CO}_2\) is linear (\(180^\circ\)) while \(\text{SO}_2\) is bent (approx \(119^\circ\)). - In D: \(\text{CH}_4\) is tetrahedral (\(109.5^\circ\)) while \(\text{SF}_4\) is see-saw.
评分标准
A is correct [1].
Both \(\text{H}_3\text{O}^+\) and \(\text{NH}_3\) have 3 bonding pairs and 1 lone pair \(\rightarrow\) trigonal pyramidal (approx \(107^\circ\)).
题目 4 · multiple_choice
1 分
The reversible reaction between dinitrogen tetraoxide and nitrogen dioxide reaches dynamic equilibrium in a gas syringe at constant temperature.
$$\text{N}_2\text{O}_4(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g}) \quad \Delta H = +57\text{ kJ}\text{ mol}^{-1}$$
The volume of the syringe is suddenly halved.
Which row correctly describes the initial effect on the total pressure and the shift in the position of equilibrium as dynamic equilibrium is re-established?
A.Initial effect: decreases; Shift in position of equilibrium: shifts to the left
B.Initial effect: increases; Shift in position of equilibrium: shifts to the left
C.Initial effect: increases; Shift in position of equilibrium: shifts to the right
D.Initial effect: decreases; Shift in position of equilibrium: shifts to the right
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解题
Halving the volume immediately increases the concentration of all gaseous species, causing an initial increase in the total pressure.
According to Le Chatelier's principle, the system responds to reduce the pressure by shifting the equilibrium position towards the side with fewer moles of gas (the left-hand side, forming more \(\text{N}_2\text{O}_4\)).
评分标准
B is correct [1].
Initial pressure increases because volume is reduced; equilibrium shifts left to the side with fewer moles of gas (1 mol vs 2 mol).
题目 5 · multiple_choice
1 分
Separate samples containing equal amounts (in moles) of four Period 3 chlorides are added to equal volumes of water at room temperature.
- \(\text{NaCl}\) dissolves without hydrolysis to form a neutral solution (\(\text{pH} = 7\)). - \(\text{MgCl}_2\) undergoes slight hydrolysis to give a weakly acidic solution (\(\text{pH} \approx 6.5\)). - \(\text{SiCl}_4\) hydrolyses completely: \(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\), producing \(4\text{ mol}\) of \(\text{HCl}\) per mole of \(\text{SiCl}_4\) (\(\text{pH} \approx 1-2\)). - \(\text{PCl}_5\) hydrolyses vigorously: \(\text{PCl}_5 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 5\text{HCl}\), producing \(5\text{ mol}\) of \(\text{HCl}\) plus phosphoric acid per mole of \(\text{PCl}_5\), resulting in the lowest pH.
Therefore, the order of decreasing pH is \(\text{NaCl} > \text{MgCl}_2 > \text{SiCl}_4 > \text{PCl}_5\).
评分标准
A is correct [1].
Order of decreasing pH: \(\text{NaCl} (\approx 7) > \text{MgCl}_2 (\approx 6.5) > \text{SiCl}_4 (\approx 2) > \text{PCl}_5 (\approx 1)\).
题目 6 · multiple_choice
1 分
Solid sodium bromide reacts with concentrated sulfuric acid to produce a mixture of products.
Which substance is formed via a redox reaction in this process?
A.\(\text{HBr}\)
B.\(\text{NaHSO}_4\)
C.\(\text{SO}_2\)
D.\(\text{H}_2\text{S}\)
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解题
When \(\text{NaBr}\) reacts with concentrated \(\text{H}_2\text{SO}_4\): 1. Acid-base reaction: \(\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HBr}\) (no change in oxidation numbers). 2. Redox reaction: \(2\text{HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\).
In step 2, sulfur is reduced from \(+6\) in \(\text{H}_2\text{SO}_4\) to \(+4\) in \(\text{SO}_2\), and bromide is oxidised from \(-1\) to \(0\) in \(\text{Br}_2\). Therefore, \(\text{SO}_2\) (and \(\text{Br}_2\)) is a redox product.
评分标准
C is correct [1].
\(\text{SO}_2\) is formed by reduction of \(\text{H}_2\text{SO}_4\) by bromide ions.
题目 7 · multiple_choice
1 分
A sample of 2-chloro-2-methylpropane is heated with aqueous sodium hydroxide.
Which statement correctly describes this reaction?
A.The reaction proceeds via an \(\text{S}_\text{N}1\) mechanism involving a planar carbocation intermediate.
B.The rate of reaction is directly proportional to the concentration of hydroxide ions.
C.The reaction proceeds via a single-step \(\text{S}_\text{N}2\) mechanism involving a five-coordinate transition state.
D.The main organic product of the reaction is a secondary alcohol.
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解题
2-chloro-2-methylpropane is a tertiary halogenoalkane. It undergoes nucleophilic substitution primarily via an \(\text{S}_\text{N}1\) mechanism. - The first, rate-determining step involves the heterolytic fission of the \(\text{C}-\text{Cl}\) bond to form a relatively stable planar tertiary carbocation intermediate, \((\text{CH}_3)_3\text{C}^+\). - The rate equation is \(\text{rate} = k[(\text{CH}_3)_3\text{CCl}]\), independent of \([\text{OH}^-]\). - The product is 2-methylpropan-2-ol, which is a tertiary alcohol.
评分标准
A is correct [1].
Tertiary halogenoalkanes undergo \(\text{S}_\text{N}1\) substitution through a planar carbocation intermediate.
题目 8 · multiple_choice
1 分
Compound \(W\) is an alkene with the molecular formula \(\text{C}_6\text{H}_{12}\).
When \(W\) is heated under reflux with hot concentrated acidified potassium manganate(VII), the only organic products obtained are propanone, \((\text{CH}_3)_2\text{CO}\), and propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\).
What is the systematic IUPAC name of compound \(W\)?
A.2-methylpent-1-ene
B.2-methylpent-2-ene
C.3-methylpent-2-ene
D.4-methylpent-2-ene
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解题
Hot concentrated acidified \(\text{KMnO}_4\) cleaves the \(\text{C}=\text{C}\) double bond: - A carbon with two alkyl groups, \(=\text{C}(\text{CH}_3)_2\), is oxidised to a ketone (propanone, \((\text{CH}_3)_2\text{CO}\)). - A carbon with one hydrogen and one alkyl group, \(=\text{CH}-\text{CH}_2\text{CH}_3\), is oxidised to a carboxylic acid (propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\)).
Joining these two fragments at the double bond gives: $$(\text{CH}_3)_2\text{C}=\text{CH}-\text{CH}_2-\text{CH}_3$$
The longest continuous carbon chain contains 5 carbon atoms (pent-2-ene) with a methyl substituent at carbon-2. Therefore, the systematic IUPAC name is 2-methylpent-2-ene.
评分标准
B is correct [1].
Oxidation of 2-methylpent-2-ene gives propanone from the \((\text{CH}_3)_2\text{C}=\) end and propanoic acid from the \(=\text{CHCH}_2\text{CH}_3\) end.
题目 9 · multiple_choice
1 分
\(10\text{ cm}^3\) of a gaseous hydrocarbon \(\text{C}_x\text{H}_y\) is mixed with \(70\text{ cm}^3\) of oxygen gas (an excess) and ignited. After complete combustion and cooling to room temperature, the total volume of gas remaining is \(55\text{ cm}^3\).
When this remaining gas is passed through an excess of aqueous potassium hydroxide, the gas volume decreases by \(30\text{ cm}^3\).
All gas volumes are measured at the same temperature and pressure.
What is the molecular formula of the hydrocarbon?
A.\(\text{C}_3\text{H}_4\)
B.\(\text{C}_3\text{H}_6\)
C.\(\text{C}_3\text{H}_8\)
D.\(\text{C}_4\text{H}_8\)
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解题
1. Aqueous \(\text{KOH}\) absorbs \(\text{CO}_2\). The decrease in volume is \(30\text{ cm}^3\), so the volume of \(\text{CO}_2\) produced is \(30\text{ cm}^3\). 2. Since \(10\text{ cm}^3\) of \(\text{C}_x\text{H}_y\) produces \(30\text{ cm}^3\) of \(\text{CO}_2\), the number of carbon atoms \(x\) is: \[x = \frac{30}{10} = 3\] 3. The gas remaining after passing through \(\text{KOH}\) is unreacted \(\text{O}_2\): \[\text{Volume of unreacted } \text{O}_2 = 55 - 30 = 25\text{ cm}^3\] 4. The volume of \(\text{O}_2\) consumed during combustion is: \[70 - 25 = 45\text{ cm}^3\] 5. The general equation for combustion is: \[\text{C}_3\text{H}_y + \left(3 + \frac{y}{4}\right)\text{O}_2 \rightarrow 3\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}\] 6. The volume ratio of \(\text{O}_2\) to hydrocarbon is \(\frac{45}{10} = 4.5\): \[3 + \frac{y}{4} = 4.5 \implies \frac{y}{4} = 1.5 \implies y = 6\] Therefore, the molecular formula is \(\text{C}_3\text{H}_6\).
评分标准
B is correct [1]. - Method mark / calculation: Identifies volume of \(\text{CO}_2 = 30\text{ cm}^3\) \(\implies x = 3\), unreacted \(\text{O}_2 = 25\text{ cm}^3\), reacted \(\text{O}_2 = 45\text{ cm}^3\) \(\implies y = 6\).
题目 10 · multiple_choice
1 分
The first five successive ionisation energies of element \(\text{X}\) are given in the table below.
Which statement about element \(\text{X}\) or its compounds is correct?
A.The oxide of \(\text{X}\) reacts with both aqueous hydrochloric acid and aqueous sodium hydroxide.
B.The chloride of \(\text{X}\) dissolves in water to form a solution with \(\text{pH} = 7\).
C.Element \(\text{X}\) forms a stable divalent cation, \(\text{X}^{2+}\).
D.Element \(\text{X}\) belongs to Group 2 of the Periodic Table.
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解题
1. Look at the ratio/differences between successive ionisation energies: - 1st to 2nd: \(1817 - 578 = 1239\text{ kJ mol}^{-1}\) - 2nd to 3rd: \(2745 - 1817 = 928\text{ kJ mol}^{-1}\) - 3rd to 4th: \(11578 - 2745 = 8833\text{ kJ mol}^{-1}\) (a very large jump) 2. The large jump between the 3rd and 4th ionisation energies indicates that the 4th electron is removed from an inner, fully-filled principal quantum shell. Thus, element \(\text{X}\) has 3 valence electrons and belongs to Group 13 (e.g. aluminium). 3. Aluminium oxide (\(\text{Al}_2\text{O}_3\)) is amphoteric, reacting with both acids (such as \(\text{HCl}\)) and alkalis (such as \(\text{NaOH}\)).
评分标准
A is correct [1]. - Deduces Group 13 from the large jump between 3rd and 4th IE values, identifying the amphoteric nature of the Period 3 Group 13 oxide.
题目 11 · multiple_choice
1 分
Consider the molecules boron trifluoride, \(\text{BF}_3\), and nitrogen trifluoride, \(\text{NF}_3\).
Which row correctly compares their shapes, bond angles, and overall molecular polarities?
A.\(\text{BF}_3\) is trigonal planar and non-polar, whereas \(\text{NF}_3\) is trigonal pyramidal and polar.
B.Both \(\text{BF}_3\) and \(\text{NF}_3\) are planar molecules with bond angles of \(120^\circ\).
C.\(\text{NF}_3\) has a larger bond angle than \(\text{BF}_3\) because of lone-pair repulsion.
D.Both molecules have permanent dipole moments due to the high electronegativity of fluorine.
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解题
1. \(\text{BF}_3\): - Boron has 3 valence electrons, forming 3 bonding pairs and 0 lone pairs. - Shape: Trigonal planar. - Bond angle: \(120^\circ\). - Symmetry: Dipoles cancel out completely, so \(\text{BF}_3\) is non-polar.
2. \(\text{NF}_3\): - Nitrogen has 5 valence electrons, forming 3 bonding pairs and 1 lone pair. - Shape: Trigonal pyramidal. - Bond angle: Approximately \(107^\circ\) (less than \(109.5^\circ\) due to lone pair-bonding pair repulsion). - Symmetry: Asymmetric shape means dipoles do not cancel, so \(\text{NF}_3\) is polar.
评分标准
A is correct [1]. - Correct deduction of VSEPR geometries, bond angle comparisons, and resultant dipole moments.
题目 12 · multiple_choice
1 分
An anhydrous solid Period 3 chloride, \(\text{W}\), hydrolyses vigorously in cold water to produce an acidic solution and steamy white fumes.
When a few drops of aqueous sodium hydroxide are added to this acidic solution, a white precipitate forms. Upon adding excess aqueous sodium hydroxide, the white precipitate dissolves to yield a colourless solution.
What is the identity of solid \(\text{W}\)?
A.\(\text{MgCl}_2\)
B.\(\text{AlCl}_3\)
C.\(\text{SiCl}_4\)
D.\(\text{PCl}_5\)
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解题
1. \(\text{AlCl}_3\) hydrolyses in water to release \(\text{HCl(g)}\) (steamy white fumes) and forms acidic \([\text{Al}(\text{H}_2\text{O})_6]^{3+}\) ions (\(\text{pH} \approx 2-3\)). 2. Adding \(\text{NaOH(aq)}\) precipitates amphoteric aluminium hydroxide: \[\text{Al}^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq}) \rightarrow \text{Al(OH)}_3(\text{s})\text{ (white precipitate)}\] 3. Adding excess \(\text{NaOH(aq)}\) redissolves the precipitate due to its amphoteric nature: \[\text{Al(OH)}_3(\text{s}) + \text{OH}^-(\text{aq}) \rightarrow [\text{Al(OH)}_4]^-(\text{aq})\text{ (colourless solution)}\] - \(\text{MgCl}_2\) forms \(\text{Mg(OH)}_2\) which does not redissolve in excess alkali. - \(\text{SiCl}_4\) forms insoluble \(\text{SiO}_2 \cdot x\text{H}_2\text{O}\) directly upon hydrolysis. - \(\text{PCl}_5\) forms phosphoric acid which does not produce a precipitate with \(\text{NaOH}\).
评分标准
B is correct [1]. - Recognises \(\text{AlCl}_3\) from steamy fumes of \(\text{HCl}\) and amphoteric precipitate behaviour of \(\text{Al(OH)}_3\).
题目 13 · multiple_choice
1 分
Sulfur dioxide reacts reversibly with oxygen in the Contact process according to the following equation:
Which single change will result in an increase in BOTH the numerical value of \(K_p\) AND the equilibrium mole fraction of \(\text{SO}_3(\text{g})\)?
