An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V1) Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.
卷二 選擇題
Answer all forty multiple choice questions. Choose the best response (A, B, C or D).
40 题目 · 40 分
题目 1 · MCQ
1 分
A trolley of mass \(1.5\text{ kg}\) is pulled from rest along a horizontal frictionless surface by a constant resultant force of \(4.5\text{ N}\) for \(2.0\text{ s}\).
What is the kinetic energy of the trolley after \(2.0\text{ s}\)?
C is correct [1]. - A (9.0 J) is the momentum change \(F \times t\). - B (13.5 J) incorrectly calculates \(\frac{1}{2}mv\) or \(\frac{1}{2}Ft\). - D (54 J) omits the factor of \(\frac{1}{2}\) in the kinetic energy equation.
题目 2 · MCQ
1 分
Which combination of reactants is suitable for preparing a pure, dry sample of barium sulfate by precipitation?
A.solid barium carbonate and dilute sulfuric acid
B.aqueous barium chloride and dilute sulfuric acid
C.aqueous barium hydroxide and aqueous copper(II) sulfate
D.aqueous barium nitrate and dilute hydrochloric acid
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解题
To prepare an insoluble salt such as barium sulfate by precipitation, two soluble starting materials must be mixed. Aqueous barium chloride and dilute sulfuric acid are both soluble solutions that react to form an insoluble precipitate of barium sulfate and aqueous hydrochloric acid: \(\text{BaCl}_2\text{(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)} + 2\text{HCl(aq)}\). The solid precipitate is then separated by filtration, washed with distilled water, and dried.
评分标准
B is correct [1]. - A is incorrect because solid barium carbonate is insoluble in water, forming an insoluble crust of sulfate that prevents complete reaction. - C is incorrect because mixing barium hydroxide and copper(II) sulfate produces two insoluble precipitates (barium sulfate and copper(II) hydroxide), so pure barium sulfate cannot be isolated by simple filtration. - D is incorrect because barium chloride and nitric acid are formed, both of which are soluble, producing no precipitate.
题目 3 · MCQ
1 分
A circuit contains a \(12\text{ V}\) power supply connected to three resistors. Two \(6.0\ \Omega\) resistors are connected in parallel with each other, and this combination is connected in series with a \(3.0\ \Omega\) resistor.
What is the total current drawn from the power supply?
A.\(0.80\text{ A}\)
B.\(1.3\text{ A}\)
C.\(2.0\text{ A}\)
D.\(4.0\text{ A}\)
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解题
1. Combined resistance of the two \(6.0\ \Omega\) parallel resistors: \(R_p = \frac{6.0 \times 6.0}{6.0 + 6.0} = 3.0\ \Omega\). 2. Total resistance of the circuit: \(R_{\text{total}} = R_p + 3.0\ \Omega = 3.0\ \Omega + 3.0\ \Omega = 6.0\ \Omega\). 3. Total current: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\).
评分标准
C is correct [1]. - A (0.80 A) incorrectly treats all three resistors as connected in series (total resistance \(15\ \Omega\)). - B (1.3 A) incorrectly treats all three resistors in a different combination or miscalculates parallel resistance. - D (4.0 A) ignores the resistance of the parallel combination.
题目 4 · MCQ
1 分
A student carries out tests to identify a solid compound, \(\text{X}\).
1. Adding dilute nitric acid to solid \(\text{X}\) produces effervescence. The gas given off forms a white precipitate in limewater. 2. An aqueous solution of \(\text{X}\) is prepared. Adding aqueous sodium hydroxide dropwise produces a green precipitate that is insoluble in excess. 3. Adding aqueous ammonia dropwise to a separate portion of the solution produces a green precipitate that is insoluble in excess.
What is compound \(\text{X}\)?
A.iron(II) carbonate
B.iron(II) chloride
C.iron(III) carbonate
D.copper(II) carbonate
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解题
- The production of a gas that turns limewater cloudy (carbon dioxide) with acid confirms the presence of carbonate ions (\(\text{CO}_3^{2-}\)). - The formation of a green precipitate with aqueous sodium hydroxide and with aqueous ammonia, insoluble in excess of both reagents, confirms the presence of iron(II) ions (\(\text{Fe}^{2+}\)). - Therefore, the compound is iron(II) carbonate.
评分标准
A is correct [1]. - B is incorrect because chlorides do not produce carbon dioxide when treated with dilute acid. - C is incorrect because iron(III) forms a red-brown precipitate with sodium hydroxide and ammonia. - D is incorrect because copper(II) forms a light blue precipitate that dissolves in excess ammonia to give a deep blue solution.
题目 5 · MCQ
1 分
Plant cells are placed in a concentrated sucrose solution.
Which statement correctly describes the condition of the cells and the direction of the net movement of water?
A.The cells become turgid because there is a net movement of water into the cells.
B.The cells become flaccid because there is a net movement of water into the cells.
C.The cells become turgid because there is a net movement of water out of the cells.
D.The cells become flaccid because there is a net movement of water out of the cells.
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解题
A concentrated sucrose solution has a lower water potential than the cell sap inside plant cells. As a result, water moves out of the cells down the water potential gradient by osmosis. This loss of internal turgor pressure causes the cells to become flaccid (and eventually plasmolysed).
评分标准
D is correct [1]. - A is incorrect because water enters cells when the external solution has a higher water potential, making cells turgid. - B is incorrect because water moves from higher to lower water potential, not out of cells placed in a higher water potential solution. - C is incorrect because water moves out of cells when placed in a solution of lower water potential.
题目 6 · MCQ
1 分
An object of mass \(4.0\text{ kg}\) is pulled along a horizontal floor by a forward horizontal force of \(18\text{ N}\). A constant friction force of \(6.0\text{ N}\) opposes the motion. What is the acceleration of the object?
A.\(3.0\text{ m/s}^2\)
B.\(1.5\text{ m/s}^2\)
C.\(4.5\text{ m/s}^2\)
D.\(6.0\text{ m/s}^2\)
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解题
First calculate the resultant force: \(F_{\text{net}} = 18\text{ N} - 6.0\text{ N} = 12\text{ N}\). Then use Newton's second law: \(a = \frac{F_{\text{net}}}{m} = \frac{12\text{ N}}{4.0\text{ kg}} = 3.0\text{ m/s}^2\).
评分标准
A is correct [1]; resultant force is \(12\text{ N}\) and acceleration is \(12/4.0 = 3.0\text{ m/s}^2\).
题目 7 · MCQ
1 分
Which procedure is used to prepare pure, dry crystals of zinc sulfate from solid zinc oxide and dilute sulfuric acid?
A.Add dilute sulfuric acid to zinc oxide, heat until completely dry, and wash residue with water.
B.Mix equimolar volumes of zinc oxide and sulfuric acid, filter off excess liquid, and evaporate to dryness.
C.Add excess zinc oxide to warm dilute sulfuric acid, filter to remove unreacted solid, heat filtrate to crystallisation point, and leave to cool.
D.Titrate zinc oxide solution against dilute sulfuric acid using an indicator, then boil off all water.
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解题
Zinc oxide is an insoluble base. The standard preparation involves adding excess solid zinc oxide to warm dilute sulfuric acid to ensure all acid reacts, filtering off the unreacted solid zinc oxide, heating the filtrate to the point of crystallisation, and then leaving it to cool and crystallise before drying between filter papers.
评分标准
C is correct [1]; preparation of a soluble salt from an insoluble base requires excess solid, filtration, and crystallisation.
题目 8 · MCQ
1 分
A \(12\text{ V}\) power supply is connected in series with two resistors of resistance \(4.0\ \Omega\) and \(8.0\ \Omega\). What is the total current in the circuit and the potential difference across the \(8.0\ \Omega\) resistor?
Total circuit resistance \(R = 4.0\ \Omega + 8.0\ \Omega = 12.0\ \Omega\). Circuit current \(I = \frac{V}{R} = \frac{12\text{ V}}{12.0\ \Omega} = 1.0\text{ A}\). The potential difference across the \(8.0\ \Omega\) resistor is \(V_2 = I \times R_2 = 1.0\text{ A} \times 8.0\ \Omega = 8.0\text{ V}\).
评分标准
D is correct [1]; total resistance is \(12\ \Omega\), giving \(I = 1.0\text{ A}\) and \(V = 8.0\text{ V}\).
题目 9 · MCQ
1 分
A sample of an unknown solid is dissolved in distilled water to form solution X.
1. When aqueous sodium hydroxide is added gradually to solution X, a light blue precipitate forms which does not dissolve in excess sodium hydroxide. 2. When dilute nitric acid followed by aqueous barium nitrate is added to another portion of solution X, a white precipitate forms.
What is the identity of the unknown solid?
A.copper(II) chloride
B.copper(II) sulfate
C.iron(II) sulfate
D.zinc sulfate
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解题
The formation of a light blue precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of copper(II) ions, \(\text{Cu}^{2+}\). The formation of a white precipitate with acidified barium nitrate confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). Therefore, the solid is copper(II) sulfate.
评分标准
B is correct [1]; light blue precipitate insoluble in excess NaOH indicates \(\text{Cu}^{2+}\), and white precipitate with acidified \(\text{Ba(NO}_3)_2\) indicates \(\text{SO}_4^{2-}\).
题目 10 · MCQ
1 分
Strips of fresh plant tissue are placed into a concentrated sucrose solution for 45 minutes. Which row correctly identifies the direction of net water movement and the resulting condition of the plant cells?
A.Net water movement: out of the cells; Condition of cells: plasmolysed
B.Net water movement: into the cells; Condition of cells: turgid
C.Net water movement: into the cells; Condition of cells: flaccid
D.Net water movement: out of the cells; Condition of cells: turgid
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解题
A concentrated sucrose solution has a lower water potential than the cytoplasm/vacuole of the plant cells. Water leaves the cells by osmosis down a water potential gradient (net movement out of the cells). As the cytoplasm shrinks away from the cell wall, the cells become plasmolysed.
评分标准
A is correct [1]; water moves out of the cells by osmosis down the water potential gradient, causing plasmolysis.
题目 11 · MCQ
1 分
An object of mass \( 4.0\text{ kg} \) moves along a smooth horizontal surface. A forward horizontal force of \( 18\text{ N} \) and a backward resistive force of \( 6.0\text{ N} \) act on the object simultaneously.
What is the acceleration of the object?
A.\( 1.5\text{ m/s}^2 \)
B.\( 3.0\text{ m/s}^2 \)
C.\( 4.5\text{ m/s}^2 \)
D.\( 6.0\text{ m/s}^2 \)
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解题
First determine the resultant horizontal force acting on the object: \( F_{\text{net}} = 18\text{ N} - 6.0\text{ N} = 12\text{ N} \)
Using Newton's second law, \( F = ma \): \( a = \frac{F_{\text{net}}}{m} = \frac{12\text{ N}}{4.0\text{ kg}} = 3.0\text{ m/s}^2 \)
Therefore, option B is correct.
