An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V1) Cambridge IGCSE Science - Combined (0653) paper. Not affiliated with or reproduced from Cambridge.
Paper 21 (選擇題 - Extended)
There are forty questions on this paper. Answer all questions. Choose the one correct answer out of four possible choices.
40 题目 · 40 分
题目 1 · 選擇題
1 分
An object of mass \(m\) travels at a constant speed \(v\) and has kinetic energy \(E\). What is the kinetic energy of an object of mass \(\frac{1}{2}m\) traveling at a speed of \(2v\)?
A.\(\frac{1}{2}E\)
B.\(E\)
C.\(2E\)
D.\(4E\)
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解题
The kinetic energy is given by the formula \(E_k = \frac{1}{2}mv^2\). Let the initial kinetic energy be \(E = \frac{1}{2}mv^2\). For the new object, the mass is \(m' = \frac{1}{2}m\) and the speed is \(v' = 2v\). Substituting these values into the formula: \(E' = \frac{1}{2} m' (v')^2 = \frac{1}{2} \left(\frac{1}{2}m\right) (2v)^2 = \frac{1}{4}m (4v^2) = mv^2\). Since \(E = \frac{1}{2}mv^2\), we have \(mv^2 = 2E\). Thus, the new kinetic energy is \(2E\).
评分标准
Award 1 mark for the correct option (C).
题目 2 · 選擇題
1 分
Which row correctly describes the colour changes when aqueous chlorine is added to separate aqueous solutions of potassium bromide and potassium iodide?
A.potassium bromide: colourless to orange-brown | potassium iodide: colourless to brown
B.potassium bromide: remains colourless | potassium iodide: colourless to brown
C.potassium bromide: colourless to orange-brown | potassium iodide: remains colourless
Chlorine is more reactive than both bromine and iodine because it is higher up in Group VII. Consequently, chlorine displaces bromide ions from potassium bromide to produce bromine (which forms an orange-brown solution) and displaces iodide ions from potassium iodide to produce iodine (which forms a brown solution).
评分标准
Award 1 mark for the correct option (A).
题目 3 · 選擇題
1 分
A ray of light in air is incident on the surface of a glass block. Which row correctly describes the change in speed of the light ray and the direction of its bending as it enters the glass block?
A.speed of light: decreases | direction of bending: towards the normal
B.speed of light: decreases | direction of bending: away from the normal
C.speed of light: increases | direction of bending: towards the normal
D.speed of light: increases | direction of bending: away from the normal
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解题
Glass has a higher refractive index (is optically denser) than air. When light travels from air into glass, its speed decreases. Because it slows down, the ray bends towards the normal line.
评分标准
Award 1 mark for the correct option (A).
题目 4 · 選擇題
1 分
Which row correctly matches the site of fertilisation and the site of embryo implantation in the female reproductive system?
A.site of fertilisation: oviduct | site of implantation: uterus
B.site of fertilisation: ovary | site of implantation: uterus
C.site of fertilisation: oviduct | site of implantation: ovary
D.site of fertilisation: uterus | site of implantation: oviduct
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解题
In the human female reproductive system, fertilisation (the fusion of the sperm nucleus and egg nucleus) normally takes place in the oviduct (fallopian tube). The fertilised egg (zygote) divides to become an embryo, which then travels to the uterus, where it implants in the uterine wall.
评分标准
Award 1 mark for the correct option (A).
题目 5 · 選擇題
1 分
Two resistors of resistance \(R_1\) and \(R_2\) are connected in parallel to a cell. The resistance of \(R_1\) is greater than the resistance of \(R_2\). Which statement about the current in the resistors and the potential difference (p.d.) across them is correct?
A.The current in \(R_1\) is greater than the current in \(R_2\), and the p.d. across them is the same.
B.The current in \(R_1\) is less than the current in \(R_2\), and the p.d. across them is the same.
