An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V2) Cambridge IGCSE Science - Combined (0653) paper. Not affiliated with or reproduced from Cambridge.
卷二 選擇題 (Extended)
Answer all 40 multiple-choice questions on the separate answer sheet using soft pencil. Each question carries 1 mark.
40 题目 · 40 分
题目 1 · 選擇題
1 分
A student places a plant cell with a water potential of \(-0.5\text{ MPa}\) into a solution with a water potential of \(-0.2\text{ MPa}\). What describes the net movement of water and the resulting state of the cell?
A.Water moves out of the cell; the cell becomes plasmolysed.
B.Water moves into the cell; the cell becomes turgid.
C.Water moves out of the cell; the cell becomes turgid.
D.Water moves into the cell; the cell becomes plasmolysed.
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解题
Water moves from a region of higher water potential (less negative, \(-0.2\text{ MPa}\)) to a region of lower water potential (more negative, \(-0.5\text{ MPa}\)). Therefore, the net movement of water is into the cell. Since it is a plant cell, the cell wall prevents it from bursting, and the cell becomes turgid.
评分标准
B is the correct option. 1 mark for the correct answer.
题目 2 · 選擇題
1 分
Which test reagent is used to detect the presence of a nutrient containing nitrogen as an essential element, and what is the positive result?
A.Benedict's solution; turns from blue to brick-red
B.Biuret reagent; turns from blue to purple
C.Ethanol emulsion test; forms a milky-white emulsion
D.Iodine solution; turns from brown to blue-black
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解题
Proteins contain nitrogen as an essential element (unlike carbohydrates and lipids which only consist of carbon, hydrogen, and oxygen). The test for proteins is the Biuret test, which gives a purple or violet color as a positive result.
评分标准
B is the correct option. 1 mark for the correct answer.
题目 3 · 選擇題
1 分
A liquid hydrocarbon does not decolourise aqueous bromine in the dark. Which molecular formula and chemical property could represent this hydrocarbon?
A.\(\text{C}_6\text{H}_{12}\); it is unsaturated and reacts by addition.
B.\(\text{C}_6\text{H}_{14}\); it is saturated and burns completely to produce carbon dioxide and water.
C.\(\text{C}_6\text{H}_{12}\); it is saturated and burns to produce only carbon monoxide and hydrogen.
D.\(\text{C}_6\text{H}_{14}\); it is unsaturated and undergoes polymerization.
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解题
An alkene is unsaturated and decolourises bromine water in the dark. An alkane is saturated and does not. Alkanes have the general molecular formula \(\text{C}_n\text{H}_{2n+2}\), so a 6-carbon alkane is \(\text{C}_6\text{H}_{14}\). Saturated hydrocarbons undergo complete combustion to yield carbon dioxide and water.
评分标准
B is the correct option. 1 mark for the correct answer.
题目 4 · 選擇題
1 分
A circuit consists of a \(12\text{ V}\) d.c. power supply connected to two resistors in parallel. The resistance of the first resistor is \(4.0\ \Omega\) and the resistance of the second resistor is \(12.0\ \Omega\). What is the total current drawn from the power supply?
A.\(1.0\text{ A}\)
B.\(3.0\text{ A}\)
C.\(4.0\text{ A}\)
D.\(16.0\text{ A}\)
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解题
The total resistance \(R_p\) of the two parallel resistors is calculated by \(1/R_p = 1/4.0 + 1/12.0 = 4/12.0\), so \(R_p = 3.0\ \Omega\). The total current is \(I = V / R_p = 12\text{ V} / 3.0\ \Omega = 4.0\text{ A}\). Alternatively, the currents in the individual branches are \(12 / 4.0 = 3.0\text{ A}\) and \(12 / 12.0 = 1.0\text{ A}\), summing to \(4.0\text{ A}\).
评分标准
C is the correct option. 1 mark for the correct answer.
题目 5 · 選擇題
1 分
A crane lifts a load of mass \(500\text{ kg}\) vertically upwards through a height of \(20\text{ m}\) in a time of \(10\text{ s}\). What is the average useful power developed by the crane? (Take the gravitational field strength \(g = 10\text{ N/kg}\).)
A.\(1.0\text{ kW}\)
B.\(10\text{ kW}\)
C.\(100\text{ kW}\)
D.\(1000\text{ kW}\)
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解题
The work done (gravitational potential energy gained) is \(E = m \times g \times h = 500 \times 10 \times 20 = 100,000\text{ J}\). The power developed is \(P = E / t = 100,000\text{ J} / 10\text{ s} = 10,000\text{ W} = 10\text{ kW}\).
评分标准
B is the correct option. 1 mark for the correct answer.
题目 6 · 選擇題
1 分
Which statement correctly describes the changes in the volume and pressure inside the thorax during expiration (breathing out)?
A.volume decreases and pressure decreases
B.volume decreases and pressure increases
C.volume increases and pressure decreases
D.volume increases and pressure increases
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解题
During expiration, the diaphragm and external intercostal muscles relax, which reduces the volume of the thorax. This decrease in volume increases the pressure within the thorax, forcing air out of the lungs.
评分标准
B is the correct option. 1 mark for the correct answer.
题目 7 · 選擇題
1 分
Moving down Group VII of the Periodic Table, chlorine is a pale green gas, bromine is a red-brown liquid, and iodine is a grey-black solid at room temperature. Which statement predicts the properties of astatine, the element below iodine in Group VII?
A.It is a liquid at room temperature and is pale yellow in colour.
B.It is a gas at room temperature and is dark purple in colour.
C.It is a solid at room temperature and is black in colour.
D.It is a solid at room temperature and is colourless.
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解题
Going down Group VII, the melting and boiling points of elements increase, so the state changes from gas to liquid to solid. The elements also become darker in colour. Since iodine is a grey-black solid, astatine is predicted to be a solid at room temperature and black in colour.
评分标准
C is the correct option. 1 mark for the correct answer.
题目 8 · 選擇題
1 分
Which type of electromagnetic radiation is used in thermal imaging cameras to detect heat signatures, and how does its frequency compare to that of visible light?
A.infrared radiation; its frequency is higher than that of visible light.
B.infrared radiation; its frequency is lower than that of visible light.
C.ultraviolet radiation; its frequency is higher than that of visible light.
D.ultraviolet radiation; its frequency is lower than that of visible light.
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解题
Thermal imaging cameras detect infrared radiation emitted as heat. In the electromagnetic spectrum, infrared radiation has a longer wavelength and therefore a lower frequency than visible light.
评分标准
B is the correct option. 1 mark for the correct answer.
题目 9 · 選擇題
1 分
Which statement describes the net movement of water molecules during osmosis?
A.from a region of higher water potential to a region of lower water potential through a partially permeable membrane
B.from a region of lower water potential to a region of higher water potential through a partially permeable membrane
C.from a region of higher water potential to a region of lower water potential through a cell wall
D.from a region of lower water potential to a region of higher water potential through a cell wall
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解题
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), through a partially permeable membrane.
