2025 Cambridge IGCSE Science - Combined (0653) 模拟试题及答案详解

(a) \(6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2\)

(b) (i) Destarching ensures that any starch detected at the end of the experiment was produced during the experiment, rather than being stored from before the experiment started.

(ii) The entire leaf will remain orange-brown / will not turn blue-black. This is because both factors needed for photosynthesis are restricted: the white part of the leaf lacks chlorophyll, and the green part cannot photosynthesise because carbon dioxide has been absorbed by the sodium hydroxide.

(c) Magnesium ions are required for the synthesis of chlorophyll, which is essential for absorbing light energy during photosynthesis.

(a) Correct reactant and product formulas [1]
Correct balancing of the equation [1]

(b) (i) To remove all pre-existing starch [1]
To show that any starch found was produced during the 24-hour experimental period [1]

(ii) Leaf remains orange-brown / does not turn blue-black [1]
White parts lack chlorophyll to absorb light [1]
Green parts lack carbon dioxide (absorbed by sodium hydroxide) [1]

(c) Synthesis/production of chlorophyll [1]
For absorption of light energy [1]

(a) (i) Decane molecule: \(C_{10}H_{22}\). Subtracting butane (\(C_4H_{10}\)) leaves \(C_6H_{12}\). Dividing by 3 gives \(C_2H_4\) for X.

(ii) Hydrocarbon X is ethene, which belongs to the alkenes.

(b) (i) Add aqueous bromine (bromine water) to each hydrocarbon. With butane, the solution remains orange-brown. With hydrocarbon X (ethene), the solution turns colourless (decolourises).

(ii) Displayed formula of butane:
H H H H
| | | |
H-C - C - C - C-H
| | | |
H H H H

(c) (i) A monomer is a small molecule that can join together with other identical molecules to form a polymer.

(ii) Ethene is unsaturated because it contains a carbon-to-carbon double bond (\(C=C\)). Poly(ethene) is saturated because the double bonds break during polymerisation, leaving only carbon-to-carbon single bonds (\(C-C\)).

(a) (i) \(C_2H_4\) [1]
(ii) Alkene(s) [1]

(b) (i) Add bromine water / aqueous bromine [1]
Butane: remains orange/brown/yellow [1]
Hydrocarbon X: turns colourless / decolourises [1]
(ii) Displayed structure of butane showing all C and H atoms and all single bonds [1]

(c) (i) Simple/small molecule used to make a polymer [1]
(ii) Ethene contains carbon-carbon double bonds (\(C=C\)) [1]
Poly(ethene) contains only carbon-carbon single bonds (\(C-C\)) [1]

(a) (i) \(a = \frac{v - u}{t} = \frac{12 - 0}{8.0} = 1.5\text{ m/s}^2\).

(ii) \(E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 65 \times 12^2 = 0.5 \times 65 \times 144 = 4680\text{ J}\).

(b) \(\Delta E_p = mgh = 65 \times 9.8 \times 15 = 9555\text{ J}\).

(c) Work is also done against resistive forces such as air resistance and friction between the tyres and the ground. This work is dissipated as thermal energy to the surroundings.

(a) (i) Use of \(a = \frac{\Delta v}{\Delta t}\) [1]
\(1.5\) [1]
\(\text{m/s}^2\) [1]

(ii) Use of \(E_k = \frac{1}{2}mv^2\) [1]
\(4680\text{ J}\) [1]

(b) Use of \(E_p = mgh\) [1]
\(9555\text{ J}\) [1]

(c) Work is done against friction / air resistance [1]
Thermal energy is lost to the surroundings [1]

(a) Active immunity involves the body's own immune system producing antibodies in response to an antigen, which leads to the production of memory cells and long-term protection. Passive immunity involves receiving antibodies from an external source (e.g., breast milk or injection), which does not produce memory cells and only provides short-term protection.