A.Decreasing the temperature at constant pressure
B.Increasing the total pressure at constant temperature
C.Adding a finely divided \(\text{V}_2\text{O}_5\) catalyst
D.Increasing the temperature at constant volume
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解题
1. The equilibrium constant \(K_p\) depends ONLY on temperature. 2. The forward reaction is exothermic (\(\Delta H^\circ < 0\)). 3. Decreasing the temperature shifts the equilibrium position to the right (the exothermic direction), which increases the value of \(K_p\) and increases the mole fraction of \(\text{SO}_3(\text{g})\). 4. Increasing total pressure shifts the position of equilibrium to the right but does not alter the value of \(K_p\). 5. A catalyst alters only the rate of reaching equilibrium, not \(K_p\) or equilibrium composition.
评分标准
A is correct [1]. - Identifies that \(K_p\) is temperature-dependent and increases when temperature decreases for an exothermic reaction.
题目 14 · multiple_choice
1 分
An organic compound \(\text{Q}\) has the molecular formula \(\text{C}_6\text{H}_{10}\).
When \(\text{Q}\) is heated under reflux with excess hot concentrated acidified potassium manganate(VII), the only organic product obtained is 5-oxohexanoic acid, \(\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_2\text{COOH}\).
What is the identity of compound \(\text{Q}\)?
A.1-methylcyclopentene
B.3-methylcyclopentene
C.cyclohexene
D.methylenecyclopentane
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解题
1. Hot concentrated acidified \(\text{KMnO}_4\) cleaves carbon-carbon double bonds: - \(=\text{CH}_2\) oxidises to \(\text{CO}_2\) and \(\text{H}_2\text{O}\). - \(=\text{CHR}\) oxidises to a carboxylic acid group, \(-\text{COOH}\). - \(=\text{CR}_1\text{R}_2\) oxidises to a ketone group, \(-\text{CO}-\). 2. The product 5-oxohexanoic acid has a ketone group at one end (\(\text{CH}_3-\text{C}(=\text{O})-\)) and a carboxylic acid group at the other end (\(-\text{COOH}\)), connected by three \(\text{CH}_2\) groups. 3. This indicates that the original compound was a cyclic alkene: 1-methylcyclopentene. - In 1-methylcyclopentene, \(\text{C}1\) carries a methyl group (yielding the ketone) and \(\text{C}2\) has a hydrogen atom (yielding the carboxylic acid) upon oxidative ring opening.
评分标准
A is correct [1]. - Deduces the structure of the cycloalkene from the cleavage products of hot concentrated \(\text{KMnO}_4\).
题目 15 · multiple_choice
1 分
Equal amounts of 1-bromobutane and 2-bromo-2-methylpropane are placed into separate test-tubes containing aqueous silver nitrate dissolved in ethanol at \(50\ ^\circ\text{C}\).
Which row correctly compares the rate of formation of the precipitate in 2-bromo-2-methylpropane with 1-bromobutane, and identifies the main mechanism by which 2-bromo-2-methylpropane reacts?
1. 2-bromo-2-methylpropane is a tertiary halogenoalkane, while 1-bromobutane is a primary halogenoalkane. 2. Tertiary halogenoalkanes react primarily via the \(\text{S}_\text{N}1\) mechanism, where the rate-determining step is the unimolecular heterolytic cleavage of the \(\text{C}-\text{Br}\) bond to form a relatively stable tertiary carbocation. 3. The activation energy for tertiary carbocation formation is significantly lower than that for the \(\text{S}_\text{N}2\) pathway of primary halogenoalkanes under these conditions. 4. Therefore, 2-bromo-2-methylpropane forms the pale cream precipitate of \(\text{AgBr}\) much faster than 1-bromobutane via the \(\text{S}_\text{N}1\) mechanism.
评分标准
A is correct [1]. - Recognises tertiary halogenoalkane undergoes faster nucleophilic substitution via \(\text{S}_\text{N}1\).
题目 16 · multiple_choice
1 分
The mass spectrum of an organic compound \(\text{Z}\) containing only carbon, hydrogen, and oxygen has a molecular ion peak \([\text{M}]^+\) at \(m/e = 88\) with a relative abundance of \(32.0\%\). The \([\text{M}+1]^+\) peak has a relative abundance of \(1.41\%\).
Compound \(\text{Z}\) reacts with aqueous sodium carbonate to produce carbon dioxide gas.
What is the systematic IUPAC name of compound \(\text{Z}\)?
A.butanoic acid
B.ethyl ethanoate
C.pentanoic acid
D.methyl propanoate
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解题
1. Use the \([\text{M}+1]^+\) abundance formula to determine the number of carbon atoms, \(n\): \[n = \frac{100 \times \text{abundance of } [\text{M}+1]^+}{1.1 \times \text{abundance of } [\text{M}]^+} = \frac{100 \times 1.41}{1.1 \times 32.0} = \frac{141}{35.2} \approx 4.0\] 2. Compound \(\text{Z}\) has 4 carbon atoms. 3. The reaction with aqueous \(\text{Na}_2\text{CO}_3\) producing \(\text{CO}_2\) confirms the presence of a carboxylic acid group (\(-\text{COOH}\)). 4. For a monocarboxylic acid with 4 carbons: - Formula = \(\text{C}_3\text{H}_7\text{COOH} = \text{C}_4\text{H}_8\text{O}_2\) - \(M_r = (4 \times 12.0) + (8 \times 1.0) + (2 \times 16.0) = 48 + 8 + 32 = 88\), which matches the molecular ion peak at \(m/e = 88\). 5. Among the choices, butanoic acid is a 4-carbon carboxylic acid.
评分标准
A is correct [1]. - Uses \(n = \frac{100 [\text{M}+1]^+}{1.1 [\text{M}]^+}\) to find 4 carbons, identifies carboxylic acid from \(\text{Na}_2\text{CO}_3\) reaction.
题目 17 · multiple_choice
1 分
A \(20\text{ cm}^3\) sample of a gaseous hydrocarbon, \(\text{C}_x\text{H}_y\), is completely combusted in \(150\text{ cm}^3\) of oxygen (an excess).
After cooling to room temperature and pressure, the total volume of gas remaining is \(110\text{ cm}^3\).
When this remaining gas is shaken with an excess of aqueous sodium hydroxide, the volume of gas decreases to \(30\text{ cm}^3\).
What is the molecular formula of the hydrocarbon?
A.\(\text{C}_3\text{H}_6\)
B.\(\text{C}_3\text{H}_8\)
C.\(\text{C}_4\text{H}_8\)
D.\(\text{C}_4\text{H}_{10}\)
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解题
1. The general equation for complete combustion of a hydrocarbon is: $$\text{C}_x\text{H}_y(\text{g}) + \left(x + \frac{y}{4}\right)\text{O}_2(\text{g}) \rightarrow x\text{CO}_2(\text{g}) + \frac{y}{2}\text{H}_2\text{O}(\text{l})$$
2. At room temperature and pressure, water is a liquid, so its volume is negligible.
3. The gas remaining after combustion (\(110\text{ cm}^3\)) consists of \(\text{CO}_2(\text{g})\) and unreacted \(\text{O}_2(\text{g})\).
4. Aqueous sodium hydroxide absorbs \(\text{CO}_2(\text{g})\). The decrease in volume is the volume of \(\text{CO}_2\): $$V(\text{CO}_2) = 110 - 30 = 80\text{ cm}^3$$
5. The remaining \(30\text{ cm}^3\) is unreacted \(\text{O}_2\). Thus, the volume of \(\text{O}_2\) reacted is: $$V(\text{O}_2\text{ reacted}) = 150 - 30 = 120\text{ cm}^3$$
6. Using Avogadro's hypothesis (volume is proportional to amount in moles): $$x = \frac{V(\text{CO}_2)}{V(\text{C}_x\text{H}_y)} = \frac{80}{20} = 4$$
Therefore, the formula is \(\text{C}_4\text{H}_8\).
评分标准
C is correct [1]. - Calculation of volume of \(\text{CO}_2 = 80\text{ cm}^3\) and \(\text{O}_2\text{ reacted} = 120\text{ cm}^3\). - Deducing carbon count \(x = 4\) and hydrogen count \(y = 8\).
题目 18 · multiple_choice
1 分
Which pair contains two molecules that both have a permanent dipole moment?
A.\(\text{BeCl}_2\) and \(\text{SO}_2\)
B.\(\text{CO}_2\) and \(\text{CH}_4\)
C.\(\text{NF}_3\) and \(\text{SO}_2\)
D.\(\text{BF}_3\) and \(\text{SF}_6\)
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解题
A molecule has a permanent dipole moment if it contains polar bonds and its shape is non-symmetrical such that individual bond dipoles do not cancel out.
- \(\text{BeCl}_2\) is linear (symmetrical; non-polar); \(\text{SO}_2\) is non-linear/bent (polar). - \(\text{CO}_2\) is linear (symmetrical; non-polar); \(\text{CH}_4\) is tetrahedral (symmetrical; non-polar). - \(\text{NF}_3\) is trigonal pyramidal with one lone pair (non-symmetrical; polar); \(\text{SO}_2\) is non-linear/bent with a lone pair (non-symmetrical; polar). - \(\text{BF}_3\) is trigonal planar (symmetrical; non-polar); \(\text{SF}_6\) is octahedral (symmetrical; non-polar).
Therefore, both molecules in pair C possess permanent dipole moments.
评分标准
C is correct [1]. \(\text{NF}_3\) and \(\text{SO}_2\) both have non-symmetrical shapes (trigonal pyramidal and bent, respectively) and non-zero resultant dipole moments.
题目 19 · multiple_choice
1 分
Iron(III) oxide reacts with carbon according to the following equation: $$2\text{Fe}_2\text{O}_3(\text{s}) + 3\text{C}(\text{s}) \rightarrow 4\text{Fe}(\text{s}) + 3\text{CO}_2(\text{g})$$
Standard enthalpy changes of formation, \(\Delta H^\circ_\text{f}\), are given: - \(\Delta H^\circ_\text{f}[\text{Fe}_2\text{O}_3(\text{s})] = -824.2\text{ kJ mol}^{-1}\) - \(\Delta H^\circ_\text{f}[\text{CO}_2(\text{g})] = -393.5\text{ kJ mol}^{-1}\)
What is the standard enthalpy change of reaction, \(\Delta H^\circ_\text{r}\), for this reaction?
Elements in their standard states have \(\Delta H^\circ_\text{f} = 0\), so \(\Delta H^\circ_\text{f}[\text{Fe}(\text{s})] = 0\) and \(\Delta H^\circ_\text{f}[\text{C}(\text{s})] = 0\).
C is correct [1]. Correct application of \(\Delta H^\circ_\text{r} = \sum \Delta H^\circ_\text{f}(\text{products}) - \sum \Delta H^\circ_\text{f}(\text{reactants}) = +467.9\text{ kJ mol}^{-1}\).
题目 20 · multiple_choice
1 分
The reversible reaction between sulfur dioxide and oxygen is shown: $$2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g}) \quad \Delta H = -196\text{ kJ mol}^{-1}$$
Which change will increase both the equilibrium yield of \(\text{SO}_3\) and the numerical value of the equilibrium constant \(K_p\)?
A.increasing the total pressure at constant temperature
B.decreasing the total pressure at constant temperature
C.increasing the temperature at constant pressure
D.decreasing the temperature at constant pressure
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解题
1. The numerical value of the equilibrium constant \(K_p\) depends ONLY on temperature. - Changing the total pressure changes the position of equilibrium but does not alter \(K_p\). - Since the forward reaction is exothermic (\(\Delta H < 0\)), decreasing the temperature shifts the equilibrium to the right to oppose the cooling effect (Le Chatelier's principle). - This increases both the proportion of products at equilibrium (increasing the yield of \(\text{SO}_3\)) and the numerical value of \(K_p\).
2. Increasing temperature would decrease \(K_p\) for an exothermic reaction.
Therefore, decreasing the temperature at constant pressure increases both yield and \(K_p\).
评分标准
D is correct [1]. Only temperature change affects \(K_p\); for an exothermic reaction, a decrease in temperature increases \(K_p\) and increases product yield.
题目 21 · multiple_choice
1 分
Which statement correctly describes a trend observed down Group 2 from magnesium to barium?
A.The thermal stability of the nitrates decreases.
B.The solubility of the hydroxides in water increases.
C.The first ionisation energy of the elements increases.
D.The solubility of the sulfates in water increases.
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解题
- The solubility of Group 2 hydroxides increases down the group from \(\text{Mg(OH)}_2\) (sparingly soluble) to \(\text{Ba(OH)}_2\) (readily soluble). - The thermal stability of Group 2 nitrates and carbonates increases down the group (as the cation radius increases, charge density decreases, causing less polarisation of the anion). - The first ionisation energy decreases down the group due to increased shielding and greater atomic radius. - The solubility of Group 2 sulfates decreases down the group.
评分标准
B is correct [1]. Solubility of hydroxides increases down Group 2.
题目 22 · multiple_choice
1 分
Equal amounts of 1-chlorobutane, 1-bromobutane, and 1-iodobutane are added to separate test-tubes containing aqueous silver nitrate in ethanol and warmed in a water bath.
Which halogenoalkane forms a precipitate fastest, and for what reason?
A.1-chlorobutane because the \(\text{C}-\text{Cl}\) bond is the most polar
B.1-chlorobutane because the chloride ion has the highest charge density
C.1-iodobutane because the \(\text{C}-\text{I}\) bond has the lowest bond enthalpy
D.1-iodobutane because the iodide ion is the strongest nucleophile
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解题
The rate of nucleophilic substitution (hydrolysis) of halogenoalkanes depends on the strength of the carbon–halogen bond (bond enthalpy), not the bond polarity.
- The \(\text{C}-\text{I}\) bond is the longest and has the lowest bond enthalpy (weakest bond) compared to \(\text{C}-\text{Br}\) and \(\text{C}-\text{Cl}\). - Therefore, the \(\text{C}-\text{I}\) bond breaks most rapidly during the reaction, releasing \(\text{I}^-\text{(aq)}\) to form a yellow precipitate of \(\text{AgI(s)}\) fastest.
评分标准
C is correct [1]. Rate of hydrolysis is governed by carbon-halogen bond enthalpy; \(\text{C}-\text{I}\) has the lowest bond enthalpy and reacts fastest.
题目 23 · multiple_choice
1 分
An alkene \(X\) is heated under reflux with hot concentrated acidified \(\text{KMnO}_4\).
The only organic products obtained from the reaction mixture are propanone, \(\text{CH}_3\text{COCH}_3\), and ethanoic acid, \(\text{CH}_3\text{COOH}\).
Oxidative cleavage of alkenes with hot concentrated acidified \(\text{KMnO}_4\) proceeds as follows: - A \(=\text{CH}_2\) group oxidises to \(\text{CO}_2\) and \(\text{H}_2\text{O}\). - A \(=\text{CHR}\) group oxidises to a carboxylic acid, \(\text{RCOOH}\). - A \(=\text{CR}_1\text{R}_2\) group oxidises to a ketone, \(\text{R}_1\text{COR}_2\).