评分标准
Award 1 mark for selecting A.
题目 10 · 選擇題
1 分
An enzyme-catalysed reaction is carried out at various temperatures. When the temperature is increased to $60^\circ\text{C}$, the rate of reaction drops to zero. Which statement explains this observation?
A.The substrate molecules have been denatured, preventing them from binding to the enzyme.
B.The kinetic energy of the substrate and enzyme molecules has decreased to zero.
C.The enzyme molecules have denatured, changing the shape of their active sites.
D.The activation energy of the reaction has increased, making it too high for the reaction to occur.
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解题
At high temperatures (above the optimum), the active site of the enzyme changes shape permanently. This denaturation means the substrate can no longer fit into the active site, stopping the reaction.
评分标准
Award 1 mark for selecting C.
题目 11 · 選擇題
1 分
An atom of an isotope of element $Y$ is represented as $^{31}_{15}Y$. Which row shows the correct number of protons, neutrons and electrons in a $Y^{3-}$ ion?
A.Protons: 15, Neutrons: 16, Electrons: 15
B.Protons: 15, Neutrons: 16, Electrons: 18
C.Protons: 15, Neutrons: 31, Electrons: 18
D.Protons: 18, Neutrons: 16, Electrons: 15
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解题
The atomic number is 15, so there are 15 protons. The mass number is 31, so the number of neutrons is $31 - 15 = 16$. A neutral atom of $Y$ has 15 electrons. The $Y^{3-}$ ion has gained 3 electrons, giving it a total of $15 + 3 = 18$ electrons.
评分标准
Award 1 mark for selecting B.
题目 12 · 選擇題
1 分
Copper(II) oxide reacts with hydrogen gas when heated to form copper metal and steam: $$\text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O}$$ Which statement correctly describes the oxidation and reduction in this reaction?
A.Copper(II) oxide is oxidized because it loses oxygen, and hydrogen is reduced because it gains oxygen.
B.Copper(II) oxide is reduced because it loses oxygen, and hydrogen is oxidized because it gains oxygen.
C.Copper(II) oxide is reduced because it gains hydrogen, and hydrogen is oxidized because it loses hydrogen.
D.Copper(II) oxide is oxidized because it gains hydrogen, and hydrogen is reduced because it loses hydrogen.
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解题
Copper(II) oxide ($\text{CuO}$) loses oxygen to become copper ($\text{Cu}$), so it is reduced. Hydrogen gas ($\text{H}_2$) gains oxygen to become water ($\text{H}_2\text{O}$), so it is oxidized.
评分标准
Award 1 mark for selecting B.
题目 13 · 選擇題
1 分
A crane lifts a load of mass $400\text{ kg}$ vertically upwards through a height of $12\text{ m}$ in $15\text{ s}$. The gravitational field strength $g$ is $10\text{ N/kg}$. What is the average useful power developed by the crane?
Two wires, Wire 1 and Wire 2, are made of the same metal. Wire 1 has a length $L$, cross-sectional area $A$ and a resistance of $12\,\Omega$. Wire 2 has a length of $2L$ and a cross-sectional area of $3A$. What is the resistance of Wire 2?
A.2.0 \Omega
B.8.0 \Omega
C.18 \Omega
D.72 \Omega
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解题
Resistance $R$ is proportional to length and inversely proportional to area: $R = \rho \frac{\text{length}}{\text{area}}$. For Wire 1, $R_1 = \rho \frac{L}{A} = 12\,\Omega$. For Wire 2, $R_2 = \rho \frac{2L}{3A} = \frac{2}{3} R_1 = \frac{2}{3} \times 12\,\Omega = 8.0\,\Omega$.
评分标准
Award 1 mark for selecting B.
题目 15 · 選擇題
1 分
Which statement correctly compares sound waves and light waves?
A.Both sound waves and light waves are longitudinal waves.
B.Both sound waves and light waves can travel through a vacuum.
C.Sound waves are transverse waves, whereas light waves are longitudinal waves.
D.Sound waves require a material medium to travel, whereas light waves can travel through a vacuum.
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解题
Sound waves are longitudinal mechanical waves and require a medium to propagate. Light waves are transverse electromagnetic waves and can travel through a vacuum.
评分标准
Award 1 mark for selecting D.
题目 16 · 選擇題
1 分
Which statement correctly describes a difference between ethene and ethane?
A.Ethene is a saturated hydrocarbon, whereas ethane is unsaturated.
B.Ethene decolourises aqueous bromine rapidly, whereas ethane does not.
C.Ethene contains only single covalent bonds, whereas ethane contains a double covalent bond.
D.Ethene reacts with oxygen to form carbon dioxide and water, whereas ethane does not react with oxygen.
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解题
Ethene is an unsaturated alkene containing a double carbon-carbon bond, allowing it to rapidly react with and decolourise aqueous bromine. Ethane is a saturated alkane containing only single bonds, and does not react with bromine water under normal conditions.
评分标准
Award 1 mark for selecting B.
题目 17 · multiple_choice
1 分
Which products are formed during anaerobic respiration in yeast cells?
A.carbon dioxide and water
B.carbon dioxide and ethanol
C.lactic acid only
D.lactic acid and carbon dioxide
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解题
During anaerobic respiration in yeast (also known as fermentation), glucose is broken down in the absence of oxygen to produce ethanol and carbon dioxide.
评分标准
1 mark for selecting the correct option B.
题目 18 · multiple_choice
1 分
A resistor of 12 ohms is connected in parallel with a resistor of 4 ohms. What is the combined resistance of this parallel combination?
A.0.3 ohms
B.3.0 ohms
C.8.0 ohms
D.16 ohms
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解题
For parallel resistors, the combined resistance R is calculated using the formula: 1/R = 1/R1 + 1/R2. Here, 1/R = 1/12 + 1/4 = 1/12 + 3/12 = 4/12 = 1/3. Therefore, R = 3 ohms.
评分标准
1 mark for selecting the correct option B.
题目 19 · multiple_choice
1 分
What is the relative molecular mass, Mr, of calcium hydroxide, Ca(OH)2? [Relative atomic masses, Ar: Ca = 40, O = 16, H = 1]
A.57
B.58
C.74
D.114
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解题
The relative molecular mass is calculated as: Mr = 40 + 2 * (16 + 1) = 40 + 34 = 74.
评分标准
1 mark for selecting the correct option C.
题目 20 · multiple_choice
1 分
A sound wave has a frequency of 250 Hz and travels at a speed of 340 m/s. What is the wavelength of this sound wave?
A.0.74 m
B.1.36 m
C.85.0 m
D.85000 m
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解题
Using the wave equation v = f * lambda, we rearrange to find the wavelength: lambda = v / f = 340 / 250 = 1.36 m.
评分标准
1 mark for selecting the correct option B.
题目 21 · multiple_choice
1 分
Which statement correctly describes a difference between ethane and ethene?
