2024 Cambridge IGCSE Science - Combined (0653) 模擬試題連答案詳解

(a) \(\Delta E_p = mgh = 1.5\text{ kg} \times 10\text{ N/kg} \times 4.0\text{ m} = 60\text{ J}\). (b)(i) \(\text{Energy} = P \times t = 15\text{ W} \times 5.0\text{ s} = 75\text{ J}\). (b)(ii) \(\text{Efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\% = \frac{60\text{ J}}{75\text{ J}} \times 100\% = 80\%\). (c) Energy is dissipated as thermal energy (heat) and sound to the surroundings, which is caused by friction between the moving parts of the motor.

Part (a): [2 marks] - Correct formula \(mgh\) or substitution (1 mark) - Correct final answer with unit, 60 J (1 mark). Part (b)(i): [2 marks] - Correct formula \(E = P \times t\) or substitution (1 mark) - Correct final answer with unit, 75 J (1 mark). Part (b)(ii): [2 marks] - Correct efficiency formula / substitution (1 mark) - Correct final answer, 80% (1 mark). Part (c): [2 marks] - Identifying energy is dissipated as thermal energy/heat/sound (1 mark) - Reference to friction in the moving parts of the motor (1 mark).

(a) Hydrocarbon Y (\(\text{C}_3\text{H}_6\)) is the alkene. Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond (\(\text{C}=\text{C}\)). (b) The standard test is adding aqueous bromine (bromine water), which is orange/yellow. With Hydrocarbon X (propane, an alkane), no reaction occurs under normal conditions, so the mixture remains orange. With Hydrocarbon Y (propene, an alkene), an addition reaction occurs across the double bond, decoloring the bromine water to colorless. (c) Industrial catalytic cracking requires a catalyst (silica, alumina, or zeolite) and a high temperature (typically within the range of 450 to 800 degrees Celsius).

Part (a): [3 marks] - Identifies Y as the alkene (1 mark) - Identifies carbon-carbon double bond (1 mark) - Mentions the presence of fewer hydrogens per carbon compared to alkanes or unsaturated nature (1 mark). Part (b): [3 marks] - Reagent: Bromine water / aqueous bromine with starting color orange/yellow/brown (1 mark) - Observation with X: remains orange / yellow / no change (1 mark) - Observation with Y: decolors / turns colorless (1 mark) (reject: 'turns clear'). Part (c): [2 marks] - High temperature (accept 450 to 800 degrees Celsius) (1 mark) - Catalyst (accept alumina / silica / zeolite) (1 mark).

(a) The balanced chemical equation for photosynthesis is: \(6\text{CO}_2 + 6\text{H}_2\text{O} \xrightarrow{\text{light}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\). (b) Oxygen is produced as a byproduct of photosynthesis. Light energy is absorbed by chlorophyll. Increasing light intensity provides more energy to drive the chemical reaction, resulting in a faster rate of photosynthesis and thus more oxygen released per unit time. (c) Other limiting factors include carbon dioxide concentration and temperature. (d) Palisade mesophyll cells are long, column-shaped, and arranged vertically near the upper epidermis to capture maximum sunlight. They also contain a very high density of chloroplasts to absorb light energy efficiently.

Part (a): [3 marks] - Correct reactant formulae (\(\text{CO}_2\) and \(\text{H}_2\text{O}\)) (1 mark) - Correct product formulae (\(\text{C}_6\text{H}_{12}\text{O}_6\) and \(\text{O}_2\)) (1 mark) - Correct balancing (1 mark). Part (b): [2 marks] - Light energy is absorbed by chlorophyll / is needed for photosynthesis (1 mark) - More light energy increases the rate of reaction, which releases more oxygen as a byproduct per unit time (1 mark). Part (c): [2 marks] - Temperature (1 mark) - Carbon dioxide concentration (1 mark). Part (d): [2 marks] - Cells are closely packed / vertically arranged near the upper surface of the leaf to absorb maximum light (1 mark) - Cells contain a large number of chloroplasts / high concentration of chlorophyll (1 mark).

(a) Work done = Force \times distance = weight \times height = m \times g \times h = 4.5 \text{ kg} \times 10 \text{ m/s}^2 \times 8.0 \text{ m} = 360 \text{ J}.
(b) Efficiency = (useful energy output / total energy input) \times 100 = (360 \text{ J} / 450 \text{ J}) \times 100 = 80\% (or 0.8).
(c) The useful energy transfer is electrical energy to gravitational potential energy of the parcel. The wasted energy is transferred to thermal energy of the motor and surroundings (and sound energy), dispersing/dissipating into the air.

Total Marks: 8.88

(a) [3 Marks]
- State formula \(W = mgh\) or \(F \times d\) [1 mark]
- Correct calculation: \(360\) [1 mark]
- Correct unit: \(\text{J}\) or \(\text{Joules}\) [1 mark]

(b) [3 Marks]
- State formula \(\text{Efficiency} = \frac{\text{useful energy}}{\text{total energy}} \times 100\%\) [1 mark]
- Correct calculation: \(80\%\) or \(0.8\) [2 marks]

(c) [2.88 Marks]
- Identifies useful transfer: electrical to gravitational potential energy (GPE) [1.44 marks]
- Explains that wasted energy is dissipated/lost to the surroundings as thermal/heat (or sound) energy [1.44 marks]

(a) Decane cracks to form propene and heptane: \(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_3\text{H}_6 + \text{C}_7\text{H}_{16}\).
(b) To distinguish alkanes from alkenes, use aqueous bromine (bromine water). Alkenes are unsaturated and react readily, decolourizing the bromine. Alkanes are unreactive and do not react with bromine in the absence of UV light, so the solution remains orange/yellow.
(c) A hydrocarbon is a compound containing only carbon and hydrogen. 'Saturated' means all carbon-carbon bonds are single covalent bonds. Alkenes are unsaturated because they contain a double covalent bond (\(\text{C}=\text{C}\)) between carbon atoms.