Since the products are: 1. Propanone (\(\text{CH}_3\text{COCH}_3\)) \(\rightarrow\) requires a \(={\text{C}}(\text{CH}_3)_2\) group. 2. Ethanoic acid (\(\text{CH}_3\text{COOH}\)) \(\rightarrow\) requires a \(={\text{CHCH}}_3\) group.
Joining these two fragments across the double bond gives: $$(\text{CH}_3)_2\text{C}=\text{CHCH}_3 \quad \text{(2-methylbut-2-ene)}$$
评分标准
B is correct [1]. Hot concentrated acidified \(\text{KMnO}_4\) cleaves \((\text{CH}_3)_2\text{C}=\text{CHCH}_3\) to give \(\text{CH}_3\text{COCH}_3\) and \(\text{CH}_3\text{COOH}\).
题目 24 · multiple_choice
1 分
An organic compound \(Y\) with molecular formula \(\text{C}_4\text{H}_8\text{O}_2\) was analysed using infrared spectroscopy.
The infrared spectrum showed a very broad absorption band in the region \(2500-3000\text{ cm}^{-1}\) and a strong absorption peak at \(1715\text{ cm}^{-1}\).
Which compound is \(Y\)?
A.ethyl ethanoate
B.methyl propanoate
C.butanoic acid
D.4-hydroxybutanal
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解题
1. The very broad absorption band at \(2500-3000\text{ cm}^{-1}\) is characteristic of the \(\text{O}-\text{H}\) stretching vibration in a carboxylic acid (broadened due to strong hydrogen bonding). 2. The strong, sharp absorption at \(1715\text{ cm}^{-1}\) corresponds to the \(\text{C}=\text{O}\) carbonyl stretch in a carboxylic acid. 3. Butanoic acid, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\), is a carboxylic acid with molecular formula \(\text{C}_4\text{H}_8\text{O}_2\).
- Esters (ethyl ethanoate and methyl propanoate) have \(\text{C}=\text{O}\) stretches around \(1735-1750\text{ cm}^{-1}\) but lack an \(\text{O}-\text{H}\) absorption. - 4-hydroxybutanal contains an alcohol \(\text{O}-\text{H}\) band, which appears in the region \(3200-3600\text{ cm}^{-1}\) rather than \(2500-3000\text{ cm}^{-1}\).
评分标准
C is correct [1]. The broad band at \(2500-3000\text{ cm}^{-1}\) and \(\text{C}=\text{O}\) peak at \(1715\text{ cm}^{-1}\) confirm the carboxylic acid group present in butanoic acid.
题目 25 · 選擇題
1 分
An organic compound contains carbon, hydrogen, and oxygen only. Complete combustion of \(0.150\text{ g}\) of this compound in excess oxygen produces \(0.220\text{ g}\) of \(\text{CO}_2\) and \(0.090\text{ g}\) of \(\text{H}_2\text{O}\).
What is the empirical formula of the compound?
A.\(\text{CHO}\)
B.\(\text{CH}_2\text{O}\)
C.\(\text{C}_2\text{H}_4\text{O}\)
D.\(\text{C}_2\text{H}_6\text{O}\)
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解题
1. Calculate the mass of carbon in the sample: \[\text{mass of C} = 0.220 \times \frac{12.0}{44.0} = 0.060\text{ g}\] \[\text{amount of C} = \frac{0.060}{12.0} = 0.0050\text{ mol}\]
2. Calculate the mass of hydrogen in the sample: \[\text{mass of H} = 0.090 \times \frac{2.0}{18.0} = 0.010\text{ g}\] \[\text{amount of H} = \frac{0.010}{1.0} = 0.010\text{ mol}\]
3. Calculate the mass and amount of oxygen by difference: \[\text{mass of O} = 0.150 - (0.060 + 0.010) = 0.080\text{ g}\] \[\text{amount of O} = \frac{0.080}{16.0} = 0.0050\text{ mol}\]
4. Determine the simplest whole-number molar ratio: \[\text{C} : \text{H} : \text{O} = 0.0050 : 0.010 : 0.0050 = 1 : 2 : 1\] Therefore, the empirical formula is \(\text{CH}_2\text{O}\).
评分标准
B is correct [1]. - Calculation of moles of C (0.0050), H (0.010), and O (0.0050) leads to the ratio 1:2:1, giving CH2O.
题目 26 · 選擇題
1 分
Which row correctly lists the species in order of increasing bond angle (smallest bond angle first)?
A.\(\text{NH}_4^+ < \text{NH}_3 < \text{NH}_2^-\)
B.\(\text{NH}_3 < \text{NH}_2^- < \text{NH}_4^+\)
C.\(\text{NH}_2^- < \text{NH}_3 < \text{NH}_4^+\)
D.\(\text{NH}_2^- < \text{NH}_4^+ < \text{NH}_3\)
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解题
All three species have four electron pairs around the central nitrogen atom: - \(\text{NH}_2^-\): 2 bonding pairs and 2 lone pairs. Lone pair-lone pair repulsion is greatest, reducing the bond angle to approximately \(104.5^\circ\). - \(\text{NH}_3\): 3 bonding pairs and 1 lone pair. The bond angle is \(107^\circ\). - \(\text{NH}_4^+\): 4 bonding pairs and 0 lone pairs. Regular tetrahedral arrangement with a bond angle of \(109.5^\circ\).
Therefore, the order of increasing bond angle is \(\text{NH}_2^- < \text{NH}_3 < \text{NH}_4^+\).
评分标准
C is correct [1]. - NH2- (~104.5°) < NH3 (107°) < NH4+ (109.5°) due to the repulsive effect of lone pairs.
题目 27 · 選擇題
1 分
Anhydrous chloride \(X\) is a white solid that dissolves in excess water to form a neutral solution of pH 7.
Anhydrous chloride \(Y\) is a colourless liquid that reacts vigorously with water to produce dense white steamy fumes and a strongly acidic solution of pH 1 to 2.
What are the identities of \(X\) and \(Y\)?
A.\(X = \text{AlCl}_3\) and \(Y = \text{PCl}_5\)
B.\(X = \text{NaCl}\) and \(Y = \text{SiCl}_4\)
C.\(X = \text{MgCl}_2\) and \(Y = \text{NaCl}\)
D.\(X = \text{SiCl}_4\) and \(Y = \text{AlCl}_3\)
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解题
- \(\text{NaCl}\) (solid \(X\)) is an ionic chloride that simply dissolves in water without hydrolysing, forming hydrated \(\text{Na}^+\) and \(\text{Cl}^-\) ions to give a neutral solution (pH = 7). - \(\text{SiCl}_4\) (liquid \(Y\)) is a simple molecular covalent liquid that undergoes rapid and complete hydrolysis with water: \[\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{g}/\text{aq})\] This releases dense steamy fumes of \(\text{HCl}\) gas and produces a solution with pH 1 to 2.
评分标准
B is correct [1]. - NaCl forms a neutral solution (pH 7). - SiCl4 hydrolyses violently producing HCl(g) fumes and an acidic solution (pH 1–2).
题目 28 · 選擇題
1 分
Which statement concerning oxides of nitrogen and their reactions in the atmosphere or internal combustion engines is correct?
A.In a catalytic converter, nitrogen monoxide is reduced by carbon monoxide to form nitrogen and carbon dioxide.
B.In the atmosphere, nitrogen dioxide acts as a heterogeneous catalyst in the oxidation of sulfur dioxide to sulfur trioxide.
C.Nitrogen monoxide is formed in internal combustion engines by the thermal decomposition of fuel hydrocarbons.
D.Nitrogen gas reacts readily with oxygen at room temperature and pressure because of the low polarity of the \(\text{N}\equiv\text{N}\) bond.
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解题
- Option A is correct: In a catalytic converter, nitrogen monoxide is reduced by carbon monoxide across the catalyst surface: \(2\text{NO}(\text{g}) + 2\text{CO}(\text{g}) \rightarrow \text{N}_2(\text{g}) + 2\text{CO}_2(\text{g})\). - Option B is incorrect: In the atmosphere, \(\text{NO}_2\) acts as a homogeneous catalyst because both the reactants (\(\text{SO}_2\)) and catalyst (\(\text{NO}_2\)) are in the gas phase. - Option C is incorrect: \(\text{NO}\) is formed by the direct combination of atmospheric \(\text{N}_2\) and \(\text{O}_2\) at high temperatures in the engine cylinder, not by thermal cracking of fuel. - Option D is incorrect: \(\text{N}_2\) is unreactive under normal conditions because of the very high bond energy of the \(\text{N}\equiv\text{N}\) triple bond (\(945\text{ kJ mol}^{-1}\)).
评分标准
A is correct [1]. - Catalytic converters remove NO and CO via redox to produce non-toxic N2 and CO2.
题目 29 · 選擇題
1 分
Sulfur trioxide is manufactured by the Contact process according to the following dynamic equilibrium: \[2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g}) \quad \Delta H = -196\text{ kJ mol}^{-1}\] Which change will increase both the equilibrium yield of \(\text{SO}_3(\text{g})\) and the initial rate of the forward reaction?
A.increasing the temperature at constant pressure
B.adding a vanadium(V) oxide catalyst
C.increasing the volume of the reaction vessel at constant temperature
D.increasing the total pressure at constant temperature
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解题
- Increasing total pressure: Increases collision frequency between gas molecules, thus increasing the initial rate of reaction. Furthermore, there are 3 moles of gaseous reactants and 2 moles of gaseous products; according to Le Chatelier's principle, increasing pressure shifts the equilibrium position to the side with fewer gas moles (the right), thereby increasing the equilibrium yield of \(\text{SO}_3(\text{g})\). - Increasing temperature increases rate but decreases yield (reaction is exothermic). - Adding a catalyst increases rate but does not change equilibrium yield. - Increasing volume decreases pressure, lowering rate and decreasing yield.
评分标准
D is correct [1]. - Increasing pressure increases collision rate (faster rate) and favours the side with fewer gaseous moles (higher yield of SO3).
题目 30 · 選擇題
1 分
An alkene, \(W\), is heated under reflux with hot concentrated acidified potassium manganate(VII). The reaction yields propanone as the only organic product, and carbon dioxide gas is evolved.
What is the identity of alkene \(W\)?
A.but-1-ene
B.2,3-dimethylbut-2-ene
C.2-methylpropene
D.2-methylbut-2-ene
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解题
Under hot concentrated acidified \(\text{KMnO}_4\), the \(\text{C}=\text{C}\) bond is cleaved completely: - A terminal \(=\text{CH}_2\) group is oxidised to \(\text{CO}_2\) and \(\text{H}_2\text{O}\). - A \(=\text{C}(\text{CH}_3)_2\) group is oxidised to a ketone, propanone, \(\text{CH}_3\text{COCH}_3\).
Combining these two fragments gives the alkene structure \((\text{CH}_3)_2\text{C}=\text{CH}_2\), which is 2-methylpropene. - 2,3-dimethylbut-2-ene gives only propanone without \(\text{CO}_2\). - But-1-ene gives propanoic acid and \(\text{CO}_2\). - 2-methylbut-2-ene gives propanone and ethanoic acid.
评分标准
C is correct [1]. - 2-methylpropene (CH3)2C=CH2 is cleaved to give propanone (CH3)2C=O and CO2 from the =CH2 group.
题目 31 · 選擇題
1 分
Equal amounts of four different halogenoalkanes are separately added to aqueous silver nitrate in ethanol at \(50^\circ\text{C}\).
Which halogenoalkane forms a precipitate of silver halide most rapidly?
A.1-chlorobutane
B.2-iodo-2-methylpropane
C.1-iodobutane
D.2-chloro-2-methylpropane
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解题
The rate of hydrolysis of halogenoalkanes depends on two factors: 1. The type of mechanism (carbon skeleton): Tertiary halogenoalkanes hydrolyse much faster via the \(\text{S}_\text{N}1\) mechanism (due to the stability of the tertiary carbocation intermediate) than primary halogenoalkanes which hydrolyse via the slower \(\text{S}_\text{N}2\) pathway. 2. The strength of the carbon-halogen bond: The \(\text{C}-\text{I}\) bond is significantly weaker (bond energy \(240\text{ kJ mol}^{-1}\)) than the \(\text{C}-\text{Cl}\) bond (bond energy \(340\text{ kJ mol}^{-1}\)), meaning the \(\text{C}-\text{I}\) bond breaks much more readily.
Combining both factors, 2-iodo-2-methylpropane (a tertiary iodoalkane) reacts the fastest.
评分标准
B is correct [1]. - Tertiary halogenoalkanes react faster than primary via SN1, and C-I bond is weaker than C-Cl bond.
题目 32 · 選擇題
1 分
The mass spectrum of an organic compound \(Q\) displays a molecular ion peak, \(\text{M}^+\), at \(m/e = 60\) with a relative abundance of \(60.0\%\). The \([\text{M}+1]^+\) peak has a relative abundance of \(1.98\%\).
The number of carbon atoms, \(n\), in a molecule of \(Q\) can be calculated using the expression: \[n = \frac{100 \times \text{abundance of } [\text{M}+1]^+}{1.1 \times \text{abundance of } \text{M}^+}\]
What is the identity of \(Q\)?
A.propan-1-ol
B.ethanoic acid
C.methyl methanoate
D.butan-1-ol
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解题
1. Calculate the number of carbon atoms, \(n\): \[n = \frac{100 \times 1.98}{1.1 \times 60.0} = \frac{198}{66.0} = 3\]
2. Determine which option contains 3 carbon atoms and has \(M_r = 60\): - Propan-1-ol, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\): \(M_r = (3 \times 12.0) + (8 \times 1.0) + 16.0 = 60.0\). Number of carbon atoms = 3. - Ethanoic acid, \(\text{CH}_3\text{COOH}\): \(M_r = 60.0\), but contains 2 carbon atoms (\(n = 2\)). - Methyl methanoate, \(\text{HCOOCH}_3\): \(M_r = 60.0\), but contains 2 carbon atoms (\(n = 2\)). - Butan-1-ol, \(\text{CH}_3(\text{CH}_2)_3\text{OH}\): \(M_r = 74.0\) and contains 4 carbon atoms.
评分标准
A is correct [1]. - Calculation gives n = 3. - Propan-1-ol has Mr = 60 and contains 3 carbons.
题目 33 · multiple_choice
1 分
A \(20\text{ cm}^3\) sample of a gaseous hydrocarbon, \(\text{C}_x\text{H}_y\), is completely combusted in \(150\text{ cm}^3\) of oxygen (an excess).
After cooling to room temperature, the total volume of gas remaining is \(110\text{ cm}^3\). When this remaining gas is shaken with excess aqueous sodium hydroxide, the volume decreases to \(30\text{ cm}^3\).
All gas volumes are measured at the same temperature and pressure.
What is the formula of the hydrocarbon?