(b) Physical barriers physically prevent the entry of pathogens. An example is the skin, which blocks pathogens from entering tissues. Chemical barriers destroy or inhibit pathogens using chemical substances. An example is hydrochloric acid in the stomach, which kills pathogens ingested with food.

(c) Antibiotics work by disrupting specific cell structures or metabolic pathways found in bacteria (such as cell wall synthesis or ribosome function). Viruses do not have these structures or their own metabolism, as they live and reproduce inside host cells, making them unaffected by antibiotics.

(a) Active immunity involves the production of antibodies by the host's own body [1]
Active immunity produces memory cells / provides long-term immunity [1]
Passive immunity is temporary / does not produce memory cells / antibodies are imported [1]

(b) Physical barrier description: traps or blocks pathogens [1]
Example of physical barrier: skin / nasal hairs / mucus [1]
Chemical barrier description: kills or destroys pathogens chemically [1]
Example of chemical barrier: stomach acid / tears / saliva [1]

(c) Antibiotics target cell structures / cell walls / metabolic machinery of bacteria [1]
Viruses lack cell walls / cell structures / metabolic machinery / reproduce inside host cells [1]

(a) (i) For parallel resistors: \(\frac{1}{R_p} = \frac{1}{15} + \frac{1}{15} = \frac{2}{15}\), so \(R_p = 7.5\ \Omega\).

(ii) Using Ohm's Law: \(I = \frac{V}{R_p} = \frac{12}{7.5} = 1.6\text{ A}\).

(b) (i) The parallel combination is in series with the 10 \(\Omega\) resistor: \(R_{total} = R_p + R_3 = 7.5 + 10 = 17.5\ \Omega\).

(ii) First find the new total current: \(I_{new} = \frac{V}{R_{total}} = \frac{12}{17.5} \approx 0.686\text{ A}\). The potential difference across the series resistor is: \(V_{10} = I_{new} \times 10 = 0.686 \times 10 \approx 6.9\text{ V}\).

(a) (i) Formula for parallel resistors used correctly [1]
\(7.5\ \Omega\) [1]

(ii) Use of \(I = \frac{V}{R}\) [1]
\(1.6\text{ A}\) [1]

(b) (i) Adding series resistor to parallel resistance [1]
\(17.5\ \Omega\) [1]

(ii) Use of \(I = \frac{V_{total}}{R_{total}}\) to find new current (\(0.686\text{ A}\)) [1]
Use of \(V = IR\) for 10 \(\Omega\) resistor [1]
\(6.9\text{ V}\) (allow 6.8 to 6.9 V) [1]

(a) Balanced equation: \(2CH_3OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 4H_2O(l)\).

(b) Bond breaking absorbs energy, which is an endothermic process. Bond making releases energy, which is an exothermic process. Since more energy is released during the formation of new bonds in \(CO_2\) and \(H_2O\) than is absorbed to break the bonds in \(CH_3OH\) and \(O_2\), the reaction is exothermic.

(c) The reaction pathway diagram must show reactants on a higher energy level than the products. The curve rises from the reactants to a peak (representing the transition state) and then drops down to the product level. The activation energy \(E_a\) is the upward arrow from the reactants to the peak. The overall energy change \(\Delta H\) is represented by a downward arrow from the reactant level to the product level.

(a) Correct balancing coefficients: 2, 3, 2, 4 [2]
(Award 1 mark if coefficients are balanced but not in lowest whole-number ratio, e.g. 1, 1.5, 1, 2)

(b) Bond breaking absorbs energy AND bond making releases energy [1]
Energy released in bond making is greater than energy absorbed in bond breaking [1]
Result is overall loss of heat / energy transfer to surroundings [1]

(c) Reactants at higher energy than products [1]
Curve showing peak between reactants and products [1]
Activation energy (\(E_a\)) correctly labelled from reactant level to peak [1]
Overall energy change (\(\Delta H\)) correctly labelled with downward arrow [1]

(a) Balanced reduction equation: \(Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2\).

