An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V3) Cambridge International A Level Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.
卷二 Extended 選擇題
Answer all 40 multiple-choice questions on the answer sheet. Each correct answer scores 1 mark.
40 題目 · 40 分
題目 1 · 選擇題
1 分
A 0.120 g sample of an unknown Group II metal reacts completely with excess hydrochloric acid to produce 120 cm3 of hydrogen gas, measured at room temperature and pressure (r.t.p.).
Which metal is it?
(1 mol of any gas occupies 24.0 dm3 at r.t.p.)
A.Beryllium
B.Magnesium
C.Calcium
D.Strontium
查看答案詳解收起答案詳解
解題
The equation for the reaction of a Group II metal (M) with hydrochloric acid is: \( \text{M(s)} + 2\text{HCl(aq)} \rightarrow \text{MCl}_2\text{(aq)} + \text{H}_2\text{(g)} \)
First, calculate the number of moles of hydrogen gas produced: \( \text{Moles of H}_2 = \frac{120 \text{ cm}^3}{24000 \text{ cm}^3/\text{mol}} = 0.0050 \text{ mol} \)
From the stoichiometry of the reaction, 1 mole of M produces 1 mole of \( \text{H}_2 \). Therefore, moles of metal M reacted = 0.0050 mol.
Now, calculate the relative atomic mass (\( A_r \)) of metal M: \( A_r = \frac{\text{mass}}{\text{moles}} = \frac{0.120 \text{ g}}{0.0050 \text{ mol}} = 24.0 \)
Looking at the Periodic Table, the Group II metal with a relative atomic mass of 24.0 is magnesium.
評分準則
1 mark for the correct option. - Award 1 mark for identifying Magnesium (B) through correct calculation of moles of gas (0.005 mol) and atomic mass (24). - Reject other options: A (Ar = 9), C (Ar = 40), D (Ar = 88).
題目 2 · 選擇題
1 分
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes.
Which row correctly identifies the product at each electrode and the overall change in pH of the electrolyte?
A.Anode product: chlorine; Cathode product: hydrogen; pH of electrolyte: increases
B.Anode product: oxygen; Cathode product: sodium; pH of electrolyte: decreases
D.Anode product: oxygen; Cathode product: hydrogen; pH of electrolyte: increases
查看答案詳解收起答案詳解
解題
During the electrolysis of concentrated aqueous sodium chloride: - At the anode (positive electrode), halide ions are in high concentration, so chloride ions (\( \text{Cl}^- \)) are preferentially discharged to produce chlorine gas: \( 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^- \). - At the cathode (negative electrode), hydrogen ions (\( \text{H}^+ \)) from water are preferentially discharged instead of sodium ions (\( \text{Na}^+ \)) because hydrogen is less reactive than sodium: \( 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2 \). - As \( \text{H}^+ \) ions are discharged, the concentration of hydroxide ions (\( \text{OH}^- \)) increases. The remaining solution contains sodium hydroxide (\( \text{NaOH} \)), which is alkaline, so the overall pH of the electrolyte increases.
評分準則
1 mark for the correct option. - Award 1 mark for A. - Reject B, C, and D because they contain incorrect electrode products (sodium metal is not formed in aqueous solutions, and oxygen is only produced in dilute solutions) and incorrect pH changes.
題目 3 · 選擇題
1 分
Three metals, X, Y, and Z, were reacted with aqueous solutions of their nitrates. The results of the experiments are shown.
- Metal X + Nitrate of Y \( \rightarrow \) No reaction - Metal X + Nitrate of Z \( \rightarrow \) Metal Z formed - Metal Y + Nitrate of X \( \rightarrow \) Metal X formed
What is the order of reactivity of the metals, starting with the most reactive?
A.X > Y > Z
B.Y > X > Z
C.Z > X > Y
D.Y > Z > X
查看答案詳解收起答案詳解
解題
Let's analyse the displacement reactions: 1. Metal X does not react with the nitrate of Y, which means X is less reactive than Y (Y > X). 2. Metal X reacts with the nitrate of Z to form metal Z, which means X is more reactive than Z (X > Z). 3. Metal Y reacts with the nitrate of X to form metal X, which confirms Y is more reactive than X (Y > X).
Combining these relationships: Y > X > Z.
評分準則
1 mark for the correct option. - Award 1 mark for B. - Reject A, C, and D because they do not correctly represent the order of reactivity derived from the displacement observations.
題目 4 · 選擇題
1 分
An organic compound has the structural formula shown:
Which two functional groups are present in this compound?
A.carboxylic acid and alcohol
B.ester and alcohol
C.carboxylic acid and alkene
D.ester and alkene
查看答案詳解收起答案詳解
解題
We can identify the functional groups from the structural formula: - The group \( -\text{OH} \) attached to the second carbon atom (\( \text{CH}(\text{OH}) \)) is an alcohol (hydroxyl) group. - The group \( -\text{COO}- \) (\( \text{COOCH}_2\text{CH}_3 \)) is an ester linkage.
Therefore, the compound contains both an ester group and an alcohol group.
評分準則
1 mark for the correct option. - Award 1 mark for B. - Reject A, C, and D because they contain functional groups not present in the given formula (the formula contains no \( -\text{COOH} \) for a carboxylic acid or \( \text{C}=\text{C} \) double bond for an alkene).
題目 5 · 選擇題
1 分
A student investigates the rate of reaction between excess calcium carbonate (marble chips) and dilute hydrochloric acid:
In Experiment 1, 10 g of large marble chips and 50 cm3 of 1.0 mol/dm3 HCl are used at 25 °C.
Which change in a second experiment, keeping all other variables constant, will produce the same volume of carbon dioxide but at a slower initial rate?
A.Using 10 g of powdered marble chips instead of large chips.
B.Using 50 cm3 of 2.0 mol/dm3 HCl.
C.Using 100 cm3 of 0.5 mol/dm3 HCl.
D.Carrying out the reaction at 35 °C instead of 25 °C.
查看答案詳解收起答案詳解
解題
To obtain the same volume of \( \text{CO}_2 \), the total number of moles of the limiting reactant, HCl, must remain the same because the calcium carbonate is in excess.
Since the number of moles of HCl is identical, the volume of gas produced will be the same. Since the concentration of HCl is lower (0.5 mol/dm3 compared to 1.0 mol/dm3), the rate of reaction will be slower.
- Option A increases the rate by increasing surface area. - Option B increases both rate (due to higher concentration) and the volume of gas (moles of HCl doubles to 0.10 mol). - Option D increases the rate by increasing temperature.
評分準則
1 mark for the correct option. - Award 1 mark for C. - Reject A and D because they increase the rate of reaction. - Reject B because it increases both the rate and the volume of gas produced.
題目 6 · 選擇題
1 分
The equation shows a reversible reaction in the industrial production of methanol:
Which set of conditions would produce the maximum yield of methanol at equilibrium?
A.High temperature and high pressure
B.High temperature and low pressure
C.Low temperature and high pressure
D.Low temperature and low pressure
查看答案詳解收起答案詳解
解題
To maximise the yield of methanol, we want to shift the equilibrium position to the right: 1. Temperature: The forward reaction is exothermic (\( \Delta H = -91 \text{ kJ/mol} \)). According to Le Chatelier's principle, lowering the temperature shifts the equilibrium in the exothermic direction (to the right) to release heat, thereby increasing the yield of methanol. 2. Pressure: There are 3 moles of gaseous reactants on the left (1 mol of CO and 2 mol of \( \text{H}_2 \)) and only 1 mole of gaseous product on the right (1 mol of \( \text{CH}_3\text{OH} \)). According to Le Chatelier's principle, increasing the pressure shifts the equilibrium towards the side with fewer gas moles (to the right), thereby increasing the yield of methanol.
Therefore, low temperature and high pressure will produce the maximum yield.
評分準則
1 mark for the correct option. - Award 1 mark for C. - Reject A, B, and D because they include high temperature (which decreases yield for exothermic reactions) or low pressure (which decreases yield when there are fewer moles of gas on the product side).
題目 7 · 選擇題
1 分
An aqueous solution of salt X is tested:
- Addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess sodium hydroxide. - Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate.
What is the identity of salt X?
A.Chromium(III) sulfate
B.Iron(II) sulfate
C.Iron(III) sulfate
D.Iron(II) chloride
查看答案詳解收起答案詳解
解題
Let's identify the cation and the anion from the test results: 1. Cation test: Aqueous sodium hydroxide produces a green precipitate of \( \text{Fe(OH)}_2 \) which is insoluble in excess. This indicates the presence of Iron(II) ions (\( \text{Fe}^{2+} \)). Note: Chromium(III) also forms a green precipitate, but it dissolves in excess sodium hydroxide to form a green solution. 2. Anion test: The addition of dilute nitric acid followed by aqueous barium nitrate yields a white precipitate of barium sulfate (\( \text{BaSO}_4 \)), which is insoluble in acid. This confirms the presence of sulfate ions (\( \text{SO}_4^{2-} \)).
Combining these, salt X is Iron(II) sulfate.
評分準則
1 mark for the correct option. - Award 1 mark for B. - Reject A because chromium(III) hydroxide is soluble in excess NaOH. - Reject C because iron(III) forms a red-brown precipitate. - Reject D because chloride ions are tested with silver nitrate, not barium nitrate.
題目 8 · 選擇題
1 分
An ion \( \text{Y}^{2-} \) has a nucleon number (mass number) of 34 and contains 18 electrons.
How many protons and neutrons are in the nucleus of this ion?
A.16 protons and 16 neutrons
B.16 protons and 18 neutrons
C.18 protons and 16 neutrons
D.18 protons and 18 neutrons
查看答案詳解收起答案詳解
解題
Let's calculate the subatomic particles: 1. Protons: The ion has a 2- charge (\( \text{Y}^{2-} \)), which means it has gained 2 extra electrons. \( \text{Number of electrons} = 18 \) \( \text{Number of protons} = 18 - 2 = 16 \)
2. Neutrons: The nucleon number (mass number) is the total number of protons and neutrons in the nucleus. \( \text{Nucleon number} = 34 \) \( \text{Number of neutrons} = \text{Nucleon number} - \text{Number of protons} = 34 - 16 = 18 \)
Thus, the nucleus has 16 protons and 18 neutrons.
