An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V1) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.
Paper 11 (選擇題)
Answer all 40 multiple choice questions by choosing the correct option (A, B, C or D).
40 題目 · 40 分
題目 1 · 選擇題
1 分
10 cm³ of a gaseous hydrocarbon, \(C_xH_y\), was exploded with 100 cm³ of oxygen (an excess). After cooling to room temperature, the residual gas volume was 85 cm³. On passing this residual gas through aqueous potassium hydroxide, the volume decreased to 55 cm³. What is the molecular formula of the hydrocarbon?
A.C3H4
B.C3H6
C.C3H8
D.C4H10
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解題
1. Analyze the residual gas components: After cooling to room temperature, the water vapor condenses to liquid of negligible volume. Thus, the 85 cm³ of residual gas consists of excess \(O_2\) and produced \(CO_2\). 2. Carbon dioxide absorption: Passing the gas through aqueous KOH absorbs \(CO_2\). The decrease in volume is the volume of \(CO_2\) produced: \(85 \text{ cm}^3 - 55 \text{ cm}^3 = 30 \text{ cm}^3\). 3. Excess oxygen: The remaining 55 cm³ of gas is the unreacted excess \(O_2\). 4. Reacted oxygen: The volume of \(O_2\) consumed during combustion is \(100 \text{ cm}^3 - 55 \text{ cm}^3 = 45 \text{ cm}^3\). 5. Stoichiometric ratio: The combustion equation of a hydrocarbon is: \(C_xH_y + (x + \frac{y}{4}) O_2 \rightarrow x CO_2 + \frac{y}{2} H_2O\) Ratio of hydrocarbon to \(CO_2\) = \(10 : 30 = 1 : 3\), so \(x = 3\). Ratio of hydrocarbon to reacted \(O_2\) = \(10 : 45 = 1 : 4.5\), so \(x + \frac{y}{4} = 4.5\). Substituting \(x = 3\) gives \(3 + \frac{y}{4} = 4.5 \Rightarrow \frac{y}{4} = 1.5 \Rightarrow y = 6\). Therefore, the formula is \(C_3H_6\).
評分準則
Award 1 mark for the correct option (B). - Reject: other options due to incorrect stoichiometric analysis of gas contraction.
題目 2 · 選擇題
1 分
A sample of hydrated magnesium sulfate, \(MgSO_4 \cdot xH_2O\), has a mass of 4.93 g. After heating to constant mass to remove all water of crystallisation, the mass of the anhydrous residue, \(MgSO_4\), is 2.41 g.
What is the value of \(x\)? [Relative atomic masses, \(A_r\): H, 1.0; O, 16.0; Mg, 24.3; S, 32.1]
A.1
B.5
C.7
D.10
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解題
1. Calculate the mass of water of crystallisation lost: \(4.93 \text{ g} - 2.41 \text{ g} = 2.52 \text{ g}\). 2. Calculate the molar mass of anhydrous \(MgSO_4\): \(M_r = 24.3 + 32.1 + (4 \times 16.0) = 120.4 \text{ g mol}^{-1}\). 3. Calculate the number of moles of anhydrous \(MgSO_4\): \(n(MgSO_4) = \frac{2.41 \text{ g}}{120.4 \text{ g mol}^{-1}} \approx 0.0200 \text{ mol}\). 4. Calculate the molar mass of \(H_2O\): \(M_r = (2 \times 1.0) + 16.0 = 18.0 \text{ g mol}^{-1}\). 5. Calculate the number of moles of water lost: \(n(H_2O) = \frac{2.52 \text{ g}}{18.0 \text{ g mol}^{-1}} = 0.1400 \text{ mol}\). 6. Find the mole ratio: \(x = \frac{n(H_2O)}{n(MgSO_4)} = \frac{0.1400}{0.0200} = 7\). Thus, the formula is \(MgSO_4 \cdot 7H_2O\).
評分準則
Award 1 mark for the correct option (C). - Reject: A, B, D due to incorrect calculation of water/anhydrous salt mole ratio.
題目 3 · 選擇題
1 分
Which statement best explains why the first ionisation energy of sulfur is lower than that of phosphorus?
A.Sulfur has a greater nuclear charge, which increases the shielding effect of the inner shell electrons.
B.The outermost electron in a sulfur atom is in a 3p orbital, whereas in phosphorus it is in a 3s orbital.
C.The sulfur atom has spin-paired electrons in one of its 3p orbitals, resulting in inter-electron repulsion.
D.Phosphorus has a stable half-filled 3p subshell, and its nuclear charge is larger than that of sulfur.
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解題
The electronic configurations of phosphorus and sulfur are: - P: \(1s^2 2s^2 2p^6 3s^2 3p^3\) - S: \(1s^2 2s^2 2p^6 3s^2 3p^4\)
In phosphorus, the 3p subshell is half-filled with one electron in each of the three 3p orbitals (no pairing). In sulfur, there are four electrons in the 3p subshell, meaning one of the 3p orbitals contains a pair of spin-paired electrons. The repulsion between these two paired electrons in the same orbital makes it easier to remove the outer electron from sulfur, resulting in a lower first ionisation energy than phosphorus.
評分準則
Award 1 mark for the correct option (C). - Reject: A (incorrect science regarding shielding and nuclear charge interaction), - Reject: B (both are in 3p orbitals), - Reject: D (sulfur has a larger nuclear charge than phosphorus).
題目 4 · 選擇題
1 分
Four separate 0.1 mol samples of Period 3 oxides are added to 100 cm³ of water:
Which of the resulting mixtures will have a pH of less than 3?
A.3 and 4 only
B.1, 3 and 4 only
C.2 and 3 only
D.4 only
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解題
1. \(Na_2O\) reacts with water to form strongly alkaline sodium hydroxide (pH around 13-14). 2. \(Al_2O_3\) is insoluble in water and does not react or dissolve, so the water's pH remains close to 7 (neutral). 3. \(P_4O_{10}\) reacts vigorously with water to form phosphoric(V) acid, \(H_3PO_4\), which is a moderately strong acid. In this concentration, the pH is well below 3. 4. \(SO_3\) reacts violently with water to form sulfuric(VI) acid, \(H_2SO_4\), which is a strong acid that completely dissociates, giving a pH far below 3. Therefore, only 3 and 4 yield solutions with pH < 3.
評分準則
Award 1 mark for the correct option (A). - Reject: options including 1 (basic) or 2 (insoluble/neutral).
題目 5 · 選擇題
1 分
In a calorimetry experiment, 50.0 cm³ of 1.00 mol dm⁻³ aqueous hydrochloric acid, \(HCl\), at 21.5 °C is mixed with 50.0 cm³ of 1.00 mol dm⁻³ aqueous sodium hydroxide, \(NaOH\), also at 21.5 °C in a polystyrene cup. The temperature rises to a maximum of 28.2 °C.
Assume that the density of the solution is 1.00 g cm⁻³ and the specific heat capacity of the solution is 4.18 J g⁻¹ K⁻¹.
What is the enthalpy change of neutralisation, \(\Delta H_{neut}\), for this reaction?
A.-56.0 kJ mol⁻¹
B.-28.0 kJ mol⁻¹
C.+56.0 kJ mol⁻¹
D.-1.12 x 10³ kJ mol⁻¹
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解題
1. Calculate temperature rise: \(\Delta T = 28.2 - 21.5 = 6.7 \text{ °C} = 6.7 \text{ K}\). 2. Calculate the total mass of the mixture: \(m = 50.0 \text{ cm}^3 + 50.0 \text{ cm}^3 = 100.0 \text{ cm}^3 \rightarrow 100.0 \text{ g}\) (since density is 1.00 g cm⁻³). 3. Calculate heat released: \(q = m c \Delta T = 100.0 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times 6.7 \text{ K} = 2800.6 \text{ J} = 2.801 \text{ kJ}\). 4. Calculate the moles of water formed during neutralisation: \(n(H^+) = c \times V = 1.00 \text{ mol dm}^{-3} \times 0.0500 \text{ dm}^3 = 0.0500 \text{ mol}\). 5. Calculate standard enthalpy change of neutralisation per mole of water formed: \(\Delta H_{neut} = -\frac{q}{n} = -\frac{2.801 \text{ kJ}}{0.0500 \text{ mol}} = -56.02 \text{ kJ mol}^{-1}\) (exothermic, so negative). Thus, \(\Delta H_{neut} = -56.0 \text{ kJ mol}^{-1}\).
評分準則
Award 1 mark for the correct option (A). - Reject: B (incorrect division by total moles instead of moles of water formed, or arithmetic error), - Reject: C (positive sign instead of negative), - Reject: D (wrong multiplier).
題目 6 · 選擇題
1 分
What is the standard enthalpy change of formation of liquid ethanol, \(C_2H_5OH(l)\), given the following standard enthalpies of combustion?
Which change would increase the value of the equilibrium constant, \(K_c\)?
A.Adding a catalyst
B.Decreasing the temperature
C.Increasing the pressure
D.Increasing the concentration of SO2(g)
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解題
1. The equilibrium constant, \(K_c\), of a reaction is only altered by a change in temperature. 2. The forward reaction is exothermic (\(\Delta H = -197 \text{ kJ mol}^{-1}\)). 3. According to Le Chatelier's principle, lowering the temperature will shift the equilibrium position to the right (the exothermic direction) to release heat and counteract the change. 4. Shifting to the right increases the equilibrium concentration of the products and decreases that of the reactants, which increases the value of \(K_c\).
評分準則
Award 1 mark for the correct option (B). - Reject: A (catalysts do not change \(K_c\)), - Reject: C and D (pressure and concentration shift the position but do not change the value of \(K_c\)).
題目 8 · 選擇題
1 分
In an experiment, 1.00 mol of ethyl ethanoate and 1.00 mol of water are mixed together and allowed to reach equilibrium at a constant temperature:
At equilibrium, it is found that 0.33 mol of ethanoic acid has been formed.
What is the value of the equilibrium constant, \(K_c\), under these conditions?
A.0.24
B.0.49
C.2.0
D.4.1
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解題
1. Set up an ICE table (Initial, Change, Equilibrium) in terms of moles: - Initial: \(n(CH_3COOCH_2CH_3) = 1.00\), \(n(H_2O) = 1.00\), \(n(CH_3COOH) = 0\), \(n(CH_3CH_2OH) = 0\). - Change: Since 0.33 mol of \(CH_3COOH\) is formed, \(-0.33\) mol of each reactant is consumed, and \(+0.33\) mol of each product is formed. - Equilibrium: \(n(CH_3COOCH_2CH_3) = 0.67\), \(n(H_2O) = 0.67\), \(n(CH_3COOH) = 0.33\), \(n(CH_3CH_2OH) = 0.33\).
