An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V1) Cambridge IGCSE Science - Combined (0653) paper. Not affiliated with or reproduced from Cambridge.
Paper 41 (Theory - Extended)
Answer all questions. Use a black or dark blue pen. You may use a calculator. You must show all your working and use appropriate units.
9 题目 · 81 分
题目 1 · structured
9 分
(a) A student studies the microscopic structure of human blood vessels. A cross-section of three different types of blood vessels, A, B, and C, is shown under a microscope (not to scale).
(i) Identify which vessel, A, B, or C, represents a vein. State two structural features visible in such a cross-section that justify your choice. [3]
(ii) Explain how the wall of capillary C is structurally adapted to allow the efficient exchange of substances between the blood and tissue cells. [2]
(b) Mammals have a double circulatory system. Explain what is meant by double circulation and why it is an advantage for active warm-blooded animals. [3]
(c) State the name of the blood vessel that carries deoxygenated blood from the right ventricle of the heart to the lungs. [1]
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解题
Detailed breakdown of the parts: (a)(i) Vessel B is a vein. It has a relatively thin muscular wall, a wide lumen, and may contain valves to prevent backflow. (ii) Capillaries (C) have walls that are only one-cell thick, which reduces the diffusion distance for gases and nutrients, making the exchange highly efficient. (b) Double circulation involves two separate circuits: the pulmonary circuit (to and from the lungs) and the systemic circuit (to and from the rest of the body). It ensures blood travels to the tissues under high pressure, delivering oxygen and glucose rapidly. (c) The pulmonary artery is the vessel that transports deoxygenated blood from the right ventricle to the lungs.
评分标准
(a)(i) Identifies B as the vein [1] Any two visible features: thin wall / wide lumen / presence of valve [2] (ii) (Wall is) one-cell thick / very thin [1] Provides a short diffusion distance / pathway [1] (b) Blood flows through the heart twice for each complete circuit of the body [1] Blood can be pumped to the body tissues at a higher pressure [1] Oxygen/nutrients delivered more rapidly to support active metabolism [1] (c) Pulmonary artery [1]
题目 2 · structured
9 分
(a) Table 2.1 is an incomplete comparison of the structural features of insect-pollinated and wind-pollinated flowers.
Complete the table by writing the correct descriptions for the missing features (i), (ii), and (iii). [3]
(b) Describe the pathway of a male gamete from the moment a pollen grain lands on the stigma until fertilization of the ovule occurs. [4]
(c) State whether offspring produced by sexual reproduction are genetically identical or genetically different from the parents, and explain why. [2]
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解题
Detailed breakdown: (a) For wind-pollinated flowers, petals are small/dull to avoid wasting energy. Anthers in insect-pollinated flowers are positioned inside the flower where insects brush against them. Pollen grains in insect-pollinated flowers are sticky or spiky to adhere to insects. (b) The pollen grain germinates on the stigma, growing a pollen tube down through the style into the ovary. The male nucleus travels down the tube and enters the ovule via the micropyle, where it fuses with the female gamete nucleus. (c) The offspring are genetically different because sexual reproduction combines gametes from two parents, each carrying unique genetic combinations due to meiosis and random fertilization.
评分标准
(a) (i) Small / dull / green / absent petals [1] (ii) Inside the flower / firmly attached (to rub against insects) [1] (iii) Large / sticky / spiky pollen grains [1] (b) Pollen grain germinates on the stigma [1] Pollen tube grows down [1] Through the style [1] Enters the ovary / micropyle of the ovule where male gamete fuses with female nucleus [1] (c) Genetically different [1] Combination of genetic material from two parents / fusion of gametes / meiosis [1]
题目 3 · structured
9 分
Decane, \(C_{10}H_{22}\), is a long-chain alkane that can be cracked to produce octane, \(C_8H_{18}\), and ethene, \(C_2H_4\).
(a) (i) Write a balanced chemical equation for this cracking reaction. [2]
(ii) State two reaction conditions required for this industrial cracking process. [2]
(b) Ethene is described as an unsaturated hydrocarbon.
