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

(a) The palisade mesophyll layer contains the most chloroplasts. The cells are long, narrow, and arranged vertically to maximize the path length of light passing through. They are tightly packed together to absorb as much light as possible. (b) The gas produced is oxygen. The test is to insert a glowing splint into the gas; if it is oxygen, the splint will relight. (c) As carbon dioxide concentration increases, the rate of photosynthesis initially increases rapidly because carbon dioxide is a reactant and acts as a limiting factor. Eventually, the rate plateaus because carbon dioxide is no longer the limiting factor; light intensity or temperature becomes the limiting factor.

(a) [3 marks] Palisade mesophyll (1 mark). Cells are vertically elongated or tightly packed (1 mark). High density of chloroplasts (1 mark). (b) [2 marks] Oxygen identified (1 mark). Glowing splint relights (1 mark). (c) [3.88 marks] Rate increases initially (1 mark). Rate plateaus/levels off (1 mark). Carbon dioxide is the limiting factor initially (1 mark). Another factor becomes limiting at high concentrations (0.88 marks).

(a) Bond breaking is endothermic and bond making is exothermic. A reaction is exothermic when the total energy released when new bonds form in the products is greater than the total energy absorbed to break existing bonds in the reactants. (b) In an exothermic reaction pathway diagram, the reactants are at a higher energy level than the products. The curve rises to a peak and then drops. Activation energy \(E_a\) is the energy barrier to start the reaction (reactants to peak), and the overall energy change \(\Delta H\) is the net energy difference (reactants to products). (c) \(\Delta H = 1450\text{ kJ/mol} - 1850\text{ kJ/mol} = -400\text{ kJ/mol}\). Since the value is negative, the reaction is exothermic.

(a) [3 marks] Bond breaking absorbs energy AND bond making releases energy (1 mark). Energy released in bond making is greater than energy absorbed in bond breaking (1 mark). Net transfer of thermal energy to surroundings (1 mark). (b) [3 marks] Reactants at higher energy level than products (1 mark). \(E_a\) correctly described as peak energy minus reactant energy (1 mark). \(\Delta H\) correctly described as product energy minus reactant energy (1 mark). (c) [2.88 marks] Correct subtraction: \(1450 - 1850\) (1 mark). Correct negative sign and value: \(-400\text{ kJ/mol}\) (0.88 marks). Concludes reaction is exothermic (1 mark).

(a) In a longitudinal wave, particles vibrate parallel to the direction of wave travel, while in a transverse wave, they vibrate perpendicular to it. Sound is a longitudinal wave. (b) Using Snell's Law: \(n = \frac{\sin(i)}{\sin(r)} = \frac{\sin(38^\circ)}{\sin(24^\circ)} = \frac{0.6157}{0.4067} \approx 1.51\). (c) Using the wave equation: \(v = f \lambda \Rightarrow \lambda = \frac{v}{f} = \frac{3.0 \times 10^8\text{ m/s}}{4.6 \times 10^{14}\text{ Hz}} = 6.52 \times 10^{-7}\text{ m}\).

(a) [3 marks] Longitudinal wave: vibrations parallel to energy transfer (1 mark). Transverse wave: vibrations perpendicular to energy transfer (1 mark). Sound is longitudinal (1 mark). (b) [3 marks] State formula: \(n = \frac{\sin(i)}{\sin(r)}\u200b\) (1 mark). Correct substitution: \(\frac{\sin(38)}{\sin(24)}\u200b\) (1 mark). Final value: \(1.5\) or \(1.51\) (1 mark). (c) [2.88 marks] State formula: \(v = f \lambda\) (1 mark). Substitution and calculation: \(6.52 \times 10^{-7}\) (1 mark). Correct unit: \(\text{m}\) or meters (0.88 marks).

(a) The active site of an enzyme has a highly specific 3D shape that is complementary only to a specific substrate. Like a key fitting into a lock, only this substrate can bind to the active site to form an enzyme-substrate complex. (b) As temperature rises, molecules gain kinetic energy and move faster. This increases the frequency of collisions between enzyme and substrate molecules, and a higher proportion of these collisions have sufficient energy to overcome the activation energy, increasing the reaction rate. (c) At high temperatures, the heat breaks the weak chemical bonds holding the enzyme's specific 3D structure together. This alters the shape of the active site so that the substrate can no longer bind. The enzyme is denatured.