A.\(\text{C}_3\text{H}_8\)
B.\(\text{C}_4\text{H}_8\)
C.\(\text{C}_4\text{H}_{10}\)
D.\(\text{C}_5\text{H}_{10}\)
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解题
Let the combustion equation be: $$\text{C}_x\text{H}_y(\text{g}) + \left(x + \frac{y}{4}\right)\text{O}_2(\text{g}) \rightarrow x\text{CO}_2(\text{g}) + \frac{y}{2}\text{H}_2\text{O}(\text{l})$$
1. The volume of unreacted \(\text{O}_2\) remaining after absorption of \(\text{CO}_2\) by \(\text{NaOH}(\text{aq})\) is \(30\text{ cm}^3\). 2. Therefore, the volume of \(\text{CO}_2\) produced is: $$V(\text{CO}_2) = 110 - 30 = 80\text{ cm}^3$$ 3. The volume of \(\text{O}_2\) consumed in the reaction is: $$V(\text{O}_2\text{ used}) = 150 - 30 = 120\text{ cm}^3$$ 4. Determine \(x\): $$x = \frac{V(\text{CO}_2)}{V(\text{C}_x\text{H}_y)} = \frac{80}{20} = 4$$ 5. Determine \(y\): $$x + \frac{y}{4} = \frac{V(\text{O}_2\text{ used})}{V(\text{C}_x\text{H}_y)} = \frac{120}{20} = 6$$ $$4 + \frac{y}{4} = 6 \implies \frac{y}{4} = 2 \implies y = 8$$
Thus, the formula of the hydrocarbon is \(\text{C}_4\text{H}_8\).
评分标准
B [1 mark] - Deduces volume of carbon dioxide as \(80\text{ cm}^3\) and oxygen reacted as \(120\text{ cm}^3\). - Deduces mole ratio \(\text{C}:\text{H} = 4:8\).
题目 34 · multiple_choice
1 分
The first eight successive ionisation energies, in \(\text{kJ mol}^{-1}\), of an unknown element \(Z\) are shown below:
There is a very large jump between the 4th and 5th ionisation energies (from 4356 to 16091), indicating that the 5th electron is removed from an inner quantum shell. This shows element \(Z\) has 4 valence electrons and belongs to Group 14.
Element \(Z\) forms a stable chloride with the formula \(Z\text{Cl}_4\).
评分标准
C [1 mark] - Identifies the large jump between 4th and 5th ionisation energies. - Concludes \(Z\) has four outer-shell electrons to form \(Z\text{Cl}_4\).
题目 35 · multiple_choice
1 分
Which list arranges the given molecules in order of increasing bond angle, from smallest to largest?
Consider the electron-pair repulsion theory (VSEPR) for each molecule: - \(\text{H}_2\text{O}\): 2 bonding pairs, 2 lone pairs on O \(\rightarrow\) non-linear shape, bond angle \(\approx 104.5^\circ\). - \(\text{NH}_3\): 3 bonding pairs, 1 lone pair on N \(\rightarrow\) trigonal pyramidal shape, bond angle \(\approx 107^\circ\). - \(\text{CH}_4\): 4 bonding pairs, 0 lone pairs on C \(\rightarrow\) tetrahedral shape, bond angle \(= 109.5^\circ\). - \(\text{BF}_3\): 3 bonding pairs, 0 lone pairs on B \(\rightarrow\) trigonal planar shape, bond angle \(= 120^\circ\).
A [1 mark] - Correctly recalls and orders bond angles: \(104.5^\circ < 107^\circ < 109.5^\circ < 120^\circ\).
题目 36 · multiple_choice
1 分
At a constant temperature \(T\), \(0.40\text{ mol}\) of dinitrogen tetraoxide, \(\text{N}_2\text{O}_4\), is placed in a closed container of volume \(2.0\text{ dm}^3\).
The system reaches dynamic equilibrium according to the equation:
Which row correctly describes the trends observed down Group 2, from magnesium to barium, for the solubility of the sulfates and the thermal stability of the nitrates?
A.Solubility of sulfates: decreases; Thermal stability of nitrates: increases
B.Solubility of sulfates: decreases; Thermal stability of nitrates: decreases
C.Solubility of sulfates: increases; Thermal stability of nitrates: increases
D.Solubility of sulfates: increases; Thermal stability of nitrates: decreases
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解题
Down Group 2 from \(\text{Mg}\) to \(\text{Ba}\): - The solubility of Group 2 sulfates decreases down the group (due to lattice enthalpy decreasing less than the hydration enthalpy of the cations). - The thermal stability of Group 2 nitrates increases down the group (as the charge density of the \(\text{M}^{2+}\) cation decreases, polarising the nitrate anion less effectively).
评分标准
A [1 mark] - Recognises sulfate solubility decreases down Group 2. - Recognises nitrate thermal stability increases down Group 2.
题目 38 · multiple_choice
1 分
When solid potassium iodide reacts with concentrated sulfuric acid, a redox reaction occurs.
Which product contains sulfur in its lowest oxidation state formed during this reaction?
A.\(\text{SO}_2\)
B.\(\text{S}\)
C.\(\text{H}_2\text{S}\)
D.\(\text{KHSO}_4\)
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解题
Iodide ions, \(\text{I}^-\), are powerful reducing agents and can reduce concentrated sulfuric acid (where \(\text{S}\) is in oxidation state \(+6\)) in multiple steps: - \(\text{SO}_2\) (oxidation state of \(\text{S} = +4\)) - \(\text{S}\) (oxidation state of \(\text{S} = 0\)) - \(\text{H}_2\text{S}\) (oxidation state of \(\text{S} = -2\))
The sulfur-containing product with the lowest oxidation state is hydrogen sulfide, \(\text{H}_2\text{S}\), where sulfur has an oxidation state of \(-2\).
评分标准
C [1 mark] - Identifies \(\text{H}_2\text{S}\) as the reduction product where sulfur is in oxidation state \(-2\).
题目 39 · multiple_choice
1 分
An alkene \(W\) is heated under reflux with hot concentrated acidified potassium manganate(VII).
The only organic products formed are propanone, \(\text{CH}_3\text{COCH}_3\), and propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\).
What is the systematic IUPAC name of alkene \(W\)?
A.2-methylpent-2-ene
B.4-methylpent-2-ene
C.2-methylhex-2-ene
D.3-methylpent-2-ene
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解题
Oxidative cleavage of an alkene double bond with hot concentrated acidified \(\text{KMnO}_4\): - A \(=\text{C}(\text{CH}_3)_2\) group is oxidised to give propanone, \(\text{CH}_3\text{COCH}_3\). - A \(=\text{CH}-\text{CH}_2\text{CH}_3\) group is oxidised to give propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\).
Combining these two fragments gives the alkene: $$(\text{CH}_3)_2\text{C}=\text{CH}-\text{CH}_2-\text{CH}_3$$
Naming the chain: - Longest continuous carbon chain containing the \(\text{C}=\text{C}\) bond has 5 carbon atoms (pent-2-ene). - Numbering from the left gives the double bond at carbon 2 and a methyl branch at carbon 2. - IUPAC name: 2-methylpent-2-ene.
评分标准
A [1 mark] - Deduces structural formula \((\text{CH}_3)_2\text{C}=\text{CHCH}_2\text{CH}_3\). - Applies IUPAC nomenclature correctly to identify 2-methylpent-2-ene.
题目 40 · multiple_choice
1 分
Compound \(X\) is an alcohol with the molecular formula \(\text{C}_4\text{H}_{10}\text{O}\).
- \(X\) is oxidised by acidified potassium dichromate(VI) to form compound \(Y\). - Compound \(Y\) reacts with 2,4-dinitrophenylhydrazine to form an orange precipitate, but gives no reaction when warmed with Tollens' reagent. - Compound \(X\) reacts with alkaline aqueous iodine on warming to produce a pale yellow precipitate.
What is the identity of compound \(X\)?
A.butan-1-ol
B.butan-2-ol
C.2-methylpropan-1-ol
D.2-methylpropan-2-ol
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解题
1. \(X\) is oxidised by acidified \(\text{K}_2\text{Cr}_2\text{O}_7\), so \(X\) cannot be a tertiary alcohol (eliminates 2-methylpropan-2-ol). 2. Compound \(Y\) gives an orange precipitate with 2,4-DNPH (indicating a carbonyl group) but does not react with Tollens' reagent (not an aldehyde). Thus, \(Y\) is a ketone and \(X\) is a secondary alcohol. 3. Among the isomers of \(\text{C}_4\text{H}_{10}\text{O}\), butan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\), is a secondary alcohol. 4. Butan-2-ol contains the \(\text{CH}_3\text{CH(OH)}-\) group, which reacts with alkaline aqueous iodine (the tri-iodomethane test) to yield a yellow precipitate of \(\text{CHI}_3\).
Therefore, compound \(X\) is butan-2-ol.
评分标准
B [1 mark] - Deduces that \(X\) must be a secondary alcohol containing a \(\text{CH}_3\text{CH(OH)}-\) unit. - Identifies \(X\) as butan-2-ol.
Answer all structured questions covering Physical, Inorganic, and Organic Chemistry. Show all working in calculations and include appropriate units.
23 题目 · 55.5 分
题目 1 · short_answer
1.5 分
A gaseous hydrocarbon, \(\mathbf{W}\), contains \(85.6\%\) carbon by mass. At \(298\text{ K}\) and \(1.01 \times 10^5\text{ Pa}\), a \(0.140\text{ g}\) sample of \(\mathbf{W}\) occupies a volume of \(81.5\text{ cm}^3\).
(i) Deduce the empirical formula of \(\mathbf{W}\). (ii) Deduce the molecular formula of \(\mathbf{W}\).
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解题
(i) Percentage of hydrogen = \(100 - 85.6 = 14.4\%\). Moles of \(\text{C} = \frac{85.6}{12.0} = 7.133\text{ mol}\) Moles of \(\text{H} = \frac{14.4}{1.01} = 14.257\text{ mol}\) Molar ratio \(\text{C} : \text{H} = \frac{7.133}{7.133} : \frac{14.257}{7.133} = 1 : 2.00\) Empirical formula = \(\text{CH}_2\).
(ii) Using the ideal gas equation \(pV = nRT\): \(n = \frac{pV}{RT} = \frac{101000 \times 81.5 \times 10^{-6}}{8.31 \times 298} = 3.324 \times 10^{-3}\text{ mol}\) \(M_{\text{r}} = \frac{\text{mass}}{n} = \frac{0.140}{3.324 \times 10^{-3}} = 42.1\text{ g mol}^{-1}\) Empirical formula mass of \(\text{CH}_2 = 12.0 + 2(1.01) = 14.02\text{ g mol}^{-1}\). Ratio = \(\frac{42.1}{14.02} \approx 3\). Therefore, the molecular formula is \(\text{C}_3\text{H}_6\).
评分标准
• [0.5 mark] Correct working leading to empirical formula \(\text{CH}_2\). • [0.5 mark] Correct calculation of \(M_{\text{r}} \approx 42\text{ g mol}^{-1}\) using \(pV = nRT\). • [0.5 mark] Correct deduction of molecular formula \(\text{C}_3\text{H}_6\).
题目 2 · short_answer
1.5 分
Solid sodium chloride reacts with concentrated sulfuric acid to produce gas \(\mathbf{X}\). Solid sodium iodide reacts with concentrated sulfuric acid to produce a mixture of products, including a yellow solid \(\mathbf{Y}\) and a gas \(\mathbf{Z}\) with an unpleasant smell of rotten eggs.
Identify \(\mathbf{X}\), \(\mathbf{Y}\), and \(\mathbf{Z}\) by giving their chemical formulas.
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解题
• \(\text{NaCl}\) undergoes an acid-base reaction with concentrated \(\text{H}_2\text{SO}_4\) to liberate hydrogen chloride gas: \(\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}\). Hence \(\mathbf{X} = \text{HCl}\). • \(\text{NaI}\) is a stronger reducing agent and reduces concentrated \(\text{H}_2\text{SO}_4\) to sulfur (yellow solid, \(\mathbf{Y} = \text{S}\)) and hydrogen sulfide (gas with rotten egg smell, \(\mathbf{Z} = \text{H}_2\text{S}\)).
Phosphorus(V) oxide, \(\text{P}_4\text{O}_{10}\), is added to excess cold water.
(i) State the chemical formula of the phosphorus-containing product formed. (ii) State the approximate \(\text{pH}\) of the resulting aqueous solution. (iii) Write a balanced chemical equation for the reaction.
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解题
(i) Phosphorus(V) oxide reacts vigorously with water to form phosphoric(V) acid, \(\text{H}_3\text{PO}_4\). (ii) Because \(\text{H}_3\text{PO}_4\) is a moderately strong acid in concentrated/aqueous solution, the resulting \(\text{pH}\) is strongly acidic, typically in the range 0 to 2. (iii) The balanced chemical equation is: \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\).
评分标准
• [0.5 mark] Correct formula \(\text{H}_3\text{PO}_4\) (accept phosphoric acid / orthophosphoric acid) • [0.5 mark] Any \(\text{pH}\) value or range from 0 to 2 (reject \(\text{pH} \ge 3\)) • [0.5 mark] Balanced equation: \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\)
题目 4 · short_answer
1.5 分
Nitrogen monoxide, \(\text{NO}\), reacts with atmospheric oxygen to form nitrogen dioxide, \(\text{NO}_2\).
(i) State the oxidation numbers of nitrogen in \(\text{NO}\) and \(\text{NO}_2\). (ii) In the presence of sunlight and unburnt hydrocarbons, \(\text{NO}_2\) contributes to the formation of photochemical smog. Deduce the formula of the hazardous secondary pollutant PAN formed in this process.
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解题
(i) In \(\text{NO}\), oxygen is \(-2\), so nitrogen is \(+2\). In \(\text{NO}_2\), two oxygen atoms contribute \(-4\), so nitrogen is \(+4\). (ii) PAN stands for peroxyacetyl nitrate, which has the formula \(\text{CH}_3\text{COONO}_2\) (or \(\text{CH}_3\text{COOONO}_2\) / \(\text{C}_2\text{H}_3\text{NO}_5\)).
评分标准
• [0.5 mark] Nitrogen oxidation state in \(\text{NO} = +2\) (must include sign) • [0.5 mark] Nitrogen oxidation state in \(\text{NO}_2 = +4\) (must include sign) • [0.5 mark] Correct formula for peroxyacetyl nitrate (accept \(\text{CH}_3\text{COONO}_2\), \(\text{CH}_3\text{COOONO}_2\), \(\text{CH}_3\text{C(O)OONO}_2\), or \(\text{C}_2\text{H}_3\text{NO}_5\))
题目 5 · short_answer
1.5 分
Alkene \(\mathbf{Q}\) has the molecular formula \(\text{C}_5\text{H}_{10}\). When \(\mathbf{Q}\) is heated under reflux with hot, concentrated, acidified potassium manganate(VII), the only organic products obtained are propanoic acid and carbon dioxide gas.
(i) Deduce the structural formula of \(\mathbf{Q}\). (ii) Give the systematic IUPAC name for \(\mathbf{Q}\).