评分标准
B [1 mark] - \( 3.0\text{ m/s}^2 \)
题目 12 · MCQ
1 分
A student prepares a pure, dry sample of hydrated zinc sulfate crystals by reacting excess insoluble zinc carbonate with dilute sulfuric acid.
Which sequence of experimental steps is correct?
A.filter to remove excess solid \(\rightarrow\) heat filtrate to crystallization point \(\rightarrow\) cool to form crystals \(\rightarrow\) filter and dry between filter papers
B.evaporate all water to dryness \(\rightarrow\) filter the solid \(\rightarrow\) wash with distilled water \(\rightarrow\) dry in an oven
C.filter to remove excess solid \(\rightarrow\) add universal indicator \(\rightarrow\) evaporate to complete dryness \(\rightarrow\) cool
D.heat the mixture to boiling \(\rightarrow\) cool immediately without filtering \(\rightarrow\) collect crystals by decanting \(\rightarrow\) dry in an oven
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解题
To prepare a soluble salt from an acid and an insoluble carbonate: 1. Filter the mixture to remove the unreacted excess zinc carbonate. 2. Heat the filtrate until the crystallization point (point of saturation) is reached. 3. Allow the hot saturated solution to cool and form crystals. 4. Filter off the crystals and dry them carefully between sheets of filter paper.
Option A provides this exact correct sequence.
评分标准
A [1 mark] - filter to remove excess solid \(\rightarrow\) heat filtrate to crystallization point \(\rightarrow\) cool to form crystals \(\rightarrow\) filter and dry between filter papers
题目 13 · MCQ
1 分
Two fixed resistors of resistance \( 6.0\,\Omega \) and \( 12\,\Omega \) are connected in parallel across a \( 12\text{ V} \) direct current (d.c.) power supply.
What is the total current supplied by the power source?
A.\( 0.67\text{ A} \)
B.\( 1.5\text{ A} \)
C.\( 3.0\text{ A} \)
D.\( 4.0\text{ A} \)
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解题
Method 1: Find branch currents. - Current through the \( 6.0\,\Omega \) resistor: \( I_1 = \frac{V}{R_1} = \frac{12\text{ V}}{6.0\,\Omega} = 2.0\text{ A} \) - Current through the \( 12\,\Omega \) resistor: \( I_2 = \frac{V}{R_2} = \frac{12\text{ V}}{12\,\Omega} = 1.0\text{ A} \) - Total current: \( I_{\text{total}} = I_1 + I_2 = 2.0\text{ A} + 1.0\text{ A} = 3.0\text{ A} \)
A sample of an unknown ionic compound \( \text{X} \) is dissolved in water to make an aqueous solution. Two tests are performed:
1. When aqueous sodium hydroxide is added, a green precipitate forms that is insoluble in excess sodium hydroxide. 2. When dilute nitric acid is added followed by aqueous barium nitrate, a white precipitate forms.
What is the identity of compound \( \text{X} \)?
A.copper(II) chloride
B.iron(II) sulfate
C.iron(III) sulfate
D.zinc sulfate
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解题
- The formation of a green precipitate with aqueous \( \text{NaOH} \) that remains insoluble in excess identifies the presence of iron(II) ions, \( \text{Fe}^{2+} \). - The formation of a white precipitate upon adding dilute nitric acid followed by aqueous barium nitrate confirms the presence of sulfate ions, \( \text{SO}_4^{2-} \).
Combining these ions gives iron(II) sulfate, \( \text{FeSO}_4 \). Thus, option B is correct.
评分标准
B [1 mark] - iron(II) sulfate
题目 15 · MCQ
1 分
Strips of fresh potato tissue of identical initial length were placed into test-tubes containing sucrose solutions of different concentrations.
After 45 minutes: - The potato strip in \( 0.1\text{ mol/dm}^3 \) sucrose solution increased in length by \( 4\% \). - The potato strip in \( 0.6\text{ mol/dm}^3 \) sucrose solution decreased in length by \( 7\% \).
Which statement correctly explains the observation in the \( 0.6\text{ mol/dm}^3 \) sucrose solution?
A.Sucrose molecules entered the potato cells by diffusion down a concentration gradient.
B.Sucrose was actively transported out of the cells, causing the cell walls to contract.
C.Water entered the potato cells by osmosis because the cells had a lower water potential than the external solution.
D.Water left the cells by osmosis because the external solution had a lower water potential than the cell contents.
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解题
In the \( 0.6\text{ mol/dm}^3 \) sucrose solution, the external solution has a lower water potential (higher solute concentration) than the cytoplasm inside the potato cells. Consequently, water moves out of the potato cells by osmosis down the water potential gradient, causing cells to lose turgor/volume and the tissue to decrease in length. Thus, option D is correct.
评分标准
D [1 mark] - Water left the cells by osmosis because the external solution had a lower water potential than the cell contents.
题目 16 · MCQ
1 分
A vehicle of mass \(1200\text{ kg}\) accelerates uniformly along a horizontal road from rest to a speed of \(15\text{ m/s}\) in \(6.0\text{ s}\). A constant frictional force of \(400\text{ N}\) opposes the motion.
What is the driving force produced by the engine?
A.\(2600\text{ N}\)
B.\(3000\text{ N}\)
C.\(3400\text{ N}\)
D.\(7600\text{ N}\)
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解题
First, calculate the acceleration of the vehicle: \(a = \frac{v - u}{t} = \frac{15\text{ m/s} - 0\text{ m/s}}{6.0\text{ s}} = 2.5\text{ m/s}^2\)
Next, determine the resultant (net) force using Newton's second law: \(F_{\text{net}} = m \times a = 1200\text{ kg} \times 2.5\text{ m/s}^2 = 3000\text{ N}\)
The resultant force is the difference between the driving force (\(F_{\text{engine}}\)) and the opposing friction (\(F_{\text{friction}}\)): \(F_{\text{net}} = F_{\text{engine}} - F_{\text{friction}}\) \(3000\text{ N} = F_{\text{engine}} - 400\text{ N}\) \(F_{\text{engine}} = 3000\text{ N} + 400\text{ N} = 3400\text{ N}\)
评分标准
C ; [1]
题目 17 · MCQ
1 分
Which row correctly classifies aluminium oxide (\(\text{Al}_2\text{O}_3\)), calcium oxide (\(\text{CaO}\)), and sulfur dioxide (\(\text{SO}_2\))?
Aluminium oxide (\(\text{Al}_2\text{O}_3\)) is an amphoteric oxide because it reacts with both acids and bases to form salts. Calcium oxide (\(\text{CaO}\)) is a basic oxide because it is a metallic oxide that reacts with acids. Sulfur dioxide (\(\text{SO}_2\)) is an acidic oxide because it is a non-metallic oxide that dissolves in water to form an acid and reacts with bases.
评分标准
A ; [1]
题目 18 · MCQ
1 分
Two identical resistors, each of resistance \(R\), are connected in parallel with each other. This combination is then connected in series with a third identical resistor of resistance \(R\).
The total combined resistance of the entire arrangement is \(18\ \Omega\).
What is the value of \(R\)?
A.\(6.0\ \Omega\)
B.\(12\ \Omega\)
C.\(27\ \Omega\)
D.\(36\ \Omega\)
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解题
The equivalent resistance of two identical resistors of resistance \(R\) connected in parallel is: \(R_{\text{parallel}} = \frac{R \times R}{R + R} = \frac{R}{2} = 0.5R\)
When this combination is connected in series with a third identical resistor of resistance \(R\), the total resistance is: \(R_{\text{total}} = R_{\text{parallel}} + R = 0.5R + R = 1.5R\)
Given that \(R_{\text{total}} = 18\ \Omega\): \(1.5R = 18\ \Omega \implies R = \frac{18}{1.5} = 12\ \Omega\)
评分标准
B ; [1]
题目 19 · MCQ
1 分
An aqueous solution of compound \(X\) is tested as follows:
1. When aqueous sodium hydroxide is added dropwise, a light blue precipitate forms that does not dissolve in excess sodium hydroxide. 2. When dilute nitric acid followed by aqueous barium nitrate is added, a white precipitate forms.
What is the identity of compound \(X\)?
A.copper(II) chloride
B.copper(II) sulfate
C.iron(II) sulfate
D.iron(III) chloride
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解题
1. Addition of aqueous sodium hydroxide producing a light blue precipitate that is insoluble in excess confirms the presence of copper(II) ions (\(\text{Cu}^{2+}\)). 2. Addition of dilute nitric acid followed by aqueous barium nitrate yielding a white precipitate of barium sulfate confirms the presence of sulfate ions (\(\text{SO}_4^{2-}\)).
Therefore, compound \(X\) is copper(II) sulfate.
评分标准
B ; [1]
题目 20 · MCQ
1 分
Cylinders of potato tissue of equal initial mass were placed in sucrose solutions of different concentrations. After two hours, the percentage change in mass of each potato cylinder was determined.
The percentage change in mass was \(0\%\) at a sucrose concentration of \(0.35\text{ mol/dm}^3\).
Which statement is correct?
A.Potato cells placed in a \(0.20\text{ mol/dm}^3\) sucrose solution lose water and become plasmolysed.
B.Potato cells placed in a \(0.60\text{ mol/dm}^3\) sucrose solution gain water and become turgid.
C.The water potential inside the potato cells is equal to the water potential of a \(0.35\text{ mol/dm}^3\) sucrose solution.
D.Active transport of sucrose into the potato tissue stops at a concentration of \(0.35\text{ mol/dm}^3\).
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解题
When there is no net change in mass (percentage change in mass is \(0\%\)), there is no net movement of water into or out of the cells. This indicates that the water potential inside the potato cells is equal to the water potential of the surrounding \(0.35\text{ mol/dm}^3\) sucrose solution.
评分标准
C ; [1]
题目 21 · MCQ
1 分
A trolley of mass \(2.0\text{ kg}\) is pulled along a horizontal frictionless track by a constant horizontal force of \(6.0\text{ N}\) over a distance of \(3.0\text{ m}\).
The trolley starts from rest.
What is the kinetic energy of the trolley after moving \(3.0\text{ m}\)?
A.\(6.0\text{ J}\)
B.\(9.0\text{ J}\)
C.\(18\text{ J}\)
D.\(36\text{ J}\)
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解题
The work done on the trolley by the resultant force is transferred entirely to its kinetic energy (since the track is frictionless and the trolley starts from rest).
Therefore, the kinetic energy of the trolley is \(18\text{ J}\).
评分标准
C [1]
题目 22 · MCQ
1 分
Which method is used to prepare a pure, dry sample of the insoluble salt barium sulfate from aqueous solutions of barium chloride and sodium sulfate?