(i) Explain the meaning of the term *unsaturated hydrocarbon*. [2]
(ii) Describe a chemical test to distinguish between octane and ethene. Include the test reagent and the results for both compounds. [3]
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解题
Detailed solution: (a)(i) Cracking breaks decane down: \(C_{10}H_{22} \rightarrow C_8H_{18} + C_2H_4\). (ii) Industrial cracking requires high temperatures (around 500-600 °C) and a catalyst (such as zeolite, alumina, or silica). (b)(i) Unsaturated means the compound contains a carbon-carbon double bond (\(C=C\)). Hydrocarbon means it contains only carbon and hydrogen atoms. (ii) Bromine water is added to both liquids. With ethene (unsaturated), the orange-brown bromine water is decolourised (turns colourless). With octane (saturated), the bromine water remains orange-brown.
评分标准
(a)(i) Correct formulas for reactants and products [1] Correctly balanced equation: \(C_{10}H_{22} \rightarrow C_8H_{18} + C_2H_4\) [1] (ii) High temperature [1] Catalyst (e.g., zeolite / alumina / silica) [1] (b)(i) Unsaturated: contains a carbon-carbon double bond / \(C=C\) [1] Hydrocarbon: compound containing hydrogen and carbon only [1] (ii) Add bromine water / aqueous bromine [1] Result with ethene: decolourises / turns colourless [1] Result with octane: no change / remains orange-brown [1]
题目 4 · structured
9 分
A toy car of mass 0.50 kg is driven along a straight, horizontal track. The car accelerates uniformly from rest to a speed of 6.0 m/s in 4.0 s, and then continues at this constant speed of 6.0 m/s for another 4.0 s.
(a) (i) Calculate the acceleration of the car during the first 4.0 s of its motion. State the unit of your answer. [3]
(ii) Calculate the total distance travelled by the car during the entire 8.0 s journey. [2]
(b) The car is then lifted vertically by a small electric motor to a shelf 1.2 m above the track. The gravitational field strength \(g\) is 10 N/kg.
(i) Calculate the work done in lifting the car to the shelf. [2]
(ii) The motor takes 3.0 s to lift the car. Calculate the useful power output of the motor. State the unit of your answer. [2]
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解题
Detailed solution: (a)(i) Acceleration is the gradient of the speed-time graph or the change in speed divided by time: \(a = \frac{v - u}{t} = \frac{6.0 - 0}{4.0} = 1.5\text{ m/s}^2\). (ii) Distance is the area under the speed-time graph. From 0 to 4 s: Area of triangle = \(\frac{1}{2} \times 4.0 \times 6.0 = 12.0\text{ m}\). From 4 to 8 s: Area of rectangle = \(4.0 \times 6.0 = 24.0\text{ m}\). Total distance = \(12.0 + 24.0 = 36.0\text{ m}\). (b)(i) Work done is equal to the increase in gravitational potential energy: \(W = mgh = 0.50 \times 10 \times 1.2 = 6.0\text{ J}\). (ii) Power is work done divided by time: \(P = \frac{W}{t} = \frac{6.0}{3.0} = 2.0\text{ W}\).
评分标准
(a)(i) \(a = \frac{\Delta v}{t}\) or evidence of correct substitution [1] Correct calculation: 1.5 [1] Unit: \(\text{m/s}^2\) [1] (ii) Calculates triangular area (12 m) OR rectangular area (24 m) [1] Correct total distance: 36 m [1] (b)(i) \(W = mgh\) or \(F = mg = 5.0\text{ N}\) [1] Correct work done: 6.0 J [1] (ii) \(P = \frac{W}{t}\) [1] Correct power with unit: 2.0 W / Watts [1]
题目 5 · structured
9 分
Methane, \(CH_4\), undergoes complete combustion in oxygen according to the equation:
(a) (i) State whether this reaction is exothermic or endothermic, and explain your choice in terms of heat transfer to the surroundings. [2]
(ii) Sketch an energy level diagram for this combustion reaction. Label the reactants, products, activation energy (\(E_a\)), and the overall energy change (\(\Delta H\)). [3]
(b) Explain, in terms of bond breaking and bond making, why this combustion reaction releases energy. [4]
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解题
Detailed solution: (a)(i) Combustion of methane is exothermic because thermal energy (heat) is released to the surroundings, causing the temperature of the surroundings to increase. (ii) An energy level diagram for an exothermic reaction has the reactants at a higher energy level than the products. A curve rises from the reactants level to a peak (activation energy) and then drops down to the products level. \(E_a\) is the arrow from the reactants energy level to the peak. \(\Delta H\) is the downward arrow from the reactants level to the products level. (b) Bond breaking requires energy (endothermic process) while bond making releases energy (exothermic process). In this reaction, the total energy released during the formation of new bonds (C=O and H-O bonds in the products) is greater than the total energy absorbed to break the existing bonds (C-H and O=O bonds in the reactants).
