Cambridge IGCSE · Thinka 原创模拟试题

2024 Cambridge IGCSE Chemistry (0620) 模拟试题及答案详解

Thinka Nov 2024 (V3) Cambridge International A Level-Style Mock — Chemistry (0620)

160 分180 分钟2024

An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V3) Cambridge International A Level Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.

卷一 (Core - 選擇題)

Answer all forty questions on the multiple choice answer sheet. Choose one correct option A, B, C, or D.

40 题目 · 40 分

题目 1 · 選擇題

1 分

A solid sample of a pure substance is heated at a constant rate. The temperature is recorded at regular intervals:

- 0 minutes: 20 °C
- 1 minute: 45 °C
- 2 minutes: 80 °C
- 3 minutes: 80 °C
- 4 minutes: 80 °C
- 5 minutes: 120 °C
- 6 minutes: 160 °C

Which statement about this substance is correct?

A.Its melting point is 20 °C and it is completely liquid at 3 minutes.
B.Its melting point is 80 °C and it is a mixture of solid and liquid at 3 minutes.
C.Its boiling point is 80 °C and it is completely gas at 3 minutes.
D.Its melting point is 80 °C and it is completely solid at 3 minutes.

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解题

At 0 minutes, the temperature of the solid is 20 °C. As it is heated, the temperature rises to 80 °C at 2 minutes. From 2 to 4 minutes (including 3 minutes), the temperature remains constant at 80 °C while heating continues. This constant temperature represents a change of state (melting), meaning the melting point is 80 °C. During melting, both solid and liquid states are present simultaneously. Therefore, at 3 minutes, the substance is a mixture of solid and liquid.

评分标准

1 mark for the correct option B.
- Reject A: at 3 minutes, melting is not yet complete, so it is not liquid only.
- Reject C: 20 °C is the starting temperature, not the melting point.
- Reject D: it cannot be completely solid while undergoing melting at 80 °C.

题目 2 · 選擇題

1 分

Dilute sulfuric acid is added to three separate test-tubes containing sodium carbonate, zinc, and sodium hydroxide respectively. In which of the test-tubes are bubbles of gas observed?

A.sodium carbonate and zinc only
B.sodium carbonate and sodium hydroxide only
C.zinc and sodium hydroxide only
D.sodium carbonate, zinc and sodium hydroxide

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解题

Acids react with carbonates (like sodium carbonate) to produce a salt, water, and carbon dioxide gas, which is seen as bubbles. Acids react with metals above hydrogen in the reactivity series (like zinc) to produce a salt and hydrogen gas, which is also seen as bubbles. Acids react with metal hydroxides (alkalis, like sodium hydroxide) in a neutralisation reaction to produce a salt and water, but no gas is evolved. Therefore, bubbles of gas are only observed with sodium carbonate and zinc.

评分标准

1 mark for the correct option A.
- Reject B, C, and D because the neutralisation of sodium hydroxide with sulfuric acid does not produce any gas.

题目 3 · 選擇題

1 分

A potassium atom reacts with a chlorine atom to form the ionic compound potassium chloride. Which statement describes how this bond is formed?

A.Each potassium atom shares one electron with a chlorine atom.
B.Each chlorine atom transfers one electron to a potassium atom.
C.Each potassium atom transfers one electron to a chlorine atom.
D.Each potassium atom transfers two electrons to a chlorine atom.

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解题

Potassium is a metal in Group I and has 1 electron in its outer shell. Chlorine is a non-metal in Group VII and has 7 electrons in its outer shell. To achieve a stable noble gas electronic configuration, the potassium atom transfers its 1 outer electron to the chlorine atom. This forms a potassium ion $K^{+}$ and a chloride ion $Cl^{-}$ which are held together by strong electrostatic attraction.

评分标准

1 mark for the correct option C.
- Reject A: ionic bonds involve the transfer of electrons, not sharing (which would be covalent bonding).
- Reject B: the electron is transferred from the metal (potassium) to the non-metal (chlorine), not the other way around.
- Reject D: potassium only has 1 outer electron to lose, so it cannot transfer two electrons.

题目 4 · 選擇題

1 分

Two isotopes of carbon are carbon-12 ($^{12}\text{C}$) and carbon-14 ($^{14}\text{C}$). Which statement about these two isotopes is correct?

A.They have different numbers of electrons in their outer shell, so they have different chemical properties.
B.They have the same number of protons and neutrons, so they have the same chemical properties.
C.They have the same number of electrons in their outer shell, so they have the same chemical properties.
D.They have different numbers of protons, so they are placed in different groups of the Periodic Table.

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解题

Isotopes of the same element have the same number of protons and electrons, and thus identical chemical properties because they have the same electronic configuration (specifically the number of outer shell electrons). Carbon-12 and carbon-14 have different numbers of neutrons (6 and 8 respectively), so their mass numbers (nucleon numbers) are different, but they possess the exact same outer shell electrons (4) and have identical chemical properties.

评分标准

1 mark for the correct option C.

题目 5 · 選擇題

1 分

A student dissolves an unknown ionic compound in water. Aqueous sodium hydroxide is added to the solution, and the mixture is gently warmed. A gas is released that turns damp red litmus paper blue. Which ion is present in the solution?

A.ammonium, $\text{NH}_4^+$
B.copper(II), $\text{Cu}^{2+}$
C.iron(III), $\text{Fe}^{3+}$
D.zinc, $\text{Zn}^{2+}$

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解题

Ammonium ions ($\text{NH}_4^+$) react with hydroxide ions from the sodium hydroxide on heating to produce ammonia gas ($\text{NH}_3$). Ammonia is an alkaline gas, which turns damp red litmus paper blue. None of the other listed ions (copper(II), iron(III), zinc) produce an alkaline gas when treated with sodium hydroxide.

评分标准

1 mark for the correct option A.

题目 6 · 選擇題

1 分

Which option correctly describes the physical state and color of chlorine at room temperature and pressure, and the observation when chlorine gas is bubbled into aqueous potassium iodide?

A.pale green gas; the solution turns brown
B.pale green gas; the solution remains colorless
C.red-brown liquid; the solution turns brown
D.red-brown liquid; the solution remains colorless

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解题

At room temperature and pressure, chlorine is a pale green gas. Chlorine is more reactive than iodine, so when chlorine is bubbled into aqueous potassium iodide, a displacement reaction occurs: $\text{Cl}_2 + 2\text{KI} \rightarrow 2\text{KCl} + \text{I}_2$. The displaced iodine dissolves in the solution, causing it to turn from colorless to brown.

评分标准

1 mark for the correct option A.

题目 7 · 選擇題

1 分

A mixture contains solid sand and solid salt (sodium chloride). Which sequence of steps should be used to obtain separate, dry samples of both sand and salt?

A.Add water, stir, filter, evaporate the filtrate, and dry the residue.
B.Add water, stir, evaporate the mixture to dryness, and then filter.
C.Filter the dry mixture, add water to the residue, and evaporate.
D.Heat the mixture to melt the salt, filter, and dry the residue.

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解题

To separate sand (insoluble in water) and salt (soluble in water):
1. Add water and stir: salt dissolves, while sand remains as an insoluble solid.
2. Filter: the sand remains on the filter paper as the residue, and the salt solution passes through as the filtrate.
3. Evaporate the filtrate to dryness: this removes the water, leaving behind dry salt.
4. Dry the residue (sand): this removes any remaining water, leaving behind dry sand.

Therefore, option A is the correct sequence.

评分标准

1 分

The table shows the melting points and boiling points of four substances: W, X, Y and Z.

$$\begin{array}{|c|c|c|} \hline \text{Substance} & \text{Melting point } / ^\circ\text{C} & \text{Boiling point } / ^\circ\text{C} \\ \hline \text{W} & -114 & 78 \\ \hline \text{X} & -182 & -161 \\ \hline \text{Y} & 180 & 1340 \\ \hline \text{Z} & 0 & 100 \\ \hline \end{array}$$

Which substance is a liquid at
$-100~^\circ\text{C}$ and a gas at $100~^\circ\text{C}$?

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解题

To determine the state of each substance, we compare the given temperatures with their melting and boiling points:

1. At $-100~^\circ\text{C}$:
- Substance W has a melting point of $-114~^\circ\text{C}$. Since $-100~^\circ\text{C}$ is above the melting point but below the boiling point ($78~^\circ\text{C}$), W is a **liquid**.
- Substance X has a boiling point of $-161~^\circ\text{C}$. Since $-100~^\circ\text{C}$ is above the boiling point, X is a **gas**.
- Substances Y and Z have melting points of $180~^\circ\text{C}$ and $0~^\circ\text{C}$ respectively. At $-100~^\circ\text{C}$, both are below their melting points and thus are **solids**.

2. At $100~^\circ\text{C}$:
- Substance W has a boiling point of $78~^\circ\text{C}$. Since $100~^\circ\text{C}$ is above $78~^\circ\text{C}$, W is a **gas**.

