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

Bile plays two critical roles in fat digestion. First, it is alkaline and neutralizes the acidic mixture of food and gastric juice entering the duodenum from the stomach. This creates the optimum pH for the enzyme lipase to function. Second, bile emulsifies fats, breaking down large fat droplets into much smaller droplets. This physical process significantly increases the surface area of the fats, allowing lipase to digest them more rapidly.

Award 1 mark for neutralising the acid / providing alkaline conditions.
Award 1 mark for providing the optimum pH for lipase action.
Award 1 mark for emulsifying fats / breaking large droplets to small droplets to increase surface area.

During a chemical reaction, energy is required to break existing chemical bonds, which is an endothermic process. In this reaction, energy is absorbed to break the \(\text{H}-\text{H}\) and \(\text{Cl}-\text{Cl}\) bonds. Conversely, energy is released when new chemical bonds are formed, which is an exothermic process. Here, energy is released during the formation of the new \(\text{H}-\text{Cl}\) bonds. Since the reaction is exothermic overall, the amount of energy released when forming the new bonds is greater than the amount of energy taken in to break the original bonds.

Award 1 mark for stating that bond breaking takes in energy.
Award 1 mark for stating that bond making releases energy.
Award 1 mark for explaining that more energy is released when making bonds than is absorbed when breaking bonds.

A vaccine contains weakened or harmless pathogens that carry specific antigens. When introduced into the body, these antigens are detected by lymphocytes. The lymphocytes are triggered to produce specific antibodies that bind to the antigens. During this immune response, memory cells are also produced. If the person is infected by the actual live pathogen in the future, these memory cells recognize the antigens immediately and produce specific antibodies much more rapidly and in greater quantity, preventing disease.

Award 1 mark for mentioning that the vaccine / weakened pathogens contain antigens.
Award 1 mark for stating that lymphocytes are stimulated to produce specific antibodies.
Award 1 mark for stating that memory cells are produced (providing rapid future response).

When ethene gas is bubbled through aqueous bromine (bromine water), the solution changes colour from orange/brown to colourless (it is decolourised). This is because ethene is an unsaturated hydrocarbon, meaning it contains a reactive double carbon-to-carbon bond (\(\text{C}=\text{C}\)) which easily undergoes an addition reaction with bromine. Ethane, on the other hand, is a saturated hydrocarbon containing only single carbon-to-carbon bonds (\(\text{C}-\text{C}\)), so it does not react with bromine water under standard conditions.

Award 1 mark for the correct colour change: orange / yellow / brown to colourless (do not accept 'clear').
Award 1 mark for identifying ethene as unsaturated / having a carbon-to-carbon double bond.
Award 1 mark for explaining that ethane is saturated / contains only single bonds (and thus does not undergo addition easily).

In concentrated aqueous sodium chloride, both sodium ions (\(\text{Na}^+\)) and hydrogen ions (\(\text{H}^+\)) are attracted to the cathode (the negative electrode). Because hydrogen is lower in the reactivity series than sodium, the hydrogen ions are preferentially discharged. Each hydrogen ion gains one electron at the cathode (reduction) to form hydrogen atoms, which then pair up to form hydrogen gas molecules (\(\text{H}_2\)).

Award 1 mark for identifying the product as hydrogen (gas).
Award 1 mark for stating that hydrogen ions (\(\text{H}^+\)) are attracted to the negative electrode / cathode.
Award 1 mark for explaining that hydrogen ions gain electrons / are reduced (to form hydrogen molecules).

Once stable fusion ends, a high-mass star expands into a red supergiant, whereas a low-mass star like our Sun expands into a red giant. A high-mass star then undergoes a violent explosion called a supernova, whereas a low-mass star sheds its outer layers as a planetary nebula, leaving behind a white dwarf. Finally, the remains of the supernova of a high-mass star collapse into either an extremely dense neutron star or a black hole, depending on the remaining core mass.

Award 1 mark for stating that a high-mass star becomes a red supergiant (compared to red giant for low-mass).
Award 1 mark for stating that a high-mass star ends in a supernova explosion (compared to planetary nebula / white dwarf for low-mass).
Award 1 mark for stating that a high-mass star leaves behind a neutron star or black hole.

First, calculate the change in kinetic energy (useful work done) of the toy car:
\(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.50\text{ kg} \times (4.0\text{ m/s})^2 = 0.25 \times 16 = 4.0\text{ J}\).
Next, calculate the useful power using the formula:
\(P = \frac{E_k}{t} = \frac{4.0\text{ J}}{2.0\text{ s}} = 2.0\text{ W}\).

Award 1 mark for correct recall of the kinetic energy formula and substitution: \(\frac{1}{2} \times 0.50 \times (4.0)^2\).
Award 1 mark for calculating kinetic energy as \(4.0\text{ J}\).
Award 1 mark for calculating power as \(2.0\text{ W}\) (including the correct unit).