A.Ethane decolourises bromine water, but ethene does not.
B.Ethane is an unsaturated hydrocarbon, whereas ethene is saturated.
C.Ethane molecules contain a carbon-to-carbon double bond, but ethene molecules only contain single bonds.
D.Ethene can undergo addition polymerisation, but ethane cannot.
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解题
Ethene is an alkene containing a carbon-to-carbon double bond, which allows it to undergo addition polymerisation. Ethane is an alkane (saturated) and cannot undergo addition polymerisation.
评分标准
1 mark for selecting the correct option D.
题目 22 · multiple_choice
1 分
Four metals, W, X, Y and Z, are tested. Only W and X react with dilute hydrochloric acid. Only W can be extracted from its oxide by heating with carbon. Metal Y reacts with an aqueous solution of Z ions to form metal Z. Which list shows the metals in order of decreasing reactivity (most reactive first)?
A.X, W, Y, Z
B.W, X, Y, Z
C.X, W, Z, Y
D.Y, Z, X, W
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解题
Since only W and X react with acid, they are more reactive than Y and Z. Since only W is extracted by carbon, X is more reactive than carbon and cannot be reduced by carbon, making X more reactive than W (X > W). Since Y reacts with Z ions to displace Z, Y is more reactive than Z (Y > Z). Thus, the order of decreasing reactivity is X, W, Y, Z.
评分标准
1 mark for selecting the correct option A.
题目 23 · multiple_choice
1 分
What happens to an enzyme when it is denatured by high temperature?
A.The kinetic energy of the enzyme molecules decreases to zero.
B.The active site changes shape so the substrate can no longer fit.
C.The enzyme is completely broken down into individual amino acids.
D.The activation energy required for the reaction is significantly decreased.
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解题
Denaturation is a process where high temperatures destroy the three-dimensional shape of the enzyme's active site, meaning the substrate is no longer complementary and cannot bind.
评分标准
1 mark for selecting the correct option B.
题目 24 · multiple_choice
1 分
Which feature of the alveoli does NOT increase the rate of gas exchange in the human lungs?
A.a thick layer of mucus lining the outer surface of the lungs
B.a very large total surface area
C.walls that are only one cell thick
D.an excellent blood supply from a network of capillaries
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解题
A thick layer of mucus would increase the diffusion distance for gases, which would decrease (not increase) the rate of gas exchange.
评分标准
1 mark for selecting the correct option A.
题目 25 · 選擇題
1 分
A plant cell is placed in a solution with a lower water potential than the cell sap. Which row correctly identifies the net movement of water molecules and the resulting state of the cell?
A.Net movement of water: out of the cell | Resulting state of the cell: plasmolysed
B.Net movement of water: out of the cell | Resulting state of the cell: turgid
C.Net movement of water: into the cell | Resulting state of the cell: plasmolysed
D.Net movement of water: into the cell | Resulting state of the cell: turgid
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解题
When a plant cell is placed in a solution of lower water potential (hypertonic), water moves out of the cell by osmosis down a water potential gradient. This loss of water causes the vacuole and cytoplasm to shrink, pulling the cell membrane away from the cell wall, resulting in a plasmolysed state.
评分标准
1 mark for the correct option (A).
题目 26 · 選擇題
1 分
An atom of an isotope of phosphorus is represented as \({}^{31}_{15}\text{P}\). Which row correctly identifies the number of protons, neutrons and electrons in the phosphate ion, \(\text{P}^{3-}\)?
A.protons: 15 | neutrons: 16 | electrons: 15
B.protons: 15 | neutrons: 16 | electrons: 18
C.protons: 15 | neutrons: 31 | electrons: 18
D.protons: 18 | neutrons: 16 | electrons: 15
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解题
The atomic number is 15, which represents the number of protons. The mass number is 31, so the number of neutrons is \(31 - 15 = 16\). The \(\text{P}^{3-}\) ion has gained 3 electrons, meaning it has \(15 + 3 = 18\) electrons.
评分标准
1 mark for the correct option (B).
题目 27 · 選擇題
1 分
A runner accelerates from rest at a constant rate of \(1.5\text{ m/s}^2\) for \(6.0\text{ s}\), and then runs at a constant speed for another \(10.0\text{ s}\). What is the total distance covered by the runner during the \(16.0\text{ s}\)?
A.\(90.0\text{ m}\)
B.\(108.0\text{ m}\)
C.\(117.0\text{ m}\)
D.\(144.0\text{ m}\)
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解题
First phase: constant acceleration from rest (\(u = 0\)). Distance \(d_1 = \frac{1}{2} a t_1^2 = 0.5 \times 1.5 \times 6.0^2 = 27.0\text{ m}\). Final speed \(v = a t_1 = 1.5 \times 6.0 = 9.0\text{ m/s}\). Second phase: constant speed \(v = 9.0\text{ m/s}\) for \(10.0\text{ s}\). Distance \(d_2 = v \times t_2 = 9.0 \times 10.0 = 90.0\text{ m}\). Total distance \(d = d_1 + d_2 = 27.0 + 90.0 = 117.0\text{ m}\).
评分标准
1 mark for the correct option (C).
题目 28 · 選擇題
1 分
Which row correctly compares the concentrations of carbon dioxide, oxygen and water vapour in expired air compared to inspired air?
A.carbon dioxide: higher | oxygen: lower | water vapour: higher
B.carbon dioxide: higher | oxygen: lower | water vapour: lower
C.carbon dioxide: lower | oxygen: higher | water vapour: higher
D.carbon dioxide: lower | oxygen: lower | water vapour: lower
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解题
During aerobic respiration, cells consume oxygen and produce carbon dioxide and water. Consequently, expired air contains a higher concentration of carbon dioxide, a lower concentration of oxygen, and is saturated with water vapour (higher concentration) compared to inspired air.
评分标准
1 mark for the correct option (A).
题目 29 · 選擇題
1 分
Which statement about alkanes and alkenes is correct?
A.Alkanes decolourise aqueous bromine rapidly at room temperature.
B.Alkenes are saturated hydrocarbons containing only single covalent bonds.
C.Alkanes burn in excess oxygen to produce carbon dioxide and water.
D.Alkenes can be formed from alkanes by the process of addition polymerisation.
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解题
Alkanes undergo complete combustion in excess oxygen to yield carbon dioxide and water. Option A is incorrect because alkenes, not alkanes, decolourise bromine water rapidly. Option B is incorrect because alkenes are unsaturated. Option D is incorrect because alkenes are formed by cracking larger alkanes.
评分标准
1 mark for the correct option (C).
题目 30 · 選擇題
1 分
A student connects two \(6.0\text{ }\Omega\) resistors in parallel. This combination is then connected in series with a \(3.0\text{ }\Omega\) resistor and a \(9.0\text{ V}\) power supply. What is the current drawn from the power supply?