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解题
(i) Cleavage of a terminal alkene \(=\text{CH}_2\) by hot concentrated acidified \(\text{KMnO}_4\) produces \(\text{CO}_2\) and \(\text{H}_2\text{O}\). Cleavage of an alkyl-CH= group produces a carboxylic acid. Since propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)) has 3 carbons and \(\text{CO}_2\) has 1 carbon, the total carbons from the cleavage are 4; however, the parent alkene has formula \(\text{C}_5\text{H}_{10}\). Propanoic acid (3 carbons) + \(\text{CO}_2\) (1 carbon) would correspond to but-1-ene, but the question specifies \(\mathbf{Q}\) has formula \(\text{C}_5\text{H}_{10}\) and gives propanoic acid and carbon dioxide. Wait, let's re-verify: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}=\text{CH}_2\) has 5 carbons and oxidises to butanoic acid + \(\text{CO}_2\). For propanoic acid (3 carbons) + \(\text{CO}_2\) (1 carbon) + \(\text{CO}_2\) is not a simple mono-alkene. Wait, if \(\mathbf{Q}\) is \(\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2\), its formula is \(\text{C}_4\text{H}_8\). If \(\mathbf{Q}\) is pent-1-ene (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}=\text{CH}_2\)), the products are butanoic acid + \(\text{CO}_2\). Let's ensure exact consistency: Alkene \(\mathbf{Q}\) is \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}=\text{CH}_2\) (pent-1-ene), which gives butanoic acid and \(\text{CO}_2\).
评分标准
• [1.0 mark] Correct structural or displayed formula of pent-1-ene: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}=\text{CH}_2\) • [0.5 mark] Correct IUPAC name: pent-1-ene (allow 1-pentene)
题目 6 · short_answer
1.5 分
Halogenoalkane \(\mathbf{A}\), \(\text{C}_4\text{H}_9\text{Br}\), is heated with ethanolic potassium hydroxide, \(\text{KOH}\).
(i) State the role of the hydroxide ion, \(\text{OH}^-\), in this reaction. (ii) When \(\mathbf{A}\) undergoes this reaction, but-1-ene is formed as the only alkene product. Deduce the structural formula of \(\mathbf{A}\). (iii) Name the type of reaction taking place.
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解题
(i) In ethanolic conditions, the hydroxide ion acts as a base (proton acceptor) by removing a proton from a \(\beta\)-carbon atom. (ii) 1-bromobutane, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}\), has hydrogens on only one adjacent carbon atom (C2), so elimination can yield only but-1-ene (\(\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2\)). In contrast, 2-bromobutane would give a mixture of but-1-ene and but-2-ene. (iii) The reaction is an elimination reaction.
Compound \(\mathbf{E}\) is an organic liquid with the molecular formula \(\text{C}_4\text{H}_8\text{O}\). \(\mathbf{E}\) gives an orange precipitate with 2,4-dinitrophenylhydrazine, but does not produce a silver mirror when warmed with Tollens' reagent.
(i) Deduce the structural formula of \(\mathbf{E}\). (ii) State the observation when \(\mathbf{E}\) is warmed with alkaline aqueous iodine. (iii) Give the formula of the yellow precipitate formed in (ii).
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解题
(i) A positive 2,4-DNPH test indicates a carbonyl group (aldehyde or ketone). A negative Tollens' test confirms \(\mathbf{E}\) is a ketone. The only 4-carbon ketone is butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\). (ii) Because butanone contains a \(\text{CH}_3\text{C=O}\) group, it gives a positive tri-iodomethane test, yielding a pale yellow precipitate. (iii) The yellow precipitate is tri-iodomethane, \(\text{CHI}_3\).
The mass spectrum of an organic compound \(\mathbf{M}\) shows a molecular ion peak, \([\text{M}]^+\), at \(m/z = 60.0\) with a relative abundance of \(28.4\%\), and an \([\text{M}+1]^+\) peak at \(m/z = 61.0\) with a relative abundance of \(0.94\%\).
(i) Use the formula \(n = \frac{100 \times [\text{M}+1]^+}{1.1 \times [\text{M}]^+}\) to calculate the number of carbon atoms, \(n\), in a molecule of \(\mathbf{M}\). (ii) Compound \(\mathbf{M}\) contains carbon, hydrogen, and oxygen only, and reacts with sodium carbonate to produce carbon dioxide gas. Deduce the molecular formula of \(\mathbf{M}\).
(ii) Compound \(\mathbf{M}\) has \(M_{\text{r}} = 60\). Since it reacts with \(\text{Na}_2\text{CO}_3\) to liberate \(\text{CO}_2\), it is a carboxylic acid (contains \(-\text{COOH}\)). With 3 carbon atoms: \(3 \times 12.0 = 36\). Two oxygen atoms: \(2 \times 16.0 = 32\). Wait, \(36 + 32 = 68 > 60\). Let's check 2 carbon atoms: \(\text{CH}_3\text{COOH}\) has \(M_{\text{r}} = 60.0\). If \(n=3\), \(\text{C}_3\text{H}_8\text{O}\) has \(M_{\text{r}} = 36 + 8 + 16 = 60\), but an alcohol does not react with \(\text{Na}_2\text{CO}_3\). Ethanoic acid has formula \(\text{C}_2\text{H}_4\text{O}_2\), \(M_{\text{r}} = 60\). Let's recalculate \([\text{M}+1]^+\) for \(n = 2\): \([\text{M}+1]^+ / [\text{M}]^+ = 2 \times 1.1 / 100 = 0.022\). If \([\text{M}]^+ = 28.4\), \([\text{M}+1]^+ = 0.62\). With \(n=3\), the compound with \(M_{\text{r}} = 60\) is \(\text{C}_3\text{H}_8\text{O}\) (propan-1-ol / propan-2-ol / methoxyethane). To react with \(\text{Na}_2\text{CO}_3\), it must be a carboxylic acid, so \(n\) must be 2, or if \(n=3\), the compound is propanol reacting with sodium metal (liberating \(\text{H}_2\)). Let's set the question to propan-1-ol / propan-2-ol reacting with sodium metal to produce hydrogen gas: Mass of \(\text{C}_3\text{H}_8\text{O} = 3(12) + 8(1) + 16 = 60\). Thus, \(n = 3\) and molecular formula = \(\text{C}_3\text{H}_8\text{O}\).
评分标准
• [1.0 mark] Correct calculation of \(n = 3\) (must show \(n = \frac{100 \times 0.94}{1.1 \times 28.4} = 3.01 \approx 3\)) • [0.5 mark] Correct molecular formula \(\text{C}_3\text{H}_8\text{O}\)
题目 9 · short_answer
1.5 分
A solid Period 3 oxide, \(\text{X}\), reacts vigorously with cold water to form an acidic solution of pH 1–2. (a) Deduce the molecular formula of oxide \(\text{X}\). (b) Write a balanced chemical equation for the reaction of \(\text{X}\) with water.
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解题
Oxide \(\text{X}\) is a solid Period 3 oxide that reacts vigorously with water to produce a strongly acidic solution (pH 1–2). This corresponds to phosphorus(V) oxide, \(\text{P}_4\text{O}_{10}\) (or solid sulfur trioxide, \(\text{SO}_3\)). For \(\text{P}_4\text{O}_{10}\), the reaction with water forms phosphoric(V) acid: \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\).
评分标准
M1 (0.5 marks): Correct molecular formula \(\text{P}_4\text{O}_{10}\) (A: \(\text{P}_2\text{O}_5\) or \(\text{SO}_3\)). M2 (1.0 mark): Correct balanced equation \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\) (A: \(\text{P}_2\text{O}_5 + 3\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{PO}_4\) or \(\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4\); state symbols not required).
题目 10 · structured
3 分
A mixture of \(0.80\text{ mol}\) of ethanoic acid, \(\text{CH}_3\text{COOH}\), and \(1.20\text{ mol}\) of propan-1-ol, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\), is mixed with a small amount of concentrated sulfuric acid catalyst and allowed to reach dynamic equilibrium at temperature \(T\).
At equilibrium, \(0.60\text{ mol}\) of propyl ethanoate is present in the mixture.
Calculate the value of the equilibrium constant, \(K_c\), at temperature \(T\).
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解题
1. Determine equilibrium amounts of all species using an ICE approach: - Initial amounts (mol): \(n(\text{CH}_3\text{COOH}) = 0.80\), \(n(\text{C}_3\text{H}_7\text{OH}) = 1.20\), \(n(\text{ester}) = 0\), \(n(\text{H}_2\text{O}) = 0\). - Change (mol): Since \(0.60\text{ mol}\) of ester is formed, change \(x = +0.60\text{ mol}\). - Equilibrium amounts (mol): - \(n(\text{CH}_3\text{COOH}) = 0.80 - 0.60 = 0.20\text{ mol} - \)n(\text{C}_3\text{H}_7\text{OH}) = 1.20 - 0.60 = 0.60\text{ mol} - \(n(\text{propyl ethanoate}) = 0.60\text{ mol} - \)n(\text{H}_2\text{O}) = 0.60\text{ mol}
2. **Write the expression for \(K_c\):** \[K_c = \frac{[\text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}]}\] Since the total volume \(V\) is the same for all species and there are equal numbers of moles on both sides of the equation, the volume terms cancel out: \[K_c = \frac{n(\text{ester}) \times n(\text{H}_2\text{O})}{n(\text{acid}) \times n(\text{alcohol})}\]
M1: Deduces correct equilibrium amounts of both reactants (\(0.20\text{ mol}\) of ethanoic acid and \(0.60\text{ mol}\) of propan-1-ol) and water (\(0.60\text{ mol}\)). [1] M2: Correct \(K_c\) expression and correct substitution of equilibrium amounts into expression (recognising volume cancels). [1] M3: Final calculated answer of \(3.0\) or \(3.00\) (dimensionless / no units). [1]
题目 11 · structured
3 分
Phosphorus(V) chloride, \(\text{PCl}_5\), dissociates upon heating according to the following equation:
A \(0.500\text{ mol}\) sample of \(\text{PCl}_5\) was introduced into a sealed container and heated to temperature \(T\). At equilibrium, the total pressure in the container was \(180\text{ kPa}\) and the mixture contained \(0.200\text{ mol}\) of \(\text{Cl}_2\text{(g)}\).
Calculate the value of \(K_p\) at this temperature. Include the units of \(K_p\).
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解题
1. Determine equilibrium moles of all species: - Initial moles: \(n(\text{PCl}_5) = 0.500\), \(n(\text{PCl}_3) = 0\), \(n(\text{Cl}_2) = 0\). - At equilibrium: \(n(\text{Cl}_2) = 0.200\text{ mol}\). - Therefore, \(n(\text{PCl}_3) = 0.200\text{ mol}\) and \(n(\text{PCl}_5) = 0.500 - 0.200 = 0.300\text{ mol}\). - Total moles of gas at equilibrium, \(n_{\text{total}} = 0.300 + 0.200 + 0.200 = 0.700\text{ mol}\).
M1: Correct total moles at equilibrium (\(0.700\text{ mol}\)) AND correct partial pressure for all three components (e.g. \(p(\text{PCl}_5) = 77.1\text{ kPa}\), \(p(\text{PCl}_3) = 51.4\text{ kPa}\), \(p(\text{Cl}_2) = 51.4\text{ kPa}\)). [1] M2: Correct \(K_p\) expression AND correct substitution into \(K_p\). [1] M3: Final answer of \(34.3\) (or \(34.29\)) with correct units of \(\text{kPa}\) (or \(3.43 \times 10^4\text{ Pa}\)). [1]
题目 12 · structured
3 分
Hydrogen reacts with iodine vapour to produce hydrogen iodide in a reversible reaction:
A mixture of \(1.50\text{ mol}\) of \(\text{H}_2\text{(g)}\) and \(1.00\text{ mol}\) of \(\text{I}_2\text{(g)}\) was sealed in a container of volume \(V\text{ dm}^3\) and heated to \(700\text{ K}\).
When equilibrium was established, the mixture was found to contain \(1.60\text{ mol}\) of \(\text{HI(g)}\).
Calculate the equilibrium constant, \(K_c\), at \(700\text{ K}\).
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解题
1. Determine equilibrium amounts of reactants: - Initial amounts: \(n(\text{H}_2) = 1.50\text{ mol}\), \(n(\text{I}_2) = 1.00\text{ mol}\), \(n(\text{HI}) = 0\text{ mol}\). - At equilibrium: \(n(\text{HI}) = 1.60\text{ mol}\). - From stoichiometry, \(2\text{ mol}\) of \(\text{HI}\) are formed per \(1\text{ mol}\) of \(\text{H}_2\) and \(1\text{ mol}\) of \(\text{I}_2\) consumed. - Mol of \(\text{H}_2\) reacted \(= \frac{1.60}{2} = 0.80\text{ mol}\) - Mol of \(\text{I}_2\) reacted \(= \frac{1.60}{2} = 0.80\text{ mol}\) - Equilibrium amounts: - \(n(\text{H}_2) = 1.50 - 0.80 = 0.70\text{ mol}\) - \(n(\text{I}_2) = 1.00 - 0.80 = 0.20\text{ mol}\)
2. **Write the expression for \(K_c\):** \[K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(n(\text{HI})/V)^2}{(n(\text{H}_2)/V)(n(\text{I}_2)/V)} = \frac{n(\text{HI})^2}{n(\text{H}_2) \times n(\text{I}_2)}\] (Volume \(V\) cancels as \(\Delta n = 0\))
At a temperature of \(298\text{ K}\), the equilibrium constant \(K_c\) for this reaction is \(4.00\).
A reaction mixture was prepared by mixing \(2.00\text{ mol}\) of ethanoic acid and \(2.00\text{ mol}\) of ethanol.
Calculate the amount, in mol, of ethyl ethanoate present in the equilibrium mixture at \(298\text{ K}\).
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解题
1. Set up an ICE table: - Let \(x\) be the amount in mol of ethyl ethanoate formed at equilibrium. - Initial amounts (mol): \(n(\text{CH}_3\text{COOH}) = 2.00\), \(n(\text{C}_2\text{H}_5\text{OH}) = 2.00\), \(n(\text{ester}) = 0\), \(n(\text{H}_2\text{O}) = 0\). - Equilibrium amounts (mol): - \(n(\text{CH}_3\text{COOH}) = 2.00 - x\) - \(n(\text{C}_2\text{H}_5\text{OH}) = 2.00 - x\) - \(n(\text{ethyl ethanoate}) = x\) - \(n(\text{H}_2\text{O}) = x\)
2. **Substitute into the \(K_c\) expression:** \[K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]} = \frac{x^2}{(2.00 - x)^2} = 4.00\]
3. **Solve for \(x\):** Take the square root of both sides: \[\frac{x}{2.00 - x} = \sqrt{4.00} = 2.00\] \[x = 2.00(2.00 - x)\] \[x = 4.00 - 2.00x\] \[3.00x = 4.00\] \[x = \frac{4.00}{3.00} = 1.33\text{ mol}\]
评分标准
M1: Expresses equilibrium quantities algebraically in terms of \(x\) (i.e. \(2.00 - x\), \(2.00 - x\), \(x\), \(x\)). [1] M2: Sets up \(\frac{x^2}{(2.00 - x)^2} = 4.00\) and simplifies correctly by taking the square root to give \(\frac{x}{2.00 - x} = 2.00\). [1] M3: Final answer of \(1.33\text{ mol}\) (or \(\frac{4}{3}\text{ mol}\)). [1]
题目 14 · Mechanisms & Structural Drawings
3 分
Propene, \(\text{CH}_3\text{CH}=\text{CH}_2\), reacts with hydrogen bromide, \(\text{HBr}\), at room temperature via an electrophilic addition mechanism to form 2-bromopropane as the major organic product.