2. Write the expression for \(K_c\): \(K_c = \frac{[CH_3COOH][CH_3CH_2OH]}{[CH_3COOCH_2CH_3][H_2O]}\)
3. Since the total volume \(V\) cancels out in the numerator and denominator: \(K_c = \frac{(0.33 / V)(0.33 / V)}{(0.67 / V)(0.67 / V)} = \frac{0.33 \times 0.33}{0.67 \times 0.67} \approx 0.24\).
評分準則
Award 1 mark for the correct option (A). - Reject: B (calculated \(0.33 / 0.67\) without squaring), - Reject: C (calculated reactant/product or other inverse ratios), - Reject: D (calculated \(1 / K_c = 4.1\)).
題目 9 · 選擇題
1 分
A mixture of \(0.10\text{ mol}\) of anhydrous calcium carbonate, \(\text{CaCO}_3\), and \(0.10\text{ mol}\) of anhydrous sodium carbonate, \(\text{Na}_2\text{CO}_3\), is heated strongly in a crucible until there is no further change in mass. What is the total volume of gas released, measured at room temperature and pressure (r.t.p.)?
[Assume \(1\text{ mol}\) of gas occupies \(24.0\text{ dm}^3\) at r.t.p.]
A.\(2.40\text{ dm}^3\)
B.\(4.80\text{ dm}^3\)
C.\(1.20\text{ dm}^3\)
D.\(0.24\text{ dm}^3\)
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解題
Upon heating strongly, calcium carbonate decomposes according to the equation: \(\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)\) Therefore, \(0.10\text{ mol}\) of \(\text{CaCO}_3\) produces \(0.10\text{ mol}\) of \(\text{CO}_2\) gas. Sodium carbonate, \(\text{Na}_2\text{CO}_3\), is thermally stable at these temperatures and does not decompose. Thus, the total volume of gas released is: \(0.10\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 2.40\text{ dm}^3\).
評分準則
1 mark for identifying that only calcium carbonate decomposes, calculating the moles of gas as 0.10 mol, and converting this to a volume of 2.40 dm³.
題目 10 · 選擇題
1 分
Which statement correctly explains the difference in physical properties between elements in Period 3?
A.Silicon has the highest electrical conductivity of all the elements in Period 3 because of its giant covalent structure.
B.The melting points of the elements decrease continuously from sodium to argon across Period 3.
C.Phosphorus exists as \(\text{P}_4\) molecules and has a higher melting point than sulfur, which exists as \(\text{S}_8\) molecules.
D.Chlorine has a lower boiling point than sulfur because chlorine molecules (\(\text{Cl}_2\)) have fewer electrons than sulfur molecules (\(\text{S}_8\)).
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解題
Chlorine exists as diatomic \(\text{Cl}_2\) molecules, which contain 34 electrons per molecule. Sulfur exists as octatomic \(\text{S}_8\) molecules, which contain 128 electrons per molecule. The larger number of electrons in sulfur molecules leads to stronger instantaneous dipole-induced dipole (London dispersion) forces, resulting in a higher boiling point for sulfur and a lower boiling point for chlorine.
評分準則
1 mark for selecting option D, which correctly identifies that chlorine has weaker intermolecular forces than sulfur due to fewer electrons per molecule.
題目 11 · 選擇題
1 分
What is the standard enthalpy change of formation, \(\Delta H_f^\ominus\), of propan-1-ol, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}(l)\), given the following standard enthalpy changes of combustion, \(\Delta H_c^\ominus\)?
The equation for the formation of propan-1-ol is: \(3\text{C}(s) + 4\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}(l)\)
Using Hess's Law and combustion data: \(\Delta H_f^\ominus = \sum \Delta H_c^\ominus (\text{reactants}) - \sum \Delta H_c^\ominus (\text{products})\)
If the temperature of the container is increased while the volume is kept constant, which row correctly describes the effect on the concentration of \(\text{SO}_3(g)\) and the value of the equilibrium constant, \(K_c\)?
A.Concentration of \(\text{SO}_3(g)\) increases; \(K_c\) increases
B.Concentration of \(\text{SO}_3(g)\) decreases; \(K_c\) decreases
C.Concentration of \(\text{SO}_3(g)\) decreases; \(K_c\) remains constant
D.Concentration of \(\text{SO}_3(g)\) increases; \(K_c\) remains constant
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解題
According to Le Chatelier's principle, if the temperature of a system in dynamic equilibrium is increased, the equilibrium position shifts in the endothermic direction to absorb heat. Since the forward reaction is exothermic (\(\Delta H = -197\text{ kJ mol}^{-1}\)), the reverse reaction is endothermic. Thus, the equilibrium shifts to the left, which decreases the concentration of \(\text{SO}_3(g)\). Since temperature is the only variable that changes the value of \(K_c\), and the equilibrium has shifted to the reactants side, the value of \(K_c\) decreases.
評分準則
1 mark for identifying both that the concentration of sulfur trioxide decreases and the value of Kc decreases with an increase in temperature.
題目 13 · 選擇題
1 分
A sample of \(1.20\text{ g}\) of an unknown Group 2 metal, \(M\), reacts completely with excess hydrochloric acid to produce \(1.20\text{ dm}^3\) of hydrogen gas, measured at room temperature and pressure (r.t.p.).
What is the identity of the metal \(M\)?
[Assume \(1\text{ mol}\) of gas occupies \(24.0\text{ dm}^3\) at r.t.p.]
A.Beryllium
B.Magnesium
C.Calcium
D.Strontium
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解題
The reaction of a Group 2 metal with hydrochloric acid is: \(M(s) + 2\text{HCl}(aq) \rightarrow M\text{Cl}_2(aq) + \text{H}_2(g)\)
Moles of \(\text{H}_2\) gas produced: \(n(\text{H}_2) = \frac{1.20\text{ dm}^3}{24.0\text{ dm}^3\text{ mol}^{-1}} = 0.050\text{ mol}\)
Since the molar ratio of \(M\) to \(\text{H}_2\) is 1:1, moles of \(M\) reacted = \(0.050\text{ mol}\).
Molar mass of \(M\): \(A_r = \frac{\text{mass}}{\text{moles}} = \frac{1.20\text{ g}}{0.050\text{ mol}} = 24.0\text{ g mol}^{-1}\)
This corresponds to magnesium (\(\text{Mg}\)), which has an relative atomic mass of approximately \(24.3\text{ g mol}^{-1}\).
評分準則
1 mark for calculating the moles of hydrogen gas, using the stoichiometry to find the moles of metal, calculating the relative atomic mass of the metal, and correctly matching it to Magnesium.
題目 14 · 選擇題
1 分
A solid white oxide of a Period 3 element dissolves in water to form a strongly acidic solution with a pH of approximately 2.
What is the formula of the oxide?
A.\(\text{MgO}\)
B.\(\text{Al}_2\text{O}_3\)
C.\(\text{SiO}_2\)
D.\(\text{P}_4\text{O}_{10}\)
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解題
\(\text{P}_4\text{O}_{10}\) is a solid white molecular oxide of Period 3 that reacts vigorously with water to form phosphoric(V) acid, \(\text{H}_3\text{PO}_4\), which is a moderately strong acid yielding a pH around 1–2. - \(\text{MgO}\) is a basic oxide that is only slightly soluble in water, forming a weakly alkaline solution (pH ~ 9). - \(\text{Al}_2\text{O}_3\) is amphoteric and completely insoluble in water. - \(\text{SiO}_2\) is a giant covalent acidic oxide but is completely insoluble in water.
評分準則
1 mark for identifying phosphorus(V) oxide as the only Period 3 oxide that is a white solid and dissolves in water to form a strongly acidic solution.
題目 15 · 選擇題
1 分
In a calorimetry experiment, \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) aqueous sodium hydroxide, \(\text{NaOH}\), is mixed with \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) hydrochloric acid, \(\text{HCl}\). The temperature of the mixture increases by \(6.6\ ^\circ\text{C}\).
Assume the density of the final mixture is \(1.00\text{ g cm}^{-3}\) and its specific heat capacity is \(4.18\text{ J g}^{-1}\,^\circ\text{C}^{-1}\).
What is the enthalpy change of neutralisation, \(\Delta H_{\text{neu}}\), for this reaction?
A.\(-55.2\text{ kJ mol}^{-1}\)
B.\(-27.6\text{ kJ mol}^{-1}\)
C.\(-5.52\text{ kJ mol}^{-1}\)
D.\(+55.2\text{ kJ mol}^{-1}\)
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解題
Total volume of mixture = \(50.0 + 50.0 = 100.0\text{ cm}^3\) Total mass of mixture (\(m\)) = \(100.0\text{ g}\)
Heat energy released (\(q\)): \(q = m \times c \times \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\,^\circ\text{C}^{-1} \times 6.6\ ^\circ\text{C} = 2758.8\text{ J} = 2.76\text{ kJ}\)
Moles of water formed: \(n(\text{NaOH}) = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\) \(n(\text{HCl}) = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\) Since \(\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}\), the moles of \(\text{H}_2\text{O}\) formed is \(0.0500\text{ mol}\).
Enthalpy change of neutralisation (per mole of water formed): \(\Delta H_{\text{neu}} = -\frac{2.7588\text{ kJ}}{0.0500\text{ mol}} = -55.2\text{ kJ mol}^{-1}\).
評分準則
1 mark for calculating the correct heat energy released (2.76 kJ), finding the moles of water formed (0.0500 mol), and calculating the exothermic enthalpy change of neutralisation as -55.2 kJ/mol.
題目 16 · 選擇題
1 分
In a reaction flask of volume \(1.00\text{ dm}^3\), \(1.00\text{ mol}\) of ethyl ethanoate and \(1.00\text{ mol}\) of water are allowed to reach dynamic equilibrium at a constant temperature.
At equilibrium, the mixture is analyzed and found to contain \(0.25\text{ mol}\) of ethanoic acid. What is the value of the equilibrium constant, \(K_c\), at this temperature?
A.\(0.11\)
B.\(0.33\)
C.\(3.0\)
D.\(9.0\)
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解題
Let's set up the ICE (Initial, Change, Equilibrium) table: - **Initial moles**: Ester = 1.00 mol, Water = 1.00 mol, Acid = 0 mol, Alcohol = 0 mol
- **Change in moles**: Since 0.25 mol of acid is present at equilibrium, change is +0.25 mol. This means: Acid: +0.25 mol Alcohol: +0.25 mol Ester: -0.25 mol Water: -0.25 mol
1 mark for calculating correct equilibrium amounts of all four components, substituting them into the Kc expression, and obtaining 0.11.
