An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V3) Cambridge International A Level Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.
Paper 23 (選擇題 - Extended)
Forty multiple-choice questions. Answer all questions. Choose the correct option A, B, C or D.
40 題目 · 40 分
題目 1 · 選擇題
1 分
Which statement correctly defines excretion as a characteristic of living organisms?
A.the taking in of materials for energy, growth and development
B.the permanent increase in size and dry mass by an increase in cell number or cell size or both
C.the removal of toxic materials, the waste products of metabolism and substances in excess of requirements
D.the ability to detect or sense stimuli in the internal or external environment and to make appropriate responses
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解題
Excretion is defined as the removal from organisms of toxic materials, the waste products of metabolism (chemical reactions in cells including respiration) and substances in excess of requirements. Option A defines nutrition, option B defines growth, and option D defines sensitivity.
評分準則
1 mark for the correct choice C. No marks for other options.
題目 2 · 選擇題
1 分
Which statement explains why the rate of an enzyme-catalysed reaction decreases rapidly at temperatures above the optimum temperature?
A.The kinetic energy of the substrate and enzyme molecules decreases.
B.The active site of the enzyme changes shape so the substrate can no longer fit.
C.The substrate molecules are denatured and can no longer fit the active site.
D.The enzyme molecules break down completely into individual amino acids.
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解題
At high temperatures, the thermal energy causes the bonds maintaining the three-dimensional structure of the enzyme's active site to break. This changes the shape of the active site (denaturation) so that the substrate is no longer complementary and cannot fit into it. The substrate itself does not denature, kinetic energy actually increases rather than decreases, and the enzyme does not break down into free amino acids.
評分準則
1 mark for the correct choice B. No marks for other options.
題目 3 · 選擇題
1 分
Which method is used to prepare a pure, dry sample of the insoluble salt, barium sulfate?
A.Mix barium chloride solution with dilute sulfuric acid, filter the mixture, wash the residue with distilled water, and dry.
B.React barium carbonate solid with dilute sulfuric acid, filter the mixture, evaporate the filtrate, and crystallise.
C.Heat barium metal with dilute sulfuric acid, filter, crystallise, and dry the crystals.
D.Titrate barium hydroxide solution with dilute sulfuric acid using an indicator, then evaporate the solution to dryness.
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解題
Barium sulfate is an insoluble salt. Insoluble salts are prepared by precipitation, which involves mixing two soluble reactants (barium chloride solution and dilute sulfuric acid). The barium sulfate forms a solid precipitate, which is obtained by filtration as the residue. It is then washed with distilled water to remove soluble impurities and dried. Methods B, C, and D are not suitable because barium sulfate is insoluble and would either stop the reaction prematurely (in B and C) or not be obtained via crystallisation of a filtrate.
評分準則
1 mark for identifying the precipitation method (A). Other options are incorrect.
題目 4 · 選擇題
1 分
An ion of chlorine is represented as \({}^{37}\text{Cl}^{-}\). Chlorine has an atomic number of 17. How many protons, neutrons, and electrons are present in this ion?
A.17 protons, 20 neutrons, 18 electrons
B.17 protons, 20 neutrons, 16 electrons
C.17 protons, 37 neutrons, 18 electrons
D.18 protons, 20 neutrons, 17 electrons
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解題
The atomic number of chlorine is 17, which means it has 17 protons. The mass number is 37, so the number of neutrons is \(37 - 17 = 20\). Since it is a negatively charged ion with a charge of \(1-\), it has one more electron than the number of protons: \(17 + 1 = 18\) electrons. Therefore, there are 17 protons, 20 neutrons, and 18 electrons.
評分準則
1 mark for the correct choice A. No marks for incorrect proton, neutron, or electron counts.
題目 5 · 選擇題
1 分
Aqueous copper(II) sulfate is electrolysed using inert carbon electrodes. What is observed at each electrode during this reaction?
A.Anode: bubbles of a colourless gas; Cathode: a pink-brown solid is deposited.
B.Anode: a pink-brown solid is deposited; Cathode: bubbles of a colourless gas.
C.Anode: bubbles of a pale green gas; Cathode: bubbles of a colourless gas.
D.Anode: a pink-brown solid is deposited; Cathode: bubbles of a pale green gas.
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解題
At the cathode (negative electrode), copper ions (\(\text{Cu}^{2+}\)) are reduced to copper metal, which is deposited as a pink-brown solid. At the anode (positive electrode), hydroxide ions (\(\text{OH}^{-}\)) from the water are oxidised to form oxygen gas, which is observed as bubbles of a colourless gas.
評分準則
1 mark for the correct combination of anode and cathode observations (A).
題目 6 · 選擇題
1 分
An object of mass \(4.0\text{ kg}\) is dropped from rest from a height of \(20\text{ m}\). Air resistance is negligible. The acceleration of free fall, \(g\), is \(10\text{ m/s}^2\). What is the kinetic energy of the object when it is at a height of \(5.0\text{ m}\) above the ground?
A.\(200\text{ J}\)
B.\(600\text{ J}\)
C.\(800\text{ J}\)
D.\(1200\text{ J}\)
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解題
Due to the conservation of energy, the loss in gravitational potential energy (\(\Delta E_p\)) is equal to the gain in kinetic energy (\(E_k\)). The change in height, \(\Delta h = 20\text{ m} - 5.0\text{ m} = 15\text{ m}\). Therefore, \(\Delta E_p = m g \Delta h = 4.0\text{ kg} \times 10\text{ m/s}^2 \times 15\text{ m} = 600\text{ J}\). Since the initial kinetic energy was zero, the kinetic energy at \(5.0\text{ m}\) is \(600\text{ J}\).
評分準則
1 mark for calculating the correct kinetic energy of 600 J (B).
題目 7 · 選擇題
1 分
A ray of light is travelling inside a glass block with a refractive index of \(1.50\). The light reaches the boundary between the glass and air. What is the critical angle for light in this glass block?
A.\(30^\circ\)
B.\(42^\circ\)
C.\(48^\circ\)
D.\(56^\circ\)
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解題
The relationship between the critical angle \(c\) and the refractive index \(n\) is given by \(\sin(c) = 1/n\). Substituting \(n = 1.50\) gives \(\sin(c) = 1/1.50 = 0.667\). Calculating \(c = \sin^{-1}(0.667)\) yields approximately \(41.8^\circ\), which rounds to \(42^\circ\).
評分準則
1 mark for using the correct equation and obtaining 42 degrees (B).
題目 8 · 選擇題
1 分
A circuit consists of a \(12\text{ V}\) power supply connected to three resistors. Two resistors, each of resistance \(6.0\ \Omega\), are connected in parallel. This parallel combination is connected in series with a third resistor of resistance \(3.0\ \Omega\). What is the 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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解題
First, calculate the equivalent resistance of the parallel combination: \(R_p = (6.0 \times 6.0) / (6.0 + 6.0) = 3.0\ \Omega\). Next, calculate the total resistance of the circuit: \(R_{\text{total}} = R_p + 3.0\ \Omega = 3.0\ \Omega + 3.0\ \Omega = 6.0\ \Omega\). Using Ohm's law, the current drawn from the power supply is \(I = V / R_{\text{total}} = 12\text{ V} / 6.0\ \Omega = 2.0\text{ A}\).
評分準則
1 mark for calculating the correct total current of 2.0 A (C).
題目 9 · 選擇題
1 分
An ideal step-up transformer has a primary coil with 150 turns and a secondary coil with 600 turns. The primary coil is connected to a \(12\text{ V}\) alternating current (a.c.) supply. The secondary coil is connected to a resistor of resistance \(8.0\ \Omega\). What is the current in the primary coil?
A.\(1.5\text{ A}\)
B.\(6.0\text{ A}\)
C.\(24\text{ A}\)
D.\(96\text{ A}\)
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解題
First, calculate the secondary voltage \(V_s\) using the transformer equation: \(V_s = V_p \times \frac{N_s}{N_p}\). Substituting the given values: \(V_s = 12\text{ V} \times \frac{600}{150} = 48\text{ V}\). Next, calculate the current in the secondary coil \(I_s\) using Ohm's law: \(I_s = \frac{V_s}{R} = \frac{48\text{ V}}{8.0\ \Omega} = 6.0\text{ A}\). Since the transformer is ideal, the input power equals the output power: \(V_p \times I_p = V_s \times I_s\). Therefore, \(12\text{ V} \times I_p = 48\text{ V} \times 6.0\text{ A}\), which gives \(I_p = 24\text{ A}\).
評分準則
1 mark for the correct answer C. Award 1 mark for correct application of transformer equations and power conservation.
題目 10 · 選擇題
1 分
A trolley of mass \(2.0\text{ kg}\) travels at a speed of \(6.0\text{ m/s}\) to the right. It collides with a stationary trolley of mass \(4.0\text{ kg}\). After the collision, the two trolleys stick together and move off. What is the loss in total kinetic energy of the trolleys as a result of the collision?
A.\(12\text{ J}\)
B.\(18\text{ J}\)
C.\(24\text{ J}\)
D.\(36\text{ J}\)
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解題
First, find the initial total kinetic energy: \(E_{ki} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} (2.0\text{ kg}) (6.0\text{ m/s})^2 = 36\text{ J}\). Next, find the final speed \(v\) of the combined trolleys using conservation of momentum: \(m_1 u_1 = (m_1 + m_2)v \Rightarrow (2.0\text{ kg} \times 6.0\text{ m/s}) = (2.0\text{ kg} + 4.0\text{ kg})v \Rightarrow 12 = 6.0v \Rightarrow v = 2.0\text{ m/s}\). Find the final total kinetic energy: \(E_{kf} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} (6.0\text{ kg}) (2.0\text{ m/s})^2 = 12\text{ J}\). The loss in kinetic energy is \(E_{ki} - E_{kf} = 36\text{ J} - 12\text{ J} = 24\text{ J}\).
評分準則
1 mark for the correct answer C. Award 1 mark for calculating kinetic energy before and after the collision using momentum conservation.
題目 11 · 選擇題
1 分
The chemical equation for the synthesis of ammonia is: \(\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}\). The reaction is exothermic. The bond energies are: \(\text{N}\equiv\text{N}\) is \(945\text{ kJ/mol}\); \(\text{H}-\text{H}\) is \(436\text{ kJ/mol}\); \(\text{N}-\text{H}\) is \(391\text{ kJ/mol}\). What is the overall energy change for the forward reaction?