A.Crystallisation after evaporating half the water, followed by filtration
B.Precipitation followed by filtration, washing the residue with distilled water, and drying
C.Neutralisation followed by heating the filtrate to dryness
D.Titration using an indicator followed by simple distillation
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解题
Barium sulfate is an insoluble salt formed via a precipitation reaction between aqueous solutions containing barium ions and sulfate ions.
To obtain a pure, dry sample: 1. Mix the two aqueous solutions to form a precipitate. 2. Filter the mixture to collect the insoluble residue. 3. Wash the residue on the filter paper with distilled water to remove soluble impurities. 4. Dry the residue between filter papers or in a warm oven.
评分标准
B [1]
题目 23 · MCQ
1 分
A \(12\text{ V}\) power supply is connected across a series combination of a \(4.0\,\Omega\) resistor and an \(8.0\,\Omega\) resistor.
What is the potential difference across the \(8.0\,\Omega\) resistor?
A.\(4.0\text{ V}\)
B.\(6.0\text{ V}\)
C.\(8.0\text{ V}\)
D.\(12\text{ V}\)
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解题
1. Find total resistance: \(R_{\text{total}} = 4.0\,\Omega + 8.0\,\Omega = 12.0\,\Omega\). 2. Calculate circuit current: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{12.0\,\Omega} = 1.0\text{ A}\). 3. Calculate potential difference across the \(8.0\,\Omega\) resistor: \(V = I \times R = 1.0\text{ A} \times 8.0\,\Omega = 8.0\text{ V}\).
评分标准
C [1]
题目 24 · MCQ
1 分
A student adds aqueous sodium hydroxide dropwise to an aqueous solution of an unknown salt until the sodium hydroxide is in excess.
A green precipitate forms which is insoluble in excess sodium hydroxide.
Which metal ion is present in the solution?
A.\(\text{Cu}^{2+}\)
B.\(\text{Fe}^{2+}\)
C.\(\text{Fe}^{3+}\)
D.\(\text{Cr}^{3+}\)
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解题
\(\text{Fe}^{2+}\) ions react with aqueous sodium hydroxide to form a green precipitate of iron(II) hydroxide, \(\text{Fe(OH)}_2\), which is insoluble in excess aqueous sodium hydroxide. - \(\text{Cu}^{2+}\) gives a light blue precipitate. - \(\text{Fe}^{3+}\) gives a red-brown precipitate. - \(\text{Cr}^{3+}\) gives a green precipitate that dissolves in excess to give a green solution.
评分标准
B [1]
题目 25 · MCQ
1 分
A variegated leaf (having green and white areas) on a destarched plant is partially covered with a strip of black paper. The plant is placed in bright sunlight for several hours.
The leaf is then removed, decolourised in boiling ethanol, and tested with iodine solution.
Which part of the leaf turns blue-black?
A.The white part that was covered with black paper
B.The white part that was exposed to light
C.The green part that was exposed to light
D.The green part that was covered with black paper
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解题
Photosynthesis requires both light and chlorophyll to produce glucose, which is then stored as starch. - The green areas contain chlorophyll, while the white areas do not. - The uncovered areas receive light, while the covered areas do not.
Therefore, only the green parts exposed to sunlight can carry out photosynthesis and synthesise starch, turning blue-black when tested with iodine solution.
评分标准
C [1]
题目 26 · MCQ
1 分
A crane lifts a load of mass \(450\text{ kg}\) vertically upwards through a height of \(16\text{ m}\) in a time of \(12\text{ s}\).
The gravitational field strength \(g = 10\text{ N/kg}\).
What is the average power developed by the crane?
A.\(600\text{ W}\)
B.\(3800\text{ W}\)
C.\(6000\text{ W}\)
D.\(72\,000\text{ W}\)
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解题
1. Calculate the work done in lifting the load: \[ \text{Work done} = mgh = 450\text{ kg} \times 10\text{ N/kg} \times 16\text{ m} = 72\,000\text{ J} \] 2. Calculate the power developed: \[ \text{Power} = \frac{\text{Work done}}{\text{time}} = \frac{72\,000\text{ J}}{12\text{ s}} = 6\,000\text{ W} \]
评分标准
C ; [1]
题目 27 · MCQ
1 分
Which method is the most suitable for preparing a pure, dry sample of barium sulfate?
A.reacting barium carbonate with dilute hydrochloric acid and evaporating the solution to dryness
B.mixing aqueous barium chloride with dilute sulfuric acid, filtering, and washing the residue
C.adding excess barium metal to dilute sulfuric acid and crystallising the filtrate
D.titrating aqueous barium hydroxide with dilute sulfuric acid using an indicator
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解题
Barium sulfate is an insoluble salt. Insoluble salts are prepared by precipitation: mixing two soluble salt solutions (e.g., aqueous barium chloride and dilute sulfuric acid) to form an insoluble precipitate, which is then separated by filtration, washed with distilled water to remove impurities, and dried.
评分标准
B ; [1]
题目 28 · MCQ
1 分
A circuit contains a \(12\text{ V}\) power supply connected across two resistors in parallel: a \(20\,\Omega\) resistor and a \(30\,\Omega\) resistor.
What is the total current drawn from the power supply?
An aqueous solution of compound \(X\) undergoes two separate qualitative tests:
1. Adding aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. 2. Adding dilute nitric acid followed by aqueous silver nitrate produces a yellow precipitate.
What is the identity of compound \(X\)?
A.chromium(III) chloride
B.copper(II) bromide
C.iron(II) iodide
D.iron(III) iodide
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解题
- The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions (\(\text{Fe}^{2+}\)). - The formation of a yellow precipitate when acidified silver nitrate is added confirms the presence of iodide ions (\(\text{I}^-\)). Therefore, compound \(X\) is iron(II) iodide.
评分标准
C ; [1]
题目 30 · MCQ
1 分
Which row correctly describes the changes that occur as the temperature of an enzyme-catalysed reaction is increased from \(25\,^\circ\text{C}\) to the optimum temperature of \(40\,^\circ\text{C}\)?
A.Kinetic energy of molecules decreases; Frequency of effective collisions decreases
B.Kinetic energy of molecules increases; Frequency of effective collisions decreases
C.Kinetic energy of molecules increases; Frequency of effective collisions increases
D.Kinetic energy of molecules stays constant; Frequency of effective collisions increases
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解题
As temperature increases towards the optimum temperature, the kinetic energy of both the enzyme and substrate molecules increases. This causes them to move faster, increasing the frequency of collisions and the frequency of successful/effective collisions between substrate molecules and the enzyme's active site.
评分标准
C ; [1]
题目 31 · MCQ
1 分
A trolley of mass \(4.0\text{ kg}\) travels along a smooth, horizontal surface at an initial speed of \(2.0\text{ m/s}\). A constant forward horizontal force of \(12\text{ N}\) acts on the trolley over a distance of \(5.0\text{ m}\).
What is the final kinetic energy of the trolley?
A.\(60\text{ J}\)
B.\(68\text{ J}\)
C.\(80\text{ J}\)
D.\(128\text{ J}\)
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解题
1. Calculate the initial kinetic energy of the trolley: \[E_{k,\text{initial}} = \frac{1}{2}mv^2 = \frac{1}{2} \times 4.0\text{ kg} \times (2.0\text{ m/s})^2 = 8.0\text{ J}\]
2. Calculate the work done by the forward force: \[W = F \times d = 12\text{ N} \times 5.0\text{ m} = 60\text{ J}\]
3. The work done on the trolley increases its kinetic energy: \[E_{k,\text{final}} = E_{k,\text{initial}} + W = 8.0\text{ J} + 60\text{ J} = 68\text{ J}\]
Therefore, the correct option is B.
评分标准
B [1 mark]
题目 32 · MCQ
1 分
Oxides can be classified based on their acid-base properties.
- \(\text{SO}_2\) (sulfur dioxide) is a non-metal oxide that dissolves in water to form an acid; it is an acidic oxide. - \(\text{CaO}\) (calcium oxide) is a basic metal oxide that reacts with acids to form a salt and water. - \(\text{Al}_2\text{O}_3\) (aluminium oxide) reacts with both acids and bases; it is an amphoteric oxide.
Therefore, row A is correct.
评分标准
A [1 mark]
题目 33 · MCQ
1 分
A circuit contains three identical resistors, each with a resistance of \(6.0\,\Omega\).
Two of the resistors are connected in parallel with each other. This parallel pair is then connected in series with the third resistor.
What is the total combined resistance of this circuit arrangement?
A.\(2.0\,\Omega\)
B.\(4.0\,\Omega\)
C.\(9.0\,\Omega\)
D.\(18\,\Omega\)
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解题
1. Calculate the equivalent resistance \(R_p\) of the two \(6.0\,\Omega\) resistors in parallel: \[\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{6.0} = \frac{2}{6.0} \implies R_p = 3.0\,\Omega\]
2. Add the resistance of the third resistor in series: \[R_{\text{total}} = R_p + 6.0\,\Omega = 3.0\,\Omega + 6.0\,\Omega = 9.0\,\Omega\]
Therefore, the correct option is C.
评分标准
C [1 mark]
题目 34 · MCQ
1 分
An unknown solid is dissolved in distilled water and tested as follows:
1. Addition of aqueous sodium hydroxide produces a light blue precipitate that remains insoluble when excess sodium hydroxide is added. 2. Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate.
Which compound is present?
A.copper(II) chloride
B.copper(II) sulfate
C.iron(II) sulfate
D.zinc sulfate
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解题
- Formation of a light blue precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of copper(II) ions (\(\text{Cu}^{2+}\)). - Formation of a white precipitate upon adding dilute nitric acid followed by aqueous barium nitrate confirms the presence of sulfate ions (\(\text{SO}_4^{2-}\)).
Thus, the compound is copper(II) sulfate. The correct option is B.
评分标准
B [1 mark]
题目 35 · MCQ
1 分
Plant cells with a cell sap concentration equivalent to a \(0.4\text{ mol/dm}^3\) sucrose solution are placed into a beaker containing a \(0.8\text{ mol/dm}^3\) sucrose solution.
Which row correctly describes the net direction of water movement and the resulting state of the plant cells?
A.Net movement of water: into the cells | State of cells: turgid
B.Net movement of water: into the cells | State of cells: plasmolysed
C.Net movement of water: out of the cells | State of cells: plasmolysed
D.Net movement of water: out of the cells | State of cells: turgid
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解题
- The external solution has a higher solute concentration (\(0.8\text{ mol/dm}^3\)) than the cell sap (\(0.4\text{ mol/dm}^3\)), meaning the external solution has a lower water potential. - Water moves out of the plant cells down the water potential gradient by osmosis. - As water leaves, the volume of the vacuole and cytoplasm decreases, pulling the cell membrane away from the cell wall, causing the cells to become plasmolysed.
Therefore, row C is correct.
