An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V2) Cambridge IGCSE Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.
Paper 22 選擇題 (Extended)
There are forty questions on this paper. Answer all questions. For each question, choose the correct option from the four options A, B, C, and D.
40 题目 · 40 分
题目 1 · 選擇題
1 分
A mixture of dyes is analysed using paper chromatography. The solvent front travels a distance of $16.0\text{ cm}$ from the baseline. A spot of yellow dye, Y, travels $10.4\text{ cm}$ from the baseline. What is the $R_{\text{f}}$ value of dye Y?
A.0.54
B.0.60
C.0.65
D.0.81
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解题
The $R_{\text{f}}$ value is calculated using the formula: $R_{\text{f}} = \frac{\text{distance travelled by the substance}}{\text{distance travelled by the solvent front}}$. Substituting the given values: $R_{\text{f}} = \frac{10.4\text{ cm}}{16.0\text{ cm}} = 0.65$.
评分标准
1 mark: Correct calculation of 0.65 (Option C).
题目 2 · 選擇題
1 分
What is the total number of atoms present in $4.4\text{ g}$ of carbon dioxide, $\text{CO}_2$? [Relative atomic masses: $A_{\text{r}}(\text{C}) = 12$, $A_{\text{r}}(\text{O}) = 16$. Avogadro constant, $L = 6.0 \times 10^{23}\text{/mol}$]
A.$6.0 \times 10^{22}$
B.$1.2 \times 10^{23}$
C.$1.8 \times 10^{23}$
D.$1.8 \times 10^{24}$
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解题
1. Calculate the relative formula mass ($M_{\text{r}}$) of $\text{CO}_2$: $12 + 2(16) = 44\text{ g/mol}$. 2. Calculate the moles of $\text{CO}_2$: $\frac{4.4\text{ g}}{44\text{ g/mol}} = 0.1\text{ mol}$. 3. Calculate the number of molecules of $\text{CO}_2$: $0.1 \times 6.0 \times 10^{23} = 6.0 \times 10^{22}$ molecules. 4. Calculate the total number of atoms: Each molecule of $\text{CO}_2$ contains 3 atoms (1 carbon atom and 2 oxygen atoms). Therefore, total atoms = $3 \times 6.0 \times 10^{22} = 1.8 \times 10^{23}$ atoms.
评分标准
1 mark: Correctly identifies the number of atoms as $1.8 \times 10^{23}$ (Option C).
题目 3 · 選擇題
1 分
Concentrated aqueous sodium chloride is electrolysed using platinum electrodes. What are the products at the electrodes and the pH change of the remaining electrolyte?
During the electrolysis of concentrated aqueous sodium chloride: - At the anode, chloride ions ($\text{Cl}^-$) are selectively discharged because they are in high concentration, producing chlorine gas. - At the cathode, hydrogen ions ($\text{H}^+$) are selectively discharged because hydrogen is lower in the reactivity series than sodium, producing hydrogen gas. - Sodium ions ($\text{Na}^+$) and hydroxide ions ($\text{OH}^-$) remain in solution, forming sodium hydroxide, which is alkaline. This causes the pH of the remaining electrolyte to increase.
评分标准
1 mark: Correctly identifies products as chlorine at the anode, hydrogen at the cathode, and an increase in pH (Option A).
题目 4 · 選擇題
1 分
The equation for the reaction of ethene with hydrogen is shown: $\text{C}_2\text{H}_4(\text{g}) + \text{H}_2(\text{g}) \rightarrow \text{C}_2\text{H}_6(\text{g})$. The bond energies are: $\text{C}=\text{C}$: $610\text{ kJ/mol}$; $\text{C}-\text{C}$: $348\text{ kJ/mol}$; $\text{H}-\text{H}$: $436\text{ kJ/mol}$; $\text{C}-\text{H}$: $412\text{ kJ/mol}$. What is the energy change for this reaction?
A.$-126\text{ kJ/mol}$
B.$-326\text{ kJ/mol}$
C.$+126\text{ kJ/mol}$
D.$-1172\text{ kJ/mol}$
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解题
Energy change = (energy needed to break bonds) - (energy released when bonds are formed). Bonds broken: $1 \times \text{C}=\text{C} = 610\text{ kJ/mol}$, $4 \times \text{C}-\text{H} = 4 \times 412 = 1648\text{ kJ/mol}$, $1 \times \text{H}-\text{H} = 436\text{ kJ/mol}$. Total energy to break bonds = $610 + 1648 + 436 = 2694\text{ kJ/mol}$. Bonds formed: $1 \times \text{C}-\text{C} = 348\text{ kJ/mol}$, $6 \times \text{C}-\text{H} = 6 \times 412 = 2472\text{ kJ/mol}$. Total energy released = $348 + 2472 = 2820\text{ kJ/mol}$. Energy change = $2694 - 2820 = -126\text{ kJ/mol}$.
评分标准
1 mark: Correctly calculates the energy change of $-126\text{ kJ/mol}$ (Option A).
题目 5 · 選擇題
1 分
An excess of marble chips (calcium carbonate) is reacted with $50\text{ cm}^3$ of $1.0\text{ mol/dm}^3$ hydrochloric acid. Which change will increase both the initial rate of reaction and the total volume of carbon dioxide gas collected?
A.using $50\text{ cm}^3$ of $2.0\text{ mol/dm}^3$ hydrochloric acid at the same temperature
B.powdering the marble chips while keeping the acid volume and concentration the same
C.using $100\text{ cm}^3$ of $0.5\text{ mol/dm}^3$ hydrochloric acid at the same temperature
D.increasing the temperature of the mixture while keeping the acid volume and concentration the same
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解题
1. Initial rate of reaction: Increasing the concentration of hydrochloric acid increases the number of acid particles per unit volume, which increases the collision frequency and thus the initial rate of reaction. 2. Total volume of gas produced: Since calcium carbonate is in excess, the hydrochloric acid is the limiting reactant. The moles of acid initially present are $50\text{ cm}^3 \times 1.0\text{ mol/dm}^3 = 0.05\text{ mol}$. Using $50\text{ cm}^3$ of $2.0\text{ mol/dm}^3$ hydrochloric acid increases the moles of acid to $0.10\text{ mol}$, which produces double the amount of carbon dioxide gas.
评分标准
1 mark: Correctly identifies Option A as increasing both the initial rate and the total volume of gas.
题目 6 · 選擇題
1 分
Two separate aqueous solutions, P and Q, both have a concentration of $0.1\text{ mol/dm}^3$. Solution P is hydrochloric acid (a strong acid) and solution Q is ethanoic acid (a weak acid). Which statement comparing P and Q is correct?
A.Solution P has a higher pH than solution Q.
B.Solution P is a better electrical conductor than solution Q.
C.Solution Q reacts more rapidly with magnesium ribbon than solution P.
D.Solution Q requires a larger volume of $0.1\text{ mol/dm}^3$ sodium hydroxide for complete neutralisation than solution P.
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解题
Hydrochloric acid (P) is a strong acid and fully dissociates in water, producing a high concentration of $\text{H}^+$ ions. Ethanoic acid (Q) is a weak acid and only partially dissociates, producing a low concentration of $\text{H}^+$ ions. Because solution P contains a higher concentration of free, mobile ions, it conducts electricity much better than solution Q. P has a lower pH (more acidic) and reacts faster with magnesium than Q. Both require the exact same volume of sodium hydroxide for neutralisation because they contain the same total moles of acidic protons per decimetre cubed.
评分标准
1 mark: Correctly identifies that the strong acid P is a better electrical conductor (Option B).