Therefore, substance W is a liquid at $-100~^\circ\text{C}$ and a gas at $100~^\circ\text{C}$.

评分标准

1 mark for the correct option A.
- Reject B, C, D as they do not meet both state requirements at the specified temperatures.

题目 20 · 選擇題

1 分

A student investigates four different liquid food colourings, P, Q, R and S, using paper chromatography.

The results show:
- P separates into two spots.
- Q remains as a single spot that travels the same distance as the upper spot of P.
- R remains as a single spot that travels a shorter distance than the spot of Q.
- S separates into three spots, one of which travels the exact same distance as the spot of R.

Which statement is correct?

A.P is a pure substance.
B.Q and R are mixtures.
C.S contains substance R.
D.S contains substance Q.

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解题

Let's analyze the chromatography results:
- **P** separates into two spots, meaning it is a **mixture** of at least two substances.
- **Q** remains as a single spot, meaning it is a **pure substance**.
- **R** remains as a single spot, meaning it is a **pure substance**.
- **S** separates into three spots, meaning it is a **mixture** of at least three substances. One of these spots travels the same distance as the spot of R, which indicates that the mixture S **contains substance R**.

Therefore, statement C is correct.

评分标准

卷二 (Extended - 選擇題)

Answer all forty questions on the multiple choice answer sheet. Choose one correct option A, B, C, or D.

35 题目 · 35 分

题目 1 · 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 electrolyte?

A.Anode product: chlorine; Cathode product: hydrogen; pH: increases
B.Anode product: chlorine; Cathode product: sodium; pH: decreases
C.Anode product: oxygen; Cathode product: hydrogen; pH: increases
D.Anode product: oxygen; Cathode product: sodium; pH: no change

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解题

During the electrolysis of concentrated aqueous sodium chloride: (1) At the anode (positive electrode), chloride ions ($Cl^-$) are discharged in preference to hydroxide ions because they are in high concentration, forming chlorine gas. (2) At the cathode (negative electrode), hydrogen ions ($H^+$) are discharged in preference to sodium ions because hydrogen is less reactive than sodium, forming hydrogen gas. (3) As $H^+$ and $Cl^-$ ions are selectively discharged and removed, sodium ($Na^+$) and hydroxide ($OH^-$) ions remain in the solution. This forms alkaline sodium hydroxide solution, causing the pH of the electrolyte to increase.

评分标准

1 mark for the correct option.
- Accept A.
- Reject B, C, D.

题目 2 · 選擇題

1 分

In Experiment 1, excess large marble chips are reacted with $50\text{ cm}^3$ of $1.0\text{ mol/dm}^3$ hydrochloric acid at $25^\circ\text{C}$. Experiment 2 is carried out under different conditions to produce a faster initial rate of reaction but the same final volume of carbon dioxide gas. Which set of conditions is used for Experiment 2?

A.Excess large marble chips with $100\text{ cm}^3$ of $0.5\text{ mol/dm}^3$ hydrochloric acid at $25^\circ\text{C}$
B.Excess large marble chips with $50\text{ cm}^3$ of $1.0\text{ mol/dm}^3$ hydrochloric acid at $35^\circ\text{C}$
C.Excess large marble chips with $25\text{ cm}^3$ of $1.0\text{ mol/dm}^3$ hydrochloric acid at $35^\circ\text{C}$
D.Excess large marble chips with $50\text{ cm}^3$ of $2.0\text{ mol/dm}^3$ hydrochloric acid at $25^\circ\text{C}$

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解题

To obtain the same final volume of carbon dioxide, the total number of moles of the limiting reactant (hydrochloric acid) must remain the same: In Experiment 1: $\text{moles of } HCl = 0.050\text{ dm}^3 \times 1.0\text{ mol/dm}^3 = 0.050\text{ mol}$. Option A gives $0.100\text{ dm}^3 \times 0.5\text{ mol/dm}^3 = 0.050\text{ mol}$, but the lower concentration decreases the initial rate. Option B gives $0.050\text{ dm}^3 \times 1.0\text{ mol/dm}^3 = 0.050\text{ mol}$. The higher temperature ($35^\circ\text{C}$ compared to $25^\circ\text{C}$) increases the rate, giving a faster initial rate and the same volume of gas. Option C halves the moles of $HCl$, and Option D doubles the moles of $HCl$, changing the final volume of gas.

评分标准

1 mark for identifying the correct conditions that maintain the moles of the limiting reactant while increasing the temperature to increase the reaction rate.
- Accept B.
- Reject A, C, D.

题目 3 · 選擇題

1 分

Ammonia is manufactured by the Haber process according to the following reversible reaction: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \quad \Delta H = -92\text{ kJ/mol}$. Which row correctly describes the conditions that produce the highest equilibrium yield of ammonia and the conditions that produce the fastest rate of reaction?

A.Highest equilibrium yield: low temperature and high pressure; Fastest rate: high temperature and high pressure
B.Highest equilibrium yield: high temperature and high pressure; Fastest rate: high temperature and high pressure
C.Highest equilibrium yield: low temperature and low pressure; Fastest rate: low temperature and high pressure
D.Highest equilibrium yield: high temperature and low pressure; Fastest rate: high temperature and low pressure

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解题

For equilibrium yield: The forward reaction is exothermic ($\Delta H < 0$), so a lower temperature shifts the position of equilibrium to the right to increase the yield of ammonia. There are fewer moles of gas on the right (2 moles) than on the left (4 moles), so higher pressure shifts the equilibrium to the right. Thus, the highest equilibrium yield is obtained at low temperature and high pressure. For rate of reaction: Both increasing temperature and increasing pressure increase the frequency of effective collisions, resulting in a faster rate of reaction. Thus, the fastest rate of reaction is obtained at high temperature and high pressure.

评分标准

1 mark for the correct combination of conditions for maximum yield (low temperature, high pressure) and maximum rate (high temperature, high pressure).
- Accept A.
- Reject B, C, D.

题目 4 · 選擇題

1 分

When 10.0 g of an impure sample of calcium carbonate is heated strongly, it decomposes according to the equation: $\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})$. If 1.92 $\text{dm}^3$ of carbon dioxide gas is collected at r.t.p., what is the percentage purity of the calcium carbonate? [Relative atomic masses: $\text{Ca} = 40$, $\text{C} = 12$, $\text{O} = 16$. One mole of any gas occupies 24.0 $\text{dm}^3$ at r.t.p.]

A.8.0%
B.19.2%
C.80.0%
D.92.0Op00037nxo3KiqRDOKhAHryjz_purity_q1_fixed_D_corrected_80.0%

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解题

First, calculate the number of moles of $\text{CO}_2$ gas collected: $\text{moles of CO}_2 = 1.92 / 24.0 = 0.08\text{ mol}$. Since the molar ratio of $\text{CaCO}_3$ to $\text{CO}_2$ is 1:1, the moles of pure $\text{CaCO}_3$ that reacted is also 0.08 mol. Next, calculate the molar mass of $\text{CaCO}_3$: $40 + 12 + (3 \times 16) = 100\text{ g/mol}$. Find the mass of pure $\text{CaCO}_3$ in the sample: $0.08\text{ mol} \times 100\text{ g/mol} = 8.0\text{ g}$. Finally, calculate the percentage purity: $(8.0\text{ g} / 10.0\text{ g}) \times 100\% = 80.0\%$.

评分标准

1 mark for the correct option C. Reject other options: A represents the mass of pure substance, B uses incorrect volume scaling, and D is the percentage of impurity.

题目 5 · 選擇題

1 分

A concentrated aqueous solution of sodium chloride is electrolysed using inert carbon electrodes. Which row correctly describes the substances formed at the electrodes and the substance left in the solution?

A.Cathode: hydrogen; Anode: chlorine; Left in solution: sodium hydroxide
B.Cathode: sodium; Anode: chlorine; Left in solution: water
C.Cathode: hydrogen; Anode: oxygen; Left in solution: sodium chloride
D.Cathode: sodium; Anode: oxygen; Left in solution: hydrochloric acid

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解题

During the electrolysis of concentrated aqueous sodium chloride: 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. 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. The remaining sodium ions ($\text{Na}^+$) and hydroxide ions ($\text{OH}^-$) combine, leaving sodium hydroxide in the solution.

评分标准

1 mark for the correct option A. Correctly identifies products of cathode, anode, and solution species.

题目 6 · 選擇題

1 分

Methanol is manufactured by the reversible reaction: $\text{CO}(\text{g}) + 2\text{H}_2(\text{g}) \rightleftharpoons \text{CH}_3\text{OH}(\text{g})$ where $\Delta H = -91\text{ kJ/mol}$. Which set of conditions will produce the highest yield of methanol at equilibrium?

