An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 AQA GCSE Chemistry 8462 paper. Not affiliated with or reproduced from AQA.
卷一 Foundation
Answer all questions. Show your working in all calculations.
10 題目 · 100 分
題目 1 · 選擇題
10 分
A student wants to prepare a pure, dry sample of crystals of soluble copper(II) sulfate. They react copper(II) oxide (an insoluble base) with dilute sulfuric acid. Which of the following describes the correct, complete sequence of steps to safely obtain the pure, dry crystals?
A.Add excess copper(II) oxide to the acid, warm the mixture, filter to remove unreacted oxide, heat the filtrate to evaporate some water, leave to crystallise, and dry the crystals with filter paper.
B.Add excess sulfuric acid to copper(II) oxide, warm the mixture, evaporate all the water until dry, filter the remaining crystals, and wash with hot water.
C.Add equal amounts of copper(II) oxide and acid, boil the mixture until it turns green, filter, discard the filtrate, and dry the residue in an oven.
D.Add excess copper(II) oxide to the acid, filter the mixture immediately, discard the filtrate, heat the residue until it melts, and then cool to form crystals.
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解題
The correct answer is a. To prepare a soluble salt from an insoluble base, we add excess base (copper(II) oxide) to ensure all the acid is completely neutralized. Heating the mixture gently increases the rate of reaction. Filtration is then carried out to remove the unreacted, excess copper(II) oxide. The filtrate (copper(II) sulfate solution) is heated in an evaporating basin to remove some of the water until the crystallization point is reached. It is then left to cool so that crystals can form slowly (as solubility decreases with temperature). Finally, the crystals are separated and gently dried using filter paper.
評分準則
Award up to 10 marks as follows: 2 marks for stating that excess copper(II) oxide is added to ensure all acid is fully reacted; 2 marks for heating the acid/mixture to increase the reaction rate; 2 marks for filtering the mixture to remove the unreacted insoluble copper(II) oxide; 2 marks for heating the filtrate to evaporate water partially until crystallization point (do not accept evaporating to dryness); 2 marks for leaving to crystallise and drying the crystals with filter paper.
題目 2 · 選擇題
10 分
A student investigates the rate of reaction between sodium thiosulfate solution and dilute hydrochloric acid. They measure the time taken for a cross drawn on paper under the flask to disappear. If the student increases the concentration of the sodium thiosulfate solution, how and why does the rate of reaction change?
A.The rate increases because there are more reacting particles per unit volume, which increases the frequency of successful collisions.
B.The rate decreases because the particles have less kinetic energy and collide less frequently.
C.The rate increases because the activation energy of the reaction is lowered, allowing more particles to react.
D.The rate remains the same because concentration only affects the yield of the sulfur precipitate, not the speed of the reaction.
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解題
The correct answer is a. Increasing the concentration of reactants in a solution means there are more reacting particles in the same volume (per unit volume). Because the particles are packed more closely together, they will collide with each other more frequently. Consequently, the frequency of successful collisions increases, which increases the overall rate of reaction.
評分準則
Award up to 10 marks as follows: 2 marks for identifying that the rate of reaction increases; 2 marks for explaining that higher concentration means more reacting particles in a given volume; 2 marks for stating that the particles are closer together; 2 marks for explaining that the frequency of collisions increases; 2 marks for stating that the frequency of successful collisions increases.
題目 3 · 選擇題
10 分
An atom of an isotope of chlorine is represented as \(^{37}_{17}\text{Cl}\). Which option correctly identifies the number of protons, neutrons, and electrons in this neutral atom, and describes how it differs from an atom of chlorine-35 (\(^{35}_{17}\text{Cl}\))?
A.It has 17 protons, 20 neutrons, and 17 electrons. It differs from chlorine-35 because chlorine-35 has 18 neutrons.
B.It has 17 protons, 17 neutrons, and 20 electrons. It differs from chlorine-35 because chlorine-35 has 2 fewer protons.
C.It has 37 protons, 17 neutrons, and 37 electrons. It differs from chlorine-35 because chlorine-35 has 35 protons.
D.It has 20 protons, 17 neutrons, and 20 electrons. It differs from chlorine-35 because chlorine-35 has the same number of neutrons but different protons.
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解題
The correct answer is a. For the chlorine isotope \(^{37}_{17}\text{Cl}\), the atomic number (the subscript, 17) represents the number of protons. Since the atom is neutral, the number of electrons is also equal to 17. The mass number (the superscript, 37) is the sum of protons and neutrons, so the number of neutrons is calculated as 37 - 17 = 20. For chlorine-35 (\(^{35}_{17}\text{Cl}\)), the number of neutrons is 35 - 17 = 18. Both isotopes have the same number of protons (17), meaning they represent the same element but have different physical masses due to the differing number of neutrons.
評分準則
Award up to 10 marks as follows: 2 marks for identifying that the number of protons is 17; 2 marks for identifying that the number of electrons in a neutral atom is 17; 2 marks for calculating that the number of neutrons in chlorine-37 is 37 - 17 = 20; 2 marks for calculating that the number of neutrons in chlorine-35 is 35 - 17 = 18; 2 marks for explaining that isotopes have the same proton number but different neutron numbers.
題目 4 · 選擇題
10 分
In the UK, potable (drinking) water is produced from fresh water sources like lakes and rivers. In arid countries, potable water is often produced from seawater. Which option correctly identifies the processes used to treat fresh water in the UK, and the main disadvantage of obtaining potable water from seawater by distillation?
A.Fresh water treatment: filtration and sterilisation (using chlorine, ozone, or UV light). Disadvantage of distillation: requires high amounts of energy, making it expensive.
