2025 Cambridge IGCSE Science - Combined (0653) 模擬試題連答案詳解

Part (a): The balanced chemical equation for photosynthesis is: \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\).

Part (b): 1. Place the leaf in boiling water for about 30 seconds to kill the cells and stop chemical reactions. 2. Place the leaf in hot ethanol, heated in a water bath (rather than over a direct flame because ethanol is highly flammable) to remove chlorophyll and decolorize the leaf. 3. Rinse the leaf in cold water to soften it. 4. Spread the leaf on a white tile and add a few drops of iodine solution. Positive result (starch present): color changes from yellow-brown to blue-black. Negative result (starch absent): color remains yellow-brown.

Part (c): Boiling the leaf in water breaks down the cell walls and denatures cell membranes. This increases permeability, allowing the ethanol to penetrate and extract chlorophyll, and later allowing the iodine solution to reach starch grains inside the chloroplasts.

(a) [3 marks]
- 1 mark for correct reactants: \(\text{CO}_2 + \text{H}_2\text{O}\)
- 1 mark for correct products: \(\text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2\)
- 1 mark for correct balancing: \(6, 6 \rightarrow 1, 6\)

(b) [4 marks]
- 1 mark for boiling the leaf in ethanol using a water bath (safety precaution must be specified: no naked flame because ethanol is flammable).
- 1 mark for rinsing leaf in water and adding iodine solution.
- 1 mark for positive result color change (yellow-brown to blue-black).
- 1 mark for negative result color change (remains yellow-brown).

(c) [1.88 marks]
- 1 mark for stating that boiling kills the leaf cells.
- 0.88 marks for explaining that this breaks down cell membranes, making the cells permeable to the iodine solution.

Part (a): A pathogen is defined as a disease-causing organism.

Part (b): A vaccine contains dead, weakened, or harmless forms of a pathogen. This pathogen has specific antigens on its surface. When introduced into the body, white blood cells called lymphocytes recognize these antigens as foreign. The lymphocytes divide and produce specific antibodies that bind to the antigens. After the pathogen is cleared, some of these lymphocytes remain in the blood as memory cells. If the real, active pathogen enters the body in the future, these memory cells can rapidly produce large quantities of antibodies to destroy the pathogen before illness occurs.

Part (c): Two differences between active and passive immunity are:
1. Active immunity involves the production of antibodies by the person's own immune system (and results in memory cells), whereas passive immunity involves receiving antibodies from an external source (no memory cells made).
2. Active immunity provides long-term protection, whereas passive immunity provides only short-term/temporary protection.

(a) [1 mark]
- 1 mark for stating: disease-causing organism.

(b) [5 marks]
- 1 mark for stating that a vaccine contains dead, weakened, or harmless pathogens/antigens.
- 1 mark for stating that lymphocytes recognize the foreign antigens.
- 1 mark for stating that lymphocytes produce specific antibodies.
- 1 mark for stating that memory cells are produced.
- 1 mark for stating that memory cells allow a faster, larger antibody response if reinfected.

(c) [2.88 marks]
- 1 mark for stating that active immunity produces memory cells, whereas passive immunity does not.
- 1 mark for stating that active immunity is long-lasting, whereas passive immunity is short-lived.
- 0.88 marks for stating that active immunity requires time to develop, whereas passive immunity is immediate.

Part (a): An exothermic reaction is a chemical reaction that transfers thermal energy to its surroundings. This causes the temperature of the surroundings to increase.

Part (b): In an energy level diagram for an exothermic reaction:
1. The reactants are placed at a higher energy level than the products.
2. The activation energy (\(E_a\)) is represented by an arrow pointing upwards from the reactant level to the peak of the energy curve.
3. The enthalpy change (\(\Delta H\)) is represented by a downward-pointing arrow from the reactant energy level to the product energy level.

Part (c): Chemical reactions involve two stages: bond breaking and bond making. Breaking bonds is an endothermic process (it requires an input of energy), whereas making new bonds is an exothermic process (it releases energy). A reaction is endothermic if the energy required to break the existing bonds in the reactants is greater than the energy released when new bonds are formed in the products.