Total Marks: 8.88

(a) [2 Marks]
- Correct formula for heptane \(\text{C}_7\text{H}_{16}\) [1 mark]
- Fully balanced equation [1 mark]

(b) [3.88 Marks]
- Correct reagent: aqueous bromine / bromine water [1 mark]
- Decane observation: remains orange/yellow/brown / no change [1.44 marks]
- Propene observation: decolourizes / becomes colourless [1.44 marks]

(c) [3 Marks]
- Hydrocarbon defined as containing carbon and hydrogen only [1 mark]
- Saturated defined as containing only single carbon-carbon bonds [1 mark]
- Alkenes contain at least one double carbon-carbon (\(\text{C}=\text{C}\)) bond [1 mark]

(a) The balanced equation is \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\).
(b) Photosynthesis is controlled by enzymes, which are temperature-sensitive. If temperature varies, the rate of photosynthesis will change due to temperature variation rather than light intensity alone, meaning temperature must be controlled as a constant variable to ensure valid results.
(c) Palisade mesophyll cells are situated at the upper surface of the leaf to receive maximum light. They are column-shaped and tightly packed to maximize absorption area, and contain a very high density of chloroplasts (containing chlorophyll) to trap light energy.

Total Marks: 8.88

(a) [3 Marks]
- Correct chemical symbols for reactants and products [1.5 marks]
- Correct balancing [1.5 marks]

(b) [2.88 Marks]
- Explaining that temperature is a limiting factor / affects enzyme activity [1.44 marks]
- Explaining that controlling it ensures the investigation is a fair test / light intensity is the only independent variable affecting the rate [1.44 marks]

(c) [3 Marks]
- Location: upper layer of leaf to access sunlight [1 mark]
- High concentration/density of chloroplasts [1 mark]
- Vertically elongated / tightly packed to intercept maximum light [1 mark]

a) Acceleration is the change in velocity divided by time: \(a = \frac{8.0 \text{ m/s} - 0 \text{ m/s}}{4.0 \text{ s}} = 2.0 \text{ m/s}^2\). b) Total distance is the area under the speed-time graph. Distance in the first 4.0 s is \(\frac{1}{2} \times 4.0 \text{ s} \times 8.0 \text{ m/s} = 16 \text{ m}\). Distance in the next 10.0 s is \(10.0 \text{ s} \times 8.0 \text{ m/s} = 80 \text{ m}\). Total distance is \(16 \text{ m} + 80 \text{ m} = 96 \text{ m}\). c) Work done = force * distance. During the constant speed phase, the distance is \(80 \text{ m}\). Force is \(3.0 \text{ N}\). Work done = \(3.0 \text{ N} \times 80 \text{ m} = 240 \text{ J}\).

a) 2.22 marks: 1.11 marks for correct acceleration calculation (2.0) and 1.11 marks for correct units (m/s^2). b) 3.33 marks: 1.11 marks for distance in accelerating phase (16 m), 1.11 marks for distance in constant speed phase (80 m), and 1.11 marks for correct total (96 m). c) 3.33 marks: 1.11 marks for correct formula (Work = force * distance), 1.11 marks for using the correct distance of 80 m, and 1.11 marks for final answer of 240 J.

a) Because copper(II) oxide is added in excess, when all the acid has reacted, some black copper(II) oxide solid will remain unreacted at the bottom of the beaker. b) Filtration is used to separate the insoluble copper(II) oxide residue from the soluble copper(II) sulfate filtrate. c) To obtain crystals: Heat the filtrate to evaporate some water until a saturated solution is formed (crystallisation point), leave to cool so crystals form, filter the mixture to collect the crystals, and wash with a little cold distilled water, then dry the crystals between sheets of filter paper. d) The balanced chemical equation is \(\text{CuO} + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O}\).

a) 2.22 marks: 1.11 marks for mentioning addition of excess CuO, and 1.11 marks for stating that black solid remains undissolved. b) 1.11 marks for naming filtration. c) 3.33 marks: 1.11 marks for heating/evaporating to point of crystallisation, 1.11 marks for cooling to crystallise and filtering, and 1.11 marks for drying crystals with filter paper. d) 2.22 marks: 1.11 marks for correct reactants and products, and 1.11 marks for correct balancing.

a) The word equation for photosynthesis is: carbon dioxide + water -> glucose + oxygen (in the presence of light and chlorophyll). b) Boiling the leaf in ethanol breaks down cell membranes and extracts chlorophyll from the leaf. Decolourising the leaf is necessary so that the green colour of the leaf does not mask the blue-black colour change of the iodine test. c) The uncovered green region turns blue-black because chlorophyll was present, allowing photosynthesis to occur and starch to be produced. The uncovered white region remains orange-brown (the colour of iodine) because it lacks chlorophyll, meaning no photosynthesis could occur, so no starch was produced.

a) 2.22 marks: 1.11 marks for correct reactants, 1.11 marks for correct products. b) 2.22 marks: 1.11 marks for stating it removes chlorophyll/decolourises the leaf, and 1.11 marks for explaining this allows the color change with iodine to be clearly visible. c) 4.44 marks: 1.11 marks for stating the green region turns blue-black, 1.11 marks for explaining that chlorophyll is present for photosynthesis to make starch, 1.11 marks for stating the white region remains orange-brown, and 1.11 marks for explaining that it lacks chlorophyll so no photosynthesis/starch occurred.