B.Fresh water treatment: distillation and electrolysis. Disadvantage of distillation: does not remove dissolved toxic salts.
C.Fresh water treatment: chromatography and sedimentation. Disadvantage of distillation: produces water that is too acidic to drink safely.
D.Fresh water treatment: chlorination and crystallization. Disadvantage of distillation: removes all mineral ions, making the water toxic to human cells.
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解題
The correct answer is a. Fresh water treatment in the UK involves passing the water through filter beds (filtration) to remove insoluble suspended solids, and then sterilising it (using chlorine, ozone, or UV light) to kill harmful bacteria and microbes. Distillation of seawater involves heating the water to boil it, which requires a huge amount of thermal energy. This high energy requirement makes distillation very expensive and unsustainable unless cheap energy sources are available.
評分準則
Award up to 10 marks as follows: 2 marks for identifying filtration as the step to remove insoluble solids; 2 marks for identifying sterilisation as the step to destroy microbes; 2 marks for naming a correct sterilising agent (such as chlorine, ozone, or UV light); 2 marks for identifying distillation as a method to desalinate seawater; 2 marks for explaining that distillation is expensive due to its high thermal energy demands.
題目 5 · 選擇題
10 分
A student adds dilute hydrochloric acid to solid calcium carbonate in a flask. Which option correctly states the balanced chemical equation for this reaction and the chemical test used to identify the gas produced?
A.Equation: \(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\). Test: Bubble the gas through limewater; it turns cloudy.
B.Equation: \(\text{CaCO}_3(\text{s}) + \text{HCl}(\text{aq}) \rightarrow \text{CaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\). Test: Hold a burning splint near the gas; it makes a squeaky pop.
C.Equation: \(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2(\text{g})\). Test: Hold a glowing splint in the gas; it relights.
D.Equation: \(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{Ca(HCl)}_2(\text{aq}) + \text{CO}_3(\text{g})\). Test: Expose the gas to damp blue litmus paper; it turns white.
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解題
The correct answer is a. When an acid reacts with a metal carbonate, the products are a salt, water, and carbon dioxide. Hydrochloric acid reacting with calcium carbonate produces calcium chloride, water, and carbon dioxide. The balanced equation is \(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\). The gas produced is carbon dioxide, which is tested by bubbling it through limewater; the limewater turns cloudy (milky) due to the formation of a white precipitate of calcium carbonate.
評分準則
Award up to 10 marks as follows: 2 marks for identifying the correct products (calcium chloride, water, and carbon dioxide); 2 marks for using the correct chemical formula for calcium chloride (\(\text{CaCl}_2\)); 2 marks for balancing the equation correctly with a 1:2:1:1:1 ratio; 2 marks for identifying the gas produced as carbon dioxide; 2 marks for describing the correct gas test (bubbled through limewater, which turns cloudy/milky).
題目 6 · practical-and-calculation
10 分
A student investigated the preparation of pure, dry crystals of copper sulfate. The student added excess black copper oxide powder (\(\text{CuO}\)) to hot dilute sulfuric acid (\(\text{H}_2\text{SO}_4\)).
(a) Explain why the student added excess copper oxide rather than excess sulfuric acid. [1 mark]
(b) State how the student can remove the unreacted copper oxide from the reaction mixture. [1 mark]
(c) Describe how the student can obtain dry, pure crystals of copper sulfate from the copper sulfate solution obtained. [3 marks]
(d) In a different experiment, the theoretical yield of copper sulfate crystals is \(12.5\text{ g}\). The student actually obtained \(9.0\text{ g}\) of crystals. Calculate the percentage yield of copper sulfate. Show your working. [3 marks]
(e) Suggest one safety precaution the student should take when heating the sulfuric acid, and explain why this is necessary. [2 marks]
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解題
(a) Excess copper oxide is added to ensure that all of the sulfuric acid reacts completely, so that no unreacted acid remains in the final product.
(b) The unreacted copper oxide can be removed by filtration (using a filter funnel and filter paper).
(c) To obtain dry, pure crystals: 1. Heat the copper sulfate solution in an evaporating basin to evaporate some water until the crystallization point is reached. 2. Leave the solution to cool slowly so that crystals form. 3. Filter the crystals to separate them from the remaining liquid, and dry them by patting with filter paper or leaving them in a warm oven.
(e) Safety precaution: Wear safety goggles because dilute sulfuric acid is corrosive/an irritant and can damage eyes. Alternatively, heat the acid gently using a water bath rather than a direct Bunsen burner flame to prevent the acid from boiling over or spitting.
評分準則
(a) [1 mark]: to ensure all sulfuric acid reacts / is neutralised / is used up.
(b) [1 mark]: filtration / filtering.
(c) [3 marks]: 1 mark for heating/evaporating some water (to crystallization point); 1 mark for leaving to cool / crystallise; 1 mark for filtering crystals AND drying (with filter paper / warm oven).
(d) [3 marks]: 1 mark for correct substitution: \(9.0 / 12.5\); 1 mark for multiplying by 100; 1 mark for correct final answer of 72 (accept 72%).
(e) [2 marks]: 1 mark for suitable precaution (e.g., wear safety goggles / heat gently / use water bath); 1 mark for correct linked reason (e.g., acid is corrosive / prevents spitting or boiling over).
題目 7 · practical-and-calculation
10 分
A student investigated the rate of reaction between calcium carbonate lumps and dilute hydrochloric acid. The reaction produces calcium chloride, water, and carbon dioxide gas.
