2025 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) 模拟试题及答案详解

(a) (i) The shoots will grow / bend upwards.

(ii) Negative gravitropism (or negative geotropism).

(iii) Auxin is produced in the shoot tip and moves downwards due to gravity, accumulating on the lower side of the horizontal shoot. A higher concentration of auxin on the lower side promotes faster cell elongation compared to the upper side, causing the shoot to bend upwards.

(b) (i) A hormone is a chemical substance, produced by a gland and carried by the blood, which alters the activity of one or more specific target organs.

(ii) Any three of:
- increased heart rate
- increased breathing rate
- dilated pupils
- increased blood glucose concentration
- diversion of blood flow to the muscles / away from the digestive system.

(a) (i)
- shoots grow / bend upwards [1] (Reject: grow towards light)

(ii)
- negative gravitropism / negative geotropism [1]

(iii)
- auxin is made in the shoot tip [1]
- auxin moves / diffuses to the lower side of the shoot [1]
- high concentration of auxin on the lower side stimulates cell elongation, causing the shoot to bend upwards [1]

(b) (i) Any two from:
- chemical substance produced by a gland [1]
- transported in the blood / plasma [1]
- alters the activity of one or more specific target organs [1]

(ii) Any three from:
- increased heart rate [1]
- increased breathing rate [1]
- widened / dilated pupils [1]
- increased blood glucose level [1]
- diversion of blood to muscles / away from gut [1]

(a) (i) The percentage of oxygen is lower in expired air because oxygen diffuses from the alveoli into the blood capillaries in the lungs, as it is needed by cells for aerobic respiration.

(ii) The percentage of carbon dioxide is higher in expired air because carbon dioxide is produced as a waste product of cellular respiration and diffuses from the blood capillaries into the alveoli to be exhaled.

(b) (i) Alveoli (or air sacs).

(ii) Three of:
- thin walls (one cell thick) which provide a short diffusion distance
- very large surface area for diffusion
- surrounded by a dense network of capillaries to maintain a steep concentration gradient
- moist lining which allows gases to dissolve before diffusing.

(c) Goblet cells secrete sticky mucus that traps inhaled dust, dirt, and pathogens. Ciliated cells have tiny hair-like projections (cilia) that beat in unison to sweep the mucus up and out of the trachea and airways toward the throat to be swallowed.

(a) (i)
- oxygen diffuses into the blood / is absorbed in the lungs [1]
- because it is used for respiration [1]

(ii)
- carbon dioxide is produced during respiration [1]
- and diffuses out of the blood into the lungs / alveoli to be exhaled [1]

(b) (i)
- alveoli / air sacs [1] (Accept: alveolus)

(ii) Any three from:
- thin walls / one cell thick [1]
- large surface area [1]
- surrounded by a dense network of capillaries / good blood supply [1]
- moist lining [1]

(c)
- goblet cells secrete / produce mucus which traps pathogens / dust [1]
- ciliated cells have cilia that sweep / beat the mucus upwards / away from the lungs [1]

(a)
- **A**: ovary
- **B function**: site of fertilisation / transport of egg from ovary to uterus
- **C**: uterus
- **D function**: ring of muscle at opening of uterus / holds developing fetus in place during pregnancy

(b) (i) The placenta allows oxygen, glucose, amino acids, and antibodies to diffuse from the maternal blood to the fetal blood. It also allows carbon dioxide and urea to diffuse from the fetal blood to the maternal blood for excretion, and it secretes hormones (like progesterone) to maintain the uterine lining during pregnancy.

(ii) The bloods do not mix to prevent high maternal blood pressure from damaging delicate fetal capillaries, and to protect the fetus from maternal immune responses (or mismatched blood groups causing agglutination).

(c) 46 (or 23 pairs).