(a) [3 marks] Active site has a specific shape (1 mark). Substrate shape is complementary (1 mark). Only specific substrate fits to form enzyme-substrate complex (1 mark). (b) [3 marks] Molecules gain kinetic energy / move faster (1 mark). More frequent collisions between enzyme and substrate (1 mark). More successful/productive collisions per unit time (1 mark). (c) [2.88 marks] High temperature breaks bonds within enzyme (1 mark). Active site shape changes / substrate cannot fit (1 mark). Enzyme is denatured (0.88 marks) (reject 'killed').

(a) Solid copper(II) oxide is added in excess to a beaker of warm dilute sulfuric acid until no more dissolves. Excess is added to ensure that all of the acid is completely reacted. The mixture is filtered to remove the unreacted, insoluble copper(II) oxide. (b) The balanced equation is: \(\text{CuO(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{CuSO}_4\text{(aq)} + \text{H}_2\text{O(l)}\). (c) Heating hydrated copper(II) sulfate causes it to lose its water of crystallization. The blue crystals turn into a white anhydrous powder, and water droplets condense near the top of the test tube. Word equation: hydrated copper(II) sulfate \(\rightarrow\) anhydrous copper(II) sulfate + water.

(a) [3 marks] Warm acid and add excess CuO to neutralize all acid (1 mark). Filter to remove excess insoluble CuO (1 mark). Heat filtrate to crystallize and dry crystals (1 mark). (b) [3 marks] Correct formulae for reactants: \(\text{CuO}\) and \(\text{H}_2\text{SO}_4\) (1 mark). Correct formulae for products: \(\text{CuSO}_4\) and \(\text{H}_2\text{O}\) (1 mark). Correct state symbols: \(\text{(s), (aq), (aq), (l)}\) (1 mark). (c) [2.88 marks] Blue crystals turn to white powder AND water/condensation seen (1 mark). Hydrated copper(II) sulfate (1 mark). Anhydrous copper(II) sulfate + water (0.88 marks) (accept reversible arrow).

(a) Acceleration is given by: \(a = \frac{v - u}{t} = \frac{18\text{ m/s} - 0}{12\text{ s}} = 1.5\text{ m/s}^2\). (b) Kinetic energy is given by: \(KE = \frac{1}{2} m v^2 = 0.5 \times 25000\text{ kg} \times (18\text{ m/s})^2 = 12500 \times 324 = 4,050,000\text{ J}\) (or \(4.05\text{ MJ}\)). (c) Power is the rate of doing work, so: \(\text{Work Done} = \text{Power} \times \text{time} = 375,000\text{ W} \times 12\text{ s} = 4,500,000\text{ J}\) (or \(4.5\text{ MJ}\)). The work done by the motor (\(4.5\text{ MJ}\)) is greater than the kinetic energy gained (\(4.05\text{ MJ}\)) because in any real system, some energy is lost as heat/sound due to friction/air resistance.

(a) [2.5 marks] Correct formula or substitution: \(18 / 12\) (1 mark). Answer: \(1.5\) (1 mark). Unit: \(\text{m/s}^2\) (0.5 marks). (b) [3.38 marks] Correct formula: \(KE = \frac{1}{2}mv^2\) (1 mark). Substitution: \(0.5 \times 25000 \times 18^2\) (1 mark). Answer: \(4,050,000\) or \(4.05 \times 10^6\) (1 mark). Unit: \(\text{J}\) or Joules (0.38 marks). (c) [3 marks] Correct formula: \(W = P \times t\) (1 mark). Calculation: \(375000 \times 12 = 4.5 \times 10^6\text{ J}\) (1 mark). Explains difference due to resistive forces / energy lost as heat (1 mark).

(a) The test is bromine water (aqueous bromine). When added to an alkene, the orange-brown solution becomes colorless. When added to an alkane, there is no color change (remains orange-brown). (b) Catalytic cracking involves heating long-chain alkanes to vaporize them, then passing them over a hot catalyst (such as silica or alumina) at high temperatures (approx. \(600-700^\circ\text{C}\)). It is done to meet the high demand for shorter-chain alkanes (used as fuels) and alkenes (used to make polymers). (c) Complete combustion equation: \(\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\). Propene structure contains 3 carbon atoms, with one double bond: \(\text{H}_2\text{C}=\text{CH}-\text{CH}_3\).