(a) Describe a chemical test to show that the gas produced is carbon dioxide. Include the result of the test. [2 marks]
(b) The student measured the volume of gas collected every 10 seconds. Explain why the rate of reaction decreases as the reaction proceeds. [2 marks]
(c) In the first 20 seconds, the student collected \(30\text{ cm}^3\) of carbon dioxide gas. Calculate the mean rate of reaction in this time interval. State the units. [3 marks]
(d) The student repeated the experiment using the same mass of calcium carbonate, but as a fine powder instead of large lumps. Describe and explain, in terms of particles, the effect of using powder on the rate of reaction. [3 marks]
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解題
(a) Test: Bubble the gas through limewater. Result: The limewater turns cloudy/milky.
(b) The rate of reaction decreases because the reactant particles (acid particles and calcium carbonate) are being used up. This decreases the concentration of acid, leading to less frequent collisions between reactant particles per second.
(c) Mean rate of reaction calculation: \(\text{Mean Rate} = \frac{\text{Volume of Gas}}{\text{Time}} = \frac{30\text{ cm}^3}{20\text{ s}} = 1.5\text{ cm}^3/\text{s}\).
(d) Effect: The rate of reaction increases. Explanation: Fine powder has a larger surface area (to volume ratio) than lumps. This means there are more exposed particles, leading to more frequent collisions (more collisions per second) between the acid particles and calcium carbonate.
評分準則
(a) [2 marks]: 1 mark for bubbling gas through limewater; 1 mark for limewater turning cloudy/milky.
(b) [2 marks]: 1 mark for saying reactant particles / acid is used up (or concentration decreases); 1 mark for fewer collisions per unit time / less frequent collisions.
(c) [3 marks]: 1 mark for dividing volume by time (\(30 / 20\)); 1 mark for correct value: 1.5; 1 mark for correct unit: \(\text{cm}^3/\text{s}\) (accept cm3/s or \(\text{cm}^3\text{ s}^{-1}\)).
(d) [3 marks]: 1 mark for stating that rate increases; 1 mark for stating that powder has a larger surface area; 1 mark for stating there are more frequent collisions / more collisions per second.
題目 8 · practical-and-calculation
10 分
Salty water can be processed to produce potable water.
(a) Define the term 'potable water'. [2 marks]
(b) Pure water can be obtained from salty water by distillation. Describe how distillation separates pure water from a salt solution. You should refer to state changes in your answer. [3 marks]
(c) A student analyzed a \(100\text{ cm}^3\) sample of river water. The student evaporated all the water from the sample and weighed the dry residue left behind. The mass of the dry residue was \(0.08\text{ g}\). Calculate the concentration of dissolved solids in the river water in \(\text{g/dm}^3\). Remember: \(1\text{ dm}^3 = 1000\text{ cm}^3\). [3 marks]
(d) Before water is piped to homes, it is treated. State one substance added to water during treatment to sterilize it, and explain why sterilization is necessary. [2 marks]
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解題
(a) Potable water is water that is safe to drink. It is not pure water because it contains low levels of dissolved salts and microbes, but these levels are safe.
(b) Distillation process: 1. The salty water is heated, causing the water to evaporate (change state from liquid to gas) to form steam. 2. The salt is left behind in the flask because it has a much higher boiling point than water. 3. The steam enters a condenser where it cools and condenses (changes state from gas to liquid) to form pure liquid water.
(d) Substance: Chlorine (or ozone / UV light). Reason: To sterilize the water by killing harmful bacteria, microbes, or pathogens, making it safe to drink.
評分準則
(a) [2 marks]: 1 mark for water that is safe to drink; 1 mark for mentioning it contains low/safe levels of dissolved salts or microbes.
(b) [3 marks]: 1 mark for heating/evaporating the water to form steam; 1 mark for stating that salt is left behind; 1 mark for cooling/condensing the steam back into liquid water.
(c) [3 marks]: 1 mark for converting volume to \(\text{dm}^3\) (\(100 / 1000 = 0.1\)); 1 mark for dividing mass by volume (\(0.08 / 0.1\)); 1 mark for correct final value: 0.8 (accept 0.8 with unit \(\text{g/dm}^3\)).
(d) [2 marks]: 1 mark for identifying a sterilizing agent (chlorine / ozone / UV); 1 mark for linked explanation (to kill bacteria / microbes / pathogens).
題目 9 · practical-and-calculation
10 分
Atoms are made of protons, neutrons, and electrons.
(a) Complete the sentences to describe the properties of subatomic particles: (i) The relative electrical charge of an electron is _______. [1 mark] (ii) The relative mass of a neutron is _______. [1 mark]
(b) An atom of lithium has a mass number of 7 and an atomic number of 3. State the number of protons, neutrons, and electrons in this lithium atom. [3 marks] - Number of protons: _______ - Number of neutrons: _______ - Number of electrons: _______
(c) Neon has two main isotopes: Neon-20 (\(90\%\) abundance) and Neon-22 (\(10\%\) abundance). Calculate the relative atomic mass (\(A_r\)) of this sample of neon. Show your working. [3 marks]
(d) Explain why isotopes of the same element have the same chemical properties. [2 marks]
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解題
(a) (i) The relative charge of an electron is \(-1\). (ii) The relative mass of a neutron is \(1\).
(b) Lithium atom with atomic number 3 and mass number 7: Number of protons = 3 (equal to atomic number). Number of electrons = 3 (equal to number of protons in a neutral atom). Number of neutrons = \(7 - 3 = 4\) (mass number minus atomic number).
(d) Chemical properties depend on the electronic structure. Isotopes of the same element have the same number of electrons (and the same outer shell electronic configuration), which means they react in the exact same way.
評分準則
(a) [2 marks]: 1 mark for (i) -1 (accept minus 1); 1 mark for (ii) 1.