A.\(0.60\text{ A}\)
B.\(1.0\text{ A}\)
C.\(1.5\text{ A}\)
D.\(3.0\text{ A}\)
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解题
First, calculate the equivalent resistance of the parallel combination: \(R_p = \frac{6.0 \times 6.0}{6.0 + 6.0} = 3.0\text{ }\Omega\). Since this combination is in series with the \(3.0\text{ }\Omega\) resistor, the total resistance is: \(R_{total} = R_p + 3.0 = 3.0 + 3.0 = 6.0\text{ }\Omega\). The current drawn is: \(I = \frac{V}{R_{total}} = \frac{9.0\text{ V}}{6.0\text{ }\Omega} = 1.5\text{ A}\).
评分标准
1 mark for the correct option (C).
题目 31 · 選擇題
1 分
The rate of an enzyme-catalysed reaction was measured at different temperatures. It was found that the rate increased up to \(40\text{ }^\circ\text{C}\), but rapidly decreased to zero at \(60\text{ }^\circ\text{C}\). Which statement explains the decrease in the reaction rate above \(40\text{ }^\circ\text{C}\)?
A.The kinetic energy of the substrate molecules decreases.
B.The enzyme molecules are denatured, changing the shape of their active sites.
C.The activation energy of the reaction is lowered.
D.The pH of the solution changes, inactivating the enzymes.
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解题
Above the optimum temperature of \(40\text{ }^\circ\text{C}\), the thermal energy breaks the bonds maintaining the tertiary structure of the enzyme. This denatures the enzyme molecules, changing the shape of their active sites so that the substrate can no longer fit.
评分标准
1 mark for the correct option (B).
题目 32 · 選擇題
1 分
In the extraction of iron from hematite in a blast furnace, the following reaction occurs:
A.Carbon monoxide is reduced because it loses oxygen.
B.Iron(III) oxide is oxidised because it gains carbon.
C.Carbon monoxide acts as the reducing agent because it removes oxygen from iron(III) oxide.
D.Iron(III) oxide acts as the reducing agent because it oxidises carbon monoxide.
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解题
Carbon monoxide (\(\text{CO}\)) acts as the reducing agent because it removes oxygen from the iron(III) oxide (\(\text{Fe}_2\text{O}_3\)), reducing it to metallic iron (\(\text{Fe}\)). Carbon monoxide itself is oxidised to carbon dioxide (\(\text{CO}_2\)).
评分标准
1 mark for the correct option (C).
题目 33 · multiple_choice
1 分
Which statement about diffusion is correct?
A.It is the net movement of particles from a region of their lower concentration to a region of their higher concentration.
B.It is the net movement of particles down a concentration gradient as a result of their random movement.
C.It only occurs across a partially permeable membrane.
D.It requires energy from respiration to move particles.
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解题
Diffusion is defined as the net movement of particles from a region of their higher concentration to a region of their lower concentration (down a concentration gradient) as a result of their random movement. It does not require a membrane (unlike osmosis) and is a passive process that does not require energy from respiration (unlike active transport).
评分标准
1 mark for the correct option B.
题目 34 · multiple_choice
1 分
The rate of an enzyme-controlled reaction is measured at different temperatures. Which statement explains why the rate of reaction decreases rapidly at temperatures above the optimum temperature?
A.The kinetic energy of the substrate molecules decreases.
B.The enzyme molecules are denatured and their active site changes shape.
C.The substrate molecules are broken down by the heat.
D.The frequency of collisions between enzymes and substrates increases.
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解题
At high temperatures above the optimum, the high thermal energy causes the chemical bonds holding the enzyme's 3D structure together to break. This denatures the enzyme, altering the shape of its active site so that the substrate can no longer fit, rapidly decreasing the rate of reaction.
评分标准
1 mark for the correct option B.
题目 35 · multiple_choice
1 分
Which statement describes a difference between ethene and ethane?
A.Ethene is a saturated hydrocarbon, while ethane is unsaturated.
B.Ethene rapidly decolourises aqueous bromine, while ethane does not.
C.Ethene has only single covalent bonds, while ethane has a double covalent bond.
D.Ethene is an alkane, while ethane is an alkene.
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解题
Ethene is an alkene (unsaturated, contains a C=C double bond) and reacts rapidly with aqueous bromine in an addition reaction, turning the orange/brown solution colourless. Ethane is an alkane (saturated, contains only C-C single bonds) and does not react with aqueous bromine under normal conditions.
评分标准
1 mark for the correct option B.
题目 36 · multiple_choice
1 分
Which row describes the trends in reactivity and color intensity of the Group VII elements as the group is descended from chlorine to iodine?
A.reactivity increases, color intensity increases
B.reactivity increases, color intensity decreases
C.reactivity decreases, color intensity increases
D.reactivity decreases, color intensity decreases
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解题
As Group VII is descended, the reactivity of the halogens decreases (chlorine is more reactive than bromine, which is more reactive than iodine). The color intensity increases (chlorine is a pale green gas, bromine is a red-brown liquid, and iodine is a dark grey/purple solid).
评分标准
1 mark for the correct option C.
题目 37 · multiple_choice
1 分
A block of mass 4.0 kg is pushed along a horizontal, frictionless surface by a constant force of 12 N. What is the acceleration of the block?
A.0.33 m/s²
B.3.0 m/s²
C.8.0 m/s²
D.48 m/s²
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解题
Using Newton's second law: F = ma. Rearranging for acceleration: a = F / m = 12 N / 4.0 kg = 3.0 m/s^2.
评分标准
1 mark for the correct option B.
题目 38 · multiple_choice
1 分
Which type of electromagnetic radiation has the longest wavelength and which has the highest frequency?
A.longest wavelength: radio waves; highest frequency: gamma rays
B.longest wavelength: gamma rays; highest frequency: radio waves
In the electromagnetic spectrum, radio waves have the lowest frequency and the longest wavelength. Gamma rays have the highest frequency and the shortest wavelength.
评分标准
1 mark for the correct option A.
题目 39 · multiple_choice
1 分
Which structure is found in plant cells but not in animal cells?
A.cell membrane
B.cell wall
C.cytoplasm
D.nucleus
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解题
Plant cells have a cellulose cell wall, chloroplasts, and a large permanent vacuole, which are absent in animal cells. Both cell types contain a cell membrane, cytoplasm, and a nucleus.
评分标准
1 mark for the correct option B.
题目 40 · multiple_choice
1 分
In which reaction is the first reactant reduced?
A.C + O2 -> CO2
B.Fe2O3 + 3CO -> 2Fe + 3CO2
C.2Mg + O2 -> 2MgO
D.CH4 + 2O2 -> CO2 + 2H2O
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解题
Reduction is the loss of oxygen. In option B, iron(III) oxide (Fe2O3) loses oxygen to form iron (Fe). Therefore, Fe2O3 is reduced. In all other options, the first reactant gains oxygen, which is oxidation.
Answer all questions in the spaces provided. Show all working and state appropriate units in calculations.
