An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V3) Cambridge IGCSE Science - Combined (0653) paper. Not affiliated with or reproduced from Cambridge.
Extended Theory Paper
Answer all questions. Show your working in calculations. A calculator and Periodic Table are permitted.
10 题目 · 90 分
题目 1 · structured
9 分
(a) Define osmosis with reference to water potential. [3] (b) An experiment is conducted where fresh potato cylinders are weighed and placed into sucrose solutions of different concentrations. After two hours, the potato cylinders are weighed again. (i) Explain, in terms of water potential, why a potato cylinder placed in a high concentration sucrose solution decreases in mass. [3] (ii) Suggest the concentration of sucrose inside the potato cells if a cylinder placed in a 0.25 mol/dm3 sucrose solution shows no change in mass. [1] (c) State two factors, other than concentration gradient and temperature, that increase the rate of diffusion of a solute across a cell membrane. [2]
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解题
(a) Osmosis is the net movement of water molecules from a region of higher water potential to a region of lower water potential, through a partially permeable membrane. (b)(i) The sucrose solution has a lower water potential than the cytoplasm of the potato cells, so water moves out of the cells down a water potential gradient by osmosis, leading to a decrease in mass. (ii) If there is no change in mass, the water potential inside is equal to that outside, so the internal concentration is 0.25 mol/dm3. (c) A larger surface area of the membrane increases the rate of diffusion, and a thinner membrane (smaller diffusion distance) also increases the rate.
评分标准
(a) Net movement of water molecules [1], from higher water potential to lower water potential [1], through a partially permeable membrane [1]. (b)(i) Solution has lower water potential than potato cells [1], water moves out of the cells/cytoplasm [1], by osmosis [1]. (b)(ii) 0.25 mol/dm3 [1]. (c) Any two from: larger surface area of membrane [1], thinner membrane / shorter diffusion distance [1].
题目 2 · structured
9 分
A heavy hydrocarbon fraction containing dodecane, C12H26, is cracked at high temperatures to yield shorter-chain hydrocarbons. (a) Complete the chemical equation for this cracking reaction which produces one molecule of hexane, C6H14, and two molecules of an alkene, X: C12H26 -> C6H14 + 2 [ alkene X ]. [1] (b) Deduce the molecular formula of alkene X and state its name. [2] (c) Describe a chemical test to distinguish between hexane and alkene X, stating the reagent, observation with hexane, and observation with alkene X. [3] (d) State the conditions required for industrial cracking. [2] (e) State the general formula of the homologous series to which alkene X belongs. [1]
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解题
(a) Subtracting C6H14 from C12H26 leaves C6H12. Since two molecules of X are formed, each molecule of X has the formula C3H6. (b) The formula is C3H6, which is propene. (c) Bromine water is used. Hexane (saturated) does not react, so the solution remains orange/brown. Propene (unsaturated) reacts, decolourising the bromine water. (d) Industrial cracking requires high temperatures (approx. 500-700 degrees C) and a catalyst (silica or alumina). (e) Alkenes have the general formula CnH2n.
评分标准
(a) C3H6 [1]. (b) Molecular formula: C3H6 [1], name: propene [1]. (c) Reagent: aqueous bromine / bromine water [1]; Hexane: remains orange/brown/yellow [1] (reject: clear / no change); Alkene X: decolourises / turns colourless [1] (reject: turns clear). (d) High temperature [1], catalyst [1] (or high pressure). (e) CnH2n [1].
题目 3 · structured
9 分
A parcel of mass 12 kg is pulled up a smooth conveyor belt of a sorting facility. The parcel starts from rest and accelerates uniformly up the slope, reaching a speed of 1.5 m/s in 4.0 s. (a) Calculate the acceleration of the parcel. Show your working and state the unit. [3] (b) Calculate the distance travelled by the parcel during this acceleration phase. Show your working. [2] (c) The conveyor belt raises the parcel through a vertical height of 2.5 m. Calculate the increase in the gravitational potential energy of the parcel. (g = 10 N/kg) [2] (d) In reality, the electric motor driving the conveyor belt does 420 J of work on the parcel. Explain why the work done is greater than the increase in the gravitational potential energy of the parcel. [2]
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解题
(a) Acceleration = (v - u) / t = (1.5 - 0) / 4.0 = 0.375 m/s^2. (b) Distance = average speed x time = (1.5 / 2) x 4.0 = 3.0 m. (c) GPE = mgh = 12 x 10 x 2.5 = 300 J. (d) In reality, work must also be done to overcome frictional forces within the belt system and between the parcel and the belt, which dissipates energy as thermal energy to the surroundings.