(a) 1 mark for each correct entry:
- **A**: ovary [1]
- **B function**: site of fertilisation / transport of egg to uterus [1]
- **C**: uterus (or womb) [1]
- **D function**: keeps fetus in place / ring of muscle at entrance to uterus [1]

(b) (i) Any three from:
- allows diffusion of oxygen / nutrients / glucose / amino acids from mother to fetus [1]
- allows diffusion of carbon dioxide / urea from fetus to mother [1]
- secretes progesterone / hormones to maintain pregnancy [1]
- provides passive immunity / transfers antibodies [1]

(ii) Any two from:
- high maternal blood pressure would damage fetal capillaries [1]
- prevents mismatch of blood groups / agglutination [1]
- protects fetus from some pathogens / toxins in maternal blood [1]

(c)
- 46 (or 23 pairs) [1]

(a) (i) Runoff / leaching of chemical fertilizers from agricultural land, and discharge of untreated sewage into water bodies.

(ii)
1. Fertilizer runoff increases nutrient levels in the water, causing rapid growth of algae (an algal bloom).
2. The algal bloom covers the surface, blocking sunlight from reaching plants underneath.
3. Underwater plants die due to lack of light for photosynthesis.
4. Decomposers (bacteria) feed on the dead plant matter and multiply rapidly.
5. The bacteria respire aerobically, depleting dissolved oxygen levels in the water, which causes fish to die of suffocation.

(b) (i) Deforestation reduces the number of trees available to absorb carbon dioxide from the atmosphere during photosynthesis. Additionally, when cut trees are burned or decay, they release stored carbon back into the atmosphere as carbon dioxide.

(ii) Soil erosion (or flooding / habitat destruction / loss of biodiversity).

(a) (i) 1 mark for each correct activity:
- leaching / runoff of fertilizers [1]
- release of untreated sewage / animal waste [1]

(ii) Any five from:
- increased growth of algae / algal bloom on surface [1]
- algae block light from reaching submerged plants [1]
- submerged plants die due to lack of photosynthesis [1]
- bacteria / decomposers multiply rapidly as they feed on dead plants [1]
- bacteria respire aerobically [1]
- oxygen concentration in water decreases / depleted [1]
- fish die from lack of oxygen [1]

(b) (i)
- fewer trees to photosynthesize and absorb carbon dioxide [1]
- burning of wood / decay of cut trees releases carbon dioxide [1]

(ii) Any one from:
- soil erosion [1]
- flooding [1]
- habitat destruction / loss of biodiversity [1]

(a) (i) Fractional distillation. (ii) Crude oil is heated and vaporised. The vapour rises up a fractionating column which is hot at the bottom and cooler at the top. The fractions condense at different heights depending on their boiling points. (b) (i) High temperature and a catalyst (e.g., alumina or silica). (ii) \(C_7H_{16}\). (c) (i) It contains a carbon-carbon double bond (\(C=C\)). (ii) Add aqueous bromine (bromine water) to both hydrocarbons. Propene turns the orange-brown bromine water colourless (decolourises it). Propane does not react and the mixture remains orange-brown.

(a) (i) Fractional distillation [1]. (ii) Crude oil is heated/vaporised [1]; fractions condense at different temperatures/boiling points [1]. (b) (i) High temperature / heat (approx. 500-600 °C) [1]; catalyst (alumina/silica/zeolite) [1]. (ii) \(C_7H_{16}\) [1]. (c) (i) Contains a double bond (\(C=C\)) [1]. (ii) Test: add bromine water / aqueous bromine [1]; result with propene: turns colourless / decolourises [1]; result with propane: remains orange/brown [1].

(a) (i) Anode. (ii) Positive electrode: chlorine gas; Negative electrode: zinc. (iii) In solid zinc chloride, the ions are held in fixed positions in a giant lattice and cannot move. When molten, the giant structure breaks down and the ions are free to move and carry the electric current. (b) (i) Oxygen gas. Test: relights a glowing splint. (ii) A pink-brown solid is deposited on the electrode. (iii) The mass of the positive electrode decreases because the copper atoms lose electrons to form copper ions, which dissolve into the solution.