Xylem vessels have several adaptations. First, their cell walls are reinforced with a tough substance called lignin, which makes them very strong to withstand the negative pressure (tension) of the transpiration stream without collapsing, and also provides structural support to the entire plant. Second, xylem vessels are made of dead cells with no cytoplasm or organelles, and their end walls have completely broken down. This forms continuous, hollow tubes that allow water to flow upwards with minimal resistance.

Award 1 mark for mentioning lignin / thick walls to prevent collapse / provide support.
Award 1 mark for mentioning hollow tubes / no cytoplasm / no end walls to allow continuous water column / low resistance flow.
Award 1 mark for mentioning that cells are dead (preventing water absorption by osmosis / maintaining flow path).

A vaccine contains weakened or harmless forms of a pathogen, which still present specific antigens to the body's immune system. When administered, lymphocytes recognize these antigens as foreign and are stimulated to produce antibodies that have a complementary shape to bind and neutralize them. Crucially, this process also leads to the production of memory cells. These memory cells remain in the circulation for a long time, so if the person is exposed to the active, virulent pathogen in the future, the immune system can produce antibodies much more rapidly and in greater quantities, destroying the pathogen before it causes illness.

Any three from:
- Vaccine contains antigens / weakened pathogens (which are recognized by the immune system); [1]
- Lymphocytes are stimulated to produce antibodies (with complementary shapes to the antigens); [1]
- Memory cells are produced (which persist in the body); [1]
- (Upon future infection) antibodies are produced faster / in greater quantities to destroy the pathogen; [1]

1. At 30 units of light intensity, the rate of photosynthesis increases linearly as light intensity increases, which indicates that light intensity is the factor directly limiting the process.
2. At 80 units of light intensity, the graph levels off and the rate of photosynthesis becomes constant. Since further increases in light intensity do not increase the rate, light is no longer limiting; instead, another factor like temperature or carbon dioxide concentration has become the limiting factor.

1. Light intensity [1]
2. Rate of photosynthesis levels off / becomes constant / does not increase as light intensity increases [1]; therefore, another factor (e.g., carbon dioxide concentration or temperature) is limiting the rate [1]

For an exothermic reaction, the products have less chemical energy than the reactants. Therefore, the product line must be drawn below the reactant line. The activation energy ($E_a$) is the minimum energy required to start the reaction and is shown as an upward arrow from the reactant energy level to the highest point on the curve. The enthalpy change ($\Delta H$) is the overall energy difference and is shown as a downward arrow from the reactant level to the product level.

1. Product line drawn/described below reactant line [1]
2. Activation energy ($E_a$) represented as an upward arrow from reactant level to the peak of the curve [1]
3. Enthalpy change ($\Delta H$) represented as a downward arrow from reactant level to product level [1]

1. First, convert distance from km to meters (or use consistent units where km cancels):
$$H_0 = \frac{v}{d} = \frac{3.3 \times 10^3\text{ km/s}}{1.5 \times 10^{21}\text{ km}} = 2.2 \times 10^{-18}\text{ s}^{-1}$$
2. The age of the Universe is estimated as:
$$T = \frac{1}{H_0} = \frac{1}{2.2 \times 10^{-18}\text{ s}^{-1}} = 4.545 \times 10^{17}\text{ s} \approx 4.5 \times 10^{17}\text{ s}$$

1. Recall/use of $H_0 = v/d$ with values [1]; calculation to give $2.2 \times 10^{-18}\text{ s}^{-1}$ [1]
2. Recall/use of $T = 1/H_0$ to give $4.5 \times 10^{17}\text{ s}$ (allow ecf from part 1) [1]

At terminal velocity, the falling object is moving at a constant speed, meaning there is no acceleration and the resultant force is zero. Therefore, the upward force of air resistance must be equal in magnitude and opposite in direction to the downward force of gravity (weight). The arrows representing these two forces should point in opposite directions and be of equal length.

1. Vertically downward arrow labeled 'weight' / 'gravity' and vertically upward arrow labeled 'air resistance' / 'drag' [1]
2. Both arrows drawn with equal lengths to show balanced forces / resultant force is zero [1]

1. Recall the formula relating orbital speed, radius, and period:
\(v = \frac{2\pi r}{T}\)

2. Rearrange the formula to solve for the orbital period \(T\) in seconds:
\(T = \frac{2\pi r}{v} = \frac{2 \times \pi \times 1.1 \times 10^8 \text{ km}}{25 \text{ km/s}} \approx 2.76 \times 10^7 \text{ s}\)

3. Convert the period from seconds to days:
\(T \text{ (days)} = \frac{2.76 \times 10^7 \text{ s}}{86400 \text{ s/day}} \approx 319.98 \text{ days}\)

4. Rounding to two significant figures gives \(320 \text{ days}\).

1 mark for correct rearrangement of the formula or substitution of values to find the orbital period in seconds: \(T = \frac{2\pi \times 1.1 \times 10^8}{25}\) or \(2.76 \times 10^7 \text{ s}\) (accept \(2.8 \times 10^7 \text{ s}\)).
1 mark for dividing the calculated time in seconds by \(86400\) (or converting sequentially by dividing by \(3600\) and then \(24\)).
1 mark for the correct final value rounded to two significant figures: \(320 \text{ days}\) (accept values in the range \(318\)–\(322\)).

