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

(a) The balanced chemical 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) The rate of photosynthesis can be measured by counting the number of oxygen bubbles released by the plant per unit of time (e.g., bubbles per minute), or by measuring the volume of oxygen gas collected in a gas syringe over a set period. One control variable is the temperature of the water bath, or the concentration of carbon dioxide (using a fixed concentration of sodium hydrogencarbonate).

(c) At very high light intensities, the rate of photosynthesis stops increasing and levels off because light intensity is no longer the limiting factor. Some other factor, such as temperature or carbon dioxide concentration, is in short supply and limits the rate of the reaction.

(a) [3 marks total]:
- Correct reactants and products: \(\text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2\) (1 mark)
- Balanced correctly: \(6\text{CO}_2\) and \(6\text{H}_2\text{O}\) (1 mark)
- Balanced correctly: \(6\text{O}_2\) (1 mark)

(b) [3 marks total]:
- Method of measurement: counting bubbles or measuring gas volume in a set time (1 mark)
- Specifying the gas is oxygen (1 mark)
- One correct control variable (e.g., water temperature, concentration of dissolved carbon dioxide) (1 mark)

(c) [3 marks total]:
- Light intensity is no longer the limiting factor (1 mark)
- Another factor is now limiting (1 mark)
- Example of another limiting factor given: temperature or carbon dioxide concentration (1 mark)

(a) Arteries have thick, muscular, and elastic walls with a narrow lumen to withstand and maintain high pressure as blood is pumped away from the heart. Veins have thinner walls, a wider lumen, and contain valves to prevent the backflow of blood, which flows at a much lower pressure back to the heart.

(b) Double circulation means that for every complete circuit of the body, blood passes through the heart twice (pulmonary circulation to the lungs, and systemic circulation to the rest of the body). This is important because it maintains high pressure to body tissues, ensuring fast delivery of oxygen and nutrients, and keeps oxygenated and deoxygenated blood separate.

(c) The pulmonary vein.

(a) [4 marks total]:
- Arteries have thicker muscular/elastic walls than veins (1 mark)
- Arteries have a narrower lumen, veins have a wider lumen (1 mark)
- Arteries must withstand/maintain high blood pressure (1 mark)
- Veins contain valves to prevent backflow of low-pressure blood (1 mark)

(b) [3 marks total]:
- Double circulation definition: blood passes through the heart twice for each complete circuit of the body (1 mark)
- Importance: maintains high blood pressure / faster delivery of oxygen to respiring tissues (1 mark)
- Importance: keeps oxygenated and deoxygenated blood separate (1 mark)

(c) [2 marks total]:
- Pulmonary (1 mark)
- Vein (1 mark)

(a) Acceleration is given by:
\(a = \frac{v - u}{t} = \frac{8.0 - 0}{10} = 0.80\text{ m/s}^2\)

(b) Kinetic energy is calculated using:
\(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 65 \times (8.0)^2 = 0.5 \times 65 \times 64 = 2080\text{ J}\)

(c) Work done against resistance:
\(W = F \times d = 15\text{ N} \times 40\text{ m} = 600\text{ J}\)

(d) The kinetic energy of the bicycle and rider is transferred into thermal energy in the brakes and surroundings due to friction.

(a) [2 marks total]:
- Formula or substitution: \(\frac{8.0}{10}\) (1 mark)
- Correct value and unit: \(0.80\text{ m/s}^2\) (1 mark)

(b) [3 marks total]:
- Formula: \(\frac{1}{2}mv^2\) (1 mark)
- Correct calculation: \(2080\) (1 mark)
- Correct unit: \(\text{J}\) or Joules (1 mark)

(c) [2 marks total]:
- Formula/working: \(15 \times 40\) (1 mark)
- Correct answer: \(600\text{ J}\) (1 mark)

(d) [2 marks total]:
- Identifying kinetic energy (1 mark)
- Transferred to thermal/heat energy (1 mark)

(a) Catalytic cracking involves heating long-chain alkanes to high temperatures (approx. 500 °C) in the presence of a catalyst (such as alumina or silica). This breaks large, less useful hydrocarbon molecules into smaller, more useful molecules (shorter-chain alkanes like petrol, and alkenes like ethene which are used to make plastics).