(a) (i) Anode [1]. (ii) Chlorine / \(Cl_2\) [1]; zinc / \(Zn\) [1]. (iii) Ions in solid are in fixed positions/cannot move [1]; ions in molten state are free to move and carry charge [1]. (b) (i) Oxygen / \(O_2\) [1]; Test: relights a glowing splint [1]. (ii) Pink/brown/red-brown solid deposit [1]. (iii) Mass decreases [1]; copper atoms oxidised to copper ions / \(Cu \rightarrow Cu^{2+} + 2e^-\)[1].

(a) (i) zinc oxide + sulfuric acid -> zinc sulfate + water. (ii) To ensure all of the dilute sulfuric acid has reacted. (iii) By filtration (the excess zinc oxide remains on the filter paper as residue). (iv) Heat the solution to evaporate most of the water until a saturated solution is formed. Leave the hot solution to cool so that crystals form. Filter off the crystals and dry them between filter papers. (b) (i) Add dilute hydrochloric acid followed by aqueous barium chloride. A white precipitate is formed. (ii) A white precipitate forms with a few drops of sodium hydroxide. The white precipitate dissolves in excess sodium hydroxide to form a colourless solution.

(a) (i) zinc oxide + sulfuric acid -> zinc sulfate + water [1]. (ii) To ensure all acid is fully reacted/neutralised [1]. (iii) Filtration [1]. (iv) Heat/evaporate solution to crystallization point / saturated solution [1]; cool to allow crystal growth [1]; filter and dry crystals between filter papers [1]. (b) (i) Acidify with dilute hydrochloric acid / nitric acid AND add aqueous barium chloride / barium nitrate [1]; white precipitate [1]. (ii) White precipitate formed initially [1]; precipitate dissolves in excess to give a colourless solution [1].

(a) (i) Nitrogen is 78% (accept 78% to 80%); Oxygen is 21% (accept 20% to 21%). (ii) Methane (or water vapour). (iii) Greenhouse gases absorb infrared radiation reflected/radiated from the Earth's surface and trap heat in the atmosphere. (b) (i) Filtration: to remove insoluble solids/particles from the water; Chlorination: to kill harmful bacteria/microorganisms. (ii) Add the liquid to anhydrous copper(II) sulfate. It turns from white to blue. (Alternatively, add to anhydrous cobalt(II) chloride, which turns from blue to pink).

(a) (i) Nitrogen: 78% [1]; Oxygen: 21% [1]. (ii) Methane / water vapour [1]. (iii) Greenhouse gases absorb infrared radiation [1]; trapped heat warms the atmosphere / prevents heat from escaping into space [1]. (b) (i) Filtration [1]; to remove suspended insoluble solids [1]; Chlorination [1]; to kill bacteria/pathogens [1]. (ii) Anhydrous copper(II) sulfate turns blue (OR anhydrous cobalt(II) chloride turns pink) [1].

(a) (i) The line is a straight diagonal going upwards, representing constant acceleration (or uniform acceleration).
(ii) Acceleration is calculated using the formula:
\(a = \frac{\Delta v}{\Delta t}\)
\(a = \frac{8.0\text{ m/s} - 0\text{ m/s}}{4.0\text{ s}} = 2.0\text{ m/s}^2\)
The unit for acceleration is \(\text{m/s}^2\) (or \(\text{m s}^{-2}\)).
(iii) The total distance is the area under the speed-time graph. This can be calculated as the area of a trapezium, or the sum of the areas of a triangle (from 0 to 4 s) and a rectangle (from 4 to 10 s):
\(\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = 0.5 \times 4.0\text{ s} \times 8.0\text{ m/s} = 16.0\text{ m}\)
\(\text{Area of rectangle} = \text{base} \times \text{height} = (10.0\text{ s} - 4.0\text{ s}) \times 8.0\text{ m/s} = 6.0\text{ s} \times 8.0\text{ m/s} = 48.0\text{ m}\)
\(\text{Total distance} = 16.0\text{ m} + 48.0\text{ m} = 64\text{ m}\)

(b) (i) Kinetic energy is calculated using the formula:
\(E_k = \frac{1}{2} m v^2\)
\(E_k = 0.5 \times 65\text{ kg} \times (8.0\text{ m/s})^2 = 0.5 \times 65 \times 64 = 2080\text{ J}\) (or \(2.08\text{ kJ}\))
(ii) The body uses chemical energy from food reserves to produce muscular movement.