A.\(-93\text{ kJ/mol}\)
B.\(-231\text{ kJ/mol}\)
C.\(+93\text{ kJ/mol}\)
D.\(+2253\text{ kJ/mol}\)
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解題
Energy required to break bonds (reactants) = \(1 \times (\text{N}\equiv\text{N}) + 3 \times (\text{H}-\text{H}) = 945 + 3(436) = 2253\text{ kJ/mol}\). Energy released in forming bonds (products) = \(6 \times (\text{N}-\text{H}) = 6(391) = 2346\text{ kJ/mol}\). Overall energy change = energy input - energy output = \(2253 - 2346 = -93\text{ kJ/mol}\).
評分準則
1 mark for the correct answer A. Award 1 mark for subtracting total bond energy of products from total bond energy of reactants.
題目 12 · 選擇題
1 分
Aqueous copper(II) sulfate is electrolysed using different electrodes. Which row correctly describes what is observed at each electrode?
A.Carbon electrodes: Cathode = pink solid forms; Anode = bubbles of a colourless gas
B.Carbon electrodes: Cathode = bubbles of a colourless gas; Anode = pink solid forms
C.Copper electrodes: Cathode = electrode dissolves and decreases in mass; Anode = pink solid forms
D.Copper electrodes: Cathode = pink solid forms; Anode = bubbles of a colourless gas
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解題
Using carbon (inert) electrodes, copper ions (\(\text{Cu}^{2+}\)) are reduced at the cathode to form a pink copper solid coating. At the anode, hydroxide ions (\(\text{OH}^-\)) are oxidised to form oxygen gas, which is seen as bubbles of a colourless gas.
評分準則
1 mark for the correct answer A. Award 1 mark for correct identification of products at the cathode and anode during the electrolysis of aqueous copper(II) sulfate.
題目 13 · 選擇題
1 分
A student accidentally touches a hot object and rapidly withdraws their hand. Which sequence represents the path of the electrical impulse along the neurones in the reflex arc?
C.sensory neurone \(\rightarrow\) motor neurone \(\rightarrow\) relay neurone
D.sensory neurone \(\rightarrow\) relay neurone \(\rightarrow\) motor neurone
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解題
In a reflex arc, the stimulus is detected by receptors which generate impulses in the sensory neurone. The impulse is passed to a relay neurone in the central nervous system, which then transmits the signal to a motor neurone to stimulate the effector (muscle).
評分準則
1 mark for the correct answer D. Award 1 mark for correctly sequencing sensory, relay, and motor neurones.
題目 14 · 選擇題
1 分
Plant cells are placed in a highly concentrated salt solution. Which row correctly describes the state of the cell and the net movement of water?
A.State of cell: plasmolysed; Net movement of water: out of the cell
B.State of cell: turgid; Net movement of water: into the cell
C.State of cell: plasmolysed; Net movement of water: into the cell
D.State of cell: turgid; Net movement of water: out of the cell
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解題
A highly concentrated salt solution has a lower water potential than the plant cell vacuole/cytoplasm. Water moves out of the cell by osmosis from an area of higher water potential to an area of lower water potential, causing the cell membrane to pull away from the cell wall, which is known as plasmolysis.
評分準則
1 mark for the correct answer A. Award 1 mark for correctly identifying the net movement of water out of the cell and the resulting plasmolysed state.
題目 15 · 選擇題
1 分
Which reaction is used to distinguish between an alkane and an alkene, and what is the observation for the alkene?
A.Reaction: addition of aqueous bromine; Observation for alkene: colour change from orange to colourless
B.Reaction: addition of aqueous bromine; Observation for alkene: colour change from colourless to orange
C.Reaction: burning in air; Observation for alkene: burns with a completely blue, soot-free flame
D.Reaction: addition of steam; Observation for alkene: forms an ester with a sweet smell
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解題
Aqueous bromine (bromine water) is used to test for unsaturation (double bonds). Alkenes undergo an addition reaction with bromine, decolourising the orange bromine solution to colourless. Alkanes do not react with bromine water in the absence of UV light, so the solution remains orange.
評分準則
1 mark for the correct answer A. Award 1 mark for identifying the bromine water test and its positive observation for an alkene.
題目 16 · 選擇題
1 分
A ray of light travels from a glass block into air. The refractive index of the glass is \(1.50\). What is the critical angle for the boundary between the glass and the air?
A.\(30.0^\circ\)
B.\(41.8^\circ\)
C.\(48.6^\circ\)
D.\(56.3^\circ\)
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解題
The relationship between refractive index \(n\) and critical angle \(c\) when a ray travels into air is given by: \(\sin(c) = \frac{1}{n}\). Substituting \(n = 1.50\) yields \(\sin(c) = \frac{1}{1.50} \approx 0.6667\). Taking the inverse sine: \(c = \arcsin(0.6667) \approx 41.8^\circ\).
評分準則
1 mark for the correct answer B. Award 1 mark for the correct application of \(\sin(c) = 1/n\) to calculate the angle.
題目 17 · 選擇題
1 分
A toy car of mass \(0.5\text{ kg}\) moving at a velocity of \(4.0\text{ m/s}\) collides with a stationary toy car of mass \(1.5\text{ kg}\). After the collision, the two cars couple together and move off with a common velocity. What is their combined velocity after the collision?
A.\(0.75\text{ m/s}\)
B.\(1.0\text{ m/s}\)
C.\(1.3\text{ m/s}\)
D.\(2.0\text{ m/s}\)
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解題
According to the principle of conservation of momentum:
\(\text{Total initial momentum} = \text{Total final momentum}\)
\(v = \frac{2.0\text{ kg m/s}}{2.0\text{ kg}} = 1.0\text{ m/s}\)
Therefore, the correct option is B.
評分準則
Award 1 mark for the correct option B. - 1 mark: correct calculation of the final velocity using the conservation of momentum formula.
題目 18 · 選擇題
1 分
A radioactive isotope has a half-life of \(12\text{ minutes}\). A sample originally has a count rate of \(800\text{ counts per minute}\). Neglecting background radiation, how long will it take for the count rate of this sample to fall to \(50\text{ counts per minute}\)?
A.\(24\text{ minutes}\)
B.\(36\text{ minutes}\)
C.\(48\text{ minutes}\)
D.\(96\text{ minutes}\)
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解題
We can determine the number of half-lives that pass by halving the count rate step-by-step:
- Start: \(800\text{ counts per minute}\) - 1st half-life: \(400\text{ counts per minute}\) - 2nd half-life: \(200\text{ counts per minute}\) - 3rd half-life: \(100\text{ counts per minute}\) - 4th half-life: \(50\text{ counts per minute}\)
Award 1 mark for the correct option C. - 1 mark: identification of 4 half-lives needed and correct multiplication to find total time.
題目 19 · 選擇題
1 分
Two \(6.0\ \Omega\) resistors are connected in parallel with each other. This combination is connected in series with a \(5.0\ \Omega\) resistor and a \(12\text{ V}\) d.c. power supply. What is the current flowing through the \(5.0\ \Omega\) resistor?
A.\(1.1\text{ A}\)
B.\(1.5\text{ A}\)
C.\(2.4\text{ A}\)
D.\(4.0\text{ A}\)
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解題
First, find the equivalent resistance of the two \(6.0\ \Omega\) resistors in parallel:
Since the \(5.0\ \Omega\) resistor is in series with the parallel pair, the entire supply current of \(1.5\text{ A}\) flows through it.
Therefore, the correct option is B.
評分準則
Award 1 mark for the correct option B. - 1 mark: correct calculation of the parallel equivalent resistance, total resistance, and subsequent current calculation.
題目 20 · 選擇題
1 分
Which compound is formed by the dehydration of ethanol and will rapidly decolourise aqueous bromine in the dark?
A.Ethane
B.Ethene
C.Ethanoic acid
D.Ethyl ethanoate
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解題
Dehydration of ethanol (an alcohol) is a reaction that removes a water molecule, forming ethene (an alkene):
Because ethene contains a carbon-carbon double bond (\(\text{C}=\text{C}\)), it is unsaturated and will rapidly undergo an addition reaction with aqueous bromine, decolourising it from orange to colourless in the dark.
Therefore, the correct option is B.
評分準則
Award 1 mark for the correct option B. - 1 mark: identification of the product of dehydration of ethanol (ethene) and its reaction with bromine water.
題目 21 · 選擇題
1 分
Which statement correctly describes the energy changes that occur during an exothermic chemical reaction?
A.Energy is absorbed to break bonds, and this is greater than the energy released when new bonds are formed.
B.Energy is released when bonds are broken, and this is greater than the energy absorbed when new bonds are formed.
C.Energy is absorbed to break bonds, and this is less than the energy released when new bonds are formed.
D.Energy is released when bonds are broken, and this is less than the energy absorbed when new bonds are formed.
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解題
In any chemical reaction, bond breaking is an endothermic process (requires energy to be absorbed), and bond making is an exothermic process (releases energy).
For a reaction to be overall exothermic, the total energy released during bond making must be greater than the total energy absorbed during bond breaking. In other words, energy is absorbed to break bonds, and this is less than the energy released when new bonds are formed.
Therefore, the correct option is C.
評分準則
Award 1 mark for the correct option C. - 1 mark: correct distinction between the endothermic nature of bond breaking and the exothermic nature of bond making, and their balance in an exothermic reaction.
題目 22 · 選擇題
1 分
Healthy plant cells are placed into a highly concentrated solution of sodium chloride. Which of the following correctly describes the movement of water and the final state of the plant cells?
A.Water enters the cells by osmosis and the cells become turgid.
B.Water enters the cells by active transport and the cells become turgid.
C.Water leaves the cells by osmosis and the cells become plasmolysed.
D.Water leaves the cells by active transport and the cells become plasmolysed.
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解題
A highly concentrated solution has a lower water potential than the cytoplasm and cell sap of the plant cells. Water will move out of the cells down a water potential gradient, from a region of higher water potential to a region of lower water potential, by the passive process of osmosis. As a result of losing water, the vacuole shrinks and the cell membrane pulls away from the cell wall, leaving the cells plasmolysed.
Therefore, the correct option is C.
評分準則
Award 1 mark for the correct option C. - 1 mark: identification of the net direction of water movement (leaving the cell), the mechanism (osmosis), and the resulting cellular state (plasmolysed).
題目 23 · 選擇題
1 分
Which pathway shows the correct order of structures involved in a simple reflex arc?