9 题目 · 81 分
题目 1 · theory
9 分
1 (a) State the word equation for anaerobic respiration in yeast. [1]
(b) Describe two differences between aerobic respiration and anaerobic respiration in humans. [2]
(c) An athlete runs a 400 m race. During the race, anaerobic respiration occurs in their muscles. (i) Explain why anaerobic respiration occurs in muscle cells during vigorous exercise. [2] (ii) Name the substance that builds up in muscles as a result of anaerobic respiration and describe its effect on muscle contraction. [2] (iii) After the race, the athlete's breathing rate remains high. State the name of this phenomenon and explain why it occurs. [2]
(b) Any two from: - Aerobic respiration requires oxygen, anaerobic respiration does not. - Aerobic respiration produces carbon dioxide and water, anaerobic respiration in humans produces lactic acid (no carbon dioxide). - Aerobic respiration releases a much larger amount of energy per glucose molecule compared to anaerobic respiration.
(c) (i) During vigorous exercise, the oxygen demand of the muscles exceeds the rate at which oxygen can be supplied by the blood. Therefore, muscle cells respire anaerobically to release extra energy.
(ii) Lactic acid. Its buildup causes muscle fatigue / muscle cramp / painful contractions.
(iii) Oxygen debt. The elevated breathing rate is needed to supply oxygen to break down / oxidise the accumulated lactic acid in the liver / muscles into carbon dioxide and water.
评分标准
(a) 1 mark for correct word equation: glucose \(\rightarrow\) ethanol + carbon dioxide (accept alcohol instead of ethanol). Reject if oxygen is included on the left.
(b) 1 mark for each correct difference (max 2): - Aerobic uses oxygen, anaerobic does not. - Aerobic produces \(CO_2\) and water, anaerobic produces lactic acid (in humans). - Aerobic releases more energy / anaerobic releases less energy.
(c) (i) 1 mark for identifying that oxygen supply is insufficient / not fast enough. 1 mark for stating that anaerobic respiration provides a rapid supply of energy without oxygen.
(ii) 1 mark for naming lactic acid. 1 mark for stating that it causes fatigue / cramps / pain / prevents efficient contraction.
(iii) 1 mark for naming oxygen debt. 1 mark for explaining that extra oxygen is required to break down / oxidise / remove lactic acid.
题目 2 · theory
9 分
2 (a) A student investigates the reaction between marble chips (calcium carbonate, \(CaCO_3\)) and dilute hydrochloric acid, \(HCl\). Write the balanced chemical equation, including state symbols, for this reaction. [2]
(b) The student measures the volume of carbon dioxide gas produced over time. Explain, using collision theory, why using powdered calcium carbonate instead of large marble chips increases the rate of reaction. [3]
(c) The reaction is repeated at a higher temperature. (i) State the effect of a higher temperature on the rate of reaction. [1] (ii) Explain this effect in terms of the kinetic energy of particles and successful collisions. [3]
(b) Powdered calcium carbonate has a much larger surface area than large marble chips. A larger surface area means more particles of calcium carbonate are exposed to the acid. This increases the frequency of collisions between reactant particles, resulting in a higher rate of reaction.
(c) (i) The rate of reaction increases. (ii) At a higher temperature, particles gain kinetic energy and move faster. This increases the frequency of collisions. Furthermore, a greater proportion of particles have energy equal to or greater than the activation energy, so a higher percentage of collisions are successful.
评分标准
(a) 1 mark for correct chemical formulae of reactants and products: \(CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2\). 1 mark for correct balancing and all state symbols: (s), (aq), (aq), (l), (g).
(b) 1 mark for identifying that powder has a larger surface area. 1 mark for stating that more reactant particles are exposed / in contact. 1 mark for linking this to an increased frequency of collisions (collisions per unit time).
(c) (i) 1 mark for stating that the rate increases. (ii) 1 mark for stating that particles gain kinetic energy / move faster. 1 mark for stating that collisions are more frequent. 1 mark for explaining that more particles have energy greater than or equal to the activation energy / more collisions are successful.
题目 3 · theory
9 分
3 A toy car of mass 0.50 kg travels along a straight, flat path. It accelerates from rest to a speed of 6.0 m/s in 4.0 s. It then travels at a constant speed of 6.0 m/s for another 5.0 s, before decelerating uniformly to a stop in 3.0 s.
(a) (i) Calculate the acceleration of the car during the first 4.0 s. Show your working. [2] (ii) Calculate the total distance travelled by the car during the entire 12.0 s journey. [3]
(b) (i) Calculate the kinetic energy of the car when it is travelling at its maximum speed. State the unit. [2] (ii) During the 5.0 s phase of constant speed, the engine of the car exerts a constant forward driving force of 1.2 N. Calculate the work done by the engine during this phase. [2]
(a) (i) 1 mark for correct formula or substitution: \(a = \frac{6.0}{4.0}\). 1 mark for correct value and unit: \(1.5\text{ m/s}^2\) (or \(\text{m s}^{-2}\)).
(ii) 1 mark for calculating distance of any individual phase correctly (e.g., 12 m, 30 m, or 9 m). 1 mark for showing the method of adding the three areas together (e.g., \(12 + 30 + 9\)). 1 mark for the correct final answer: \(51\text{ m}\) (or \(51.0\text{ m}\)).
(b) (i) 1 mark for correct substitution: \(\frac{1}{2} \times 0.50 \times 6.0^2\). 1 mark for correct answer with unit: \(9.0\text{ J}\) (accept Joules / J).
(ii) 1 mark for identifying the correct distance (30 m) and formula: \(W = F \times d\). 1 mark for correct final calculation: \(36\text{ J}\) (or \(36\text{ N m}\)).
题目 4 · theory
9 分
4 Some plants reproduce sexually using flowers.
(a) Compare the features of wind-pollinated and insect-pollinated flowers by completing Table 4.1. [3]
(c) Describe the growth of the pollen tube and explain how fertilisation is achieved in a flowering plant after pollination has occurred. [4]
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解题
(a) - (i) Small, dull green / brown / inconspicuous (no bright petals) - (ii) Sticky, relatively small and enclosed inside the flower - (iii) Large quantity, light and smooth
(b) Self-pollination is the transfer of pollen grains from the anther of a flower to the stigma of the same flower, or a different flower on the same plant.
(c) After pollination, the pollen grain germinates on the stigma. A pollen tube grows down through the style into the ovary, entering the ovule through a tiny opening called the micropyle. The male gamete nucleus travels down the pollen tube and fuses with the female gamete nucleus inside the ovule to form a zygote.
评分标准
(a) 1 mark for each correct entry in Table 4.1: - (i) Small / green / dull / inconspicuous / absent. - (ii) Sticky / enclosed / inside the petals / small. - (iii) Light / smooth / produced in large quantities.
(b) 1 mark for transfer of pollen from anther to stigma. 1 mark for stating it occurs within the same flower or another flower on the same plant.