题目 7 · 選擇題
1 分
Three metals, D, E, and F, are added to separate aqueous solutions of their nitrates. Metal D displaces F but has no reaction with E. Metal E displaces both D and F. Metal F has no reaction with either D or E. What is the order of reactivity of the metals, from most reactive to least reactive?
A.E $\rightarrow$ D $\rightarrow$ F
B.F $\rightarrow$ D $\rightarrow$ E
C.E $\rightarrow$ F $\rightarrow$ D
D.D $\rightarrow$ E $\rightarrow$ F
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解题
Metal E can displace both D and F from their solutions, so E is more reactive than both D and F (E > D, E > F). Metal D can displace F, but not E, so D is more reactive than F but less reactive than E (E > D > F). Metal F cannot displace either D or E, making it the least reactive. Thus, the reactivity order from most to least reactive is E $\rightarrow$ D $\rightarrow$ F.
评分标准
1 mark: Correctly identifies the order as E $\rightarrow$ D $\rightarrow$ F (Option A).
题目 8 · 選擇題
1 分
A synthetic polymer has the repeating unit shown: $-[\text{O}-\text{CH}_2-\text{CH}_2-\text{O}-\text{CO}-\text{C}_6\text{H}_4-\text{CO}]_n-$. Which statement about this polymer is correct?
A.It is a polyamide formed by addition polymerisation.
B.It is a polyamide formed by condensation polymerisation.
C.It is a polyester formed by addition polymerisation.
D.It is a polyester formed by condensation polymerisation.
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解题
The repeating unit contains the ester linkage ($-\text{O}-\text{CO}-$), which classifies the polymer as a polyester (specifically Terylene). Polyesters are formed by condensation polymerisation, during which monomers join together with the elimination of small molecules such as water.
评分标准
1 mark: Correctly identifies the polymer as a polyester formed by condensation polymerisation (Option D).
题目 9 · 選擇題
1 分
Hydrogen gas reacts with chlorine gas according to the equation:
| Bond | Bond energy / kJ/mol | | :--- | :---: | | H–H | 436 | | Cl–Cl | 242 | | H–Cl | 431 |
What is the overall energy change for this reaction?
A.-184 kJ/mol
B.+184 kJ/mol
C.-247 kJ/mol
D.+247 kJ/mol
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解题
To calculate the overall energy change: 1. Energy needed to break reactant bonds: \[ \text{H–H} + \text{Cl–Cl} = 436\text{ kJ/mol} + 242\text{ kJ/mol} = 678\text{ kJ/mol} \] 2. Energy released when product bonds are formed: \[ 2 \times \text{H–Cl} = 2 \times 431\text{ kJ/mol} = 862\text{ kJ/mol} \] 3. Overall energy change: \[ \text{Energy change} = \text{Energy in} - \text{Energy out} = 678\text{ kJ/mol} - 862\text{ kJ/mol} = -184\text{ kJ/mol} \] Therefore, the correct option is A.
评分标准
1 mark for the correct option A.
题目 10 · 選擇題
1 分
A paper chromatography experiment is carried out on a mixture of colorless amino acids. A locating agent is sprayed on the paper to reveal the spots.
If the distance moved by the solvent front is 12.0 cm, and one of the revealed spots has an \( R_{\text{f}} \) value of 0.65, how far from the baseline did this spot travel?
A.4.2 cm
B.7.8 cm
C.11.35 cm
D.18.5 cm
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解题
Using the formula: \[ R_{\text{f}} = \frac{\text{distance moved by the spot}}{\text{distance moved by the solvent front}} \] Rearranging to find the distance moved by the spot: \[ \text{distance moved by the spot} = R_{\text{f}} \times \text{distance moved by the solvent front} \] \[ \text{distance moved by the spot} = 0.65 \times 12.0\text{ cm} = 7.8\text{ cm} \] Therefore, the spot travelled 7.8 cm from the baseline.
评分标准
1 mark for the correct option B.
题目 11 · 選擇題
1 分
What is the maximum volume of carbon dioxide gas, measured at room temperature and pressure (r.t.p.), produced when 5.0 g of calcium carbonate, \( \text{CaCO}_3 \), reacts completely with excess dilute hydrochloric acid?
During the electrolysis of concentrated aqueous sodium chloride (brine): - At the anode (positive electrode), chloride ions (\( \text{Cl}^- \)) are discharged in preference to hydroxide ions because of their high concentration, forming chlorine gas. - At the cathode (negative electrode), hydrogen ions (\( \text{H}^+ \)) are discharged in preference to sodium ions (\( \text{Na}^+ \)) because hydrogen is lower in the reactivity series, forming hydrogen gas. - Hydroxide ions (\( \text{OH}^- \)) and sodium ions (\( \text{Na}^+ \)) remain in the solution, forming sodium hydroxide (\( \text{NaOH} \)), which is alkaline. This causes the pH of the remaining solution to increase.
评分标准
1 mark for the correct option A.
题目 13 · 選擇題
1 分
The reaction between iron and dilute sulfuric acid is investigated at different temperatures.
Which statement correctly explains why increasing the temperature increases the rate of this reaction?
A.The activation energy of the reaction is lowered, making it easier for more collisions to be successful.
B.The particles move faster and collide more frequently, and a higher proportion of collisions have energy greater than or equal to the activation energy.
C.The concentration of acid particles increases, leading to a higher frequency of successful collisions.
D.The reacting particles decrease in physical size, allowing them to collide and react more rapidly.
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解题
Increasing the temperature increases the kinetic energy of the reacting particles. As a result: 1. The particles move faster and collide more frequently (increasing collision frequency). 2. A much higher proportion of colliding particles possess energy equal to or greater than the activation energy (making more collisions successful).
Option A is incorrect because temperature does not lower the activation energy (only a catalyst does). Option C is incorrect because temperature does not change the concentration of the acid. Option D is incorrect because the physical size of the iron particles is unchanged.
评分标准
1 mark for the correct option B.
题目 14 · 選擇題
1 分
An organic compound has the molecular formula \( \text{C}_3\text{H}_6\text{O}_2 \). It reacts with sodium carbonate to produce carbon dioxide gas.
What is the name and functional group of this compound?
A.ethyl methanoate, ester
B.propan-1-ol, alcohol
C.propanoic acid, carboxylic acid
D.propanone, ketone
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解题
The reaction of an organic compound with a carbonate to produce carbon dioxide gas indicates that the compound is an acid, specifically a carboxylic acid. With three carbon atoms, the carboxylic acid is propanoic acid. The functional group of carboxylic acids is the carboxyl group (\( -\text{COOH} \)).
评分标准
1 mark for the correct option C.
题目 15 · 選擇題
1 分
Three metals, P, Q, and R, are reacted with aqueous solutions of their salts. The results are shown in the table.
| Reactants | Result | | :--- | :--- | | P + salt of Q | no reaction | | Q + salt of R | reaction occurs | | R + salt of P | reaction occurs |
What is the correct order of reactivity of the metals, starting with the most reactive?
A.P, R, Q
B.Q, P, R
C.Q, R, P
D.R, Q, P
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解题
From the results: - 'P + salt of Q' gives 'no reaction': this means P cannot displace Q, so Q is more reactive than P (Q > P). - 'Q + salt of R' gives 'reaction occurs': this means Q can displace R, so Q is more reactive than R (Q > R). - 'R + salt of P' gives 'reaction occurs': this means R can displace P, so R is more reactive than P (R > P).
Combining these relationships yields: Q > R > P. Therefore, the order starting with the most reactive is Q, R, P.
评分标准
1 mark for the correct option C.
题目 16 · 選擇題
1 分
Many modern petrol cars are fitted with catalytic converters in their exhaust systems to minimize harmful emissions.