评分标准
C [1 mark]
题目 36 · MCQ
1 分
A trolley of mass \(2.0\text{ kg}\) travels along a horizontal surface with an initial velocity of \(3.0\text{ m/s}\). A constant resultant force of \(6.0\text{ N}\) acts on the trolley in the direction of motion for \(4.0\text{ s}\). What is the final kinetic energy of the trolley?
C is correct [1]. A is incorrect (incorrectly calculates kinetic energy from momentum or simple addition). B is incorrect (omits initial velocity \(u\), giving \(v = 12\text{ m/s}\) and \(E_k = 144\text{ J}\)). D is incorrect (forgets the factor of \(\frac{1}{2}\), giving \(mv^2 = 450\text{ J}\)).
题目 37 · MCQ
1 分
Excess solid zinc carbonate is added to dilute sulfuric acid to prepare a sample of pure hydrated zinc sulfate crystals. Which sequence of practical steps should be used to obtain the crystals from the reaction mixture?
A.filter to remove excess zinc carbonate \(\rightarrow\) heat filtrate to crystallisation point \(\rightarrow\) leave to cool and crystallise \(\rightarrow\) dry crystals between filter paper
B.evaporate mixture completely to dryness \(\rightarrow\) filter solid \(\rightarrow\) wash with cold water \(\rightarrow\) dry in an oven
C.heat mixture to boiling \(\rightarrow\) filter hot solution \(\rightarrow\) evaporate all water rapidly to dryness
D.filter to remove excess zinc carbonate \(\rightarrow\) add excess sulfuric acid \(\rightarrow\) evaporate to dryness
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解题
To prepare crystals of a soluble salt from an insoluble reactant: 1. Filter the mixture to remove unreacted excess zinc carbonate. 2. Heat the filtrate until the crystallisation point (saturation point) is reached. 3. Allow the hot saturated solution to cool and crystallise slowly. 4. Filter off the crystals and dry them carefully between sheets of filter paper.
评分标准
A is correct [1]. B is incorrect (evaporating to complete dryness produces anhydrous powder rather than hydrated crystals, and filtering after drying is impossible). C is incorrect (evaporating to dryness gives contaminated or anhydrous product without controlled crystallisation). D is incorrect (adding more sulfuric acid would introduce acid impurities).
题目 38 · MCQ
1 分
A circuit contains a \(12\text{ V}\) power supply connected to three resistors. A \(6.0\,\Omega\) resistor is connected in series with a parallel combination of two \(4.0\,\Omega\) resistors. What is the potential difference across the parallel combination of resistors?
A.\(2.0\text{ V}\)
B.\(3.0\text{ V}\)
C.\(6.0\text{ V}\)
D.\(9.0\text{ V}\)
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解题
1. Find equivalent resistance of the two \(4.0\,\Omega\) parallel resistors: \(R_p = \frac{4.0 \times 4.0}{4.0 + 4.0} = 2.0\,\Omega\). 2. Find total circuit resistance: \(R_{\text{total}} = 6.0\,\Omega + 2.0\,\Omega = 8.0\,\Omega\). 3. Find total circuit current: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{8.0\,\Omega} = 1.5\text{ A}\). 4. Calculate potential difference across the parallel combination: \(V_p = I \times R_p = 1.5\text{ A} \times 2.0\,\Omega = 3.0\text{ V}\).
评分标准
B is correct [1]. A is incorrect (confuses equivalent parallel resistance with voltage drop). C is incorrect (assumes equal voltage sharing across series and parallel parts). D is incorrect (calculates the potential difference across the \(6.0\,\Omega\) series resistor: \(1.5\text{ A} \times 6.0\,\Omega = 9.0\text{ V}\)).
题目 39 · MCQ
1 分
An aqueous solution of salt \(\mathbf{X}\) is tested as follows: - Addition of aqueous sodium hydroxide dropwise gives a green precipitate that is insoluble in excess sodium hydroxide. - Addition of dilute nitric acid followed by aqueous barium nitrate gives a white precipitate.
What is the identity of salt \(\mathbf{X}\)?
A.copper(II) sulfate
B.iron(II) sulfate
C.iron(II) chloride
D.iron(III) sulfate
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解题
1. The formation of a green precipitate with aqueous sodium hydroxide that remains insoluble in excess indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\). 2. The formation of a white precipitate (barium sulfate) upon adding acidified barium nitrate confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\). Therefore, salt \(\mathbf{X}\) is iron(II) sulfate.
评分标准
B is correct [1]. A is incorrect (copper(II) gives a light blue precipitate with sodium hydroxide). C is incorrect (chloride gives a white precipitate with silver nitrate, not barium nitrate). D is incorrect (iron(III) gives a red-brown precipitate with sodium hydroxide).
题目 40 · MCQ
1 分
Which row correctly states the main function of the palisade mesophyll and the tissue responsible for transporting sucrose away from a leaf?
A.Main function: site of most photosynthesis; Tissue transporting sucrose: phloem
B.Main function: site of most photosynthesis; Tissue transporting sucrose: xylem
C.Main function: main site of gas exchange; Tissue transporting sucrose: phloem
D.Main function: main site of gas exchange; Tissue transporting sucrose: xylem
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解题
Palisade mesophyll cells are packed with chloroplasts and arranged near the upper surface of the leaf to maximize light absorption, making them the primary site of photosynthesis. Sucrose and amino acids are translocated away from leaves via the phloem tissue (xylem transports water and dissolved mineral ions).
评分标准
A is correct [1]. B is incorrect (xylem transports water and minerals, not sucrose). C is incorrect (gas exchange is the primary function of spongy mesophyll and stomata). D is incorrect (wrong function for palisade mesophyll and wrong transport tissue).
Answer all questions. Show your working where required and provide appropriate units.
12 题目 · 120 分
题目 1 · Structured
10 分
(a) A toy electric cart of mass \(0.45\text{ kg}\) accelerates uniformly from rest along a straight horizontal track to a speed of \(3.6\text{ m/s}\) in a time of \(2.4\text{ s}\).
(iii) Calculate the resultant horizontal force acting on the cart during this acceleration. [2]
resultant force = .................................... \(\text{N}\)
(b) After \(2.4\text{ s}\), the cart travels at a constant speed of \(3.6\text{ m/s}\) before reaching a frictionless upward ramp.
(i) Calculate the kinetic energy of the cart when it is travelling at \(3.6\text{ m/s}\). [2]
kinetic energy = .................................... \(\text{J}\)
(ii) The cart travels up the frictionless ramp until it momentarily comes to rest at vertical height \(h\). Assuming all kinetic energy is converted into gravitational potential energy, calculate the maximum vertical height \(h\) reached by the cart. (acceleration due to gravity, \(g = 9.8\text{ m/s}^2\)) [2]
(b)(ii) (equating \(mgh = E_k\) or \(h = \frac{v^2}{2g}\)) \(h = \frac{2.916}{0.45 \times 9.8}\) or \(\frac{12.96}{19.6}\) (allow ecf from (b)(i)) [1]; \(0.66\) (\(\text{m}\)) (accept \(0.661\)) [1];
(b)(iii) energy cannot be created or destroyed / energy can only be transformed/transferred from one form to another [1];
题目 2 · Structured
10 分
Copper(II) sulfate is a soluble salt prepared by reacting an insoluble metal oxide with a dilute acid.
(a) Write a balanced chemical equation, including state symbols, for the reaction between solid copper(II) oxide, \(\text{CuO}\), and dilute sulfuric acid, \(\text{H}_2\text{SO}_4\). [2]
(b) Describe three practical steps to obtain pure, dry crystals of hydrated copper(II) sulfate starting from an excess of copper(II) oxide powder and dilute sulfuric acid. [3]
(c) Explain why an excess of copper(II) oxide is added to the sulfuric acid rather than an exact stoichiometric amount. [1]
(d) A student uses \(25.0\text{ cm}^3\) of \(1.20\text{ mol/dm}^3\) sulfuric acid in this preparation.
(i) Calculate the amount, in moles, of sulfuric acid used. [1]
(ii) Calculate the maximum mass of anhydrous copper(II) sulfate (\(\text{CuSO}_4\)) that can be produced. [\(M_r\): \(\text{CuSO}_4 = 160\)] [2]
mass = .................................... \(\text{g}\)
(e) Name an alternative copper compound that reacts with dilute sulfuric acid to form copper(II) sulfate in a neutralization reaction without producing a gas. [1]
(b) 1. Filter the mixture to remove the unreacted excess copper(II) oxide solid. 2. Heat the filtrate (aqueous copper(II) sulfate) gently in an evaporating dish to evaporate water until the crystallization point / saturation point is reached. 3. Allow the hot concentrated solution to cool slowly to form crystals, filter off the crystals, and dry them gently between sheets of filter paper (or in a warm desiccator/low-temperature oven).
(c) To ensure that all of the acid is completely reacted/neutralized so that the salt solution is pure and not contaminated with unreacted acid.
(b) filter (to remove excess copper(II) oxide) [1]; heat / evaporate filtrate to point of crystallisation / until saturated [1]; leave to cool/crystallise AND dry crystals with filter paper / in a warm oven [1];
(c) to ensure all the acid is used up / neutralised / so that no acid remains in the product [1];
(a) A circuit contains a \(12.0\text{ V}\) power supply connected to a series resistor \(R_1 = 6.0\ \Omega\) and a parallel branch containing two identical resistors, \(R_2 = 10.0\ \Omega\) and \(R_3 = 10.0\ \Omega\).
(i) Calculate the combined equivalent resistance of the two parallel resistors \(R_2\) and \(R_3\). [2]
(b)(ii) As current increases, the temperature of the filament rises. The positive metal ions vibrate faster / with greater amplitude, which increases the frequency of collisions between the delocalised conduction electrons and the vibrating lattice ions, impeding current flow.
(b)(ii) temperature of filament increases / ions vibrate more (vigorously) [1]; more frequent collisions between electrons and ions [1];
题目 4 · Structured
10 分
A student carries out chemical tests to identify the ions present in an unknown salt, compound \(\mathbf{X}\). Compound \(\mathbf{X}\) contains one cation and one anion.
(a) (i) Solid \(\mathbf{X}\) is dissolved in distilled water to make an aqueous solution. When aqueous sodium hydroxide is added dropwise, a green precipitate forms which is insoluble in excess sodium hydroxide.
Identify the cation present in compound \(\mathbf{X}\). [1]
(ii) To a fresh portion of the solution of \(\mathbf{X}\), dilute nitric acid is added followed by aqueous barium nitrate. A dense white precipitate is formed.
Identify the anion present in compound \(\mathbf{X}\). [1]
(iii) Write the ionic equation, including state symbols, for the precipitation reaction in (a)(ii). [2]
(iv) Deduce the chemical formula of compound \(\mathbf{X}\). [1]
(b) Paper chromatography is used to separate and identify dyes in a mixture.