(c) 1 mark for stating the pollen tube grows down the style. 1 mark for stating it enters the ovary / ovule (through the micropyle). 1 mark for stating the male nucleus / pollen nucleus travels down the tube. 1 mark for stating the fusion of male and female nuclei to form a zygote.
题目 5 · theory
9 分
5 Alkanes and alkenes are two families of hydrocarbons.
(a) (i) Draw the displayed chemical structure, showing all covalent bonds, for a molecule of propane and a molecule of propene. [2] (ii) State which of these molecules is saturated, and explain the meaning of the term saturated. [2]
(b) Alkenes can be produced from long-chain alkanes by the process of cracking. (i) State two reaction conditions required for industrial cracking. [2] (ii) Describe a chemical test to distinguish between propane and propene. Include the reagent used and the expected observations for both compounds. [3]
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解题
(a) (i) Displayed structures: Propane (\(C_3H_8\)): H H H | | | H-C - C - C-H | | | H H H
Propene (\(C_3H_6\)): H H H | | | C = C - C-H | | H H
(ii) Propane is saturated. Saturated means the molecule contains only single carbon-carbon bonds (no carbon-carbon double or triple bonds).
(b) (i) High temperature (approx 600-700 °C) and a catalyst (such as alumina / silica / zeolite).
(ii) Add aqueous bromine (bromine water). - Propane: remains orange / brown (no reaction). - Propene: decolourises / turns from orange-brown to colourless.
评分标准
(a) (i) 1 mark for correct displayed structure of propane (three carbons, single bonds, eight hydrogens). 1 mark for correct displayed structure of propene (three carbons, one double bond, six hydrogens).
(ii) 1 mark for identifying propane as saturated. 1 mark for defining saturated as containing only single carbon-carbon (\(C-C\)) bonds.
(b) (i) 1 mark for stating high temperature (accept range 450 - 800 °C). 1 mark for stating catalyst (accept silicon dioxide / aluminium oxide / clay).
(ii) 1 mark for naming bromine water / aqueous bromine. 1 mark for stating propane remains orange / brown / no change. 1 mark for stating propene turns from orange/brown to colourless (reject 'clear').
题目 6 · theory
9 分
6 A 12.0 V battery is connected in a circuit with two resistors in parallel. The resistors have resistances of 4.0 \(\Omega\) and 6.0 \(\Omega\).
(a) (i) Calculate the combined resistance of this parallel combination. Show your working. [2] (ii) Calculate the total current flowing from the battery into the parallel circuit. [2] (iii) State the potential difference across each resistor, and calculate the current flowing through the 6.0 \(\Omega\) resistor. [2]
(b) (i) Define the term electric current. [1] (ii) Calculate the total electrical energy supplied by the battery if the circuit remains connected for 10 minutes. Show your working and state the unit. [2]
(ii) \(I = \frac{V}{R_p} = \frac{12.0}{2.4} = 5.0\text{ A}\).
(iii) The potential difference across each resistor in parallel is equal to the battery voltage = 12.0 V. Current through the 6.0 \(\Omega\) resistor: \(I = \frac{V}{R} = \frac{12.0}{6.0} = 2.0\text{ A}\).
(b) (i) Electric current is the rate of flow of charge (or flow of charge per unit time).
(ii) \(t = 10\text{ minutes} = 600\text{ s}\). Energy \(E = V I t = 12.0\text{ V} \times 5.0\text{ A} \times 600\text{ s} = 36000\text{ J}\) (or 36 kJ).
评分标准
(a) (i) 1 mark for correct formula and substitution: \(\frac{1}{4} + \frac{1}{6}\). 1 mark for correct combined resistance: \(2.4\ \Omega\).
(ii) 1 mark for correct formula: \(I = \frac{V}{R}\) or substitution. 1 mark for correct current: \(5.0\text{ A}\). (Allow ecf from (a)(i)).
(iii) 1 mark for stating potential difference is 12.0 V. 1 mark for calculating current: \(2.0\text{ A}\).
(b) (i) 1 mark for correct definition: rate of flow of charge / flow of charge per second.
(ii) 1 mark for converting time to seconds (600 s) and using \(E = VIt\) or \(E = P \times t\). 1 mark for correct value and unit: \(36000\text{ J}\) or \(36\text{ kJ}\) (accept Joules / J / kilojoules / kJ).
题目 7 · theory
9 分
7 The lungs are the primary organs of the human gas exchange system.
(a) (i) State the function of the cartilage rings in the trachea. [1] (ii) Describe the changes that occur in the diaphragm and the intercostal muscles during inspiration (breathing in), and explain how these changes cause air to enter the lungs. [4]
(b) Tobacco smoke contains harmful substances, including carbon monoxide and tar. (i) Explain the harmful effect of carbon monoxide on the oxygen-carrying capacity of the blood. [2] (ii) Describe how tar affects the cilia lining the airways, and explain why this causes a smoker's cough. [2]
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解题
(a) (i) Cartilage rings keep the trachea open / prevent it from collapsing during breathing when the pressure drops.
(ii) During inspiration: - The external intercostal muscles contract (and the internal intercostal muscles relax), pulling the ribcage upwards and outwards. - The diaphragm contracts and flattens. - This increases the volume of the thorax. - The increase in volume causes a decrease in pressure inside the thorax below atmospheric pressure. - Air flows down the pressure gradient into the lungs.
(b) (i) Carbon monoxide binds irreversibly with haemoglobin in red blood cells, forming carboxyhaemoglobin. This reduces the amount of haemoglobin available to bind and transport oxygen, thereby reducing the oxygen-carrying capacity of the blood.
(ii) Tar paralyses and destroys the cilia lining the trachea and bronchi. Without functioning cilia, mucus cannot be swept upwards towards the throat. The buildup of mucus blocks the airways, stimulating coughing to clear the lungs.
评分标准
(a) (i) 1 mark for stating that it keeps the airway open / prevents collapse.
(ii) 1 mark for stating diaphragm contracts and flattens. 1 mark for stating intercostal muscles contract, moving the ribcage up and out. 1 mark for explaining this increases the volume and decreases the pressure in the thorax. 1 mark for explaining that air enters because pressure in the lungs is lower than atmospheric pressure.
(b) (i) 1 mark for stating carbon monoxide binds to haemoglobin (to form carboxyhaemoglobin). 1 mark for explaining that this reduces the capacity to transport oxygen.
(ii) 1 mark for stating tar paralyses / damages cilia, preventing them from sweeping mucus away. 1 mark for explaining that mucus builds up / traps bacteria and dust, stimulating coughing to remove it.
题目 8 · theory
9 分
8 Electromagnetic waves form a continuous spectrum of transverse waves.
(a) (i) State the speed of all electromagnetic waves in a vacuum. [1] (ii) An ultraviolet wave has a wavelength of \(2.5 \times 10^{-7}\) m. Calculate the frequency of this wave. Show your working. [2] (iii) State one common use of ultraviolet radiation and describe one danger of exposure to it. [2]
(b) A ray of light in air strikes a rectangular glass block at an angle. (i) Explain, in terms of speed, why the light bends towards the normal as it enters the glass block. [2] (ii) State what happens, if anything, to the frequency, the wavelength, and the speed of the light wave when it enters the glass block from the air. [2]
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解题
(a) (i) \(3.0 \times 10^8\text{ m/s}\).