C.The current in both resistors is the same, and the p.d. across \(R_1\) is greater than across \(R_2\).
D.The current in both resistors is the same, and the p.d. across \(R_1\) is less than across \(R_2\).
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解题
When components are connected in parallel, the potential difference (p.d.) across each of them is the same and equal to the voltage supplied by the cell. According to Ohm's law, \(I = \frac{V}{R}\). Since the p.d. \(V\) is identical, the branch with the larger resistance (\(R_1\)) will have a smaller current flowing through it compared to \(R_2\).
评分标准
Award 1 mark for the correct option (B).
题目 6 · 選擇題
1 分
An unknown salt solution is tested as follows: - Addition of dilute nitric acid followed by aqueous silver nitrate produces a white precipitate. - Addition of aqueous sodium hydroxide produces a light blue precipitate that is insoluble in excess. What is the identity of the salt?
A.copper(II) chloride
B.copper(II) sulfate
C.iron(II) chloride
D.zinc chloride
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解题
The reaction of the solution with nitric acid and silver nitrate results in a white precipitate, which chemically indicates the presence of chloride anions (\(Cl^-\)). The formation of a light blue precipitate that is insoluble in excess sodium hydroxide is characteristic of copper(II) cations (\(Cu^{2+}\)). Therefore, the salt is copper(II) chloride.
评分标准
Award 1 mark for the correct option (A).
题目 7 · 選擇題
1 分
Which combination of environmental conditions will result in the highest rate of transpiration in a healthy plant?
A.temperature: high | humidity: high | air movement: still air
B.temperature: high | humidity: low | air movement: moving air
C.temperature: low | humidity: low | air movement: still air
D.temperature: low | humidity: high | air movement: moving air
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解题
Transpiration rate increases when environmental conditions keep the water vapour concentration gradient between the leaf interior and the surrounding air as steep as possible. High temperatures increase water evaporation inside the mesophyll; low humidity ensures that the surrounding air remains dry; and moving air quickly carries away water vapour surrounding the stomata, maintaining a fast diffusion rate.
评分标准
Award 1 mark for the correct option (B).
题目 8 · 選擇題
1 分
One molecule of a gaseous alkane with formula \(C_{12}H_{26}\) undergoes cracking to produce one molecule of ethene, \(C_2H_4\), and one molecule of another hydrocarbon. What is the formula of this other hydrocarbon?
A.\(C_{10}H_{20}\)
B.\(C_{10}H_{22}\)
C.\(C_{10}H_{24}\)
D.\(C_{14}H_{30}\)
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解题
In a cracking reaction, both the total number of carbon atoms and hydrogen atoms must be conserved on both sides of the equation. Let the unknown product be \(C_xH_y\). - Carbon balance: \(12 = 2 + x \implies x = 10\) - Hydrogen balance: \(26 = 4 + y \implies y = 22\) Therefore, the formula is \(C_{10}H_{22}\).
评分标准
Award 1 mark for the correct option (B).
题目 9 · 選擇題
1 分
Which row correctly matches an adaptive feature of a human sperm cell to its function?
A.flagellum | stores nutrient energy for the developing embryo
B.jelly coat | protects the genetic material inside the sperm head
C.presence of enzymes in the acrosome | digests a pathway through the jelly coat of the egg cell
D.large cytoplasm | increases the probability of colliding with the egg cell
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解题
A sperm cell is adapted to reach and fertilise an egg cell. Its acrosome contains digestive enzymes that break down the outer jelly coat of the egg cell, allowing the sperm nucleus to enter. The flagellum is used for swimming (motility), not food storage.
评分标准
Award 1 mark for selecting C.
题目 10 · 選擇題
1 分
A rectangular block of concrete has a weight of \(240\text{ N}\) and dimensions \(0.50\text{ m} \times 0.40\text{ m} \times 0.20\text{ m}\). What is the maximum pressure the block can exert when resting on a flat horizontal surface?