A simple reflex arc consists of the following sequence:
1. Receptor detects a stimulus. 2. Sensory neurone transmits the impulse to the central nervous system (CNS). 3. Relay neurone in the spinal cord passes the impulse from the sensory to the motor neurone. 4. Motor neurone transmits the impulse from the CNS to the effector. 5. Effector (muscle or gland) carries out the response.
Therefore, the correct pathway is: receptor \(\to\) sensory neurone \(\to\) relay neurone \(\to\) motor neurone \(\to\) effector.
Therefore, the correct option is B.
評分準則
Award 1 mark for the correct option B. - 1 mark: identification of the correct sequence of neurons and structures in a reflex arc.
題目 24 · 選擇題
1 分
An atom of element X has a proton number of 17 and a nucleon number of 37. Which statement correctly describes the composition of an ion of X with a single negative charge, \(\text{X}^-\mathrm{?}\)
A.It contains 17 protons, 20 neutrons and 18 electrons.
B.It contains 17 protons, 37 neutrons and 18 electrons.
C.It contains 18 protons, 20 neutrons and 17 electrons.
D.It contains 17 protons, 20 neutrons and 16 electrons.
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解題
An atom of element X has 17 protons. - The nucleon number (mass number) is 37, so the number of neutrons is: \(37 - 17 = 20\text{ neutrons}\). - A neutral atom of X would have 17 electrons. - The ion has a single negative charge (\(\text{X}^-\)), meaning it has gained one electron. The number of electrons is: \(17 + 1 = 18\text{ electrons}\).
Therefore, the ion \(\text{X}^-\)\ contains 17 protons, 20 neutrons, and 18 electrons.
Therefore, the correct option is A.
評分準則
Award 1 mark for the correct option A. - 1 mark: correct determination of the number of protons, neutrons, and electrons in the specified anion.
題目 25 · 選擇題
1 分
An organism is observed changing its position using energy from respiration, detecting a change in its environment and responding to it, and producing offspring of its own kind. Which three characteristics of living organisms are described?
A.growth, nutrition and respiration
B.movement, respiration and growth
C.movement, sensitivity and reproduction
D.sensitivity, respiration and excretion
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解題
Changing position using energy is movement. Detecting and responding to environmental changes is sensitivity. Producing offspring is reproduction. Therefore, the three characteristics described are movement, sensitivity, and reproduction.
評分準則
1 mark: Identify movement, sensitivity, and reproduction as the correct characteristics (Option C).
題目 26 · 選擇題
1 分
An enzyme-controlled reaction is carried out at two different temperatures, \(25\ ^\circ\text{C}\) and \(40\ ^\circ\text{C}\). The optimum temperature for this enzyme is \(40\ ^\circ\text{C}\). Which statement correctly explains the difference in the rate of reaction at these two temperatures?
A.At \(25\ ^\circ\text{C}\), the enzyme is fully denatured, so the rate of reaction is zero.
B.At \(40\ ^\circ\text{C}\), the kinetic energy of the substrate and enzyme molecules is higher, leading to more frequent successful collisions.
C.At \(40\ ^\circ\text{C}\), the activation energy of the reaction is increased, allowing the reaction to proceed faster.
D.At \(25\ ^\circ\text{C}\), the active site of the enzyme changes shape permanently, preventing substrate binding.
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解題
At a higher temperature (up to the optimum), both the enzyme and substrate molecules have more kinetic energy. This causes them to move faster, resulting in more frequent collisions, and a higher proportion of these collisions have sufficient energy to overcome the activation energy barrier, leading to a faster rate of reaction.
評分準則
1 mark: Identify that higher kinetic energy leads to more frequent successful collisions at \(40\ ^\circ\text{C}\) (Option B).
題目 27 · 選擇題
1 分
An atom of an isotope of element \(X\) is represented by \({}_{15}^{31}X\). Which statement about an atom of this isotope is correct?
A.It contains 15 protons, 15 electrons and 16 neutrons.
B.It contains 15 protons, 16 electrons and 15 neutrons.
C.It contains 16 protons, 16 electrons and 15 neutrons.
D.It contains 31 protons, 15 electrons and 16 neutrons.
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解題
The atomic number (proton number) is 15, which means there are 15 protons. In a neutral atom, the number of electrons equals the number of protons, so there are 15 electrons. The nucleon number (mass number) is 31, so the number of neutrons is \(31 - 15 = 16\).
評分準則
1 mark: Correctly determine the number of protons, electrons, and neutrons (Option A).
題目 28 · 選擇題
1 分
Molten lead(II) bromide is electrolysed using inert carbon electrodes. Which row correctly describes the observations at each electrode?
A.Anode: brown fumes are evolved; Cathode: grey beads of metal are formed
B.Anode: grey beads of metal are formed; Cathode: brown fumes are evolved
C.Anode: bubbles of colourless gas; Cathode: brown fumes are evolved
D.Anode: brown fumes are evolved; Cathode: bubbles of colourless gas
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解題
During the electrolysis of molten lead(II) bromide: Bromide ions migrate to the positive anode, where they are oxidised to form bromine gas, seen as brown fumes. Lead(II) ions migrate to the negative cathode, where they are reduced to form molten lead metal, seen as grey beads.
評分準則
1 mark: Correctly identify the products and observations at both the anode and cathode (Option A).
題目 29 · 選擇題
1 分
Which statement describes the trends in physical properties of the Group VII elements (halogens) as the group is descended from fluorine to iodine?
A.The colour of the element becomes lighter and the boiling point increases.
B.The colour of the element becomes darker and the boiling point increases.
C.The reactivity with metals increases and the density decreases.
D.The reactivity with metals increases and the melting point decreases.
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解題
As Group VII is descended, the elements change state from gas to liquid to solid, meaning their melting and boiling points increase. Their colours also become progressively darker (fluorine is pale yellow, chlorine is pale green, bromine is red-brown, and iodine is grey-black/violet). Reactivity decreases down the group.
評分準則
1 mark: Identify that colour becomes darker and boiling point increases down the group (Option B).
題目 30 · 選擇題
1 分
A crane lifts a heavy crate of mass \(400\text{ kg}\) vertically upwards through a height of \(12\text{ m}\) in a time of \(8.0\text{ s}\). The gravitational field strength \(g\) is \(10\text{ N/kg}\). What is the useful average power developed by the crane?
A.600 W
B.4800 W
C.6000 W
D.48000 W
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解題
First, calculate the weight of the crate: \(W = m \times g = 400 \times 10 = 4000\text{ N}\). Next, calculate the work done: \(\text{Work Done} = F \times d = 4000 \times 12 = 48000\text{ J}\). Finally, calculate the power: \(\text{Power} = \frac{\text{Work Done}}{\text{time}} = \frac{48000}{8.0} = 6000\text{ W}\).
評分準則
1 mark: Show correct calculation steps for force, work, and power to arrive at 6000 W (Option C).
題目 31 · 選擇題
1 分
Two resistors, with resistances of \(4.0\ \Omega\) and \(6.0\ \Omega\), are connected in parallel to a \(12\text{ V}\) d.c. power supply. What is the total current drawn from the power supply?
A.1.2 A
B.2.0 A
C.5.0 A
D.10 A
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解題
The potential difference across each parallel branch is \(12\text{ V}\). The current through the \(4.0\ \Omega\) resistor is \(I_1 = \frac{12}{4.0} = 3.0\text{ A}\). The current through the \(6.0\ \Omega\) resistor is \(I_2 = \frac{12}{6.0} = 2.0\text{ A}\). The total current is \(I_{\text{total}} = 3.0 + 2.0 = 5.0\text{ A}\). Alternatively, using equivalent resistance: \(R_p = \frac{4.0 \times 6.0}{4.0 + 6.0} = 2.4\ \Omega\), so \(I = \frac{12}{2.4} = 5.0\text{ A}\).
評分準則
1 mark: Calculate the parallel currents or parallel equivalent resistance correctly to find the total current of 5.0 A (Option C).
題目 32 · 選擇題
1 分
A sound wave travels through water with a speed of \(1500\text{ m/s}\). If the frequency of the sound wave is \(2.5\text{ kHz}\), what is its wavelength in water?
A.0.60 m
B.1.67 m
C.3.75 m
D.600 m
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解題
The wave equation is \(v = f \lambda\). Given speed \(v = 1500\text{ m/s}\) and frequency \(f = 2.5\text{ kHz} = 2500\text{ Hz}\). Rearranging for wavelength: \(\lambda = \frac{v}{f} = \frac{1500}{2500} = 0.60\text{ m}\).
評分準則
1 mark: Correctly convert the frequency to Hz and use the wave equation to calculate the wavelength of 0.60 m (Option A).
題目 33 · multiple_choice
1 分
A car of mass \(1200\text{ kg}\) accelerates from \(10\text{ m/s}\) to \(25\text{ m/s}\) in a time of \(5.0\text{ s}\). What is the average useful power required to achieve this acceleration, ignoring any resistive forces?
A.27 kW
B.36 kW
C.63 kW
D.90 kW
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解題
First, calculate the change in kinetic energy (\(\Delta E_k\)) of the car: \(E_{ki} = \frac{1}{2} m u^2 = 0.5 \times 1200 \times 10^2 = 60\,000\text{ J}\) \(E_{kf} = \frac{1}{2} m v^2 = 0.5 \times 1200 \times 25^2 = 375\,000\text{ J}\) \(\Delta E_k = 375\,000 - 60\,000 = 315\,000\text{ J}\)
Next, calculate the average power (\(P\)): \(P = \frac{\Delta E_k}{t} = \frac{315\,000}{5.0} = 63\,000\text{ W} = 63\text{ kW}\).
評分準則
1 mark for calculating the correct change in kinetic energy and dividing by time to obtain 63 kW (Option C).
題目 34 · multiple_choice
1 分
Two resistors, with resistances of \(6.0\ \Omega\) and \(12.0\ \Omega\), are connected in parallel. This parallel combination is then connected in series with a third resistor of resistance \(4.0\ \Omega\) and a \(12.0\text{ V}\) power supply of negligible internal resistance. What is the current flowing through the \(12.0\ \Omega\) resistor?
A.0.50 A
B.0.75 A
C.1.0 A
D.1.5 A
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解題
1. Find the equivalent resistance of the parallel branch (\(R_p\)): \(1/R_p = 1/6.0 + 1/12.0 = 3/12.0 \implies R_p = 4.0\ \Omega\).