A.High pressure and high temperature
B.High pressure and low temperature
C.Low pressure and high temperature
D.Low pressure and low temperature

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解题

According to Le Chatelier's principle: 1. Pressure: The left side of the equation has 3 moles of gas, while the right side has 1 mole of gas. Increasing pressure shifts the equilibrium position to the right (fewer moles of gas), increasing the yield. Therefore, high pressure is needed. 2. Temperature: The forward reaction is exothermic (negative $\Delta H$). Decreasing the temperature shifts the equilibrium position to the right (the exothermic direction) to release heat, increasing the yield. Therefore, low temperature is needed. Thus, high pressure and low temperature produce the highest yield of methanol.

评分标准

1 mark for the correct option B. Correct application of Le Chatelier's principle to both pressure and temperature changes.

题目 7 · 選擇題

1 分

A student reacts 2.50 g of impure calcium carbonate (containing 80.0% CaCO3 by mass) with an excess of dilute hydrochloric acid. The equation for the reaction is: CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g). What is the volume of carbon dioxide gas, in dm3, produced at room temperature and pressure (r.t.p.)? [Assume the molar volume of a gas at r.t.p. is 24.0 dm3/mol, and Mr(CaCO3) = 100]

A.0.48 dm3
B.0.60 dm3
C.0.24 dm3
D.0.96 dm3

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解题

Step 1: Calculate the mass of pure CaCO3 in the sample: Mass = 2.50 g * 0.80 = 2.00 g. Step 2: Calculate the number of moles of pure CaCO3: Moles = 2.00 g / 100 g/mol = 0.020 mol. Step 3: Use the stoichiometric ratio from the balanced equation (1 mol CaCO3 produces 1 mol CO2) to find the moles of CO2: Moles of CO2 = 0.020 mol. Step 4: Calculate the volume of CO2 gas at r.t.p.: Volume = 0.020 mol * 24.0 dm3/mol = 0.48 dm3.

评分标准

A is correct. 1 mark for calculating the volume as 0.48 dm3. B is incorrect (uses 2.50 g of CaCO3 directly without accounting for 80% purity). C is incorrect (divides moles by 2). D is incorrect (multiplies moles by 2).

题目 8 · 選擇題

1 分

Concentrated aqueous sodium chloride is electrolysed using inert platinum electrodes. Which row in the table correctly identifies the product at each electrode and the change in pH of the remaining solution?

A.Anode product: chlorine; Cathode product: hydrogen; pH of remaining solution: increases
B.Anode product: chlorine; Cathode product: sodium; pH of remaining solution: decreases
C.Anode product: oxygen; Cathode product: hydrogen; pH of remaining solution: increases
D.Anode product: oxygen; Cathode product: sodium; pH of remaining solution: stays the same

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解题

During the electrolysis of concentrated aqueous sodium chloride: 1. At the anode (positive electrode), chloride ions (Cl-) are discharged in preference to hydroxide ions (OH-) because they are in high concentration, producing chlorine gas. 2. At the cathode (negative electrode), hydrogen ions (H+) are discharged in preference to sodium ions (Na+) because hydrogen is lower in the reactivity series, producing hydrogen gas. 3. As H+ and Cl- ions are removed from the solution, Na+ and OH- ions remain, forming sodium hydroxide (NaOH), which is alkaline. Therefore, the pH of the remaining solution increases.

评分标准

A is correct. 1 mark for identifying chlorine at the anode, hydrogen at the cathode, and an increase in pH. B is incorrect because sodium is not formed at the cathode in aqueous solution, and the pH increases rather than decreases. C is incorrect because chlorine, not oxygen, is formed at the anode when the solution is concentrated. D is incorrect because oxygen and sodium are not formed, and the pH does not stay the same.

题目 9 · 選擇題

1 分

A mixture of gases A and B reacts reversibly to form gas C according to the equation: 2A(g) + B(g) <=> 2C(g) (forward reaction is exothermic). Which conditions of temperature and pressure would produce the highest yield of C at equilibrium?

A.Low temperature and high pressure
B.High temperature and high pressure
C.Low temperature and low pressure
D.High temperature and low pressure

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解题

1. Effect of temperature: The forward reaction is exothermic. 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 C. 2. Effect of pressure: The left side of the equation has 3 moles of gas (2 + 1) while the right side has 2 moles of gas. Increasing the pressure shifts the equilibrium towards the side with fewer moles of gas (to the right) to reduce pressure, thereby increasing the yield of C. Therefore, low temperature and high pressure produce the highest yield of C.

评分标准

A is correct. 1 mark for choosing the combination of low temperature (favours exothermic direction) and high pressure (favours fewer gas moles). B, C, and D are incorrect because they contain at least one condition that decreases the equilibrium yield of C.

题目 10 · 選擇題

1 分

A $20\text{ cm}^3$ sample of propane ($C_3H_8$) is mixed with $120\text{ cm}^3$ of oxygen ($O_2$) and ignited. The mixture is allowed to cool to room temperature and pressure. What is the total volume of gas remaining? (Assume water is a liquid at room temperature and pressure.)

A.$60\text{ cm}^3$
B.$80\text{ cm}^3$
C.$100\text{ cm}^3$
D.$140\text{ cm}^3$

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解题

According to the balanced equation for the complete combustion of propane: $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$, 1 volume of $C_3H_8$ reacts with 5 volumes of $O_2$ to yield 3 volumes of $CO_2$. $20\text{ cm}^3$ of $C_3H_8$ requires $20 \times 5 = 100\text{ cm}^3$ of $O_2$. Since $120\text{ cm}^3$ of oxygen was provided, oxygen is in excess and $120 - 100 = 20\text{ cm}^3$ of unreacted $O_2$ remains. The volume of $CO_2$ gas produced is $20 \times 3 = 60\text{ cm}^3$. Water is a liquid at room temperature and pressure, so its volume is negligible. The total volume of gas remaining is the sum of the excess oxygen and the carbon dioxide produced: $20\text{ cm}^3 + 60\text{ cm}^3 = 80\text{ cm}^3$.

评分标准

1 mark for correct selection B. Award 1 mark for calculating the volume of excess oxygen (20 cm3), the volume of carbon dioxide produced (60 cm3), and correctly summing them to obtain 80 cm3.

题目 11 · 選擇題

1 分

Two electrolysis cells are set up. Cell 1 contains aqueous copper(II) sulfate with inert carbon electrodes. Cell 2 contains aqueous copper(II) sulfate with copper electrodes. Which row correctly describes what is observed at the anode (positive electrode) in each cell?

A.Cell 1 anode: bubbles of a colourless gas; Cell 2 anode: the electrode decreases in mass
B.Cell 1 anode: a pink-brown solid is deposited; Cell 2 anode: bubbles of a colourless gas
C.Cell 1 anode: bubbles of a colourless gas; Cell 2 anode: a pink-brown solid is deposited
D.Cell 1 anode: the electrode decreases in mass; Cell 2 anode: bubbles of a colourless gas

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解题

In Cell 1, because the carbon electrodes are inert, hydroxide ions ($OH^-$) from the water are preferentially discharged at the positive anode to produce oxygen gas, which is observed as bubbles of a colourless gas. In Cell 2, because the copper electrodes are active, the copper anode itself is oxidised, losing electrons to form copper(II) ions ($Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$), which causes the anode to dissolve and decrease in mass.

评分标准

1 mark for correct selection A. Award 1 mark for identifying that oxygen gas (bubbles) forms at an inert anode and that an active copper anode dissolves (decreases in mass).

题目 12 · 選擇題

1 分

Ammonia gas reacts with water according to the reversible reaction shown: $NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$. Which statement correctly describes the reactants in the forward reaction according to the Brønsted-Lowry theory?

A.$H_2O$ acts as an acid because it donates a proton; $NH_3$ acts as a base because it accepts a proton
B.$NH_3$ acts as an acid because it donates a proton; $H_2O$ acts as a base because it accepts a proton
C.$H_2O$ acts as an acid because it accepts a proton; $NH_3$ acts as a base because it donates a proton
D.$NH_3$ acts as an acid because it accepts a proton; $H_2O$ acts as a base because it donates a proton

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解题

According to the Brønsted-Lowry theory, an acid is a proton ($H^+$) donor and a base is a proton ($H^+$) acceptor. In the forward reaction, water ($H_2O$) transfers a hydrogen ion (proton) to ammonia ($NH_3$), forming a hydroxide ion ($OH^-$) and an ammonium ion ($NH_4^+$). Thus, water acts as the acid because it donates a proton, and ammonia acts as the base because it accepts a proton.

评分标准

1 mark for correct selection A. Award 1 mark for recalling the definitions of a Brønsted-Lowry acid (proton donor) and base (proton acceptor) and correctly applying them to the forward reaction.

题目 13 · 選擇題

1 分

When $10\text{ cm}^3$ of a gaseous hydrocarbon was reacted with $70\text{ cm}^3$ of oxygen (an excess), the total volume of gas remaining after complete combustion and cooling to room temperature and pressure was $50\text{ cm}^3$.