(a) [2 marks]
- 1 mark for stating that thermal energy is transferred to the surroundings.
- 1 mark for stating that the temperature of the surroundings increases.

(b) [4 marks]
- 1 mark for stating that reactants are drawn at a higher energy level than products.
- 1 mark for describing an energy barrier/curve rising from reactants before falling to products.
- 1 mark for describing activation energy (\(E_a\)) as the energy difference between reactants and the peak (pointing up).
- 1 mark for describing the enthalpy change (\(\Delta H\)) as the energy difference between reactants and products (pointing down / negative).

(c) [2.88 marks]
- 1 mark for stating that bond breaking requires/absorbs energy.
- 1 mark for stating that bond making releases energy.
- 0.88 marks for stating that in an endothermic reaction, more energy is absorbed during bond breaking than is released during bond making.

Part (a): The balanced equation is \(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_8\text{H}_{18} + \text{C}_2\text{H}_4\).

Part (b): The chemical test uses aqueous bromine (bromine water), which is orange/brown. When added to octane (an alkane), there is no reaction in the dark, so the solution remains orange. When added to ethene (an alkene containing a \(\text{C}=\text{C}\) double bond), an addition reaction occurs, decoloring the bromine water (it turns colorless).

Part (c): The reaction is an addition reaction. Ethene reacts with bromine to form 1,2-dibromoethane. Its molecular structure consists of two carbon atoms joined by a single covalent bond, with each carbon atom bonded to two hydrogen atoms and one bromine atom (\(\text{CH}_2\text{Br}-\text{CH}_2\text{Br}\)).

(a) [2 marks]
- 1 mark for correct formula of reactants and products (\(\text{C}_{10}\text{H}_{22}\), \(\text{C}_8\text{H}_{18}\), \(\text{C}_2\text{H}_4\)).
- 1 mark for equation being fully balanced (1:1:1 ratio).

(b) [4 marks]
- 1 mark for using bromine water / aqueous bromine as the reagent.
- 1 mark for stating that bromine water is orange/yellow/brown initially.
- 1 mark for stating that with octane, the mixture remains orange/yellow/brown (no change).
- 1 mark for stating that with ethene, the mixture turns colorless (decolorized).

(c) [2.88 marks]
- 1 mark for naming the reaction: addition reaction.
- 1 mark for identifying the product: 1,2-dibromoethane (accept dibromoethane).
- 0.88 marks for describing the structure (single bond between carbons, with each carbon having two H and one Br attached).

Part (a): The three raw materials are coke (carbon), limestone (calcium carbonate), and air (which provides oxygen).

Part (b) (i): The chemical equation is: \(\text{CO}_2 + \text{C} \rightarrow 2\text{CO}\).
Part (b) (ii): The chemical equation is: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\).

Part (c): Limestone (\(\text{CaCO}_3\)) decomposes in the high heat of the furnace to produce calcium oxide (\(\text{CaO}\)) and carbon dioxide (\(\text{CO}_2\)). Calcium oxide is a basic metal oxide, which reacts with the acidic silicon dioxide (\(\text{SiO}_2\)) impurities to form calcium silicate (\(\text{CaSiO}_3\)), commonly called slag.

(a) [3 marks]
- 1 mark for coke / carbon.
- 1 mark for limestone / calcium carbonate.
- 1 mark for air / oxygen (accept hot air).

(b) [4 marks]
- 1 mark for correct reactant and product formulas in (i) (\(\text{CO}_2 + \text{C} \rightarrow \text{CO}\)).
- 1 mark for correctly balancing (i) (\(\text{CO}_2 + \text{C} \rightarrow 2\text{CO}\)).
- 1 mark for correct reactant and product formulas in (ii) (\(\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow \text{Fe} + \text{CO}_2\)).
- 1 mark for correctly balancing (ii) (\(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\)).

(c) [1.88 marks]
- 1 mark for explaining that thermal decomposition of calcium carbonate produces calcium oxide (\(\text{CaO}\)).
- 0.88 marks for explaining that the basic oxide (\(\text{CaO}\)) reacts with acidic silica (\(\text{SiO}_2\)) to form calcium silicate / slag.