2. Find the total resistance of the circuit (\(R_{total}\)): \(R_{total} = R_p + 4.0 = 4.0 + 4.0 = 8.0\ \Omega\).
3. Find the total current supplied (\(I_{total}\)): \(I_{total} = V / R_{total} = 12.0 / 8.0 = 1.5\text{ A}\).
4. Find the potential difference across the parallel branch (\(V_p\)): \(V_p = I_{total} \times R_p = 1.5 \times 4.0 = 6.0\text{ V}\).
5. Find the current through the \(12.0\ \Omega\) resistor (\(I_{12}\)): \(I_{12} = V_p / 12.0 = 6.0 / 12.0 = 0.50\text{ A}\).
評分準則
1 mark for using parallel/series resistance rules and Ohm's law to calculate the current as 0.50 A (Option A).
題目 35 · multiple_choice
1 分
Consider the following reaction for the synthesis of ammonia: \(\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \to 2\text{NH}_3\text{(g)}\)
The average bond energies are: • \(\text{N}\equiv\text{N}\) bond: \(945\text{ kJ/mol}\) • \(\text{H}-\text{H}\) bond: \(436\text{ kJ/mol}\) • \(\text{N}-\text{H}\) bond: \(391\text{ kJ/mol}\)
What is the enthalpy change, \(\Delta H\), for this reaction?
A.-93 kJ/mol
B.+93 kJ/mol
C.-1092 kJ/mol
D.+1092 kJ/mol
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解題
Energy required to break bonds (reactants): \(1 \times (\text{N}\equiv\text{N}) + 3 \times (\text{H}-\text{H}) = 945 + 3(436) = 2253\text{ kJ/mol}\).
Energy released when forming bonds (products): \(2 \times 3 \times (\text{N}-\text{H}) = 6 \times 391 = 2346\text{ kJ/mol}\).
1 mark for calculating bond breaking energy, bond making energy, and taking the difference to yield -93 kJ/mol (Option A).
題目 36 · multiple_choice
1 分
A sample of \(2.4\text{ g}\) of magnesium ribbon is reacted with \(100\text{ cm}^3\) of \(1.0\text{ mol/dm}^3\) hydrochloric acid according to the equation:
What volume of hydrogen gas, measured at room temperature and pressure (r.t.p.), is produced in this reaction? [\(A_r\text{: Mg} = 24\); molar volume of gas at r.t.p. = \(24\text{ dm}^3/\text{mol}\)]
A.1.2 dm³
B.2.4 dm³
C.4.8 dm³
D.12 dm³
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解題
First, find the moles of each reactant: • Moles of \(\text{Mg} = 2.4 / 24 = 0.10\text{ mol}\). • Moles of \(\text{HCl} = 1.0\text{ mol/dm}^3 \times 0.100\text{ dm}^3 = 0.10\text{ mol}\).
According to the equation, \(1\text{ mol}\) of \(\text{Mg}\) requires \(2\text{ mol}\) of \(\text{HCl}\). Therefore, \(0.10\text{ mol}\) of \(\text{Mg}\) would require \(0.20\text{ mol}\) of \(\text{HCl}\). Since only \(0.10\text{ mol}\) of \(\text{HCl}\) is present, \(\text{HCl}\) is the limiting reactant.
Using the stoichiometry of the limiting reactant: • Moles of \(\text{H}_2\) produced = \(0.10 / 2 = 0.05\text{ mol}\). • Volume of \(\text{H}_2 = 0.05\text{ mol} \times 24\text{ dm}^3/\text{mol} = 1.2\text{ dm}^3\).
評分準則
1 mark for identifying HCl as the limiting reactant and calculating the volume of hydrogen gas as 1.2 dm³ (Option A).
題目 37 · multiple_choice
1 分
Which statement correctly describes how high temperatures affect the rate of an enzyme-controlled reaction?
A.They increase the rate indefinitely because the substrate molecules gain more kinetic energy and collide more frequently.
B.They decrease the rate beyond the optimum temperature because the shape of the enzyme's active site is permanently altered.
C.They decrease the rate because the activation energy barrier of the reaction is raised.
D.They keep the rate constant after the optimum because all the active sites of the enzymes remain saturated.
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解題
As temperature increases, kinetic energy increases, raising the collision rate between enzymes and substrates up to an optimum. Beyond the optimum temperature, high thermal energy breaks intermolecular bonds in the enzyme, permanently changing the shape of its active site (denaturation) so the substrate can no longer fit, and the reaction rate rapidly falls to zero.
評分準則
1 mark for identifying that high temperatures denature enzymes by changing the shape of their active site (Option B).
題目 38 · multiple_choice
1 分
Four identical healthy plant cells with standard turgor pressure are placed into four different test tubes containing sucrose solutions of concentrations \(0.1\text{ mol/dm}^3\), \(0.3\text{ mol/dm}^3\), \(0.5\text{ mol/dm}^3\), and \(0.7\text{ mol/dm}^3\). In which solution will plasmolysis occur most rapidly, and why?
A.0.1 mol/dm³ because water enters the cell down a water potential gradient.
B.0.1 mol/dm³ because sucrose molecules actively enter the cell, increasing internal pressure.
C.0.7 mol/dm³ because water leaves the cell down a water potential gradient.
D.0.7 mol/dm³ because sucrose molecules diffuse out of the cell, pulling the cell membrane with them.
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解題
Plasmolysis occurs when water leaves the vacuole of a plant cell via osmosis. Water moves down a water potential gradient (from a region of higher water potential to a region of lower water potential). The \(0.7\text{ mol/dm}^3\) sucrose solution has the highest solute concentration and thus the lowest water potential. Therefore, water leaves the plant cell most rapidly in this solution, causing the cell membrane to pull away from the cell wall.
評分準則
1 mark for identifying the correct solution (0.7 mol/dm³) and the direction/cause of water movement (Option C).
題目 39 · multiple_choice
1 分
Which statement correctly describes the properties and trends of elements within the Periodic Table?
A.Group I alkali metals become less reactive down the group, and their density decreases.
B.Group VII halogens become less reactive down the group, and their boiling points increase.
C.Group VII halogens become more reactive down the group, and their melting points decrease.
D.Transition elements have lower melting points and lower densities than Group I alkali metals.
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解題
For Group VII (halogens), reactivity decreases down the group as the atomic radius increases and it becomes harder to attract an electron. However, their melting and boiling points increase down the group because the molecules become larger, resulting in stronger intermolecular forces of attraction.
評分準則
1 mark for recognizing that Group VII elements become less reactive down the group and experience an increase in boiling points (Option B).
題目 40 · multiple_choice
1 分
A radioactive isotope has a half-life of \(6.0\text{ hours}\). A sample initially contains \(1.6 \times 10^{20}\) nuclei of this radioactive isotope. How many radioactive nuclei of this isotope will remain in the sample after \(24\text{ hours}\)?
A.4.0 x 10^19
B.2.0 x 10^19
C.1.0 x 10^19
D.1.0 x 10^18
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解題
First, find the number of half-lives that have elapsed: Number of half-lives = \(24\text{ hours} / 6.0\text{ hours} = 4\) half-lives.
After each half-life, the number of remaining nuclei is halved: • After 1 half-life: \(8.0 \times 10^{19}\) • After 2 half-lives: \(4.0 \times 10^{19}\) • After 3 half-lives: \(2.0 \times 10^{19}\) • After 4 half-lives: \(1.0 \times 10^{19}\)
1 mark for calculating the correct number of half-lives and identifying the remaining radioactive nuclei as 1.0 x 10^19 (Option C).
Paper 43 (Theory - Extended)
Twelve structured theory questions. Answer all questions. Show all working and use appropriate units.
12 題目 · 120 分
題目 1 · 結構題
10 分
A roller coaster car of mass 400 kg is moving along a track. It starts from rest at point A, at a height of 30 m above the ground, and rolls down to point B, which is at a height of 5 m above the ground.
(a) Calculate the gravitational potential energy (g.p.e.) lost by the roller coaster car between point A and point B. [Take \(g = 9.8\text{ m/s}^2\)] [3]
(b) During the descent, 15% of the initial potential energy lost is converted into thermal energy due to friction and air resistance. Calculate the kinetic energy of the car at point B. [2]
(c) Calculate the speed of the roller coaster car at point B. [3]
(d) Explain, in terms of work done, why the car eventually comes to a halt when the brakes are applied. [2]
(d) The brakes exert a frictional force on the wheels opposing the motion. Work is done by this force over a distance, transferring the kinetic energy of the car into thermal energy in the brakes and surroundings.
評分準則
(a) - Use of \(h = 25\text{ m}\) [1] - Correct formula \(mgh\) [1] - Correct calculation to get 98000 J (or 98 kJ) [1]
(b) - Multiplies lost GPE by 0.85 (or calculates 15% as 14700 J and subtracts) [1] - Correct calculation to get 83300 J (or 83.3 kJ) [1]
(c) - Correct formula \(KE = \frac{1}{2}mv^2\) rearranged or substituted [1] - Correct substitution: \(83300 = 200 \times v^2\) [1] - Correct speed value: 20.4 m/s (accept 20 m/s with correct working, 20.41 m/s) [1]
(d) - Identifies that friction/brakes apply a force opposing direction of motion, doing work [1] - Work done transfers kinetic energy into thermal energy [1]
題目 2 · 結構題
10 分
A student investigates the reaction of magnesium with dilute hydrochloric acid:
(a) The student reacts 1.20 g of magnesium ribbon with excess hydrochloric acid. Calculate the number of moles of magnesium used. [Relative atomic mass: \(A_r(\text{Mg}) = 24.0\)] [2]
(b) Calculate the volume of hydrogen gas, in \(\text{dm}^3\), produced at room temperature and pressure (r.t.p.). [One mole of any gas occupies \(24.0\text{ dm}^3\) at r.t.p.] [2]
(c) In another experiment, the student reacts \(50.0\text{ cm}^3\) of \(1.50\text{ mol/dm}^3\) hydrochloric acid with excess magnesium. (i) Calculate the number of moles of hydrochloric acid used. [2] (ii) Determine which reactant is the limiting reactant if 1.20 g of Mg (0.050 moles) was reacted with this volume of acid. Show your working. [2]
(d) Describe the chemical test to confirm the gas produced is hydrogen and state the positive result. [2]
(b) From the equation, 1 mole of \(\text{Mg}\) produces 1 mole of \(\text{H}_2\). \(\text{Moles of } \text{H}_2 = 0.050\text{ mol}\). \(\text{Volume of } \text{H}_2 = 0.050\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 1.20\text{ dm}^3\).