題目 17 · 選擇題
1 分
When \( 0.240\text{ g} \) of an impure sample of magnesium (containing only unreactive impurities) is treated with an excess of dilute hydrochloric acid, \( 216\text{ cm}^3 \) of hydrogen gas is collected at room temperature and pressure (r.t.p.).
What is the percentage purity of the magnesium sample?
[Assume 1 mole of gas occupies \( 24.0\text{ dm}^3 \) at r.t.p. and \( A_{\text{r}}(\text{Mg}) = 24.3 \)]
A.82.3%
B.90.0%
C.91.1%
D.92.7%
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解題
Step 1: Calculate the number of moles of hydrogen gas produced: \( n(\text{H}_2) = \frac{216\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.00900\text{ mol} \)
Step 2: Use the balanced equation to find the moles of magnesium that reacted: \( \text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2(\text{aq}) + \text{H}_2(\text{g}) \) Since the ratio is 1:1, \( n(\text{Mg}) = 0.00900\text{ mol} \).
Step 3: Calculate the mass of pure magnesium: \( \text{Mass} = 0.00900\text{ mol} \times 24.3\text{ g mol}^{-1} = 0.2187\text{ g} \)
1 mark: Correct calculation of moles of hydrogen, relative moles of magnesium, mass of magnesium, and final percentage purity to 3 significant figures.
題目 18 · 選擇題
1 分
A \( 10.0\text{ cm}^3 \) sample of a gaseous hydrocarbon \( \text{C}_x\text{H}_y \) is completely burned in \( 70.0\text{ cm}^3 \) of oxygen (an excess). After cooling to room temperature, the total volume of gas remaining is \( 50.0\text{ cm}^3 \). This residual gas is passed through an excess of aqueous sodium hydroxide, and the volume of gas decreases to \( 20.0\text{ cm}^3 \).
All gas volumes are measured at the same temperature and pressure.
What is the molecular formula of the hydrocarbon?
A.\( \text{CH}_4 \)
B.\( \text{C}_2\text{H}_6 \)
C.\( \text{C}_3\text{H}_6 \)
D.\( \text{C}_3\text{H}_8 \)
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解題
Step 1: Identify the volume of carbon dioxide produced. Passing the gas through aqueous \( \text{NaOH} \) absorbs the acidic oxide \( \text{CO}_2 \). \( \text{Volume of CO}_2 \text{ produced} = 50.0\text{ cm}^3 - 20.0\text{ cm}^3 = 30.0\text{ cm}^3 \). Since \( 10.0\text{ cm}^3 \) of hydrocarbon produced \( 30.0\text{ cm}^3 \) of \( \text{CO}_2 \), \( x = \frac{30.0}{10.0} = 3 \).
Step 2: Identify the volume of unreacted oxygen. The final remaining \( 20.0\text{ cm}^3 \) of gas must be the excess unreacted oxygen gas.
Step 3: Find the volume of oxygen that reacted. \( \text{Volume of O}_2 \text{ reacted} = 70.0\text{ cm}^3 - 20.0\text{ cm}^3 = 50.0\text{ cm}^3 \).
Step 4: Write the general equation for combustion and solve for \( y \): \( \text{C}_3\text{H}_y + (3 + \frac{y}{4})\text{O}_2 \rightarrow 3\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O} \) Since 1 volume of hydrocarbon reacts with 5 volumes of \( \text{O}_2 \): \( 3 + \frac{y}{4} = 5 \implies \frac{y}{4} = 2 \implies y = 8 \).
Thus, the formula is \( \text{C}_3\text{H}_8 \).
評分準則
1 mark: Correct determination of carbon number from CO2 absorption and hydrogen number from volume of oxygen reacted to yield C3H8.
題目 19 · 選擇題
1 分
An aqueous solution of a dicarboxylic acid, \( \text{H}_2\text{A} \), has a concentration of \( 0.0500\text{ mol dm}^{-3} \).
A \( 25.0\text{ cm}^3 \) sample of this acid solution requires \( 20.0\text{ cm}^3 \) of an aqueous sodium hydroxide solution, \( \text{NaOH} \), for complete neutralisation.
What is the concentration of the sodium hydroxide solution?
A.\( 0.0400\text{ mol dm}^{-3} \)
B.\( 0.0625\text{ mol dm}^{-3} \)
C.\( 0.125\text{ mol dm}^{-3} \)
D.\( 0.250\text{ mol dm}^{-3} \)
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解題
Step 1: Write the balanced equation for the reaction: \( \text{H}_2\text{A(aq)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{A(aq)} + 2\text{H}_2\text{O(l)} \)
Step 2: Calculate the moles of the dicarboxylic acid: \( n(\text{H}_2\text{A}) = 0.0500\text{ mol dm}^{-3} \times \frac{25.0}{1000}\text{ dm}^3 = 1.25 \times 10^{-3}\text{ mol} \)
Step 4: Calculate the concentration of the \( \text{NaOH} \) solution: \( \text{Concentration} = \frac{2.50 \times 10^{-3}\text{ mol}}{0.0200\text{ dm}^3} = 0.125\text{ mol dm}^{-3} \)
評分準則
1 mark: Correct stoichiometry used (1:2 ratio of acid to base) to find concentration of NaOH.
題目 20 · 選擇題
1 分
The elements in Period 3 show distinct trends in physical properties such as electrical conductivity and melting point.
Which row correctly pairs a Period 3 element with the reason for its relative physical property?
| Row | Element | Physical Property | Reason | | :--- | :--- | :--- | :--- | | A | Silicon | High melting point | Strong metallic bonding between delocalised electrons and positive ions | | B | Phosphorus | Low electrical conductivity | Giant molecular structure with localized covalent bonds | | C | Sulfur | Higher melting point than phosphorus | Greater van der Waals' forces between \( \text{S}_8 \) molecules than \( \text{P}_4 \) molecules | | D | Argon | High melting point | Strong covalent bonds within the monatomic structure |
A.A
B.B
C.C
D.D
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解題
A is incorrect: Silicon is a metalloid with a giant molecular (covalent) structure, not metallic bonding. B is incorrect: Phosphorus has a simple molecular structure (\( \text{P}_4 \)), not a giant molecular structure. C is correct: Sulfur exists as \( \text{S}_8 \) rings, which contain more electrons than \( \text{P}_4 \) molecules. This results in stronger London dispersion (van der Waals') forces between molecules, giving sulfur a higher melting point than phosphorus. D is incorrect: Argon is monatomic with weak van der Waals' forces, giving it an extremely low melting point.
評分準則
1 mark: Correctly identify that sulfur's higher melting point than phosphorus is due to stronger intermolecular forces because of its larger molecular formula (S8 vs P4).
題目 21 · 選擇題
1 分
An excess of each of the following Period 3 oxides is added separately to containers of water.
What is the order of the pH of the resulting solutions, from lowest pH (most acidic) to highest pH (most alkaline)?
A.2, 3, 1
B.3, 2, 1
C.1, 2, 3
D.1, 3, 2
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解題
- \( \text{SO}_3 \) reacts with water to form sulfuric acid, \( \text{H}_2\text{SO}_4 \), which is a strong diprotic acid (giving a very low pH, around 0-1). - \( \text{P}_4\text{O}_{10} \) reacts with water to form phosphoric acid, \( \text{H}_3\text{PO}_4 \), which is a moderately strong acid (giving a low pH, around 1-2, but higher than sulfuric acid). - \( \text{Na}_2\text{O} \) reacts with water to form sodium hydroxide, \( \text{NaOH} \), which is a strong alkali (giving a high pH, around 13-14).
Therefore, the order from lowest pH to highest pH is 3, 2, 1.
評分準則
1 mark: Correct ranking of acid-base character of the Period 3 oxides in water from most acidic to most basic.
題目 22 · 選擇題
1 分
The standard enthalpy change of combustion of butane, \( \text{C}_4\text{H}_{10}\text{(g)} \), is \( -2877\text{ kJ mol}^{-1} \).
What mass of butane must be completely burned in oxygen to raise the temperature of \( 500\text{ g} \) of water from \( 20.0\ ^\circ\text{C} \) to \( 85.0\ ^\circ\text{C} \), assuming that only \( 60.0\% \) of the heat released from combustion is absorbed by the water?
[Specific heat capacity of water, \( c = 4.18\text{ J g}^{-1}\ \text{K}^{-1} \); \( M_{\text{r}}(\text{C}_4\text{H}_{10}) = 58.0 \)]
A.\( 1.64\text{ g} \)
B.\( 2.74\text{ g} \)
C.\( 4.56\text{ g} \)
D.\( 7.61\text{ g} \)
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解題
Step 1: Calculate the heat energy absorbed by the water: \( q = m c \Delta T \) \( q = 500\text{ g} \times 4.18\text{ J g}^{-1}\text{K}^{-1} \times (85.0 - 20.0)\text{ K} = 135850\text{ J} = 135.85\text{ kJ} \)
Step 2: Account for the heating efficiency of \( 60.0\% \): \( Q_{\text{total}} = \frac{135.85\text{ kJ}}{0.600} = 226.42\text{ kJ} \)
Step 3: Calculate the moles of butane needed: \( n = \frac{226.42\text{ kJ}}{2877\text{ kJ mol}^{-1}} = 0.07870\text{ mol} \)
Step 4: Calculate the mass of butane needed: \( \text{mass} = n \times M_{\text{r}} = 0.07870\text{ mol} \times 58.0\text{ g mol}^{-1} = 4.56\text{ g} \)
評分準則
1 mark: Correctly calculates heat absorbed, total combustion energy required, moles of butane, and mass of butane.
題目 23 · 選擇題
1 分
Consider the following standard enthalpy changes of formation, \( \Delta H_{\text{f}}^\ominus \): - \( \Delta H_{\text{f}}^\ominus[\text{CO(g)}] = -110.5\text{ kJ mol}^{-1} \) - \( \Delta H_{\text{f}}^\ominus[\text{CO}_2\text{(g)}] = -393.5\text{ kJ mol}^{-1} \) - \( \Delta H_{\text{f}}^\ominus[\text{Fe}_2\text{O}_3\text{(s)}] = -824.2\text{ kJ mol}^{-1} \)
The equation for the reduction of iron(III) oxide by carbon monoxide is: \( \text{Fe}_2\text{O}_3\text{(s)} + 3\text{CO(g)} \rightarrow 2\text{Fe(s)} + 3\text{CO}_2\text{(g)} \)
What is the standard enthalpy change of this reaction?
1 mark: Correctly applies Hess's law to determine the reaction enthalpy from standard enthalpies of formation.
