2024 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) 模拟试题及答案详解

(a)(i) Structures that attract insects include brightly coloured petals and nectar/nectaries (or scent).
(ii) Pollen grains of wind-pollinated flowers are small, light, and smooth so they can easily be carried by the wind. Insect-pollinated pollen grains are larger, sticky, or spiky to adhere to insects.
(b)(i) Seeds require water (moisture), oxygen, and a suitable temperature (warmth) to germinate.
(ii) Enzymes break down stored insoluble food reserves (such as starch) into soluble nutrients (such as glucose/maltose) which are used for respiration and growth of the embryo.
(c) Vegetative propagation (or asexual reproduction/runners/tubers).

(a)(i) Award 1 mark for petals and 1 mark for nectary/nectar/scent. (Max 2 marks)
(ii) Award 1 mark for describing wind-pollinated pollen (light/smooth/small) and 1 mark for describing insect-pollinated pollen (sticky/spiky/heavy). (Max 2 marks)
(b)(i) Award 1 mark for each condition: water, oxygen, and suitable temperature/warmth. (Max 3 marks)
(ii) Award 1 mark for stating that enzymes break down insoluble stored food into soluble substances, and 1 mark for stating that these are used for growth/respiration. (Max 2 marks)
(c) Award 1 mark for vegetative propagation (or runners/tubers/asexual reproduction). (1 mark)

(a)(i) Carbon-14 has a proton number of 6. Number of protons = 6, number of electrons = 6 (in a neutral atom), and number of neutrons = mass number - proton number = 14 - 6 = 8.
(ii) They are isotopes because they have the same number of protons (6) but different numbers of neutrons (carbon-12 has 6 neutrons, carbon-14 has 8).
(b)(i) Carbon dioxide has covalent bonding, which involves the sharing of pairs of electrons between non-metal atoms (carbon and oxygen).
(ii) There is a double covalent bond between carbon and each oxygen atom, which means 4 shared electrons are involved in each carbon-oxygen double bond.
(c) Macromolecular structures like silicon(IV) oxide have very high melting and boiling points, and they do not conduct electricity (are insulators).

(a)(i) Award 1 mark for each correct sub-part: protons = 6, neutrons = 8, electrons = 6. (Max 3 marks)
(ii) Award 1 mark for stating they have the same number of protons and 1 mark for stating they have different numbers of neutrons. (Max 2 marks)
(b)(i) Award 1 mark for identifying covalent bonding and 1 mark for explaining it as the sharing of electrons. (Max 2 marks)
(ii) Award 1 mark for 4 shared electrons (or 2 pairs / double bond). (1 mark)
(c) Award 1 mark for each property: high melting/boiling point, does not conduct electricity, insoluble in water. (Max 2 marks)

(a)(i) Acceleration, \(a = \frac{v - u}{t} = \frac{8.0 - 0}{10} = 0.8\text{ m/s}^2\).
(ii) During the first 10 s (acceleration phase): \(\text{distance}_1 = \frac{1}{2} \times \text{base} \times \text{height} = 0.5 \times 10 \times 8.0 = 40\text{ m}\). During the next 20 s (constant speed phase): \(\text{distance}_2 = \text{speed} \times \text{time} = 8.0 \times 20 = 160\text{ m}\). Total distance = \(40 + 160 = 200\text{ m}\).
(b)(i) Kinetic energy, \(E_k = \frac{1}{2} m v^2 = 0.5 \times 75 \times 8.0^2 = 0.5 \times 75 \times 64 = 2400\text{ J}\).
(ii) Frictional forces, such as air resistance and friction from the road, oppose the motion. To maintain a constant speed, the cyclist must apply a forward force that is equal and opposite to these resistive forces so that the resultant force is zero.
(c) Chemical energy.

(a)(i) Award 1 mark for correct calculation (0.8) and 1 mark for the correct unit (m/s^2). (Max 2 marks)
(ii) Award 1 mark for distance during acceleration (40 m), 1 mark for distance during constant speed (160 m), and 1 mark for adding them to get 200 m. (Max 3 marks)
(b)(i) Award 1 mark for the formula or correct substitution (0.5 x 75 x 64) and 1 mark for the final answer 2400 J. (Max 2 marks)
(ii) Award 1 mark for mentioning resistive forces (friction/air resistance) and 1 mark for explaining that a forward force is needed to make the resultant force zero. (Max 2 marks)
(c) Award 1 mark for chemical energy. (1 mark)

(a)(i) The other two raw materials are coke (carbon) and limestone (calcium carbonate).
(ii) Carbon monoxide acts as a reducing agent (it reduces the iron ore to iron).
(iii) Word equation: iron(III) oxide + carbon monoxide \(\rightarrow\) iron + carbon dioxide (or chemical equation: \(Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2\)).
(b)(i) Alloy.
(ii) Pure iron consists of layers of identical atoms that slide over each other easily. Steel contains carbon/other atoms of different sizes, which disrupt the regular lattice arrangement and prevent the layers from sliding over each other.
(c) Galvanising. Zinc is more reactive than iron, so it corrodes/loses electrons preferentially (sacrificial protection) instead of the iron underneath, even if scratched.