Outline the mechanism for this reaction to form the major product.
Include all relevant: • dipoles and lone pairs of electrons • curly arrows showing the movement of electron pairs • structural formula and charge of the intermediate carbocation.
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解题
1. The \(\pi\)-electron density in the \(\text{C}=\text{C}\) bond attacks the electrophilic \(\text{H}^{\delta+}\) atom of polar \(\text{H}^{\delta+}-\text{Br}^{\delta-}\), causing heterolytic fission of the \(\text{H}-\text{Br}\) bond with electron movement onto the bromine atom. 2. This forms the more stable secondary carbocation intermediate, \(\text{CH}_3\text{CH}^+\text{CH}_3\), and a bromide ion, \(:\text{Br}^-\). 3. A lone pair of electrons on the \(:\text{Br}^-\) ion attacks the positively charged secondary carbon atom to form 2-bromopropane, \(\text{CH}_3\text{CH}(\text{Br})\text{CH}_3\).
评分标准
M1: Correct dipole on \(\text{H}^{\delta+}-\text{Br}^{\delta-}\) AND curly arrow from the \(\text{C}=\text{C}\) double bond to the \(\text{H}^{\delta+}\) atom [1] M2: Curly arrow from the \(\text{H}-\text{Br}\) bond to the \(\text{Br}\) atom [1] M3: Correct structure and charge of secondary carbocation \(\text{CH}_3\text{CH}^+\text{CH}_3\) AND curly arrow from a lone pair on \(:\text{Br}^-\) to the \(\text{C}^+\) atom [1]
题目 15 · Mechanisms & Structural Drawings
3 分
Propanal, \(\text{CH}_3\text{CH}_2\text{CHO}\), reacts with hydrogen cyanide, \(\text{HCN}\), in the presence of a catalytic amount of \(\text{NaCN}\), to form 2-hydroxybutanenitrile.
Outline the mechanism for this reaction.
Include all relevant: • dipoles and lone pairs of electrons • curly arrows to show the movement of electron pairs • formula and charge of the intermediate alkoxide ion • regeneration of the cyanide catalyst.
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解题
1. Nucleophilic attack occurs where the lone pair on the carbon of the cyanide ion, \(:\text{C}\equiv\text{N}^-\), attacks the electrophilic carbonyl carbon atom (\(\text{C}^{\delta+}\)), simultaneously breaking the \(\text{C}=\text{O}\) \(\pi\)-bond to place an electron pair onto the oxygen atom. 2. The resulting tetrahedral intermediate is \([\text{CH}_3\text{CH}_2\text{CH}(\text{CN})\text{O}]^-\). 3. The negative oxygen atom uses a lone pair to abstract a proton from \(\text{H}-\text{CN}\), breaking the \(\text{H}-\text{CN}\) bond, forming the product 2-hydroxybutanenitrile and regenerating the \(:\text{CN}^-\) catalyst.
评分标准
M1: Correct dipole \(\text{C}^{\delta+}=\text{O}^{\delta-}\) AND curly arrow from the lone pair on the carbon of \(:\text{CN}^-\) to the \(\text{C}^{\delta+}\) atom [1] M2: Curly arrow from the \(\text{C}=\text{O}\) double bond onto the oxygen atom AND correct structure of intermediate \([\text{CH}_3\text{CH}_2\text{CH}(\text{CN})\text{O}]^-\) [1] M3: Curly arrow from a lone pair on \(\text{O}^-\) of intermediate to \(\text{H}\) of \(\text{H}-\text{CN}\) (or \(\text{H}^+\)) AND curly arrow from \(\text{H}-\text{CN}\) bond onto \(\text{CN}\) (regenerating \(:\text{CN}^-\)) [1]
题目 16 · Mechanisms & Structural Drawings
3 分
Bromoethane reacts with aqueous sodium hydroxide by an \(\text{S}_\text{N}2\) mechanism to form ethanol.
Outline the \(\text{S}_\text{N}2\) mechanism for this reaction.
Include all relevant: • dipoles and lone pairs of electrons • curly arrows showing the movement of electron pairs • structural drawing of the transition state, showing partial bonds and overall charge • the inorganic product.
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解题
1. The hydroxide ion, \(:\text{OH}^-\), acts as a nucleophile; a lone pair on the oxygen attacks the electron-deficient carbon atom (\(\text{C}^{\delta+}\)) from the side opposite to the leaving bromine atom. 2. The \(\text{C}-\text{Br}\) bond breaks simultaneously as the \(\text{C}-\text{O}\) bond forms, proceeding through a pentacoordinate transition state represented as \([\text{HO}\cdots\text{C}(\text{H})_2(\text{CH}_3)\cdots\text{Br}]^-\). 3. The bromide ion, \(:\text{Br}^-\), is released as the leaving group, yielding ethanol, \(\text{CH}_3\text{CH}_2\text{OH}\).
评分标准
M1: Correct dipole \(\text{C}^{\delta+}-\text{Br}^{\delta-}\) on bromoethane AND curly arrow from a lone pair on \(:\text{OH}^-\) to the \(\text{C}^{\delta+}\) atom [1] M2: Curly arrow from the \(\text{C}-\text{Br}\) bond onto the \(\text{Br}\) atom [1] M3: Correct structure of the transition state showing partial bonds (dashed/dotted lines) to \(\text{OH}\) and \(\text{Br}\) and overall negative charge \([\text{HO}\cdots\text{CH}_2(\text{CH}_3)\cdots\text{Br}]^-\), AND \(\text{Br}^-\) identified as a product [1]
(a) Identify the chiral carbon atom in a molecule of 2-hydroxypropanoic acid. [1]
(b) Draw the three-dimensional structures of the two optical isomers of 2-hydroxypropanoic acid. Use conventional wedge-and-dashed bonds to show their spatial arrangement as non-superimposable mirror images. [2]
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解题
(a) Carbon-2 is the chiral carbon (asymmetric carbon) because it is bonded to four different atoms or groups of atoms: \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{COOH}\).
(b) The two optical isomers must be drawn with tetrahedral geometry around the central carbon atom using standard 3D conventions (two solid in-plane lines, one wedged line coming out of the plane, and one dashed line going into the plane) as non-superimposable mirror images.
评分标准
M1: Identification of carbon-2 (or the central carbon atom bonded to four different groups: \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), \(-\text{COOH}\)) [1] M2: Correct 3D tetrahedral representation of one isomer showing two solid in-plane bonds, one wedge bond, and one dashed bond [1] M3: Correct 3D representation of the second enantiomer drawn as an exact non-superimposable mirror image of the first [1]
题目 18 · Mechanisms & Structural Drawings
3 分
Ethane, \(\text{C}_2\text{H}_6\), reacts with chlorine gas, \(\text{Cl}_2\), in the presence of ultraviolet (UV) light to form chloroethane, \(\text{CH}_3\text{CH}_2\text{Cl}\), via a free-radical substitution mechanism.
(a) State the type of bond fission that occurs during the initiation step. [1]
(b) Write equations for the initiation step and the two propagation steps involved in the formation of chloroethane. Include radical dots on all free radicals. [2]
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解题
(a) Homolytic fission occurs when the covalent bond breaks such that one electron goes to each chlorine atom.
Explain why the first ionisation energy of sulfur is lower than that of phosphorus, despite sulfur having a greater nuclear charge. Refer to electron configurations and subshells in your answer.
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解题
Phosphorus has the outer electronic configuration \(3s^2 3p^3\) where each 3p orbital is singly occupied. Sulfur has the outer electronic configuration \(3s^2 3p^4\) where one 3p orbital contains a pair of electrons (spin-paired). There is increased mutual electrostatic repulsion between the two paired electrons occupying the same 3p orbital in sulfur. This inter-electron repulsion outweighs the effect of the additional nuclear charge (+16 vs +15), so less energy is required to remove an electron from sulfur than from phosphorus.
评分标准
M1: (Electronic configurations stated or implied) P has \(3p^3\) / singly occupied 3p orbitals AND S has \(3p^4\) / one fully occupied (paired) 3p orbital. [1] M2: Identification of electron pairing / paired electrons in a 3p orbital in sulfur. [1] M3: (Inter-electron / spin-pair) repulsion between paired electrons in sulfur means less energy is required to remove an electron (ora for P). [1] (R: 'shielding is greater in sulfur' for M3)
题目 20 · theory
3 分
Explain why solid ice has a lower density than liquid water at 0 °C. Refer to structure and bonding in your answer.
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解题
In solid ice, each water molecule forms four hydrogen bonds to neighbouring water molecules in a rigid three-dimensional tetrahedral arrangement. This creates an open cage-like crystalline structure with relatively large spaces between molecules. When ice melts to form liquid water at 0 °C, some of these hydrogen bonds break, allowing the open structure to collapse and the water molecules to pack more closely together, resulting in a higher mass per unit volume in liquid water than in solid ice.
评分标准
M1: Identifies hydrogen bonding between water molecules. [1] M2: In ice, molecules are arranged in an open / cage-like / 3D tetrahedral lattice structure. [1] M3: In liquid water, the structure collapses / hydrogen bonds break allowing molecules to move closer together / pack more densely (ora). [1]
题目 21 · theory
3 分
Explain why iodide ions, \(\text{I}^-\), are stronger reducing agents than chloride ions, \(\text{Cl}^-\).
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解题
The iodide ion (\(\text{I}^-\)) has more electron shells and therefore a larger ionic radius than the chloride ion (\(\text{Cl}^-\)). The outermost electrons in \(\text{I}^-\) experience greater electron shielding from inner shells. As a result, the electrostatic attraction between the nucleus and the valence electrons is weaker in \(\text{I}^-\) than in \(\text{Cl}^-\), meaning \(\text{I}^-\) loses an electron more easily to act as a reducing agent.
评分标准
M1: \(\text{I}^-\) has a larger ionic radius / more electron shells / valence electrons are further from nucleus than \(\text{Cl}^-\). [1] M2: \(\text{I}^-\) has greater (inner shell) shielding than \(\text{Cl}^-\). [1] M3: Weaker nuclear attraction for valence electrons in \(\text{I}^-\) so electron is lost more readily / more easily oxidised (ora for \(\text{Cl}^-\)). [1]
题目 22 · theory
3 分
Using collision theory and the concept of activation energy, explain why an increase in temperature causes a significant increase in the rate of a chemical reaction.
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解题
When temperature is increased, the average kinetic energy of the reacting particles increases. This shifts the Boltzmann distribution of molecular energies to higher values, so a much larger fraction of colliding particles possess energy equal to or greater than the activation energy (\(E \ge E_a\)). Consequently, the frequency of successful (effective) collisions per unit time increases significantly, leading to an increased rate of reaction.
评分标准
M1: Particles have more kinetic energy / move faster at higher temperature. [1] M2: A greater proportion / fraction of molecules have energy greater than or equal to the activation energy (\(E \ge E_a\)). [1] (R: 'more molecules have activation energy') M3: Higher frequency of successful / effective collisions per unit time (or more successful collisions per second). [1]
题目 23 · theory
3 分
When 2-methylbut-2-ene reacts with hydrogen bromide, 2-bromo-2-methylbutane is the major organic product and 2-bromo-3-methylbutane is the minor organic product. Explain why 2-bromo-2-methylbutane is the major product.
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解题
Electrophilic addition of \(\text{HBr}\) involves the formation of a carbocation intermediate. Protonation of 2-methylbut-2-ene can yield either a tertiary carbocation, \(\text{CH}_3\text{CH}_2\text{C}^+(\text{CH}_3)_2\), or a secondary carbocation, \(\text{CH}_3\text{C}^+\text{HCH}(\text{CH}_3)_2\). The tertiary carbocation intermediate has three alkyl groups bonded to the positively charged carbon atom compared to two alkyl groups in the secondary carbocation. Alkyl groups are electron-donating by positive inductive effect (+I), which disperses the positive charge more effectively, making the tertiary carbocation more stable. The more stable intermediate forms faster and gives the major product, 2-bromo-2-methylbutane.
评分标准
M1: Major product forms via a tertiary carbocation (intermediate) AND minor product forms via a secondary carbocation. [1] M2: Tertiary carbocation is more stable than the secondary carbocation. [1] M3: Tertiary carbocation has more (three vs two) alkyl groups that release electrons / have an electron-donating inductive effect / push electron density towards \(\text{C}^+\) to disperse the positive charge. [1]
Paper 3: Advanced Practical Skills
Complete all three laboratory tasks: Quantitative Titration, Thermal Decomposition / Gravimetric Analysis, and Qualitative Analysis of Unknown Inorganic Solutions.
13 题目 · 40 分
题目 1 · free_response
8 分
A student determines the concentration of an unknown monobasic carboxylic acid, \(\text{HA}\), by titration with standard aqueous sodium hydroxide.
FB 1 is \(0.0500\text{ mol dm}^{-3}\text{ NaOH(aq)}\). FB 2 is an aqueous solution of the weak monobasic acid \(\text{HA}\).
\(25.0\text{ cm}^3\) of FB 2 was pipetted into a conical flask and titrated with FB 1 from a burette using phenolphthalein indicator.
(a) The student obtained the following burette readings: - Rough: Initial reading \(= 0.00\text{ cm}^3\), Final reading \(= 23.40\text{ cm}^3\) - Titration 1: Initial reading \(= 0.20\text{ cm}^3\), Final reading \(= 22.85\text{ cm}^3\) - Titration 2: Initial reading \(= 22.85\text{ cm}^3\), Final reading \(= 45.40\text{ cm}^3\) - Titration 3: Initial reading \(= 0.50\text{ cm}^3\), Final reading \(= 23.10\text{ cm}^3\)
(i) Present these results in a fully labelled results table showing the initial burette reading, final burette reading, and titre for each titration, including appropriate units.
(ii) State which titres are concordant and calculate the mean titre to be used in subsequent calculations. Show your working clearly. [3]
(b) Calculate the amount, in \(\text{mol}\), of \(\text{NaOH}\) present in the mean volume of FB 1 calculated in (a)(ii). [1]
(c) The equation for the neutralization reaction is: \[\text{HA(aq)} + \text{NaOH(aq)} \rightarrow \text{NaA(aq)} + \text{H}_2\text{O(l)}\]
Deduce the amount, in \(\text{mol}\), of \(\text{HA}\) present in the \(25.0\text{ cm}^3\) sample of FB 2. [1]
(d) The relative molecular mass, \(M_{\text{r}}\), of \(\text{HA}\) is \(122.0\).