Which pollutant gas is chemically reduced to nitrogen gas, \( \text{N}_2 \), inside a catalytic converter?
A.carbon monoxide
B.carbon dioxide
C.nitrogen dioxide
D.sulfur dioxide
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解题
A catalytic converter facilitates the reduction of nitrogen oxides (such as nitrogen dioxide, \( \text{NO}_2 \), or nitrogen monoxide, \( \text{NO} \)) to harmless nitrogen gas, \( \text{N}_2 \). Carbon monoxide (\( \text{CO} \)) is oxidized to carbon dioxide (\( \text{CO}_2 \)). Therefore, the pollutant gas that is reduced to nitrogen gas is nitrogen dioxide (or other oxides of nitrogen).
评分标准
1 mark for the correct option C.
题目 17 · 選擇題
1 分
Which statement about the particles of a solid substance during melting is correct?
A.The average kinetic energy of the particles increases.
B.The energy absorbed is used to overcome the forces of attraction holding the particles in fixed positions.
C.The particles begin to vibrate faster about their fixed positions.
D.The particles lose energy to the surroundings.
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解题
During melting, temperature remains constant. Therefore, the average kinetic energy of the particles does not change. The heat energy absorbed is used to overcome the attractive forces (or regular lattice arrangement) holding the solid particles in fixed positions.
评分标准
Award 1 mark for the correct answer (B). Reject all other options.
题目 18 · 選擇題
1 分
A chromatogram of a colorless mixture of compounds is developed. Why is a locating agent sprayed onto the chromatogram and how is the \(R_f\) value calculated?
A.to dissolve the spots; \(R_f = \frac{\text{distance travelled by solvent}}{\text{distance travelled by substance}}\)
B.to make the spots visible; \(R_f = \frac{\text{distance travelled by substance}}{\text{distance travelled by solvent}}\)
C.to prevent the spots from evaporating; \(R_f = \frac{\text{distance travelled by substance}}{\text{distance travelled by solvent}}\)
D.to react with the mobile phase; \(R_f = \frac{\text{distance travelled by solvent}}{\text{distance travelled by substance}}\)
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解题
Locating agents react with colorless compounds (such as amino acids) to produce colored spots, making them visible. The retardation factor, \(R_f\), is calculated as the distance travelled by the substance divided by the distance travelled by the solvent front.
评分标准
Award 1 mark for the correct answer (B). Reject all other options.
题目 19 · 選擇題
1 分
A sample of element X contains two isotopes, \(^{10}\text{X}\) and \(^{11}\text{X}\). The percentage abundance of \(^{10}\text{X}\) is \(20.0\%\) and the percentage abundance of \(^{11}\text{X}\) is \(80.0\%\). What is the relative atomic mass, \(A_r\), of this sample of element X?
A.10.20
B.10.50
C.10.80
D.11.00
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解题
The relative atomic mass is the weighted average: \(A_r = \frac{(10 \times 20.0) + (11 \times 80.0)}{100} = \frac{200 + 880}{100} = 10.80\).
评分标准
Award 1 mark for the correct answer (C). Reject all other options.
题目 20 · 選擇題
1 分
A student performs a titration. They need to transfer exactly \(25.0\text{ cm}^3\) of sodium hydroxide solution into a conical flask, and then add dilute hydrochloric acid from a second piece of apparatus until the end-point is reached. Which row describes the most suitable apparatus for measuring these volumes?
A.\(25.0\text{ cm}^3\) of sodium hydroxide: volumetric pipette; volume of hydrochloric acid added: burette
B.\(25.0\text{ cm}^3\) of sodium hydroxide: measuring cylinder; volume of hydrochloric acid added: volumetric pipette
C.\(25.0\text{ cm}^3\) of sodium hydroxide: burette; volume of hydrochloric acid added: measuring cylinder
D.\(25.0\text{ cm}^3\) of sodium hydroxide: volumetric pipette; volume of hydrochloric acid added: measuring cylinder
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解题
A volumetric pipette is used to deliver a fixed volume (exactly \(25.0\text{ cm}^3\)) of alkali solution to the conical flask. A burette is used to deliver and measure the variable volume of acid required to reach the end-point.
评分标准
Award 1 mark for the correct answer (A). Reject all other options.
题目 21 · 選擇題
1 分
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. Which products are formed at each electrode and what happens to the pH of the remaining electrolyte?
During the electrolysis of concentrated aqueous \(\text{NaCl}\), chloride ions are preferentially discharged at the anode (forming chlorine gas, \(\text{Cl}_2\)) and hydrogen ions are preferentially discharged at the cathode (forming hydrogen gas, \(\text{H}_2\)). The remaining ions in solution are \(\text{Na}^+\) and \(\text{OH}^-\), which form sodium hydroxide (an alkali), causing the pH of the electrolyte to increase.
评分标准
Award 1 mark for the correct answer (B). Reject all other options.
题目 22 · 選擇題
1 分
For an endothermic chemical reaction, which statement is correct?
A.The products have more energy than the reactants, and \(\Delta H\) is positive.
B.The products have more energy than the reactants, and \(\Delta H\) is negative.
C.The reactants have more energy than the products, and \(\Delta H\) is positive.
D.The reactants have more energy than the products, and \(\Delta H\) is negative.
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解题
In an endothermic reaction, heat energy is taken in from the surroundings. Thus, the products have more chemical energy than the reactants, resulting in a positive enthalpy change (\(\Delta H > 0\)).
评分标准
Award 1 mark for the correct answer (A). Reject all other options.
题目 23 · 選擇題
1 分
The reaction between excess solid magnesium carbonate and dilute nitric acid is investigated. Which changes both increase the initial rate of the reaction?
A.increase the temperature and use larger pieces of magnesium carbonate
B.decrease the concentration of nitric acid and add a catalyst
C.increase the concentration of nitric acid and use smaller pieces of magnesium carbonate
D.decrease the temperature and increase the volume of nitric acid used
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解题
Increasing the concentration of nitric acid increases the number of acid particles per unit volume, leading to a higher frequency of successful collisions. Using smaller pieces of solid magnesium carbonate increases its surface area, which also increases the frequency of collisions.
评分标准
Award 1 mark for the correct answer (C). Reject all other options.
题目 24 · 選擇題
1 分
The results of three displacement experiments involving metals J, K, and L, and their oxides are shown: 1. Metal J reacts with the oxide of K. 2. Metal L does not react with the oxide of K. 3. Metal K reacts with the oxide of L. What is the order of reactivity of the metals, starting with the most reactive?
A.J, K, L
B.J, L, K
C.K, J, L
D.L, K, J
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解题
From experiment 1, J displaces K, so J is more reactive than K (J > K). From experiment 2, L cannot displace K, so K is more reactive than L (K > L). This is confirmed by experiment 3, where K displaces L. Therefore, the order of reactivity is J > K > L.
评分标准
Award 1 mark for the correct answer (A). Reject all other options.
题目 25 · 選擇題
1 分
Two gas jars, one containing gas X and the other containing gas Y, are separated by a porous barrier. The relative molecular mass of X is 17 and that of Y is 44.
Which statement about the diffusion of these gases is correct?
A.Gas Y diffuses faster than gas X because Y has a higher relative molecular mass.
B.Gas X diffuses faster than gas Y because X has a lower relative molecular mass.
C.Both gases diffuse at the same rate because diffusion is independent of molecular mass.
D.Gas Y diffuses faster than gas X because heavier particles travel with greater kinetic energy at the same temperature.
