An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V1) Cambridge IGCSE Science - Combined (0653) paper. Not affiliated with or reproduced from Cambridge.
Extended Theory Paper
Answer all questions. Write your answers in the spaces provided. Show all your working for calculations.
9 题目 · 81 分
题目 1 · structured
9 分
Mammals, such as humans, possess a double circulatory system, whereas fish possess a single circulatory system.
(a) (i) Describe what is meant by a double circulatory system. [2]
(ii) Explain the physiological advantages of a double circulatory system compared to a single circulatory system. [2]
(b) Coronary heart disease (CHD) is a serious condition that affects the heart's blood supply.
(i) Explain how the blockage of coronary arteries can lead to a heart attack. [3]
(ii) State two lifestyle changes a person can make to reduce their risk of developing CHD. [2]
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解题
(a) (i) In a double circulatory system, blood passes through the heart twice for each complete circuit of the body. This consists of the pulmonary circulation (to the lungs) and systemic circulation (to the rest of the body).
(ii) A double circulatory system allows blood returning from the lungs to be repressurised by the heart before being sent to the rest of the body. This maintains a high blood pressure in the systemic circulation, ensuring a faster delivery of oxygen and glucose to respiring tissues, and a faster removal of metabolic waste.
(b) (i) Coronary arteries supply the heart muscle cells with blood containing oxygen and glucose. A blockage (often due to cholesterol plaques) stops or severely reduces this blood flow. As a result, heart muscle cells are deprived of oxygen and glucose, preventing them from performing aerobic respiration. The cells cannot produce enough energy, leading to anaerobic respiration, lactic acid build-up, and ultimately cell death and tissue damage (a heart attack).
(ii) To reduce the risk of CHD, a person can: 1. Reduce their intake of saturated fats and cholesterol. 2. Exercise regularly to maintain cardiovascular fitness.
评分标准
(a) (i) - Blood passes through the heart twice [1] - For each complete circuit of the body [1]
(ii) - Blood can be pumped to the body tissues at a higher pressure [1] - Increases the rate of delivery of oxygen/glucose OR increases rate of removal of carbon dioxide/waste [1]
(b) (i) - Blockage reduces/stops blood flow to the heart muscle [1] - Prevents the delivery of oxygen/glucose to the heart muscle cells [1] - Muscle cells cannot respire (aerobically) / cells die or are damaged [1]
(b) (ii) - Any two from: reduce intake of saturated fats/cholesterol; regular exercise; stop smoking; reduce stress levels; maintain healthy weight / BMI [2] - (Accept any other sensible and valid lifestyle change. Reject medical interventions like taking aspirin or bypass surgery.)
题目 2 · structured
9 分
A student investigates the rate of photosynthesis in an aquatic plant at different light intensities and under two different concentrations of carbon dioxide ($0.03\%$ and $0.15\%$).
(a) (i) Write the balanced chemical equation for photosynthesis in symbols. [2]
(ii) Identify the form in which carbohydrates are transported through the phloem of a plant and the form in which they are stored in the leaves. [2]
(b) (i) With reference to limiting factors, explain why the rate of photosynthesis eventually reaches a maximum plateau as light intensity increases. [3]
(ii) Explain why this plateau is higher when the carbon dioxide concentration is increased from $0.03\%$ to $0.15\%$. [2]
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解题
(a) (i) The balanced chemical equation for photosynthesis is: $$6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$$
(ii) Carbohydrates are transported through the phloem as sucrose, and stored in the leaves as starch.
(b) (i) At low light intensities, light intensity is the limiting factor. As light intensity increases, the rate of photosynthesis increases proportionally. However, at high light intensities, the rate plateaus because light is no longer the limiting factor. At this point, some other factor, such as carbon dioxide concentration or temperature, has become the limiting factor and restricts further increase in the rate.
(ii) Carbon dioxide is a raw material used in photosynthesis. At $0.03\%$ concentration, carbon dioxide acts as a limiting factor. Increasing the concentration of carbon dioxide to $0.15\%$ provides more raw materials, making carbon dioxide less of a limiting factor and allowing a higher maximum rate of photosynthesis to be achieved.