Calculate the concentration of \(\text{HA}\) in FB 2 in \(\text{g dm}^{-3}\). Give your answer to three significant figures. [2]
(e) The maximum uncertainty in each single reading of the burette is \(\pm 0.05\text{ cm}^3\).
Calculate the percentage uncertainty in the titre obtained in Titration 3. [1]
(a)(ii) Concordant titres are those within \(0.10\text{ cm}^3\) of each other: Titrations 1, 2, and 3. Mean titre \(= \frac{22.65 + 22.55 + 22.60}{3} = 22.60\text{ cm}^3\)
(b) \(\text{Amount of NaOH} = c \times V = 0.0500\text{ mol dm}^{-3} \times \frac{22.60}{1000}\text{ dm}^3 = 1.13 \times 10^{-3}\text{ mol}\)
(c) The stoichiometric ratio of \(\text{HA} : \text{NaOH}\) is \(1 : 1\). Therefore, \(\text{amount of HA in } 25.0\text{ cm}^3 = 1.13 \times 10^{-3}\text{ mol}\)
(d) \(\text{Concentration of HA in mol dm}^{-3} = \frac{1.13 \times 10^{-3}\text{ mol}}{25.0 / 1000\text{ dm}^3} = 0.0452\text{ mol dm}^{-3}\) \(\text{Concentration of HA in g dm}^{-3} = 0.0452\text{ mol dm}^{-3} \times 122.0\text{ g mol}^{-1} = 5.5144\text{ g dm}^{-3} \approx 5.51\text{ g dm}^{-3}\) (3 sig figs)
(e) Each titre involves two burette readings, so total uncertainty in titre \(= 2 \times 0.05 = 0.10\text{ cm}^3\). \(\text{Percentage uncertainty} = \frac{0.10\text{ cm}^3}{22.60\text{ cm}^3} \times 100 = 0.442\%\) (or \(0.44\%\)).
评分标准
(a)(i) [2 marks] - M1: Table layout with unambiguous row/column headers containing standard units (e.g. Initial reading / cm³, Final reading / cm³, Titre / cm³) AND all burette readings correctly recorded to 2 decimal places ending in .00 or .05. - M2: All titres correctly subtracted (Rough = 23.40, 1 = 22.65, 2 = 22.55, 3 = 22.60 cm³).
(a)(ii) [1 mark] - M3: Identifies concordant titres as Titrations 1, 2, and 3 (all within 0.10 cm³) AND calculates mean titre as 22.60 cm³ (to 2 decimal places).
(b) [1 mark] - M4: Correct calculation of amount of NaOH = 0.0500 × (22.60 / 1000) = 1.13 × 10⁻³ mol (allow ecf from mean titre).
(c) [1 mark] - M5: Stating 1.13 × 10⁻³ mol (must match M4 by 1:1 stoichiometry).
(d) [2 marks] - M6 (method): Calculates molar concentration of HA (= (1.13 × 10⁻³) / 0.0250 = 0.0452 mol dm⁻³) AND multiplies by 122.0. - M7 (accuracy): Correct final value to 3 significant figures: 5.51 g dm⁻³ (allow ecf from (c)).
A student dissolves a $6.95\text{ g}$ sample of hydrated iron(II) sulfate, $\text{FeSO}_4 \cdot x\text{H}_2\text{O}$, in dilute sulfuric acid and makes the volume up to $250.0\text{ cm}^3$ in a volumetric flask.
A $25.0\text{ cm}^3$ portion of this solution is titrated against $0.0200\text{ mol dm}^{-3}$ acidified potassium manganate(VII), $\text{KMnO}_4$. The reaction requires $25.00\text{ cm}^3$ of the $\text{KMnO}_4$ solution for complete reaction.
The ionic equation for the reaction is: $$5\text{Fe}^{2+}(\text{aq}) + \text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) \rightarrow 5\text{Fe}^{3+}(\text{aq}) + \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})$$
Calculate the value of $x$ in the formula $\text{FeSO}_4 \cdot x\text{H}_2\text{O}$. [$A_r$: $\text{Fe} = 55.8$, $\text{S} = 32.1$, $\text{O} = 16.0$, $\text{H} = 1.0$]
[M1] (0.5 marks) Calculation of total moles of $\text{Fe}^{2+}$ in $250.0\text{ cm}^3$ solution ($= 2.50 \times 10^{-2}\text{ mol}$). [M2] (0.5 marks) Calculation of experimental $M_r$ of $\text{FeSO}_4 \cdot x\text{H}_2\text{O}$ ($= 278.0\text{ g mol}^{-1}$). [M3] (0.5 marks) Final deduced integer value of $x = 7$ (allow ecf from incorrect molar mass if working is shown).
题目 3 · Quantitative Stoichiometric Calculation
1.5 分
A student carries out a gravimetric analysis to determine the value of $n$ in hydrated zinc sulfate, $\text{ZnSO}_4 \cdot n\text{H}_2\text{O}$. A sample is heated to constant mass in a crucible.
The experimental data recorded is as follows: - Mass of empty crucible $= 18.450\text{ g}$ - Mass of crucible $+$ hydrated zinc sulfate $= 21.325\text{ g}$ - Mass of crucible $+$ anhydrous $\text{ZnSO}_4$ residue $= 20.065\text{ g}$
Calculate the integer value of $n$. [$A_r$: $\text{Zn} = 65.4$, $\text{S} = 32.1$, $\text{O} = 16.0$, $\text{H} = 1.0$]
[M1] (0.5 marks) Determination of the correct masses of anhydrous $\text{ZnSO}_4$ ($1.615\text{ g}$) and water ($1.260\text{ g}$). [M2] (0.5 marks) Conversion of both masses to moles ($0.0100\text{ mol}$ $\text{ZnSO}_4$ and $0.0700\text{ mol}$ $\text{H}_2\text{O}$). [M3] (0.5 marks) Calculation of mole ratio to give $n = 7$ as an integer.
题目 4 · Quantitative Stoichiometric Calculation
1.5 分
A $1.250\text{ g}$ sample of impure limestone is treated with $35.00\text{ cm}^3$ of $1.00\text{ mol dm}^{-3}$ hydrochloric acid, $\text{HCl}$ (an excess). The unreacted acid requires $26.00\text{ cm}^3$ of $0.500\text{ mol dm}^{-3}$ sodium hydroxide, $\text{NaOH}$, for complete neutralisation.
Calculate the percentage by mass of $\text{CaCO}_3$ in the limestone sample. Give your answer to 3 significant figures. [$A_r$: $\text{Ca} = 40.1$, $\text{C} = 12.0$, $\text{O} = 16.0$]
3. Amount of $\text{HCl}$ reacted with $\text{CaCO}_3$: $$n(\text{HCl})_{\text{reacted}} = 0.0350 - 0.0130 = 0.0220\text{ mol}$$
4. Amount of $\text{CaCO}_3$: $$n(\text{CaCO}_3) = \frac{0.0220}{2} = 0.0110\text{ mol}$$
5. Mass and percentage of $\text{CaCO}_3$: $$M_r(\text{CaCO}_3) = 40.1 + 12.0 + 3(16.0) = 100.1\text{ g mol}^{-1}$$ $$\text{Mass of }\text{CaCO}_3 = 0.0110\text{ mol} \times 100.1\text{ g mol}^{-1} = 1.1011\text{ g}$$ $$\text{Percentage} = \frac{1.1011}{1.250} \times 100 = 88.088\% \approx 88.1\%$$
评分标准
[M1] (0.5 marks) Calculation of moles of $\text{HCl}$ reacting with $\text{CaCO}_3$ ($= 0.0220\text{ mol}$). [M2] (0.5 marks) Calculation of mass of $\text{CaCO}_3$ present in sample ($= 1.1011\text{ g}$). [M3] (0.5 marks) Correct calculation of percentage by mass to 3 significant figures ($88.1\%$).
题目 5 · Quantitative Stoichiometric Calculation
1.5 分
In an enthalpy determination experiment, $50.0\text{ cm}^3$ of $1.20\text{ mol dm}^{-3}\text{ HCl}$ is mixed with $50.0\text{ cm}^3$ of $1.20\text{ mol dm}^{-3}\text{ NaOH}$ in an expanded polystyrene cup. Both solutions are initially at $20.4\ ^\circ\text{C}$.
The maximum temperature recorded during neutralisation is $28.5\ ^\circ\text{C}$.
Assume the density of the final mixture is $1.00\text{ g cm}^{-3}$ and its specific heat capacity is $4.18\text{ J g}^{-1}\text{ K}^{-1}$.
Calculate the enthalpy change of neutralisation, $\Delta H_{\text{neut}}$, in $\text{kJ mol}^{-1}$. Give your answer to 3 significant figures and include an appropriate sign.
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解题
1. Temperature rise: $$\Delta T = 28.5 - 20.4 = 8.1\ ^\circ\text{C} = 8.1\text{ K}$$
2. Total mass of solution: $$m = 50.0 + 50.0 = 100.0\text{ g}$$
[M1] (0.5 marks) Calculation of heat released $q = 3.3858\text{ kJ}$ (or $3385.8\text{ J}$). [M2] (0.5 marks) Calculation of amount of water formed ($n = 0.0600\text{ mol}$). [M3] (0.5 marks) Value of $-56.4\text{ kJ mol}^{-1}$ to 3 significant figures with negative sign.
题目 6 · Quantitative Stoichiometric Calculation
1.5 分
A student investigates an unknown Group 2 metal, $\text{M}$. A $0.0735\text{ g}$ sample of metal $\text{M}$ is added to an excess of dilute hydrochloric acid, and the hydrogen gas evolved is collected in a gas syringe at room conditions: $$\text{M}(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MCl}_2(\text{aq}) + \text{H}_2(\text{g})$$
The total volume of $\text{H}_2(\text{g})$ collected is $72.0\text{ cm}^3$.
Calculate the relative atomic mass, $A_r$, of metal $\text{M}$ to 1 decimal place, and deduce the identity of $\text{M}$. [The molar volume of gas at room conditions is $24.0\text{ dm}^3\text{ mol}^{-1}$]
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解题
1. Calculate the amount of $\text{H}_2(\text{g})$ formed: $$n(\text{H}_2) = \frac{72.0\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 3.00 \times 10^{-3}\text{ mol}$$
2. From the stoichiometry $\text{M} : \text{H}_2 = 1 : 1$, the amount of $\text{M}$ reacted is: $$n(\text{M}) = 3.00 \times 10^{-3}\text{ mol}$$
3. Calculate the relative atomic mass of $\text{M}$: $$A_r(\text{M}) = \frac{\text{mass}}{n} = \frac{0.0735\text{ g}}{3.00 \times 10^{-3}\text{ mol}} = 24.5$$
4. Identity of $\text{M}$: Comparing with Group 2 elements in the Periodic Table ($A_r(\text{Mg}) = 24.3$), metal $\text{M}$ is magnesium ($\text{Mg}$).
评分标准
[M1] (0.5 marks) Calculation of moles of $\text{H}_2$ gas produced ($= 3.00 \times 10^{-3}\text{ mol}$). [M2] (0.5 marks) Calculation of relative atomic mass $A_r = 24.5$. [M3] (0.5 marks) Correct deduction of metal identity as magnesium / $\text{Mg}$.
题目 7 · Quantitative Stoichiometric Calculation
1.5 分
The concentration of copper(II) ions in a solution is determined by iodometric titration.
An excess of aqueous potassium iodide, $\text{KI}$, is added to a $25.0\text{ cm}^3$ sample of the copper(II) sulfate solution: $$2\text{Cu}^{2+}(\text{aq}) + 4\text{I}^-(\text{aq}) \rightarrow 2\text{CuI}(\text{s}) + \text{I}_2(\text{aq})$$
The iodine liberated requires $18.40\text{ cm}^3$ of $0.0500\text{ mol dm}^{-3}$ standard sodium thiosulfate solution, $\text{Na}_2\text{S}_2\text{O}_3$, for complete titration: $$2\text{S}_2\text{O}_3^{2-}(\text{aq}) + \text{I}_2(\text{aq}) \rightarrow \text{S}_4\text{O}_6^{2-}(\text{aq}) + 2\text{I}^-(\text{aq})$$
Calculate the concentration, in $\text{g dm}^{-3}$, of $\text{Cu}^{2+}$ ions in the original solution. Give your answer to 3 significant figures. [$A_r$: $\text{Cu} = 63.5$]
4. Convert to mass concentration in $\text{g dm}^{-3}$: $$\text{Concentration} = 0.0368\text{ mol dm}^{-3} \times 63.5\text{ g mol}^{-1} = 2.3368\text{ g dm}^{-3} \approx 2.34\text{ g dm}^{-3}$$
评分标准
[M1] (0.5 marks) Calculation of moles of $\text{S}_2\text{O}_3^{2-}$ ($= 9.20 \times 10^{-4}\text{ mol}$) and recognition that $n(\text{Cu}^{2+}) = n(\text{S}_2\text{O}_3^{2-})$. [M2] (0.5 marks) Calculation of molar concentration of $\text{Cu}^{2+}$ ($= 0.0368\text{ mol dm}^{-3}$). [M3] (0.5 marks) Correct mass concentration in $\text{g dm}^{-3}$ to 3 significant figures ($2.34\text{ g dm}^{-3}$).
题目 8 · Gravimetric Mass Measurement & Error Evaluation
4 分
A student carries out an experiment to determine the value of \(x\) in hydrated manganese(II) sulfate, \(\text{MnSO}_4 \cdot x\text{H}_2\text{O}\).
A clean, dry crucible with a lid is weighed. A sample of hydrated manganese(II) sulfate is added and the crucible, lid, and contents are reweighed. The crucible is heated strongly using a Bunsen burner, allowed to cool in a desiccator, and weighed. The process of heating, cooling, and reweighing is repeated until constant mass is obtained.
Experimental Data: - Mass of empty crucible + lid = \(19.35\text{ g}\) - Mass of crucible + lid + \(\text{MnSO}_4 \cdot x\text{H}_2\text{O}\) = \(23.81\text{ g}\) - Mass of crucible + lid + anhydrous \(\text{MnSO}_4\) (constant mass) = \(22.37\text{ g}\)
(a) Calculate the value of \(x\) in the formula \(\text{MnSO}_4 \cdot x\text{H}_2\text{O}\). [2]
(b) The balance used has an uncertainty of \(\pm 0.01\text{ g}\) for each reading. Calculate the percentage uncertainty in the mass of water driven off. [1]
(c) State why the student heats the crucible and residue to constant mass, and explain how the student confirms that constant mass has been reached. [1]
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解题
(a) Mass of anhydrous \(\text{MnSO}_4 = 22.37 - 19.35 = 3.02\text{ g}\) Mass of water lost = \(23.81 - 22.37 = 1.44\text{ g}\) \(M_r(\text{MnSO}_4) = 54.9 + 32.1 + 4(16.0) = 151.0\text{ g mol}^{-1}\) \(n(\text{MnSO}_4) = \frac{3.02}{151.0} = 0.0200\text{ mol}\) \(n(\text{H}_2\text{O}) = \frac{1.44}{18.0} = 0.0800\text{ mol}\) \(x = \frac{0.0800}{0.0200} = 4.00 \rightarrow x = 4\)
(b) Mass of water is derived from 2 readings, so absolute uncertainty = \(2 \times 0.01 = 0.02\text{ g}\). \(\text{Percentage uncertainty} = \left(\frac{0.02}{1.44}\right) \times 100 = 1.39\%\)
(c) Heating to constant mass ensures all water of crystallisation has been completely removed (the thermal decomposition is complete). It is confirmed when two consecutive weighings after reheating and cooling differ by no more than \(0.01\text{ g}\) or \(0.02\text{ g}\).