(b) [3 marks]: 1 mark for Protons = 3; 1 mark for Neutrons = 4; 1 mark for Electrons = 3.
(c) [3 marks]: 1 mark for setting up the equation: \((20 \times 90) + (22 \times 10)\); 1 mark for dividing by 100; 1 mark for correct final calculation: 20.2 (award 3 marks for 20.2 alone).
(d) [2 marks]: 1 mark for stating they have the same number of electrons / same electronic arrangement; 1 mark for stating chemical reactions/properties depend on outer electrons.
題目 10 · practical-and-calculation
10 分
A student investigated the temperature change when different volumes of sodium hydroxide solution are added to dilute hydrochloric acid. The student added \(5\text{ cm}^3\) portions of sodium hydroxide solution to \(25\text{ cm}^3\) of hydrochloric acid in a polystyrene cup, measuring the highest temperature reached after each addition.
(a) Explain why a polystyrene cup is used instead of a glass beaker. [2 marks]
(b) In one trial, the initial temperature of the acid was \(18.4\text{ }^\circ\text{C}\). After adding sodium hydroxide, the maximum temperature reached was \(24.9\text{ }^\circ\text{C}\). Calculate the temperature increase. [1 mark]
(c) The student repeated the experiment under identical conditions. The temperature increases recorded for one specific addition were: - Trial 1: \(6.4\text{ }^\circ\text{C}\) - Trial 2: \(6.6\text{ }^\circ\text{C}\) - Trial 3: \(3.2\text{ }^\circ\text{C}\) - Trial 4: \(6.5\text{ }^\circ\text{C}\) (i) Identify the anomalous result. [1 mark] (ii) Calculate the mean temperature increase, ignoring the anomalous result. Show your working. [2 marks]
(d) Write a word equation for the reaction between hydrochloric acid and sodium hydroxide. [2 marks]
(e) What type of reaction takes place when an acid reacts with an alkali? [1 mark]
(f) State the ion present in all acidic solutions that reacts with hydroxide ions (\(\text{OH}^-\)) from the alkali. [1 mark]
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解題
(a) A polystyrene cup is a good thermal insulator, which reduces heat loss to the surroundings, making the temperature measurement more accurate.
(c) (i) The anomalous result is Trial 3 (\(3.2\text{ }^\circ\text{C}\)), as it is much lower than the other three trials. (ii) Mean = \frac{6.4 + 6.6 + 6.5}{3} = \frac{19.5}{3} = 6.5\text{ }^\circ\text{C}.
(d) The word equation is: hydrochloric acid + sodium hydroxide \(\rightarrow\) sodium chloride + water.
(e) The reaction is neutralisation (accept exothermic).
(f) The hydrogen ion (\(\text{H}^+\)) is present in all acidic solutions.
評分準則
(a) [2 marks]: 1 mark for polystyrene being an insulator / poor conductor of heat; 1 mark for reducing heat loss (to surroundings).
(b) [1 mark]: 1 mark for \(6.5\text{ }^\circ\text{C}\) (accept 6.5).
(c) [3 marks total]: (i) 1 mark for identifying Trial 3 / 3.2. (ii) 2 marks: 1 mark for summing the remaining trials and dividing by 3: \((6.4 + 6.6 + 6.5) / 3\) (or \(19.5 / 3\)); 1 mark for correct mean value of 6.5.
(d) [2 marks]: 1 mark for correct reactants (hydrochloric acid + sodium hydroxide); 1 mark for correct products (sodium chloride + water).
(e) [1 mark]: 1 mark for neutralisation / neutralization (accept exothermic).
(f) [1 mark]: 1 mark for hydrogen ion / \(\text{H}^+\).
卷二 Foundation
Answer all questions. You must have a scientific calculator and periodic table.
10 題目 · 100 分
題目 1 · Recall and data plotting
10 分
A student investigates the rate of reaction between calcium carbonate and dilute hydrochloric acid. They measure the volume of carbon dioxide gas produced every 10 seconds. The table below shows their results:
a) Recall the chemical test for carbon dioxide gas and describe the positive result. [2 marks] b) State the independent variable and the dependent variable in this experiment. [2 marks] c) The student plots these results on a grid with 'Time in seconds' on the x-axis and 'Volume of gas in \(\text{cm}^3\)' on the y-axis. i) Describe the curve of best fit that should be drawn through these points. [2 marks] ii) Use the data in the table to calculate the mean rate of reaction between 0 and 30 seconds. Include the correct unit. [3 marks] iii) Explain why the volume of gas does not increase after 50 seconds. [1 mark]
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解題
a) Bubble the gas through limewater. The positive result is that the limewater turns cloudy or milky.
b) The independent variable is Time (s). The dependent variable is the Volume of gas (\(\text{cm}^3\)).
c) i) A smooth curve of best fit should be drawn starting at \((0,0)\) that curves upwards with a decreasing gradient and levels off to become completely horizontal from 50 seconds onwards.
ii) \(\text{Mean rate} = \frac{\text{Volume of gas produced}}{\text{Time taken}} = \frac{42\text{ cm}^3}{30\text{ s}} = 1.4\text{ cm}^3/\text{s}\).
iii) The reaction has stopped because one of the reactants (calcium carbonate or hydrochloric acid) has been completely used up.
評分準則
a) 1 mark for adding/bubbling through limewater. 1 mark for turning cloudy/milky. b) 1 mark for independent variable as Time. 1 mark for dependent variable as Volume of gas. c) i) 1 mark for describing a smooth curve starting at the origin with a decreasing gradient. 1 mark for describing it leveling off/becoming horizontal at 50 seconds. ii) 1 mark for \(42 / 30\). 1 mark for correct value of 1.4. 1 mark for correct unit (\(\text{cm}^3/\text{s}\) or \(\text{cm}^3\text{ s}^{-1}\)). iii) 1 mark for stating a reactant is completely used up / reaction has finished.