Passage of this remaining gas through aqueous sodium hydroxide (which absorbs carbon dioxide) reduced the volume of gas to $20\text{ cm}^3$.

What is the formula of the hydrocarbon?

A.$\text{C}_2\text{H}_4$
B.$\text{C}_3\text{H}_6$
C.$\text{C}_3\text{H}_8$
D.$\text{C}_4\text{H}_{10}$

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解题

1. **Determine the volume of carbon dioxide ($\text{CO}_2$) produced:**
The gas remaining after combustion contains unreacted $\text{O}_2$ and product $\text{CO}_2$. Passing it through aqueous sodium hydroxide removes the $\text{CO}_2$, reducing the volume from $50\text{ cm}^3$ to $20\text{ cm}^3$.
$\text{Volume of CO}_2 = 50\text{ cm}^3 - 20\text{ cm}^3 = 30\text{ cm}^3$.

2. **Determine the volume of unreacted and reacted oxygen ($\text{O}_2$):**
- Unreacted $\text{O}_2$ remaining = $20\text{ cm}^3$.
- Volume of $\text{O}_2$ reacted = $70\text{ cm}^3 - 20\text{ cm}^3 = 50\text{ cm}^3$.

3. **Determine the mole ratio:**
- Hydrocarbon : $\text{O}_2\text{ reacted}$ : $\text{CO}_2\text{ produced}$ = $10\text{ cm}^3 : 50\text{ cm}^3 : 30\text{ cm}^3$
- This simplifies to a mole ratio of $1 : 5 : 3$.

4. **Determine the formula $\text{C}_x\text{H}_y$:**
- Since $1$ mole of hydrocarbon produces $3$ moles of $\text{CO}_2$, $x = 3$.
- The combustion equation is:
$\text{C}_3\text{H}_y + 5\text{O}_2 \rightarrow 3\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}$
- Equating oxygen atoms on both sides:
$10\text{ (from } 5\text{O}_2) = 6\text{ (from } 3\text{CO}_2) + \frac{y}{2}\text{ (from H}_2\text{O})$
$4 = \frac{y}{2} \Rightarrow y = 8$.

Therefore, the formula of the hydrocarbon is $\text{C}_3\text{H}_8$ (propane).

评分标准

Award 1 mark for the correct answer (C).

- Reject other options because they do not match the stoichiometric ratios calculated from the gas volumes.

题目 14 · 選擇題

1 分

An electrolysis experiment is set up with two different cells connected in series, so the same electric current passes through both:

- **Cell 1:** Aqueous copper(II) sulfate using inert carbon electrodes
- **Cell 2:** Aqueous copper(II) sulfate using copper electrodes

Which statement about the changes occurring at the electrodes is correct?

A.The mass of the anode in Cell 1 decreases.
B.The concentration of copper(II) ions in both cell solutions remains constant.
C.Copper metal is deposited at the cathode in both Cell 1 and Cell 2.
D.Bubbles of hydrogen gas are observed at the cathode in Cell 1.

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解题

Let's analyze the electrodes in both cells:

- **Cell 1 (inert carbon electrodes):**
- **At the cathode (-):** Copper is lower than hydrogen in the reactivity series, so $\text{Cu}^{2+}$ ions are preferentially discharged to form copper metal: $\text{Cu}^{2+}(aq) + 2\text{e}^- \rightarrow \text{Cu}(s)$. Thus, copper is deposited.
- **At the anode (+):** Hydroxide ions ($\text{OH}^-$) from water are discharged to produce oxygen gas: $4\text{OH}^-(aq) \rightarrow \text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4\text{e}^-$. The carbon anode is inert, so its mass does not change.
- **Solution concentration:** Since $\text{Cu}^{2+}$ ions are removed from solution without being replaced, the blue color fades and the concentration of $\text{Cu}^{2+}$ decreases.

- **Cell 2 (copper electrodes):**
- **At the cathode (-):** Like Cell 1, $\text{Cu}^{2+}$ ions are reduced to copper metal: $\text{Cu}^{2+}(aq) + 2\text{e}^- \rightarrow \text{Cu}(s)$. Thus, copper is deposited.
- **At the anode (+):** The copper electrode dissolves because copper atoms are oxidized to $\text{Cu}^{2+}$ ions: $\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2\text{e}^-$. The mass of this anode decreases.
- **Solution concentration:** The rate at which $\text{Cu}^{2+}$ ions are deposited at the cathode equals the rate at which they are formed at the anode. Therefore, the concentration of $\text{Cu}^{2+}$ remains constant.

Comparing the statements:
- Option A is incorrect because the anode in Cell 1 is inert carbon and does not decrease in mass.
- Option B is incorrect because $\text{Cu}^{2+}$ concentration only remains constant in Cell 2, but decreases in Cell 1.
- Option C is correct because copper is deposited at the cathode in both cells.
- Option D is incorrect because $\text{Cu}^{2+}$ is discharged in preference to $\text{H}^+$, so copper metal is deposited instead of hydrogen gas.

评分标准

Award 1 mark for the correct answer (C).

- Reject A: The anode in Cell 1 is inert.
- Reject B: Copper ion concentration decreases in Cell 1.
- Reject D: Copper, not hydrogen, is discharged at the cathode because copper is lower in the reactivity series.

题目 15 · 選擇題

1 分

The equation for a reversible reaction is shown below:

\[2\text{X}(g) + \text{Z}(g) \rightleftharpoons 2\text{Y}(g) \quad \Delta H = -92\text{ kJ/mol}\]

Which row correctly describes the conditions that maximize the rate of reaction and the conditions that maximize the equilibrium yield of Y?

| | Conditions for maximum rate of reaction | Conditions for maximum equilibrium yield of Y |
|---|---|---|
| **A** | High temperature and high pressure | Low temperature and high pressure |
| **B** | High temperature and high pressure | High temperature and low pressure |
| **C** | Low temperature and low pressure | Low temperature and high pressure |
| **D** | High temperature and low pressure | High temperature and high pressure |

A.Row A
B.Row B
C.Row C
D.Row D

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解题

To determine the correct conditions:

1. **Maximizing the Rate of Reaction:**
- **Temperature:** Increasing the temperature always increases the rate of reaction because particles have more kinetic energy, leading to more frequent collisions and a higher proportion of collisions having energy greater than or equal to the activation energy.
- **Pressure:** Increasing the pressure of a gaseous system increases the concentration of the reactant molecules, leading to more frequent collisions and thus a higher rate of reaction.
- Therefore, the **rate of reaction is maximized at high temperature and high pressure**.

2. **Maximizing the Equilibrium Yield of Y:**
- **Temperature:** The forward reaction is exothermic ($\Delta H = -92\text{ kJ/mol}$). According to Le Chatelier's principle, decreasing the temperature shifts the equilibrium in the direction of the exothermic reaction (to the right) to release heat. Therefore, the yield of Y is maximized at a **low temperature**.
- **Pressure:** There are 3 moles of gas on the reactant side ($2\text{X} + 1\text{Z}$) and only 2 moles of gas on the product side ($2\text{Y}$). Increasing the pressure shifts the equilibrium towards the side with fewer gas moles (to the right). Therefore, the yield of Y is maximized at a **high pressure**.
- Therefore, the **equilibrium yield of Y is maximized at low temperature and high pressure**.

Matching these findings to the table, Row A is the correct option.

评分标准

Award 1 mark for the correct answer (A).

- Reject B: High temperature would decrease the yield of Y because the forward reaction is exothermic, and low pressure would shift the equilibrium to the left.
- Reject C: Low temperature and low pressure would decrease the rate of reaction.
- Reject D: Low pressure decreases both rate and yield of Y.

题目 16 · 選擇題

1 分

A sample of 10.0 cm3 of a gaseous hydrocarbon, $C_xH_y$, was completely burned in 100.0 cm3 of oxygen (an excess). After cooling to room temperature, the total volume of gas remaining was 85.0 cm3. This gas mixture was passed through aqueous sodium hydroxide, which absorbed the carbon dioxide. The remaining volume of gas was 55.0 cm3. All gas volumes were measured at r.t.p. What is the formula of the hydrocarbon?

A.$CH_4$
B.$C_3H_6$
C.$C_3H_8$
D.$C_4H_8$

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解题

1. Identify the volume of gas absorbed by aqueous sodium hydroxide: The volume decrease is due to the absorption of carbon dioxide. Volume of $CO_2$ = 85.0 cm3 - 55.0 cm3 = 30.0 cm3. 2. Identify the unreacted oxygen: The remaining gas after passing through NaOH is unreacted oxygen, which is 55.0 cm3. 3. Calculate the volume of oxygen reacted: Volume of oxygen reacted = 100.0 cm3 - 55.0 cm3 = 45.0 cm3. 4. Find the mole ratio of reactants and products: Hydrocarbon : $O_2$ reacted : $CO_2$ produced = 10.0 cm3 : 45.0 cm3 : 30.0 cm3 = 1 : 4.5 : 3. Since 1 mole of $C_xH_y$ produces 3 moles of $CO_2$, we have x = 3. From the combustion equation: $C_3H_y + 4.5 O_2 \rightarrow 3 CO_2 + (y/2) H_2O$. Comparing oxygen atoms on both sides: 9 = 6 + y/2, which gives y = 6. Thus, the formula is $C_3H_6$.