(c) (i) \(\text{Moles of HCl} = \text{concentration} \times \text{volume in dm}^3 = 1.50 \times \frac{50.0}{1000} = 0.075\text{ mol}\). (ii) According to the equation, 1 mole of \(\text{Mg}\) reacts with 2 moles of \(\text{HCl}\). 0.050 moles of \(\text{Mg}\) would require \(2 \times 0.050 = 0.100\text{ mol}\) of \(\text{HCl}\). Since only 0.075 moles of \(\text{HCl}\) are available, \(\text{HCl}\) is the limiting reactant.
(d) Apply a lighted splint to the gas. If hydrogen is present, it burns with a squeaky pop sound.
(b) - Multiplies moles of hydrogen by 24 [1] - Correct answer: 1.20 \(\text{dm}^3\) (accept correct conversion to 1200 \(\text{cm}^3\) if unit stated) [1]
(c)(ii) - Explains the mole ratio (1 Mg : 2 HCl) or calculates moles of Mg needed [1] - Concludes that HCl is the limiting reactant because 0.075 mol is less than the 0.100 mol required [1]
(d) - Test: Lighted splint / flame introduced [1] - Result: Squeaky pop / pop sound heard [1]
題目 3 · 結構題
10 分
Photosynthesis is the key process by which green plants produce carbohydrates.
(a) State the balanced chemical equation for photosynthesis. [3]
(b) Describe an experimental method that can be used to measure the rate of photosynthesis in an aquatic plant, such as Elodea, at different light intensities. [3]
(c) When light intensity is increased beyond a certain point, the rate of photosynthesis no longer increases and reaches a plateau. Explain why this happens. [2]
(d) Explain how the structure of palisade mesophyll cells in a leaf is adapted to maximize photosynthesis. [2]
(b) Place the aquatic plant in a beaker of water containing sodium hydrogencarbonate (to provide carbon dioxide). Position a light source at a measured distance from the plant. Count the number of bubbles of oxygen produced per unit time (e.g. per minute) or collect the volume of gas using a gas syringe. Change the distance of the light source to vary light intensity while keeping temperature constant.
(c) At very high light intensities, light is no longer the limiting factor. Another factor, such as temperature or carbon dioxide concentration, is in short supply and limits the overall rate of photosynthesis.
(d) Palisade mesophyll cells are long, column-shaped, and tightly packed together near the upper surface of the leaf to receive maximum sunlight. They contain a high density of chloroplasts to capture as much light energy as possible.
(b) - Description of setup with aquatic plant and method to vary light intensity (e.g., moving a lamp) [1] - Measurement: Count bubbles of oxygen / measure volume of gas collected in a set time [1] - Controlled variable: Mention of keeping temperature constant or using a heat shield / keeping carbon dioxide concentration constant [1]
(c) - States that light is no longer the limiting factor [1] - Identifies another limiting factor such as carbon dioxide concentration or temperature [1]
(d) - High density / many chloroplasts to absorb maximum light [1] - Long, vertical shape / tightly packed to maximize absorption near the leaf surface [1]
題目 4 · 結構題
10 分
A circuit contains a 12 V battery connected to a resistor network. The network consists of a \(6.0\ \Omega\) resistor and a \(12.0\ \Omega\) resistor connected in parallel with each other. This parallel combination is connected in series with a third resistor of resistance \(4.0\ \Omega\).
(a) Calculate the combined resistance of the two parallel resistors. [2]
(b) Show that the total equivalent resistance of the entire circuit is \(8.0\ \Omega\). [2]
(c) Calculate the total current flowing from the battery. [2]
(d) Calculate the electrical energy transferred in the \(4.0\ \Omega\) resistor over a period of 5.0 minutes. [4]
(b) The parallel combination is in series with the \(4.0\ \Omega\) resistor: \(R_{\text{total}} = R_p + R_3 = 4.0\ \Omega + 4.0\ \Omega = 8.0\ \Omega\).
(c) Using Ohm's Law for the whole circuit: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{8.0\ \Omega} = 1.5\text{ A}\).
(d) The current through the \(4.0\ \Omega\) series resistor is the total current of \(1.5\text{ A}\). Power in this resistor: \(P = I^2 R = (1.5)^2 \times 4.0 = 2.25 \times 4.0 = 9.0\text{ W}\). Time: \(t = 5.0\text{ minutes} = 5.0 \times 60\text{ s} = 300\text{ s}\). Energy transferred: \(E = P \times t = 9.0\text{ W} \times 300\text{ s} = 2700\text{ J}\) (or 2.7 kJ).
評分準則
(a) - Correct formula for parallel resistance used [1] - Correct calculation to get 4.0 \(\Omega\) [1]
(b) - States series addition formula \(R_{\text{total}} = R_p + R_3\) [1] - Adds \(4.0 + 4.0\) to show 8.0 \(\Omega\) [1]
(c) - Correct formula \(I = V / R\) [1] - Correct calculation: 1.5 A [1]
(d) - Converts time to seconds: \(5.0 \times 60 = 300\text{ s}\) [1] - Uses correct power formula \(P = I^2 R\) or \(P = V I\) with appropriate voltage across the resistor (6.0 V) [1] - Calculates power as 9.0 W [1] - Obtains final energy value: 2700 J (or 2.7 kJ) [1]
題目 5 · 結構題
10 分
Hydrazine, \(\text{N}_2\text{H}_4\), reacts with oxygen according to the following equation:
The structural formulas of the reactants and products are shown below: - Hydrazine has four \(\text{N}-\text{H}\) single bonds and one \(\text{N}-\text{N}\) single bond. - Oxygen gas has one \(\text{O}=\text{O}\) double bond. - Nitrogen gas has one \(\text{N}\equiv\text{N}\) triple bond. - Water has two \(\text{O}-\text{H}\) single bonds.
Use the bond energy values in the table to answer the questions.
(a) Calculate the total energy required to break all the bonds in 1 mole of \(\text{N}_2\text{H}_4\) and 1 mole of \(\text{O}_2\). [3]
(b) Calculate the total energy released when new bonds are formed to produce 1 mole of \(\text{N}_2\) and 2 moles of \(\text{H}_2\text{O}\). [3]
(c) Calculate the overall energy change (\(\Delta H\)) for this reaction and state whether the reaction is exothermic or endothermic. [2]
(d) Explain, in terms of bond breaking and bond forming, why this reaction is exothermic or endothermic as identified in part (c). [2]
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解題
(a) Energy to break bonds (reactants): - Four \(\text{N}-\text{H}\) bonds: \(4 \times 391 = 1564\text{ kJ/mol}\) - One \(\text{N}-\text{N}\) bond: \(1 \times 160 = 160\text{ kJ/mol}\) - One \(\text{O}=\text{O}\) bond: \(1 \times 498 = 498\text{ kJ/mol}\) Total energy input = \(1564 + 160 + 498 = 2222\text{ kJ/mol}\).
(b) Energy released in forming bonds (products): - One \(\text{N}\equiv\text{N}\) bond: \(1 \times 945 = 945\text{ kJ/mol}\) - Four \(\text{O}-\text{H}\) bonds (2 water molecules): \(4 \times 463 = 1852\text{ kJ/mol}\) Total energy output = \(945 + 1852 = 2797\text{ kJ/mol}\).
(c) Overall energy change: \(\Delta H = \text{Energy input} - \text{Energy output} = 2222 - 2797 = -575\text{ kJ/mol}\). Since \(\Delta H\) is negative, the reaction is exothermic.
(d) The reaction is exothermic because more energy is released when forming the new bonds in the products than is absorbed/required to break the bonds in the reactants.
(d) - Mentions that energy released during bond forming is greater than energy absorbed during bond breaking [2]
題目 6 · 結構題
10 分
A student investigates osmosis using potato tissue.
(a) Define the term osmosis. [2]
(b) Cylinders of potato tissue are weighed and placed into test tubes containing either distilled water (\(0.0\text{ mol/dm}^3\)) or a concentrated salt solution (\(0.8\text{ mol/dm}^3\)). After two hours, they are weighed again. (i) Explain why the potato cylinder in distilled water increases in mass. [2] (ii) Describe the appearance and physical state of the potato cells after being in the concentrated salt solution. [2]
(c) State two variables that must be controlled in this experiment to ensure a fair comparison. [2]
(d) Explain how active transport differs from osmosis. [2]
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解題
(a) Osmosis is the net movement of water molecules from a region of higher water potential (a dilute solution) to a region of lower water potential (a concentrated solution) through a partially permeable membrane.
(b) (i) Distilled water has a higher water potential than the cytoplasm inside the potato cells. Water moves into the potato cells down a water potential gradient by osmosis, increasing the mass of the tissue. (ii) The cells lose water and become flaccid/plasmolysed. The cell membrane pulls away from the cell wall, and the potato cylinder feels soft/bendy.
(c) Variables to control: Temperature of the solution; Surface area / length / volume of the potato cylinders; Source/type of potato; Time the cylinders are left in the solution.
(d) Active transport is the movement of particles through a cell membrane from a region of lower concentration to a region of higher concentration (against a concentration gradient), which requires energy from respiration and membrane proteins. Osmosis is passive, goes down a gradient, and is specifically for water.
評分準則
(a) - Movement of water from higher to lower water potential [1] - Through a partially permeable membrane [1]
(b)(i) - Water moves into the potato cells by osmosis [1] - Because distilled water has a higher water potential than the cell sap [1]
(b)(ii) - Cells become plasmolysed / cytoplasm shrinks / membrane pulls away from wall [1] - Cells/cylinder become flaccid / soft / flexible [1]
(c) - Any two from: temperature, surface area/dimension/diameter of potato cylinder, time left in solution, species/source of potato [2]
(d) - Active transport is against a concentration gradient / osmosis is down a water potential gradient [1] - Active transport requires energy (from respiration) / osmosis is passive [1]
題目 7 · 結構題
10 分
A student analyzes an unknown compound, Salt X, using qualitative tests and chromatography.