題目 2 · Recall and data plotting
10 分
A student investigates the neutralisation of hydrochloric acid using sodium hydroxide solution. They add different volumes of sodium hydroxide solution to \(25\text{ cm}^3\) of hydrochloric acid and measure the maximum temperature of the mixture. Their results are:
a) Write a word equation for the chemical reaction between hydrochloric acid and sodium hydroxide. [2 marks] b) Recall the approximate pH of: i) The initial hydrochloric acid solution. [1 mark] ii) The neutral solution formed when the acid is completely neutralised. [1 mark] c) The student plots a graph of 'Maximum temperature in \(^\circ\text{C}\)' on the y-axis against 'Volume of NaOH added in \(\text{cm}^3\)' on the x-axis. i) Describe the pattern shown by the plotted data points as the volume of NaOH increases from 0 to \(35\text{ cm}^3\). [2 marks] ii) How can the exact volume of sodium hydroxide needed to neutralise the acid be found using a graph of these results? [2 marks] iii) Name the type of reaction that releases thermal energy to the surroundings. [1 mark] iv) State one safety precaution the student should take when handling dilute hydrochloric acid. [1 mark]
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解題
a) Hydrochloric acid reacts with sodium hydroxide to form a salt and water: \(\text{hydrochloric acid} + \text{sodium hydroxide} \rightarrow \text{sodium chloride} + \text{water}\).
b) i) Dilute hydrochloric acid is a strong acid, so its pH is in the range of 0 to 3 (accept any value in this range, e.g., pH 1 or 2). ii) Complete neutralisation results in a neutral solution, which has a pH of 7.
c) i) The temperature increases as the volume of sodium hydroxide increases up to \(20\text{ cm}^3\), and then the temperature decreases as more volume is added beyond \(20\text{ cm}^3\).
ii) Draw two straight lines of best fit: one through the rising temperature points and one through the falling temperature points. The point of intersection of these two lines corresponds to the exact volume needed for neutralisation.
iii) A reaction that releases thermal energy is an exothermic reaction.
iv) Wear safety goggles to protect eyes, or wear protective gloves to prevent skin contact.
評分準則
a) 1 mark for reactants: hydrochloric acid + sodium hydroxide. 1 mark for products: sodium chloride + water. b) i) 1 mark for pH 0, 1, 2, or 3. ii) 1 mark for pH 7. c) i) 1 mark for noting the temperature increase up to \(20\text{ cm}^3\). 1 mark for noting the temperature decrease after \(20\text{ cm}^3\). ii) 1 mark for drawing two lines of best fit. 1 mark for identifying the neutralisation point at the intersection of these lines. iii) 1 mark for exothermic. iv) 1 mark for safety goggles / protective gloves / protective clothing.
題目 3 · Recall and data plotting
10 分
An atom of neon has an atomic number of 10 and a mass number of 20.
a) Recall the names of the three subatomic particles found in this atom. [3 marks] b) Determine the number of protons, neutrons, and electrons in this atom of neon-20. [3 marks] c) Naturally occurring neon consists of three isotopes. The relative abundance of each isotope is shown in the table below:
i) Explain what is meant by the term 'isotopes'. [2 marks] ii) On a bar chart showing 'Relative abundance (%)' on the vertical (y) axis and 'Mass number' on the horizontal (x) axis, describe how the height of the bars for Neon-20 and Neon-22 compare. [2 marks]
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解題
a) The three subatomic particles are protons, neutrons, and electrons.
b) The atomic number is 10, so there are 10 protons. In a neutral atom, the number of electrons equals the number of protons, so there are 10 electrons. The number of neutrons is the mass number minus the atomic number: \(20 - 10 = 10\) neutrons.
c) i) Isotopes are atoms of the same element (same number of protons) that have different mass numbers (different numbers of neutrons).
ii) On the bar chart, the bar for Neon-20 is plotted at \(90.5\%\) and the bar for Neon-22 is plotted at \(9.2\%\). Comparing these, the bar for Neon-20 is significantly taller (about ten times taller) than the bar for Neon-22.
評分準則
a) 1 mark for protons, 1 mark for neutrons, 1 mark for electrons. b) 1 mark for 10 protons, 1 mark for 10 electrons, 1 mark for 10 neutrons. c) i) 1 mark for same number of protons / same element. 1 mark for different number of neutrons / different mass number. ii) 1 mark for identifying that Neon-20 has a much taller bar than Neon-22. 1 mark for quantifying the comparison (e.g. Neon-20 is about 10 times taller / Neon-22 is about one-tenth of the height of Neon-20).