(a) (i)
- Constant / uniform acceleration [1] (Reject: just 'acceleration' or 'speed increases')

(ii)
- \(a = \frac{\Delta v}{\Delta t}\) or \(\frac{8.0}{4.0}\) [1]
- \(2.0\) [1]
- \(\text{m/s}^2\) or \(\text{m s}^{-2}\) [1]

(iii)
- distance = area under graph (stated or shown in working) [1]
- calculation of either triangle area (\(16\text{ m}\)) or rectangle area (\(48\text{ m}\)) [1]
- \(64\text{ m}\) [1]

(b) (i)
- \(E_k = \frac{1}{2} m v^2\) or \(0.5 \times 65 \times 8.0^2\) [1]
- \(2080\text{ J}\) (or \(2.08\text{ kJ}\)) [1]

(ii)
- Chemical (energy) [1]

(a) (i) The speed of light decreases (slows down) because glass is optically denser than air.
(ii) Use the refractive index formula:
\(n = \frac{\sin i}{\sin r}\)
Rearranging for \(\sin r\):
\(\sin r = \frac{\sin i}{n} = \frac{\sin 45.0^\circ}{1.50}\)
\(\sin r = \frac{0.7071}{1.50} = 0.4714\)
\(r = \sin^{-1}(0.4714) \approx 28.1^\circ\)
(iii) Total internal reflection can only happen if:
1. The light is travelling in the more optically dense medium (glass) towards the less dense medium (air).
2. The angle of incidence is greater than the critical angle for the glass-air boundary.

(b) (i) The loudspeaker cone vibrates back and forth, pushing on nearby air molecules. These vibrations are passed along as air particles collide, creating longitudinal waves consisting of compressions (regions of high pressure where particles are close together) and rarefactions (regions of low pressure where particles are spread apart).
(ii) Use the wave equation:
\(v = f \lambda\)
Rearranging for wavelength \(\lambda\):
\(\lambda = \frac{v}{f} = \frac{340\text{ m/s}}{250\text{ Hz}} = 1.36\text{ m}\)

(a) (i)
- Decreases / slows down [1]

(ii)
- \(n = \frac{\sin i}{\sin r}\) (or rearranged) [1]
- \(\sin r = \frac{\sin 45^\circ}{1.50}\) or \(\sin r = 0.4714\) [1]
- \(28.1^\circ\) (Accept: \(28^\circ\) or \(28.13^\circ\)) [1]

(iii)
- Ray must be travelling from a more dense medium to a less dense medium [1]
- Angle of incidence must be greater than the critical angle [1]

(b) (i)
- Particles vibrate / oscillate back and forth parallel to wave direction [1]
- Creating compressions and rarefactions (or passing energy via collisions) [1]

(ii)
- \(\lambda = \frac{v}{f}\) or \(\frac{340}{250}\) [1]
- \(1.36\text{ m}\) [1]

(a) (i) For parallel resistors:
\(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\)
\(\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{12.0} = \frac{2}{12.0} + \frac{1}{12.0} = \frac{3}{12.0}\)
\(R_p = \frac{12.0}{3} = 4.0\text{ }\Omega\)
(ii) The total current \(I\) from the battery is:
\(I = \frac{V}{R_p} = \frac{12.0\text{ V}}{4.0\text{ }\Omega} = 3.0\text{ A}\)
(iii) In a parallel circuit, the voltage across each parallel branch is equal to the voltage of the supply (if there are no other series components).
Voltage across \(R_1 = 12.0\text{ V}\).