评分标准
(a) (i) - Correct formulae of all reactants and products (\(\text{CO}_2\), \(\text{H}_2\text{O}\), \(\text{C}_6\text{H}_{12}\text{O}_6\), \(\text{O}_2\)) [1] - Correct balancing (6, 6, 1, 6) [1]
(a) (ii) - Transport form: sucrose [1] - Storage form: starch [1]
(b) (i) - At lower light intensities, light is the limiting factor [1] - At higher light intensities, light is no longer limiting / rate does not increase [1] - Another factor (such as carbon dioxide concentration or temperature) has become limiting [1]
(b) (ii) - Carbon dioxide is a raw material for photosynthesis [1] - Increasing concentration reduces its limitation / allows more carbon dioxide to be fixed per unit time [1]
题目 3 · structured
9 分
A farm next to a lake applies nitrogen-based chemical fertilisers to its crops. Following heavy rain, nutrients run off from the soil into the lake, leading to eutrophication.
(a) (i) Explain how the runoff of chemical fertilisers causes rapid growth of algae at the lake surface. [2]
(ii) Describe how this rapid algal growth blocks light and affects the aquatic plants growing deeper in the lake. [2]
(b) (i) Explain the role of decomposers (bacteria) in the depletion of dissolved oxygen in the lake. [3]
(ii) State the consequence of this oxygen depletion on fish populations and explain why this consequence occurs. [2]
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解题
(a) (i) Nitrogen-based chemical fertilisers contain highly soluble nitrates. When washed into the lake, these nitrates act as nutrients that promote rapid cell division and growth of algae, leading to an algal bloom on the surface.
(ii) The thick layer of algae on the surface blocks sunlight from penetrating deeper into the water. This prevents aquatic plants growing deeper in the lake from absorbing light energy, meaning they cannot photosynthesise and consequently die.
(b) (i) As deeper plants and algae die, they sink to the bottom of the lake. Decomposers, such as aerobic bacteria, feed on the dead organic matter. The abundant food supply causes a rapid increase in the bacterial population. These bacteria respire aerobically, consuming the dissolved oxygen in the water at a rate faster than it can be replenished.
(ii) The consequence is the death of fish populations. Fish require dissolved oxygen to perform aerobic respiration to obtain energy; without sufficient dissolved oxygen, they cannot respire and die of suffocation.
评分标准
(a) (i) - Nitrates/nutrients from fertilisers are highly soluble [1] - Algae absorb these nutrients and use them for rapid growth/cell division [1]
(a) (ii) - Algae form a layer/bloom at the surface that blocks sunlight [1] - Deeper plants cannot photosynthesise and die [1]
(b) (i) - Dead plants/algae are decomposed/broken down by bacteria [1] - Bacteria population increases rapidly [1] - Bacteria respire aerobically, consuming/using up dissolved oxygen [1]
(b) (ii) - Fish die [1] - Because they cannot perform aerobic respiration / lack of oxygen for respiration [1]
题目 4 · structured
9 分
An electric current is passed through dilute sulfuric acid using inert platinum electrodes.
(a) Describe what is observed at: (i) the anode (positive electrode) [1] (ii) the cathode (negative electrode). [1]
(b) (i) Write the ionic half-equation for the reaction occurring at the cathode. [2] (ii) State a chemical test, and its positive result, to identify the gas produced at the cathode. [2]
(c) (i) Write the ionic half-equation for the reaction occurring at the anode. [2] (ii) Explain why the volume of gas collected at the cathode is twice the volume of gas collected at the anode. [1]
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解题
(a) (i) Colourless bubbles of gas are produced at the anode. (ii) Colourless bubbles of gas are produced at the cathode.
(b) (i) The reaction at the cathode is: \( 2\text{H}^+(aq) + 2\text{e}^- \rightarrow \text{H}_2(g) \). (ii) Test: Place a lighted splint near the mouth of the test tube. Result: A squeaky pop sound is heard.
(c) (i) The reaction at the anode is: \( 4\text{OH}^-(aq) \rightarrow \text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4\text{e}^- \). (ii) The half-equations show that transferring 4 moles of electrons produces 2 moles of hydrogen gas at the cathode, but only 1 mole of oxygen gas at the anode, resulting in a 2:1 volume ratio.