(i) Explain why the baseline on the chromatography paper must be drawn in pencil and not in ink. [1]
(ii) In an experiment, a spot of yellow dye travels a distance of \(4.2\text{ cm}\) from the baseline while the solvent front travels a distance of \(7.0\text{ cm}\). Calculate the \(R_f\) value of this yellow dye. [2]
\(R_f\) value = ....................................
(iii) State one factor that determines how far a dye travels up the chromatography paper. [1]
(iv) Describe how colorless substances, such as amino acids, can be located on a chromatogram after separation. [1]
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解题
(a)(i) Addition of aqueous \(\text{NaOH}\) giving a green precipitate insoluble in excess indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\).
(a)(ii) Addition of dilute nitric acid followed by aqueous barium nitrate giving a white precipitate indicates sulfate ions, \(\text{SO}_4^{2-}\).
(a)(iii) The reaction between barium ions and sulfate ions produces insoluble barium sulfate: \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\).
(a)(iv) Combining \(\text{Fe}^{2+}\) and \(\text{SO}_4^{2-}\) gives \(\text{FeSO}_4\).
(b)(i) Ink would dissolve in the mobile solvent and separate into its own dye components, interfering with the chromatogram, whereas pencil graphite is insoluble in the solvent.
(b)(ii) \(R_f = \frac{\text{distance travelled by substance}}{\text{distance travelled by solvent front}} = \frac{4.2\text{ cm}}{7.0\text{ cm}} = 0.60\).
(b)(iii) The solubility of the dye in the solvent (mobile phase) or its attraction/affinity to the paper (stationary phase).
(b)(iv) Spraying the chromatogram with a locating agent (such as ninhydrin) and then heating/drying it to develop visible coloured spots.
评分标准
(a)(i) iron(II) / \(\text{Fe}^{2+}\) [1];
(a)(ii) sulfate / \(\text{SO}_4^{2-}\) [1];
(a)(iii) \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\) correct species [1]; correct state symbols [1];
(a)(iv) \(\text{FeSO}_4\) [1];
(b)(i) pencil (graphite) is insoluble in the solvent / ink would dissolve and separate/contaminate chromatogram [1];
(b)(iii) solubility in the solvent / affinity to the stationary phase (paper) [1];
(b)(iv) spray with a locating agent (allow ninhydrin) / shine UV light [1];
题目 5 · Structured
10 分
Plants carry out photosynthesis to synthesize glucose.
(a) (i) State the balanced chemical equation for photosynthesis. [2]
(ii) Name the green pigment that absorbs light energy for photosynthesis and state the specific plant cell organelle in which this pigment is found. [2]
pigment: ....................................
organelle: ....................................
(b) An investigation was conducted into the rate of photosynthesis of an aquatic plant at different light intensities and two different concentrations of carbon dioxide (\(0.04\%\) and \(0.12\%\)), while keeping temperature constant at \(20^\circ\text{C}\).
(i) Describe the effect of increasing light intensity on the rate of photosynthesis at \(0.04\%\) carbon dioxide concentration. [2]
(ii) Explain why the rate of photosynthesis becomes constant at high light intensity when the carbon dioxide concentration is \(0.04\%\). [2]
(c) Water is an essential reactant for photosynthesis.
Name the tissue that transports water from the roots to the leaves and describe one structural adaptation of this tissue for its function. [2]
tissue: ....................................
adaptation: ....................................
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解题
(a)(i) The balanced chemical equation for photosynthesis is: \(6\text{CO}_2 + 6\text{H}_2\text{O} \xrightarrow{\text{light and chlorophyll}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\).
(a)(ii) The pigment is chlorophyll, located within chloroplasts.
(b)(i) At low to moderate light intensities, the rate of photosynthesis increases steadily / proportionally as light intensity increases. At high light intensities, the rate of photosynthesis levels off / remains constant (plateaus).
(b)(ii) Light intensity is no longer the limiting factor; carbon dioxide concentration (or temperature) has become the factor that is in shortest supply / limiting the rate of reaction.
(c) Tissue: Xylem. Structural adaptation: Xylem vessels are hollow, continuous tubes with no end walls (or no cytoplasm/cell contents) to allow unimpeded water flow; walls are reinforced with lignin to provide structural support and prevent collapse under tension.
评分标准
(a)(i) \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\) correct formulas of reactants and products [1]; correct balancing [1];
(a)(ii) chlorophyll [1]; chloroplast [1];
(b)(i) (initially) rate increases as light intensity increases [1]; levels off / reaches a plateau / becomes constant at higher light intensity [1];
(b)(ii) light intensity is no longer the limiting factor [1]; carbon dioxide (concentration) is the limiting factor / is in short supply [1];
(c) xylem [1]; hollow / dead cells / no end walls / lignified walls / waterproof walls [1];
题目 6 · Structured
10 分
**1** A motorized test cart travels along a straight, horizontal track.
**(a)** The cart accelerates uniformly from rest to a speed of \(12\text{ m/s}\) in a time of \(4.0\text{ s}\).
(i) State the type of motion shown by the cart during the first \(4.0\text{ s}\). [1]
(ii) Calculate the acceleration of the cart during this time. [2]
**(c)** After several seconds, the cart reaches a steady speed of \(18\text{ m/s}\). At this speed, the total resistive forces opposing motion are \(350\text{ N}\).
(i) Calculate the useful power output developed by the motor when moving at \(18\text{ m/s}\). State the unit. [3]
**(c)(i)** Since speed is constant, driving force = resistive force = \(350\text{ N}\). $$P = F \times v = 350\text{ N} \times 18\text{ m/s} = 6300\text{ W}$$ Unit: \(\text{W}\) or \(\text{J/s}\).
**2** Soluble salts can be prepared by reacting an insoluble base or carbonate with a dilute acid.
**(a)** A student prepares pure, dry crystals of hydrated copper(II) sulfate, \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\), using solid copper(II) oxide and dilute sulfuric acid.
(i) State why an excess of copper(II) oxide is added to the warm dilute sulfuric acid. [1]
(ii) Name the technique used to remove the unreacted copper(II) oxide from the reaction mixture. [1]
(iii) Describe how the student obtains pure, dry crystals of hydrated copper(II) sulfate from the filtrate. [3]
**(b)** Copper(II) sulfate can also be prepared by reacting solid copper(II) carbonate with dilute sulfuric acid.
Write a balanced chemical equation, including state symbols, for this reaction. [2]
**(c)** The student carries out a qualitative test to confirm the presence of sulfate ions in the copper(II) sulfate solution.
(i) State the observation made when aqueous barium nitrate is added to the acidified solution. [1]
(ii) Write the ionic equation, including state symbols, for the reaction that occurs during this test. [2]
[Total: 10]
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解题
**(a)(i)** Excess copper(II) oxide is added to ensure all the sulfuric acid is completely neutralized/used up so that unreacted acid does not contaminate the salt crystals.
**(a)(ii)** Filtration is used to separate the unreacted solid copper(II) oxide from the copper(II) sulfate solution.
**(a)(iii)** 1. Heat the copper(II) sulfate solution to evaporate water until the point of crystallization (or until a saturated solution is formed). 2. Leave the solution to cool and allow crystals to form. 3. Filter to separate the crystals and dry them between sheets of filter paper or in a low-temperature oven.
**(a)(i)** * to ensure all the (sulfuric) acid is reacted / completely neutralized ; [1]
**(a)(ii)** * filtration / filter ; [1]
**(a)(iii)** * heat / evaporate until crystallization point / saturation point ; [1] * leave to cool / crystallize ; [1] * filter (to collect crystals) AND dry with filter paper / in warm desiccator / warm oven ; [1]
**(b)** A potential divider circuit consists of a \(9.0\text{ V}\) power supply connected in series with a fixed resistor of resistance \(400\,\Omega\) and a thermistor.
(i) State what happens to the electrical resistance of the thermistor as its temperature increases. [1]
(ii) At a particular temperature, the resistance of the thermistor is \(800\,\Omega\).
Calculate the potential difference across the thermistor at this temperature. [3]
**(c)** Copper has delocalised / free electrons that can move through the structure and conduct electric charge. In plastic, all electrons are held tightly in covalent bonds and are not free to move.
**(a)(ii)** * \(I = \frac{V}{R}\) or \(\frac{12.0}{12}\) (ecf from (a)(i)) ; [1] * \(1.0\) (\(\text{A}\)) ; [1]
**(b)(i)** * resistance decreases ; [1]
**(b)(ii)** * total resistance \(= 400 + 800 = 1200\,(\Omega)\) or current \(I = \frac{9.0}{1200} = 0.0075\text{ (A)}\) ; [1] * \(V = 9.0 \times \frac{800}{1200}\) or \(0.0075 \times 800\) ; [1] * \(6.0\) (\(\text{V}\)) ; [1]
**(c)** * (copper has) delocalised / free electrons that are free to move (and carry charge) ; [1] * (plastic has) no free electrons / electrons are held in fixed positions / fixed in bonds ; [1]
题目 9 · Structured
10 分
**4** Chemical analysis and separation techniques are essential for identifying substances and purifying mixtures.
**(a)** A sample of a solid ionic salt, **X**, is tested.
(i) Solid **X** is dissolved in water and warmed with aqueous sodium hydroxide and aluminium foil. A pungent gas is given off which turns damp red litmus paper blue.
Identify the gas evolved and name the anion present in salt **X**. [2]
**(b)** Paper chromatography is used to separate and identify dyes present in a food coloring mixture.
(i) Explain why the baseline on the chromatography paper must be drawn in pencil rather than pen ink. [1]
(ii) Explain why the solvent level in the chromatography tank must be below the baseline at the start of the experiment. [1]
(iii) In a chromatography experiment, a dye travels a distance of \(5.4\text{ cm}\) from the baseline while the solvent front travels \(9.0\text{ cm}\).
Calculate the \(R_\text{f}\) value of this dye. [2]
**(c)** Name the separation technique used to obtain pure water from a mixture of sodium chloride and water. [1]
**(d)** Describe the test and positive result used to identify chlorine gas. [1]
[Total: 10]
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解题
**(a)(i)** Adding aqueous \(\text{NaOH}\) and \(\text{Al}\) foil followed by warming reduces nitrate ions (\(\text{NO}_3^-\)) to ammonia gas (\(\text{NH}_3\)), which is alkaline and turns damp red litmus paper blue.
**(a)(ii)** A green precipitate insoluble in excess sodium hydroxide indicates iron(II) ions, \(\text{Fe}^{2+}\).
**(a)(iii)** Combining \(\text{Fe}^{2+}\) and \(\text{NO}_3^-\) gives \(\text{Fe(NO}_3)_2\).
**(b)(i)** Graphite in pencil is insoluble in chromatography solvents, so it will not dissolve or produce extraneous spots that interfere with the chromatogram.