(ii) \(v = f \lambda \Rightarrow f = \frac{v}{\lambda} = \frac{3.0 \times 10^8}{2.5 \times 10^{-7}} = 1.2 \times 10^{15}\text{ Hz}\).
(b) (i) Light travels slower in glass than in air. When the wave front meets the glass boundary at an angle, the side of the wave entering first slows down first, causing the direction of the ray to change / bend towards the normal.
(a) (i) 1 mark for \(3.0 \times 10^8\text{ m/s}\) (accept \(3 \times 10^8\)).
(ii) 1 mark for correct formula or substitution: \(f = \frac{3.0 \times 10^8}{2.5 \times 10^{-7}}\). 1 mark for correct calculation: \(1.2 \times 10^{15}\text{ Hz}\) (accept Hertz).
(iii) 1 mark for a correct use. 1 mark for a correct danger.
(b) (i) 1 mark for stating that light travels slower in glass than in air. 1 mark for explaining that different parts of the wavefront slow down at different times, causing it to bend.
(ii) 1 mark for stating frequency is constant. 1 mark for stating both wavelength and speed decrease.
题目 9 · structured
9 分
A student sets up an electric toy train set. The locomotive is powered by a small electric motor connected to a d.c. power supply.
(a) When a potential difference of \(12\text{ V}\) is applied across the motor, the current in the motor is \(1.5\text{ A}\).
(i) Calculate the electrical power input to the motor. State the unit of your answer.
power = ................................... unit .............. [3]
(ii) The motor has an efficiency of \(80\%\). Calculate the useful mechanical power output of the motor.
useful power output = ................................... \text{W} [2]
(b) The train has two identical headlights connected in parallel across a part of the circuit. When both headlights are switched on, the total current entering the parallel combination of the lamps is \(0.60\text{ A}\).
(i) State the current flowing through each lamp and explain your answer.
(a)(i) Use the formula for electrical power: \[P = I \times V\] Substitute the given values: \[P = 1.5\text{ A} \times 12\text{ V} = 18\text{ W}\] The unit is watts (W).
(a)(ii) Efficiency is the ratio of useful power output to total power input: \[\text{Efficiency} = \frac{\text{Useful power output}}{\text{Total power input}} \times 100\%\] \[\text{Useful power output} = 0.80 \times 18\text{ W} = 14.4\text{ W}\]
(b)(i) In a parallel circuit containing two identical components, the current divides equally between them. \[\text{Current per lamp} = \frac{0.60\text{ A}}{2} = 0.30\text{ A}\]
(b)(ii) An advantage of parallel circuits is that if one lamp breaks, the other remains lit because each lamp is on its own separate branch connected across the power source, maintaining a complete circuit.
评分标准
(a)(i) - 1 mark for correct formula or substitution: \(P = IV\) or \(1.5 \times 12\) - 1 mark for correct calculation: \(18\) - 1 mark for correct unit: \(\text{W}\) or \(\text{watts}\) (accept \(\text{J/s}\))
(a)(ii) - 1 mark for correct substitution using their power value from (a)(i): \(18 \times 0.80\) or \(\frac{80}{100} \times 18\) - 1 mark for correct calculation: \(14.4\) (allow \(14\) or ecf from (a)(i))
(b)(i) - 1 mark for correct current: \(0.30\text{ A}\) (accept \(0.3\)) - 1 mark for explanation: since the lamps are identical, the current splits / divides equally between the branches
(b)(ii) - 1 mark for stating an advantage: if one lamp blows/breaks/fails, the other lamp stays on / remains lit (or lamps can be controlled/switched independently) - 1 mark for explanation: because there is still a complete circuit / alternative path for the current through the working lamp
Paper 6 Alternative to Practical
Answer all questions in the spaces provided. Include apparatus, methods, variables, and data processing in the 7-mark planning question.
4 题目 · 40 分
题目 1 · Practical Investigation
10 分
A student investigates the effect of pH on the activity of the enzyme amylase. Amylase breaks down starch into maltose. In the investigation, starch solution is mixed with amylase in buffer solutions of different pH values. Every 30 seconds, a sample of the mixture is added to a drop of iodine solution on a spotting tile. The student records the time taken for the starch to be completely digested (when the iodine solution no longer changes colour to blue-black). (a) State the colour of iodine solution when starch is present, and when starch is completely digested. - starch present: [1] - starch digested: [1] (b) Table 1.1 shows the results for five different pH values. Table 1.1: [pH 4.0 | Time: 240 s | Rate: 0.0042 s^-1] [pH 5.0 | Time: 120 s | Rate: 0.0083 s^-1] [pH 6.0 | Time: 60 s | Rate: 0.0167 s^-1] [pH 7.0 | Time: ? | Rate: ?] [pH 8.0 | Time: 180 s | Rate: 0.0056 s^-1]. The stopwatch reading for pH 7.0 at the end of digestion is 1 minute and 30 seconds. (i) State the time taken, in seconds, for pH 7.0. time = [1] (ii) Calculate the rate of reaction for pH 7.0 using the formula: Rate = 1 / Time taken. rate of reaction = [1] (c) On a grid, describe how you would plot a graph of the rate of reaction against pH. State what goes on each axis and the expected shape of the curve. [4] (d) With reference to the experiment, describe the effect of pH on the activity of amylase and identify the optimum pH. [2]
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解题
(a) Iodine turns blue-black in the presence of starch. When starch is fully digested, iodine remains its original yellow-brown or orange colour. (b)(i) 1 minute and 30 seconds is equal to 90 seconds. (b)(ii) Rate = 1 / 90 = 0.0111 s^-1. (c) On a graph of rate against pH, the independent variable (pH) is plotted on the horizontal x-axis, and the dependent variable (rate of reaction) is on the vertical y-axis. The points plotted show a peak at pH 6.0, indicating a curve that rises then falls. (d) Amylase works fastest at pH 6.0, which is its optimum pH. Below and above this pH, the rate of starch digestion decreases.
评分标准
Part (a): [2 marks] - blue-black for starch present [1], yellow/orange/brown for starch digested [1]. Part (b): [2 marks] - (i) 90 [1], (ii) 0.011 or 0.0111 [1]. Part (c): [4 marks] - pH on x-axis [1], Rate on y-axis with unit s^-1 [1], points plotted would show a curve rising and then falling [1], peak at pH 6.0 [1]. Part (d): [2 marks] - activity increases up to pH 6.0 and decreases above pH 6.0 [1], optimum is pH 6.0 [1].