评分标准
(a) (i) bubbles of gas / effervescence [1] (ii) bubbles of gas / effervescence [1]
(b) (i) \( \text{H}^+ \) and \( \text{e}^- \) on left and \( \text{H}_2 \) on right [1], correctly balanced [1] (ii) lighted splint [1], squeaky pop / pops [1] (reject: glowing splint)
(c) (i) \( \text{OH}^- \) on left and \( \text{O}_2 \), \( \text{H}_2\text{O} \), \( \text{e}^- \) on right [1], correctly balanced [1] (ii) 2 moles of hydrogen gas are produced for every 1 mole of oxygen gas / ratio of hydrogen to oxygen is 2:1 [1]
题目 5 · structured
9 分
In an industrial refinery, large hydrocarbon molecules can be broken down into smaller, more useful molecules by a process called cracking. Decane, \( \text{C}_{10}\text{H}_{22} \), can be cracked to produce pentane, \( \text{C}_5\text{H}_{12} \), propene, \( \text{C}_3\text{H}_6 \), and ethene, \( \text{C}_2\text{H}_4 \).
(a) Write a balanced chemical equation for this cracking reaction. [2]
(b) State two conditions required for industrial cracking to occur. [2]
(c) Ethene molecules can undergo addition polymerisation to form poly(ethene). (i) Draw the structure of the monomer, ethene, showing all atoms and all covalent bonds. [1] (ii) Draw the structure of poly(ethene) showing two repeating units. [2]
(d) Describe a chemical test using aqueous bromine to distinguish between pentane and propene. State the observations for each. [2]
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解题
(a) The balanced equation is: \( \text{C}_{10}\text{H}_{22} \rightarrow \text{C}_5\text{H}_{12} + \text{C}_3\text{H}_6 + \text{C}_2\text{H}_4 \)
(b) The two essential conditions are a high temperature (around 450–800 °C) and a catalyst (such as alumina, silica, or zeolites).
(c) (i) Ethene monomer: H H \ / C=C / \ H H (ii) Poly(ethene) macromolecule showing two repeating units: H H H H | | | | --C - C - C - C-- | | | | H H H H
(d) Add aqueous bromine to separate samples of each gas. Propene (the unsaturated alkene) decolorises the orange-brown bromine water. Pentane (the saturated alkane) does not react, and the mixture remains orange-brown.
评分标准
(a) reactant formula correct [1], products formulae correct and balanced [1] (b) high temperature / heat [1], catalyst / zeolite / silica / alumina [1] (c) (i) \( \text{C}=\text{C} \) double bond with 4 \( \text{C}-\text{H} \) single bonds correctly shown [1] (ii) single \( \text{C}-\text{C} \) bonds in the chain with extension bonds shown on both ends [1], 4 hydrogens per carbon pair [1] (d) bromine water decolorises / turns colourless with propene [1], no change / remains orange-brown with pentane [1]
题目 6 · structured
9 分
Iron is extracted from its ore, hematite, in a blast furnace. Hematite contains iron(III) oxide, \( \text{Fe}_2\text{O}_3 \).
(a) Carbon monoxide is the main reducing agent in the blast furnace. (i) Write a balanced chemical equation for the formation of carbon monoxide from carbon dioxide and coke (carbon). [1] (ii) Write a balanced chemical equation for the reduction of iron(III) oxide by carbon monoxide. [2]
(b) Limestone (calcium carbonate) is added to the furnace to remove impurities. (i) State the name of the substance added to remove acidic impurities. [1] (ii) Explain how this substance removes silicon dioxide, \( \text{SiO}_2 \), including the name of the neutralisation product. [2]
(c) (i) Explain, in terms of their structures, why alloys of iron (such as steel) are harder and stronger than pure iron. [2] (ii) State one practical use of stainless steel. [1]
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解题
(a) (i) The reaction between carbon dioxide and carbon to form carbon monoxide is: \( \text{CO}_2 + \text{C} \rightarrow 2\text{CO} \) (ii) The reaction for the reduction of iron(III) oxide is: \( \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2 \)
(b) (i) Calcium carbonate / limestone decomposes to form calcium oxide, which removes the impurities. (ii) Calcium oxide is a basic oxide. It reacts with silicon dioxide (an acidic oxide) in a neutralisation reaction to produce calcium silicate, \( \text{CaSiO}_3 \), commonly known as slag.
(c) (i) Pure iron contains regularly arranged layers of identical atoms that can slide past each other easily when force is applied. In steel (an alloy), atoms of different elements and sizes disrupt this uniform layer structure, preventing them from sliding easily. (ii) Stainless steel is used to manufacture cutlery, kitchen sinks, or surgical instruments.