評分準則
1 mark for the correct option. - Award 1 mark for B. - Reject A, C, and D due to incorrect calculation of protons (which must be 16 for a 2- ion with 18 electrons) or neutrons (which must be 18).
題目 9 · multiple_choice
1 分
A \(20\text{ cm}^3\) sample of a gaseous hydrocarbon is completely burned in \(120\text{ cm}^3\) of oxygen (an excess). After cooling to room temperature and pressure, the remaining gas volume is \(90\text{ cm}^3\). This remaining gas is shaken with excess aqueous sodium hydroxide, and the volume decreases to \(30\text{ cm}^3\).
What is the formula of the hydrocarbon?
A.\(CH_4\)
B.\(C_2H_6\)
C.\(C_3H_6\)
D.\(C_3H_8\)
查看答案詳解收起答案詳解
解題
1. Shaking the remaining gas with aqueous sodium hydroxide absorbs the carbon dioxide (\(CO_2\)). The decrease in volume is equivalent to the volume of \(CO_2\) produced: \text{Volume of } CO_2 = 90\text{ cm}^3 - 30\text{ cm}^3 = 60\text{ cm}^3\.
2. The ratio of the volume of \(CO_2\) produced to the volume of hydrocarbon reacted is: \frac{60\text{ cm}^3}{20\text{ cm}^3} = 3\. This means that 1 mole of hydrocarbon contains 3 moles of carbon atoms, so the hydrocarbon has the formula \(C_3H_y\).
3. The remaining \(30\text{ cm}^3\) of gas is the unreacted excess oxygen. Therefore, the volume of oxygen that reacted is: 120\text{ cm}^3 - 30\text{ cm}^3 = 90\text{ cm}^3\.
4. The ratio of reacted \(O_2\) to hydrocarbon is: \frac{90\text{ cm}^3}{20\text{ cm}^3} = 4.5\.
5. The general equation for the complete combustion of a hydrocarbon \(C_3H_y\) is: \(C_3H_y + (3 + \frac{y}{4})O_2 \rightarrow 3CO_2 + \frac{y}{2}H_2O\)
6. Equating the oxygen coefficient to 4.5: \(3 + \frac{y}{4} = 4.5 \implies \frac{y}{4} = 1.5 \implies y = 6\).
Therefore, the molecular formula of the hydrocarbon is \(C_3H_6\).
評分準則
Award 1 mark for the correct answer C. No partial marks.
題目 10 · multiple_choice
1 分
In an electrolysis experiment, aqueous copper(II) sulfate is electrolyzed using copper electrodes.
Which row correctly describes the changes in mass of the electrodes and the color of the electrolyte?
A.Anode mass: decreases; Cathode mass: increases; Color of electrolyte: remains blue
B.Anode mass: no change; Cathode mass: increases; Color of electrolyte: fades
C.Anode mass: decreases; Cathode mass: increases; Color of electrolyte: fades
D.Anode mass: decreases; Cathode mass: no change; Color of electrolyte: remains blue
查看答案詳解收起答案詳解
解題
During the electrolysis of aqueous copper(II) sulfate with active copper electrodes: - At the anode (positive electrode), copper atoms are oxidized to form copper(II) ions: \(Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-\). Therefore, the mass of the anode decreases. - At the cathode (negative electrode), copper(II) ions are reduced to form copper metal: \(Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)\). Therefore, the mass of the cathode increases. - The rate of copper ions leaving the solution at the cathode is exactly equal to the rate of copper ions entering the solution from the anode. Consequently, the concentration of copper(II) ions remains constant, and the blue color of the electrolyte remains unchanged.
評分準則
Award 1 mark for the correct answer A. No partial marks.
題目 11 · multiple_choice
1 分
The equation for a reversible reaction is shown:
\(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\) \(\Delta H = -197\text{ kJ/mol}\)
Which change to the conditions increases both the equilibrium yield of \(SO_3(g)\) and the rate of the forward reaction?
A.Decreasing the temperature
B.Decreasing the pressure
C.Increasing the temperature
D.Increasing the pressure
查看答案詳解收起答案詳解
解題
- Effect on equilibrium yield: There are 3 moles of gas on the reactant side and 2 moles of gas on the product side. Increasing the pressure shifts the position of equilibrium to the side with fewer gas moles (the right-hand side), thereby increasing the yield of \(SO_3(g)\). - Effect on reaction rate: Increasing the pressure increases the concentration of gas molecules, resulting in more frequent collisions and hence a faster forward reaction rate. - Decreasing the temperature increases the yield of \(SO_3(g)\) (since the forward reaction is exothermic), but it decreases the rate of the reaction. Therefore, only increasing the pressure accomplishes both desired changes.
評分準則
Award 1 mark for the correct answer D. No partial marks.
題目 12 · multiple_choice
1 分
Two separate solutions, X and Y, both have a concentration of \(0.1\text{ mol/dm}^3\).
- Solution X has a pH of 1. - Solution Y has a pH of 4.
Which statement about these solutions is correct?
A.Solution X is a weaker acid than solution Y.
B.Solution Y has a higher concentration of hydrogen ions than solution X.
C.Solution Y is only partially ionized in aqueous solution, whereas solution X is fully ionized.
D.Solution X reacts with magnesium ribbon at a slower rate than solution Y.
查看答案詳解收起答案詳解
解題
- A pH of 1 indicates a high concentration of hydrogen ions, which is characteristic of a strong acid that is fully ionized in aqueous solution (Solution X). - A pH of 4 at the same concentration indicates a much lower hydrogen ion concentration, which is characteristic of a weak acid that is only partially ionized in aqueous solution (Solution Y). - Therefore, Solution Y is only partially ionized while Solution X is fully ionized. - Since Solution X has a higher hydrogen ion concentration, it will react with magnesium ribbon at a faster rate, not slower.
評分準則
Award 1 mark for the correct answer C. No partial marks.
題目 13 · multiple_choice
1 分
The results of three displacement experiments are shown:
1. Metal J is added to an aqueous solution of L ions: a reaction occurs. 2. Metal J is added to an aqueous solution of M ions: no reaction occurs. 3. Metal K is added to an aqueous solution of M ions: a reaction occurs.
What is the order of reactivity of these metals, from most reactive to least reactive?
A.K, M, J, L
B.K, J, M, L
C.M, K, J, L
D.L, J, M, K
查看答案詳解收起答案詳解
解題
- Experiment 1 shows that metal J can displace L from its salt solution, meaning J is more reactive than L (\(J > L\)). - Experiment 2 shows that metal J cannot displace M, meaning M is more reactive than J (\(M > J\)). - Experiment 3 shows that metal K can displace M from its salt solution, meaning K is more reactive than M (\(K > M\)).
Combining these relationships: \(K > M > J > L\).
評分準則
Award 1 mark for the correct answer A. No partial marks.
題目 14 · multiple_choice
1 分
How many structural isomers are there with the molecular formula \(C_4H_9Br\)?
A.2
B.3
C.4
D.5
查看答案詳解收起答案詳解
解題
There are exactly four structural isomers with the formula \(C_4H_9Br\): 1. 1-bromobutane (straight chain: \(CH_3CH_2CH_2CH_2Br\)) 2. 2-bromobutane (straight chain: \(CH_3CH_2CH(Br)CH_3\)) 3. 1-bromo-2-methylpropane (branched chain: \((CH_3)_2CHCH_2Br\)) 4. 2-bromo-2-methylpropane (branched chain: \((CH_3)_3CBr\))
評分準則
Award 1 mark for the correct answer C. No partial marks.
題目 15 · multiple_choice
1 分
An unknown salt solution X is tested as follows:
- When aqueous sodium hydroxide is added dropwise, a green precipitate is formed which is insoluble in excess. - When dilute nitric acid followed by aqueous barium nitrate is added, a white precipitate is formed.
What is the identity of salt X?
A.Chromium(III) sulfate
B.Iron(II) sulfate
C.Iron(III) sulfate
D.Iron(II) chloride
查看答案詳解收起答案詳解
解題
- The addition of dropwise sodium hydroxide resulting in a green precipitate insoluble in excess confirms the presence of iron(II) ions, \(Fe^{2+}\). (Note: Chromium(III) also gives a green precipitate, but it is soluble in excess sodium hydroxide to form a green solution). - The addition of barium nitrate in the presence of dilute nitric acid yields a white precipitate of barium sulfate, confirming the presence of sulfate ions, \(SO_4^{2-}\). - Therefore, the identity of salt X is iron(II) sulfate.
評分準則
Award 1 mark for the correct answer B. No partial marks.
題目 16 · multiple_choice
1 分
Which statement about the catalytic converter in a car's exhaust system is correct?
A.Carbon monoxide is reduced to carbon dioxide.
B.Nitrogen oxides are reduced to nitrogen gas.
C.Hydrocarbons are oxidized to hydrogen and carbon.
D.Carbon dioxide is removed by reacting with nitrogen dioxide.
查看答案詳解收起答案詳解
解題
- In a catalytic converter, nitrogen oxides (such as \(NO\) and \(NO_2\)) are reduced to harmless nitrogen gas (\(N_2\)) by losing oxygen. - Carbon monoxide (\(CO\)) is oxidized (not reduced) to carbon dioxide (\(CO_2\)). - Unburnt hydrocarbons are oxidized to carbon dioxide and water, not to hydrogen and carbon. - Carbon dioxide is a product of these reactions, not a reactant that is removed by nitrogen dioxide.
評分準則
Award 1 mark for the correct answer B. No partial marks.
題目 17 · 選擇題
1 分
Excess calcium carbonate is added to \(50\text{ cm}^3\) of \(2.0\text{ mol/dm}^3\) hydrochloric acid. What is the maximum volume of carbon dioxide gas, measured at room temperature and pressure (r.t.p.), produced in this reaction? (Assume 1 mole of any gas occupies \(24\text{ dm}^3\) at r.t.p.)