评分标准

Award 1 mark for the correct option B. Award 1 mark for calculating the volume of $CO_2$ (30.0 cm3) and volume of reacted $O_2$ (45.0 cm3), and using these ratios to deduce x = 3 and y = 6. Reject all other options.

题目 17 · 選擇題

1 分

Concentrated aqueous sodium chloride (brine) is electrolysed using inert carbon electrodes. Which row correctly identifies the products formed at each electrode and the change in pH of the electrolyte during the process?

A.Anode product: oxygen; Cathode product: hydrogen; pH of electrolyte: decreases
B.Anode product: chlorine; Cathode product: sodium; pH of electrolyte: increases
C.Anode product: chlorine; Cathode product: hydrogen; pH of electrolyte: increases
D.Anode product: oxygen; Cathode product: sodium; pH of electrolyte: no change

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解题

During the electrolysis of concentrated aqueous sodium chloride: 1. At the anode (+): Chloride ions ($Cl^-$) and hydroxide ions ($OH^-$) migrate to the anode. Since chloride ions are highly concentrated, they are selectively discharged to form chlorine gas. 2. At the cathode (-): Sodium ions ($Na^+$) and hydrogen ions ($H^+$) migrate to the cathode. Since hydrogen is lower in the reactivity series than sodium, hydrogen ions are selectively discharged to form hydrogen gas. 3. Inside the solution: The discharge of hydrogen ions leaves an excess of hydroxide ions, which pair with the remaining sodium ions to form sodium hydroxide, a strong alkali. Thus, the pH of the electrolyte increases.

评分标准

Award 1 mark for the correct option C. Award 1 mark for identifying chlorine at the anode, hydrogen at the cathode, and an increase in the pH of the solution. Reject other options.

题目 18 · 選擇題

1 分

A gaseous reaction is represented by the equation shown: $2A(g) + B(g) \rightleftharpoons 2C(g)$ with $\Delta H = -92$ kJ/mol. Which conditions of temperature and pressure produce the highest yield of C at equilibrium, and what is the effect of adding a catalyst on the position of equilibrium?

A.Temperature: high; Pressure: high; Effect of catalyst: shifts equilibrium to the right
B.Temperature: low; Pressure: high; Effect of catalyst: no change
C.Temperature: low; Pressure: low; Effect of catalyst: shifts equilibrium to the right
D.Temperature: high; Pressure: low; Effect of catalyst: no change

查看答案详解

解题

To maximize the yield of C, we want to favor the forward reaction: 1. Temperature: The forward reaction is exothermic. 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 C. 2. Pressure: There are 3 moles of gas on the reactant side ($2A + B$) and 2 moles of gas on the product side ($2C$). Increasing the pressure shifts the equilibrium to the side with fewer gas moles (to the right) to reduce the pressure, thereby increasing the yield of C. 3. Catalyst: A catalyst increases the rates of both the forward and reverse reactions by the same factor. Therefore, it does not change the position of equilibrium, only the rate at which equilibrium is reached.

评分标准

Award 1 mark for the correct option B. Award 1 mark for identifying low temperature, high pressure, and no change from the catalyst. Reject other options.

题目 19 · 選擇題

1 分

A $10\text{ cm}^3$ sample of a gaseous hydrocarbon was completely burnt in $70\text{ cm}^3$ of oxygen (an excess). After cooling to room temperature, the total volume of gas remaining was $50\text{ cm}^3$. Shaking this remaining gas with excess aqueous sodium hydroxide reduced the volume to $20\text{ cm}^3$.

What is the molecular formula of the hydrocarbon?

A.$CH_4$
B.$C_2H_6$
C.$C_3H_6$
D.$C_3H_8$

查看答案详解

解题

1. Shaking the remaining gas with sodium hydroxide removes the carbon dioxide ($CO_2$) produced. The decrease in volume represents the volume of $CO_2$ produced:
$V(CO_2) = 50\text{ cm}^3 - 20\text{ cm}^3 = 30\text{ cm}^3$.

2. The remaining $20\text{ cm}^3$ of gas is the unreacted oxygen ($O_2$).

3. The volume of oxygen that reacted is:
$V(O_2\text{ reacted}) = 70\text{ cm}^3 - 20\text{ cm}^3 = 50\text{ cm}^3$.

4. Write the mole (volume) ratio of hydrocarbon reacted : $O_2$ reacted : $CO_2$ produced:
$10 : 50 : 30 = 1 : 5 : 3$.

5. For a hydrocarbon of formula $C_xH_y$, the combustion equation is:
$C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$
- From the carbon balance, $x = 3$.
- From the oxygen balance, $x + \frac{y}{4} = 5 \implies 3 + \frac{y}{4} = 5 \implies y = 8$.

Therefore, the molecular formula of the hydrocarbon is $C_3H_8$.

评分标准

1 mark for the correct option (D).
- Deduce the volume of CO2 is 30 cm3 (1 mark)
- Deduce the volume of O2 reacted is 50 cm3 (1 mark)
- Apply volume ratios to determine the formula C3H8 (1 mark)

题目 20 · 選擇題

1 分

A reversible reaction is shown.

$2A(g) + B(g) \rightleftharpoons 2C(g)$ $\Delta H = -92\text{ kJ/mol}$

Which set of conditions produces the highest equilibrium yield of $C(g)$ and what is the effect of adding a catalyst on this yield?

A.High temperature, high pressure, catalyst increases the yield
B.Low temperature, high pressure, catalyst has no effect on the yield
C.Low temperature, low pressure, catalyst increases the yield
D.High temperature, low pressure, catalyst has no effect on the yield

查看答案详解

解题

1. **Temperature effect:** The forward reaction is exothermic ($\Delta H = -92\text{ kJ/mol}$). 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 $C(g)$.

2. **Pressure effect:** There are 3 moles of gas on the left-hand side and 2 moles of gas on the right-hand side. Increasing the pressure shifts the equilibrium to the side with fewer moles of gas (to the right), increasing the yield of $C(g)$.

3. **Catalyst effect:** A catalyst increases the rates of both the forward and reverse reactions equally. It helps the reaction reach equilibrium faster but has no effect on the position of equilibrium or the yield of the products.

Therefore, a low temperature, high pressure, and a catalyst having no effect on the yield represent the correct combination (Option B).

评分标准

1 mark for the correct option (B).
- Low temperature increases yield of exothermic reaction (1 mark)
- High pressure increases yield of reaction with fewer gaseous product moles (1 mark)
- Catalyst has no effect on the equilibrium position/yield (1 mark)

题目 21 · 選擇題

1 分

Concentrated aqueous copper(II) chloride is electrolysed using inert carbon electrodes. Which row correctly identifies the product at each electrode and describes the color change of the electrolyte during the process?

A.Anode product: $Cl_2(g)$; Cathode product: $Cu(s)$; Electrolyte color: blue color fades
B.Anode product: $O_2(g)$; Cathode product: $Cu(s)$; Electrolyte color: blue color remains
C.Anode product: $Cl_2(g)$; Cathode product: $H_2(g)$; Electrolyte color: blue color fades
D.Anode product: $O_2(g)$; Cathode product: $H_2(g)$; Electrolyte color: blue color remains

查看答案详解

解题

1. **At the anode (positive electrode):** Chlorine gas ($Cl_2$) is produced. Chloride ions ($Cl^-$) are discharged in preference to hydroxide ions ($OH^-$) because the solution is concentrated.

2. **At the cathode (negative electrode):** Copper metal ($Cu$) is produced. Copper ions ($Cu^{2+}$) are discharged in preference to hydrogen ions ($H^+$) because copper is lower in the reactivity series.

3. **Color change of electrolyte:** Copper(II) ions are responsible for the blue color of the solution. As $Cu^{2+}$ ions are discharged and deposited as solid copper on the cathode, their concentration in the solution decreases, causing the blue color of the electrolyte to fade.