评分标准
(a)(i) Exothermic [1] Heat/thermal energy is released to the surroundings / temperature of surroundings increases [1] (ii) Reactants shown at a higher energy level than products [1] Activation energy (\(E_a\)) correctly labelled from reactants to peak of curve [1] Energy change of reaction (\(\Delta H\)) correctly labelled with downward arrow from reactants to products [1] (b) Bond breaking is endothermic / requires energy [1] Bond making is exothermic / releases energy [1] More energy is released in making the product bonds (C=O and H-O) [1] Than is taken in to break the reactant bonds (C-H and O=O) [1]
题目 6 · structured
9 分
A student sets up an electrical circuit containing three identical resistors, each with a resistance of \(12\ \Omega\), connected in parallel.
(a) (i) Calculate the total combined resistance of this parallel combination. [2]
(ii) The parallel combination is connected across a 6.0 V battery. Calculate the total current in the circuit. [2]
(b) Calculate the total electrical charge that passes through the battery if the circuit remains switched on for 5.0 minutes. State the unit of your answer. [3]
(c) Calculate the total electrical energy transferred by the battery during these 5.0 minutes. [2]
(a)(i) Correct formula used: \(\frac{1}{R_p} = \frac{3}{12}\) or equivalent [1] Correct combined resistance: 4.0 \(\Omega\) [1] (ii) Correct use of \(I = \frac{V}{R}\) [1] Correct total current: 1.5 A [1] (b) Converts minutes to seconds: \(300\text{ s}\) [1] Correct calculation: \(1.5 \times 300 = 450\) [1] Correct unit: C / Coulombs [1] (c) Correct use of \(E = VIt\) or \(E = VQ\) [1] Correct energy: 2700 J / Joules [1]
题目 7 · structured
9 分
Flowering plants have specialised vascular tissues to transport substances throughout the plant.
(a) Contrast the function of xylem vessels with the function of phloem vessels in terms of the substances they transport and the direction of transport. [3]
(b) Transpiration is the loss of water vapour from the leaves of a plant.
(i) Explain how the structures of the mesophyll cells and the stomata are involved in the process of transpiration. [3]
(ii) State and explain the effect of increasing atmospheric humidity on the rate of transpiration. [3]
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解题
Detailed solution: (a) Xylem vessels transport water and dissolved mineral ions upwards from the roots to the leaves. In contrast, phloem vessels transport organic nutrients, specifically sucrose and amino acids, from sources (where they are made, such as leaves) to sinks (where they are used or stored, such as roots or growing tips). (b)(i) Mesophyll cells have wet cell walls from which water evaporates into the internal air spaces of the leaf. Stomata are pores on the leaf surface that open to allow the water vapour in the air spaces to diffuse out into the atmosphere. (ii) Increasing atmospheric humidity decreases the rate of transpiration. This is because high humidity increases the concentration of water vapour in the air outside the leaf, reducing the water vapour concentration gradient between the inside and the outside of the leaf. Consequently, the rate of diffusion of water vapour out of the stomata slows down.
评分标准
(a) Xylem transports water and mineral ions [1] Phloem transports sucrose and amino acids [1] Transport in xylem is one-way (upwards) whereas phloem transport is two-way / from source to sink [1] (b)(i) Water evaporates from the wet cell walls of mesophyll cells into air spaces [1] Creates water vapour in the air spaces [1] Water vapour diffuses out of the leaf through the stomata [1] (ii) Rate of transpiration decreases [1] Humidity increases water vapour concentration outside the leaf / reduces the concentration gradient [1] Slower diffusion of water vapour out of the stomata [1]
题目 8 · structured
9 分
(a) A student is provided with a dry mixture containing solid copper(II) carbonate (insoluble in water) and solid sodium chloride (soluble in water).
Describe how the student can obtain a pure sample of solid copper(II) carbonate and a pure, dry sample of sodium chloride crystals from this mixture. [4]
(b) The student conducts qualitative tests on an unknown solution, X.