題目 24 · 選擇題
1 分
A \( 2.00\text{ dm}^3 \) sealed vessel is filled with \( 4.00\text{ mol} \) of gas \( \text{A} \) and \( 4.00\text{ mol} \) of gas \( \text{B} \). The mixture is allowed to reach equilibrium at a constant temperature according to the following equation:
Step 4: Calculate the value of \( K_{\text{c}} \): \( K_{\text{c}} = \frac{[\text{C}]^2 [\text{D}]}{[\text{A}] [\text{B}]^3} = \frac{(1.20)^2 \times 0.60}{1.40 \times (0.20)^3} = \frac{0.864}{1.40 \times 0.008} = 77.1\text{ dm}^6\text{ mol}^{-2} \)
評分準則
1 mark: Correctly calculates equilibrium concentrations by dividing moles by the volume, and applies the Kc formula accurately.
題目 25 · 選擇題
1 分
A sample of \( 0.120\text{ g} \) of a magnesium–aluminium alloy reacts completely with an excess of dilute hydrochloric acid. The volume of hydrogen gas collected at room temperature and pressure (r.t.p.) is \( 144\text{ cm}^3 \).
What is the percentage by mass of magnesium in the alloy?
[Assume 1 mole of gas occupies \( 24.0\text{ dm}^3 \) at r.t.p. \( A_r \) values: \( \text{Mg} = 24.3 \), \( \text{Al} = 27.0 \)]
A.38.6%
B.47.4%
C.51.4%
D.61.4%
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解題
1. Calculate the total moles of hydrogen gas produced:
2. Note that chloride ions are spectator ions and do not participate in the precipitation reaction between \(\text{Ba}^{2+}\) and \(\text{SO}_4^{2-}\). Therefore, the total moles of \(\text{Cl}^-\) in solution remain unchanged at \(0.0120\text{ mol}\).
- Method: Calculate moles of chloride ions (spectator ions) and divide by the total volume of the mixture. - Reject: 0.120 mol dm^-3 (failing to multiply by 2 for BaCl2).
題目 27 · 選擇題
1 分
An element, \( \text{X} \), in Period 3 of the Periodic Table reacts with oxygen to form an oxide, \( \text{Y} \).
Oxide \( \text{Y} \) is a white solid at room temperature. It reacts with both dilute aqueous sodium hydroxide and dilute hydrochloric acid.
Which statement about element \( \text{X} \) and its compounds is correct?
A.Element X conducts electricity only in the liquid state.
B.The chloride of X dissolves in water to form a solution with a pH of approximately 7.
C.The oxide of X has a giant covalent structure.
D.Element X has a lower first ionisation energy than the element immediately to its left.
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解題
1. Identify element \(\text{X}\): An oxide in Period 3 that is a white solid and reacts with both acids and bases (amphoteric) is aluminium oxide, \(\text{Al}_2\text{O}_3\). Therefore, \(\text{X}\) is aluminium (\(\text{Al}\)) and \(\text{Y}\) is aluminium oxide (\(\text{Al}_2\text{O}_3\)).
2. Evaluate the statements:
- Option A: Aluminium is a metal and conducts electricity in both solid and liquid states due to its delocalised valence electrons. (Incorrect)
- Option B: Aluminium chloride (\(\text{AlCl}_3\)) undergoes hydrolyses in water to form an acidic solution with a pH of about 3. (Incorrect)
- Option C: Aluminium oxide has a giant ionic lattice structure with significant covalent character, not a giant covalent structure. (Incorrect)
- Option D: The element immediately to the left of aluminium is magnesium. Magnesium (\(1s^2 2s^2 2p^6 3s^2\)) has a higher first ionisation energy than aluminium (\(1s^2 2s^2 2p^6 3s^2 3p^1\)) because the outer electron of aluminium is in a \(3p\) orbital, which is higher in energy and more shielded than the \(3s\) orbital of magnesium. Thus, aluminium has a lower first ionisation energy than magnesium. (Correct)
評分準則
Award 1 mark for the correct answer (D).
- Method: Identify X as Al, and apply principles of shielding and sub-shell energy levels to explain the anomaly in Period 3 first ionisation energies.
題目 28 · 選擇題
1 分
The table below shows the melting points and electrical conductivities of four consecutive elements in Period 3 of the Periodic Table.
| Element | Melting point / K | Electrical conductivity of the solid | |---|---|---| | P | 371 | Good | | Q | 923 | Good | | R | 933 | Good | | S | 1687 | Very poor |
Which statement about these elements is correct?
A.Element P reacts with water to produce a solution with a pH of about 9.
B.Element Q has a higher first ionisation energy than element P.
C.The oxide of element R reacts with water to form a strongly alkaline solution.
D.Element S reacts with chlorine to form a liquid chloride that does not react with water.
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解題
1. Identify the elements: Based on the consecutive melting points in Period 3, the elements are:
- \(\text{P} = \text{Sodium (Na)}\) (low melting point metal)
- Option A: Sodium (\(\text{P}\)) reacts vigorously with water to form sodium hydroxide, which is a strong base producing a pH of around 13–14. (Incorrect)
- Option B: Magnesium (\(\text{Q}\)) has a higher nuclear charge than sodium (\(\text{P}\)) with similar shielding, so its outer electrons are held more tightly, resulting in a higher first ionisation energy. (Correct)
- Option C: Aluminium oxide (\(\text{Al}_2\text{O}_3\)) is insoluble in water and does not react to form any alkaline solution. (Incorrect)
- Option D: Silicon (\(\text{S}\)) reacts with chlorine to form silicon tetrachloride (\(\text{SiCl}_4\)), which is a liquid but hydrolyses rapidly and violently in water to form \(\text{SiO}_2\) and \(\text{HCl}\). (Incorrect)
評分準則
Award 1 mark for the correct answer (B).
- Method: Correctly identify elements P to S as Na, Mg, Al, Si, and apply periodicity trends of physical and chemical properties.
題目 29 · 選擇題
1 分
The table shows the bond energies for some covalent bonds.
- Method: Correctly apply Hess's Law with stoichiometric factors 3 for C and 4 for H2. - Reject: -1540 (failing to multiply by coefficients) or +106 (incorrect sign due to reversing reactants/products).
題目 31 · 選擇題
1 分
An equilibrium is established in a closed container according to the following equation:
Which change will increase the yield of sulfur trioxide, \( \text{SO}_3 \), at equilibrium?
A.Increasing the temperature at constant volume.
B.Decreasing the volume of the container at constant temperature.
C.Adding a catalyst at constant temperature and volume.
D.Removing oxygen from the mixture.
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解題
1. Analyze each option using Le Chatelier's principle:
- Option A: The forward reaction is exothermic (\(\Delta H = -197\text{ kJ mol}^{-1}\)). Increasing the temperature shifts the equilibrium in the endothermic (reverse) direction to absorb the added thermal energy. This decreases the yield of \(\text{SO}_3\). (Incorrect)
- Option B: Decreasing the volume increases the overall pressure. The system shifts in the direction of fewer gas molecules to reduce the pressure. There are 3 moles of gaseous reactants and 2 moles of gaseous products. Therefore, the equilibrium shifts to the right, increasing the yield of \(\text{SO}_3\). (Correct)
- Option C: Adding a catalyst increases the rate of both forward and reverse reactions equally. It does not alter the position of the equilibrium or change the yield. (Incorrect)
- Option D: Removing oxygen (\(\text{O}_2\)) decreases the concentration of a reactant, causing the system to shift to the left to replace the lost reactant. This decreases the yield of \(\text{SO}_3\). (Incorrect)
評分準則
Award 1 mark for the correct answer (B).
- Method: Apply Le Chatelier's principle to temperature, pressure (volume changes), concentration, and catalysts.
題目 32 · 選擇題
1 分
A mixture of \( 2.00\text{ mol} \) of hydrogen gas and \( 1.50\text{ mol} \) of iodine vapor is allowed to reach equilibrium in a closed vessel of volume \( V\text{ dm}^3 \) at a certain temperature.
- Method: Calculate equilibrium moles, set up Kc expression, and substitute values correctly. - Reject: 10.0 (forgetting to square the [HI] term).
題目 33 · 選擇題
1 分
A 20 cm\(^3\) sample of a gaseous hydrocarbon, \(\text{C}_x\text{H}_y\), was exploded with 120 cm\(^3\) of oxygen gas (which is in excess). After cooling to room temperature, the volume of the remaining gaseous mixture was 90 cm\(^3\). After shaking this remaining gaseous mixture with concentrated aqueous sodium hydroxide, the volume decreased to 30 cm\(^3\).
What is the molecular formula of the hydrocarbon? (All gas volumes are measured at room temperature and pressure.)
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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解題
1. Shaking with concentrated aqueous \(\text{NaOH\)} removes \(\text{CO}_2\) gas by absorption. The volume of \(\text{CO}_2\) produced = \(90\text{ cm}^3 - 30\text{ cm}^3 = 60\text{ cm}^3\). 2. The remaining 30 cm\(^3\) of gas is unreacted oxygen. The volume of oxygen reacted = \(120\text{ cm}^3 - 30\text{ cm}^3 = 90\text{ cm}^3\). 3. Use the mole ratio of gas volumes (Avogadro's hypothesis): \(\text{C}_x\text{H}_y + (x + y/4)\text{O}_2 \rightarrow x\text{CO}_2 + y/2\text{H}_2\text{O}\). Ratio of volumes reacting/produced: \(\text{C}_x\text{H}_y : \text{O}_2 : \text{CO}_2 = 20 : 90 : 60 = 1 : 4.5 : 3\). 4. Therefore: \(x = 3\). From oxygen stoichiometry: \(x + y/4 = 4.5 \Rightarrow 3 + y/4 = 4.5 \Rightarrow y = 6\). The molecular formula is \(\text{C}_3\text{H}_6\).
評分準則
1 mark for the correct option B.
題目 34 · 選擇題
1 分
A student heated a sample of hydrated iron(II) sulfate, \(\text{FeSO}_4 \cdot x\text{H}_2\text{O}\), in a crucible to constant mass to determine the value of \(x\).
The following experimental masses were recorded: - Mass of empty crucible and lid = 25.12 g - Mass of crucible, lid, and hydrated salt = 27.90 g - Mass of crucible, lid, and anhydrous salt = 26.64 g
What is the value of \(x\)? [\(A_r\): H = 1.0; O = 16.0; S = 32.1; Fe = 55.8]
A.5
B.6
C.7
D.8
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解題
1. Calculate the mass of anhydrous \(\text{FeSO}_4\) remaining: \(26.64\text{ g} - 25.12\text{ g} = 1.52\text{ g}\). 2. Calculate the mass of water lost: \(27.90\text{ g} - 26.64\text{ g} = 1.26\text{ g}\). 3. Calculate the molar masses: \(M_r(\text{FeSO}_4) = 55.8 + 32.1 + (16.0 \times 4) = 151.9\text{ g mol}^{-1}\), and \(M_r(\text{H}_2\text{O}) = 18.0\text{ g mol}^{-1}\). 4. Convert masses to moles: \(\text{Moles of FeSO}_4 = 1.52 / 151.9 = 0.0100\text{ mol}\); \(\text{Moles of H}_2\text{O} = 1.26 / 18.0 = 0.0700\text{ mol}\). 5. Determine the ratio: \(0.0700 / 0.0100 = 7\). Therefore, \(x = 7\).