A.\(0.6\text{ dm}^3\)
B.\(1.2\text{ dm}^3\)
C.\(2.4\text{ dm}^3\)
D.\(4.8\text{ dm}^3\)
查看答案詳解收起答案詳解
解題
Step 1: Write the balanced chemical equation for the reaction: \(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\)
Step 2: Calculate the number of moles of \(\text{HCl}\) used: \(\text{moles of HCl} = \text{concentration} \times \text{volume} = 2.0\text{ mol/dm}^3 \times 0.050\text{ dm}^3 = 0.10\text{ mol}\)
Step 3: Determine the moles of \(\text{CO}_2\) produced using the stoichiometric ratio (\(2\text{HCl} : 1\text{CO}_2\)): \(\text{moles of CO}_2 = \frac{0.10\text{ mol}}{2} = 0.050\text{ mol}\)
Step 4: Calculate the volume of \(\text{CO}_2\) gas at r.t.p.: \(\text{volume} = 0.050\text{ mol} \times 24\text{ dm}^3/\text{mol} = 1.2\text{ dm}^3\)
評分準則
Correct Answer: B 1 mark: Awarded for the correct calculation showing a gas volume of 1.2 dm³.
題目 18 · 選擇題
1 分
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. Which row correctly identifies the products formed at each electrode and the change in the pH of the remaining electrolyte?
During the electrolysis of concentrated aqueous sodium chloride (brine): - At the anode (+), chloride ions (\(\text{Cl}^-\)) are discharged in preference to hydroxide ions (\(\text{OH}^-\)) because they are in high concentration, forming chlorine gas (\(\text{Cl}_2\)). - At the cathode (-), hydrogen ions (\(\text{H}^+\)) are discharged in preference to sodium ions (\(\text{Na}^+\)) because hydrogen is lower in the reactivity series, forming hydrogen gas (\(\text{H}_2\)). - As \(\text{H}^+\) and \(\text{Cl}^-\) ions are discharged and removed, \(\text{Na}^+\) and \(\text{OH}^-\)
評分準則
Correct Answer: C 1 mark: Correctly identifies chlorine at the anode, hydrogen at the cathode, and an increase in pH.
題目 19 · 選擇題
1 分
Three metals, X, Y and Z, are tested to compare their chemical reactivities: - Metal X reacts rapidly with cold water. - Metal Y does not react with dilute hydrochloric acid, but reacts when heated with copper(II) oxide to displace copper. - Metal Z does not react when heated with copper(II) oxide.
What is the order of reactivity of these metals from most reactive to least reactive?
A.X \(\rightarrow\) Y \(\rightarrow\) Z
B.Y \(\rightarrow\) X \(\rightarrow\) Z
C.Z \(\rightarrow\) Y \(\rightarrow\) X
D.X \(\rightarrow\) Z \(\rightarrow\) Y
查看答案詳解收起答案詳解
解題
1. Metal X is the most reactive because it is highly reactive enough to react with cold water (e.g., an alkali metal). 2. Metal Y does not react with dilute acid, meaning it is below hydrogen in the reactivity series, but it is reactive enough to reduce copper(II) oxide (meaning Y is more reactive than copper). 3. Metal Z is less reactive than copper because it cannot reduce copper(II) oxide. Therefore, the order of reactivity from most to least reactive is: X \(\rightarrow\) Y \(\rightarrow\) Z.
評分準則
Correct Answer: A 1 mark: Correctly deduces the order of reactivity as X -> Y -> Z.
題目 20 · 選擇題
1 分
Which statement correctly explains why an increase in temperature increases the rate of a chemical reaction?
A.The activation energy of the reaction is lowered, so a greater proportion of colliding particles have energy greater than the activation energy.
B.The particles have more kinetic energy, so they collide more frequently and a larger proportion of the collisions have energy greater than the activation energy.
C.The particles are closer together, resulting in more frequent collisions.
D.The reaction becomes more exothermic, which releases more heat to speed up the process.
查看答案詳解收起答案詳解
解題
When the temperature is increased, the reactant particles gain kinetic energy and move faster. This results in more frequent collisions. More importantly, a much higher proportion of the colliding particles now possess energy equal to or greater than the activation energy, which leads to a higher frequency of successful collisions.
評分準則
Correct Answer: B 1 mark: Correctly explains the effect of temperature in terms of collision frequency and the proportion of particles with energy exceeding the activation energy.
題目 21 · 選擇題
1 分
How many of the following compounds are structural isomers of butanoic acid? 1. Methyl propanoate 2. Ethyl ethanoate 3. Propyl methanoate 4. Butan-1-ol
A.1
B.2
C.3
D.4
查看答案詳解收起答案詳解
解題
Structural isomers have the same molecular formula but different structural arrangements. Butanoic acid is a carboxylic acid with the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\). - Methyl propanoate is an ester with the formula \(\text{C}_2\text{H}_5\text{COOCH}_3\) (molecular formula \(\text{C}_4\text{H}_8\text{O}_2\)) - Isomer. - Ethyl ethanoate is an ester with the formula \(\text{CH}_3\text{COOCH}_2\text{CH}_3\) (molecular formula \(\text{C}_4\text{H}_8\text{O}_2\)) - Isomer. - Propyl methanoate is an ester with the formula \(\text{HCOOCH}_2\text{CH}_2\text{CH}_3\) (molecular formula \(\text{C}_4\text{H}_8\text{O}_2\)) - Isomer. - Butan-1-ol is an alcohol with the formula \(\text{C}_4\text{H}_9\text{OH}\) (molecular formula \(\text{C}_4\text{H}_{10}\text{O}\)) - Not an isomer. Therefore, 3 of the compounds are structural isomers.
評分準則
Correct Answer: C 1 mark: Correctly identifies that exactly 3 of the given structures are isomers of butanoic acid.
題目 22 · 選擇題
1 分
The equation shows a reversible reaction in the Haber process:
\(\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})\) where \(\Delta H = -92\text{ kJ/mol}\)
Which set of conditions will produce the highest equilibrium yield of ammonia, \(\text{NH}_3\)?
A.High temperature and high pressure
B.High temperature and low pressure
C.Low temperature and high pressure
D.Low temperature and low pressure
查看答案詳解收起答案詳解
解題
According to Le Chatelier's principle: - Temperature: The forward reaction is exothermic (\(\Delta H < 0\)). Lowering the temperature will shift the equilibrium position to the right to produce more heat, thereby increasing the yield of ammonia. - Pressure: There are 4 moles of gas on the left side and 2 moles of gas on the right side. Increasing the pressure will shift the equilibrium position to the side with fewer moles of gas (the right side), thereby increasing the yield of ammonia. Thus, low temperature and high pressure give the highest equilibrium yield.
評分準則
Correct Answer: C 1 mark: Deduces that low temperature and high pressure maximize the equilibrium yield.
題目 23 · 選擇題
1 分
A student wants to measure the rate of reaction between zinc granules and dilute sulfuric acid by collecting the hydrogen gas produced and measuring its volume at regular intervals. Which list of apparatus is most suitable for this experiment?
A.beaker, Bunsen burner, gas syringe, balance
B.conical flask, delivery tube, gas syringe, stopwatch
To carry out this experiment successfully, the student needs: - A reaction vessel (conical flask) where the zinc and acid can react. - A delivery tube to securely transport the gas from the reaction vessel to the gas syringe without leaking. - A gas syringe (or a measuring cylinder inverted over water) to measure the exact volume of gas produced. - A stopwatch to measure the time intervals.
評分準則
Correct Answer: B 1 mark: Correctly selects the complete and appropriate set of apparatus for gas collection and rate measurement.
題目 24 · 選擇題
1 分
A solid salt, W, is dissolved in water to make an aqueous solution. - Addition of aqueous sodium hydroxide to a portion of the solution produces a green precipitate that is insoluble in excess. - Addition of dilute nitric acid followed by aqueous barium nitrate to another portion of the solution produces a white precipitate.
What is the chemical formula of salt W?
A.\(\text{Fe}_2(\text{SO}_4)_3\)
B.\(\text{FeSO}_4\)
C.\(\text{Cr}_2(\text{SO}_4)_3\)
D.\(\text{FeCl}_2\)
查看答案詳解收起答案詳解
解題
1. The reaction with aqueous sodium hydroxide produces a green precipitate that is insoluble in excess, which is the characteristic test confirming the presence of iron(II) ions, \(\text{Fe}^{2+}\). (Note: Chromium(III) also gives a green precipitate, but it is soluble in excess sodium hydroxide to form a green solution). 2. The reaction with aqueous barium nitrate after acidification with nitric acid produces a white precipitate of barium sulfate, which confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). 3. Combining these ions gives Iron(II) sulfate, \(\text{FeSO}_4\).
評分準則
Correct Answer: B 1 mark: Correctly identifies the cation as Fe²⁺ and the anion as SO₄²⁻, yielding FeSO₄.
題目 25 · 選擇題
1 分
An aqueous solution of copper(II) sulfate is electrolysed using inert carbon electrodes.
Which row correctly describes the observations at each electrode and the change in the electrolyte?
A.Cathode: Pink-brown solid forms; Anode: Colourless gas bubbles form; Electrolyte: Blue colour fades
B.Cathode: Colourless gas bubbles form; Anode: Pink-brown solid forms; Electrolyte: Blue colour fades
C.Cathode: Pink-brown solid forms; Anode: Green gas bubbles form; Electrolyte: Blue colour stays the same
D.Cathode: Colourless gas bubbles form; Anode: Colourless gas bubbles form; Electrolyte: Blue colour stays the same
查看答案詳解收起答案詳解
解題
At the cathode (negative electrode), copper ions (\( \text{Cu}^{2+} \)) are attracted and reduced to copper metal, which is seen as a pink-brown solid depositing on the electrode.
At the anode (positive electrode), hydroxide ions (\( \text{OH}^- \)) from water are discharged to form oxygen gas, which is seen as colourless bubbles.
As \( \text{Cu}^{2+} \) ions are removed from the solution, the blue colour of the electrolyte gradually fades.
評分準則
A - 1 mark: Correctly identifies the cathode product (copper), the anode product (oxygen), and the fading of the blue colour of the electrolyte.
題目 26 · 選擇題
1 分
How many structural isomers are there with the molecular formula \( \text{C}_4\text{H}_9\text{Cl} \)?