(a) [3 marks] Aqueous bromine / bromine water (1 mark). Alkene de-colorizes / turns colorless (1 mark). Alkane remains orange/brown (1 mark). (b) [3 marks] High temperature AND catalyst (1 mark). Breaks long-chain alkanes to shorter-chain hydrocarbons (1 mark). Shorter chains are more useful as fuels/polymers (1 mark). (c) [2.88 marks] Balanced equation: \(\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\) (1.88 marks) (deduct 1 mark if unbalanced). Correct displayed structure of propene with carbon-carbon double bond (1 mark).

(a) Electric current is the rate of flow of charge. Current \(I = \frac{Q}{t} = \frac{36\text{ C}}{12\text{ s}} = 3.0\text{ A}\). (b) For resistors in parallel: \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3.0} + \frac{1}{6.0} = \frac{2}{6.0} + \frac{1}{6.0} = \frac{3}{6.0} = \frac{1}{2.0}\). Thus, \(R_p = 2.0\ \Omega\). (c) Total current \(I = \frac{V}{R_p} = \frac{9.0\text{ V}}{2.0\ \Omega} = 4.5\text{ A}\). Total power supplied \(P = V \times I = 9.0\text{ V} \times 4.5\text{ A} = 40.5\text{ W}\).

(a) [3 marks] Definition: rate of flow of charge (1 mark). Calculation: \(3.0\) (1 mark). Unit: \(\text{A}\) or Amperes (1 mark). (b) [3 marks] Formula: \(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\) (1 mark). Substitution: \(\frac{1}{3} + \frac{1}{6}\) (1 mark). Answer: \(2.0\ \Omega\) (1 mark). (c) [2.88 marks] Correct current calculation: \(4.5\text{ A}\) (1 mark). Power formula: \(P = VI\) or \(I^2 R\) (1 mark). Answer: \(40.5\text{ W}\) (0.88 marks).

(a)(i)
Using the formula for acceleration:
\(a = \frac{v - u}{t}\)
\(a = \frac{6.0\text{ m/s} - 0\text{ m/s}}{3.0\text{ s}} = 2.0\text{ m/s}^2\)

(a)(ii)
Using Newton's second law:
\(F = m \times a\)
\(F = 0.50\text{ kg} \times 2.0\text{ m/s}^2 = 1.0\text{ N}\)

(b)
The resultant force is \(0\text{ N}\).
According to Newton's first law, if an object travels at a constant velocity, the forces acting on it are balanced, meaning there is no resultant force (acceleration is zero).

(c)(i)
The kinetic energy of the car before braking is:
\(E_k = \frac{1}{2}mv^2\)
\(E_k = \frac{1}{2} \times 0.50\text{ kg} \times (6.0\text{ m/s})^2 = 9.0\text{ J}\)
Since the car is brought to a complete stop, all of its kinetic energy is dissipated. Therefore, the work done by the braking force is equal to the initial kinetic energy, which is \(9.0\text{ J}\).

(c)(ii)
Using the formula for work done:
\(W = F \times d\)
\(9.0\text{ J} = F \times 9.0\text{ m}\)
\(F = \frac{9.0\text{ J}}{9.0\text{ m}} = 1.0\text{ N}\)

(a)(i) [2 marks total]
- Correct formula or substitution: \(a = \frac{6.0}{3.0}\) [1 mark]
- Correct final value with unit: \(2.0\text{ m/s}^2\) (accept \(\text{m/s}^2\) or \(\text{m s}^{-2}\)) [1 mark]

(a)(ii) [2 marks total]
- Correct substitution using \(F = ma\): \(0.50 \times 2.0\) (allow ecf from (a)(i)) [1 mark]
- Correct value with unit: \(1.0\text{ N}\) [1 mark]

(b) [2 marks total]
- State that resultant force is \(0\text{ N}\) (or zero) [1 mark]
- Explain that constant velocity implies zero acceleration / balanced forces [1 mark]

(c)(i) [2 marks total]
- Correct substitution into kinetic energy formula: \(\frac{1}{2} \times 0.50 \times 6.0^2\) [1 mark]
- Correct calculation leading to \(9.0\text{ J}\) AND equating this to work done to stop [1 mark]

(c)(ii) [1 mark total]
- Correct calculation and unit: \(1.0\text{ N}\) (derived from \(F = \frac{W}{d} = \frac{9.0}{9.0}\)) [1 mark]