評分準則
1 mark for the correct option C.
題目 35 · 選擇題
1 分
Under suitable conditions, an element \(X\) in Period 3 of the Periodic Table reacts with excess oxygen to form an oxide \(Y\).
Oxide \(Y\) reacts vigorously with water to form a solution that turns blue litmus paper red. Oxide \(Y\) also reacts with aqueous sodium hydroxide but is unreactive towards dilute hydrochloric acid.
Which element is \(X\)?
A.aluminium
B.silicon
C.phosphorus
D.sodium
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解題
1. Since the oxide \(Y\) reacts with water to form an acidic solution (turns blue litmus red), \(Y\) must be an acidic oxide of a non-metal that is soluble in water. 2. Silicon dioxide (\(\text{SiO}_2\)) is an acidic oxide but is insoluble in water, so it does not affect litmus. 3. Aluminium oxide (\(\text{Al}_2\text{O}_3\)) is amphoteric, insoluble, and reacts with both acids and bases. 4. Sodium oxide (\(\text{Na}_2\text{O}\)) is basic, reacting with water to form an alkaline solution (\(\text{NaOH}\)) which would turn red litmus blue. 5. Phosphorus reacts with oxygen to form phosphorus(V) oxide (\(\text{P}_4\text{O}_{10}\)), which is a soluble acidic oxide. It reacts vigorously with water to form phosphoric(V) acid, turning blue litmus red, and reacts with bases but not acids. Thus, element \(X\) is phosphorus.
評分準則
1 mark for the correct option C.
題目 36 · 選擇題
1 分
Four elements in Period 3 are sodium, magnesium, aluminium, and silicon.
Which statement about the physical properties of these elements is correct?
A.Silicon has the highest electrical conductivity among these elements because of its giant covalent structure.
B.Aluminium has a higher melting point than sodium because its cations have a higher charge density and contribute more delocalised electrons per atom to the metallic lattice.
C.Sodium has a greater electrical conductivity than magnesium because sodium has more valence electrons free to move.
D.Magnesium has a lower melting point than sodium because of its stronger metallic bonding.
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解題
1. Silicon is a semiconductor and has a low electrical conductivity compared to the metals Na, Mg, and Al, so option A is incorrect. 2. Aluminium has a higher melting point than sodium because \(\text{Al}^{3+}\) ions have a higher charge density and contribute 3 delocalised electrons per atom, leading to much stronger metallic bonding than \(\text{Na}^+\) (which has a lower charge density and contributes only 1 electron per atom). This makes option B correct. 3. Magnesium has higher electrical conductivity than sodium because it has more delocalised electrons per atom available for conduction, so option C is incorrect. 4. Magnesium has a higher melting point than sodium due to its stronger metallic bonding, so option D is incorrect.
評分準則
1 mark for the correct option B.
題目 37 · 選擇題
1 分
In a calorimetry experiment, 50.0 cm\(^3\) of 1.00 mol dm\(^{-3}\) hydrochloric acid, \(\text{HCl(aq)}\), was mixed with 50.0 cm\(^3\) of 1.00 mol dm\(^{-3}\) sodium hydroxide solution, \(\text{NaOH(aq)}\), in a polystyrene cup.
The initial temperature of both solutions was 18.5 °C and the maximum temperature reached was 25.0 °C.
Assume that the density of the final mixture is 1.00 g cm\(^{-3}\) and its specific heat capacity is 4.18 J g\(^{-1\)} K\(^{-1\)}.
What is the enthalpy change of neutralisation, \(\Delta H\), for this reaction?
A.\(-5.4\text{ kJ mol}^{-1}\)
B.\(-27.2\text{ kJ mol}^{-1}\)
C.\(-54.3\text{ kJ mol}^{-1}\)
D.\(-108.7\text{ kJ mol}^{-1}\)
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解題
1. Calculate heat released (\(q\)) using \(q = m c \Delta T\): - Total volume of mixture = \(50.0 + 50.0 = 100.0\text{ cm}^3\). - Mass of mixture (\(m\)) = \(100.0\text{ g}\). - Temperature change (\(\Delta T\)) = \(25.0 - 18.5 = 6.5\text{ K}\). - \(q = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.5\text{ K} = 2717\text{ J} = 2.717\text{ kJ}\). 2. Calculate the moles of water formed: - \(\text{Moles of HCl} = \text{Moles of NaOH} = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\). - Since they react in a 1:1 ratio, 0.0500 mol of water is formed. 3. Calculate \(\Delta H\): - \(\Delta H = -q / n = -2.717\text{ kJ} / 0.0500\text{ mol} = -54.34\text{ kJ mol}^{-1} \approx -54.3\text{ kJ mol}^{-1}\).
評分準則
1 mark for the correct option C.
題目 38 · 選擇題
1 分
The standard enthalpy changes of combustion, \(\Delta H_c^\ominus\), for three substances are given: - \(\text{C(s)}\): \(-393.5\text{ kJ mol}^{-1}\) - \(\text{H}_2\text{(g)}\): \(-285.8\text{ kJ mol}^{-1}\) - \(\text{C}_2\text{H}_4\text{(g)}\): \(-1411.0\text{ kJ mol}^{-1}\)
What is the standard enthalpy change of formation of ethene, \(\text{C}_2\text{H}_4\text{(g)}\)?
A.\(-731.7\text{ kJ mol}^{-1}\)
B.\(-52.4\text{ kJ mol}^{-1}\)
C.\(+52.4\text{ kJ mol}^{-1}\)
D.\(+731.7\text{ kJ mol}^{-1}\)
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解題
1. Write the equation for the standard enthalpy change of formation of ethene: \(2\text{C(s)} + 2\text{H}_2\text{(g)} \rightarrow \text{C}_2\text{H}_4\text{(g)}\). 2. Using combustion data: \(\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\). 3. Substitute the values: \(\Delta H_f^\ominus = 2 \times (-393.5) + 2 \times (-285.8) - (-1411.0)\). 4. Evaluate: \(\Delta H_f^\ominus = -787.0 - 571.6 + 1411.0 = +52.4\text{ kJ mol}^{-1}\).
評分準則
1 mark for the correct option C.
題目 39 · 選擇題
1 分
Consider the following gaseous equilibrium established in a closed container of fixed volume:
Which of the following changes will increase the value of the equilibrium constant, \(K_c\), for this reaction?
A.adding a suitable catalyst to the mixture
B.decreasing the volume of the container
C.increasing the temperature of the system
D.introducing more \(\text{PCl}_5\text{(g)}\) into the container
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解題
1. The value of the equilibrium constant \(K_c\) is only changed by temperature. Changes in concentration, pressure, or the addition of a catalyst have no effect on the value of \(K_c\). Thus, options A, B, and D are incorrect. 2. Since the forward reaction is endothermic (\(\Delta H > 0\)), increasing the temperature of the system shifts the position of equilibrium to the right to absorb the extra thermal energy. 3. This increase in the concentration of products (\(\text{PCl}_3\) and \(\text{Cl}_2\)) relative to the reactant (\(\text{PCl}_5\)) results in a larger value of \(K_c\).
評分準則
1 mark for the correct option C.
題目 40 · 選擇題
1 分
A mixture of 2.00 mol of gas A and 1.50 mol of gas B is allowed to reach equilibrium in a sealed container of volume 2.00 dm\(^3\) at a constant temperature.
At equilibrium, the concentration of C is found to be 0.50 mol dm\)^{-3}\).
What is the value of the equilibrium constant, \(K_c\), for this reaction at this temperature?
A.0.083
B.0.167
C.0.333
D.0.500
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解題
1. Find equilibrium moles: Equilibrium concentration of C = \(0.50\text{ mol dm}^{-3}\). Since \(V = 2.00\text{ dm}^3\), equilibrium moles of C = \(0.50 \times 2.00 = 1.00\text{ mol}\). 2. From the stoichiometry: \(\text{A} + \text{B} \rightleftharpoons 2\text{C} + \text{D}\). - Moles of A reacted = \(1.00 / 2 = 0.50\text{ mol}\). Remaining A = \(2.00 - 0.50 = 1.50\text{ mol}\). - Moles of B reacted = \(1.00 / 2 = 0.50\text{ mol}\). Remaining B = \(1.50 - 0.50 = 1.00\text{ mol}\). - Moles of D formed = \(1.00 / 2 = 0.50\text{ mol}\). 3. Find equilibrium concentrations (divided by 2.00): - \([\text{A}] = 1.50 / 2.00 = 0.75\text{ mol dm}^{-3}\). - \([\text{B}] = 1.00 / 2.00 = 0.50\text{ mol dm}^{-3}\). - \([\text{C}] = 1.00 / 2.00 = 0.50\text{ mol dm}^{-3}\). - \([\text{D}] = 0.50 / 2.00 = 0.25\text{ mol dm}^{-3}\). 4. Calculate \(K_c = \frac{[\text{C}]^2[\text{D}]}{[\text{A}][\text{B}]} = \frac{(0.50)^2 \times 0.25}{0.75 \times 0.50} = \frac{0.25 \times 0.25}{0.375} = \frac{0.0625}{0.375} \approx 0.167\text{ dm}^3\text{ mol}^{-1}\).
評分準則
1 mark for the correct option B.
Paper 21 (AS Level Structured)
Answer all six structured theoretical questions in the space provided on the question paper.
6 題目 · 60 分
題目 1 · structured-theory
10 分
(a) A liquid organic compound, **X**, contains carbon, hydrogen, and oxygen only. A \(1.20\text{ g}\) sample of compound **X** was completely burned in excess oxygen. The combustion products were passed through a drying agent, which increased in mass by \(1.44\text{ g}\). The remaining gases were then passed through concentrated aqueous sodium hydroxide, which increased in mass by \(2.64\text{ g}\).
(i) Calculate the mass of carbon and the mass of hydrogen present in the \(1.20\text{ g}\) sample of **X**. [2]
(ii) Use your answers from (a)(i) to determine the empirical formula of compound **X**. [3]
(b) In a separate experiment, a sample of an anhydrous metal carbonate, \(\text{MCO}_3\), was analyzed. A \(1.18\text{ g}\) sample of \(\text{MCO}_3\) reacted completely with excess dilute hydrochloric acid to produce carbon dioxide gas.