A.\(1200\text{ Pa}\)
B.\(2400\text{ Pa}\)
C.\(3000\text{ Pa}\)
D.\(4800\text{ Pa}\)
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解题
Pressure is calculated using the formula \(p = \frac{F}{A}\). To find the maximum pressure, we must use the minimum contact area. The minimum area of contact is calculated using the two smallest dimensions: \(A_{\text{min}} = 0.40\text{ m} \times 0.20\text{ m} = 0.080\text{ m}^2\). Thus, the maximum pressure is \(p_{\text{max}} = \frac{240\text{ N}}{0.080\text{ m}^2} = 3000\text{ Pa}\).
评分标准
Award 1 mark for selecting C.
题目 11 · 選擇題
1 分
A student performs paper chromatography on a food dye. The solvent front travels \(8.0\text{ cm}\) from the baseline. One of the coloured spots has an \(R_f\) value of \(0.35\). What is the distance travelled by this spot from the baseline?
A.\(2.3\text{ cm}\)
B.\(2.8\text{ cm}\)
C.\(5.2\text{ cm}\)
D.\(22.9\text{ cm}\)
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解题
The retardation factor \(R_f\) is calculated as: \(R_f = \frac{\text{distance travelled by spot}}{\text{distance travelled by solvent front}}\). Rearranging the formula to find the distance travelled by the spot gives: \(\text{distance} = R_f \times \text{distance travelled by solvent front} = 0.35 \times 8.0\text{ cm} = 2.8\text{ cm}\).
评分标准
Award 1 mark for selecting B.
题目 12 · 選擇題
1 分
A sound wave of frequency \(1200\text{ Hz}\) travels through a liquid at a speed of \(1500\text{ m/s}\). What is the wavelength of this sound wave in the liquid?
A.\(0.80\text{ m}\)
B.\(1.25\text{ m}\)
C.\(1.80\text{ m}\)
D.\(1.8 \times 10^6\text{ m}\)
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解题
Wavelength \(\lambda\) is related to speed \(v\) and frequency \(f\) by the wave equation: \(v = f\lambda\). Rearranging for wavelength gives: \(\lambda = \frac{v}{f} = \frac{1500\text{ m/s}}{1200\text{ Hz}} = 1.25\text{ m}\).
评分标准
Award 1 mark for selecting B.
题目 13 · 選擇題
1 分
A battery delivers a constant current of \(0.40\text{ A}\) to a circuit for \(5.0\text{ minutes}\). How much charge passes through the circuit during this time?
A.\(2.0\text{ C}\)
B.\(12\text{ C}\)
C.\(120\text{ C}\)
D.\(750\text{ C}\)
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解题
Electric charge \(Q\) is calculated using \(Q = I \times t\), where \(I\) is current in amperes and \(t\) is time in seconds. First, convert time to seconds: \(t = 5.0\text{ minutes} \times 60\text{ s/minute} = 300\text{ s}\). Then, calculate the charge: \(Q = 0.40\text{ A} \times 300\text{ s} = 120\text{ C}\).
评分标准
Award 1 mark for selecting C.
题目 14 · 選擇題
1 分
Which statement correctly describes a difference between transition elements, such as copper, and alkali metals, such as sodium?
A.Transition elements have lower densities than alkali metals.
B.Transition elements have lower melting points than alkali metals.
C.Transition elements form coloured compounds, whereas alkali metals form white compounds.
D.Transition elements are highly reactive with cold water, whereas alkali metals are unreactive.
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解题
Transition elements are hard, dense metals with high melting points, and they are characterised by forming coloured compounds (e.g., blue copper(II) salts). Alkali metals like sodium are soft, have low melting points, low densities, react vigorously with cold water, and form white/colourless compounds.
评分标准
Award 1 mark for selecting C.
题目 15 · 選擇題
1 分
A student wants to prepare a pure, dry sample of the insoluble salt lead(II) sulfate. Which pair of aqueous solutions should be mixed together to produce this salt by precipitation?