评分标准

1 mark for the correct option (A).
- Correct anode product: Cl2(g) (1 mark)
- Correct cathode product: Cu(s) (1 mark)
- Correct explanation for the fading blue color due to loss of Cu2+ ions (1 mark)

题目 22 · 選擇題

1 分

A student reacts $10.0\text{ g}$ of a mixture containing calcium carbonate, $\text{CaCO}_3$, and sand. Sand does not react with acids. The mixture is reacted with excess dilute hydrochloric acid, producing $1.2\text{ dm}^3$ of carbon dioxide gas measured at room temperature and pressure (r.t.p.). What is the percentage by mass of calcium carbonate in the mixture? [$M_r$ of $\text{CaCO}_3 = 100$; molar volume of gas at r.t.p. = $24\text{ dm}^3\text{/mol}$]

A.12%
B.24%
C.50%
D.83?

查看答案详解

解题

1. Find the moles of carbon dioxide gas produced: moles = $1.2\text{ dm}^3 / 24\text{ dm}^3\text{/mol} = 0.05\text{ mol}$. 2. Write the balanced chemical equation: $\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})$. The mole ratio of calcium carbonate to carbon dioxide is 1:1, so moles of calcium carbonate reacted = 0.05 mol. 3. Calculate the mass of calcium carbonate: mass = $0.05\text{ mol} \times 100\text{ g/mol} = 5.0\text{ g}$. 4. Calculate the percentage by mass: $(5.0\text{ g} / 10.0\text{ g}) \times 100 = 50\%$.

评分标准

Award 1 mark for correct selection of C. Correct calculation steps: moles of CO2 = 0.05 mol; mass of CaCO3 = 5.0 g; percentage by mass = 50%.

题目 23 · 選擇題

1 分

Aqueous copper(II) sulfate is electrolysed in two separate experiments. In Experiment 1, inert carbon (graphite) electrodes are used. In Experiment 2, copper electrodes are used. Which statement correctly describes the changes that occur during these electrolyses?

A.In Experiment 1, the pH of the electrolyte decreases. In Experiment 2, the blue colour of the electrolyte remains unchanged.
B.In Experiment 1, the pH of the electrolyte increases. In Experiment 2, the blue colour of the electrolyte fades.
C.In Experiment 1, the pH of the electrolyte remains unchanged. In Experiment 2, the blue colour of the electrolyte remains unchanged.
D.In Experiment 1, the pH of the electrolyte decreases. In Experiment 2, the blue colour of the electrolyte fades.

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解题

In Experiment 1, water is oxidized at the anode, releasing oxygen gas and leaving hydrogen ions behind. This increases the acidity of the solution and therefore the pH of the electrolyte decreases. In Experiment 2, copper is oxidized at the anode and copper is deposited at the cathode at the same rate. This means the concentration of blue copper(II) ions in the electrolyte remains constant, so the blue colour of the electrolyte remains unchanged.

评分标准

Award 1 mark for correct identification that the pH decreases in Experiment 1 and the blue colour remains unchanged in Experiment 2.

题目 24 · 選擇題

1 分

A reversible reaction is shown: $\text{A}(\text{g}) + 2\text{B}(\text{g}) \rightleftharpoons 2\text{C}(\text{g})$ where the forward reaction is exothermic. Which changes to temperature and pressure will both increase the equilibrium yield of $\text{C}(\text{g})$?

A.decrease temperature, decrease pressure
B.decrease temperature, increase pressure
C.increase temperature, decrease pressure
D.increase temperature, increase pressure

查看答案详解

解题

To increase the yield of C, the equilibrium must shift to the right. Since the forward reaction is exothermic, decreasing the temperature will favour the exothermic forward reaction. Since there are 3 moles of gas on the left side and 2 moles of gas on the right side, increasing the pressure will shift the equilibrium to the side with fewer gas moles (the right side). Therefore, decreasing temperature and increasing pressure will both increase the yield of C.

评分标准

Award 1 mark for correct identification that decreasing temperature and increasing pressure favour the forward reaction.

题目 25 · 選擇題

1 分

A sample of 4.4 g of a gaseous hydrocarbon occupies a volume of $2.4\text{ dm}^3$ at room temperature and pressure (r.t.p.).

Which hydrocarbon is it?

(Molar volume of a gas at r.t.p. = $24.0\text{ dm}^3/\text{mol}$)

A.methane, $CH_4$
B.ethane, $C_2H_6$
C.propane, $C_3H_8$
D.butane, $C_4H_{10}$

查看答案详解

解题

1. Calculate the number of moles of the hydrocarbon:
$\text{Moles} = \frac{\text{Volume}}{\text{Molar volume}} = \frac{2.4\text{ dm}^3}{24.0\text{ dm}^3/\text{mol}} = 0.1\text{ mol}$.

2. Calculate the relative molecular mass ($M_r$) of the hydrocarbon:
$M_r = \frac{\text{Mass}}{\text{Moles}} = \frac{4.4\text{ g}}{0.1\text{ mol}} = 44$.

3. Determine which hydrocarbon has a relative molecular mass of 44:
- Methane, $CH_4$: $12.0 + 4 \times 1.0 = 16.0$
- Ethane, $C_2H_6$: $2 \times 12.0 + 6 \times 1.0 = 30.0$
- Propane, $C_3H_8$: $3 \times 12.0 + 8 \times 1.0 = 44.0$
- Butane, $C_4H_{10}$: $4 \times 12.0 + 10 \times 1.0 = 58.0$

Therefore, the hydrocarbon is propane (C).

评分标准

Award 1 mark for selecting correct option C.

题目 26 · 選擇題

1 分

Concentrated aqueous sodium chloride is electrolysed using inert platinum electrodes.

Which row correctly identifies the product at the anode, the product at the cathode, and the change in pH of the electrolyte?

A.Anode: oxygen; Cathode: hydrogen; pH decreases
B.Anode: chlorine; Cathode: hydrogen; pH increases
C.Anode: chlorine; Cathode: sodium; pH increases
D.Anode: oxygen; Cathode: hydrogen; pH increases

查看答案详解

解题

During the electrolysis of concentrated aqueous sodium chloride:
- At the anode (+), chloride ions ($Cl^-$) are selectively discharged in preference to hydroxide ions ($OH^-$) due to their high concentration, forming chlorine gas.
- At the cathode (-), hydrogen ions ($H^+$) are selectively discharged in preference to sodium ions ($Na^+$) because hydrogen is lower in the reactivity series, forming hydrogen gas.
- As hydrogen ions and chloride ions are removed from the solution, sodium ions ($Na^+$) and hydroxide ions ($OH^-$) remain in the solution, forming sodium hydroxide. This causes the solution to become alkaline and the pH to increase.

Therefore, the correct row is B.

评分标准

Award 1 mark for selecting correct option B.

题目 27 · 選擇題

1 分

Sulfur dioxide reacts with oxygen to form sulfur trioxide in a reversible reaction:

$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \quad \Delta H = -197\text{ kJ/mol}$

Which change in conditions will increase the equilibrium yield of sulfur trioxide?

A.adding more vanadium(V) oxide catalyst
B.decreasing the temperature
C.decreasing the pressure
D.removing some oxygen

查看答案详解

解题

- Option A is incorrect: A catalyst increases the rate of both forward and reverse reactions equally, which helps to reach equilibrium faster but has no effect on the equilibrium yield.
- Option B is correct: The forward reaction is exothermic (indicated by $\Delta H = -197\text{ kJ/mol}$). Decreasing the temperature shifts the equilibrium position to the right (the exothermic direction) to release heat, thereby increasing the yield of $SO_3$.
- Option C is incorrect: Decreasing the pressure shifts the equilibrium to the left (the side with more moles of gas, 3 moles of reactants vs 2 moles of products), decreasing the yield of $SO_3$.
- Option D is incorrect: Removing a reactant like oxygen shifts the equilibrium to the left to replace the lost reactant, decreasing the yield.

评分标准

Award 1 mark for selecting correct option B.

题目 28 · 選擇題

1 分

A sample of $ 0.120\text{ g} $ of magnesium ribbon is reacted completely with excess dilute hydrochloric acid at room temperature and pressure (r.t.p.). The equation for the reaction is: $ \text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)} $ What is the volume of hydrogen gas, in $ \text{cm}^3 $, produced? [$ A_r(\text{Mg}) = 24.0 $; 1 mole of any gas occupies $ 24.0\text{ dm}^3 $ at r.t.p.]

A.1.20
B.12.0
C.120
D.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解题

First, calculate the number of moles of magnesium: $ \text{moles of Mg} = \frac{0.120\text{ g}}{24.0\text{ g/mol}} = 0.0050\text{ mol} $. According to the balanced equation, $ 1\text{ mol} $ of $ \text{Mg} $ produces $ 1\text{ mol} $ of $ \text{H}_2 $ gas. Thus, $ 0.0050\text{ mol} $ of $ \text{H}_2 $ gas is produced. The volume of $ \text{H}_2 $ gas at r.t.p. is $ 0.0050\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 0.120\text{ dm}^3 $. To convert this volume to $ \text{cm}^3 $, multiply by 1000: $ 0.120\text{ dm}^3 \times 1000 = 120\text{ cm}^3 $.

评分标准

1 mark for the correct answer C. Award 0 marks for incorrect options.