The volume of carbon dioxide gas collected was \(342\text{ cm}^3\) at a temperature of \(298\text{ K}\) and a pressure of \(101\text{ kPa}\).
(i) Calculate the number of moles of carbon dioxide gas produced. Assume carbon dioxide behaves as an ideal gas. (The gas constant \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)). [3]
(ii) Use your answer to (b)(i) to calculate the relative formula mass, \(M_{\text{r}}\), of \(\text{MCO}_3\), and identify the metal \(\text{M}\). [2]
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解題
(a) (i) - The increase in mass of the drying agent is due to water: \(m(\text{H}_2\text{O}) = 1.44\text{ g}\). - Mass of H in the sample = \(1.44 \times \frac{2 \times 1.008}{18.016} = 0.161\text{ g}\) (or \(1.44 \times \frac{2}{18} = 0.160\text{ g}\)). - The increase in mass of the sodium hydroxide is due to carbon dioxide: \(m(\text{CO}_2) = 2.64\text{ g}\). - Mass of C in the sample = \(2.64 \times \frac{12.011}{44.01} = 0.720\text{ g}\).
(ii) - Mass of oxygen in the sample = \(1.20 - (0.720 + 0.160) = 0.320\text{ g}\). - Moles of C = \(\frac{0.720}{12.0} = 0.060\text{ mol}\). - Moles of H = \(\frac{0.160}{1.0} = 0.160\text{ mol}\). - Moles of O = \(\frac{0.320}{16.0} = 0.020\text{ mol}\). - Dividing by the smallest number of moles (0.020): \(\text{C} : \text{H} : \text{O} = 3 : 8 : 1\). - Empirical formula is \(\text{C}_3\text{H}_8\text{O}\).
(ii) - Moles of \(\text{MCO}_3 = \text{moles of } \text{CO}_2 = 0.01395\text{ mol}\). - \(M_{\text{r}}\text{ of } \text{MCO}_3 = \frac{1.18\text{ g}}{0.01395\text{ mol}} = 84.6\text{ g mol}^{-1}\) (or \(84.3\text{ g mol}^{-1}\) if using \(0.0140\text{ mol}\)). - \(A_{\text{r}}\text{ of M} = 84.6 - (12.0 + 3 \times 16.0) = 84.6 - 60.0 = 24.6\). - The metal M is magnesium (\(\text{Mg}\), actual \(A_{\text{r}} = 24.3\)).
評分準則
Part (a)(i): - M1: Calculation of mass of H: \(0.160\text{ g}\) (or \(0.161\text{ g}\)) [1] - M2: Calculation of mass of C: \(0.720\text{ g}\) [1]
Part (a)(ii): - M1: Calculation of mass of O: \(0.320\text{ g}\) (by subtracting C and H from 1.20 g) [1] - M2: Conversion of masses of C, H, and O to moles [1] - M3: Deducing the correct empirical formula ratio (3:8:1) to give \(\text{C}_3\text{H}_8\text{O}\) [1]
Part (b)(i): - M1: Correct conversions: \(V = 3.42 \times 10^{-4}\text{ m}^3\) AND \(p = 101000\text{ Pa}\) [1] - M2: Correct rearrangement of ideal gas equation: \(n = \frac{pV}{RT}\) [1] - M3: Calculation of moles: \(0.01395\text{ mol}\) (accept \(0.0140\text{ mol}\)) [1]
Part (b)(ii): - M1: Calculation of \(M_{\text{r}}\): \(84.3 - 84.6\text{ g mol}^{-1}\) [1] - M2: Identifying metal M as Magnesium / \(\text{Mg}\) [1]
題目 2 · structured-theory
10 分
(a) The elements in Period 3 (sodium to chlorine) show distinct trends in their chemical reactions.
(i) Describe the differences in the reactions of sodium and magnesium with cold water. Include observations and a balanced chemical equation for the reaction of sodium with cold water. [3]
(ii) Describe how magnesium reacts with steam, stating the observations and writing a balanced chemical equation for this reaction. [2]
(b) The oxides of the elements in Period 3 also show characteristic reactions with water.
(i) Describe the reaction of sodium oxide, \(\text{Na}_2\text{O}\), with water. State the pH of the resulting solution and write a balanced equation for the reaction. [2]
(ii) Aluminium oxide, \(\text{Al}_2\text{O}_3\), and sulfur dioxide, \(\text{SO}_2\), are added to separate test tubes of water. Describe what is observed in each test tube, and state the pH (if any) of the resulting mixture in each test tube. Write an equation for any reaction that occurs. [3]
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解題
(a) (i) - Sodium reacts vigorously / rapidly with cold water, melting into a sphere and moving across the surface, fizzing and dissolving. - Magnesium reacts extremely slowly with cold water, producing only a few bubbles of gas over a long period. - Equation for sodium: \(2\text{Na}(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{NaOH}(\text{aq}) + \text{H}_2(\text{g})\).
(ii) - Magnesium reacts vigorously / burns with a bright white flame with steam, producing a white solid. - Equation: \(\text{Mg}(\text{s}) + \text{H}_2\text{O}(\text{g}) \rightarrow \text{MgO}(\text{s}) + \text{H}_2(\text{g})\).
(b) (i) - Sodium oxide reacts vigorously / readily dissolves in water to form a strongly alkaline solution of pH 13–14. - Equation: \(\text{Na}_2\text{O}(\text{s}) + \text{H}_2\text{O}(\text{l}) \rightarrow 2\text{NaOH}(\text{aq})\).
(ii) - For \(\text{Al}_2\text{O}_3\): There is no reaction / it does not dissolve (remains as a white solid) and the pH is 7. - For \(\text{SO}_2\): It dissolves to form a colorless solution / reacts to form a strongly acidic solution with pH of 1–2. - Equation for \(\text{SO}_2\): \(\text{SO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_2\text{SO}_3(\text{aq})\).
評分準則
Part (a)(i): - M1: Describes sodium reaction: vigorous reaction / melting/floating / fizzing AND describes magnesium reaction: very slow reaction [1] - M2: Mentions gas (hydrogen) evolved or alkaline solution formed [1] - M3: Correct balanced equation for sodium: \(2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\) (state symbols not required unless specified) [1]
Part (a)(ii): - M1: Observation: bright white light / white solid formed [1] - M2: Correct equation: \(\text{Mg} + \text{H}_2\text{O} \rightarrow \text{MgO} + \text{H}_2\) with \(\text{H}_2\text{O}\) as steam/gas [1]
Part (b)(i): - M1: Description of reaction / dissolution and pH = 13 or 14 [1] - M2: Correct equation: \(\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}\) [1]
Part (b)(ii): - M1: Aluminium oxide: observation (insoluble solid/no change) and pH = 7 [1] - M2: Sulfur dioxide: observation (dissolves to a clear solution) and pH = 1 or 2 [1] - M3: Correct equation for sulfur dioxide: \(\text{SO}_2 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_3\) [1]
題目 3 · structured-theory
10 分
(a) Define the term *standard enthalpy change of combustion*. [2]
(b) Propan-1-ol, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\), can be used as a liquid fuel. The standard enthalpy changes of combustion, \(\Delta H_{\text{c}}^\ominus\), for carbon, hydrogen, and propan-1-ol are given in the table below.
(i) Write a balanced chemical equation, including state symbols, for the reaction representing the standard enthalpy change of formation of propan-1-ol, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH(l)}\). [2]
(ii) Construct an enthalpy cycle (Hess’s law cycle) to connect the formation of propan-1-ol from its elements with the standard enthalpy changes of combustion. [2]
(iii) Calculate the standard enthalpy change of formation, \(\Delta H_{\text{f}}^\ominus\), of propan-1-ol. Show your working. [2]
(c) A student estimated the standard enthalpy change of combustion of gaseous propan-1-ol using bond energies. Explain why the value obtained using bond energies differs from the actual standard enthalpy change of combustion of liquid propan-1-ol, even if all calculations are performed correctly. [2]
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解題
(a) Enthalpy change when 1 mole of a substance is burned completely in excess oxygen under standard conditions (298 K and 101 kPa / 1 bar).
(ii) Constructing the cycle: - Elements: \(3\text{C}(\text{s}) + 4\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g})\) at top left. - Compound: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}(\text{l})\) at top right. - Combustion Products: \(3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l})\) at bottom. - Arrow from Elements to Compound is \(\Delta H_{\text{f}}^\ominus\). - Arrow from Elements to Products is \(3 \times (-394) + 4 \times (-286)\). - Arrow from Compound to Products is \(-2021\).
(c) - Bond energies are average values (derived from a range of different compounds rather than the specific environment in propan-1-ol). - Bond energies apply only to gaseous molecules, whereas propan-1-ol in its standard state is a liquid, so intermolecular forces / enthalpy change of vaporisation must be accounted for.
評分準則
Part (a): - M1: Enthalpy change when one mole of a substance is completely burned in oxygen [1] - M2: Mention of standard conditions of 298 K and 101 kPa / 1 bar [1]
Part (b)(i): - M1: Correct species and balancing: \(3\text{C} + 4\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\) [1] - M2: Correct state symbols: \(\text{C}(\text{s})\), \(\text{H}_2(\text{g})\), \(\text{O}_2(\text{g})\), and \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}(\text{l})\) [1]
Part (b)(ii): - M1: Elements, Compound and Combustion Products correctly placed in cycle [1] - M2: All arrows pointing in the correct direction with correct stoichiometry factors (e.g., \(3\Delta H_{\text{c}}^\ominus\text{[C]} + 4\Delta H_{\text{c}}^\ominus\text{[H}_2\text{]}\)) [1]
Part (c): - M1: Average bond energies are used which are not specific to the actual molecule [1] - M2: Bond energies apply only to gases / standard combustion of propan-1-ol involves liquid which requires energy to vaporize / has intermolecular forces [1]
題目 4 · structured-theory
10 分
The synthesis of methanol is a reversible reaction carried out in the gas phase:
(b) Initially, \(1.00\text{ mol}\) of \(\text{CO}\) and \(2.00\text{ mol}\) of \(\text{H}_2\) were sealed in a \(5.00\text{ dm}^3\) reaction vessel and allowed to reach equilibrium at a constant temperature, \(T\). At equilibrium, the mixture was found to contain \(0.40\text{ mol}\) of \(\text{CH}_3\text{OH}\).