(b) (i) Power \(P\) dissipated in a resistor can be calculated using \(P = \frac{V^2}{R}\) or \(P = I^2 R\).
Using \(P = \frac{V^2}{R}\) for resistor \(R_2\):
\(P = \frac{(12.0\text{ V})^2}{12.0\text{ }\Omega} = \frac{144}{12.0} = 12.0\text{ W}\)
Alternatively, current in \(R_2\) is \(I_2 = \frac{12.0\text{ V}}{12.0\text{ }\Omega} = 1.0\text{ A}\), so \(P = I_2 \times V = 1.0\text{ A} \times 12.0\text{ V} = 12.0\text{ W}\).
(ii) Connecting a resistor in series with the parallel branch:
1. Increases the total resistance of the circuit (from \(4.0\text{ }\Omega\) to \(4.0 + 12.0 = 16.0\text{ }\Omega\)).
2. This decreases the total current in the circuit.
3. Because current flows through the new series resistor, there is a potential drop across it, leaving less voltage across the parallel branch (and thus less voltage across \(R_1\)).
4. Since \(I_1 = \frac{V_{\text{parallel}}}{R_1}\), a lower voltage across \(R_1\) results in a decreased current through \(R_1\).

(a) (i)
- \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\) or \(\frac{1}{6} + \frac{1}{12}\) [1]
- \(\frac{3}{12}\) leading to \(R_p = 4.0\text{ }\Omega\) [1]

(ii)
- \(I = \frac{V}{R}\) or \(\frac{12.0}{4.0}\) [1]
- \(3.0\text{ A}\) [1]

(iii)
- \(12.0\text{ V}\) [1]

(b) (i)
- \(P = \frac{V^2}{R}\) or \(P = I V\) with \(I_2 = 1.0\text{ A}\) [1]
- \(12.0\text{ W}\) [1]

(ii)
- Current through \(R_1\) decreases [1]
- Total resistance of the circuit increases (which decreases total current) [1]
- Voltage across parallel branch / resistor \(R_1\) decreases [1]

(a) (i) An alpha particle is a helium nucleus, consisting of 2 protons and 2 neutrons. A beta-minus particle is a high-energy, high-speed electron emitted from the nucleus.
(ii) In a nuclear reaction, the mass numbers and atomic numbers must balance on both sides of the equation.
For mass numbers: \(210 = A + 4 \Rightarrow A = 210 - 4 = 206\).
For atomic numbers: \(84 = Z + 2 \Rightarrow Z = 84 - 2 = 82\).
(iii) Number of neutrons = \(\text{nucleon number } (A) - \text{proton number } (Z) = 206 - 82 = 124\).

(b) (i) Half-life is defined as the time taken for half the nuclei of a radioactive isotope in a sample to decay (or the time taken for the activity of a sample to decrease to half of its initial value).
(ii) Number of half-lives elapsed:
\(n = \frac{414\text{ days}}{138\text{ days}} = 3\text{ half-lives}\)
After 1 half-life: \(800\text{ Bq} \rightarrow 400\text{ Bq}\)
After 2 half-lives: \(400\text{ Bq} \rightarrow 200\text{ Bq}\)
After 3 half-lives: \(200\text{ Bq} \rightarrow 100\text{ Bq}\)
(iii) Alpha radiation is the most strongly ionising because of its relatively large mass and \(+2\) charge. Ionising radiation is hazardous because it can damage DNA, leading to mutations, causing cancers, or directly destroying cells.

(a) (i)
- Alpha: helium nucleus / 2 protons and 2 neutrons [1]
- Beta: high-speed / high-energy electron [1]

(ii)
- \(A = 206\) [1]
- \(Z = 82\) [1]

(iii)
- \(124\) [1]

(b) (i)
- Time taken for half the radioactive nuclei in a sample to decay (or for activity / count rate to halve) [1]

(ii)
- Shows that 3 half-lives have elapsed (\(\frac{414}{138} = 3\)) [1]
- \(100\text{ Bq}\) [1]

(iii)
- Alpha (radiation) [1]
- Cell mutation / cancer / cell damage / cell death [1]