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解题
According to the kinetic particle theory, gases with a lower relative molecular mass ($M_r$) diffuse faster than gases with a higher relative molecular mass at the same temperature. Since gas X has an $M_r$ of 17 (which is lower than gas Y with an $M_r$ of 44), gas X will diffuse faster.
评分标准
1 mark for selecting option B.
题目 26 · 選擇題
1 分
The symbol for a phosphide ion is $_{15}^{31}\text{P}^{3-}$.
What is the number of protons, neutrons and electrons in this ion?
A.15 protons, 16 neutrons, 12 electrons
B.15 protons, 16 neutrons, 18 electrons
C.15 protons, 31 neutrons, 18 electrons
D.16 protons, 15 neutrons, 18 electrons
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解题
The atomic number (bottom number) is 15, which represents the number of protons. The mass number (top number) is 31. The number of neutrons is mass number minus atomic number: $31 - 15 = 16$. The charge is $3-$, meaning the phosphorus atom has gained 3 electrons. The number of electrons is $15 + 3 = 18$.
评分标准
1 mark for selecting option B.
题目 27 · 選擇題
1 分
What is the total number of shared pairs of electrons (bonding pairs) in one molecule of ethene, $\text{C}_2\text{H}_4$?
A.4
B.5
C.6
D.8
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解题
Ethene ($\text{CH}_2=\text{CH}_2$) contains: - Four single $\text{C}-\text{H}$ covalent bonds (4 shared pairs) - One double $\text{C}=\text{C}$ covalent bond (2 shared pairs)
Adding these together gives $4 + 2 = 6$ shared pairs of electrons.
评分标准
1 mark for selecting option C.
题目 28 · 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 change in pH of the electrolyte?
C.Anode: chlorine; Cathode: sodium; pH: stays the same
D.Anode: oxygen; Cathode: sodium; pH: increases
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解题
During the electrolysis of concentrated aqueous sodium chloride (brine): - Chloride ions ($\text{Cl}^-$) are discharged at the positive electrode (anode) to produce chlorine gas ($\text{Cl}_2$). - Hydrogen ions ($\text{H}^+$) are discharged at the negative electrode (cathode) to produce hydrogen gas ($\text{H}_2$). - Sodium ions ($\text{Na}^+$) and hydroxide ions ($\text{OH}^-$) remain in the solution, forming sodium hydroxide, which is alkaline, so the pH increases.
评分标准
1 mark for selecting option B.
题目 29 · 選擇題
1 分
The reaction between nitrogen and hydrogen to form ammonia is shown. $$\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}$$
The bond energies are: - $\text{N}\equiv\text{N}$: $945\text{ kJ/mol}$ - $\text{H}-\text{H}$: $436\text{ kJ/mol}$ - $\text{N}-\text{H}$: $391\text{ kJ/mol}$
What is the energy change, $\Delta H$, for this reaction?
A.$-93\text{ kJ/mol}$
B.$+93\text{ kJ/mol}$
C.$-1092\text{ kJ/mol}$
D.$+1092\text{ kJ/mol}$
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解题
Energy required to break bonds (reactants): $1 \times (\text{N}\equiv\text{N}) + 3 \times (\text{H}-\text{H}) = 945 + 3(436) = 2253\text{ kJ/mol}$
Energy released in making bonds (products): $6 \times (\text{N}-\text{H}) = 6(391) = 2346\text{ kJ/mol}$
Which statement explains why increasing the temperature increases the rate of a chemical reaction?
A.The activation energy of the reaction is lowered.
B.The concentration of the reacting particles is increased.
C.The particles have more energy, so a greater proportion of collisions have energy greater than the activation energy.
D.The particles are closer together, so they collide more frequently.
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解题
Increasing the temperature increases the kinetic energy of the reacting particles. As a result, the particles move faster, colliding more frequently, and a much greater proportion of these collisions have energy greater than the activation energy required for a successful reaction to occur.
评分标准
1 mark for selecting option C.
题目 31 · 選擇題
1 分
Which substance, when added in excess to dilute sulfuric acid, does NOT produce a gas?
A.copper(II) carbonate
B.magnesium ribbon
C.sodium hydroxide solution
D.zinc metal
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解题
Option A: Copper(II) carbonate reacts with sulfuric acid to produce carbon dioxide gas ($\text{CO}_2$). Option B: Magnesium reacts with sulfuric acid to produce hydrogen gas ($\text{H}_2$). Option C: Sodium hydroxide is an alkali. Its reaction with sulfuric acid is a neutralisation reaction producing only a soluble salt (sodium sulfate) and water ($\text{H}_2\text{O}$); no gas is produced. Option D: Zinc reacts with sulfuric acid to produce hydrogen gas ($\text{H}_2$).
评分标准
1 mark for selecting option C.
题目 32 · 選擇題
1 分
An organic compound has the structural formula $\text{CH}_3\text{CH}_2\text{COOCH}_3$.
To which homologous series does this compound belong and what is its name?
A.carboxylic acid, butanoic acid
B.ester, methyl propanoate
C.ester, ethyl ethanoate
D.alcohol, butanol
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解题
The compound contains the ester functional group ($-\text{COO}-$). The acid part ($\text{CH}_3\text{CH}_2\text{COO}-$) has three carbons, so it is derived from propanoic acid (propanoate). The alcohol part ($-\text{CH}_3$) is derived from methanol (methyl). Therefore, the ester is named methyl propanoate.
评分标准
1 mark for selecting option B.
题目 33 · 選擇題
1 分
Why is a locating agent used in the paper chromatography of colorless amino acids?
A.To increase the solubility of the amino acids in the mobile phase.
B.To make the colorless spots visible on the chromatogram.
C.To prevent the solvent front from evaporating during the run.
D.To decrease the retention factor \(R_f\) value of the amino acids.
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解题
Amino acids are colorless and cannot be seen on the chromatogram directly. A locating agent is sprayed onto the paper to react with the amino acids, forming colored spots that are visible.
评分标准
1 mark for option B.
题目 34 · 選擇題
1 分
Hydrogen reacts with bromine to form hydrogen bromide as shown: \(\text{H}_2\text{(g)} + \text{Br}_2\text{(g)} \rightarrow 2\text{HBr(g)}\). The bond energies are: \(\text{H–H}\) is \(436\text{ kJ/mol}\), \(\text{Br–Br}\) is \(193\text{ kJ/mol}\), and \(\text{H–Br}\) is \(366\text{ kJ/mol}\). What is the energy change for this reaction?
A.\(-103\text{ kJ/mol}\)
B.\(-263\text{ kJ/mol}\)
C.\(+103\text{ kJ/mol}\)
D.\(+263\text{ kJ/mol}\)
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解题
Energy to break bonds = \(436 + 193 = 629\text{ kJ/mol}\). Energy released in forming bonds = \(2 \times 366 = 732\text{ kJ/mol}\). Energy change = \(629 - 732 = -103\text{ kJ/mol}\).
评分标准
1 mark for option A.
题目 35 · 選擇題
1 分
A student wishes to prepare a pure sample of sodium sulfate crystals by titrating dilute sulfuric acid with aqueous sodium hydroxide. Which piece of apparatus is most suitable for adding exactly \(25.0\text{ cm}^3\) of the aqueous sodium hydroxide to the conical flask?
A.A \(50\text{ cm}^3\) burette.
B.A \(25\text{ cm}^3\) volumetric pipette.
C.A \(50\text{ cm}^3\) measuring cylinder.
D.A \(25\text{ cm}^3\) dropper.
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解题
A volumetric pipette is specifically designed to measure and deliver a single fixed volume, such as exactly \(25.0\text{ cm}^3\), with very high precision.
评分标准
1 mark for option B.