评分标准
(a) Formula or working: 1.5 / 4.0 [1], value: 0.375 [1], unit: m/s^2 [1]. (b) Working: average speed x time or 1/2 x a x t^2 [1], answer: 3.0 (m) [1]. (c) Working: 12 x 10 x 2.5 [1], answer: 300 (J) [1]. (d) Friction exists / work is done against friction [1], some energy is lost as heat/sound to the surroundings [1].
题目 4 · structured
9 分
A laboratory technician sets up an electrolysis cell to decompose molten lead(II) bromide, PbBr2, using carbon electrodes. (a) Explain why lead(II) bromide must be molten to undergo electrolysis. [2] (b) State the observation at the negative electrode (cathode) and write the ionic half-equation for the reaction that occurs. [3] (c) State the observation at the positive electrode (anode) and write the ionic half-equation for the reaction that occurs. [3] (d) State the type of chemical reaction that takes place at the positive electrode. [1]
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解题
(a) Lead(II) bromide is an ionic compound. In the solid state, its ions are held tightly in a fixed lattice and cannot move. When molten, the lattice breaks down and the ions are free to move to the electrodes to conduct electricity. (b) Lead ions (Pb2+) gain electrons at the cathode, forming silver-coloured liquid lead metal. The half-equation is Pb2+ + 2e- -> Pb. (c) Bromide ions (Br-) lose electrons at the anode, forming reddish-brown bromine gas. The half-equation is 2Br- -> Br2 + 2e-. (d) Loss of electrons occurs at the anode, which is oxidation.
评分标准
(a) Solid ions are held in a fixed lattice [1], molten state allows ions to move and carry charge [1] (reject: electrons move). (b) Observation: silver/grey liquid/metal [1], half-equation: Pb2+ + 2e- -> Pb [2] (1 mark for symbols, 1 mark for balancing). (c) Observation: brown gas/fumes [1], half-equation: 2Br- -> Br2 + 2e- [2] (1 mark for symbols, 1 mark for balancing). (d) Oxidation [1].
题目 5 · structured
9 分
Photosynthesis is a vital process occurring in green plants. (a) Write the balanced chemical equation for photosynthesis. [3] (b) Describe two structural adaptations of a leaf's palisade mesophyll layer that maximize the rate of photosynthesis. [2] (c) Explain how carbon dioxide in the air enters the photosynthesising cells inside the leaf. [3] (d) State one mineral ion required by plants to produce chlorophyll. [1]
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解题
(a) The balanced equation is 6CO2 + 6H2O -> C6H12O6 + 6O2 (in the presence of light and chlorophyll). (b) Palisade cells are elongated and packed closely together near the upper surface of the leaf to absorb maximum light. They also contain a very high concentration of chloroplasts. (c) Carbon dioxide diffuses from the atmosphere into the leaf through open stomata. It then diffuses through the air spaces in the spongy mesophyll layer down a concentration gradient, dissolves in the moist film on the cell walls, and enters the cells. (d) Magnesium ions are essential for the synthesis of chlorophyll.
评分标准
(a) Reactants: 6CO2 + 6H2O [1], products: C6H12O6 + 6O2 [1], correct balancing [1]. (b) Palisade cells packed tightly together/perpendicularly near top surface [1], contain many/large numbers of chloroplasts [1]. (c) Diffuses through stomata [1], moves through intercellular/spongy air spaces [1], dissolves in water/moisture on cell walls before entering cells [1]. (d) Magnesium / Mg2+ [1] (accept: nitrogen).
题目 6 · structured
9 分
A battery of potential difference 12.0 V is connected to a parallel circuit containing two resistors with values of 3.0 ohms and 6.0 ohms respectively. (a) Describe the correct connection of these resistors, an ammeter (to measure total current), and a voltmeter (to measure potential difference across the 6.0 ohms resistor). [3] (b) Calculate the combined resistance of the two resistors in parallel. Show your working. [2] (c) Calculate the total current in the main circuit. Show your working and state the unit. [2] (d) Calculate the electrical energy transformed by the 3.0 ohms resistor if the circuit remains closed for 5.0 minutes. Show your working. [2]
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解题
(a) The two resistors are connected across each other in parallel branches. The ammeter is connected in series in the main branch (next to the battery) to measure total current. The voltmeter is connected in parallel directly across the terminals of the 6.0 ohm resistor. (b) 1 / Rp = 1 / 3.0 + 1 / 6.0 = 2 / 6.0 + 1 / 6.0 = 3 / 6.0, therefore Rp = 6.0 / 3 = 2.0 ohms. (c) Total current I = V / Rp = 12.0 V / 2.0 ohms = 6.0 A. (d) Current through the 3.0 ohm resistor is I1 = V / R1 = 12 V / 3.0 ohms = 4.0 A. Power = V x I1 = 12 x 4.0 = 48 W. Energy = Power x time = 48 W x (5.0 x 60 s) = 48 x 300 = 14400 J (or 14.4 kJ).