题目 29 · 選擇題

1 分

A 0.10 mol sample of a hydrocarbon is completely burned in oxygen. Under these conditions, 7.2 dm$^3$ of carbon dioxide gas (measured at r.t.p.) and 7.2 g of water are produced. What is the molecular formula of the hydrocarbon?

[Assume 1 mol of any gas occupies 24 dm$^3$ at r.t.p.; relative atomic masses: H = 1, O = 16]

A.$\text{C}_3\text{H}_6$
B.$\text{C}_3\text{H}_8$
C.$\text{C}_4\text{H}_8$
D.$\text{C}_4\text{H}_{10}$

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解题

1. Calculate the number of moles of carbon dioxide ($\text{CO}_2$) produced:
$\text{moles of CO}_2 = \frac{7.2\text{ dm}^3}{24\text{ dm}^3/\text{mol}} = 0.30\text{ mol}$

2. Determine the number of carbon atoms per molecule of hydrocarbon:
Since 0.10 mol of the hydrocarbon produces 0.30 mol of $\text{CO}_2$, 1.0 mol of the hydrocarbon must contain 3.0 mol of carbon atoms. Thus, there are 3 carbon atoms per molecule.

3. Calculate the number of moles of water ($\text{H}_2\text{O}$) produced:
$\text{moles of H}_2\text{O} = \frac{7.2\text{ g}}{18\text{ g}/\text{mol}} = 0.40\text{ mol}$

4. Determine the number of hydrogen atoms per molecule of hydrocarbon:
Each mole of $\text{H}_2\text{O}$ contains 2 moles of hydrogen atoms, so 0.40 mol of $\text{H}_2\text{O}$ contains $0.40 \times 2 = 0.80\text{ mol}$ of hydrogen atoms.
Since 0.10 mol of the hydrocarbon produces 0.80 mol of hydrogen atoms, 1.0 mol of the hydrocarbon must contain 8.0 mol of hydrogen atoms. Thus, there are 8 hydrogen atoms per molecule.

Therefore, the molecular formula is $\text{C}_3\text{H}_8$.

评分标准

- 1 mark for the correct option B.
- Award marks for finding the moles of carbon dioxide (0.30 mol) and water (0.40 mol), then deducing the mole ratio of C:H as 3:8.

题目 30 · 選擇題

1 分

The reversible gaseous reaction shown is at dynamic equilibrium.

$2\text{A}(\text{g}) + \text{B}(\text{g}) \rightleftharpoons 2\text{C}(\text{g})$ $\Delta H = -115\text{ kJ/mol}$

Which pair of changes will both shift the position of equilibrium to the right and increase the rate of the forward reaction?

A.Decreasing the temperature and adding a catalyst
B.Increasing the temperature and increasing the pressure
C.Increasing the pressure and increasing the concentration of $\text{A}$
D.Decreasing the pressure and decreasing the concentration of $\text{A}$

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解题

Let us evaluate the effects of the changes in the correct option (C):
1. Increasing the pressure:
- Since there are 3 moles of gas on the left-hand side and 2 moles of gas on the right-hand side, an increase in pressure shifts the equilibrium to the side with fewer gas moles (the right-hand side).
- It also increases the concentration of gas particles, leading to more frequent collisions and a faster rate of reaction.

2. Increasing the concentration of reactant $\text{A}$:
- According to Le Chatelier's principle, the equilibrium position shifts to the right to consume the added reactant.
- It also increases the collision frequency between reactant particles, increasing the rate of reaction.

Thus, both changes in option C shift the equilibrium to the right and increase the reaction rate.

Other options analysis:
- Option A: Decreasing the temperature shifts the equilibrium to the right (exothermic), but decreases the reaction rate. Adding a catalyst increases the rate but does not affect the equilibrium position.
- Option B: Increasing the temperature increases the rate, but shifts the equilibrium to the left.
- Option D: Decreasing the pressure and reactant concentration both shift the equilibrium to the left and decrease the reaction rate.

评分标准

- 1 mark for the correct option C.
- Under physical chemistry principles, check both the thermodynamic equilibrium position (Le Chatelier's principle) and the kinetic rate (collision theory) for each proposed change.

题目 31 · 選擇題

1 分

An aqueous solution is electrolysed using inert carbon electrodes.

- At the anode, a colourless gas is evolved that relights a glowing splint.
- At the cathode, a pink-brown metal is deposited.

Which substance is being electrolysed?

A.dilute hydrochloric acid
B.concentrated aqueous sodium chloride
C.aqueous copper(II) sulfate
D.aqueous copper(II) chloride

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解题

During the electrolysis of aqueous copper(II) sulfate ($\text{CuSO}_4$(aq)) using carbon electrodes:

1. At the anode (+):
Both hydroxide ions ($\text{OH}^-$) and sulfate ions ($\text{SO}_4^{2-}$) are attracted to the anode. $\text{OH}^-$ ions are more easily oxidised than $\text{SO}_4^{2-}$ ions, producing oxygen gas:
$4\text{OH}^-(\text{aq}) \rightarrow \text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + 4\text{e}^-$
Oxygen is a colourless gas that relights a glowing splint.

2. At the cathode (-):
Both copper(II) ions ($\text{Cu}^{2+}$) and hydrogen ions ($\text{H}^+$) are attracted to the cathode. Copper is less reactive than hydrogen in the reactivity series, so $\text{Cu}^{2+}$ ions are preferentially reduced to form copper metal:
\(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})
This forms a pink-brown metallic deposit on the cathode.

Therefore, the correct substance is aqueous copper(II) sulfate (option C).

评分标准

- 1 mark for the correct option C.
- Option A is incorrect as hydrogen gas (colorless, flammable) is produced at the cathode.
- Options B and D are incorrect as chlorine gas (pale green, pungent) is produced at the anode.

题目 32 · 選擇題

1 分

A 10 cm$^{3}$ sample of a gaseous hydrocarbon was completely combusted in 100 cm$^{3}$ of oxygen (an excess). After cooling to room temperature and pressure, the total volume of gas remaining was 85 cm$^{3}$. After passing this remaining gas through concentrated aqueous sodium hydroxide, the volume of gas decreased to 55 cm$^{3}$. What is the formula of the hydrocarbon?

A.$C_{2}H_{4}$
B.$C_{3}H_{6}$
C.$C_{3}H_{8}$
D.$C_{4}H_{8}$

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解题

1. The decrease in gas volume when passed through aqueous NaOH is due to the absorption of carbon dioxide: $\text{Volume of } CO_{2} = 85\text{ cm}^{3} - 55\text{ cm}^{3} = 30\text{ cm}^{3}$. Since 10 cm$^{3}$ of hydrocarbon produced 30 cm$^{3}$ of $CO_{2}$, 1 mole of hydrocarbon contains 3 moles of carbon atoms (so $x = 3$).
2. The remaining 55 cm$^{3}$ of gas is unreacted oxygen. Since we started with 100 cm$^{3}$ of oxygen, the volume of oxygen reacted is $100\text{ cm}^{3} - 55\text{ cm}^{3} = 45\text{ cm}^{3}$.
3. For the complete combustion of a hydrocarbon $C_{3}H_{y}$: $C_{3}H_{y} + (3 + y/4)O_{2} \rightarrow 3CO_{2} + (y/2)H_{2}O$. The ratio of hydrocarbon to reacted oxygen is 10 : 45, which is 1 : 4.5. Therefore, $3 + y/4 = 4.5 \Rightarrow y/4 = 1.5 \Rightarrow y = 6$. The molecular formula is $C_{3}H_{6}$.

评分标准

Award 1 mark for the correct option B.
- Identify that the volume of carbon dioxide produced is 30 cm$^{3}$ (indicating 3 carbon atoms per molecule).
- Identify that 45 cm$^{3}$ of oxygen reacted.
- Deduce the formula of the hydrocarbon as $C_{3}H_{6}$ based on the combustion stoichiometry.

题目 33 · 選擇題

1 分

An electrolysis experiment is set up using inert platinum electrodes and an aqueous solution containing a mixture of copper(II) sulfate and dilute sulfuric acid. Which statement correctly describes the observations at the electrodes during the first few minutes of electrolysis?

A.A pink solid deposits at the anode, and a colourless gas is produced at the cathode.
B.A pink solid deposits at the cathode, and a colourless gas is produced at the anode.
C.A colourless gas is produced at both electrodes.
D.A pink solid deposits at the cathode, and a pale green gas is produced at the anode.

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解题

During the electrolysis of a mixture of aqueous copper(II) sulfate and sulfuric acid:
1. At the cathode (negative electrode), both $Cu^{2+}$ and $H^{+}$ ions are attracted. Since copper is lower than hydrogen in the reactivity series, $Cu^{2+}$ is preferentially discharged and reduced to copper metal: $Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)$, forming a pink/brown solid.
2. At the anode (positive electrode), both $SO_{4}^{2-}$ and $OH^{-}$ (from water) ions are attracted. $OH^{-}$ is preferentially discharged and oxidised to oxygen gas: $4OH^{-}(aq) \rightarrow O_{2}(g) + 2H_{2}O(l) + 4e^{-}$, which is seen as a colourless gas.
Therefore, option B is correct.