Part (a)(i): Acceleration \(a = \frac{v - u}{t} = \frac{20\text{ m/s} - 0}{8.0\text{ s}} = 2.5\text{ m/s}^2\).

Part (a)(ii): Distance traveled is equal to the area under the speed-time graph: \(d = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8.0\text{ s} \times 20\text{ m/s} = 80\text{ m}\).

Part (b): Kinetic energy \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 1200\text{ kg} \times (20\text{ m/s})^2 = 0.5 \times 1200 \times 400 = 240\,000\text{ J}\) (or \(240\text{ kJ}\)).

Part (c): Gravitational potential energy \(\Delta E_p = m g \triangle h = 1200\text{ kg} \times 9.8\text{ m/s}^2 \times 15\text{ m} = 176\,400\text{ J}\) (or \(176.4\text{ kJ}\)).

(a) [4 marks]
- 1 mark for formula: \(a = \frac{\Delta v}{t}\) or substituting values correctly.
- 1 mark for calculation of acceleration: \(2.5\) and correct unit: \(\text{m/s}^2\).
- 1 mark for formula or method for distance: \(d = \frac{1}{2} v t\) or \(d = u t + \frac{1}{2} a t^2\).
- 1 mark for calculation of distance: \(80\text{ m}\).

(b) [2 marks]
- 1 mark for correct formula: \(E_k = \frac{1}{2} m v^2\) and correct substitution.
- 1 mark for correct answer: \(240\,000\text{ J}\) (or \(240\text{ kJ}\)).

(c) [2.88 marks]
- 1 mark for correct formula: \(E_p = m g h\).
- 1 mark for substituting correct values: \(1200 \times 9.8 \times 15\) (accept \(1200 \times 10 \times 15 = 180\,000\text{ J}\) if student explicitly states \(g=10\text{ m/s}^2\)).
- 0.88 marks for correct calculation of GPE: \(176\,400\text{ J}\) (or \(176.4\text{ kJ}\)) (or \(180\,000\text{ J}\) with \(g=10\)).

Part (a): First, find the combined resistance of the two \(6.0\ \Omega\) resistors in parallel:
\(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6.0} + \frac{1}{6.0} = \frac{2}{6.0} = \frac{1}{3.0}\ \Omega^{-1}\)
Therefore, \(R_p = 3.0\ \Omega\).

Next, add this to the series resistor of \(4.0\ \Omega\):
\(R_{\text{total}} = R_s + R_p = 4.0\ \Omega + 3.0\ \Omega = 7.0\ \Omega\).

Part (b): Using Ohm's Law: \(I = \frac{V}{R_{\text{total}}} = \frac{6.0\text{ V}}{7.0\ \Omega} \approx 0.86\text{ A}\) (or \(0.857\text{ A}\)).

Part (c): Power supplied is given by \(P = I \times V = 0.857\text{ A} \times 6.0\text{ V} \approx 5.14\text{ W}\) (or \(5.1\text{ W}\)). Alternatively, \(P = \frac{V^2}{R} = \frac{36}{7} \approx 5.14\text{ W}\). The unit of power is the Watt (W).

(a) [4 marks]
- 1 mark for correct formula for parallel resistors: \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\).
- 1 mark for calculating the parallel resistance: \(3.0\ \Omega\).
- 1 mark for adding the series resistor to get \(R_{\text{total}} = R_1 + R_p\).
- 1 mark for calculating total resistance: \(7.0\ \Omega\).

(b) [2 marks]
- 1 mark for correct formula: \(I = \frac{V}{R}\).
- 1 mark for correct current calculation: \(0.86\text{ A}\) (accept \(0.857\text{ A}\)).

(c) [2.88 marks]
- 1 mark for correct formula: \(P = I V\) or \(P = I^2 R\) or \(P = \frac{V^2}{R}\).
- 1 mark for correct calculation: \(5.14\text{ W}\) (allow ecf from (b)).
- 0.88 marks for the correct unit: \(\text{W}\) or Watts.