(i) Addition of aqueous sodium hydroxide to solution X produces a green precipitate that is insoluble in excess. Identify the cation present in solution X. [1]
(ii) Acidifying solution X with dilute nitric acid followed by the addition of aqueous silver nitrate produces a white precipitate. Identify the anion present in solution X. [1]
(iii) Heating a mixture of solution X with aqueous sodium hydroxide produces a gas. Describe a chemical test to identify this gas, and state the positive result. [3]
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解题
Detailed solution: (a) First, add water to the mixture and stir so that the sodium chloride dissolves completely, while the copper(II) carbonate remains as an insoluble solid. Filter the mixture; the copper(II) carbonate is collected as the residue on the filter paper. Wash this residue with distilled water and leave it to dry to obtain pure copper(II) carbonate. Heat the filtrate (sodium chloride solution) in an evaporating basin until the crystallisation point is reached, then allow it to cool so that sodium chloride crystals form. (b)(i) A green precipitate with sodium hydroxide indicates the presence of iron(II) ions, \(Fe^{2+}\). (ii) A white precipitate with silver nitrate after acidification indicates the presence of chloride ions, \(Cl^-\). (iii) When solution X (containing ammonium ions) is heated with sodium hydroxide, ammonia gas is evolved. The test for ammonia is to hold damp red litmus paper in the gas; the positive result is that the paper turns blue.
评分标准
(a) Add water and stir to dissolve the sodium chloride [1] Filter the mixture to obtain copper(II) carbonate as the residue [1] Wash the residue with distilled water and dry it [1] Heat the filtrate (sodium chloride solution) to crystallisation point / evaporate water and leave to cool [1] (b)(i) Iron(II) / \(Fe^{2+}\) [1] (ii) Chloride / \(Cl^-\) [1] (iii) Test: Damp red litmus paper [1] Result: turns blue [1] Identifies gas as ammonia / \(NH_3\) [1]
题目 9 · structured
9 分
Fig. 6.1 represents the speed-time graph for a toy car of mass \(1.2\text{ kg}\) as it moves along a straight horizontal track.
The graph starts at \((0\text{ s}, 0\text{ m/s})\), rises linearly to \((4.0\text{ s}, 3.0\text{ m/s})\), and then remains horizontal at \(3.0\text{ m/s}\) until \(t = 10\text{ s}\).
(a) (i) Calculate the acceleration of the toy car during the first \(4.0\text{ s}\) of its motion. Show your working. [2]
(ii) Calculate the total distance travelled by the toy car during the entire \(10\text{ s}\) shown on the graph. Show your working. [3]
(b) (i) From \(t = 4.0\text{ s}\) to \(t = 10\text{ s}\), the car travels at a constant speed of \(3.0\text{ m/s}\). The forward driving force provided by the motor is \(4.5\text{ N}\). State the size of the total resistive force acting on the car during this time, and explain your answer in terms of resultant force. [2]
(ii) Calculate the kinetic energy of the toy car when it is travelling at its constant speed of \(3.0\text{ m/s}\). State the unit. [2]
(ii) Distance = area under the speed-time graph. Area of triangle (0 to 4 s) = \(0.5 \times 4.0 \times 3.0 = 6.0\text{ m}\) Area of rectangle (4 to 10 s) = \((10 - 4.0) \times 3.0 = 18.0\text{ m}\) Total distance = \(6.0 + 18.0 = 24\text{ m}\) (Alternatively, using the area of a trapezium formula: \(0.5 \times (6.0 + 10) \times 3.0 = 24\text{ m}\))
(b) (i) Resistive force = \(4.5\text{ N}\). At constant speed, acceleration is zero, which means the resultant force is zero. The backward resistive force must exactly balance the forward driving force.
**(a)(i)** - Formula or calculation process seen: \(\frac{3.0}{4.0}\) [1] - Correct answer with unit: \(0.75\text{ m/s}^2\) [1]
**(a)(ii)** - Evidence of using area under the graph (e.g., calculating triangular area as \(6\text{ m}\) or rectangular area as \(18\text{ m}\)) [1] - Summing the two areas correctly: \(6.0 + 18.0\) [1] - Final correct distance: \(24\text{ m}\) [1]
**(b)(i)** - Correct resistive force value: \(4.5\text{ N}\) [1] - Correct explanation (constant speed means zero acceleration / zero resultant force) [1]
**(b)(ii)** - Correct substitution into kinetic energy formula: \(0.5 \times 1.2 \times 3^2\) [1] - Correct calculation and unit: \(5.4\text{ J}\) [1]