评分标准

Award 1 mark for the correct option B.
- Identify that a pink solid (copper metal) deposits at the cathode.
- Identify that a colourless gas (oxygen) is produced at the anode.

题目 34 · 選擇題

1 分

The equation for a reversible gas-phase reaction is shown.

$2X(g) + Y(g) \rightleftharpoons 2Z(g) \quad \Delta H = -115\text{ kJ/mol}$

Which changes in temperature and pressure will both increase the equilibrium yield of $Z$?

A.decrease in temperature and decrease in pressure
B.decrease in temperature and increase in pressure
C.increase in temperature and decrease in pressure
D.increase in temperature and increase in pressure

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解题

1. Temperature: The forward reaction is exothermic ($\Delta H = -115\text{ kJ/mol}$). According to Le Chatelier's principle, decreasing the temperature shifts the equilibrium in the exothermic direction (to the right) to oppose the change, thereby increasing the yield of $Z$.
2. Pressure: There are 3 moles of gaseous reactants on the left-hand side and 2 moles of gaseous products on the right-hand side. Increasing the pressure shifts the equilibrium towards the side with fewer moles of gas (to the right) to decrease the pressure, increasing the yield of $Z$.
Therefore, both a decrease in temperature and an increase in pressure will increase the yield of $Z$.

评分标准

Award 1 mark for the correct option B.
- Correctly apply Le Chatelier's principle to predict that decreasing temperature increases the yield of an exothermic reaction.
- Correctly apply Le Chatelier's principle to predict that increasing pressure increases the yield of a reaction with fewer moles of gas on the product side.

题目 35 · 選擇題

1 分

Two electrolysis experiments are set up using inert platinum electrodes. Experiment 1 involves the electrolysis of dilute sulfuric acid. Experiment 2 involves the electrolysis of concentrated aqueous sodium chloride. Which row correctly identifies the products formed at each electrode?

A.Experiment 1: anode = oxygen, cathode = hydrogen; Experiment 2: anode = chlorine, cathode = hydrogen
B.Experiment 1: anode = oxygen, cathode = hydrogen; Experiment 2: anode = oxygen, cathode = sodium
C.Experiment 1: anode = hydrogen, cathode = oxygen; Experiment 2: anode = chlorine, cathode = hydrogen
D.Experiment 1: anode = hydrogen, cathode = oxygen; Experiment 2: anode = oxygen, cathode = sodium

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解题

In Experiment 1 (electrolysis of dilute sulfuric acid):
- At the anode (positive electrode), $OH^-$– ions from water are preferentially discharged over $SO_4^{2-}$– ions to produce oxygen gas.
- At the cathode (negative electrode), $H^+$– ions are discharged to produce hydrogen gas.

In Experiment 2 (electrolysis of concentrated aqueous sodium chloride):
- At the anode (positive electrode), $Cl^-$– ions are in high concentration and are preferentially discharged over $OH^-$– ions to produce chlorine gas.
- At the cathode (negative electrode), $H^+$– ions are preferentially discharged over $Na^+$– ions because hydrogen is lower in the reactivity series, producing hydrogen gas.

评分标准

Award 1 mark for selecting the correct option A.
- Correctly identify anode and cathode products for dilute sulfuric acid (oxygen and hydrogen respectively).
- Correctly identify anode and cathode products for concentrated aqueous sodium chloride (chlorine and hydrogen respectively).

Paper 3 (Core - Written Theory)

Answer all structured and short-answer questions in the spaces provided on the question paper.

(a)(i) [1 mark] Iron(II) / $Fe^{2+}$. Reject 'iron' / $Fe^{3+}$.
(a)(ii) [2 marks]
- Reagent: Aqueous ammonia [1]
- Observation: Green precipitate, insoluble in excess [1]
(b)(i) [1 mark] Sulfate (ion). Reject 'sulfite'.
(b)(ii) [1 mark] $SO_4^{2-}$. Charges must be correct.
(c) [4 marks]
- (i) Turns blue [1]
- (ii) Bleached / turns white [1] (Accept turns red then bleaches)
- (iii) Glowing splint [1] (Reject 'burning/lighted splint')
- (iv) Turns cloudy / milky / white precipitate [1]
(d) [1 mark] $FeSO_4$

Paper 4 (Extended - Written Theory)

Answer all structured and short-answer questions in the spaces provided on the question paper.

Part (a) [4 marks]:
- (i) Correct structural formula of propan-1-ol with all atoms and bonds shown explicitly (including the O-H bond) [2 marks]. (Deduct 1 mark if O-H bond is written simply as -OH without a single line bond between O and H).
- (ii) Acidified potassium manganate(VII) or acidified potassium dichromate(VI) [1 mark]. Heat / warm / reflux [1 mark].

Part (b) [3 marks]:
- Correct formulas of reactants and products: $\text{CH}_3\text{CH}_2\text{COOH}$, $\text{Mg}$, $(\text{CH}_3\text{CH}_2\text{COO})_2\text{Mg}$, and $\text{H}_2$ [1 mark].
- Equation balanced correctly: coefficient of 2 in front of propanoic acid [1 mark].
- All correct state symbols: $\text{(aq)}$, $\text{(s)}$, $\text{(aq)}$, $\text{(g)}$ [1 mark].

Part (c) [6 marks]:
- (i) Esters [1 mark].
- (ii) Concentrated sulfuric acid [1 mark] (Accept: $\text{H}_2\text{SO}_4$, reject dilute sulfuric acid).
- (iii) IUPAC name: ethyl propanoate [1 mark].
- Correct structure showing all bonds, including the ester linkage $\text{C}(=\text{O})-\text{O}$ [3 marks]. (Deduct 1 mark for each missing bond or atom, e.g. writing $-\text{CH}_3$ instead of showing all H-C bonds).

Paper 5 (Practical Test)

Carry out experimental tasks, record raw results to the required precision, construct plots, and complete the written qualitative/planning sections.

Part (a): [6 marks]
- 1 mark for specifying a fixed mass of baking powder measured with a balance.
- 1 mark for specifying a fixed volume of hydrochloric acid measured with a measuring cylinder.
- 1 mark for describing the connection to a gas syringe.
- 1 mark for starting the stopwatch immediately upon mixing.
- 1 mark for recording volume at specified time intervals (e.g., every 10 s) OR measuring the time to collect a fixed volume of gas.
- 1 mark for repeating the experiment for the other two brands AND identifying controlled variables (e.g., temperature, concentration of acid).

Part (b): [2 marks]
- 1 mark for stating that carbon dioxide is soluble in water / dissolves in water.
- 1 mark for explaining that this results in a lower/underestimated volume of gas being measured if collected over water.

Part (c): [2 marks]
- 1 mark for explaining that temperature affects the rate of reaction / particle collision energy.
- 1 mark for stating that keeping it constant ensures a fair test / makes sure only the brand is being tested.

Part (d): [3 marks]
- 1 mark for stating that both curves start at the origin and level off at the same final volume of gas.
- 1 mark for stating that the fast-reacting brand has a steeper initial slope/gradient.
- 1 mark for stating that the fast-reacting brand reaches the plateau (ends) sooner than the slow-reacting brand.

Paper 6 (Alternative to Practical)

Answer written questions testing experimental design, observation, identification tables, graphing, and planning parameters.

(a) Burette / pipette / measuring cylinder [1 mark]

(b) To remove the magnesium oxide layer / oxide coating [1 mark]

(c) Bubbling / effervescence stops OR magnesium ribbon completely dissolves [1 mark]

(d) Planning (Max 7 marks):
- MP1 (Dilution Method): Clearly describes how to make at least three different concentrations of acid keeping the total volume at 25 cm3 (e.g., giving specific volumes of acid and water such as 25 cm3 acid + 0 cm3 water, 12.5 cm3 acid + 12.5 cm3 water, etc.) [1 mark]
- MP2 (Insulation): Uses a polystyrene cup / insulated container (placed in a beaker for stability) [1 mark]
- MP3 (Initial Temperature): Measures and records the initial temperature of the acid before adding magnesium [1 mark]
- MP4 (Control Variable): Adds a constant mass / length of magnesium ribbon [1 mark]
- MP5 (Maximum Temperature): Stirs the mixture and measures the maximum temperature reached [1 mark]
- MP6 (Calculation): Calculates temperature rise by subtracting initial temperature from maximum temperature [1 mark]
- MP7 (Reliability and Analysis): Repeats the experiments (to check reliability/find average) AND plots a graph of temperature rise against concentration [1 mark]

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