题目 36 · 選擇題
1 分
A student investigates the reaction between marble chips (calcium carbonate) and dilute hydrochloric acid: \(\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}\). Which change increases the rate of reaction by increasing both the collision frequency and the proportion of successful collisions?
A.Increasing the concentration of hydrochloric acid.
B.Increasing the temperature of the reaction.
C.Decreasing the size of the marble chips.
D.Increasing the pressure of the system.
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解题
Increasing the temperature increases the kinetic energy of the particles, which increases the frequency of collisions. It also increases the proportion of particles with energy greater than or equal to the activation energy, thus increasing the proportion of successful collisions.
评分标准
1 mark for option B.
题目 37 · 選擇題
1 分
What is the total number of ions present in \(0.25\text{ mol}\) of magnesium chloride, \(\text{MgCl}_2\)? (Take the Avogadro constant as \(6.02 \times 10^{23}\text{ /mol}\).)
A.\(1.51 \times 10^{23}\)
B.\(3.01 \times 10^{23}\)
C.\(4.52 \times 10^{23}\)
D.\(7.53 \times 10^{23}\)
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解题
One formula unit of \(\text{MgCl}_2\) contains 3 ions (one \(\text{Mg}^{2+}\) ion and two \(\text{Cl}^-\) ions). Therefore, \(0.25\text{ mol}\) of \(\text{MgCl}_2\) contains \(0.25 \times 3 = 0.75\text{ mol}\) of ions. Number of ions = \(0.75 \times 6.02 \times 10^{23} = 4.52 \times 10^{23}\).
评分标准
1 mark for option C.
题目 38 · 選擇題
1 分
An element \(M\) forms a stable nitride with the formula \(M_3\text{N}_2\). What is the electronic configuration of an atom of \(M\) if it is in Period 3 of the Periodic Table?
A.2, 2
B.2, 8, 2
C.2, 8, 3
D.2, 8, 6
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解题
The nitride ion has a charge of \(3-\), so two nitride ions have a total charge of \(6-\). Thus, three ions of \(M\) must have a total charge of \(6+\), meaning each ion of \(M\) is \(M^{2+}\). Since \(M\) forms a \(2+\) ion, it is in Group II. An element in Period 3 and Group II has 3 occupied shells and 2 outer electrons, which corresponds to the electronic configuration 2, 8, 2 (magnesium).
评分标准
1 mark for option B.
题目 39 · 選擇題
1 分
Propene reacts with steam in the presence of an acid catalyst to produce an alcohol. Which statement about this reaction is correct?
A.The reaction is a substitution reaction.
B.The product formed has the molecular formula \(\text{C}_3\text{H}_8\text{O}\).
C.The product formed is a saturated hydrocarbon.
D.The reaction occurs at room temperature and pressure.
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解题
Propene (\(\text{C}_3\text{H}_6\)) reacts with steam (\(\text{H}_2\text{O}\)) in an addition reaction to produce propanol, which has the molecular formula \(\text{C}_3\text{H}_8\text{O}\). This reaction is an addition reaction, produces a saturated alcohol (not a hydrocarbon), and requires high temperature (\(300^\circ\text{C}\)) and pressure (\(60\text{ atm}\)).
评分标准
1 mark for option B.
题目 40 · 選擇題
1 分
Aqueous copper(II) sulfate is electrolysed using inert carbon electrodes. Which row correctly identifies the products formed at each electrode?
At the cathode, \(\text{Cu}^{2+}\) ions are discharged in preference to \(\text{H}^+\) because copper is lower in the reactivity series, forming copper metal. At the anode, \(\text{OH}^-\) ions are discharged in preference to \(\text{SO}_4^{2-}\) ions, releasing oxygen gas.
Answer all questions. Write your answers in the spaces provided on the question paper. You may use a calculator.
6 题目 · 80 分
题目 1 · structured
13 分
Sulfur has several isotopes.
(a) Define the term isotopes. [2]
(b) An isotope of sulfur is represented as \(^{34}_{16}S\). Complete the table to show the number of protons, neutrons and electrons in a \(^{34}_{16}S^{2-}\) ion. [3]
(c) A sample of sulfur contains two isotopes, \(^{32}S\) and \(^{34}S\). The relative atomic mass of the sample is 32.1. Calculate the percentage abundance of each isotope in this sample. [3]
(d) Sulfur reacts with fluorine to form sulfur hexafluoride, \(SF_6\).
(i) Describe covalent bonding in terms of electrostatic attraction. [2]
(ii) Draw the dot-and-cross diagram to show the arrangement of outer-shell electrons in a molecule of nitrogen trifluoride, \(NF_3\). [3]
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解题
(a) Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons.
(b) Protons = 16, Neutrons = 18, Electrons = 18.
(c) Let the percentage abundance of \(^{32}S\) be \(x\)% and \(^{34}S\) be \((100 - x)\)%. \(32.1 = \frac{32x + 34(100 - x)}{100}\) \(3210 = 32x + 3400 - 34x\) \(2x = 190\) \(x = 95\) Abundance of \(^{32}S = 95\)%, Abundance of \(^{34}S = 5\)%.
(d) (i) The strong electrostatic attraction between the shared pair of electrons and the nuclei of the bonded atoms.
(ii) Nitrogen is in Group V, so it has 5 outer-shell electrons. Fluorine is in Group VII, so it has 7 outer-shell electrons. Three single covalent bonds are formed (three shared pairs of electrons, one from N and one from each of the three F atoms). One lone pair remains on the nitrogen atom, and six non-bonding outer electrons remain on each fluorine atom.
评分标准
(a) [1] atoms of the same element / same proton number / same atomic number [1] different numbers of neutrons / different nucleon number / different mass number
(d) (i) [1] electrostatic attraction between shared pair(s) of electrons [1] and the positive nuclei of both atoms
(ii) [1] Three shared pairs of electrons (one pair between N and each F) [1] One lone pair of electrons on the nitrogen atom [1] Six non-bonding outer electrons on each of the three fluorine atoms
题目 2 · structured
13 分
Titanium is a transition element. It is extracted from its ore, rutile (impure titanium(IV) oxide, \(TiO_2\)).
(a) State two physical properties of titanium that are characteristic of transition metals but not Group I metals. [2]
(b) The extraction of titanium involves two main stages.
Stage 1: Titanium(IV) oxide is reacted with chlorine and coke (carbon) to produce titanium(IV) chloride, \(TiCl_4\), and carbon monoxide. Write the balanced symbol equation for this reaction. [2]
Stage 2: Titanium(IV) chloride is reduced by molten magnesium at \(900\text{ }^\circ\text{C}\) in an atmosphere of argon. \(TiCl_4 + 2Mg \to Ti + 2MgCl_2\)
(i) Explain, in terms of the reactivity of metals, why magnesium is used in this reaction. [1]
(ii) Explain why this reaction is carried out in an atmosphere of argon rather than air. [2]
(iii) State, with a reason, which substance is reduced in this reaction. [2]
(c) Recycling titanium is important. Give two reasons why recycling metals is beneficial to the environment. [2]
(d) Titanium alloys are used in aircraft engines. Explain why alloys are harder than pure metals. [2]
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解题
(a) Titanium has high density and a high melting point.
(b)(i) Magnesium is more reactive than titanium and can therefore displace it from titanium(IV) chloride.
(b)(ii) Argon is an inert noble gas. Air contains oxygen and nitrogen, which would react with molten magnesium or titanium at \(900\text{ }^\circ\text{C}\).
(b)(iii) Titanium(IV) chloride (or titanium ions) is reduced because it loses chlorine / gains electrons / the oxidation state of titanium decreases from +4 to 0.