(a)(i) Award 1 mark for coke and 1 mark for limestone (or air/oxygen). (Max 2 marks)
(ii) Award 1 mark for reducing agent / reduces iron oxide. (1 mark)
(iii) Award 1 mark for correct reactants and 1 mark for correct products in either word or balanced symbol equation. (Max 2 marks)
(b)(i) Award 1 mark for alloy. (1 mark)
(ii) Award 1 mark for stating that different sized atoms disrupt the structure and 1 mark for stating that this prevents layers of atoms from sliding over each other. (Max 2 marks)
(c) Award 1 mark for galvanising (or sacrificial protection) and 1 mark for explaining that zinc is more reactive than iron and reacts preferentially. (Max 2 marks)

(a)(i) Combined resistance in series: \(R = R_1 + R_2 = 4.0 + 8.0 = 12.0\text{ }\Omega\).
(ii) Current \(I = \frac{V}{R} = \frac{12}{12} = 1.0\text{ A}\).
(b)(i) In parallel, the potential difference across the 4.0 \(\Omega\) resistor increases to the full supply voltage of 12 V (whereas in series it was only 4.0 V).
(ii) Combined resistance in parallel: \(\frac{1}{R} = \frac{1}{4.0} + \frac{1}{8.0} = \frac{2}{8.0} + \frac{1}{8.0} = \frac{3}{8.0}\text{ }\Omega^{-1}\). Therefore, \(R = \frac{8.0}{3} \approx 2.7\text{ }\Omega\).
(c)(i) When the current becomes too high, the thin wire inside the fuse heats up and melts. This breaks the circuit, stopping the flow of electricity and preventing damage/overheating of other components.
(ii) Variable resistor (or rheostat).
(d) The magnetic field strength increases.

(a)(i) Award 1 mark for 12.0 \(\Omega\). (1 mark)
(ii) Award 1 mark for calculation (1.0) and 1 mark for the unit (A). (Max 2 marks)
(b)(i) Award 1 mark for stating that it becomes 12 V / increases. (1 mark)
(ii) Award 1 mark for parallel formula setup and 1 mark for the final calculated value (2.7 \(\Omega\)). (Max 2 marks)
(c)(i) Award 1 mark for stating that the fuse wire melts and 1 mark for explaining that this breaks the circuit / stops current flow. (Max 2 marks)
(ii) Award 1 mark for variable resistor/rheostat. (1 mark)
(d) Award 1 mark for stating that the magnetic field strength increases/strengthens. (1 mark)

(a)(i) Root hair cells have a long, narrow extension (the 'hair') which greatly increases the surface area for the absorption of water and mineral ions. They also have thin cell walls.
(ii) Osmosis.
(b)(i) Xylem.
(ii) Water evaporates from the wet surfaces of the spongy mesophyll cells into the air spaces within the leaf. This creates water vapour, which then diffuses out of the leaf into the atmosphere through the stomata down a concentration gradient.
(c)(i) Phloem.
(ii) Translocation; sucrose and amino acids are transported from sources (where they are produced/leaves) to sinks (where they are stored or used for growth/roots/flowers).

(a)(i) Award 1 mark for long extension/hair and 1 mark for explaining that this increases surface area. (Max 2 marks)
(ii) Award 1 mark for osmosis. (1 mark)
(b)(i) Award 1 mark for xylem. (1 mark)
(ii) Award 1 mark for stating that water evaporates from mesophyll cell surfaces, 1 mark for stating that water vapour diffuses out of the leaf, and 1 mark for mentioning the stomata. (Max 3 marks)
(c)(i) Award 1 mark for phloem. (1 mark)
(ii) Award 1 mark for the term translocation and 1 mark for explaining from source to sink / from leaves to roots/growing regions. (Max 2 marks)

(a)(i) At the cathode: lead is formed. At the anode: bromine is formed.
(ii) A red-brown gas (or vapour) is observed at the positive electrode (anode).
(iii) \(Pb^{2+} + 2e^- \rightarrow Pb\).
(b)(i) Hydrogen gas is produced at the cathode. Sodium is not formed because sodium is more reactive than hydrogen, so hydrogen ions are preferentially discharged (reduced) at the cathode.
(ii) Chlorine gas.
(c) Electroplating is done to prevent rust/corrosion (or to improve the appearance of the object). To copper-plate a steel key, a piece of pure copper is used as the anode.