1. Calculate the final velocity \(v\) of the motorbike:
\(v = u + at = 0 + (2.5 \text{ m/s}^2 \times 6.0 \text{ s}) = 15 \text{ m/s}\)

2. Calculate the kinetic energy \(E_k\) gained by the motorbike:
\(E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 180 \text{ kg} \times (15 \text{ m/s})^2 = 20250 \text{ J}\)

3. Calculate the average power \(P\):
\(P = \frac{E_k}{t} = \frac{20250 \text{ J}}{6.0 \text{ s}} = 3375 \text{ W}\)

1 mark for calculating the final velocity of \(15 \text{ m/s}\) using \(v = at\).
1 mark for calculating the kinetic energy gained of \(20250 \text{ J}\) (or \(20 \text{ kJ}\)) using \(E_k = \frac{1}{2}mv^2\).
1 mark for dividing kinetic energy by time to obtain the average power: \(3375 \text{ W}\) (accept \(3380 \text{ W}\), \(3.4 \text{ kW}\), or \(3400 \text{ W}\)).

Method 1: Using parallel resistance
1. Find the combined resistance \(R\) of the two parallel resistors:
\(\frac{1}{R} = \frac{1}{20} + \frac{1}{30} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60} \implies R = 12\ \Omega\)

2. Convert the time to seconds:
\(t = 5.0 \text{ minutes} \times 60 \text{ s/minute} = 300 \text{ s}\)

3. Calculate the total energy using \(E = \frac{V^2 t}{R}\):
\(E = \frac{12^2 \times 300}{12} = 12 \times 300 = 3600 \text{ J}\)

Method 2: Using individual currents
1. Calculate the current in each branch:
\(I_{20} = \frac{12 \text{ V}}{20\ \Omega} = 0.6 \text{ A}\)
\(I_{30} = \frac{12 \text{ V}}{30\ \Omega} = 0.4 \text{ A}\)

2. Determine the total current:
\(I_{\text{total}} = 0.6 \text{ A} + 0.4 \text{ A} = 1.0 \text{ A}\)

3. Convert time to seconds:
\(t = 300 \text{ s}\)

4. Calculate total energy transferred:
\(E = V I_{\text{total}} t = 12 \text{ V} \times 1.0 \text{ A} \times 300 \text{ s} = 3600 \text{ J}\)

1 mark for calculating the combined resistance of \(12\ \Omega\) (using \(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\)) OR for calculating individual currents in the branches: \(0.6 \text{ A}\) and \(0.4 \text{ A}\).
1 mark for converting time to seconds: \(5.0 \text{ minutes} = 300 \text{ s}\).
1 mark for calculating the correct electrical energy: \(3600 \text{ J}\) (accept \(3.6 \text{ kJ}\)).

1. A massive star has much greater gravitational forces compressing its core.
2. This results in a significantly higher core temperature and pressure.
3. Consequently, the rate of nuclear fusion (hydrogen converting to helium) is vastly higher, consuming the hydrogen fuel at an extremely rapid rate that outweighs the extra fuel available.

1 mark: Identify that a massive star has much stronger gravitational compression, leading to higher core temperature or pressure.
1 mark: State that the rate of hydrogen fusion (or nuclear fusion) is much faster in more massive stars.
1 mark: Explain that the increased fusion rate consumes the fuel supply much faster, outweighing the larger initial mass of hydrogen.

1. The vaccine contains harmless antigens of the pathogen.
2. Lymphocytes recognize these antigens and produce specific antibodies against them.
3. Memory cells are formed and remain in the blood.
4. If the live pathogen enters the body in the future, these memory cells quickly recognize it and produce a rapid, large-scale antibody response to destroy it before symptoms develop.

1 mark: Vaccine contains antigens / dead or weakened pathogens that stimulate lymphocytes.
1 mark: Lymphocytes produce specific antibodies and memory cells.
1 mark: On subsequent infection by live pathogen, memory cells produce a rapid / large-scale antibody response.