(c) Recycling metals conserves limited mineral resources (ore deposits) and reduces the energy consumption compared to extracting the metal from its ore.
(d) In pure metals, atoms are arranged in regular layers that can easily slide over each other when a force is applied. In alloys, atoms of different sizes disrupt this regular arrangement, making it much more difficult for the layers to slide.
评分标准
(a) [1] high density / denser [1] high melting point / boiling point (Accept: hard / strong)
(b) [1] Correct formulae of reactants and products [1] Correct balancing: \(TiO_2 + 2Cl_2 + 2C \to TiCl_4 + 2CO\)
(b)(i) [1] Magnesium is more reactive than titanium
(b)(ii) [1] Argon is inert / unreactive [1] To prevent titanium/magnesium reacting with oxygen/nitrogen/air
(b)(iii) [1] Titanium(IV) chloride / titanium ions / \(TiCl_4\) [1] Titanium gains electrons / oxidation state of Ti decreases / Ti loses chlorine (Reject: Titanium loses oxygen)
(c) [1] Conserves metal ores [1] Uses less energy / reduces carbon emissions / reduces landfill waste
(d) [1] Atoms of different sizes disrupt regular layers [1] This prevents layers from sliding over each other
题目 3 · structured
14 分
Hydrated nickel(II) sulfate crystals have the formula \(NiSO_4 \cdot xH_2O\). A student heated a sample of these crystals to find the value of \(x\).
(a) Define the term anhydrous. [1]
(b) The student heated a crucible containing hydrated nickel(II) sulfate. \(\bullet\) Mass of empty crucible = \(24.30\text{ g}\) \(\bullet\) Mass of crucible + hydrated nickel(II) sulfate = \(29.56\text{ g}\) \(\bullet\) Mass of crucible + anhydrous nickel(II) sulfate (after heating to constant mass) = \(27.40\text{ g}\)
(i) Explain why the student heated the sample to constant mass. [1]
(ii) Calculate the mass of anhydrous \(NiSO_4\) formed. [1]
(iii) Calculate the mass of water of crystallisation lost. [1]
(iv) Calculate the number of moles of anhydrous \(NiSO_4\) formed. [Relative formula mass, \(M_r\): \(NiSO_4 = 155\)] [2]
(v) Calculate the number of moles of water lost. [Relative formula mass, \(M_r\): \(H_2O = 18\)] [2]
(vi) Deduce the value of \(x\) to the nearest whole number. [2]
(c) In a separate experiment, a hydrocarbon was found to contain \(82.8\%\) carbon and \(17.2\%\) hydrogen by mass.
(i) Determine the empirical formula of this hydrocarbon. [3]
(ii) The relative molecular mass of this hydrocarbon is 58. Deduce its molecular formula. [1]
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解题
(a) Anhydrous means containing no water of crystallisation.
(b)(i) To ensure all the water of crystallisation had been completely removed.
(b)(ii) Mass of anhydrous \(NiSO_4 = 27.40\text{ g} - 24.30\text{ g} = 3.10\text{ g}\).
(b)(iii) Mass of water lost = \(29.56\text{ g} - 27.40\text{ g} = 2.16\text{ g}\).
(b)(iv) Moles of \(NiSO_4 = \frac{3.10}{155} = 0.02\text{ mol}\).
(b)(v) Moles of \(H_2O = \frac{2.16}{18} = 0.12\text{ mol}\).
(b)(vi) Ratio of moles of water to moles of anhydrous salt = \(0.12 / 0.02 = 6\). Thus, \(x = 6\).
(c)(i) Moles of C = \(82.8 / 12 = 6.90\). Moles of H = \(17.2 / 1 = 17.2\). Divide by the smallest (6.90): C = \(6.90 / 6.90 = 1\) H = \(17.2 / 6.90 = 2.49 \approx 2.5\). Multiply by 2 to get whole numbers: C = 2, H = 5. Empirical formula is \(C_2H_5\).
(c)(ii) Empirical formula mass of \(C_2H_5 = (2 \times 12) + (5 \times 1) = 29\). Since \(58 / 29 = 2\), the molecular formula is \(C_4H_{10}\).
评分标准
(a) [1] a substance containing no water (of crystallisation)
(b)(i) [1] to ensure all the water is lost/removed / reaction is complete
(b)(vi) [1] working: ratio of \(0.12 / 0.02\) [1] 6
(c)(i) [1] working out moles of C and H: \(82.8 / 12 = 6.90\) and \(17.2 / 1 = 17.2\) [1] simplifying ratio: \(1 : 2.5\) [1] empirical formula: \(C_2H_5\)
(c)(ii) [1] \(C_4H_{10}\)
题目 4 · structured
13 分
Hydrogen gas reacts with chlorine gas in a photochemical reaction: \(H_2\text{(g)} + Cl_2\text{(g)} \to 2HCl\text{(g)}\)
(a) State the condition required for this photochemical reaction to occur. [1]
(b) The reaction is exothermic.
(i) Explain, in terms of bond breaking and bond making, why this reaction is exothermic. [2]
(ii) A table of bond energies is shown below: \(\bullet\) \(H-H\): \(436\text{ kJ/mol}\) \(\bullet\) \(Cl-Cl\): \(242\text{ kJ/mol}\) \(\bullet\) \(H-Cl\): \(431\text{ kJ/mol}\) Calculate the energy change, \(\Delta H\), for this reaction. [3]
(c) The rate of the reaction between zinc and dilute hydrochloric acid was investigated. \(Zn\text{(s)} + 2HCl\text{(aq)} \to ZnCl_2\text{(aq)} + H_2\text{(g)}\)
(i) Describe how the rate of this reaction changes over time as the reaction proceeds. Explain your answer in terms of collision theory. [4]
(ii) State and explain the effect of increasing the temperature on the rate of this reaction. [3]
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解题
(a) Ultraviolet (UV) light.
(b)(i) More energy is released when making bonds (in \(HCl\)) than is absorbed when breaking bonds (in \(H_2\) and \(Cl_2\)).
(b)(ii) Energy needed to break bonds = \(436 + 242 = 678\text{ kJ/mol}\). Energy released making bonds = \(2 \times 431 = 862\text{ kJ/mol}\). \(\Delta H = 678 - 862 = -184\text{ kJ/mol}\).
(c)(i) The rate of reaction is fastest at the start because the concentration of hydrochloric acid is at its highest. As the reaction proceeds, the rate decreases because acid particles are used up, so their concentration decreases. This leads to a lower frequency of successful collisions. Eventually, the rate becomes zero when the limiting reactant is completely used up.
(c)(ii) Increasing the temperature increases the rate of reaction. This is because particles gain kinetic energy and move faster, which increases the frequency of collisions. Furthermore, a greater proportion of particles now possess energy equal to or greater than the activation energy, leading to a much higher frequency of successful collisions.
评分标准
(a) [1] ultraviolet light / UV light / sunlight
(b)(i) [1] bond breaking is endothermic / absorbs energy AND bond making is exothermic / releases energy [1] more energy is released making bonds than is absorbed breaking bonds
(b)(ii) [1] energy in = \(436 + 242 = 678\text{ (kJ)}\) [1] energy out = \(2 \times 431 = 862\text{ (kJ)}\) [1] \(-184\text{ kJ/mol}\) (value and sign required)
(c)(i) [1] rate decreases over time / reaction slows down [1] concentration of acid / reactants decreases [1] collision frequency decreases / fewer collisions per unit time [1] reaction stops when reactant used up
(c)(ii) [1] rate increases [1] particles have more kinetic energy / move faster (increasing collision frequency) [1] more particles have energy greater than or equal to activation energy (increasing successful collision frequency)
题目 5 · structured
14 分
Lead(II) sulfate is an insoluble salt.