(i) Write the expression for the equilibrium constant, \(K_{\text{c}}\), for this reaction, including its units. [2]
(ii) Calculate the equilibrium concentrations, in \(\text{mol dm}^{-3}\), of \(\text{CO}\) and \(\text{H}_2\). [3]
(iii) Calculate the value of \(K_{\text{c}}\) at this temperature. [2]
(c) Predict the effect, if any, of increasing the temperature on the value of \(K_{\text{c}}\). Explain your answer. [1]
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解題
(a) If a system at dynamic equilibrium is subjected to a change in conditions, the position of equilibrium shifts in the direction that tends to minimize/oppose the change.
(c) - Since the forward reaction is exothermic, an increase in temperature shifts the equilibrium to the left (favours the endothermic reaction), which decreases the concentration of methanol and increases the reactants, thus decreasing the value of \(K_{\text{c}}\).
評分準則
Part (a): - M1: Dynamic equilibrium / shifts to minimize/oppose the change [1] - M2: Explicitly mentions change in conditions [1]
Part (b)(i): - M1: Correct expression for \(K_{\text{c}}\): \(\frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}\) [1] - M2: Correct units: \(\text{dm}^6\text{ mol}^{-2}\) [1]
Part (b)(ii): - M1: Calculates equilibrium moles of \(\text{CO}\) (\(0.60\text{ mol}\)) and \(\text{H}_2\) (\(1.20\text{ mol}\)) [1] - M2: Correct concentration of \(\text{CO}\) (\(0.120\text{ mol dm}^{-3}\)) [1] - M3: Correct concentration of \(\text{H}_2\) (\(0.240\text{ mol dm}^{-3}\)) [1]
Part (b)(iii): - M1: Correct substitute of values into their \(K_{\text{c}}\) expression [1] - M2: Calculates value: \(11.6\) (accept \(11.57\)) [1]
Part (c): - M1: Decreases AND forward reaction is exothermic / shifts to endothermic direction [1]
題目 5 · structured-theory
10 分
(a) Propan-1-ol and propan-2-ol are structural isomers.
(i) State the type of structural isomerism shown by these two alcohols. [1]
(ii) Both alcohols can be oxidized when heated with acidified potassium dichromate(VI). Identify the organic product formed in each case when the reaction is carried out under reflux.
- Product from propan-1-ol under reflux: - Product from propan-2-ol under reflux: [2]
(iii) State the color change observed in the acidified potassium dichromate(VI) solution during these oxidation reactions. [1]
(iv) Describe a simple chemical test, using a reagent other than acidified potassium dichromate(VI) or Fehling’s/Tollens’ reagents, that can distinguish between the two organic products identified in (a)(ii). State the observation for each product. [3]
(b) Propan-1-ol can undergo a dehydration reaction to form an alkene.
(i) State the reagent and conditions required for this dehydration reaction. [2]
(ii) Write a balanced chemical equation for the dehydration of propan-1-ol, showing the structural formula of the organic product. [1]
(ii) - Product from propan-1-ol under reflux: Propanoic acid (accept \(\text{CH}_3\text{CH}_2\text{COOH}\)). - Product from propan-2-ol under reflux: Propanone (accept \(\text{CH}_3\text{COCH}_3\)).
(iii) - Orange to green.
(iv) - Reagent: Add aqueous sodium carbonate (\(\text{Na}_2\text{CO}_3\)) or sodium hydrogencarbonate (\(\text{NaHCO}_3\)). - Observation with propanoic acid: Effervescence / bubbles of gas produced (which turn limewater cloudy). - Observation with propanone: No visible change / no effervescence. - *(Alternative: Universal indicator: propanoic acid turns it orange/red, propanone leaves it green. Iodoform test: iodine + aqueous NaOH: propanone forms a yellow precipitate, propanoic acid does not).*
(b) (i) - Reagent: Concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) OR concentrated phosphoric acid (\(\text{H}_3\text{PO}_4\)) (or alumina catalyst, \(\text{Al}_2\text{O}_3\)). - Conditions: Heat / temperature of \(170^\circ\text{C}\) (or pass vapor over hot alumina).
(ii) - \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{CH}=\text{CH}_2 + \text{H}_2\text{O}\)
評分準則
Part (a)(i): - M1: Positional / position (isomerism) [1]
Part (a)(iv): - M1: Suitable reagent: e.g. \(\text{Na}_2\text{CO}_3\) / \(\text{NaHCO}_3\) OR Universal Indicator OR Iodine + NaOH [1] - M2: Observation with product 1 (propanoic acid): e.g. fizzing / turns red [1] - M3: Observation with product 2 (propanone): e.g. no reaction / turns yellow ppt (if using iodoform) [1]
Part (b)(i): - M1: Concentrated sulfuric acid / conc. \(\text{H}_2\text{SO}_4\) (or conc. \(\text{H}_3\text{PO}_4\) / alumina / \(\text{Al}_2\text{O}_3\)) [1] - M2: Heat / temperature \(170^\circ\text{C}\) (or \(300\text{--}350^\circ\text{C}\) for alumina) [1]
Part (b)(ii): - M1: Correct equation showing structural/displayed formula of propene: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{CH}=\text{CH}_2 + \text{H}_2\text{O}\) [1]
題目 6 · structured-theory
10 分
(a) The relative reducing power of the halide ions can be demonstrated by reacting solid sodium halides with concentrated sulfuric acid.
(i) Describe the observation when concentrated sulfuric acid is added to solid sodium chloride, and write an equation for the reaction that occurs. [2]
(ii) When concentrated sulfuric acid is added to solid sodium iodide, several reactions occur, and a variety of products are observed. Describe two different observations that can be made in this reaction, and identify the product responsible for each observation. [2]
(iii) Explain, in terms of the relative reducing powers of the halide ions, why sodium iodide reacts differently from sodium chloride when treated with concentrated sulfuric acid. [2]
(b) A student wanted to determine the percentage by mass of sodium chloride in a sample of impure table salt. A \(1.50\text{ g}\) sample of the impure table salt was dissolved in distilled water to make a \(250.0\text{ cm}^3\) solution. A \(25.0\text{ cm}^3\) portion of this solution was titrated against \(0.100\text{ mol dm}^{-3}\) aqueous silver nitrate, \(\text{AgNO}_3\). The average titre of \(\text{AgNO}_3(\text{aq})\) required to reach the end point was \(22.50\text{ cm}^3\).
(i) Write the ionic equation, with state symbols, for the precipitation reaction that occurs during the titration. [1]
(ii) Calculate the number of moles of \(\text{Ag}^+\) ions that reacted. [1]
(iii) Calculate the percentage by mass of sodium chloride in the impure sample. (Assume the impurities do not react with silver nitrate). [2]
(ii) Any two of the following pairs: - Purple vapor / dark grey solid: Iodine (\(\text{I}_2\)) - Yellow solid: Sulfur (\(\text{S}\)) - Rotten-egg smell gas: Hydrogen sulfide (\(\text{H}_2\text{S}\)) - Choking gas / misty fumes: Sulfur dioxide (\(\text{SO}_2\)) or hydrogen iodide (\(\text{HI}\))
(iii) - Iodide ions (\(\text{I}^-\)) are stronger reducing agents than chloride ions (\(\text{Cl}^-\)). - This is because iodide ions are larger / have a larger ionic radius, meaning outer electrons are further from the nucleus, experience more shielding, and are more easily lost.
Part (a)(ii): - M1: One correct observation linked with the correct product (e.g., purple vapor / \(\text{I}_2\)) [1] - M2: A second distinct observation linked with the correct product (e.g., yellow solid / \(\text{S}\) or bad egg smell / \(\text{H}_2\text{S}\)) [1]
Part (a)(iii): - M1: Iodide is a stronger reducing agent than chloride / iodide is more easily oxidized [1] - M2: Reason: Iodide has larger radius / more shielding / less nuclear attraction on outer electrons [1]
Part (b)(i): - M1: Correct ionic equation with state symbols: \(\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})\) [1]
Part (b)(ii): - M1: Correct calculation of moles: \(2.25 \times 10^{-3}\text{ mol}\) [1]
Part (b)(iii): - M1: Scales moles up to 250 cm3: \(0.0225\text{ mol}\) AND calculates mass of NaCl: \(1.316\text{ g}\) [1] - M2: Calculates percentage: \(87.7\%\) (accept \(87.7\text{--}87.8\%\)) [1]
Paper 31 (Practical Skills)
Carry out three practical quantitative/qualitative laboratory tasks and complete related evaluation sheets.
3 題目 · 39.900000000000006 分
題目 1 · practical-investigation
13.3 分
**Determination of the concentration of a diprotic acid, FB 1**
FB 1 is an aqueous solution of a diprotic acid, \(\text{H}_2\text{A}\). FB 2 is \(0.110\text{ mol dm}^{-3}\) sodium hydroxide, \(\text{NaOH}\). In this experiment, you will standardise FB 1 by titrating it against FB 2 using phenolphthalein indicator.
**(a) Method** 1. Pipette \(25.0\text{ cm}^3\) of FB 1 into a clean conical flask. 2. Add a few drops of phenolphthalein indicator. 3. Fill the burette with FB 2. 4. Perform a rough titration, then carry out sufficient accurate titrations to obtain concordant results (within \(0.10\text{ cm}^3\)). 5. Record your titration results in a suitable, single table in the space below.
**(b) Calculation** 1. Show your working and use your concordant results to calculate the mean volume of FB 2 used. 2. Calculate the number of moles of \(\text{NaOH}\) in this mean volume. 3. Determine the number of moles of \(\text{H}_2\text{A}\) that reacted. 4. Calculate the concentration of \(\text{H}_2\text{A}\) in FB 1, in \(\text{mol dm}^{-3}\). Give your answers to 3 significant figures.
**(c) Evaluation** The pipette used has an uncertainty of \(\pm 0.06\text{ cm}^3\). Calculate the percentage uncertainty in the volume of FB 1 delivered by this pipette.
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解題
**(a) Method and Results Table:** The candidate must present results in a single, clear table with all readings to 2 decimal places (ending in .00 or .05). Two concordant titres (within 0.10 cm³) must be obtained and identified. Sample data: - Rough titre = \(23.10\text{ cm}^3\) - Titre 1 = \(22.75\text{ cm}^3\) - Titre 2 = \(22.65\text{ cm}^3\) Mean volume = \(\frac{22.75 + 22.65}{2} = 22.70\text{ cm}^3\).
**(b) Calculations:** 1. Mean titre = \(22.70\text{ cm}^3\). 2. Moles of \(\text{NaOH}\) = \(\frac{22.70}{1000} \times 0.110 = 2.497 \times 10^{-3}\text{ mol}\). 3. Moles of \(\text{H}_2\text{A}\) = \(\frac{2.497 \times 10^{-3}}{2} = 1.249 \times 10^{-3}\text{ mol}\). 4. Concentration of \(\text{H}_2\text{A}\) in FB 1 = \(\frac{1.249 \times 10^{-3}}{0.0250} = 0.0499\text{ mol dm}^{-3}\) (or \(0.0500\text{ mol dm}^{-3}\) if using rounded sample values).