(b) This is an addition (or hydration) reaction.
The displayed formula of ethanol showing all bonds is:
```
H H
| |
H - C - C - O - H
| |
H H
```

(c) Reagent: Aqueous bromine (bromine water).
With ethane: The solution remains orange-brown (no reaction in the dark).
With ethene: The solution turns from orange-brown to colourless (it decolourises).

(a) [3 marks total]:
- High temperature and catalyst (e.g., alumina/silica) (1 mark)
- Breaks down long-chain molecules into shorter-chain molecules (1 mark)
- Reason: to produce more useful products (e.g., alkenes/petrol) to meet demand (1 mark)

(b) [3 marks total]:
- Reaction type: addition or hydration (1 mark)
- Correct carbon-carbon single bond and five C-H bonds (1 mark)
- Correct C-O and O-H single bonds (1 mark)

(c) [3 marks total]:
- Reagent: Bromine water / aqueous bromine (1 mark)
- Ethane: stays orange/brown/yellow (1 mark)
- Ethene: decolourises / turns colourless (1 mark; reject 'clear')

(a) At the anode, chlorine gas is produced. At the cathode, hydrogen gas is produced.

(b) At the anode:
\(2\text{Cl}^-(\text{aq}) \rightarrow \text{Cl}_2(\text{g}) + 2\text{e}^-\)
At the cathode:
\(2\text{H}^+(\text{aq}) + 2\text{e}^- \rightarrow \text{H}_2(\text{g})\) (or \(2\text{H}_2\text{O}(\text{l}) + 2\text{e}^- \rightarrow \text{H}_2(\text{g}) + 2\text{OH}^-(\text{aq})\))

(c) Water dissociates into \(\text{H}^+\) and \(\text{OH}^-\) ions, while sodium chloride dissociates into \(\text{Na}^+\) and \(\text{Cl}^-\) ions. As \(\text{H}^+\) is discharged at the cathode and \(\text{Cl}^-\) is discharged at the anode, the concentration of \(\text{Na}^+\) and \(\text{OH}^-\) ions remaining in the solution increases. This forms sodium hydroxide solution, which is highly alkaline.

(a) [2 marks total]:
- Anode: chlorine (1 mark)
- Cathode: hydrogen (1 mark)

(b) [4 marks total]:
- Anode equation reactant and product correct: \(2\text{Cl}^- \rightarrow \text{Cl}_2\) (1 mark)
- Anode equation fully balanced with charges and state symbols: \(2\text{Cl}^-(\text{aq}) \rightarrow \text{Cl}_2(\text{g}) + 2\text{e}^-\) (1 mark)
- Cathode equation reactant and product correct: \(2\text{H}^+ \rightarrow \text{H}_2\) (1 mark)
- Cathode equation fully balanced with charges and state symbols: \(2\text{H}^+(\text{aq}) + 2\text{e}^- \rightarrow \text{H}_2(\text{g})\) (1 mark)

(c) [3 marks total]:
- Identifies that \(\text{H}^+\) and \(\text{Cl}^-\) ions are removed/discharged from solution (1 mark)
- Leaves behind \(\text{Na}^+\) and \(\text{OH}^-\) ions in solution (1 mark)
- Identifies that these ions form sodium hydroxide, which is basic/alkaline (1 mark)

(a) First, find the resistance of the parallel combination of the two \(6\ \Omega\) resistors:
\(\frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \implies R_p = 3\ \Omega\)
Next, add the series \(4\ \Omega\) resistor:
\(R_{\text{total}} = 4\ \Omega + 3\ \Omega = 7\ \Omega\)

(b) The total current is found using Ohm's Law:
\(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{7\ \Omega} \approx 1.71\text{ A}\) (or \(1.7\text{ A}\))