评分标准
(a) Resistors in parallel [1], ammeter in series in main branch [1], voltmeter connected in parallel across 6.0 ohm resistor [1]. (b) Working: 1/3 + 1/6 or (3x6)/(3+6) [1], answer: 2.0 (ohms) [1]. (c) Working: 12 / Rp [1], answer: 6.0 A [1] (must include unit 'A' or 'Amperes'). (d) Working: Power = 48 W or Time = 300 s [1], answer: 14400 J or 14.4 kJ [1].
题目 7 · structured
9 分
The human circulatory system is responsible for the transport of materials throughout the body. (a) Explain, in terms of blood pressure and destination, why the muscle wall of the left ventricle is much thicker than that of the right ventricle. [2] (b) Describe the function of the septum in the human heart and explain why a hole in the septum would reduce a person's athletic performance. [3] (c) Define coronary heart disease (CHD) and state two dietary modifications that can reduce the risk of developing this condition. [4]
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解题
(a) The left ventricle pumps blood to the entire body, which is a long distance with high resistance, requiring high pressure. The right ventricle only pumps blood to the lungs, which are nearby and delicate, requiring much lower pressure. (b) The septum separates the left and right chambers of the heart, preventing the mixing of oxygenated and deoxygenated blood. A hole in the septum would allow deoxygenated blood to mix with oxygenated blood, reducing the oxygen concentration of the blood sent to body tissues, leading to less aerobic respiration in muscles during exercise. (c) Coronary heart disease is the narrowing or blockage of the coronary arteries, which supply the heart muscle with oxygenated blood. Risk can be reduced by lowering the intake of saturated fats/cholesterol and increasing the intake of soluble fiber/fresh vegetables.
评分标准
(a) Left ventricle pumps to body / at high pressure [1], right ventricle pumps to lungs / at lower pressure [1]. (b) Septum prevents oxygenated and deoxygenated blood from mixing [1], hole causes mixing / lowers oxygen level in systemic circulation [1], less oxygen is delivered to muscles/cells for aerobic respiration [1]. (c) Coronary heart disease: narrowing/blockage of coronary arteries [1] which reduces blood/oxygen supply to heart muscle cells [1]. Dietary modifications: reduce saturated fats/cholesterol [1], increase fiber/fruits/vegetables [1].
题目 8 · structured
9 分
Waves can be classified into different categories based on their physical characteristics. (a) Distinguish between longitudinal and transverse waves in terms of the direction of particle vibration and the direction of wave travel. [2] (b) Electromagnetic waves travel at a speed of 3.0 x 10^8 m/s in a vacuum. (i) State two features that all electromagnetic waves have in common, other than their speed in a vacuum. [2] (ii) Ultra-high frequency (UHF) radio waves are used for television broadcasting. Calculate the frequency of a television signal with a wavelength of 0.60 m. Show your working and state the unit. [3] (c) Describe how the frequency and wavelength of a light wave change, if at all, when it passes from air into a glass block. [2]
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解题
(a) In a longitudinal wave, particles vibrate parallel to the direction of wave propagation. In a transverse wave, particles vibrate perpendicular to the direction of wave propagation. (b)(i) All electromagnetic waves are transverse waves and can travel through a vacuum (do not require a medium). (ii) f = v / lambda = (3.0 x 10^8 m/s) / 0.60 m = 5.0 x 10^8 Hz (or 500 MHz). (c) When entering a glass block, the frequency of the light wave remains constant. However, since the speed of light is lower in glass than in air, the wavelength must decrease to satisfy the wave equation v = f x lambda.
评分标准
(a) Longitudinal: vibration is parallel to the direction of wave propagation [1]; Transverse: vibration is perpendicular to the direction of wave propagation [1]. (b)(i) Any two from: they are transverse waves [1], they can travel through a vacuum [1], they transfer energy [1] (accept: they can reflect/refract). (b)(ii) Working: (3.0 x 10^8) / 0.60 [1], value: 5.0 x 10^8 [1], unit: Hz (or hertz) [1]. (c) Frequency: remains constant/unchanged [1], Wavelength: decreases [1].
题目 9 · structured
9 分
An electric current is passed through concentrated aqueous sodium chloride (brine) using inert carbon electrodes.