A.2
B.3
C.4
D.5
查看答案詳解收起答案詳解
解題
To find the number of structural isomers with the formula \( \text{C}_4\text{H}_9\text{Cl} \):
1. Draw the isomers based on a straight four-carbon chain (butane): - 1-chlorobutane: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} \) - 2-chlorobutane: \( \text{CH}_3\text{CH}_2\text{CH(Cl)CH}_3 \)
2. Draw the isomers based on a branched three-carbon chain (methylpropane): - 1-chloro-2-methylpropane: \( \text{(CH}_3\text{)}_2\text{CHCH}_2\text{Cl} \) - 2-chloro-2-methylpropane: \( \text{(CH}_3\text{)}_3\text{CCl} \)
There are no other possible arrangements, meaning there are exactly 4 structural isomers.
評分準則
C - 1 mark: Correctly identifies that there are 4 structural isomers.
題目 27 · 選擇題
1 分
Three metals, X, Y and Z, were studied to determine their relative reactivity.
- Metal X reacts rapidly with cold water to release hydrogen gas. - Metal Y does not react with steam, but its metal oxide can be reduced by heating with carbon powder. - Metal Z reacts with steam to form hydrogen gas, but does not react with cold water.
What is the correct order of reactivity of these metals, from most reactive to least reactive?
A.X → Z → Y
B.X → Y → Z
C.Z → X → Y
D.Y → Z → X
查看答案詳解收起答案詳解
解題
- Metal X reacts with cold water, which is characteristic of highly reactive metals (such as Group I metals or calcium). - Metal Z reacts with steam but not cold water, which is characteristic of moderately reactive metals (such as magnesium, zinc, or iron). - Metal Y is very unreactive: it does not react with water or steam, and its oxide can be reduced by carbon, placing it below carbon in the reactivity series.
Therefore, the reactivity order from most to least reactive is X \(\rightarrow\) Z \(\rightarrow\) Y.
評分準則
A - 1 mark: Correctly deduces the reactivity order of X, Y, and Z based on their reactions with water, steam, and carbon.
題目 28 · 選擇題
1 分
What is the concentration of sulfate ions, \( \text{SO}_4^{2-} \), in \( 250\text{ cm}^3 \) of an aqueous solution containing \( 14.2\text{ g} \) of dissolved sodium sulfate, \( \text{Na}_2\text{SO}_4 \)?
[\( M_r \text{ of Na}_2\text{SO}_4 = 142 \)]
A.0.10 mol/dm3
B.0.20 mol/dm3
C.0.40 mol/dm3
D.0.80 mol/dm3
查看答案詳解收起答案詳解
解題
1. Calculate the number of moles of \( \text{Na}_2\text{SO}_4 \): \( \text{moles} = \frac{\text{mass}}{M_r} = \frac{14.2\text{ g}}{142\text{ g/mol}} = 0.10\text{ mol} \)
2. Determine the moles of \( \text{SO}_4^{2-} \) ions in solution: Each formula unit of \( \text{Na}_2\text{SO}_4 \) dissociates to yield one \( \text{SO}_4^{2-} \) ion: \( \text{Na}_2\text{SO}_4(aq) \rightarrow 2\text{Na}^+(aq) + \text{SO}_4^{2-}(aq) \) Therefore, moles of \( \text{SO}_4^{2-} = 0.10\text{ mol} \).
3. Convert the volume of the solution to \( \text{dm}^3 \): \( 250\text{ cm}^3 = 0.25\text{ dm}^3 \)
4. Calculate the concentration of \( \text{SO}_4^{2-} \): \( \text{concentration} = \frac{\text{moles}}{\text{volume}} = \frac{0.10\text{ mol}}{0.25\text{ dm}^3} = 0.40\text{ mol/dm}^3 \).
評分準則
C - 1 mark: Calculates the correct concentration of sulfate ions, accounting for both the stoichiometry of dissolution and unit conversion.
題目 29 · 選擇題
1 分
A student reacts \( 5.0\text{ g} \) of zinc powder with an excess of \( 1.0\text{ mol/dm}^3 \) sulfuric acid.
Which change decreases the initial rate of reaction but does not change the total volume of hydrogen gas collected at the end of the reaction?
A.Using 5.0 g of zinc pieces instead of powder.
B.Using 2.5 g of zinc powder instead of 5.0 g.
C.Using 50 cm3 of 0.5 mol/dm3 sulfuric acid instead of 100 cm3 of 1.0 mol/dm3 sulfuric acid.
D.Increasing the temperature of the sulfuric acid.
查看答案詳解收起答案詳解
解題
- Using zinc pieces instead of powder decreases the surface area of the reactant. This decreases the initial rate of reaction. - Since the mass of the limiting reactant (zinc) is still \( 5.0\text{ g} \) and the acid remains in excess, the total moles (and therefore volume) of hydrogen gas produced will remain exactly the same. - Option B decreases the total yield. Option C reduces the total yield if the acid becomes the limiting reactant. Option D increases the initial rate of reaction.
評分準則
A - 1 mark: Correctly identifies that larger pieces reduce the surface area and hence rate, without changing the final amount of limiting reactant (zinc).
題目 30 · 選擇題
1 分
Which statement about weak acids and strong acids in aqueous solution is correct?
A.A weak acid has a lower pH than a strong acid of the same concentration.
B.A strong acid is completely dissociated into ions, whereas a weak acid is only partially dissociated.
C.Weak acids react with reactive metals to produce hydrogen gas, but strong acids do not.
D.Strong acids can be concentrated, whereas weak acids are always dilute.
查看答案詳解收起答案詳解
解題
- Strong acids are defined by their complete dissociation into hydrogen ions in aqueous solution (e.g., \( \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \)). - Weak acids only partially dissociate in solution (e.g., \( \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \)). - Therefore, option B is correct.
評分準則
B - 1 mark: Identifies the correct definition and distinction of strong versus weak acids in terms of their degree of dissociation.
題目 31 · 選擇題
1 分
An ion of element X is represented as \( ^{31}_{15}\text{X}^{3-} \).
Which row correctly identifies the number of protons, neutrons and electrons in this ion?
A.Protons: 15; Neutrons: 16; Electrons: 18
B.Protons: 15; Neutrons: 16; Electrons: 12
C.Protons: 15; Neutrons: 31; Electrons: 18
D.Protons: 16; Neutrons: 15; Electrons: 15
查看答案詳解收起答案詳解
解題
- The atomic number (lower left) is 15, which represents the number of protons. - The mass number (upper left) is 31. The number of neutrons is calculated as \( \text{Mass Number} - \text{Atomic Number} = 31 - 15 = 16 \). - The charge is 3-, which means the ion has 3 more electrons than protons. Therefore, the number of electrons is \( 15 + 3 = 18 \).
This matches row A.
評分準則
A - 1 mark: Correctly determines protons (15), neutrons (16), and electrons (18).
題目 32 · 選擇題
1 分
A student tests an unknown aqueous solution and records the following observations:
- When aqueous sodium hydroxide is added, a green precipitate is formed that is insoluble in excess. - When dilute nitric acid followed by aqueous barium nitrate is added, a white precipitate is formed.
What is the identity of the solute in the unknown solution?
A.iron(II) sulfate
B.iron(III) sulfate
C.iron(II) chloride
D.chromium(III) sulfate
查看答案詳解收起答案詳解
解題
- The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess is the characteristic test for iron(II) ions (\( \text{Fe}^{2+} \)). Chromium(III) also forms a green precipitate but it dissolves in excess sodium hydroxide to form a green solution. - The formation of a white precipitate with aqueous barium nitrate after acidification is the characteristic test for sulfate ions (\( \text{SO}_4^{2-} \)). - Combining these, the compound is iron(II) sulfate.
評分準則
A - 1 mark: Correctly identifies the cation as iron(II) and the anion as sulfate.
題目 33 · 選擇題
1 分
When 5.00 g of an impure sample of calcium carbonate reacts with an excess of dilute hydrochloric acid, 1.02 dm\(^{3}\) of carbon dioxide gas is collected at room temperature and pressure (r.t.p.). What is the percentage purity of the calcium carbonate? [Assume the impurities do not react with acid. \(M_{\text{r}}(\text{CaCO}_3) = 100\); 1 mole of gas occupies 24.0 dm\(^{3}\) at r.t.p.]
A.42.5%
B.85.0%
C.90.0%
D.95.8%
查看答案詳解收起答案詳解
解題
First, calculate the number of moles of carbon dioxide gas collected: moles of \(\text{CO}_2 = 1.02\text{ dm}^3 / 24.0\text{ dm}^3/\text{mol} = 0.0425\text{ mol}\). Since the reaction stoichiometry is \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\), the mole ratio of \(\text{CaCO}_3\) to \(\text{CO}_2\) is 1:1. Therefore, there are 0.0425 mol of pure \(\text{CaCO}_3\) in the sample. Next, calculate the mass of this pure \(\text{CaCO}_3\): \(\text{mass} = 0.0425\text{ mol} \times 100\text{ g/mol} = 4.25\text{ g}\). Finally, determine the percentage purity: \(\text{percentage purity} = (4.25\text{ g} / 5.00\text{ g}) \times 100\% = 85.0\%\).
評分準則
1 mark for calculating the moles of carbon dioxide (0.0425 mol). 1 mark for calculating the mass of pure calcium carbonate (4.25 g). 1 mark for calculating the percentage purity (85.0%).
題目 34 · 選擇題
1 分
An aqueous solution of concentrated copper(II) chloride is electrolysed using inert carbon electrodes. Which row correctly describes the observations at the anode and cathode?
A.Anode: bubbles of a pale green gas; Cathode: red-brown solid deposited
B.Anode: bubbles of a colourless gas; Cathode: red-brown solid deposited
C.Anode: red-brown solid deposited; Cathode: bubbles of a pale green gas
D.Anode: bubbles of a pale green gas; Cathode: bubbles of a colourless gas
查看答案詳解收起答案詳解
解題
During the electrolysis of concentrated aqueous copper(II) chloride: At the anode (+), chloride ions (\(\text{Cl}^-\)) are discharged in preference to hydroxide ions (\(\text{OH}^-\)) due to their high concentration, forming chlorine gas, which is observed as bubbles of a pale green gas. At the cathode (-), copper(II) ions (\(\text{Cu}^{2+}\)) are discharged in preference to hydrogen ions (\(\text{H}^+\)) because copper is lower in the reactivity series, forming copper metal which is observed as a red-brown solid deposit. Therefore, option A is correct.