(c) The potential difference across the \(4\ \Omega\) resistor is:
\(V_4 = I \times 4\ \Omega = 1.714\text{ A} \times 4\ \Omega \approx 6.86\text{ V}\) (or \(6.9\text{ V}\))
The potential difference across the parallel combination is:
\(V_p = V_{\text{total}} - V_4 = 12\text{ V} - 6.86\text{ V} = 5.14\text{ V}\) (or \(5.1\text{ V}\))
Alternatively, using the parallel resistance:
\(V_p = I \times R_p = 1.714\text{ A} \times 3\ \Omega \approx 5.14\text{ V}\)

(a) [3 marks total]:
- Correct formula/calculation for parallel part: \(R_p = 3\ \Omega\) (1 mark)
- Correct formula for series addition: \(R_{\text{total}} = R_1 + R_p\) (1 mark)
- Correct answer: \(7\ \Omega\) (1 mark)

(b) [2 marks total]:
- Correct formula: \(I = V/R\) (1 mark)
- Correct calculation: \(1.71\text{ A}\) or \(1.7\text{ A}\) (1 mark)

(c) [4 marks total]:
- Correct calculation for PD across the \(4\ \Omega\) resistor: \(1.71\text{ A} \times 4 = 6.86\text{ V}\) (2 marks, allow 1 mark for formula/method)
- Correct method to find the remaining PD (either subtraction from 12 V or \(I \times R_p\)) (1 mark)
- Correct calculation for PD across parallel branch: \(5.14\text{ V}\) or \(5.1\text{ V}\) (1 mark)

(a) Sound is a longitudinal wave. It is transmitted when a vibrating source causes air molecules to vibrate back and forth parallel to the direction of wave travel. This creates alternating regions of high pressure where molecules are close together (compressions) and low pressure where they are spread apart (rarefactions).

(b) Using the wave equation:
\(v = f \lambda \implies \lambda = \frac{v}{f}\)
\(\lambda = \frac{1500\text{ m/s}}{250\text{ Hz}} = 6.0\text{ m}\)

(c) Two differences:
1. Sound waves are longitudinal, whereas light waves are transverse.
2. Sound waves require a medium (matter) to travel through and cannot travel through a vacuum, whereas light waves can travel through a vacuum.
3. Sound waves travel much slower (approx. 330 m/s in air) than light waves (approx. \(3 \times 10^8\text{ m/s}\)).

(a) [3 marks total]:
- Longitudinal wave identified (1 mark)
- Compressions described as regions of high pressure/particles close together (1 mark)
- Rarefactions described as regions of low pressure/particles spread out (1 mark)

(b) [3 marks total]:
- Correct formula: \(v = f \lambda\) or rearrange \(\lambda = v/f\) (1 mark)
- Correct substitution: \(1500 / 250\) (1 mark)
- Correct answer with unit: \(6.0\text{ m}\) (1 mark)

(c) [3 marks total]:
- States any one valid difference (e.g., longitudinal vs transverse) (1 mark)
- States a second valid difference (e.g., sound requires medium, light does not; or speed difference) (1 mark)
- Clear contrast made for both waves (1 mark)

(a) Increasing the concentration of hydrochloric acid increases the rate of reaction. This is because there are more reactant particles per unit volume, which increases the frequency of collisions between the acid particles and the calcium carbonate.

(b) Increasing the temperature increases the kinetic energy of the particles, so they move faster and collide more frequently. Crucially, a much larger proportion of the colliding particles have energy equal to or greater than the activation energy, leading to a higher rate of successful collisions.

(c) The gas produced is carbon dioxide (\(\text{CO}_2\)). The test is to bubble the gas through limewater (aqueous calcium hydroxide). The positive result is that the limewater turns cloudy or milky.

(a) [3 marks total]:
- Rate increases (1 mark)
- More particles per unit volume (1 mark)
- Frequency of collisions increases (1 mark; reject 'more collisions' without time/frequency context)

(b) [3 marks total]:
- Particles have more kinetic energy / move faster (1 mark)
- More particles have energy greater than or equal to activation energy (1 mark)
- Increased frequency of successful/effective collisions (1 mark)

(c) [3 marks total]:
- Gas identified as carbon dioxide (1 mark)
- Test: bubble through limewater (1 mark)
- Observation: turns cloudy / milky (1 mark)