(a)(i) Award 1 mark for lead (cathode) and 1 mark for bromine (anode). (Max 2 marks)
(ii) Award 1 mark for red-brown gas / vapour. (1 mark)
(iii) Award 1 mark for correct reactants and products (\(Pb^{2+}\) and \(Pb\)) and 1 mark for correct balancing with \(2e^-\). (Max 2 marks)
(b)(i) Award 1 mark for hydrogen gas and 1 mark for explaining that sodium is more reactive than hydrogen. (Max 2 marks)
(ii) Award 1 mark for chlorine. (1 mark)
(c) Award 1 mark for a correct reason (prevent corrosion / appearance) and 1 mark for identifying pure copper as the anode. (Max 2 marks)

(a)(i) Speed of light in a vacuum, \(v = 3.0 \times 10^8\text{ m/s}\).
(ii) Using \(v = f \lambda\), we have \(f = \frac{v}{\lambda} = \frac{3.0 \times 10^8}{1.5 \times 10^{-5}} = 2.0 \times 10^{13}\text{ Hz}\).
(b)(i) Microwaves.
(ii) Hazards of ultraviolet radiation include sunburn, premature aging of skin, skin cancer, or damage to eyes (cataracts).
(c)(i) The ray of light slows down when it enters the glass because glass is optically denser than air. This change in speed causes the wave to bend towards the normal.
(ii) Refractive index, \(n = \frac{\sin i}{\sin r} = \frac{\sin 40^\circ}{\sin 25^\circ} = \frac{0.6428}{0.4226} \approx 1.52\).
(iii) Critical angle.

(a)(i) Award 1 mark for \(3.0 \times 10^8\text{ m/s}\). (1 mark)
(ii) Award 1 mark for correct formula or substitution (\(\frac{3.0 \times 10^8}{1.5 \times 10^{-5}}\)) and 1 mark for the correct answer \(2.0 \times 10^{13}\text{ Hz}\). (Max 2 marks)
(b)(i) Award 1 mark for microwaves. (1 mark)
(ii) Award 1 mark for a valid hazard of UV (skin cancer/sunburn/cataracts). (1 mark)
(c)(i) Award 1 mark for stating that light slows down in glass and 1 mark for stating that this change in speed causes refraction/bending. (Max 2 marks)
(ii) Award 1 mark for correct formula/substitution (\(\frac{\sin 40}{\sin 25}\)) and 1 mark for the calculated refractive index (1.52). (Max 2 marks)
(iii) Award 1 mark for critical angle. (1 mark)

(a) (i) A potometer measures transpiration rate by tracking the movement of an air bubble along a capillary tube [1] over a measured period of time [1].

(ii) Any two from:
- increased temperature
- increased wind speed / air movement
- decreased humidity
- increased light intensity [2]

(iii) Any one explanation corresponding to part (ii):
- Temperature: water molecules gain more kinetic energy [1], increasing the rate of evaporation from mesophyll cell surfaces [1].
- Wind speed: blows water vapour away from the stomata [1], maintaining a steep concentration/diffusion gradient [1].
- Humidity: dry air decreases the water potential outside the leaf [1], increasing the diffusion gradient between the inside and outside of the leaf [1].
- Light intensity: causes stomata to open wider [1], allowing water vapour to diffuse out of the leaf more rapidly [1].

(b) (i) Any two from:
- made of dead cells joined end-to-end with no cross walls to form a continuous hollow tube [1]
- walls are thickened with lignin to provide strength and prevent collapse [1]
- no cytoplasm or organelles to obstruct water flow [1]

(ii) osmosis [1]

(iii) phloem [1]

(a) (i)
- distance moved by bubble is recorded; [1]
- per unit time / within a specified time limit; [1]

(ii) Any two for [1] mark each:
- temperature / wind speed / air flow / light intensity / low humidity; [2]

(iii)
- correct factor identified; [1]
- correct scientific explanation of its effect on evaporation rate or diffusion gradient; [1]

(b) (i) Any two for [1] mark each:
- empty cells / no cytoplasm; [1]
- no end walls / form continuous tubes; [1]
- walls thickened with lignin; [1]

(ii)
- osmosis; [1] (reject: diffusion / active transport)

(iii)
- phloem; [1]

(a) (i) Thermal decomposition [1]

(ii) \(\text{ZnCO}_3 \rightarrow \text{ZnO} + \text{CO}_2\) (all formulas correct [1], balanced [1])

(b) (i) zinc oxide + carbon \(\rightarrow\) zinc + carbon monoxide (or carbon dioxide) [1]

(ii) Zinc oxide is reduced because it loses oxygen [1], and carbon is oxidised because it gains oxygen [1].