评分标准
(a) (i) \( \text{CO}_2 + \text{C} \rightarrow 2\text{CO} \) [1] (ii) reactants and products correct [1], balancing correct [1]
(b) (i) calcium carbonate / limestone / calcium oxide [1] (ii) basic calcium oxide reacts with acidic silicon dioxide [1], to form calcium silicate / slag [1]
(c) (i) pure iron has regularly arranged layers of atoms that slide over each other [1], alloy atoms have different sizes which disrupt layers and prevent sliding [1] (ii) cutlery / surgical instruments / chemical plant / kitchen sinks [1] (reject: construction / bridges)
题目 7 · Structured
9 分
**1** (a) Fig. 1.1 shows a circuit diagram with a 9.0 V d.c. power supply connected to a parallel combination of two resistors. One resistor has a resistance of 15 \(\Omega\) and the other has a resistance of 30 \(\Omega\).
(i) Calculate the combined resistance of the two resistors in parallel. Show your working.
(ii) Determine the total current supplied by the power supply. Show your working.
Current = .................... A [2]
(b) A student sets up a temperature-sensing circuit. It contains a thermistor and a fixed resistor of 100 \(\Omega\) connected in series with a 6.0 V d.c. power supply.
(i) State and explain what happens to the potential difference across the 100 \(\Omega\) resistor as the temperature of the thermistor decreases.
(b) (i) As the temperature decreases, the resistance of the thermistor increases. Because the thermistor and the 100 \(\Omega\) resistor are in series, the thermistor takes a larger share of the total potential difference. Consequently, the potential difference across the 100 \(\Omega\) resistor decreases.
(ii) The symbol is a standard rectangular resistor with a diagonal line through it, with a flat horizontal section at the bottom-left end.
评分标准
(a) (i) - use of parallel resistor formula: \(\frac{1}{R_p} = \frac{1}{15} + \frac{1}{30}\) or \(\frac{15 \times 30}{15 + 30}\) [1] - correct value: \(10\) (\(\Omega\)) [1]
(ii) - use of \(I = \frac{V}{R}\) with their parallel resistance [1] - correct value: \(0.90\) (A) (accept \(0.9\)) [1]
(b) (i) - resistance of the thermistor increases as temperature decreases [1] - total circuit resistance increases / circuit current decreases [1] - potential difference across the 100 \(\Omega\) resistor decreases [1]
(b) (ii) - correct rectangle for a resistor [1] - correct diagonal line with flat horizontal hook at the bottom left [1]
题目 8 · Structured
9 分
**2** A small electric toy car of mass 1.5 kg starts from rest and accelerates uniformly to a speed of 4.0 m/s in 5.0 seconds. It then travels at a constant speed of 4.0 m/s for another 10.0 seconds.
(a) Describe the motion of the toy car between 5.0 s and 15.0 s.
(c) Distance travelled can be found from the area under a speed-time graph: - Distance from 0 to 5.0 s (triangle): \(\frac{1}{2} \times 5.0\text{ s} \times 4.0\text{ m/s} = 10.0\text{ m}\). - Distance from 5.0 s to 15.0 s (rectangle): \(10.0\text{ s} \times 4.0\text{ m/s} = 40.0\text{ m}\). - Total distance = \(10.0\text{ m} + 40.0\text{ m} = 50.0\text{ m}\).
(a) - constant speed / velocity or zero acceleration [1]
(b) - use of \(a = \frac{v-u}{t}\) or \(\frac{4.0}{5.0}\) [1] - correct value \(0.8\) and unit \(\text{m/s}^2\) [1]
(c) - recognition that distance is the area under the speed-time graph [1] - correct calculation of one section (either 10 m or 40 m) [1] - correct total distance: \(50\) (m) [1]
(d) - use of \(\text{K.E.} = \frac{1}{2} m v^2\) [1] - correct substitution: \(0.5 \times 1.5 \times 4.0^2\) [1] - correct value: \(12\) (J) [1]
题目 9 · Structured
9 分
**3** (a) A ray of light in air is incident on the flat surface of a transparent glass block.
(i) State what is meant by the *normal* line at the point of incidence.
(ii) When the light enters the glass block, it undergoes refraction. State and explain the direction of bending of the ray of light as it enters the glass.
(a) (i) A line drawn perpendicular (at \(90^\circ\)) to the boundary surface at the point where the incident ray strikes. (ii) The ray bends towards the normal, because glass is more optically dense than air (light travels slower in glass than in air).