(a) To identify the dyes present in X, the student performs paper chromatography. (i) State why the start line on the chromatography paper must be drawn in pencil rather than ink. [1] (ii) The solvent front travels 8.0 cm from the start line. A blue spot from Salt X travels 5.2 cm. Calculate the \(R_f\) value of this blue spot. [2]
(b) The student makes an aqueous solution of Salt X and carries out the following analytical tests: (i) Addition of dilute nitric acid followed by aqueous silver nitrate produces a cream precipitate. Identify the anion present in Salt X. [1] (ii) Addition of aqueous sodium hydroxide produces a light blue precipitate. Identify the cation present in Salt X and write the ionic equation (including state symbols) for this precipitation reaction. [3] (iii) Describe what is observed when excess aqueous ammonia is added to the mixture containing the light blue precipitate from part (b)(ii). [3]
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解題
(a) (i) Pencil graphite is insoluble in the chromatography solvent and will not dissolve or run to interfere with the results, whereas ink would dissolve and separate. (ii) \(R_f = \frac{\text{Distance travelled by solute}}{\text{Distance travelled by solvent}} = \frac{5.2\text{ cm}}{8.0\text{ cm}} = 0.65\).
(b) (i) A cream precipitate with silver nitrate indicates the presence of bromide ions (\(\text{Br}^-\)). (ii) A light blue precipitate with sodium hydroxide indicates copper(II) ions (\(\text{Cu}^{2+}\)). Ionic equation: \(\text{Cu}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Cu(OH)}_2\text{(s)}\). (iii) When excess aqueous ammonia is added to the light blue precipitate, the precipitate dissolves (solubilizes) to form a deep blue / dark blue solution.
評分準則
(a)(i) - Pencil lead/graphite is insoluble / will not dissolve and smudge/run [1]
(a)(ii) - Correct formula or working: \(5.2 / 8.0\) [1] - Correct answer: 0.65 (no units) [1]
(b)(i) - Bromide (accept \(\text{Br}^-\)) [1]
(b)(ii) - Cation: Copper(II) / \(\text{Cu}^{2+}\) [1] - Ionic equation formulae: \(\text{Cu}^{2+} + 2\text{OH}^- \rightarrow \text{Cu(OH)}_2\) [1] - State symbols correct: (aq) for reactants, (s) for product [1]
(b)(iii) - Precipitate dissolves / is soluble [1] - Forms a deep blue / dark blue [1] - Solution [1]
題目 8 · 結構題
10 分
Carbon-14 (\(^{14}_{6}\text{C}\)) is a radioisotope used in carbon dating.
(a) Define the term isotopes. [2]
(b) Carbon-14 decays into nitrogen-14 (\(\text{N}\)) by emitting a beta-minus (\(\beta^-\)\) particle (an electron). Complete the nuclear decay equation below by writing the correct proton and nucleon numbers in the empty boxes:
(c) A sample of ancient wood contains \(8.0 \times 10^{10}\) atoms of Carbon-14. The half-life of Carbon-14 is 5700 years. Calculate the number of Carbon-14 atoms remaining in this sample after 17100 years. Show your working. [3]
(d) Explain why beta radiation is more penetrating than alpha radiation. [2]
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解題
(a) Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons (or same atomic number, different mass number).
(b) The completed equation is: $$^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + ^{0}_{-1}\text{e}$$ - For Nitrogen: nucleon number = 14, proton number = 7. - For the beta particle: nucleon number = 0, proton number = -1.
(c) Calculate the number of half-lives that have passed: $$\text{Number of half-lives} = \frac{17100\text{ years}}{5700\text{ years}} = 3\text{ half-lives}$$ After 1 half-life: \(8.0 \times 10^{10} \div 2 = 4.0 \times 10^{10}\) atoms After 2 half-lives: \(4.0 \times 10^{10} \div 2 = 2.0 \times 10^{10}\) atoms After 3 half-lives: \(2.0 \times 10^{10} \div 2 = 1.0 \times 10^{10}\) atoms There are \(1.0 \times 10^{10}\) atoms of Carbon-14 remaining.
(d) Beta particles have a smaller mass and a smaller charge (\(-1\)) compared to alpha particles (which have a mass of 4 and a charge of \(+2\)). This makes beta particles less ionizing, allowing them to pass through more matter before losing all their energy.
評分準則
(a) - Atoms of same element / same proton number [1] - Different number of neutrons / different mass number [1]
(b) - Nitrogen-14 shown as \(^{14}_{7}\text{N}\) [1] - Beta particle shown as \(^{0}_{-1}\text{e}\) (or \(^{0}_{-1}\beta\)) [1] - Both sides of the equation balance (nucleon sum and proton sum are equal) [1]
(c) - Calculates number of half-lives = 3 [1] - Shows process of halving three times [1] - Correct final answer: \(1.0 \times 10^{10}\) atoms (accept \(1 \times 10^{10}\)) [1]
(d) - Beta has a lower charge (\(-1\) compared to \(+2\)) or is much smaller/lighter [1] - Beta is less strongly ionizing, so it interacts less and travels further through materials [1]
題目 9 · 結構題
10 分
A student investigates the reaction between methane and chlorine to form chloromethane and hydrogen chloride:
(a) The bond energies of some covalent bonds are shown below: - \( \text{C}-\text{H} \): \( 413 \text{ kJ/mol} \) - \( \text{Cl}-\text{Cl} \): \( 242 \text{ kJ/mol} \) - \( \text{C}-\text{Cl} \): \( 339 \text{ kJ/mol} \) - \( \text{H}-\text{Cl} \): \( 431 \text{ kJ/mol} \)
(i) Calculate the energy required to break all the bonds in 1 mole of \( \text{CH}_4 \) and 1 mole of \( \text{Cl}_2 \). [2]
(ii) Calculate the energy released when new bonds are formed to make 1 mole of \( \text{CH}_3\text{Cl} \) and 1 mole of \( \text{HCl} \). [2]
(iii) Use your answers to (i) and (ii) to calculate the overall energy change (\( \Delta H \)) for this reaction, and state whether the reaction is exothermic or endothermic. [2]
(b) Draw an energy level diagram for this reaction. Label the reactants, products, activation energy (\( E_a \)) and enthalpy change (\( \Delta H \)). [4]
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解題
(a)(i) Bonds broken: 4 moles of \( \text{C}-\text{H} \) bonds: \( 4 \times 413 = 1652 \text{ kJ} \) 1 mole of \( \text{Cl}-\text{Cl} \) bonds: \( 1 \times 242 = 242 \text{ kJ} \) Total energy to break bonds = \( 1652 + 242 = 1894 \text{ kJ} \).
(a)(iii) Enthalpy change \( \Delta H = \text{energy broken} - \text{energy formed} = 1894 - 2009 = -115 \text{ kJ/mol} \). Since \( \Delta H \) is negative, the reaction is exothermic.
(b) The energy level diagram should show: - Reactants (\( \text{CH}_4 + \text{Cl}_2 \)) on a horizontal line higher than products. - Products (\( \text{CH}_3\text{Cl} + \text{HCl} \)) on a horizontal line lower than reactants. - A smooth curve rising from the reactants line to a peak and then falling to the products line. - An arrow pointing upwards from the reactants level to the peak of the curve, labelled as activation energy \( E_a \). - An arrow pointing downwards from the reactants level to the products level, labelled as enthalpy change \( \Delta H = -115 \text{ kJ/mol} \).
(a)(iii) - Correct calculation of \( 1894 - 2009 = -115 \text{ kJ/mol} \) (allow ecf from (a)(i) and (a)(ii)) (1) - Identifies reaction as exothermic due to negative value (1)
(b) - Y-axis labelled Energy / Enthalpy AND X-axis labelled Progress of reaction / Reaction pathway (1) - Reactants line higher than products line, with formulas correctly placed (1) - Upward-pointing arrow from reactants to curve peak labelled \( E_a \) (1) - Downward-pointing arrow from reactants to products labelled \( \Delta H \) (1)
題目 10 · 結構題
10 分
An electric vehicle of mass \( 1200 \text{ kg} \) accelerates from rest along a straight, horizontal test track.
(a) The car accelerates uniformly from rest to a speed of \( 24 \text{ m/s} \) in \( 8.0 \text{ s} \).
(i) Calculate the acceleration of the car. [2]
(ii) Calculate the horizontal distance travelled by the car during these \( 8.0 \text{ s} \). [2]
(iii) Calculate the useful work done by the car's motor during this acceleration, assuming no energy losses due to air resistance or friction. [2]
(b) After \( 8.0 \text{ s} \), the car continues to travel at a constant speed of \( 24 \text{ m/s} \).
(i) Name the two non-zero forces acting horizontally on the car when it is travelling at a constant speed, and describe their relationship. [2]
(ii) The motor of the car now provides a forward driving force of \( 600 \text{ N} \) to maintain this constant speed of \( 24 \text{ m/s} \). Calculate the useful power output of the motor. [2]
(a)(iii) Work done = change in kinetic energy \( E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 24^2 = 345600 \text{ J} \) or \( 3.46 \times 10^5 \text{ J} \) (or \( 346 \text{ kJ} \))
(b)(i) The two forces are the driving force (forward force) and friction/air resistance/drag (resistive/opposing force). They are equal in magnitude and opposite in direction (so net force is zero).
(b)(ii) \( P = F \times v = 600 \times 24 = 14400 \text{ W} \) (or \( 14.4 \text{ kW} \))
評分準則
(a)(i) - Formula \( a = \frac{\Delta v}{t} \) used correctly (1) - Correct calculation with units: \( 3.0 \text{ m/s}^2 \) (1)
(a)(ii) - Suitable formula used, e.g., \( s = \frac{(u+v)}{2}t \) or \( s = \frac{1}{2}at^2 \) (1) - Correct calculation: \( 96 \text{ m} \) (1)
(a)(iii) - Recognising work done = change in kinetic energy AND formula \( E_k = \frac{1}{2}mv^2 \) (1) - Correct calculation: \( 345600 \text{ J} \) (or \( 3.46 \times 10^5 \text{ J} \) / \( 346 \text{ kJ} \)) (1)
(b)(i) - Identifies driving force (or thrust) AND friction (or drag / air resistance) (1) - States they are equal and opposite / balanced (1)
(b)(ii) - Formula \( P = F \times v \) used (1) - Correct calculation: \( 14400 \text{ W} \) (or \( 14.4 \text{ kW} \)) (1)
題目 11 · 結構題
10 分
A student investigates osmosis in plant cells using cylindrical pieces of potato. Six identical potato cylinders are weighed and placed into different test-tubes. Each test-tube contains a different concentration of sucrose solution: from \( 0.0 \text{ mol/dm}^3 \) (distilled water) to \( 1.0 \text{ mol/dm}^3 \). After 4 hours, the cylinders are removed, dried, and reweighed.