(a) State the name of the gas produced at: (i) the positive electrode (anode) [1] (ii) the negative electrode (cathode) [1]
(b) Concentrated aqueous sodium chloride contains four different ions: \(H^+\), \(OH^-\), \(Na^+\), and \(Cl^-\). Explain, in terms of the reactivity of the elements involved, why hydrogen gas is discharged at the cathode instead of sodium metal. [2]
(c) Write the ionic half-equation, including state symbols, for the reaction that occurs at the positive electrode (anode). [2]
(d) Describe a chemical test, and its positive result, that can be used to identify the gas produced at the anode. [2]
(e) After the electrolysis has run for some time, universal indicator is added to the solution near the cathode. State and explain the color change observed. [1]
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解题
(a) (i) The gas produced at the anode is chlorine. (ii) The gas produced at the cathode is hydrogen.
(b) Hydrogen ions (\(H^+\)) and sodium ions (\(Na^+\)) both migrate to the negative cathode. Because hydrogen is lower in the reactivity series (less reactive) than sodium, hydrogen ions are more easily reduced (gain electrons) and are preferentially discharged as hydrogen gas, leaving sodium ions in the solution.
(c) At the anode, chloride ions (\(Cl^-\)) lose electrons to form chlorine molecules: \(2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-\)
(d) To test for chlorine gas, place damp blue litmus paper into the mouth of the test tube. The litmus paper will turn red momentarily (due to acidity) and then quickly bleach white.
(e) The universal indicator turns blue or purple. This is because hydrogen ions are discharged at the cathode, leaving an excess of hydroxide ions (\(OH^-\)) in the solution, which makes the area around the cathode highly alkaline (forming aqueous sodium hydroxide).
评分标准
(a) (i) chlorine [1] (ii) hydrogen [1]
(b) - Hydrogen / hydrogen ions are less reactive than sodium / sodium ions [1] - Hydrogen ions (\(H^+\)) gain electrons / are reduced preferentially (or sodium ions are more stable in solution) [1]
(c) - \(2Cl^- \rightarrow Cl_2 + 2e^-\) - Correct formula for reactants, products, and balancing [1] - Correct state symbols: \((aq)\) for \(Cl^-\), \((g)\) for \(Cl_2\) [1]
(d) - Use damp blue litmus paper [1] (Reject: dry litmus paper) - The paper turns bleached / white [1]
(e) - Indicator turns blue/purple because the solution becomes alkaline / contains excess hydroxide ions (\(OH^-\)) [1]
题目 10 · structured
9 分
An electric current is passed through concentrated aqueous sodium chloride (brine) using inert carbon electrodes.
(a) State the name of the gas produced at: (i) the positive electrode (anode) [1] (ii) the negative electrode (cathode) [1]
(b) Concentrated aqueous sodium chloride contains four different ions: \(H^+\), \(OH^-\), \(Na^+\), and \(Cl^-\). Explain, in terms of the reactivity of the elements involved, why hydrogen gas is discharged at the cathode instead of sodium metal. [2]
(c) Write the ionic half-equation, including state symbols, for the reaction that occurs at the positive electrode (anode). [2]
(d) Describe a chemical test, and its positive result, that can be used to identify the gas produced at the anode. [2]
(e) After the electrolysis has run for some time, universal indicator is added to the solution near the cathode. State and explain the color change observed. [1]
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解题
(a) (i) The gas produced at the anode is chlorine. (ii) The gas produced at the cathode is hydrogen.
(b) Hydrogen ions (\(H^+\)) and sodium ions (\(Na^+\)) both migrate to the negative cathode. Because hydrogen is lower in the reactivity series (less reactive) than sodium, hydrogen ions are more easily reduced (gain electrons) and are preferentially discharged as hydrogen gas, leaving sodium ions in the solution.
(c) At the anode, chloride ions (\(Cl^-\)) lose electrons to form chlorine molecules: \(2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-\)
(d) To test for chlorine gas, place damp blue litmus paper into the mouth of the test tube. The litmus paper will turn red momentarily (due to acidity) and then quickly bleach white.
(e) The universal indicator turns blue or purple. This is because hydrogen ions are discharged at the cathode, leaving an excess of hydroxide ions (\(OH^-\)) in the solution, which makes the area around the cathode highly alkaline (forming aqueous sodium hydroxide).
评分标准
(a) (i) chlorine [1] (ii) hydrogen [1]
(b) - Hydrogen / hydrogen ions are less reactive than sodium / sodium ions [1] - Hydrogen ions (\(H^+\)) gain electrons / are reduced preferentially (or sodium ions are more stable in solution) [1]
(c) - \(2Cl^- \rightarrow Cl_2 + 2e^-\) - Correct formula for reactants, products, and balancing [1] - Correct state symbols: \((aq)\) for \(Cl^-\), \((g)\) for \(Cl_2\) [1]
(d) - Use damp blue litmus paper [1] (Reject: dry litmus paper) - The paper turns bleached / white [1]
(e) - Indicator turns blue/purple because the solution becomes alkaline / contains excess hydroxide ions (\(OH^-\)) [1]