評分準則
1 mark for identifying that bubbles of a pale green gas (chlorine) are produced at the anode and a red-brown solid (copper) is deposited at the cathode.
題目 35 · 選擇題
1 分
The reaction between excess magnesium ribbon and dilute hydrochloric acid is investigated. Which change increases the rate of the reaction because the reacting particles are closer together, resulting in a higher collision frequency?
A.Using a larger volume of the same hydrochloric acid
B.Increasing the concentration of the hydrochloric acid
C.Increasing the temperature of the hydrochloric acid
D.Using a catalyst
查看答案詳解收起答案詳解
解題
Increasing the concentration of the acid increases the number of reacting particles per unit volume, which means the particles are closer together, directly increasing the collision frequency. Increasing temperature increases collision frequency and energy. A catalyst lowers the activation energy. Changing the volume of the same acid does not change concentration or rate.
評分準則
1 mark for identifying that increasing concentration is the change that increases collision frequency by bringing particles closer together.
題目 36 · 選擇題
1 分
An organic compound Y has the structure shown: \(\text{CH}_2=\text{CH}-\text{CH}_2-\text{COOH}\). Which statements about compound Y are correct? 1. It decolourises aqueous bromine. 2. It reacts with sodium carbonate to produce carbon dioxide. 3. It reacts with ethanol to form an ester. 4. Its empirical formula is \(\text{C}_2\text{H}_3\text{O}\).
A.1, 2, 3 and 4
B.1 and 2 only
C.1, 2 and 3 only
D.2, 3 and 4 only
查看答案詳解收起答案詳解
解題
1. Correct: Compound Y contains a \(\text{C}=\text{C}\) double bond (unsaturated), so it decolourises aqueous bromine. 2. Correct: It contains a carboxylic acid group (\(-\text{COOH}\)), so it reacts with metal carbonates to produce \(\text{CO}_2\). 3. Correct: The \(-\text{COOH}\) group reacts with alcohols like ethanol to form an ester. 4. Correct: The molecular formula is \(\text{C}_4\text{H}_6\text{O}_2\), which simplifies to the empirical formula \(\text{C}_2\text{H}_3\text{O}\). Thus, all statements are correct.
評分準則
1 mark for identifying that statements 1, 2, 3 and 4 are all correct.
題目 37 · 選擇題
1 分
The reaction below is at equilibrium in a closed container: \(2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g})\) (where the forward reaction is exothermic). Which row describes the conditions that maximise the equilibrium yield of \(\text{SO}_3(\text{g})\)?
A.Temperature: High; Pressure: High; Effect of catalyst: Increases yield
B.Temperature: High; Pressure: Low; Effect of catalyst: Increases yield
C.Temperature: Low; Pressure: High; Effect of catalyst: No change in yield
D.Temperature: Low; Pressure: Low; Effect of catalyst: No change in yield
查看答案詳解收起答案詳解
解題
Since the forward reaction is exothermic, a lower temperature shifts the equilibrium to the right, increasing the yield of \(\text{SO}_3\). There are 3 moles of gas on the reactant side and 2 moles of gas on the product side; thus, high pressure shifts the equilibrium to the right to increase the yield. A catalyst increases the rate of reaction but does not affect the yield or the position of equilibrium. Therefore, low temperature, high pressure, and no change in yield with a catalyst is correct (Option C).
評分準則
1 mark for identifying that low temperature and high pressure increase the yield of product, and that a catalyst has no effect on equilibrium yield.
題目 38 · 選擇題
1 分
Four metals, W, X, Y, and Z, are added separately to aqueous solutions of their nitrates. Metal W displaces X and Z, but does not react with Y. Metal X does not react with any of the other nitrate solutions. Metal Y displaces W, X, and Z. Metal Z displaces X, but does not react with W or Y. What is the order of reactivity of the metals, from most reactive to least reactive?
A.Y > Z > W > X
B.X > Z > W > Y
C.W > Y > Z > X
D.Y > W > Z > X
查看答案詳解收起答案詳解
解題
Y is the most reactive because it displaces W, X, and Z. W is more reactive than X and Z because it displaces both of them, but less reactive than Y because it cannot displace Y. Z is more reactive than X because it displaces X, but less reactive than W because it cannot displace W. X is the least reactive as it cannot displace any other metal. The correct order from most to least reactive is: Y > W > Z > X.
評分準則
1 mark for deducing the correct order of reactivity as Y > W > Z > X based on the given displacement reactions.
題目 39 · 選擇題
1 分
An unknown salt, Q, is dissolved in water to make an aqueous solution. When aqueous sodium hydroxide is added, a green precipitate is formed which is insoluble in excess. When dilute nitric acid followed by aqueous barium nitrate is added to another portion of the solution, a white precipitate is formed. What is the identity of salt Q?
A.iron(III) sulfate
B.iron(II) sulfate
C.chromium(III) sulfate
D.iron(II) chloride
查看答案詳解收起答案詳解
解題
The cation test: Aqueous sodium hydroxide gives a green precipitate insoluble in excess, which confirms the presence of iron(II) ions (\(\text{Fe}^{2+}\)). Note that chromium(III) also gives a green precipitate, but it is soluble in excess sodium hydroxide. Anion test: Dilute nitric acid followed by aqueous barium nitrate gives a white precipitate (barium sulfate), confirming the presence of sulfate ions (\(\text{SO}_4^{2-}\)). Therefore, Q is iron(II) sulfate.
評分準則
1 mark for identifying the cation as iron(II) from the insoluble green precipitate with NaOH. 1 mark for identifying the anion as sulfate from the white precipitate with barium nitrate.
題目 40 · 選擇題
1 分
Which statement correctly describes the greenhouse effect of carbon dioxide and methane in the atmosphere?
A.They react with ozone in the stratosphere to produce heat, warming the Earth's surface.
B.They absorb short-wavelength ultraviolet radiation directly from the Sun and warm the atmosphere.
C.They absorb thermal energy (infrared radiation) re-emitted from the Earth's surface and trap it in the atmosphere.
D.They reflect incoming solar radiation back into space, preventing the cooling of the atmosphere.
查看答案詳解收起答案詳解
解題
The greenhouse effect occurs because short-wavelength solar radiation passes through the atmosphere to warm the Earth. The Earth then re-emits this thermal energy as longer-wavelength infrared radiation. Greenhouse gases like carbon dioxide and methane absorb this infrared radiation and trap the heat in the atmosphere, leading to global warming.
評分準則
1 mark for identifying that greenhouse gases absorb thermal/infrared radiation re-emitted from the Earth's surface and trap it in the atmosphere.
Paper 4 Extended Theory
Answer all structured questions in the spaces provided on the question paper. Show working where appropriate.
9 題目 · 90 分
題目 1 · Structured Theory
10 分
A student electrolyses aqueous copper(II) sulfate, \(\text{CuSO}_4(aq)\), using two different sets of electrodes.
(a) In the first experiment, inert carbon (graphite) electrodes are used. (i) Describe what is observed at the anode (positive electrode) during this electrolysis. (ii) Write the ionic half-equation, including state symbols, for the reaction occurring at the anode.
(b) Explain why the blue color of the aqueous copper(II) sulfate slowly fades during this electrolysis.
(c) In the second experiment, the carbon electrodes are replaced with copper electrodes. Describe how the observations at the anode and the cathode differ from the first experiment.
(d) Give one major industrial application of the electrolysis of aqueous copper(II) sulfate using copper electrodes.
查看答案詳解收起答案詳解
解題
For (a), inert electrodes allow hydroxide ions from water to be oxidized to oxygen gas: \(4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4e^-\). For (b), copper ions cause the blue color; their reduction to solid copper at the cathode decreases their concentration in solution. For (c), active copper electrodes mean the anode itself oxidized (dissolves): \(\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-\). This replenishes the \(\text{Cu}^{2+}\) ions, keeping the blue intensity of the solution constant. For (d), this electrolysis is used for industrial purification of copper or electroplating items with copper.
評分準則
(a) (i) Bubbles / effervescence of a colorless gas [1 mark]. (ii) Correct formulae of reactants and products: \(\text{OH}^-\), \(\text{O}_2\), \(\text{H}_2\text{O}\), and \(e^-\)[1 mark]. Correct balancing and state symbols: \(4\text{OH}^-(aq) \rightarrow \text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^-\)[1 mark].
(b) Copper(II) ions / \(\text{Cu}^{2+}\) are discharged/reduced at the cathode [1 mark]. Their concentration in the solution decreases [1 mark].
(c) The copper anode dissolves/decreases in mass (no gas bubbles) [1 mark]. The cathode gains mass with a pink/brown deposit [1 mark]. The blue color of the solution remains constant/does not fade [1 mark].
(d) Purification/refining of copper OR electroplating [1 mark].
題目 2 · Structured Theory
10 分
A student investigates the reaction between zinc metal and dilute hydrochloric acid.
(a) Write a balanced chemical equation, including state symbols, for the reaction of solid zinc with aqueous hydrochloric acid.
(b) The student reacts excess zinc with \(50.0\text{ }cm^3\) of \(2.00\text{ }mol/dm^3\) hydrochloric acid. Calculate the volume, in \(dm^3\), of hydrogen gas produced at room temperature and pressure (r.t.p.). [Molar gas volume at r.t.p. is \(24.0\text{ }dm^3/mol\)].
(c) In a separate experiment, \(3.25\text{ }g\) of zinc is completely reacted with excess hydrochloric acid. Calculate the mass, in grams, of anhydrous zinc chloride, \(\text{ZnCl}_2\), produced. [Relative atomic masses: \(A_r(\text{Zn}) = 65\), \(A_r(\text{Cl}) = 35.5\)].