A.lead(II) carbonate and dilute sulfuric acid
B.lead(II) chloride and sodium sulfate
C.lead(II) nitrate and sodium sulfate
D.lead(II) oxide and dilute sulfuric acid
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解题
Precipitation requires mixing two soluble salt solutions. Lead(II) nitrate is soluble (all nitrates are soluble) and sodium sulfate is soluble (all sodium salts are soluble). Mixing them produces insoluble lead(II) sulfate and soluble sodium nitrate.
评分标准
Award 1 mark for selecting C.
题目 16 · 選擇題
1 分
A hydrocarbon has the formula \(\text{C}_4\text{H}_8\). Which statement about this hydrocarbon is correct?
A.It belongs to the homologous series of alkanes.
B.It undergoes an addition reaction with aqueous bromine, turning it from orange to colourless.
C.It contains only carbon-carbon single covalent bonds.
D.It is a saturated hydrocarbon.
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解题
The formula \(\text{C}_4\text{H}_8\) fits the general formula \(\text{C}_n\text{H}_{2n}\) for alkenes. Alkenes are unsaturated hydrocarbons containing a carbon-carbon double covalent bond. They undergo addition reactions with aqueous bromine, causing the orange bromine water to decolourise (turn colourless).
评分标准
Award 1 mark for selecting B.
题目 17 · 選擇題
1 分
A crate of weight \(400\text{ N}\) is lifted vertically upwards at a constant speed of \(0.50\text{ m/s}\) by an electric motor. What is the useful power output of the motor?
A.\(20\text{ W}\)
B.\(80\text{ W}\)
C.\(200\text{ W}\)
D.\(800\text{ W}\)
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解题
Power is defined as work done per unit time. When an object is lifted at a constant speed, power can be calculated using the formula: \(P = F \times v\). Here, the force \(F\) is equal to the weight of the crate, \(400\text{ N}\), and the speed \(v\) is \(0.50\text{ m/s}\). Therefore, \(P = 400\text{ N} \times 0.50\text{ m/s} = 200\text{ W}\).
评分标准
1 mark: Correct calculation of power using \(P = F \times v\) to find \(200\text{ W}\).
题目 18 · 選擇題
1 分
In a paper chromatography experiment, a dye is analysed. The baseline is at \(0\text{ cm}\) and the solvent front reaches the \(12.0\text{ cm}\) mark. A component of the dye has an \(R_f\) value of \(0.75\). What is the distance of this spot from the baseline?
A.\(3.0\text{ cm}\)
B.\(4.0\text{ cm}\)
C.\(9.0\text{ cm}\)
D.\(16.0\text{ cm}\)
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解题
The retention factor (\(R_f\)) is defined as: \(R_f = \frac{\text{distance travelled by the substance}}{\text{distance travelled by the solvent front}}\). Rearranging this formula gives: \(\text{distance travelled by the substance} = R_f \times \text{distance travelled by the solvent front}\). Substituting the given values: \(\text{distance} = 0.75 \times 12.0\text{ cm} = 9.0\text{ cm}\).
评分标准
1 mark: Correct application of the \(R_f\) formula to find the distance of \(9.0\text{ cm}\).
题目 19 · 選擇題
1 分
A sound wave travels through air with a speed of \(340\text{ m/s}\). The frequency of the sound is \(850\text{ Hz}\). What is the wavelength of this sound wave?
A.\(0.40\text{ m}\)
B.\(2.5\text{ m}\)
C.\(289\text{ m}\)
D.\(2.89 \times 10^{5}\text{ m}\)
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解题
The wave equation relates wave speed (\(v\)), frequency (\(f\)), and wavelength (\(\lambda\)) as follows: \(v = f \times \lambda\). Rearranging for wavelength: \(\lambda = \frac{v}{f}\). Substituting the given values: \(\lambda = \frac{340\text{ m/s}}{850\text{ Hz}} = 0.40\text{ m}\).