題目 4 · Recall and data plotting
10 分
A student wanted to investigate the dissolved solids in three different water samples: A, B, and C. They evaporated \(100\text{ cm}^3\) of each sample to dryness and weighed the solid residue remaining in the evaporating basin. Their results are shown below:
| Sample | Mass of evaporating basin (g) | Mass of basin + dry residue (g) | | :--- | :--- | :--- | | A | 45.21 | 45.21 | | B | 44.85 | 45.65 | | C | 46.10 | 46.25 |
a) Recall the definition of 'potable water'. [2 marks] b) Calculate the mass of dissolved solids in \(100\text{ cm}^3\) of Sample B and Sample C. [2 marks] c) To present their data, the student decides to plot a bar chart. i) State the labels that should be on the horizontal (x) axis and the vertical (y) axis of this bar chart. [2 marks] ii) Identify which sample (A, B, or C) is pure water. Explain your answer using the data. [2 marks] d) Describe how pure water can be obtained from seawater using simple distillation. [2 marks]
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解題
a) Potable water is water that is safe/fit for humans to drink.
b) Mass of dissolved solids: - Sample B: \(45.65\text{ g} - 44.85\text{ g} = 0.80\text{ g}\). - Sample C: \(46.25\text{ g} - 46.10\text{ g} = 0.15\text{ g}\).
c) i) The horizontal (x) axis should be labeled 'Water Sample' (with columns for A, B, and C). The vertical (y) axis should be labeled 'Mass of dissolved solids in grams' or 'Mass of dissolved solids (g)'.
ii) Sample A is pure water. Pure water contains no dissolved solids, and Sample A left no residue (the mass of the basin remained exactly \(45.21\text{ g}\), showing \(0.00\text{ g}\) of dissolved solids).
d) In simple distillation, the seawater is heated in a flask so that the water boils and turns into water vapour (steam). The water vapour passes into a condenser where it is cooled, condensing back into pure liquid water which is collected in a separate beaker, leaving the dissolved salt behind in the flask.
評分準則
a) 2 marks for stating that potable water is water that is safe to drink. b) 1 mark for Sample B calculated as 0.80 g. 1 mark for Sample C calculated as 0.15 g. c) i) 1 mark for x-axis as Water Sample (A, B, C). 1 mark for y-axis as Mass of dissolved solids (with unit 'g' or 'grams'). ii) 1 mark for identifying Sample A. 1 mark for explaining that it left no dissolved solid/residue / mass did not change. d) 1 mark for heating/boiling to evaporate the water (forming steam/vapour). 1 mark for cooling/condensing the vapour back into liquid water.
題目 5 · Recall and data plotting
10 分
A student investigated the effect of temperature on the rate of reaction between sodium thiosulfate solution and dilute hydrochloric acid. They measured the time taken for a cross drawn on paper under the flask to disappear. The results are shown below:
| Temperature (\(^\circ\text{C}\)) | Time taken for cross to disappear (s) | | :--- | :--- | | 20 | 80 | | 30 | 40 | | 40 | 20 | | 50 | 10 | | 60 | 5 |
a) Recall the name of the insoluble solid (precipitate) formed in this reaction that causes the solution to go cloudy. [1 mark] b) State the independent variable in this experiment. [1 mark] c) The student wants to plot a graph of these results. i) State which variable should be plotted on the horizontal (x) axis. [1 mark] ii) Describe the trend shown by the data: as the temperature increases, what happens to the time taken for the cross to disappear? [1 mark] iii) Calculate the rate of reaction at \(40^\circ\text{C}\) using the formula: \(\text{Rate} = \frac{1}{\text{time}}\). Show your working. [2 marks] d) Explain, in terms of particles and collisions, why increasing the temperature increases the rate of this reaction. [4 marks]
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解題
a) The insoluble precipitate that forms is sulfur.
b) The independent variable is the temperature.
c) i) Temperature should be plotted on the horizontal (x) axis. ii) As the temperature increases, the time taken for the cross to disappear decreases (which means the reaction becomes faster). iii) At \(40^\circ\text{C}\), the time taken is \(20\text{ s}\). \(\text{Rate} = \frac{1}{20} = 0.05\text{ s}^{-1}\).
d) When the temperature increases: 1. The reactant particles gain more kinetic energy and move faster. 2. This results in more frequent collisions between particles. 3. A greater proportion of the colliding particles have energy equal to or greater than the activation energy. 4. This leads to a higher frequency of successful collisions (more successful collisions per second), which increases the rate of reaction.
評分準則
a) 1 mark for sulfur.) 1 mark for temperature. c) i) 1 mark for temperature. ii) 1 mark for identifying that the time taken decreases as temperature increases. iii) 1 mark for substituting the correct time value (20 s) into the formula. 1 mark for the correct answer of 0.05 (s⁻¹). d) 1 mark for stating that particles gain kinetic energy / move faster. 1 mark for stating that they collide more frequently. 1 mark for stating that more particles have energy greater than or equal to the activation energy. 1 mark for stating there is a higher frequency of successful collisions (or more successful collisions per second).
題目 6 · Analysis and level of response planning
10 分
A student wants to investigate how the concentration of hydrochloric acid affects the rate of reaction when it reacts with calcium carbonate chips. Part 1 (6 marks): Write a plan for the student to carry out this investigation. Your plan should include: the apparatus required, the independent, dependent, and one control variable, and how the measurements will be taken to determine the rate. Part 2 (4 marks): In one experiment, the student recorded the volume of gas collected every 10 seconds: 0 s: 0 cm3, 10 s: 15 cm3, 20 s: 28 cm3, 30 s: 49 cm3, 40 s: 42 cm3, 50 s: 45 cm3, 60 s: 45 cm3. Identify the anomalous result, suggest one practical error that could have caused this, and calculate the mean rate of reaction from 0 to 20 seconds. State the unit.
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解題
Part 1: The plan should involve reacting a fixed mass of calcium carbonate chips with a set volume of hydrochloric acid of a known concentration in a conical flask connected to a gas syringe. The volume of carbon dioxide gas produced is recorded at regular time intervals (e.g., every 10 seconds) for a set duration. The experiment is repeated with different concentrations of hydrochloric acid, keeping the temperature, mass and surface area of calcium carbonate, and the volume of acid constant. Independent variable: concentration of acid; Dependent variable: volume of gas produced per unit time; Control variables: mass of calcium carbonate, temperature. Part 2: The anomalous result is 49 cm3 at 30 s because the volume of gas cannot decrease from 30 s (49 cm3) to 40 s (42 cm3). A practical error could be misreading the scale on the gas syringe or recording the time incorrectly. The mean rate from 0 to 20 s is calculated as: volume of gas at 20 s divided by time = 28 cm3 / 20 s = 1.4 cm3/s.