(c)
- \(M_r\) of \(\text{ZnCO}_3 = 65 + 12 + (16 \times 3) = 125\) [1]
- \(M_r\) of \(\text{ZnO} = 65 + 16 = 81\)
- Moles of \(\text{ZnCO}_3 = \frac{25.0}{125} = 0.20\text{ mol}\) [1]
- Mass of \(\text{ZnO} = 0.20 \times 81 = 16.2\text{ g}\) [1]

(d) Galvanising iron / steel (to prevent rusting) OR making alloys (such as brass) [1]

(a) (i)
- thermal decomposition; [1]

(ii)
- \(\text{ZnCO}_3 \rightarrow \text{ZnO} + \text{CO}_2\);
- formulae correct; [1]
- equation balanced; [1]

(b) (i)
- zinc oxide + carbon \(\rightarrow\) zinc + carbon monoxide / carbon dioxide; [1]

(ii)
- (reduction) zinc oxide loses oxygen; [1]
- (oxidation) carbon gains oxygen; [1]

(c)
- \(M_r\) values calculated correctly (\(\text{ZnCO}_3 = 125\), \(\text{ZnO} = 81\)); [1]
- moles of zinc carbonate = \(25.0 / 125 = 0.20\text{ mol}\); [1]
- mass of zinc oxide = \(0.20 \times 81 = 16.2\text{ g}\); [1]

(d)
- galvanising / sacrificial protection / making brass; [1]

(a) (i) Weight \(W = m \times g = 150 \times 10 = 1500\text{ N}\) [1]

(ii) Work \(W_d = \text{force} \times \text{distance} = 1500 \times 8.0 = 12000\text{ J}\) (value [1], unit [1])

(iii) Power \(P = \frac{\text{work done}}{\text{time}} = \frac{12000}{12} = 1000\text{ W}\) (formula [1], calculation [1])

(b) (i) \(\text{efficiency} = \frac{\text{useful power output}}{\text{total power input}} \times 100\%\) [1]

(ii) \(\text{total power input} = \frac{1000}{0.75} = 1333\text{ W}\) (or \(1.3\text{ kW}\)) (formula rearranged [1], calculation [1])

(c) (i) Gravitational potential energy is transferred to kinetic energy. [1]

(ii) \(12000\text{ J}\) (or \(12\text{ kJ}\)) [1]

(a) (i)
- \(1500\text{ N}\); [1]

(ii)
- \(12000\); [1]
- \(\text{J}\) / Joules; [1]

(iii)
- \(P = W / t\) in any form; [1]
- \(1000\text{ W}\) / \(1.0\text{ kW}\); [1]

(b) (i)
- \(\text{efficiency} = \frac{\text{useful power output}}{\text{total power input}} (\times 100\%)\); [1]

(ii)
- \(1000 / 0.75\); [1]
- \(1333\text{ W}\) (allow \(1300\text{ W}\) to 2 s.f.); [1]

(c) (i)
- gravitational potential (energy) to kinetic (energy); [1]

(ii)
- \(12000\text{ J}\) / \(12\text{ kJ}\) (allow ecf from (a)(ii)); [1]

(a) (i) Sodium chloride has ionic bonding, which involves the transfer of electrons / electrostatic attraction between oppositely charged ions [1]. Glucose has covalent bonding, which involves the sharing of pairs of electrons [1].

(ii) Solid sodium chloride has ions held in a fixed lattice structure and cannot move [1]. In molten sodium chloride, the lattice structure breaks down, allowing the ions to be free to move and carry the charge [1].

(b) (i)
- negative electrode (cathode): shiny grey liquid/metal forms [1]
- positive electrode (anode): reddish-brown gas/vapour evolved [1]

(ii)
- cathode: \(\text{Pb}^{2+} + 2
\text{e}^- \rightarrow \text{Pb}\) [1]
- anode: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\text{ (or }2\text{Br}^- - 2\text{e}^- \rightarrow \text{Br}_2\)\) [1]

(iii) To allow the ions to be free to move (and carry the electrical current). [1]

(iv) Platinum [1]

(a) (i)
- sodium chloride is ionic / electrostatic attraction between oppositely charged ions; [1]
- glucose is covalent / shared pairs of electrons; [1]

(ii)
- in solid, ions are in fixed positions / lattice; [1]
- in molten, ions are free to move; [1] (reject: electrons are free to move)

(b) (i)
- (negative electrode / cathode) silvery/grey liquid or metal; [1]
- (positive electrode / anode) brown/red-brown gas or vapour; [1]

(ii)
- \(\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}\); [1]
- \(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\); [1]

(iii)
- so that ions are free to move / carry current; [1]

(iv)
- platinum; [1]