(a) Describe how a pure, dry sample of lead(II) sulfate can be prepared from aqueous lead(II) nitrate and aqueous sodium sulfate. Include the names of the techniques used. [5]
(b) Write an ionic equation, with state symbols, for the precipitation of lead(II) sulfate. [3]
(c) An unknown solid, **Y**, is analysed.
(i) A flame test on **Y** produces a lilac flame. Identify the metal ion present in **Y**. [1]
(ii) To an aqueous solution of **Y**, dilute nitric acid followed by aqueous silver nitrate is added. A yellow precipitate is formed. Identify the anion present in **Y**. [1]
(iii) Write the ionic equation, with state symbols, for the formation of this yellow precipitate. [3]
(iv) Describe the observation when excess aqueous ammonia is added to a solution containing copper(II) ions. [1]
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解题
(a) Mix lead(II) nitrate solution and sodium sulfate solution together in a beaker to form a precipitate. Filter the mixture using a filter funnel and filter paper to obtain the lead(II) sulfate residue. Wash the residue with distilled water to remove any soluble impurities (sodium nitrate). Dry the residue in a warm oven or between sheets of filter paper.
(a) [1] Mix the two solutions together [1] Filter (to obtain residue) [1] Wash residue / precipitate with distilled water [1] Dry residue between filter papers / in warm oven / warm place [1] Names of techniques: precipitation / filtration
(b) [1] Correct reactants: \(Pb^{2+}\) and \(SO_4^{2-}\) [1] Correct product: \(PbSO_4\) [1] Correct state symbols: (aq) + (aq) -> (s)
(c)(i) [1] Potassium / \(K^+\)
(c)(ii) [1] Iodide / \(I^-\)
(c)(iii) [1] Correct reactants: \(Ag^+\) and \(I^-\) [1] Correct product: \(AgI\) [1] Correct state symbols: (aq) + (aq) -> (s)
(c)(iv) [1] dark blue solution (accept: blue precipitate dissolves to form a dark blue solution)
题目 6 · structured
13 分
Organic compounds can contain different functional groups.
(a) Compound **P** is a carboxylic acid with the molecular formula \(C_3H_6O_2\).
(i) State the name of compound **P**. [1]
(ii) Draw the displayed formula of compound **P**, showing all atoms and all bonds. [2]
(b) Compound **P** reacts with methanol in the presence of an acid catalyst to form an ester, **Q**, and water.
(i) State the name of ester **Q**. [1]
(ii) Draw the displayed formula of ester **Q**. [2]
(iii) State the name of the acid catalyst commonly used in this reaction. [1]
(c) Polyesters can be formed by condensation polymerisation.
(i) Describe the difference between addition polymerisation and condensation polymerisation. [2]
(ii) A polyester is made from a dicarboxylic acid and a diol. Draw the structure of the repeating unit of the polyester formed from: \(HOOC-CH_2-COOH\) and \(HO-CH_2CH_2-OH\). Show all the atoms and bonds in the linkages. [4]
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解题
(a)(i) Propanoic acid.
(a)(ii) Displayed formula of propanoic acid: \(CH_3-CH_2-COOH\) fully drawn out: H-C(H)(H)-C(H)(H)-C(=O)-O-H.
(b)(i) Methyl propanoate.
(b)(ii) Displayed formula of methyl propanoate: \(CH_3-CH_2-COO-CH_3\) fully drawn out with all atoms and bonds shown.
(b)(iii) Concentrated sulfuric acid.
(c)(i) Addition polymerisation involves only one monomer with a double bond and produces only the polymer. Condensation polymerisation involves monomers with two functional groups and produces the polymer plus a small molecule like water.
(c)(ii) The repeating unit is: \(-O-CH_2-CH_2-O-CO-CH_2-CO-\). The ester linkage is shown as \(-O-CO-\).
评分标准
(a)(i) [1] Propanoic acid
(a)(ii) [2] Correct displayed structure showing all bonds (including \(O-H\) bond) (Award [1] if structure is correct but shows \(-OH\) rather than \(-O-H\))
(b)(i) [1] Methyl propanoate
(b)(ii) [2] Correct displayed structure of methyl propanoate showing all bonds (Award [1] for correct structural formula but not fully displayed)
(c)(i) [1] Addition polymerisation forms only one product (the polymer) whereas condensation polymerisation forms two products (the polymer and a small molecule / water / hydrogen chloride) [1] Addition polymerisation requires a \(C=C\) double bond / unsaturated monomer whereas condensation polymerisation requires two different functional groups per monomer (or equivalent)
(c)(ii) [1] Correct ester linkage drawn out \(-O-C(=O)-\) [1] Correct diol section \(-O-CH_2-CH_2-O-\) [1] Correct dicarboxylic acid section \(-CO-CH_2-CO-\) [1] Correct repeating unit showing open continuation bonds at the ends
Paper 62 Alternative to Practical
Answer all questions. Write your answers in the spaces provided on the question paper. Notes for qualitative analysis are provided.
4 题目 · 40 分
题目 1 · structured
10 分
A student investigates the rate of reaction between marble chips, \( \text{CaCO}_3 \), and dilute nitric acid by measuring the mass loss over time. A conical flask containing the reaction mixture is placed on a balance.
(a) Name the apparatus used to measure the mass of the flask and its contents. [1]
(b) A cotton wool plug is placed in the neck of the flask. Suggest why cotton wool is used instead of a rubber bung. [1]
(c) The mass of the flask and contents decreases during the reaction. (i) State why the mass decreases. [1] (ii) The reaction finishes when the mass stops changing. Deduce the time when the reaction finished using the following data: [1] - \( t = 0 \text{ s} \), mass = \( 200.00 \text{ g} \) - \( t = 30 \text{ s} \), mass = \( 199.65 \text{ g} \) - \( t = 60 \text{ s} \), mass = \( 199.42 \text{ g} \) - \( t = 90 \text{ s} \), mass = \( 199.30 \text{ g} \) - \( t = 120 \text{ s} \), mass = \( 199.25 \text{ g} \) - \( t = 150 \text{ s} \), mass = \( 199.25 \text{ g} \) - \( t = 180 \text{ s} \), mass = \( 199.25 \text{ g} \)
(iii) Calculate the total mass of gas lost during the experiment. [1]
(d) State and explain the effect of repeating the experiment using the same mass of smaller marble chips. [2]
(e) Describe a chemical test to confirm that the gas produced is carbon dioxide. State the observation. [2]
(f) State one safety precaution the student must take when handling dilute nitric acid. [1]
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解题
(a) A top-pan balance is used to measure mass. (b) Cotton wool allows carbon dioxide gas to escape while retaining liquid spray. A rubber bung would seal the flask, causing pressure to build up. (c)(i) The escape of carbon dioxide gas causes the mass to decrease. (c)(ii) The mass becomes constant at 120 seconds (199.25 g). (c)(iii) Total mass loss = \( 200.00 \text{ g} - 199.25 \text{ g} = 0.75 \text{ g} \). (d) Smaller marble chips have a larger surface area per unit volume, which increases the frequency of collisions, thus increasing the rate of reaction. (e) Carbon dioxide is tested by bubbling it through limewater (aqueous calcium hydroxide). The limewater turns milky/cloudy. (f) Dilute nitric acid is corrosive/irritant, so safety goggles or gloves should be worn.