Part (a): When one end of the metal rod is heated, the metal ions at that end gain thermal energy and vibrate more vigorously. They collide with neighboring ions, transferring kinetic energy along the rod. In addition, metals contain free (delocalized) electrons. These electrons gain kinetic energy, move rapidly through the metal structure, and collide with ions and other electrons further down the rod, transferring thermal energy much faster than vibration alone.

Part (b): 1. Temperature: A higher temperature means that water molecules have a higher average kinetic energy. Consequently, a larger proportion of molecules have sufficient energy to overcome the intermolecular attractive forces and escape from the liquid surface. 2. Wind: Moving air (wind) carries away the water vapor molecules that have just escaped from the puddle's surface. This prevents them from returning to the liquid and maintains a steep concentration gradient above the surface, speeding up evaporation.

Part (c): Two differences between evaporation and boiling are:
1. Evaporation occurs at any temperature (below the boiling point), whereas boiling occurs only at a specific temperature (the boiling point).
2. Evaporation occurs only at the surface of the liquid, whereas boiling occurs throughout the volume of the liquid (producing bubbles).

(a) [3 marks]
- 1 mark for describing that ions at the hot end vibrate more vigorously.
- 1 mark for stating that these vibrations are passed on to neighboring ions via collisions.
- 1 mark for describing how free/delocalized electrons move rapidly through the metal, transferring kinetic energy to distant ions.

(b) [3.88 marks]
- 1 mark for stating that temperature increases the kinetic energy of the water molecules.
- 1 mark for explaining that more molecules have enough energy to escape the intermolecular forces at the surface.
- 1 mark for stating that wind carries away the evaporated water vapor molecules from the surface.
- 0.88 marks for explaining that this prevents molecules from returning / increases net evaporation rate.

(c) [2 marks]
- 1 mark for difference 1 (e.g., surface vs. throughout liquid / bubbles vs. no bubbles).
- 1 mark for difference 2 (e.g., any temperature vs. specific boiling point).

(a) Work done by the cable is calculated using the formula:
\(W = F \times d\)
\(W = 200\text{ N} \times 120\text{ m} = 24\,000\text{ J}\) (or \(24\text{ kJ}\))

(b) The change in gravitational potential energy is given by:
\(\Delta E_p = mgh\)
\(\Delta E_p = 65\text{ kg} \times 9.8\text{ m/s}^2 \times 30\text{ m} = 19\,110\text{ J}\) (or \(19.11\text{ kJ}\))

(c) Explanation: The work done by the cable is greater than the gained GPE because some energy is transferred to the surroundings as thermal energy due to friction between the skis and the snow and air resistance.
Calculation of work done against friction:
\(W_{\text{friction}} = W_{\text{total}} - \Delta E_p\)
\(W_{\text{friction}} = 24\,000\text{ J} - 19\,110\text{ J} = 4\,890\text{ J}\)

(d) Power output is calculated as:
\(P = \frac{W}{t}\)
\(P = \frac{24\,000\text{ J}}{40\text{ s}} = 600\text{ W}\) (or \(0.6\text{ kW}\))

(a) [2 marks total]:
- 1 mark for the correct formula \(W = F \times d\) or correct substitution \(200 \times 120\).
- 1 mark for the correct calculation of \(24\,000\text{ J}\) (or \(24\text{ kJ}\)) with appropriate unit.

(b) [2.88 marks total]:
- 1 mark for the correct formula \(\Delta E_p = mgh\).
- 1 mark for correct substitution \(65 \times 9.8 \times 30\).
- 0.88 marks for the correct answer of \(19\,110\text{ J}\) (allow \(19\,500\text{ J}\) if \(g = 10\text{ m/s}^2\) is used).

(c) [2 marks total]:
- 1 mark for explaining that energy is lost to heat/thermal energy due to work being done against resistive forces/friction.
- 1 mark for subtracting GPE from total work to get \(4\,890\text{ J}\) (allow \(4\,500\text{ J}\) if using \(19\,500\text{ J}\)).

(d) [2 marks total]:
- 1 mark for the correct power formula \(P = \frac{W}{t}\) or substitution \(\frac{24\,000}{40}\).
- 1 mark for correct calculation of \(600\text{ W}\) (or \(0.6\text{ kW}\)) with correct unit.