(a) Define the term *osmosis*. [3]
(b) The potato cylinder in distilled water (\( 0.0 \text{ mol/dm}^3 \)) increased in mass by \( 12\% \). Explain this increase in mass, using terms related to water potential and the structure of plant cells. [4]
(c) The potato cylinder placed in the \( 0.8 \text{ mol/dm}^3 \) sucrose solution decreased in mass. When observed under a light microscope, the cells showed *plasmolysis*.
(i) Describe the appearance of a plasmolysed plant cell. [2]
(ii) State the name of the cell structure that is fully permeable and prevents the plant cell from bursting when placed in distilled water. [1]
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解題
(a) Osmosis is defined as the net movement of water molecules from a region of higher water potential (a dilute solution) to a region of lower water potential (a concentrated solution), down a water potential gradient, across a partially permeable membrane.
(b) Distilled water has a higher water potential than the cell sap/cytoplasm inside the potato cells. Water therefore enters the cells by osmosis down a water potential gradient, through the partially permeable cell membrane. This causes the vacuole to increase in volume, pushing the cytoplasm against the cell wall. The cell becomes turgid, leading to an increase in overall mass of the cylinder.
(c)(i) In a plasmolysed plant cell, the cell membrane has pulled away from the cell wall, and the vacuole is shrunken/reduced in size. There is a visible gap between the cell wall and the cytoplasm.
(c)(ii) The cell wall.
評分準則
(a) - net movement of water molecules (1) - from (a region of) higher water potential to (a region of) lower water potential / down a water potential gradient (1) - across/through a partially permeable membrane (1)
(b) - distilled water has a higher water potential than the vacuole/cell sap/cytoplasm (or vice versa) (1) - water moves into the cells by osmosis (1) - down a water potential gradient / across the partially permeable membrane (1) - cells become turgid / vacuole swells and exerts turgor pressure against cell wall (1)
(c)(i) - cell membrane/cytoplasm has pulled away/detached from the cell wall (1) - vacuole is shrunk / smaller (1)
(c)(ii) - cell wall (1)
題目 12 · 結構題
10 分
A simple step-down transformer is used to reduce a mains voltage of \( 240 \text{ V} \) a.c. to a safer operating voltage for an electric device.
(a) Explain how a changing current in the primary coil of the transformer induces an electromotive force (e.m.f.) in the secondary coil. [4]
(b) The primary coil of the transformer has \( 1200 \) turns and is connected to a \( 240 \text{ V} \) a.c. supply. The secondary coil has \( 60 \text{ turns} \).
(i) Calculate the output voltage of the secondary coil. Show your working. [2]
(ii) The secondary coil is connected to a resistor of resistance \( 4.0\ \Omega \). Calculate the current in the secondary circuit. [2]
(iii) Assuming the transformer is \( 100\% \) efficient, calculate the current in the primary coil when the secondary circuit is operating. [2]
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解題
(a) An alternating current in the primary coil generates a continuously changing (alternating) magnetic field. This magnetic field is guided through the soft iron core to the secondary coil. Because the magnetic field in the core is constantly changing, it cuts through the secondary coil. This changing magnetic flux linkage induces an alternating electromotive force (e.m.f.) across the secondary coil.
(a) - alternating/changing current in primary coil creates a changing/varying magnetic field (1) - the (soft) iron core guides / concentrates the magnetic field to the secondary coil (1) - the changing magnetic field cuts the secondary coil / there is a changing magnetic flux linkage (1) - this induces an electromotive force (e.m.f.) / voltage in the secondary coil (1)
(b)(ii) - Correct formula used: \( I = \frac{V}{R} \) (1) - Correct calculation with unit: \( 3.0 \text{ A} \) (allow ecf from (b)(i)) (1)
(b)(iii) - Correct formula used: \( V_p I_p = V_s I_s \) (or equivalent ratio formula) (1) - Correct calculation with unit: \( 0.15 \text{ A} \) (allow ecf from (b)(i) and (b)(ii)) (1)
Paper 63 (Alternative to Practical)
Six alternative to practical questions. Answer all questions. Write your answers in the spaces provided.
6 題目 · 60 分
題目 1 · Practical / Experimental Question
10 分
A student investigates the rate of photosynthesis in an aquatic plant, Elodea (waterweed), under different light intensities.
(a) The gas released by the plant is collected in a gas syringe. Over a period of \(5.0\text{ minutes}\), the plunger of the gas syringe moves from \(0.0\text{ cm}^3\) to \(4.5\text{ cm}^3\). (i) State the volume of gas collected. (ii) Calculate the rate of gas production in \(\text{cm}^3/\text{min}\).
(b) The light intensity is varied by placing a lamp at different distances from the beaker containing the waterweed. (i) State the independent variable in this experiment. (ii) Suggest one way the student could keep the temperature of the water around the weed constant during this experiment. (iii) State one other variable that must be kept constant to ensure a fair test.
(c) Terrestrial plants store starch in their leaves as a product of photosynthesis. Describe a method to test a leaf for the presence of starch safely. In your answer, explain the reason for each key step.
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解題
(a)(i) Volume of gas collected = \(4.5\text{ cm}^3\). (ii) Rate of gas production = \(4.5\text{ cm}^3 / 5.0\text{ min} = 0.90\text{ cm}^3/\text{min}\).
(b)(i) Independent variable: Light intensity (which is changed by moving the lamp to different distances). (ii) To keep the temperature constant, place a transparent glass block (heat shield) between the lamp and the beaker, or use a water bath, or use an LED bulb which generates very little thermal energy. (iii) Controlled variable: The concentration of carbon dioxide (maintained by adding a fixed mass of sodium hydrogencarbonate to the water) or using the exact same piece of waterweed.
(c) Starch test procedure: 1. Place the leaf in boiling water for about 30 seconds (to kill the leaf cells and break down cell membranes). 2. Place the leaf in a tube of ethanol and heat it in a water bath (ethanol extracts the green chlorophyll to decolorize the leaf so color changes are clear; a water bath must be used because ethanol is highly flammable and must not be heated with a direct Bunsen flame). 3. Rinse the leaf in warm water (to soften it). 4. Spread the leaf on a white tile and add a few drops of iodine solution. If starch is present, the color changes from orange-brown to blue-black.
評分準則
- (a)(i) \(4.5\text{ cm}^3\) [1 mark] - (a)(ii) \(0.90\text{ cm}^3/\text{min}\) (allow error carried forward from (a)(i)) [1 mark] - (b)(i) Light intensity / distance of the lamp [1 mark] - (b)(ii) Use of a transparent heat shield / water bath / LED light source [1 mark] - (b)(iii) Concentration of sodium hydrogencarbonate / carbon dioxide concentration [1 mark] - (c) Method detail 1: Boil leaf in water to halt metabolic reactions [1 mark] - Method detail 2: Heat leaf in ethanol using a water bath [1 mark] - Safety precaution: Explicitly state that ethanol is flammable and must be heated without a naked flame / in a water bath [1 mark] - Method detail 3: Rinse leaf in water and add iodine solution [1 mark] - Color change: Blue-black indicates presence of starch [1 mark]
題目 2 · Practical / Experimental Question
10 分
A student is provided with a solid sample, salt Y, which is a light green crystalline solid. They perform qualitative chemical tests to identify the cation and anion in Y.
(a) The student dissolves a sample of Y in distilled water to make a solution. They add aqueous sodium hydroxide dropwise until in excess. (i) State the observation when a few drops of aqueous sodium hydroxide are added. (ii) State the observation when excess aqueous sodium hydroxide is added. (iii) Identify the cation present in Y based on these observations.
(b) To a second portion of the solution of Y, the student adds dilute nitric acid followed by aqueous silver nitrate. A white precipitate is formed. (i) Identify the anion present in Y. (ii) Write the ionic equation, including state symbols, for the formation of this precipitate.
(c) State why the student must use distilled water rather than tap water to prepare the solution of Y.
(d) Describe how the student can perform a flame test on the solid sample Y and state the expected observation if copper(II) ions were present instead of the cation in Y.
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解題
(a)(i) Adding a few drops of aqueous sodium hydroxide to an iron(II) solution produces a green precipitate. (ii) The green precipitate remains insoluble when excess sodium hydroxide is added. (iii) The cation is therefore iron(II), \(\text{Fe}^{2+}\).
(b)(i) The formation of a white precipitate with silver nitrate after adding dilute nitric acid indicates the presence of chloride ions (\(\text{Cl}^-\)). (ii) The ionic equation is: \(\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})\).
(c) Tap water contains dissolved mineral ions (such as chloride, sulfate, or calcium) which would react with the testing reagents, leading to false-positive results.
(d) Dip a clean nichrome or platinum wire into concentrated hydrochloric acid, touch it to the solid sample Y, and place it into the hot, blue (non-luminous) flame of a Bunsen burner. If copper(II) ions were present, a blue-green flame would be observed.
評分準則
- (a)(i) Green precipitate [1 mark] - (a)(ii) Precipitate is insoluble in excess [1 mark] - (a)(iii) Iron(II) / \(\text{Fe}^{2+}\) [1 mark] - (b)(i) Chloride / \(\text{Cl}^-\)[1 mark] - (b)(ii) Equation: \(\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}\) [1 mark]; Correct state symbols: \(\text{(aq)}\) and \(\text{(s)}\) [1 mark] - (c) Tap water contains impurities/ions (e.g. chloride) which could interfere with analytical results [1 mark] - (d) Use a clean nichrome/platinum wire and concentrated acid [1 mark]; heat in non-luminous (blue) flame [1 mark]; blue-green color observed [1 mark]
題目 3 · Practical / Experimental Question
10 分
A student carries out an experiment to investigate the relationship between the load applied to a helical spring and its extension.
(a) Figure 3.1 shows the spring next to a millimeter scale ruler. - In Diagram A (unstretched spring), the lower end of the spring is aligned with the \(24.5\text{ mm}\) mark on the ruler. - In Diagram B (spring stretched with a load of \(3.0\text{ N}\)), the lower end of the spring is aligned with the \(60.5\text{ mm}\) mark on the ruler. (i) Record the unstretched length, \(L_0\), and the stretched length, \(L_1\), of the spring. (ii) Calculate the extension, \(e\), produced by the \(3.0\text{ N}\) load.