**Total Marks: 13.3** - **Table of results (4 marks):** - [1 mark]: Organised table with initial and final burette readings and volume of FB 2 added for rough and accurate titrations. - [1 mark]: Correct units in headings for all columns: / cm³ or (cm³). - [1 mark]: All accurate burette readings to 2 decimal places (ending in .00 or .05). - [1 mark]: Two accurate titrations within \(0.10\text{ cm}^3\). - **Accuracy (3 marks):** - [3 marks] awarded if candidate's titre is within \(\pm 0.20\text{ cm}^3\) of supervisor's titre. Deduct 1 mark for each extra \(0.10\text{ cm}^3\) difference up to \(0.40\text{ cm}^3\). - **Calculations (4 marks):** - [1 mark] for correct calculation of mean titre from concordant titres, clearly shown. - [1 mark] for calculating moles of \(\text{NaOH}\) to 3 significant figures. - [1 mark] for dividing moles of \(\text{NaOH}\) by 2 to get moles of \(\text{H}_2\text{A}\). - [1 mark] for dividing moles of \(\text{H}_2\text{A}\) by 0.0250 to find concentration, given to 3 significant figures. - **Evaluation (2.3 marks):** - [1 mark] for correct calculation of percentage uncertainty: \(\frac{0.06}{25.0} \times 100 = 0.24\%\). - [1.3 marks] for explaining that a pipette has a smaller percentage uncertainty than a burette because it is calibrated to deliver a single fixed volume and has only one reading point, reducing potential experimental errors.
題目 2 · practical-investigation
13.3 分
**Determination of the enthalpy change of displacement, \(\Delta H\), for the reaction between Zinc and Copper(II) sulfate**
FB 3 is \(1.00\text{ mol dm}^{-3}\) copper(II) sulfate solution, \(\text{CuSO}_4\). FB 4 is zinc powder, \(\text{Zn}\) (excess). In this experiment, you will determine the temperature rise when FB 4 is added to FB 3, and use this to calculate the enthalpy change of the reaction:
**(a) Method** 1. Support a plastic cup in a \(250\text{ cm}^3\) beaker. 2. Using a measuring cylinder, transfer \(25.0\text{ cm}^3\) of FB 3 into the plastic cup. 3. Place a thermometer in the solution and record its temperature every minute for 3 minutes. 4. At the 4th minute, do NOT record the temperature. At this exact time, add all the FB 4 provided and stir thoroughly. 5. Record the temperature of the mixture every minute from the 5th minute to the 10th minute, stirring continuously. 6. Present all your temperature measurements in a clear table below.
**(b) Graphical Representation** Plot a graph of temperature (y-axis) against time (x-axis) on a grid. Extrapolate your lines to determine the theoretical maximum temperature rise, \(\Delta T\), at the 4th minute. Show your extrapolation clearly on the graph. Determine: - Initial temperature at 4th minute (from extrapolation) = ............ \(^\circ\text{C}\) - Maximum temperature at 4th minute (from extrapolation) = ............ \(^\circ\text{C}\) - Temperature rise, \(\Delta T\) = ............ \(^\circ\text{C}\)
**(c) Calculations** 1. Calculate the heat energy, \(q\) (in J), released during the reaction. (Assume the specific heat capacity of the mixture is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\) and its density is \(1.00\text{ g cm}^{-3}\)). 2. Calculate the number of moles of \(\text{CuSO}_4\) in \(25.0\text{ cm}^3\) of FB 3. 3. Calculate the enthalpy change, \(\Delta H\), in \(\text{kJ mol}^{-1}\), for this reaction. Include the correct sign.
**(d) Evaluation** Suggest one modification to the experimental apparatus that would reduce heat loss, and explain how it achieves this.
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解題
**(a) Method and Results:** Sample temperature readings: - Minute 0: 21.0 °C - Minute 1: 21.0 °C - Minute 2: 21.0 °C - Minute 3: 21.0 °C - Minute 4: [Addition of Zn] - Minute 5: 39.5 °C - Minute 6: 41.5 °C - Minute 7: 41.0 °C - Minute 8: 40.5 °C - Minute 9: 40.0 °C - Minute 10: 39.5 °C
**(b) Graph and Extrapolation:** Plot temperature vs time. Draw a horizontal best-fit line through the points from 0 to 3 minutes. Draw a straight line of cooling through points from 5 to 10 minutes, and extrapolate both lines to the 4th minute. Using sample data: - Extrapolated initial temp at 4 min = \(21.0\ ^\circ\text{C}\) - Extrapolated maximum temp at 4 min = \(42.5\ ^\circ\text{C}\) - \(\Delta T = 42.5 - 21.0 = 21.5\ ^\circ\text{C}\).
**(d) Evaluation:** Modification: Place a lid on the plastic cup OR wrap the plastic cup in insulating material like cotton wool or use nested plastic cups. Explanation: This reduces heat transfer by convection/conduction to the surroundings, making the measured temperature rise closer to the theoretical value.
評分準則
**Total Marks: 13.3** - **Method & Table (3 marks):** - [1 mark] for complete table showing temperatures at regular 1-minute intervals, with units specified. - [1 mark] for recording all temperatures to the nearest \(0.5\ ^\circ\text{C}\) or \(0.1\ ^\circ\text{C}\) consistently. - [1 mark] for leaving the 4-minute reading blank/noted as addition. - **Graph & Extrapolation (4 marks):** - [1 mark] for correct plotting of all recorded points with temperature on y-axis and time on x-axis. - [1 mark] for drawing two appropriate best-fit straight lines (pre-addition and post-maximum cooling). - [1 mark] for showing clear extrapolation of both lines to the 4th minute. - [1 mark] for correct calculation of \(\Delta T\) from extrapolated values. - **Calculations (4 marks):** - [1 mark] for calculating \(q\) using \(mc\Delta T\) with candidate's \(\Delta T\). - [1 mark] for correct moles of \(\text{CuSO}_4 = 0.0250\text{ mol}\). - [1 mark] for division of \(q\) by moles to find \(\Delta H\). - [1 mark] for stating the final answer with a negative sign (-) and to 3 significant figures. - **Evaluation (2.3 marks):** - [1 mark] for a valid modification (e.g., adding a lid or nested cups). - [1.3 marks] for a correct explanation showing how this reduces heat loss to the environment (conduction/convection).
題目 3 · practical-investigation
13.3 分
**Qualitative analysis of unknown aqueous solutions FB 5 and FB 6**
FB 5 and FB 6 are aqueous solutions each containing one cation and one anion. Carry out the following tests to identify the ions present in each solution.
**(a) Tests on FB 5 and FB 6** Complete the table below by describing your observations. If no reaction is observed, write 'no change' or 'no reaction'.
| Test | Observations with FB 5 | Observations with FB 6 | | :--- | :--- | :--- | | **Test 1:** To a \(1\text{ cm}\) depth of sample in a test-tube, add aqueous sodium hydroxide, \(\text{NaOH}\), dropwise until in excess. | .................................... | .................................... | | **Test 2:** To a \(1\text{ cm}\) depth of sample in a test-tube, add aqueous ammonia, \(\text{NH}_3\), dropwise until in excess. | .................................... | .................................... | | **Test 3:** To a \(1\text{ cm}\) depth of sample in a test-tube, add dilute nitric acid followed by aqueous silver nitrate, \(\text{AgNO}_3\). | .................................... | .................................... | | **Test 4:** To the mixture from Test 3, add dilute aqueous ammonia. | .................................... | .................................... | | **Test 5:** To a \(1\text{ cm}\) depth of sample in a test-tube, add dilute hydrochloric acid followed by aqueous barium chloride, \(\text{BaCl}_2\). | .................................... | .................................... |
**(b) Identification** 1. Identify the cation and anion in **FB 5**. - Cation: .................... - Anion: .................... 2. Identify the cation and anion in **FB 6**. - Cation: .................... - Anion: ....................
**(c) Write an ionic equation** Write the ionic equation for the precipitation reaction that occurs when FB 6 reacts with aqueous silver nitrate in Test 3. Include state symbols.
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解題
**(a) Observations:** - **FB 5 (magnesium sulfate, \(\text{MgSO}_4\)):** - **Test 1 (NaOH):** White precipitate, insoluble in excess. - **Test 2 (\(\text{NH}_3\)):** White precipitate, insoluble in excess. - **Test 3 (\(\text{AgNO}_3\)):** No change (or no precipitate). - **Test 4 (\(\text{NH}_3\)):** No change. - **Test 5 (\(\text{BaCl}_2\)):** White precipitate. - **FB 6 (zinc chloride, \(\text{ZnCl}_2\)):** - **Test 1 (NaOH):** White precipitate, soluble in excess to form a colourless solution. - **Test 2 (\(\text{NH}_3\)):** White precipitate, soluble in excess to form a colourless solution. - **Test 3 (\(\text{AgNO}_3\)):** White precipitate. - **Test 4 (\(\text{NH}_3\)):** White precipitate dissolves/is soluble to form a colourless solution. - **Test 5 (\(\text{BaCl}_2\)):** No reaction.
**(b) Identification:** - **FB 5:** Cation is \(\text{Mg}^{2+}\); Anion is \(\text{SO}_4^{2-}\). - **FB 6:** Cation is \(\text{Zn}^{2+}\); Anion is \(\text{Cl}^-\).
**Total Marks: 13.3** - **Observations for FB 5 (3 marks):** - [1 mark] for white precipitate with \(\text{NaOH}\) and \(\text{NH}_3\), both insoluble in excess. - [1 mark] for no reaction / no change in Test 3. - [1 mark] for white precipitate with \(\text{BaCl}_2\) in Test 5. - **Observations for FB 6 (3 marks):** - [1 mark] for white precipitate with \(\text{NaOH}\) and \(\text{NH}_3\), both dissolving/soluble in excess. - [1 mark] for white precipitate in Test 3. - [1 mark] for white precipitate dissolving in dilute \(\text{NH}_3\) in Test 4. - **Identification (4 marks):** - [1 mark] for identifying \(\text{Mg}^{2+}\) in FB 5. - [1 mark] for identifying \(\text{SO}_4^{2-}\) in FB 5. - [1 mark] for identifying \(\text{Zn}^{2+}\) in FB 6. - [1 mark] for identifying \(\text{Cl}^-\) in FB 6. - **Ionic Equation (3.3 marks):** - [2 marks] for correct formula and balance: \(\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}\). - [1.3 marks] for correct state symbols: \(\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})\).
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