**(b)(ii)** If the solvent level is above the baseline, the dye samples will dissolve directly into the solvent reservoir at the bottom of the beaker instead of traveling up the paper.
**(b)(iii)** $$R_\text{f} = \frac{\text{distance moved by spot}}{\text{distance moved by solvent front}} = \frac{5.4\text{ cm}}{9.0\text{ cm}} = 0.60$$
**(c)** Simple distillation.
**(d)** Chlorine bleaches damp litmus paper (or turns damp blue litmus paper red and then bleaches it white).
**(b)** The internal structure of a leaf is adapted to maximize the efficiency of photosynthesis.
Describe how each of the following leaf structures is adapted for its function in photosynthesis:
(i) palisade mesophyll layer [1]
(ii) spongy mesophyll layer [1]
(iii) xylem tissue [1]
**(c)** A student investigates the rate of photosynthesis in an aquatic plant by counting the number of oxygen bubbles released per minute at different light intensities.
(i) State two variables that must be kept constant during this investigation to ensure valid results. [2]
(ii) As light intensity is increased, the rate of photosynthesis initially rises, but eventually levels off and remains constant.
Explain why the rate of photosynthesis does not increase further at very high light intensities. [2]
**(d)** State the main function of stomata in a plant leaf. [1]
[Total: 10]
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解题
**(a)** $$6\text{CO}_2 + 6\text{H}_2\text{O} \xrightarrow{\text{light and chlorophyll}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$$
**(b)(i)** Palisade mesophyll cells are column-shaped, closely packed, and contain numerous chloroplasts situated near the upper epidermis to absorb the maximum amount of light.
**(b)(ii)** Spongy mesophyll cells are loosely packed with large intercellular air spaces to facilitate rapid diffusion of gases (carbon dioxide in, oxygen out) through the leaf.
**(b)(iii)** Xylem vessels consist of hollow, continuous tubes with lignified walls that transport water and dissolved mineral ions from roots to photosynthetic cells in the leaf.
**(c)(i)** Two variables to keep constant: 1. Water temperature (e.g., using a thermostatically controlled water bath or heat shield). 2. Carbon dioxide concentration (e.g., by adding a fixed concentration of sodium hydrogencarbonate).
**(c)(ii)** At high light intensity, light is no longer the factor in shortest supply (limiting factor). The rate of photosynthesis is now restricted by another limiting factor, such as the concentration of carbon dioxide or temperature (which limits enzyme activity in the light-independent reactions).
**(d)** Stomata allow gas exchange (entry of \(\text{CO}_2\) and release of \(\text{O}_2\)) and allow the loss of water vapour during transpiration.
**(b)(i)** * packed / dense with chloroplasts OR located near upper surface (to absorb maximum sunlight) ; [1]
**(b)(ii)** * presence of air spaces / loosely packed (to allow diffusion / gas exchange of \(\text{CO}_2\) / \(\text{O}_2\)) ; [1]
**(b)(iii)** * hollow / continuous tubes (or lignified walls) to transport water (and mineral ions) to leaf cells ; [1]
**(c)(i)** * temperature (of water / surroundings) ; [1] * carbon dioxide concentration (or concentration of \(\text{NaHCO}_3\)) ; [1] *(accept: wavelength / color of light, mass / length of plant piece)*
**(c)(ii)** * light intensity is no longer the limiting factor ; [1] * another factor is limiting / named limiting factor (e.g. \(\text{CO}_2\) concentration / temperature / number of chloroplasts) ; [1]
**(d)** * allows gas exchange (diffusion of \(\text{CO}_2\) in / \(\text{O}_2\) out) / transpiration / water vapor release ; [1]
题目 11 · Structured
10 分
A student prepares pure, dry crystals of hydrated zinc sulfate, \(\text{ZnSO}_4 \cdot 7\text{H}_2\text{O}\), by reacting solid zinc carbonate with dilute sulfuric acid.
(a) (i) State two observations that indicate a chemical reaction is taking place when zinc carbonate is added to dilute sulfuric acid. [2]
(ii) Explain why an excess of zinc carbonate is used in this preparation. [1]
(iii) Describe how the student obtains pure, dry crystals of hydrated zinc sulfate from the reaction mixture. [3]
(b) (i) Write the balanced chemical equation, including state symbols, for the reaction between solid zinc carbonate and dilute sulfuric acid. [2]
(ii) Describe a chemical test and state the expected observation to confirm the presence of sulfate ions, \(\text{SO}_4^{2-}\), in an aqueous solution of zinc sulfate. [2]
[Total: 10]
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解题
(a) (i) When a metal carbonate reacts with an acid, carbon dioxide gas and a soluble salt are produced. Visible signs include fizzing / effervescence / bubbling and the solid carbonate dissolving / decreasing in amount.
(ii) Adding excess solid ensures that all of the sulfuric acid is used up, so the resulting salt solution is neutral and unpolluted by leftover acid.
(iii) Step 1: Filter the mixture to remove the unreacted solid zinc carbonate residue. Step 2: Heat the filtrate gently in an evaporating dish until the crystallisation point is reached (or when crystals start to form on a glass rod). Step 3: Allow the concentrated solution to cool slowly to form crystals. Step 4: Filter off the crystals and dry them gently between sheets of filter paper or in a low-temperature desiccator/oven.
(b) (i) The reaction between solid zinc carbonate and aqueous sulfuric acid produces aqueous zinc sulfate, carbon dioxide gas, and liquid water: \(\text{ZnCO}_3(\text{s}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{ZnSO}_4(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\)
(ii) Acidify the solution with dilute nitric acid (or dilute hydrochloric acid) to prevent interference from carbonate ions, then add aqueous barium nitrate (or aqueous barium chloride). Sulfate ions react with barium ions to form an insoluble white precipitate of barium sulfate (\(\text{BaSO}_4\)).
评分标准
(a) (i) Any two from: • effervescence / fizzing / bubbles (of gas); [1] • solid disappears / dissolves; [1] • mixture warms up / temperature rises; [1] (max [2])
(ii) to ensure all the (sulfuric) acid is used up / reacted / neutralised; [1]
(iii) filter (to remove excess zinc carbonate); [1] heat / evaporate filtrate to crystallisation point / saturation / reduce volume; [1] leave to cool and crystallise AND dry crystals between filter papers / in a warm oven (below \(100^{\circ}\text{C}\)); [1]
(ii) add dilute nitric acid AND aqueous barium nitrate / add dilute hydrochloric acid AND aqueous barium chloride; [1] white precipitate (formed); [1]
题目 12 · Structured
10 分
A delivery drone carries a parcel vertically upwards from rest. The mass of the drone is \(2.4\text{ kg}\) and the mass of the parcel is \(0.60\text{ kg}\). Take the acceleration due to gravity, \(g\), to be \(9.8\text{ N/kg}\).
(a) (i) Calculate the combined weight of the drone and parcel. [1]
(ii) The drone accelerates vertically upwards from rest with an acceleration of \(1.5\text{ m/s}^2\). Calculate the upward thrust force exerted by the drone's rotors during this acceleration. [3]
thrust force = ..................................................... \(\text{N}\)
(b) After accelerating, the drone climbs at a constant vertical velocity of \(4.0\text{ m/s}\).
(i) Calculate the kinetic energy of the parcel alone when moving at \(4.0\text{ m/s}\). [2]
kinetic energy = ..................................................... \(\text{J}\)
(ii) The drone lifts the drone and parcel through a vertical height of \(25\text{ m}\) at this constant speed. Calculate the increase in gravitational potential energy of the combined drone and parcel. [2]
increase in potential energy = ..................................................... \(\text{J}\)
(iii) Calculate the minimum useful power output of the drone's motors while lifting the combined mass at \(4.0\text{ m/s}\). [2]
power = ..................................................... \(\text{W}\)
(ii) Resultant force \(F = ma = 3.0\text{ kg} \times 1.5\text{ m/s}^2 = 4.5\text{ N}\). Since the drone is accelerating upwards: Resultant force = Thrust \(- W\) \(\text{Thrust} = F + W = 4.5\text{ N} + 29.4\text{ N} = 33.9\text{ N}\).
(b) (i) Kinetic energy of the parcel alone (\(m = 0.60\text{ kg}\)): \(E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.60 \times (4.0)^2 = 0.30 \times 16 = 4.8\text{ J}\).
(ii) Increase in gravitational potential energy of combined mass (\(m = 3.0\text{ kg}\)): \(\Delta E_p = mg\Delta h = 3.0 \times 9.8 \times 25 = 735\text{ J}\).
(iii) At constant speed, upward thrust equals total weight = \(29.4\text{ N}\). Power \(P = F \times v = 29.4\text{ N} \times 4.0\text{ m/s} = 117.6\text{ W}\) (or \(P = \frac{W}{t} = \frac{735\text{ J}}{25/4.0\text{ s}} = \frac{735}{6.25} = 117.6\text{ W}\)).
Answer all experimental and planning questions based on laboratory observations.
6 题目 · 60 分
题目 1 · Practical/Structured
10 分
A student investigates the effect of sucrose concentration on the mass of potato cylinders.
(a) The student uses a cork borer to cut cylinders from a large potato. (i) State how the student should prepare the potato cylinders to ensure fair testing before putting them into the solutions. .................................................................................................................................. [1] (ii) Name a piece of apparatus suitable for measuring the initial length of the cylinders. .................................................................................................................................. [1]
(b) Five potato cylinders are placed into test-tubes containing sucrose solutions of different concentrations: \(0.0\text{ mol/dm}^3\) (distilled water), \(0.2\text{ mol/dm}^3\), \(0.4\text{ mol/dm}^3\), \(0.6\text{ mol/dm}^3\), and \(0.8\text{ mol/dm}^3\).
After 60 minutes, the student removes the cylinders, blots them gently with a paper towel, and measures their final masses.
Table 1.1 shows the student's results.
**Table 1.1** | concentration of sucrose / \(\text{mol/dm}^3\) | initial mass / g | final mass / g | change in mass / g | percentage change in mass / % | | :--- | :--- | :--- | :--- | :--- | | 0.0 | 3.20 | 3.68 | +0.48 | +15.0 | | 0.2 | 3.15 | 3.37 | +0.22 | +7.0 | | 0.4 | 3.25 | 3.12 | -0.13 | -4.0 | | 0.6 | 3.10 | 2.79 | -0.31 | .................... | | 0.8 | 3.30 | 2.81 | .................... | -14.8 |
(i) Complete Table 1.1 by calculating: - the percentage change in mass for \(0.6\text{ mol/dm}^3\) sucrose solution - the change in mass for \(0.8\text{ mol/dm}^3\) sucrose solution. [2]
(ii) Explain why calculating the percentage change in mass is necessary rather than comparing only the change in mass. [1]
(c) (i) Describe the relationship between the concentration of sucrose solution and the percentage change in mass of the potato cylinders. [1]
(ii) Use the results in Table 1.1 to estimate the concentration of sucrose solution that is isotonic (equal in concentration) to the potato cell sap. Explain your answer. [2]
(d) State two variables that must be kept constant during this investigation. [2]
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解题
(a) (i) Ensure the outer skin is removed and all cylinders are trimmed to the same length or surface area. (ii) A ruler graduated in millimetres or Vernier callipers.