查看答案詳解收起答案詳解
解題
For (a), zinc is a reactive metal that displaces hydrogen from acids, forming aqueous zinc chloride and hydrogen gas. For (b), calculate moles of HCl using \(n = C \times V\). Use stoichiometric ratio (2:1) to find moles of hydrogen. Multiply by 24 to find volume in \(dm^3\). For (c), calculate moles of Zn using \(n = m / A_r\). Use stoichiometric ratio (1:1) to find moles of zinc chloride. Multiply by the molar mass of \(\text{ZnCl}_2\) (136) to get the mass.
評分準則
(a) Reactants and products correct [1 mark]. Balancing correct [1 mark]. All state symbols correct [1 mark].
(b) Calculation of moles of HCl (0.100 mol) [1 mark]. Calculation of moles of H2 (0.0500 mol) [1 mark]. Calculation of volume of hydrogen gas (1.20 dm3) [1 mark].
(c) Calculation of moles of Zn (0.0500 mol) [1 mark]. Calculation of Mr of ZnCl2 (136) [1 mark]. Correct multiplication to get mass (6.80 g) [2 marks; 1 mark for method, 1 mark for accuracy].
題目 3 · Structured Theory
10 分
Hydrogen peroxide decomposes slowly at room temperature to form water and oxygen gas.
(a) Write a balanced chemical equation for the decomposition of aqueous hydrogen peroxide, \(\text{H}_2\text{O}_2(aq)\).
(b) The rate of this reaction can be increased by adding a small amount of manganese(IV) oxide as a catalyst. Explain, in terms of collision theory and activation energy, how a catalyst increases the rate of reaction.
(c) A student performs the decomposition reaction at \(20^\circ\text{C}\) and plots a graph of the volume of oxygen gas produced against time. The student then repeats the experiment at \(40^\circ\text{C}\) using the exact same volume and concentration of hydrogen peroxide and the same mass of catalyst. Describe two ways the graph for the reaction at \(40^\circ\text{C}\) would differ from or compare to the graph at \(20^\circ\text{C}\), and explain why.
(d) State the effect of doubling the concentration of the hydrogen peroxide on: (i) the initial rate of the reaction, (ii) the final volume of oxygen gas collected.
查看答案詳解收起答案詳解
解題
For (a), hydrogen peroxide decomposes into water and oxygen. Ensure the equation is balanced. For (b), use key terms: 'alternative pathway', 'lower activation energy', and 'frequency of successful collisions'. For (c), temperature increases particle velocity, making the initial slope steeper, but has no effect on total yield of product since the reactant amounts are constant. For (d), concentration directly influences collision rate (doubling rate) and stoichiometry dictates the yield (doubling the amount of gas produced).
(b) Alternative reaction pathway with lower activation energy [1 mark]. More particles have energy greater than/equal to activation energy [1 mark]. Higher frequency of successful collisions [1 mark].
(c) Graph curve is steeper at the start at 40 °C [1 mark]. Because higher temperature increases kinetic energy / speed of particles, leading to more frequent collisions [1 mark]. Both graphs level off at the same final volume of oxygen [1 mark].
(d) (i) Rate doubles / increases [1 mark]. (ii) Final volume of oxygen doubles [1 mark].
題目 4 · Structured Theory
10 分
Esters are organic compounds with characteristic sweet smells. Propyl ethanoate is an ester.
(a) Define the term structural isomer.
(b) (i) Draw the fully displayed structure of propyl ethanoate, showing all atoms and all bonds. (ii) Name the carboxylic acid and the alcohol that react together to form propyl ethanoate. (iii) State the reaction conditions, including any catalyst, required for this esterification to take place.
(c) Draw the structural formula of an isomer of propyl ethanoate that is a carboxylic acid.
查看答案詳解收起答案詳解
解題
For (a), the definition must mention same molecular formula but different structural formula. For (b)(i), draw the ester linkage -COO- connected to a methyl group on the carbonyl carbon and a propyl group on the single-bonded oxygen. For (b)(ii), the 'ethanoate' part comes from ethanoic acid, and 'propyl' comes from propan-1-ol. For (b)(iii), concentrated sulfuric acid acts as a catalyst and dehydrating agent, and heating increases the reaction rate. For (c), propyl ethanoate has 5 carbons, 10 hydrogens, and 2 oxygens (\(\text{C}_5\text{H}_{10}\text{O}_2\)). An isomeric carboxylic acid must have 5 carbons: pentanoic acid.
評分準則
(a) Same molecular formula [1 mark] but different structural formula / arrangement [1 mark].
(b) (i) Displayed ester group -COO- with C=O and C-O shown [1 mark]. Correct propyl group and methyl group attached with all atoms and single bonds drawn out [1 mark]. (ii) Ethanoic acid [1 mark] and propan-1-ol / propanol [1 mark]. (iii) Concentrated sulfuric acid [1 mark] and heating / heat / reflux [1 mark].
(c) Correct structural formula of pentanoic acid (or structural isomers with 5 carbons, e.g., methylbutanoic acid) [2 marks; 1 mark for correct carboxylic acid functional group -COOH, 1 mark for correct total of 5 carbon atoms in the structure].
題目 5 · Structured Theory
10 分
Ammonia is manufactured industrially by the Haber Process using a reversible reaction between nitrogen and hydrogen.
(a) Write the balanced chemical equation, including state symbols, for the synthesis of ammonia in the Haber Process.
(b) State the reaction conditions used in the industrial Haber Process: (i) temperature, (ii) pressure, (iii) identity of the catalyst.
(c) Use Le Chatelier's principle to explain: (i) why a high pressure of 200 atmospheres is used to increase the yield of ammonia, (ii) why a low temperature would theoretically increase the yield of ammonia, and why a temperature of \(450^\circ\text{C}\) is used in practice instead.
查看答案詳解收起答案詳解
解題
For (a), write the balanced equation with the reversible reaction arrow and gas state symbols. For (b), memorize the Haber process parameters. For (c)(i), note the stoichiometry: 4 moles of reactants produce 2 moles of products. For (c)(ii), state that the reaction is exothermic, so cooling shifts equilibrium to the right. Explain the kinetic-thermodynamic compromise of using a moderate temperature (450 °C).
評分準則
(a) Correct formulas and reversible arrow [1 mark]. Correct balancing [1 mark]. Correct state symbols [1 mark].
(b) (i) 450 °C [1 mark]. (ii) 200 atm [1 mark]. (iii) Iron / Fe [1 mark].
(c) (i) Fewer moles of gas on the product/right side (2 vs 4) [1 mark]. Equilibrium shifts to the right to oppose pressure increase, increasing the yield [1 mark]. (ii) Forward reaction is exothermic, so lower temperature shifts equilibrium to the right [1 mark]. 450 °C is a compromise because at lower temperatures the rate of reaction is too slow [1 mark].
題目 6 · Structured Theory
10 分
The reactivity series of metals determines how they are extracted and how they react.
(a) Copper is a relatively unreactive metal. It can be extracted by heating solid copper(II) oxide with carbon powder. Write a balanced chemical equation for this reduction reaction.
(b) Explain why aluminium cannot be extracted from aluminium oxide using carbon, and state the method and key electrolyte components used to extract aluminium.
(c) A student places a piece of zinc ribbon into an aqueous solution of copper(II) sulfate. (i) Describe two physical observations that would be made during this reaction. (ii) Write the ionic equation, including state symbols, for this displacement reaction.
查看答案詳解收起答案詳解
解題
For (a), carbon reduces copper(II) oxide to copper metal and carbon dioxide (or carbon monoxide). For (b), metals higher than carbon in the reactivity series cannot be reduced by carbon and require electrolysis. Cryolite is used to lower the melting point of alumina. For (c)(i), zinc displaces copper. Solid copper is pink-brown, and \(\text{Cu}^{2+}\) ions (blue) are replaced by \(\text{Zn}^{2+}\) ions (colorless). For (c)(ii), write the ionic equation showing spectator sulfate ions omitted.
(b) Aluminium is more reactive than carbon / carbon cannot reduce it [1 mark]. Extracted by electrolysis [1 mark] of aluminium oxide dissolved in molten cryolite [1 mark].
(c) (i) Red-brown/pink solid deposit [1 mark]. Solution turns colorless / blue color fades [1 mark]. (ii) Correct species: \(\text{Zn}\) and \(\text{Cu}^{2+}\) on left, \(\text{Zn}^{2+}\) and \(\text{Cu}\) on right [1 mark]. Correct balancing and charges [1 mark]. Correct state symbols [1 mark].
題目 7 · Structured Theory
10 分
Qualitative chemical analysis can be used to identify unknown substances.
(a) A student is given a green crystalline salt, X. (i) To a solution of X, the student adds aqueous sodium hydroxide dropwise until in excess. A green precipitate is formed which is insoluble in excess sodium hydroxide. Identify the cation present in X. (ii) To another portion of the solution of X, the student adds dilute nitric acid followed by aqueous silver nitrate. A cream precipitate is formed. Identify the anion present in X. (iii) Deduce the chemical formula of salt X.
(b) Describe a chemical test to confirm the presence of carbonate ions, \(\text{CO}_3^{2-}\), in a solid sample of a different salt, including the expected observation for a positive result.
(c) Describe a chemical test to show that a colorless liquid is water, specifying the reagent used, the initial color, and the final color.
查看答案詳解收起答案詳解
解題
For (a)(i), \(\text{Fe}^{2+}\) forms a green precipitate with NaOH that does not dissolve in excess. For (a)(ii), silver nitrate forms a cream precipitate of silver bromide with bromide ions. For (a)(iii), combining \(\text{Fe}^{2+}\) and \(\text{Br}^-\) yields \(\text{FeBr}_2\). For (b), acid reacts with carbonates to liberate carbon dioxide gas, which is confirmed with limewater. For (c), chemical tests for water use anhydrous salts. Anhydrous copper(II) sulfate turns from white to blue when hydrated.
(b) Add dilute acid / HCl [1 mark]. Effervescence/bubbles observed [1 mark]. Bubble gas through limewater which turns cloudy / milky [1 mark].
(c) Reagent: anhydrous copper(II) sulfate [1 mark] (must specify anhydrous). Initial color: white [1 mark]. Final color: blue [1 mark]. Correct description of adding the liquid [1 mark]. (Accept anhydrous cobalt(II) chloride [1 mark], changing from blue [1 mark] to pink [1 mark]).