1. Long-chain alkanes are in low demand but high abundance, whereas short-chain alkanes are in high demand as fuels.
2. Cracking requires high temperature (thermal cracking) and a catalyst (catalytic cracking).
3. The reaction produces shorter-chain alkanes, which make excellent fuels, and alkenes.
4. Alkenes are highly useful because they contain a double carbon-to-carbon bond, making them highly reactive starting materials to manufacture plastics and polymers.

1 mark: State that cracking converts long-chain hydrocarbons (low demand/surplus) to short-chain hydrocarbons (high demand).
1 mark: State the conditions required: high temperature and a catalyst (or steam).
1 mark: Explain that shorter-chain alkanes are more useful as fuels (petrol/gasoline).
1 mark: Explain that alkenes are produced, which are used as monomers to make polymers/plastics.

1. Energy is taken in (absorbed) to break the bonds in the reactants (H-I bonds).
2. Energy is given out (released) when new bonds are formed in the products (H-H and I-I bonds).
3. Since the reaction is endothermic, the energy absorbed for bond breaking is greater than the energy released during bond forming.

1 mark: State that bond breaking is an endothermic process (takes in energy).
1 mark: State that bond making is an exothermic process (releases energy).
1 mark: Explain that the energy required to break the bonds in the reactants is greater than the energy released when making the bonds in the products.

1. Bile is alkaline, so it neutralizes the acidic mixture of food (chyme) arriving from the stomach, providing an optimum pH (slightly alkaline) for the enzyme lipase to work in the small intestine.
2. Bile emulsifies fats, which means it breaks large globules of fat into many tiny droplets.
3. This physical change greatly increases the surface area of the fat droplets, allowing the lipase enzyme to break down the fats into fatty acids and glycerol much faster.

1 mark: Neutralizes acidic food / stomach acid to provide the optimum pH for enzymes (lipase).
1 mark: Emulsifies fats / breaks large fat droplets into smaller droplets.
1 mark: Increases the surface area of the fats so lipase can digest/hydrolyse them faster.

1. Magnesium ions are needed to synthesize chlorophyll, the green pigment in chloroplasts. A deficiency leads to yellow leaves (chlorosis) because chlorophyll cannot be formed, reducing the plant's ability to absorb light and photosynthesise.
2. Nitrate ions are required to make amino acids, which are built up into proteins.
3. Proteins are essential for cell division, growth, and repair.
4. Therefore, a deficiency in nitrate ions severely restricts protein synthesis, leading to stunted plant growth.

1 mark: Identify that magnesium ions are required for making chlorophyll.
1 mark: Explain that lack of magnesium causes yellow leaves (chlorosis) and reduced photosynthesis.
1 mark: Identify that nitrate ions are required for making amino acids / proteins.
1 mark: Explain that lack of nitrates restricts protein synthesis, leading to stunted growth / poor cell division.

1. Water enters root hair cells and moves across the root cortex down a water potential gradient by osmosis into the xylem.
2. Inside the xylem vessels, water forms a continuous column from the roots to the leaves.
3. Transpiration (evaporation of water from leaves) creates a suction force called transpiration pull that draws the water column upwards.
4. Cohesion (attraction between water molecules) keeps the column from breaking, while water ultimately evaporates from mesophyll cell walls into air spaces and diffuses out through open stomata.

1 mark: Water moves across the root cortex into the xylem by osmosis.
1 mark: Water is transported upwards through xylem vessels as a continuous column.
1 mark: Evaporation of water from the leaves / transpiration creates a tension or transpiration pull.
1 mark: Cohesion of water molecules keeps the continuous column of water moving upwards.

1. Each lamp can be controlled independently by its own switch, whereas in a series circuit, one switch turns all lights on or off.
2. Every parallel branch receives the full power supply voltage, ensuring all lamps shine at their designed/maximum brightness. In series, voltage is shared, making them very dim.
3. If one lamp burns out or is removed, it does not break the entire circuit, so the other lamps stay lit. In a series circuit, any break causes all lights to go out.

1 mark: Each lamp can be switched on/off independently.
1 mark: Each lamp receives the full voltage / shines at maximum/normal brightness (unlike sharing voltage in series).
1 mark: If one lamp fails/breaks, the others remain operational (the circuit is not broken).

1. Inward force: Gravitational force (gravity) acts to pull all the gas and dust inwards towards the center of the star. 2. Outward force: Radiation pressure (or thermal pressure) created by the energy released during nuclear fusion reactions in the core pushes outwards. 3. Stability: Because these two opposing forces are balanced (equal and opposite), the star remains at a constant diameter and does not collapse or expand.

Award up to 3 marks as follows: - 1 mark for identifying the inward force as gravity / gravitational pull. - 1 mark for identifying the outward force as radiation pressure / thermal pressure (resulting from nuclear fusion / energy release). - 1 mark for explaining that the forces are balanced / equal (and opposite) / in equilibrium.