(b) The student obtains extensions for loads up to \(6.0\text{ N}\) and plots a graph of Load (on the y-axis) against Extension (on the x-axis). (i) State two features of the graph that confirm the spring obeys Hooke's Law. (ii) Explain how the student can use the graph to find the spring constant, \(k\).
(c) Describe how the student should position their eye to avoid parallax error when reading the ruler scale.
(d) Describe a safety precaution the student should take when performing this experiment, explaining why it is necessary.
(b)(i) Hooke's Law is obeyed because the graph is a straight line and it passes through the origin, representing direct proportionality. (ii) Since \(F = k e\), the spring constant \(k\) is equal to \(\text{Load} / \text{Extension}\), which is the gradient (slope) of the Load-Extension graph.
(c) The student should position their eye horizontally level with the bottom of the spring and read perpendicular (at a right angle) to the ruler scale to avoid parallax error.
(d) When carrying out experiments with loaded springs, the student should wear eye protection (safety goggles) in case the spring snaps or slips off the clamp. Alternatively, place a padded mat/box on the floor below the masses to catch any falling weights.
評分準則
- (a)(i) \(L_0 = 24.5\text{ mm}\) [1 mark] and \(L_1 = 60.5\text{ mm}\) [1 mark] - (a)(ii) \(e = 36.0\text{ mm}\) (allow error carried forward from (a)(i)) [1 mark] - (b)(i) Graph is a straight line [1 mark] AND passes through the origin / (0,0) [1 mark] - (b)(ii) Calculate the gradient of the graph [1 mark] - (c) Eye positioned level with the scale mark [1 mark]; view perpendicular / at right angles to the ruler [1 mark] - (d) Wear safety goggles [1 mark]; to protect eyes from a snapping spring or falling mass [1 mark]
題目 4 · Practical / Experimental Question
10 分
A student investigates the effect of temperature on the rate of starch breakdown by amylase. Starch is mixed with amylase at different temperatures, and a drop of the reaction mixture is tested with iodine solution every 30 seconds until the iodine no longer changes color.
(a) State the initial color of the iodine solution and the color observed when starch is completely broken down.
(b) Explain why the student kept both the amylase solution and the starch solution in separate tubes in a water bath for 5 minutes before mixing them together.
(c) Identify two variables that must be controlled in this investigation to ensure valid results.
(d) Table 4.1 shows the results of the experiment: - At \(10\text{ }^{\circ}\text{C}\): time taken is \(480\text{ seconds}\) - At \(40\text{ }^{\circ}\text{C}\): time taken is \(60\text{ seconds}\) - At \(70\text{ }^{\circ}\text{C}\): starch is not broken down after \(600\text{ seconds}\) (i) Explain the difference in times taken between \(10\text{ }^{\circ}\text{C}\) and \(40\text{ }^{\circ}\text{C}\). (ii) Explain the result obtained at \(70\text{ }^{\circ}\text{C}\).
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解題
(a) The iodine solution is initially yellow-brown (or orange-brown). When starch is completely broken down (into maltose and glucose), no starch is present to react with iodine, so the iodine solution remains yellow-brown (it does not turn blue-black).
(b) Keeping the separate solutions in the water bath for 5 minutes ensures that both solutions equilibrate and reach the exact, desired target temperature before they are mixed to start the reaction.
(c) Controlled variables include: the concentration of the amylase solution, the volume of amylase, the concentration of the starch solution, the volume of starch, and the pH of the mixture (using a buffer).
(d)(i) At \(40\text{ }^{\circ}\text{C}\), the enzyme and substrate molecules have significantly more kinetic energy than at \(10\text{ }^{\circ}\text{C}\). This leads to more frequent and successful collisions between the amylase active site and the starch molecules, resulting in a faster rate of reaction and a shorter time taken (\(60\text{ s}\) vs \(480\text{ s}\)). (ii) At \(70\text{ }^{\circ}\text{C}\), the high temperature has caused the amylase enzyme to denature. The specific shape of its active site is permanently lost, so the starch substrate can no longer bind to it, and no reaction occurs.
評分準則
- (a) Yellow-brown / orange-brown [1 mark]; remains yellow-brown / orange-brown (no change / does not turn blue-black) [1 mark] - (b) To allow both reactants to reach the desired target temperature before the reaction begins [2 marks] - (c) Any two from: volume of starch / volume of amylase / concentration of starch / concentration of amylase / pH [2 marks] (1 mark for each) - (d)(i) Molecules have higher kinetic energy at \(40\text{ }^{\circ}\text{C}\) than at \(10\text{ }^{\circ}\text{C}\) [1 mark]; leads to more frequent collisions between active site and substrate [1 mark] - (d)(ii) Enzyme is denatured [1 mark]; active site shape changed so substrate can no longer bind [1 mark]
題目 5 · Practical / Experimental Question
10 分
A student investigates the temperature change during the displacement reaction between zinc powder and copper(II) sulfate solution.
(a) Figure 5.1 shows the thermometer readings for the reaction. - The initial temperature of the copper(II) sulfate solution is \(19.5\text{ }^{\circ}\text{C}\). - The maximum temperature reached after adding zinc powder is \(34.2\text{ }^{\circ}\text{C}\). (i) Write down these temperatures and calculate the temperature increase, \(\Delta T\). (ii) State whether this reaction is exothermic or endothermic, giving a reason for your answer.
(b) The experiment was conducted in a plastic polystyrene cup instead of a glass beaker. (i) Explain why a polystyrene cup is a better choice for this experiment. (ii) Suggest one improvement to the apparatus to further reduce thermal energy transfer to the surroundings.
(c) Calculate the thermal energy released, \(Q\), in this reaction using the formula: \(Q = m c \Delta T\) where \(m = 25.0\text{ g}\) (mass of the solution), \(c = 4.2\text{ J}/(\text{g}^{\circ}\text{C})\) (specific heat capacity), and \(\Delta T\) is your calculated value from (a)(i). Show your working.
(d) Explain why the student stirred the mixture continuously throughout the reaction.
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解題
(a)(i) Initial temperature \(T_i = 19.5\text{ }^{\circ}\text{C}\) Maximum temperature \(T_f = 34.2\text{ }^{\circ}\text{C}\) Temperature increase \(\Delta T = 34.2 - 19.5 = 14.7\text{ }^{\circ}\text{C}\). (ii) The reaction is exothermic because the temperature of the reaction mixture increased, indicating that heat energy was released to the surroundings.
(b)(i) Polystyrene is an excellent thermal insulator compared to glass. It reduces heat transfer from the reaction mixture to the surroundings, ensuring that the measured temperature change is as close to the theoretical maximum as possible. (ii) Place a lid over the polystyrene cup, or place the cup inside a larger glass beaker packed with insulating cotton wool.
(d) Continuous stirring ensures that the solid zinc is thoroughly mixed with the copper(II) sulfate solution for a rapid and complete reaction, and that the heat released is distributed evenly throughout the liquid so the thermometer records an accurate maximum temperature.
評分準則
- (a)(i) Correct recording of initial and maximum temperatures [1 mark]; calculation of \(\Delta T = 14.7\text{ }^{\circ}\text{C}\) [1 mark] - (a)(ii) Exothermic [1 mark]; temperature increased [1 mark] - (b)(i) Polystyrene is a poor conductor of heat / good thermal insulator [1 mark]; reduces heat loss to the surroundings [1 mark] - (b)(ii) Add a lid / place cup inside a beaker packed with cotton wool [1 mark] - (c) Correct substitution of numbers [1 mark]; correct final value of \(1543.5\text{ J}\) (or \(1.54\text{ kJ}\)) with appropriate unit [1 mark] - (d) To ensure heat is evenly distributed / to ensure complete mixing for a rapid reaction [1 mark]
題目 6 · Practical / Experimental Question
10 分
A student investigates how the electrical resistance of a constantan wire varies with its length.
(a) In the space provided, draw a circuit diagram containing a cell, a switch, an ammeter, a voltmeter, and a length of resistance wire. The voltmeter must be connected to measure the potential difference across the resistance wire, and the ammeter must measure the current through it.
(b) The student sets the length of the wire to \(50.0\text{ cm}\). - The ammeter reading is \(0.35\text{ A}\). - The voltmeter reading is \(1.75\text{ V}\). (i) State the potential difference, \(V\), and current, \(I\). (ii) Calculate the resistance, \(R\), of the \(50.0\text{ cm}\) wire. Include the unit.
(c) Suggest why the student should open the switch between taking measurements at different lengths.
(d) The student tests a \(100.0\text{ cm}\) length of the same wire and finds its resistance is \(10.0\text{ }\Omega\). (i) Predict the resistance of a \(20.0\text{ cm}\) length of this wire. (ii) State the relationship between the resistance of a wire and its length.
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解題
(a) The circuit diagram must show: 1. A cell/battery, a switch, an ammeter, and the resistance wire all connected in a single series loop. 2. A voltmeter connected in parallel across the resistance wire only. 3. Standard circuit symbols used correctly.
(b)(i) Potential difference \(V = 1.75\text{ V}\), Current \(I = 0.35\text{ A}\). (ii) Resistance \(R = V / I = 1.75\text{ V} / 0.35\text{ A} = 5.0\text{ }\Omega\).
(c) When current flows through a wire, it generates thermal energy. If the wire heats up, its temperature rises, which increases its resistance and affects the accuracy of the experiment. Opening the switch between readings keeps the temperature of the wire constant.
(d)(i) If a \(100.0\text{ cm}\) wire has a resistance of \(10.0\text{ }\Omega\), a \(20.0\text{ cm}\) wire (which is \(1/5\) of the length) will have \(1/5\) of the resistance: \(R = 10.0\text{ }\Omega \times (20.0 / 100.0) = 2.0\text{ }\Omega\). (ii) Resistance is directly proportional to the length of the wire.
評分準則
- (a) Cell/power supply, switch, and ammeter in series with the test wire [1 mark] - Voltmeter connected in parallel across the test wire [1 mark] - Correct symbols for all components (cell, switch, ammeter, voltmeter, resistor/wire) [1 mark] - (b)(i) \(V = 1.75\text{ V}\) and \(I = 0.35\text{ A}\) recorded correctly [1 mark] - (b)(ii) \(R = 5.0\) [1 mark] ohms / \(\Omega\) [1 mark] (allow error carried forward) - (c) To prevent the wire from heating up [1 mark]; because temperature changes affect resistance [1 mark] - (d)(i) \(2.0\text{ }\Omega\) [1 mark] (allow error carried forward) - (d)(ii) Resistance is directly proportional to length [1 mark]
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