題目 8 · structured
10 分
A student carries out an experiment to determine the formula of a hydrated metal carbonate, \(M_2\text{CO}_3 \cdot x\text{H}_2\text{O}\), where \(M\) is a Group 1 metal.
They dissolve \(3.575\text{ g}\) of the hydrated carbonate in distilled water and make the solution up to exactly \(250\text{ cm}^3\) in a volumetric flask.
The student then titrates \(25.0\text{ cm}^3\) of this solution against \(0.100\text{ mol/dm}^3\) hydrochloric acid, \(\text{HCl}\). The volume of dilute hydrochloric acid required for complete neutralisation is \(25.0\text{ cm}^3\).
The chemical equation for the reaction is: \(M_2\text{CO}_3(\text{aq}) + 2\text{HCl}(\text{aq}) \rightarrow 2M\text{Cl}(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\)
(a) Calculate the number of moles of \(\text{HCl}\) used in the titration. [1]
(b) Use the chemical equation to determine the number of moles of \(M_2\text{CO}_3\) present in the \(25.0\text{ cm}^3\) sample. [1]
(c) Calculate the number of moles of \(M_2\text{CO}_3\) present in the \(250\text{ cm}^3\) volumetric flask. [1]
(d) Calculate the relative formula mass (\(M_r\)) of the hydrated carbonate, \(M_2\text{CO}_3 \cdot x\text{H}_2\text{O}\). [2]
(e) The Group 1 metal \(M\) is sodium, \(\text{Na}\). (i) Calculate the relative formula mass (\(M_r\)) of anhydrous sodium carbonate, \(\text{Na}_2\text{CO}_3\). [Relative atomic masses, \(A_r\): \(\text{Na} = 23\); \(\text{C} = 12\); \(\text{O} = 16\)] [1] (ii) Calculate the mass of water of crystallisation present in one mole of \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}\), and hence determine the value of \(x\). [2]
(f) Calculate the volume of carbon dioxide gas, in \(\text{dm}^3\), produced at room temperature and pressure (r.t.p.) when \(0.0125\text{ mol}\) of \(M_2\text{CO}_3\) reacts completely with excess hydrochloric acid. [One mole of any gas occupies \(24\text{ dm}^3\) at r.t.p.] [2]
(a) 0.00250 mol [1] (b) 0.00125 mol (or half of candidate's value in (a)) [1] (c) 0.0125 mol (or ten times candidate's value in (b)) [1] (d) Method: mass / candidate's (c) (e.g. 3.575 / 0.0125) [1]; Accuracy: 286 [1] (e)(i) 106 [1] (e)(ii) Method: candidate's (d) - candidate's (e)(i) (e.g. 286 - 106 = 180) [1]; Value: x = 10 (from 180 / 18) [1] (f) Method: candidate's (c) x 24 [1]; Accuracy: 0.300 dm^3 (accept 0.3 or 0.30, and accept 300 cm^3 if unit is clearly stated) [1]
題目 9 · structured
10 分
Placeholder
查看答案詳解收起答案詳解
解題
Placeholder
評分準則
Placeholder
Paper 6 Alternative to Practical
Answer all questions. Show measurements to the appropriate decimal resolution and plan experimental procedures.
5 題目 · 46 分
題目 1 · practical
12 分
A student investigated the rate of reaction between dilute sulfuric acid and zinc granules. Five experiments were carried out using different concentrations of sulfuric acid. In each experiment, the volume of hydrogen gas collected in a gas syringe was recorded after 3 minutes.
Table of Results: ExperimentConcentration of sulfuric acid / \(\text{mol/dm}^3\)Initial syringe reading / \(\text{cm}^3\)Final syringe reading / \(\text{cm}^3\)Volume of hydrogen gas collected / \(\text{cm}^3\)10.52.018.0(i)21.01.533.5(ii)31.54.052.0(iii)42.03.067.0(iv)52.51.081.0(v) Questions: (a) Complete the table by calculating the volume of hydrogen gas collected for each of the five experiments. [5] (b) Describe the relationship between the concentration of sulfuric acid and the volume of hydrogen gas collected. [2] (c) Identify the independent variable and the dependent variable in this experiment. [2] (d) Suggest why the gas syringe plunger should be moved in and out before starting the experiment. [1] (e) State one variable that should be kept constant in this investigation to ensure a fair test, other than the volume of sulfuric acid used. [1] (f) The student repeated Experiment 2 but used zinc powder instead of zinc granules. State and explain the effect this change would have on the volume of gas collected in 3 minutes. [1]
查看答案詳解收起答案詳解
解題
For part (a), calculate the volume of gas collected by subtracting the initial syringe reading from the final syringe reading for each experiment: - Experiment 1: \(18.0 - 2.0 = 16.0\ \text{cm}^3\) - Experiment 2: \(33.5 - 1.5 = 32.0\ \text{cm}^3\) - Experiment 3: \(52.0 - 4.0 = 48.0\ \text{cm}^3\) - Experiment 4: \(67.0 - 3.0 = 64.0\ \text{cm}^3\) - Experiment 5: \(81.0 - 1.0 = 80.0\ \text{cm}^3\)
For part (b), as the concentration increases by equal increments (0.5 mol/dm³), the volume of gas also increases by a constant amount (16.0 cm³), demonstrating a direct linear relationship.
For part (c), the independent variable is the one intentionally changed (concentration of acid), and the dependent variable is the one being measured (volume of hydrogen gas collected).
For part (d), pushing the plunger in and out ensures there is no friction or sticking that could distort the volume readings during gas collection.
For part (e), control variables include the temperature of the acid, the total mass of zinc, or the initial size of the granules.
For part (f), using powder increases the surface area exposed to the acid. This increases the frequency of successful collisions, causing a faster rate of reaction and therefore a greater volume of gas collected within the specified 3-minute timeframe.
評分準則
(a) Award 1 mark for each correct volume (must be written to 1 decimal place): - (i) 16.0 [1] - (ii) 32.0 [1] - (iii) 48.0 [1] - (iv) 64.0 [1] - (v) 80.0 [1] (b) Award 1 mark for stating that volume increases as concentration increases [1] and 1 mark for noting that they are directly proportional / have a linear relationship [1]. (c) Award 1 mark for identifying the independent variable as concentration of sulfuric acid [1] and 1 mark for the dependent variable as volume of hydrogen gas collected [1]. (d) Award 1 mark for stating that it checks the plunger moves smoothly / does not stick [1]. (e) Award 1 mark for temperature of acid / mass of zinc / same starting size/shape of zinc granules [1]. Reject: 'amount' of zinc. (f) Award 1 mark for stating that the volume of gas collected increases because zinc powder has a larger surface area / faster rate of reaction [1].
題目 2 · practical
12 分
A student determined the concentration of a solution of sodium hydroxide by titrating it against standard dilute hydrochloric acid. The student used a volumetric pipette to transfer 25.0 \(\text{cm}^3\) of the sodium hydroxide solution into a conical flask and added a few drops of methyl orange indicator. Acid was added from a burette until the end-point was reached.
Table of Results: Burette reading / \(\text{cm}^3\)Rough titrationTitration 1Titration 2Titration 3Final burette reading24.823.947.724.2Initial burette reading0.00.224.00.6Titre (volume of acid added)(i)(ii)(iii)(iv) Questions: (a) Complete the table by calculating the titre (volume of acid added) for each of the four titrations. [4] (b) State which titrations are concordant (consistent) and should be used to calculate the mean titre. Explain your choice. [2] (c) Calculate the mean titre of hydrochloric acid using your chosen concordant titrations. Show your working. [2] (d) State the color change of the methyl orange indicator at the end-point of this titration. [2] (e) Suggest why a volumetric pipette is used to measure the sodium hydroxide solution instead of a measuring cylinder. [1] (f) Suggest one safety precaution, other than wearing eye protection, that should be taken when filling a burette with dilute acid. [1]
查看答案詳解收起答案詳解
解題
For part (a), the volume of acid added (titre) is calculated by subtracting the initial burette reading from the final burette reading: - Rough titration: \(24.8 - 0.0 = 24.8\ \text{cm}^3\) - Titration 1: \(23.9 - 0.2 = 23.7\ \text{cm}^3\) - Titration 2: \(47.7 - 24.0 = 23.7\ \text{cm}^3\) - Titration 3: \(24.2 - 0.6 = 23.6\ \text{cm}^3\)
For part (b), concordant titres are those within \(0.1\ \text{cm}^3\) (or \(0.2\ \text{cm}^3\)) of each other. Titrations 1, 2, and 3 meet this criterion (23.7, 23.7, and 23.6). The rough titration is an approximation and must be excluded.
For part (c), calculate the average using only the concordant values: \(\text{Mean} = \frac{23.7 + 23.7 + 23.6}{3} = 23.67\ \text{cm}^3\) (or \(\frac{23.7 + 23.7}{2} = 23.7\ \text{cm}^3\)).
For part (d), methyl orange is yellow in alkaline solutions (sodium hydroxide) and turns orange at the neutralisation end-point (before turning red in excess acid). Therefore, the color change at the end-point is yellow to orange.
For part (e), a volumetric pipette is used because it has a highly precise, fixed volume with much lower tolerance (percentage error) compared to a graduated measuring cylinder.
For part (f), filling a burette above eye level can lead to acid splashing into the eyes or face. Therefore, filling it below eye level (e.g., placing it on a low stand or on the floor) is a standard safety measure.
評分準則
(a) Award 1 mark for each correct value written to 1 decimal place: - (i) 24.8 [1] - (ii) 23.7 [1] - (iii) 23.7 [1] - (iv) 23.6 [1] (b) Award 1 mark for identifying Titrations 1, 2, and 3 (or Titrations 1 and 2) [1] and 1 mark for explaining that they are within 0.1 cm³ of each other [1]. Reject: inclusion of rough titration. (c) Award 1 mark for showing correct working (summing the chosen concordant values and dividing by the count) [1] and 1 mark for correct calculation: 23.7 cm³ (or 23.67 cm³) [1]. (Include correct unit). (d) Award 1 mark for the initial color 'yellow' [1] and 1 mark for the final color 'orange' (accept pink, reject red) [1]. (e) Award 1 mark for stating that a volumetric pipette has higher precision / lower percentage error / is more accurate [1]. (f) Award 1 mark for: fill the burette below eye level / use a funnel / place burette on the floor when filling [1].