評分準則
Part 1 (6 marks Level of Response): Level 3 (5-6 marks): A detailed and coherent plan describing suitable apparatus (conical flask, delivery tube, gas syringe/measuring cylinder), identification of variables (independent: concentration, dependent: volume of gas, controls: mass/size of chips, temperature, volume of acid), and systematic measurements of volume over time. Level 2 (3-4 marks): A logical plan identifying some variables and apparatus, but lacking detail on how rate is calculated or specific controls. Level 1 (1-2 marks): Basic method with major omissions, e.g., missing how gas is collected or variables. Part 2 (4 marks): 1 mark for identifying the anomalous result at 30 s (49 cm3), 1 mark for suggesting a valid cause (e.g., scale reading error, delay in taking reading, gas syringe sticking), 1 mark for the calculation 28 / 20 = 1.4, 1 mark for the correct unit (cm3/s or cm3 s-1).
題目 7 · Analysis and level of response planning
10 分
A student uses paper chromatography to analyze the food colorings in a brown sweet. Part 1 (6 marks): Plan an experiment using paper chromatography to compare the pigments in the brown food coloring with three known pure food dyes (Red, Yellow, Blue). Explain how chromatography separates substances and how to identify if the brown coloring contains any of the known dyes. Part 2 (4 marks): A chromatogram is obtained. The start line is 1.5 cm from the bottom of the paper. The solvent front travels 8.0 cm from the start line. One dye spot travels 5.2 cm from the start line. (i) Explain why the start line must be drawn in pencil and not ink. (ii) Calculate the Rf value of the dye spot. Give your answer to 2 significant figures.
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解題
Part 1: Draw a horizontal pencil line about 1.5 cm from the bottom of the chromatography paper. Place small, concentrated spots of the brown food coloring and the three known pure dyes (Red, Yellow, Blue) on the line, leaving gaps between them. Pour a small volume of solvent (e.g., water) into a beaker, ensuring the level is below the pencil line. Suspend the paper in the beaker, cover with a lid, and allow the solvent to rise. Remove the paper before the solvent reaches the top and mark the solvent front. Separation occurs because different pigments have different solubilities in the mobile phase (solvent) and different attractions to the stationary phase (paper). The brown coloring contains a known dye if a spot from the brown coloring travels the same distance (has the same Rf value) as one of the pure dyes. Part 2: (i) Pencil lead is graphite, which is insoluble in the solvent and will not run or bleed, whereas ink is soluble and would separate, interfering with the results. (ii) Rf = distance moved by substance / distance moved by solvent = 5.2 cm / 8.0 cm = 0.65.
評分準則
Part 1 (6 marks Level of Response): Level 3 (5-6 marks): Fully coherent plan including drawing a pencil line, applying spots, solvent level below line, letting solvent run, marking front, explaining separation via stationary/mobile phases, and explaining identification by comparing spot heights or Rf values. Level 2 (3-4 marks): Clear method but may lack detailed explanation of the chromatography process (phases) or precise details like solvent level. Level 1 (1-2 marks): Simple description of chromatography without clear steps or explanations. Part 2 (4 marks): (i) 1 mark for stating pencil is insoluble, 1 mark for stating ink is soluble/would run and interfere. (ii) 1 mark for formula or working (5.2 / 8.0), 1 mark for the correct answer 0.65 (must be to 2 significant figures).
題目 8 · Analysis and level of response planning
10 分
A coastal town needs to obtain potable water. Part 1 (6 marks): Compare the processes of obtaining potable water from fresh water sources (like rivers) and from salty water sources (like sea water). In your answer, you should describe the steps involved in both methods, and discuss the ease and energy demands of each. Part 2 (4 marks): A sample of water is tested to see if it is pure water. (i) Describe a chemical test to show that water is present. (ii) Describe a physical test to show that the water is pure, and state the expected result.
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解題
Part 1: To obtain potable water from fresh water, the steps are: filtration (passing water through sand and gravel beds to remove insoluble solids) followed by sterilisation (adding chlorine, ozone, or using ultraviolet light to kill harmful microbes). This process is easy and has very low energy demands. To obtain potable water from sea water, desalination is required, which can be done by distillation (heating water to evaporate it, then condensing the steam) or reverse osmosis (passing water through membranes at high pressure). This process is difficult and has very high energy demands because of the heat or high pressure required. Part 2: (i) To show water is present, add anhydrous copper(II) sulfate. If water is present, it turns from white to blue. Alternatively, cobalt chloride paper turns from blue to pink. (ii) To show water is pure, measure its boiling point. Pure water boils at exactly 100 °C (or measure freezing point, which is exactly 0 °C). Any impurities would raise the boiling point or lower the freezing point.
評分準則
Part 1 (6 marks Level of Response): Level 3 (5-6 marks): Detailed comparison of both processes. Correctly identifies filtration and sterilisation for fresh water with chemicals/UV. Correctly identifies distillation or reverse osmosis for sea water. Compares energy demands (low for fresh, high for sea water) and ease. Level 2 (3-4 marks): Describes both methods but lacks detail on either the sterilising agents or the high energy requirements of desalination. Level 1 (1-2 marks): Simple description of filtering fresh water or boiling salty water, with little comparison. Part 2 (4 marks): (i) 1 mark for naming a suitable reagent (anhydrous copper(II) sulfate or cobalt chloride paper), 1 mark for the correct colour change (white to blue, or blue to pink). (ii) 1 mark for measuring boiling point or melting/freezing point, 1 mark for stating the correct value (100 °C or 0 °C).