(b) (i) For \(0.6\text{ mol/dm}^3\): \(\text{percentage change} = \frac{-0.31}{3.10} \times 100\% = -10.0\%\) For \(0.8\text{ mol/dm}^3\): \(\text{change in mass} = 2.81 - 3.30 = -0.49\text{ g}\)
(ii) The potato cylinders have different initial starting masses, so percentage change allows a valid comparison.
(c) (i) As the concentration of sucrose increases, the mass change decreases from gaining mass to losing mass. (ii) The estimated concentration is between \(0.2\) and \(0.4\text{ mol/dm}^3\) (approximately \(0.30\text{ mol/dm}^3\)) because this is where the graph/line crosses zero change in mass, meaning no net movement of water into or out of cells.
(d) Temperature of the water bath/room; volume of sucrose solution used in each tube; duration of immersion (time).
评分标准
(a)(i) cut to identical initial lengths / remove all skin / cut with same borer diameter; [1] (a)(ii) ruler / Vernier calipers; [1] (b)(i) \(-10.0\) (allow \(-10\)); [1] \(-0.49\) (allow \(-0.49\text{ g}\), minus sign required); [1] (b)(ii) initial masses of potato cylinders were different / enables valid comparison; [1] (c)(i) as sucrose concentration increases, (percentage) change in mass decreases / ORA; [1] (c)(ii) value between \(0.25\) and \(0.35\text{ mol/dm}^3\) (inclusive); [1] explanation: point where there is no change in mass / \(0\%\) change / no net movement of water; [1] (d) any two from: temperature; volume of solution; surface area / diameter of cylinder; immersion time; type/source of potato; [2]
题目 2 · Practical/Structured
10 分
A student carries out tests to identify the ions present in solid **E** and to analyse dilute acid **F**.
(a) Solid **E** is a pale green salt. The student dissolves solid **E** in distilled water to make an aqueous solution.
(i) To a \(2\text{ cm}^3\) portion of solution **E**, aqueous sodium hydroxide is added dropwise until in excess. A green precipitate is formed, which slowly turns brown at the surface. State the name of the cation present in solid **E**. .................................................................................................................................. [1]
(ii) To another \(2\text{ cm}^3\) portion of solution **E**, dilute nitric acid followed by aqueous barium nitrate is added. A white precipitate is formed. State the name of the anion present in solid **E**. .................................................................................................................................. [1]
(iii) Give the chemical formula of solid **E**. .................................................................................................................................. [1]
(b) Solution **F** is a colourless dilute mineral acid.
(i) A piece of universal indicator paper is dipped into solution **F** and turns red. State an approximate pH value for solution **F**. \(\text{pH} = \)................................................................................................................... [1]
(ii) A small piece of magnesium ribbon is added to \(5\text{ cm}^3\) of solution **F** in a test-tube. Rapid effervescence is observed and gas **G** is produced. Describe the test to confirm that gas **G** is hydrogen. test ............................................................................................................................. result ........................................................................................................................... [2]
(c) The student titrates \(25.0\text{ cm}^3\) of solution **F** against \(0.100\text{ mol/dm}^3\) sodium hydroxide solution using a burette.
(i) The initial burette reading is \(1.30\text{ cm}^3\) and the final burette reading is \(23.80\text{ cm}^3\). Calculate the titre volume of sodium hydroxide used. \(\text{titre volume} = \).................................................................................... \(\text{cm}^3\) [1]
(ii) Name the piece of volumetric apparatus used to transfer precisely \(25.0\text{ cm}^3\) of solution **F** into the conical flask. .................................................................................................................................. [1]
(iii) Name a suitable indicator for this titration and state the colour change observed at the end-point. indicator ..................................................................................................................... colour change from ..................................................... to ..................................... [2]
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解题
(a) (i) The formation of a green precipitate turning brown on exposure to air confirms the presence of iron(II) ions (\(\text{Fe}^{2+}\)). (ii) The addition of dilute nitric acid followed by aqueous barium nitrate forming a white precipitate of barium sulfate confirms sulfate ions (\(\text{SO}_4^{2-}\)). (iii) The salt contains \(\text{Fe}^{2+}\) and \(\text{SO}_4^{2-}\), so the formula is \(\text{FeSO}_4\).
(b) (i) Red colour on universal indicator corresponds to a strong acid, pH 1 to 2. (ii) Test: Introduce a lighted splint to the mouth of the test-tube. Result: Burns with a squeaky 'pop'.
(c) (i) \(\text{Titre volume} = 23.80 - 1.30 = 22.50\text{ cm}^3\). (ii) A volumetric pipette (pipette) is used to accurately measure \(25.0\text{ cm}^3\). (iii) Phenolphthalein indicator: changes from colourless (in acid) to pale pink (at neutralization) / OR methyl orange: changes from red to orange/yellow.
评分标准
(a)(i) iron(II) / \(\text{Fe}^{2+}\) (reject iron / iron(III)); [1] (a)(ii) sulfate / \(\text{SO}_4^{2-}\) (reject sulfite / sulfide); [1] (a)(iii) \(\text{FeSO}_4\); [1] (b)(i) \(1\) or \(2\) (allow \(0\) to \(2\)); [1] (b)(ii) lighted / burning splint; [1] 'pop' sound / squeaky pop; [1] (c)(i) \(22.50\) (allow \(22.5\)); [1] (c)(ii) (volumetric) pipette; [1] (c)(iii) phenolphthalein: colourless to (pale) pink / methyl orange: red to orange (or yellow) / thymolphthalein: colourless to blue; [2] (1 mark for named indicator, 1 mark for correct corresponding colour change)
题目 3 · Practical/Structured
10 分
A student investigates how the electrical resistance of a metallic wire depends on its length.
(a) The student sets up a circuit with a power source, an ammeter, a voltmeter, and a length \(l\) of resistance wire connected between two crocodile clips. State the meter connected in series with the wire and the meter connected in parallel across the wire. in series: ................................................................................................................. in parallel: .............................................................................................................. [1]
(b) The student records the potential difference \(V\) across different lengths \(l\) of wire and the current \(I\) flowing through the wire. Table 3.1 shows the measurements.
(i) Calculate the resistance \(R\) of the wire when \(l = 60.0\text{ cm}\) using the formula: \[ R = \frac{V}{I} \] Write your value in Table 3.1. [1]
(ii) On the grid, plot a graph of resistance \(R\) (y-axis) against length \(l\) (x-axis). Draw the straight line of best fit. [3]
(c) (i) Determine the gradient of your line of best fit. Show clearly on your graph how you obtained the values needed. \(\text{gradient} = \).................................................................................... \(\Omega/\text{cm}\) [2]
(ii) State what your graph shows about the relationship between the resistance \(R\) and length \(l\) of the wire. [1]
(d) State two precautions the student should take during the experiment to avoid heating effects in the wire that could affect the accuracy of the resistance measurements. [2]
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解题
(a) An ammeter is connected in series with the component to measure current. A voltmeter is connected in parallel across the component to measure potential difference.
(b) (i) For \(l = 60.0\text{ cm}\): \[ R = \frac{1.44\text{ V}}{0.40\text{ A}} = 3.60\text{ }\Omega \]
(ii) Plotting points on axes: - x-axis: \(l\) from 0 to 100 cm (scales: 1 large division = 20 cm) - y-axis: \(R\) from 0 to 6.0 \(\Omega\) (scales: 1 large division = 1.0 \(\Omega\)) Points plotted accurately to within half a small square, with a single straight best-fit line passing through the origin \((0,0)\).
(c) (i) Gradient calculation using triangle > 50% of the line: \[ \text{gradient} = \frac{\Delta R}{\Delta l} = \frac{6.00 - 0}{100.0 - 0} = 0.060\text{ }\Omega/\text{cm} \]
(ii) Since the graph is a straight line that passes through the origin, resistance \(R\) is directly proportional to length \(l\).
(d) 1. Switch off the power supply between taking readings so the wire does not heat up. 2. Keep current small by using a variable resistor or low supply voltage.
评分标准
(a) in series: ammeter AND in parallel: voltmeter; [1] (b)(i) \(3.60\) / \(3.6\); [1] (b)(ii) suitable linear scales occupying \(\ge 50\%\) of grid in both directions; [1] all 5 points accurately plotted to within half a small square; [1] single straight best-fit line drawn with a ruler passing through origin; [1] (c)(i) correct coordinates chosen from line with large triangle (hypotenuse \(> 50\%\) of line); [1] calculation evaluated correctly giving \(0.060\) (allow \(0.058\)–\(0.062\)); [1] (c)(ii) (resistance is) directly proportional to length / straight line passing through origin; [1] (d) any two from: switch off circuit / open switch between readings; use small currents / low voltage; keep room temperature constant; [2]
题目 4 · Practical/Structured
10 分
A student investigates the extension of a helical spring and uses it to determine the density of a small irregular stone.
(a) The student measures the unstretched length \(l_0\) of the spring. The reading on the metre ruler is \(4.5\text{ cm}\). (i) State one technique the student should use when reading the ruler to avoid parallax errors. .................................................................................................................................. [1] (ii) State one other precaution to take to ensure accurate measurement of length. .................................................................................................................................. [1]
(b) Various loads \(F\) are suspended from the spring and the new length \(l\) is measured. Table 4.1 shows the results.
(i) Complete Table 4.1 by calculating the extension \(e\) for \(F = 4.0\text{ N}\). [1]
(ii) Calculate the spring constant \(k\) of the spring in \(\text{N/cm}\) using the formula: \[ k = \frac{F}{e} \] \(k = \).................................................................................... \(\text{N/cm}\) [1]
(c) The student attaches the irregular stone to the spring. The measured length of the spring becomes \(10.5\text{ cm}\). (i) Calculate the extension caused by the stone. \(\text{extension} = \).................................................................................... \(\text{cm}\) [1] (ii) Using your answer to (b)(ii) and (c)(i), determine the weight \(W\) of the stone. \(W = \).................................................................................... N [1]
(d) The student now determines the volume \(V\) of the stone. (i) Describe how the student can measure the volume of the stone using a measuring cylinder and water. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. [2]
(ii) The mass of the stone is found to be \(60.0\text{ g}\) and its volume is \(24.0\text{ cm}^3\). Calculate the density \(\rho\) of the stone. State the unit. \(\rho = \).................................................................... unit ............................... [2]
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解题
(a) (i) View the scale perpendicular to the line of sight / at eye-level to avoid line of sight displacement (parallax error). (ii) Use a set-square against the ruler and the bottom of the spring / ensure ruler is vertical.