題目 3 · practical
10 分
A student investigated solid **A**, which is a water-soluble salt containing two cations and one anion.
The table shows the tests carried out and some of the observations.
| Test | Observation | | :--- | :--- | | **Test 1** Solid **A** was heated in a dry test-tube. The gas evolved was tested with damp red litmus paper. | Condensation formed on the cooler parts of the test-tube. The damp red litmus paper turned blue. | | **Test 2** Solid **A** was dissolved in distilled water to form solution **A**. To about 2 cm\(^3\) of solution **A**, aqueous sodium hydroxide was added dropwise and then in excess. | **(b)** ............................................................ [2] | | **Test 3** To about 2 cm\(^3\) of solution **A**, dilute nitric acid followed by aqueous silver nitrate was added. | **(c)** ............................................................ [1] | | **Test 4** To about 2 cm\(^3\) of solution **A**, dilute nitric acid followed by aqueous barium nitrate was added. | **(d)** ............................................................ [1] |
**(a)** (i) Identify the gas evolved in **Test 1**. [1] (ii) Identify the cation responsible for this gas. [1]
**(e)** Identify the anion present in solid **A** based on the result of **Test 4**. [1]
**(f)** Explain why dilute nitric acid is added before aqueous barium nitrate in **Test 4**. [1]
**(g)** State the chemical formulae of the two cations present in solid **A**. [2]
查看答案詳解收起答案詳解
解題
**(a)** (i) A gas that turns damp red litmus paper blue is alkaline. The only alkaline gas tested in the IGCSE syllabus is ammonia (\(\text{NH}_3\)). (ii) The presence of ammonia gas indicates that the cation responsible is the ammonium ion (\(\text{NH}_4^+\)).
**(b)** Solid **A** contains iron(II) ions (as shown by the cations in part g). Adding aqueous sodium hydroxide to a solution containing \(\text{Fe}^{2+}\) ions yields a dirty green precipitate of iron(II) hydroxide, which is insoluble in excess sodium hydroxide.
**(c)** Solid **A** does not contain any halide ions (it is a sulfate), so addition of dilute nitric acid followed by aqueous silver nitrate yields no reaction (the solution remains clear).
**(d)** Solid **A** contains sulfate ions. Adding dilute nitric acid followed by aqueous barium nitrate yields a white precipitate of insoluble barium sulfate (\(\text{BaSO}_4\)).
**(e)** The white precipitate with acidified barium nitrate confirms the presence of the sulfate ion (\(\text{SO}_4^{2-}\)).
**(f)** Dilute nitric acid is added to acidify the solution and remove (react with) any carbonate ions (\(\text{CO}_3^{2-}\)) present. If carbonate ions were present without acid, they would react with barium ions to form a white precipitate of barium carbonate (\(\text{BaCO}_3\)), giving a false-positive result for sulfate.
**(g)** The two cations present in solid **A** are ammonium (\(\text{NH}_4^+\)) and iron(II) (\(\text{Fe}^{2+}\)).
**Part (b)** - Green precipitate [1] - Insoluble in excess (accept: turns brown near surface / on standing) [1]
**Part (c)** - No change / no reaction / no precipitate (reject: nothing / clear solution unless 'remains' is specified) [1]
**Part (d)** - White precipitate [1]
**Part (e)** - Sulfate / \(\text{SO}_4^{2-}\) [1]
**Part (f)** - To react with / destroy / remove carbonate ions (which would otherwise produce a white precipitate of barium carbonate) [1]
**Part (g)** - \(\text{NH}_4^+\) [1] - \(\text{Fe}^{2+}\) [1] (allow 1 mark for both names instead of formulae if correctly given as 'ammonium' and 'iron(II)')
題目 4 · planning
6 分
Hydrogen peroxide, \(\text{H}_2\text{O}_2\), decomposes slowly at room temperature to form water and oxygen gas. This reaction can be catalysed by transition metal oxides.
Plan an investigation to find out which of three transition metal oxides—manganese(IV) oxide, copper(II) oxide, and iron(III) oxide—is the most effective catalyst for the decomposition of hydrogen peroxide.
You are provided with: - the three metal oxides as fine powders - a solution of hydrogen peroxide - common laboratory apparatus.
Your plan should include: - the apparatus you would use - the measurements you would make - the variables you would keep constant and how you would control them - how you would use your results to determine the most effective catalyst.
查看答案詳解收起答案詳解
解題
1. **Apparatus setup:** Set up a conical flask connected via a delivery tube to a gas syringe (or to an inverted measuring cylinder filled with water in a trough). 2. **Reactants and quantities:** Use a measuring cylinder to measure a fixed volume (e.g. \(25\text{ cm}^3\)) of hydrogen peroxide solution and pour it into the conical flask. Use a digital balance to weigh out a specific mass (e.g. \(0.5\text{ g}\)) of the first metal oxide catalyst (manganese(IV) oxide). 3. **Procedure:** Add the weighed metal oxide to the flask, immediately insert the stopper connected to the gas syringe, and start a stopwatch. 4. **Measurements:** Record the volume of oxygen gas collected in the syringe after a fixed time period (e.g., 2 minutes). 5. **Control and repeats:** Repeat the entire experiment under identical conditions (using the exact same volume and concentration of hydrogen peroxide, and the exact same mass and powder particle size of the catalyst) for the copper(II) oxide and the iron(III) oxide. 6. **Conclusion:** Compare the volumes of oxygen collected. The most effective catalyst is the one that produces the largest volume of gas in the set time (indicating the fastest rate of reaction).
評分準則
Award up to 6 marks based on the following points:
- **MP1 (Apparatus):** Mention of suitable apparatus to measure gas volume (e.g., gas syringe or inverted measuring cylinder over water) connected to a reaction vessel (e.g., conical flask or test-tube). - **MP2 (Method - Quantities):** Measure a specific volume of hydrogen peroxide solution (using a measuring cylinder or pipette) AND a specific mass of the metal oxide catalyst (using a balance). - **MP3 (Measurements):** Measure the volume of gas collected in a specified time interval OR measure the time taken to collect a specified volume of gas (using a stopwatch). - **MP4 (Comparison / Fair test):** Repeat the exact procedure for the other two metal oxides (copper(II) oxide and iron(III) oxide). - **MP5 (Control Variables):** State at least two variables that must be kept constant (e.g., concentration of hydrogen peroxide solution, initial temperature of the solution, or the surface area / particle size of the catalysts). - **MP6 (Conclusion):** Explain how the results identify the most effective catalyst (e.g., the catalyst that yields the greatest volume of oxygen in the specified time, or takes the shortest time to collect a fixed volume, is the most effective).
題目 5 · planning
6 分
Hydrogen peroxide, \(\text{H}_2\text{O}_2\), decomposes slowly at room temperature to form water and oxygen gas. This reaction can be catalysed by transition metal oxides.
Plan an investigation to find out which of three transition metal oxides—manganese(IV) oxide, copper(II) oxide, and iron(III) oxide—is the most effective catalyst for the decomposition of hydrogen peroxide.
You are provided with: - the three metal oxides as fine powders - a solution of hydrogen peroxide - common laboratory apparatus.
Your plan should include: - the apparatus you would use - the measurements you would make - the variables you would keep constant and how you would control them - how you would use your results to determine the most effective catalyst.
查看答案詳解收起答案詳解
解題
1. **Apparatus setup:** Set up a conical flask connected via a delivery tube to a gas syringe (or to an inverted measuring cylinder filled with water in a trough). 2. **Reactants and quantities:** Use a measuring cylinder to measure a fixed volume (e.g. \(25\text{ cm}^3\)) of hydrogen peroxide solution and pour it into the conical flask. Use a digital balance to weigh out a specific mass (e.g. \(0.5\text{ g}\)) of the first metal oxide catalyst (manganese(IV) oxide). 3. **Procedure:** Add the weighed metal oxide to the flask, immediately insert the stopper connected to the gas syringe, and start a stopwatch. 4. **Measurements:** Record the volume of oxygen gas collected in the syringe after a fixed time period (e.g., 2 minutes). 5. **Control and repeats:** Repeat the entire experiment under identical conditions (using the exact same volume and concentration of hydrogen peroxide, and the exact same mass and powder particle size of the catalyst) for the copper(II) oxide and the iron(III) oxide. 6. **Conclusion:** Compare the volumes of oxygen collected. The most effective catalyst is the one that produces the largest volume of gas in the set time (indicating the fastest rate of reaction).
評分準則
Award up to 6 marks based on the following points:
- **MP1 (Apparatus):** Mention of suitable apparatus to measure gas volume (e.g., gas syringe or inverted measuring cylinder over water) connected to a reaction vessel (e.g., conical flask or test-tube). - **MP2 (Method - Quantities):** Measure a specific volume of hydrogen peroxide solution (using a measuring cylinder or pipette) AND a specific mass of the metal oxide catalyst (using a balance). - **MP3 (Measurements):** Measure the volume of gas collected in a specified time interval OR measure the time taken to collect a specified volume of gas (using a stopwatch). - **MP4 (Comparison / Fair test):** Repeat the exact procedure for the other two metal oxides (copper(II) oxide and iron(III) oxide). - **MP5 (Control Variables):** State at least two variables that must be kept constant (e.g., concentration of hydrogen peroxide solution, initial temperature of the solution, or the surface area / particle size of the catalysts). - **MP6 (Conclusion):** Explain how the results identify the most effective catalyst (e.g., the catalyst that yields the greatest volume of oxygen in the specified time, or takes the shortest time to collect a fixed volume, is the most effective).
想知道自己有幾分把握？
Thinka 是 DSE 學生用的 AI 練習應用程式，有無限量練習題、即時自動批改和詳細解題步驟。逾 100,000 名學生用它確認自己真的識，而不只是「以為識」。