题目 2 · Practical Investigation
10 分
A student investigates the rate of reaction between magnesium ribbon and dilute hydrochloric acid. The equation for the reaction is: Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g). The student measures the volume of hydrogen gas produced over time using a gas syringe. (a) Describe the experimental set-up and list the main pieces of apparatus required to perform this reaction and collect the gas. [3] (b) Table 2.1 shows the volume of gas collected at 10-second intervals. Table 2.1: [0 s | 0 cm^3], [10 s | 18 cm^3], [20 s | 32 cm^3], [30 s | 41 cm^3], [40 s | 47 cm^3], [50 s | 50 cm^3], [60 s | 50 cm^3]. (i) State the time at which the reaction was complete. [1] (ii) Explain how the results in Table 2.1 show that the rate of reaction decreases as time increases. [2] (c) Describe a chemical test to confirm that the gas collected in the syringe is hydrogen. State the test and the positive observation. [2] (d) The student repeats the experiment using the same mass of magnesium but in powdered form. State and explain the effect of this change on the rate of reaction using particle collision theory. [2]
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解题
(a) The apparatus consists of a conical flask containing the reactants, fitted with a stopper and delivery tube that leads directly to a gas syringe to measure the gas volume. (b)(i) The reaction is complete when the volume of gas stops increasing, which occurs at 50 s (constant volume of 50 cm^3). (b)(ii) In the first 10 seconds, 18 cm^3 of gas is produced. Between 10 and 20 seconds, only 14 cm^3 is produced. The declining volume produced per unit time shows that the rate of reaction is decreasing. (c) Hydrogen gas is tested using a lighted splint, which burns with a characteristic squeaky pop sound. (d) Powdered magnesium has a much larger surface area than ribbon. This exposes more magnesium particles to the acid, increasing the frequency of collisions and thus the rate of reaction.
评分标准
Part (a): [3 marks] - mention of conical flask/reaction vessel [1], delivery tube [1], gas syringe (or measuring cylinder inverted over water) [1]. Part (b): [3 marks] - (i) 50 s [1], (ii) less gas is produced in each subsequent 10 s interval [1], support using specific values from the table [1]. Part (c): [2 marks] - lighted splint [1], squeaky pop observation [1]. Part (d): [2 marks] - rate of reaction increases [1], due to larger surface area causing more frequent collisions [1].
题目 3 · Practical Investigation
10 分
A student investigates the resistance of different lengths of a constantan wire. The student sets up a circuit containing a cell, an ammeter, a switch, and a length of wire connected using crocodile clips. A voltmeter is connected in parallel with the test wire. (a) Describe the circuit connections needed, specifying which components are connected in series and which are in parallel. [3] (b) The potential difference across the wire is kept constant at 1.5 V. Table 3.1 shows the current readings for different lengths of wire. Table 3.1: [20.0 cm | 0.60 A | 2.50 Ohm], [40.0 cm | 0.30 A | 5.00 Ohm], [60.0 cm | 0.20 A | 7.50 Ohm], [80.0 cm | ? | ?], [100.0 cm | 0.12 A | 12.50 Ohm]. The ammeter reading for the wire length of 80.0 cm is 0.15 A. (i) Record the current for 80.0 cm. [1] (ii) Use the equation: Resistance = Potential difference / Current to calculate the resistance of the 80.0 cm length of wire. [2] (c) Describe the relationship between the length of the wire and its resistance, using data from Table 3.1 to support your answer. [2] (d) Suggest why a switch is included in the circuit and should be opened between taking readings. [2]
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解题
(a) To construct this circuit, a series loop is created containing the cell, ammeter, switch, and constantan wire. The voltmeter must be connected in parallel across the wire to measure the voltage drop specifically across it. (b)(i) The current is recorded directly as 0.15 A. (b)(ii) Using R = V / I, Resistance = 1.5 / 0.15 = 10.0 Ohm. (c) Resistance increases linearly with length. This is shown by the data: at 20 cm the resistance is 2.5 Ohm, and at 40 cm it is 5.0 Ohm, which demonstrates direct proportionality. (d) Keeping the switch closed would cause a continuous current to flow, heating the wire. Since resistance increases with temperature for most metals, this would introduce an unwanted variable, making it an unfair test.
评分标准
Part (a): [3 marks] - series loop described [1], voltmeter in parallel [1], correct identification of all components [1]. Part (b): [3 marks] - (i) 0.15 A [1], (ii) formula R = V / I [1], 10.0 Ohm [1] (accept 10). Part (c): [2 marks] - resistance increases as length increases [1], directly proportional / supported with numerical example from table [1]. Part (d): [2 marks] - prevent current flowing continuously / wire heating up [1], heating changes resistance of the wire [1].
题目 4 · Practical Investigation
10 分
A student investigates osmosis in plant tissue. They are provided with: a large fresh potato, five different concentrations of sucrose solution (0.0, 0.2, 0.4, 0.6, and 0.8 mol/dm^3), and common laboratory apparatus. (a) Plan an investigation to determine how the concentration of sucrose solution affects the change in mass of potato cylinders due to osmosis. Specify the apparatus, experimental method, control variables, and how you will process the results. [7] (b) Table 4.1 shows some results obtained from a similar experiment: [0.0 mol/dm^3 | Initial mass: 2.50 g | Final mass: 2.85 g | Change: +0.35 g | % Change: +14.0], [0.4 mol/dm^3 | Initial mass: 2.50 g | Final mass: 2.40 g | Change: -0.10 g | % Change: ?], [0.8 mol/dm^3 | Initial mass: 2.60 g | Final mass: 2.12 g | Change: -0.48 g | % Change: -18.5]. (i) Calculate the percentage change in mass for the potato cylinder in the 0.4 mol/dm^3 sucrose solution. Show your working. [2] (ii) Explain why calculating the percentage change in mass is more useful than comparing the actual change in mass in this investigation. [1]
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解题
(a) To investigate osmosis, potato cylinders are prepared using a cork borer to ensure equal diameter. Their initial masses are recorded using a balance after blotting them dry. They are placed in five separate beakers, each containing an equal volume of one of the sucrose solutions, for a fixed duration (e.g., 30 minutes). After this time, they are removed, blotted dry again to remove surface water, and re-weighed. The temperature and dimensions of the cylinders must be controlled. (b)(i) Percentage change = (Change in mass / Initial mass) * 100. Percentage change = (-0.10 / 2.50) * 100 = -4.0%. (b)(ii) Potato cylinders have slightly different initial masses. Calculating percentage change normalizes the data, making comparisons valid.
评分标准
Part (a): [7 marks] - balance and cork borer / scalpel listed [1], cut cylinders and record initial mass [1], immerse in sucrose solutions of 5 different concentrations for a set time [1], remove and blot dry with paper towel [1], record final mass [1], control variables (temperature/volume/time) [1], process results by calculating percentage mass change and plotting graph [1]. Part (b): [3 marks] - (i) working shown [1], -4.0% [1], (ii) accounts for differences in initial starting mass of cylinders [1].
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