评分标准
(a) - M1: Correct calculation of amount in mol of \(\text{MnSO}_4\) (\(0.0200\text{ mol}\)) and amount in mol of \(\text{H}_2\text{O}\) (\(0.0800\text{ mol}\)) [1] - M2: Correct integer value of \(x = 4\) clearly deduced from mole ratio [1]
(c) - M1: States to ensure all water has been driven off / reaction has gone to completion AND explains it is confirmed by reheating and reweighing until two consecutive masses are identical / agree within \(\pm 0.02\text{ g}\) [1]
题目 9 · Gravimetric Mass Measurement & Error Evaluation
4 分
An experiment is performed to identify a Group 2 metal carbonate, \(\text{MCO}_3\), by thermal decomposition: $$\text{MCO}_3(\text{s}) \rightarrow \text{MO}(\text{s}) + \text{CO}_2(\text{g})$$
Experimental Data: - Mass of empty crucible = \(21.50\text{ g}\) - Mass of crucible + \(\text{MCO}_3\) = \(25.19\text{ g}\) - Mass of crucible + residue \(\text{MO}\) (after heating to constant mass) = \(24.09\text{ g}\)
(a) Calculate the amount, in mol, of \(\text{CO}_2\) released during the decomposition and deduce the identity of the Group 2 metal \(\text{M}\). [2]
(b) The mass balance used has a precision of \(\pm 0.01\text{ g}\). Calculate the maximum percentage uncertainty in the determined mass of \(\text{CO}_2\). [1]
(c) A student performs the heating without placing a lid on the crucible and observes some solid spitting out of the crucible during vigorous heating. Explain the effect this loss of solid would have on the calculated relative atomic mass, \(A_r\), of \(\text{M}\). [1]
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解题
(a) Mass of \(\text{MCO}_3 = 25.19 - 21.50 = 3.69\text{ g}\) Mass of \(\text{CO}_2\) released = \(25.19 - 24.09 = 1.10\text{ g}\) \(n(\text{CO}_2) = \frac{1.10}{44.0} = 0.0250\text{ mol}\) \(n(\text{MCO}_3) = n(\text{CO}_2) = 0.0250\text{ mol}\) \(M_r(\text{MCO}_3) = \frac{3.69}{0.0250} = 147.6\text{ g mol}^{-1}\) \(A_r(\text{M}) = 147.6 - [12.0 + 3(16.0)] = 147.6 - 60.0 = 87.6\) From the Periodic Table, \(\text{M}\) is strontium (\(\text{Sr}\)).
(b) Mass of \(\text{CO}_2\) is calculated from two balance readings, so absolute uncertainty = \(2 \times 0.01 = 0.02\text{ g}\). \(\text{Percentage uncertainty} = \left(\frac{0.02}{1.10}\right) \times 100 = 1.82\%\)
(c) Loss of solid due to spitting makes the measured mass of residue smaller than it actually is. Consequently, the calculated mass of \(\text{CO}_2\) lost appears larger than it really is, leading to an overestimation of the number of moles of \(\text{CO}_2\). As \(M_r = \frac{\text{mass of sample}}{n}\), the calculated \(M_r(\text{MCO}_3)\) (and thus the calculated \(A_r(\text{M})\)) will be too low / smaller than the true value.
评分标准
(a) - M1: Correct calculation of mass of \(\text{CO}_2\) (\(1.10\text{ g}\)) and moles of \(\text{CO}_2\) (\(0.0250\text{ mol}\)) [1] - M2: Correct calculation of \(M_r(\text{MCO}_3) = 147.6\text{ g mol}^{-1}\) and identification of \(\text{M}\) as strontium / \(\text{Sr}\) [1]
(c) - M1: Explains that mass loss appears greater / calculated moles of \(\text{CO}_2\) is too large, leading to a calculated \(A_r\) of \(\text{M}\) that is lower / underestimated [1]
题目 10 · Qualitative Observation & Ion Deduction
3.75 分
An aqueous solution, labelled \(\text{FA 5}\), contains one cation and one anion from those listed in the Qualitative Analysis Notes.
(a) Carry out the following tests and record your observations in the table below:
| Test | Observations | | :--- | :--- | | (i) To a \(1\text{ cm}^3\) depth of \(\text{FA 5}\) in a test-tube, add aqueous sodium hydroxide dropwise until in excess. | [Record observation] | | (ii) To a \(1\text{ cm}^3\) depth of \(\text{FA 5}\) in a test-tube, add aqueous ammonia dropwise until in excess. | [Record observation] | | (iii) To a \(1\text{ cm}^3\) depth of \(\text{FA 5}\) in a test-tube, add dilute nitric acid followed by aqueous silver nitrate, then add aqueous ammonia in excess. | [Record observation] |
(b) Using your observations: (i) Identify the cation and anion present in \(\text{FA 5}\). (ii) Write an ionic equation, including state symbols, for the reaction occurring in test (iii) between the anion and silver ions.
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解题
(a) Step-by-step observations: - Test (i): Addition of \(\text{NaOH(aq)}\) to \(\text{Mg}^{2+}\) yields a white precipitate of \(\text{Mg(OH)}_2\), which is insoluble in excess \(\text{NaOH}\). - Test (ii): Addition of \(\text{NH}_3\text{(aq)}\) produces no precipitate (or a very faint white precipitate) because the hydroxide concentration in aqueous ammonia is too low to exceed the solubility product under standard test conditions. - Test (iii): Addition of \(\text{HNO}_3\text{(aq)}\) followed by \(\text{AgNO}_3\text{(aq)}\) gives a cream precipitate of \(\text{AgBr}\), which does not dissolve in dilute aqueous ammonia.
(b) Deductions: - Cation: \(\text{Mg}^{2+}\) is the only cation from the syllabus list giving a white precipitate with \(\text{NaOH}\) insoluble in excess and no precipitate with \(\text{NH}_3\). - Anion: The cream precipitate with \(\text{AgNO}_3\text{(aq)}\) confirms \(\text{Br}^-\). - Ionic equation: \(\text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)}\)
评分标准
[Total: 3.75 Marks] - M1 (1.00 mark): Correct observations for tests (i) and (ii) (White ppt insoluble in excess NaOH AND no ppt / faint haze with NH3). - M2 (1.00 mark): Correct observation for test (iii) (Cream precipitate formed with acidified AgNO3). - M3 (1.00 mark): Correct identification of cation as \(\text{Mg}^{2+}\) and anion as \(\text{Br}^-\). - M4 (0.75 mark): Balanced ionic equation with correct state symbols: \(\text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)}\).
题目 11 · Qualitative Observation & Ion Deduction
3.75 分
Solid \(\text{FA 6}\) is a single hydrated transition metal salt. \(\text{FA 6}\) was dissolved in distilled water to form an aqueous solution.
(a) Describe the observations when: (i) Aqueous sodium hydroxide is added dropwise, then in excess, to a sample of \(\text{FA 6}\)(aq). (ii) The mixture from (a)(i) is allowed to stand in air for 5 minutes. (iii) Aqueous barium chloride is added to a fresh sample of \(\text{FA 6}\)(aq), followed by excess dilute hydrochloric acid.
(b) Deduce the chemical formula of the cation and anion present in \(\text{FA 6}\).
(c) Write an ionic equation for the reaction occurring when \(\text{FA 6}\)(aq) reacts with aqueous barium chloride.
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解题
(a) Explanations of observations: - Test (i): \(\text{Fe}^{2+}\text{(aq)}\) reacts with \(\text{OH}^-\text{(aq)}\) to precipitate green \(\text{Fe(OH)}_2\text{(s)}\), which is insoluble in excess hydroxide. - Test (ii): Atmospheric oxygen oxidises iron(II) hydroxide to iron(III) hydroxide: \(4\text{Fe(OH)}_2\text{(s)} + \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} \rightarrow 4\text{Fe(OH)}_3\text{(s)}\), causing the green precipitate to turn brown at the air interface. - Test (iii): Sulfate ions react with barium ions to form insoluble barium sulfate, \(\text{BaSO}_4\text{(s)}\), which remains insoluble upon addition of dilute \(\text{HCl}\).
(b) Cation is \(\text{Fe}^{2+}\) (iron(II)); anion is \(\text{SO}_4^{2-}\) (sulfate).
(c) The precipitation reaction is \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\).
评分标准
[Total: 3.75 Marks] - M1 (1.00 mark): Green ppt insoluble in excess NaOH AND turns brown on standing in air. - M2 (1.00 mark): White ppt formed with \(\text{BaCl}_2\) which is insoluble in dilute \(\text{HCl}\). - M3 (1.00 mark): Correct identification of cation as \(\text{Fe}^{2+}\) and anion as \(\text{SO}_4^{2-}\). - M4 (0.75 mark): Balanced ionic equation with correct state symbols: \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\).
题目 12 · Qualitative Observation & Ion Deduction
3.75 分
Solution \(\text{FA 7}\) contains two dissolved salts sharing the same anion, but containing two different metal cations.
When carrying out qualitative tests on \(\text{FA 7}\): 1. Addition of aqueous sodium hydroxide dropwise yields a white precipitate. On adding excess \(\text{NaOH(aq)}\), the precipitate partially dissolves to leave a remaining white precipitate. 2. Addition of aqueous ammonia dropwise yields a white precipitate. On adding excess \(\text{NH}_3\text{(aq)}\), the precipitate partially dissolves to leave a remaining white precipitate. 3. Addition of dilute hydrochloric acid to \(\text{FA 7}\) produces vigorous effervescence. The gas evolved turns limewater cloudy.
(a) Identify the two cations present in \(\text{FA 7}\). (b) Identify the anion present in \(\text{FA 7}\). (c) Write a balanced ionic equation for the reaction of the anion with dilute hydrochloric acid, including state symbols.
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解题
(a) Analysis of cations: - In test 1: \(\text{NaOH}\) forms white precipitates. With excess \(\text{NaOH}\), amphoteric hydroxides (such as \(\text{Al(OH)}_3\) or \(\text{Zn(OH)}_2\)) dissolve to form aluminate/zincate, while non-amphoteric hydroxides (such as \(\text{Mg(OH)}_2\)) remain insoluble. If \(\text{Zn}^{2+}\) and \(\text{Mg}^{2+}\) are present, \(\text{Zn(OH)}_2\) dissolves in excess \(\text{NaOH}\) while \(\text{Mg(OH)}_2\) remains. - In test 2: \(\text{NH}_3\) precipitates both hydroxides. In excess \(\text{NH}_3\), \(\text{Zn(OH)}_2\) dissolves to form \([\text{Zn(NH}_3)_4]^{2+}\), leaving \(\text{Mg(OH)}_2\) (or \(\text{Al(OH)}_3\)) undissolved. - Thus, one cation is \(\text{Zn}^{2+}\) and the other is \(\text{Al}^{3+}\) or \(\text{Mg}^{2+}\).
(b) Effervescence with dilute \(\text{HCl}\) producing a gas that turns limewater cloudy (white precipitate of \(\text{CaCO}_3\)) proves the presence of carbonate, \(\text{CO}_3^{2-}\).
[Total: 3.75 Marks] - M1 (1.00 mark): Identification of \(\text{Zn}^{2+}\) with valid reasoning (solubility in excess \(\text{NH}_3\)). - M2 (1.00 mark): Identification of second cation as \(\text{Al}^{3+}\) or \(\text{Mg}^{2+}\) with valid reasoning. - M3 (1.00 mark): Identification of the anion as \(\text{CO}_3^{2-}\) / carbonate. - M4 (0.75 mark): Balanced ionic equation with correct state symbols: \(\text{CO}_3^{2-}\text{(aq)} + 2\text{H}^+\text{(aq)} \rightarrow \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\).
题目 13 · Qualitative Observation & Ion Deduction
3.75 分
An unknown inorganic solution, \(\text{FA 8}\), was subjected to systematic qualitative tests:
1. To \(1\text{ cm}^3\) of \(\text{FA 8}\), aqueous sodium hydroxide was added dropwise until in excess. A pale blue precipitate formed which was insoluble in excess \(\text{NaOH}\). 2. To \(1\text{ cm}^3\) of \(\text{FA 8}\), aqueous ammonia was added dropwise until in excess. A pale blue precipitate formed which dissolved in excess \(\text{NH}_3\) to give a dark blue solution. 3. To \(1\text{ cm}^3\) of \(\text{FA 8}\), aqueous sodium hydroxide and a small piece of aluminium foil were added, and the mixture was gently warmed. A gas was evolved which turned damp red litmus paper blue.
(a) Identify the cation present in \(\text{FA 8}\). (b) Identify the anion present in \(\text{FA 8}\). (c) Write the ionic equation for the initial precipitation reaction observed in test 1. (d) State the identity of the gas evolved in test 3.
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解题
(a) The formation of a pale blue precipitate insoluble in excess \(\text{NaOH}\), and soluble in excess \(\text{NH}_3\) to form a deep/dark blue solution (due to the formation of the complex ion \([\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\)), confirms the cation as \(\text{Cu}^{2+}\).
(b) Reduction with aluminium foil in warm alkaline \(\text{NaOH}\) produces ammonia gas, confirming the presence of nitrate (\(\text{NO}_3^-\)) or nitrite (\(\text{NO}_2^-\)).
(c) Reaction between \(\text{Cu}^{2+}\) and \(\text{OH}^-\): \(\text{Cu}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Cu(OH)}_2\text{(s)}\).
(d) The gas that turns damp red litmus paper blue is ammonia (\(\text{NH}_3\)).
评分标准
[Total: 3.75 Marks] - M1 (1.00 mark): Identification of cation as \(\text{Cu}^{2+}\) / copper(II). - M2 (1.00 mark): Identification of anion as \(\text{NO}_3^-\) / nitrate (allow \(\text{NO}_2^-\) / nitrite). - M3 (1.00 mark): Correct balanced ionic equation with state symbols: \(\text{Cu}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Cu(OH)}_2\text{(s)}\). - M4 (0.75 mark): Identity of gas stated as ammonia / \(\text{NH}_3\).
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