題目 9 · Analysis and level of response planning
10 分
Earth's atmosphere has changed significantly over geological time. Part 1 (6 marks): Describe how the Earth’s atmosphere has changed over the last 4.6 billion years. Explain the roles of volcanic activity, oceans, and living organisms in causing these changes. Part 2 (4 marks): In a laboratory experiment, a student measured the percentage of oxygen in dry air by heating excess copper in a sealed system of gas syringes containing 100 cm3 of air. The air was passed over the hot copper until no further change occurred. (i) Explain why the apparatus must be cooled back to room temperature before taking the final volume reading. (ii) The final volume of gas measured was 79 cm3. Calculate the percentage of oxygen in this air sample. Show your working.
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解題
Part 1: The early atmosphere was formed by intense volcanic activity, releasing gases such as carbon dioxide, water vapour, and nitrogen, with little or no oxygen. As the Earth cooled, water vapour condensed to form the oceans. Carbon dioxide dissolved in the newly formed oceans and formed carbonate precipitates, which became sedimentary rocks, decreasing atmospheric carbon dioxide. Algae and green plants evolved and photosynthesised, taking in carbon dioxide and releasing oxygen. Over time, oxygen built up to around 21%, while carbon dioxide decreased to a trace amount, and nitrogen became the most abundant gas (around 78%) because it is unreactive. Part 2: (i) Gases expand when heated. If measured while hot, the gas volume would be higher than it actually is at room temperature, leading to an inaccurate result. (ii) Volume of oxygen reacted = starting volume - final volume = 100 cm3 - 79 cm3 = 21 cm3. Percentage of oxygen = (21 cm3 / 100 cm3) * 100 = 21%.
評分準則
Part 1 (6 marks Level of Response): Level 3 (5-6 marks): Detailed explanation of atmospheric evolution. Explains early volcanic atmosphere (mostly CO2, water vapour, N2), formation of oceans, dissolution of CO2, photosynthesis by plants/algae producing O2 and reducing CO2, and nitrogen accumulation. Level 2 (3-4 marks): Explains some stages, such as plants producing oxygen and oceans dissolving CO2, but lacks complete narrative or volcanic details. Level 1 (1-2 marks): Simple statements that oxygen increased and carbon dioxide decreased, without clear scientific mechanisms. Part 2 (4 marks): (i) 1 mark for noting that gases expand when hot, 1 mark for stating that cooling ensures a fair comparison of volumes at the same temperature. (ii) 1 mark for calculating the volume of oxygen reacted (100 - 79 = 21 cm3), 1 mark for calculating the percentage (21%).
題目 10 · Analysis and level of response planning
10 分
This question is about life cycle assessments (LCAs) of consumer products. Part 1 (6 marks): Evaluate the use of plastic carrier bags compared with paper carrier bags by carrying out a life cycle assessment (LCA). Compare them in terms of: raw materials used, manufacturing and packaging, use and reuse, and disposal at the end of their useful life. Draw a conclusion on which bag is more environmentally friendly. Part 2 (4 marks): A waste management report provides data on the energy required to make bags: To manufacture 10,000 plastic bags requires 150 MJ of energy. To manufacture 10,000 paper bags requires 550 MJ of energy. (i) Calculate the percentage increase in energy required to manufacture paper bags compared to plastic bags. (ii) Suggest one limitation of using energy values alone to decide which bag is better for the environment.
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解題
Part 1: Raw materials: Plastic bags are made from crude oil, which is a finite, non-renewable resource, whereas paper bags are made from wood, which is renewable but causes deforestation and habitat loss. Manufacturing: Plastic manufacturing requires energy-intensive cracking and polymerisation but uses little water. Paper manufacturing requires huge volumes of water and toxic chemicals, and is highly energy-intensive. Use/reuse: Plastic bags are strong and durable, meaning they can be reused multiple times. Paper bags tear easily when wet and are rarely reused. Disposal: Plastic is non-biodegradable and accumulates in landfills or as marine litter, though it can be recycled. Paper is biodegradable and easily recycled, causing fewer long-term disposal issues. Conclusion: Plastic bags may have lower manufacturing energy and higher reuse potential, but paper is superior in terms of renewability and biodegradability. Part 2: (i) Increase in energy = 550 - 150 = 400 MJ. Percentage increase = (400 / 150) * 100 = 266.67% which rounds to 267% (or 270% to 2 s.f.). (ii) Energy is only one aspect of environmental impact; it ignores resource depletion (crude oil vs trees), toxic emissions during production, and long-term disposal/pollution impacts such as plastic persistence in oceans.
評分準則
Part 1 (6 marks Level of Response): Level 3 (5-6 marks): Balanced and detailed comparison covering all four stages (raw materials, manufacturing, use, disposal) with clear pros and cons for both plastic and paper, concluding with a reasoned judgment. Level 2 (3-4 marks): Compares some of the stages (at least two or three) but lacks depth or omits a justified conclusion. Level 1 (1-2 marks): Simple comparison of one or two aspects (e.g., just biodegradeability vs crude oil) without a structured LCA approach. Part 2 (4 marks): (i) 1 mark for finding the difference (400 MJ), 1 mark for correct calculation of percentage increase (267% or 266.7%). (ii) 1 mark for identifying that energy is only one part of an LCA, 1 mark for specifying another critical factor such as resource depletion, biodegradability, toxicity, or wildlife hazard.
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