(c) (i) Extension \(e = 10.5 - 4.5 = 6.0\text{ cm}\). (ii) Weight \(W = k \times e = 0.40\text{ N/cm} \times 6.0\text{ cm} = 2.4\text{ N}\).
(d) (i) Partially fill a measuring cylinder with water and record the initial volume \(V_1\). Lower the stone gently into the water until completely submerged and record the new volume \(V_2\). The volume of the stone is \(V = V_2 - V_1\). (ii) \(\text{Density } \rho = \frac{\text{mass}}{\text{volume}} = \frac{60.0\text{ g}}{24.0\text{ cm}^3} = 2.50\text{ g/cm}^3\).
评分标准
(a)(i) line of sight perpendicular to ruler / viewing at eye level; [1] (a)(ii) use of set square / ensure ruler is held vertically / place fiducial marker at base; [1] (b)(i) \(10.0\); [1] (b)(ii) \(0.40\) (allow \(0.4\)); [1] (c)(i) \(6.0\); [1] (c)(ii) \(2.4\) (allow ecf from (b)(ii) and (c)(i)); [1] (d)(i) pour water into cylinder and record initial volume; [1] submerge stone completely and find difference between new volume and initial volume; [1] (d)(ii) \(2.5\) / \(2.50\); [1] \(\text{g/cm}^3\) (or \(\text{g cm}^{-3}\)); [1]
题目 5 · Practical/Structured
10 分
Calcium carbonate reacts with dilute hydrochloric acid to produce carbon dioxide gas: \[ \text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)} \]
Plan an experiment to investigate how the concentration of hydrochloric acid affects the rate of this reaction by collecting and measuring the volume of gas produced.
You are provided with: - large excess of calcium carbonate chips of uniform size - dilute hydrochloric acid of concentration \(2.0\text{ mol/dm}^3\) - distilled water - standard laboratory apparatus and glassware.
You are not required to carry out this experiment.
In your plan, you should: - state any additional apparatus needed - describe the experimental procedure, including how at least four different concentrations of acid are prepared - state the measurements to be taken - state two variables that must be kept constant to ensure a fair test - explain how the results are processed to determine the rate of reaction. [7]
(b) In an alternative method, a student monitors the reaction rate by measuring the loss in mass of the flask and contents over time on an electronic balance.
(i) Explain why a plug of cotton wool is placed in the neck of the conical flask during this investigation. .................................................................................................................................. .................................................................................................................................. [1]
(ii) Sketch a graph on the axes below to show how the total mass of the flask and contents changes from the start until the reaction is complete.
(a) Experimental Plan: - **Apparatus:** Conical flask, delivery tube, gas syringe (or inverted measuring cylinder in a water trough), stopwatch, measuring cylinder, balance. - **Method & Dilutions:** Measure a fixed volume (e.g. \(50\text{ cm}^3\)) of acid. Dilute the \(2.0\text{ mol/dm}^3\) acid with distilled water to prepare at least four concentrations (e.g., \(2.0\text{ mol/dm}^3\) [50 cm³ acid], \(1.5\text{ mol/dm}^3\) [37.5 cm³ acid + 12.5 cm³ water], \(1.0\text{ mol/dm}^3\) [25 cm³ acid + 25 cm³ water], and \(0.5\text{ mol/dm}^3\) [12.5 cm³ acid + 37.5 cm³ water]). Add a fixed mass of marble chips to the flask, quickly insert stopper, and start stopwatch. - **Measurements:** Measure the volume of gas collected in the gas syringe every 10 seconds for 2 minutes (or record time to collect a fixed volume of gas, e.g. 50 cm³). - **Control Variables:** Temperature of acid/room, mass of marble chips, surface area/size of chips, total volume of acid solution. - **Data Processing:** Plot a graph of volume of gas (y-axis) against time (x-axis) for each concentration. Find the initial gradient of each curve to determine the initial rate of reaction (or calculate \(\text{rate} = \text{volume} / \text{time}\)). Compare the rates across concentrations.
(b) (i) Cotton wool prevents loss of liquid acid spray caused by effervescence, while still allowing \(\text{CO}_2\) gas to escape. (ii) The graph starts at an initial positive mass value, decreases rapidly with a decreasing slope, and eventually flattens out horizontally (as all the limiting reactant is consumed and mass remains constant).
评分标准
(a) 7 marks total awarded across the following categories: **Apparatus:** (gas) syringe OR inverted measuring cylinder in water trough AND stopwatch / timer; [1] **Method (Dilution):** details of preparing at least 4 different concentrations by mixing \(2.0\text{ mol/dm}^3\) acid with distilled water; [1] **Method (Procedure):** adding chips to acid, immediately closing with bung/stopper and starting timer; [1] **Measurements:** measure volume of gas at set time intervals OR measure time to collect a fixed volume of gas; [1] **Control variables (any two):** same mass / number of chips; same surface area / size of chips; same temperature; same total volume of acid solution; [2] (1 mark each, max 2) **Data processing:** plot volume vs time graph and find initial gradient OR calculate \(\text{rate} = \frac{\text{volume}}{\text{time}}\) OR \(\frac{1}{\text{time}}\); [1]
(b)(i) allows gas / \(\text{CO}_2\) to escape but prevents loss of acid spray / liquid splashing out; [1] (b)(ii) curve starting on y-axis above zero with negative slope; [1] slope becomes less steep and levels off to a horizontal line above the x-axis; [1]
题目 6 · Practical/Structured
10 分
A student investigates the temperature change during the neutralisation of dilute sulfuric acid using aqueous potassium hydroxide.
**Method** 1. Measure \(25\text{ cm}^3\) of aqueous potassium hydroxide using a measuring cylinder and pour it into a polystyrene cup. 2. Measure and record the initial temperature \(T_1\) of the aqueous potassium hydroxide. 3. Add \(5.0\text{ cm}^3\) of dilute sulfuric acid from a burette to the polystyrene cup. 4. Stir the mixture thoroughly and measure the maximum temperature \(T_2\) reached. 5. Empty, rinse, and dry the polystyrene cup. 6. Repeat steps 1 to 5 using different volumes \(V\) of dilute sulfuric acid.
**(a)** **(i)** Name a piece of apparatus that is more accurate than a measuring cylinder for measuring \(25\text{ cm}^3\) of aqueous potassium hydroxide. [1]
**(ii)** State why a polystyrene cup is used instead of a glass beaker. [1]
**(b)** **Fig. 1.1** shows the thermometer readings for \(T_2\) when \(V = 10.0\text{ cm}^3\) and \(V = 20.0\text{ cm}^3\).
Read the thermometers shown in **Fig. 1.1** to the nearest \(0.5\ ^\circ\text{C}\) and complete **Table 1.1** for \(V = 10.0\text{ cm}^3\) and \(V = 20.0\text{ cm}^3\). Calculate the temperature rise \(\Delta T\) using the equation: \[\Delta T = T_2 - T_1\]
**(c)** **(i)** On a grid, plot a graph of \(\Delta T\) on the vertical axis against \(V\) on the horizontal axis. Draw two straight lines of best fit: one through the increasing points and one through the decreasing points. Extend both lines until they cross (intersect). [3]
**(ii)** Use your graph to determine the volume \(V_{\text{neutral}}\) of dilute sulfuric acid needed to completely neutralise \(25\text{ cm}^3\) of aqueous potassium hydroxide. Show clearly on your graph how you obtained your answer.
**(d)** State one addition to the apparatus that would reduce heat loss to the surroundings during each test. [1]
[Total: 10]
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解题
**(a)(i)** A volumetric pipette (or burette) provides a much more precise fixed volume measurement than a measuring cylinder.
**(a)(ii)** Polystyrene is an insulator of heat, which minimizes heat loss to the surroundings and ensures the recorded temperature rise is closer to the true value.
**(b)** - At \(V = 10.0\text{ cm}^3\), the liquid level reaches \(28.0\ ^\circ\text{C}\). Thus, \(\Delta T = 28.0 - 21.0 = 7.0\ ^\circ\text{C}\). - At \(V = 20.0\text{ cm}^3\), the liquid level reaches halfway between \(30\) and \(31\), so \(T_2 = 30.5\ ^\circ\text{C}\). Thus, \(\Delta T = 30.5 - 21.0 = 9.5\ ^\circ\text{C}\).
**(c)(i)** - Suitable linear axes: \(V\) from \(0\) to \(30\text{ cm}^3\) on the horizontal axis and \(\Delta T\) from \(0\) to \(12\ ^\circ\text{C}\) on the vertical axis. - Correctly plotted points: \((0.0, 0.0)\), \((5.0, 3.5)\), \((10.0, 7.0)\), \((15.0, 10.5)\), \((20.0, 9.5)\), \((25.0, 8.0)\), \((30.0, 6.5)\). - Two straight lines drawn: Line 1 passes through the first four points; Line 2 passes through the last three points. Lines extend to meet at the intersection point.
**(c)(ii)** Intersection occurs at \(V = 15.5\text{ cm}^3\). Working is shown with dashed lines extending down to the horizontal axis.
**(c)(iii)** Reading the vertical axis at the intersection point gives \(\Delta T_{\text{max}} = 10.85\ ^\circ\text{C}\) (acceptable range \(10.7 - 11.0\ ^\circ\text{C}\)).
**(d)** Adding an insulated lid (or cover) to the polystyrene cup reduces convective and evaporative heat loss.
**(a)(ii)** - (polystyrene is a) `good thermal insulator` / `poor conductor of heat` / `reduces heat loss` (to the surroundings); [1]
**(b)** - both \(T_2\) values correct: `28.0` and `30.5`; [1] - both \(\Delta T\) values correct: `7.0` and `9.5` (allow ecf from incorrect \(T_2\)); [1]
**(c)(i)** - axes labelled with quantity and unit (\(V\) in \(\text{cm}^3\) and \(\Delta T\) in \(^\circ\text{C}\)) and appropriate scales using more than half of the grid in both directions; [1] - all points plotted accurately to within half a small square; [1] - two straight, continuous best-fit lines drawn with a ruler and extended to intersect cleanly; [1]
**(c)(ii)** - \(V_{\text{neutral}}\) read correctly from intersection on candidate's graph \(\pm 0.2\text{ cm}^3\) (`15.5\text{ cm}^3` \([15.0 - 16.0]\)) AND construction lines/markings shown on graph; [1]
**(c)(iii)** - \(\Delta T_{\text{max}}\) read correctly from intersection on candidate's graph \(\pm 0.1\ ^\circ\text{C}\) (`10.8\text{ to }10.9\ ^\circ\text{C}`); [1]
**(d)** - place a `lid` / `cover` on the cup (or add extra insulation around the cup); [1]
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