评分标准
(a) 1 mark: Balance / top-pan balance. (b) 1 mark: To allow carbon dioxide/gas to escape but prevent loss of liquid spray. (c)(i) 1 mark: Carbon dioxide gas escapes from the flask. (c)(ii) 1 mark: 120 (seconds) / 120 s. (c)(iii) 1 mark: 0.75 (g). (d) 2 marks: Rate of reaction increases / reaction is faster (1 mark); due to a larger surface area of smaller chips (1 mark). (e) 2 marks: Bubble/test gas with limewater (1 mark); turns cloudy / milky / white precipitate forms (1 mark). (f) 1 mark: Wear safety goggles / protective gloves.
题目 2 · structured
10 分
A student prepares pure, dry crystals of hydrated copper(II) sulfate, \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \), starting from black copper(II) oxide powder and dilute sulfuric acid.
(a) Describe how copper(II) oxide is added to the warm acid, and explain why it must be added until some remains unreacted. [2]
(b) Name the process and the apparatus used to remove the unreacted copper(II) oxide from the mixture. [2]
(c) Describe the remaining steps required to obtain pure, dry crystals of copper(II) sulfate from the filtrate. [4]
(d) The student heats the blue crystals strongly in a dry test-tube. State the observation and explain the chemical change that occurs. [2]
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解题
(a) Copper(II) oxide is added in small portions with stirring to ensure it reacts completely. It is added in excess (until some unreacted solid remains) to ensure that all of the sulfuric acid is fully neutralised and used up. (b) The mixture is filtered using a filter funnel lined with filter paper to remove the excess insoluble copper(II) oxide. (c) The copper(II) sulfate filtrate is heated in an evaporating basin to evaporate some water until the crystallisation point is reached. The solution is then left to cool and form crystals. The crystals are filtered or decanted, washed with a small amount of cold distilled water, and dried between sheets of filter paper. (d) Hydrated copper(II) sulfate is blue. When heated strongly, the water of crystallisation is lost as steam, turning the salt into white anhydrous copper(II) sulfate.
评分标准
(a) 2 marks: Add in portions with stirring/heating (1 mark); added in excess to ensure all sulfuric acid is neutralised/reacted (1 mark). (b) 2 marks: Filtration (1 mark); filter paper and filter funnel (1 mark). (c) 4 marks: Heat the filtrate/solution to crystallisation point/evaporate some water (1 mark); leave to cool/crystallise (1 mark); filter/decant crystals from the remaining solution (1 mark); dry crystals between filter papers / in a warm oven (1 mark). (d) 2 marks: Blue crystals turn white (1 mark); water of crystallisation is lost/dehydration/steam is produced (1 mark).
题目 3 · structured
10 分
Paper chromatography is used to separate and identify food colourings in three commercial fruit drinks: X, Y, and Z.
(a) Explain why the starting line (baseline) must be drawn in pencil rather than ink. [1]
(b) Suggest why the solvent level in the chromatography tank must be below the starting line. [1]
(c) The developed chromatogram is shown: (i) Drink X separates into three spots. State what this indicates about the composition of drink X. [1] (ii) Drink Y shows a single spot that has travelled 5.4 cm from the baseline. The solvent front travelled 7.2 cm. Calculate the \( R_f \) value of this spot. Show your working. [2] (iii) Drink Z contains a dye with an \( R_f \) value of 0.50. Describe the position of this spot on the developed chromatogram. [1]
(d) Some chromatography spots are colourless. Describe how these colourless spots can be made visible on the chromatogram. [2]
(e) Suggest why ethanol is sometimes used as the solvent instead of water. [2]
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解题
(a) Pencil lead (graphite) is insoluble in the solvent and will not dissolve or run, whereas ink contains soluble dyes that would run and interfere with the chromatogram. (b) If the solvent level is above the baseline, the spotted samples will dissolve directly into the solvent pool at the bottom of the beaker instead of travelling up the paper. (c)(i) Drink X is a mixture containing three distinct dyes. (c)(ii) \( R_f = \frac{\text{distance travelled by spot}}{\text{distance travelled by solvent front}} = \frac{5.4}{7.2} = 0.75 \). (c)(iii) An \( R_f \) of 0.50 means the spot is located exactly halfway between the starting line and the solvent front. (d) Exposing the chromatogram to ultraviolet (UV) light or spraying it with a suitable locating agent (chemical reagent) makes the colourless spots visible. (e) Different solvents are chosen based on solubility; some pigments/dyes are insoluble in water but dissolve readily in organic solvents like ethanol.
评分标准
(a) 1 mark: Pencil (graphite) is insoluble (in the solvent) / will not run or interfere with results. (b) 1 mark: To prevent the spots/samples from dissolving/washing into the solvent pool. (c)(i) 1 mark: It is a mixture of (at least) three different dyes/substances. (c)(ii) 2 marks: Correct working shown: \( 5.4 / 7.2 \) (1 mark); answer = 0.75 (1 mark). (c)(iii) 1 mark: Halfway between the baseline/starting line and the solvent front. (d) 2 marks: Use of a locating agent/chemical spray (1 mark); or expose to ultraviolet/UV light (1 mark). (e) 2 marks: Dyes are insoluble in water (1 mark); but soluble in ethanol (1 mark).
题目 4 · structured
10 分
A student is provided with a green solid, salt W, containing one cation and one anion.
(a) Describe how the student would carry out a flame test on W. State the observation expected if the cation is copper(II). [2]
(b) Salt W is dissolved in water to make solution W. Describe the observations when: (i) aqueous sodium hydroxide is added dropwise and then in excess to solution W. [2] (ii) aqueous ammonia is added dropwise and then in excess to solution W. [2]
(c) To a portion of solution W, dilute nitric acid and aqueous barium nitrate are added. No precipitate forms. State what this negative result indicates. [1]
(d) To another portion of solution W, dilute nitric acid and aqueous silver nitrate are added. A cream precipitate is formed. (i) Name the anion present in W. [1] (ii) Write the chemical formula of the cream precipitate. [2]
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解题
(a) A flame test is performed by dipping a clean nichrome or platinum wire into concentrated hydrochloric acid, touching it to the solid sample W, and placing it into the hot, blue (non-luminous) Bunsen flame. A blue-green flame color confirms the presence of copper(II) ions. (b)(i) Copper(II) ions react with aqueous sodium hydroxide to form a light blue precipitate of copper(II) hydroxide, which does not dissolve in excess NaOH. (b)(ii) Adding dropwise aqueous ammonia produces a light blue precipitate of copper(II) hydroxide. Upon adding excess ammonia, the precipitate dissolves to form a clear, deep blue solution of a copper-ammonia complex. (c) Barium nitrate is used to test for sulfate ions. No precipitate indicates that sulfate ions (\( \text{SO}_4^{2-} \)) are absent. (d)(i) Silver nitrate reacts with halide ions. A cream precipitate indicates bromide ions (\( \text{Br}^- \)). (d)(ii) The cream precipitate is silver bromide, which has the chemical formula \( \text{AgBr} \).
评分标准
(a) 2 marks: Clean wire (nichrome/platinum) dipped in acid/HCl placed in Bunsen flame (1 mark); blue-green flame observed (1 mark). (b)(i) 2 marks: Light blue precipitate / ppt. (1 mark); insoluble in excess (1 mark). (b)(ii) 2 marks: Light blue precipitate / ppt. (1 mark); dissolves in excess to form a deep blue solution (1 mark). (c) 1 mark: Sulfate ions (\( \text{SO}_4^{2-} \)) are absent/not present. (d)(i) 1 mark: Bromide (accept \( \text{Br}^- \)). (d)(ii) 2 marks: \( \text{AgBr} \) (2 marks); award 1 mark for incorrect formula with correct elements (e.g. \( \text{AgBr}_2 \) or \( \text{Ag}_2\text{Br} \)).
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