(b) (i) Silicon dioxide (sand) is the main acidic impurity. Calcium carbonate decomposes to form calcium oxide, which is a basic oxide. This basic oxide reacts with the acidic silicon dioxide in a neutralisation reaction to form calcium silicate (slag), which can then be easily removed.

(ii) Thermal decomposition: \(CaCO_3 \rightarrow CaO + CO_2\)
Formation of slag: \(CaO + SiO_2 \rightarrow CaSiO_3\)

(a) Correct reactants and products: \(Fe_2O_3 + CO \rightarrow Fe + CO_2\) [1]
Correct balancing: 1, 3, 2, 3 [2]

(b) (i) Silicon dioxide is an acidic impurity [1]
Calcium oxide (formed from limestone) is basic [1]
Neutralisation reaction occurs to form slag / calcium silicate [1]

(ii) Thermal decomposition equation: \(CaCO_3 \rightarrow CaO + CO_2\) [1.5]
Slag formation equation: \(CaO + SiO_2 \rightarrow CaSiO_3\) [1.5]

(a) Conduction occurs when the particles at the hotter end of the metal wall gain kinetic energy and vibrate more. These particles collide with neighbouring, cooler particles, transferring energy. Additionally, metals contain free (delocalised) electrons which move rapidly through the metal structure, colliding with ions and quickly transferring energy throughout the material.

(b) (i) Within the liquid, particles are moving at different speeds. The most energetic particles near the surface of the liquid can overcome the attractive forces of surrounding molecules and escape into the air as a gas.

(ii) The rate of evaporation can be increased by increasing the temperature of the liquid, increasing the surface area of the liquid, or increasing the wind speed (draught) across the surface.

(a) Vibration of particles/atoms passed on by collisions [1]
Presence of free / delocalised electrons [1]
Electrons diffuse / move through the metal, transferring energy [1]

(b) (i) Particles have a distribution of kinetic energies [1]
More energetic particles are located near the surface [1]
These particles overcome attractive forces and escape as gas [1]

(ii) Increase temperature [1]
Increase surface area [1]
Increase draft / wind speed across surface [1]

(a) The reactants are hydrogen gas ($2\text{H}_2\text{(g)}$) and oxygen gas ($\text{O}_2\text{(g)}$), and the product is gaseous water vapour ($2\text{H}_2\text{O(g)}$). Balancing the equation gives:
$$2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(g)}$$

(b) An exothermic reaction is a chemical change during which thermal energy is released to the surroundings, resulting in an increase in temperature.

(c) Breaking chemical bonds requires energy input (endothermic process), whereas making new bonds releases energy (exothermic process). The reaction is exothermic because the total energy released when forming the new $\text{O-H}$ bonds in water is greater than the total energy absorbed to break the $\text{H-H}$ and $\text{O=O}$ bonds in the reactants.

(d) On a reaction pathway diagram for this exothermic reaction:
- The energy level of the products is drawn lower than that of the reactants.
- The reaction pathway rises to a peak representing the transition state.
- The activation energy is shown as the vertical distance from the reactants' energy level up to the peak of the curve.

(a)
- Correct formulae and balancing: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ [1]
- Correct state symbols: $\text{(g)}$ for all three substances [1]

(b)
- Idea of transferring/releasing thermal/heat energy to the surroundings [1]

(c)
- State that energy is taken in/absorbed for bond breaking and released/given out during bond making [1]
- State that energy released during bond making is greater than the energy taken in during bond breaking [1]
- Identify that $\text{H-H}$ and/or $\text{O=O}$ bonds are broken, and $\text{H-O}$ bonds are formed [1]

(d)
- State/show that the products are at a lower energy level than the reactants [1]
- Describe/draw a curve that rises to a peak (representing the energy barrier) before dropping to the products [1]
- Define activation energy as the vertical distance from the reactants' level to the peak [1]