评分标准
1 mark: Correct use of the wave equation \(v = f \times \lambda\) to calculate the wavelength as \(0.40\text{ m}\).
题目 20 · 選擇題
1 分
Which row correctly describes the features of a wind-pollinated flower?
A.petals: large and brightly coloured; stigmas: sticky and inside the flower
B.petals: large and dull green; stigmas: feathery and inside the flower
C.petals: small and dull green; stigmas: feathery and outside the flower
D.petals: small and brightly coloured; stigmas: sticky and outside the flower
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解题
Wind-pollinated flowers do not need to attract insects, so they typically have small, dull green petals. They have feathery stigmas that hang outside the flower to easily catch wind-borne pollen grains.
评分标准
1 mark: Correct identification of both petal and stigma adaptations for wind-pollinated flowers (Option C).
题目 21 · 選擇題
1 分
A \(6.0\text{ V}\) power supply is connected across two resistors in parallel. The resistances of the resistors are \(12\ \Omega\) and \(6.0\ \Omega\). What is the total current in the circuit?
A.\(0.33\text{ A}\)
B.\(0.50\text{ A}\)
C.\(1.0\text{ A}\)
D.\(1.5\text{ A}\)
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解题
First, find the total equivalent resistance (\(R_p\)) of the parallel combination: \(\frac{1}{R_p} = \frac{1}{12\ \Omega} + \frac{1}{6.0\ \Omega} = \frac{1}{12} + \frac{2}{12} = \frac{3}{12}\), which gives \(R_p = 4.0\ \Omega\). Next, use Ohm's law to find the total current (\(I\)): \(I = \frac{V}{R_p} = \frac{6.0\text{ V}}{4.0\ \Omega} = 1.5\text{ A}\). Alternatively, find the current through each branch: \(I_1 = \frac{6.0}{12} = 0.5\text{ A}\) and \(I_2 = \frac{6.0}{6.0} = 1.0\text{ A}\). The total current is \(I_1 + I_2 = 0.5 + 1.0 = 1.5\text{ A}\).
评分标准
1 mark: Correct calculation of total current as \(1.5\text{ A}\).
题目 22 · 選擇題
1 分
As we move down Group VII (the halogens) of the Periodic Table, how do the colour intensity and the reactivity of the elements change?
In Group VII, as we go down the group: 1. The colours of the elements become darker (pale yellow fluorine, greenish-yellow chlorine, red-brown bromine, grey-black iodine), meaning the colour intensity increases. 2. The reactivity decreases because the outer shell is further from the nucleus, making it harder to attract an electron.
评分标准
1 mark: Correct trend of increasing colour intensity and decreasing reactivity down Group VII.
题目 23 · 選擇題
1 分
A student tests a solution of compound Y. Addition of aqueous sodium hydroxide produces a blue precipitate. Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate. What is the identity of compound Y?
A.copper(II) chloride
B.copper(II) sulfate
C.iron(II) sulfate
D.iron(III) chloride
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解题
The blue precipitate with aqueous sodium hydroxide indicates the presence of copper(II) ions, \(\text{Cu}^{2+}\). The white precipitate formed with aqueous barium nitrate after acidifying with nitric acid indicates the presence of sulfate ions, \(\text{SO}_4^{2-}\). Therefore, the compound is copper(II) sulfate.
评分标准
1 mark: Correct deduction of copper(II) and sulfate ions to identify the compound as copper(II) sulfate.
题目 24 · 選擇題
1 分
What is the correct pathway taken by water as it moves through a plant from the soil?
Water is absorbed from the soil by root hair cells. It then travels across the root through root cortex cells to the xylem vessels in the centre of the root. From there, it is transported upwards to the leaves, where it enters the mesophyll cells.
评分标准
1 mark: Correct sequence of water movement through root hair cells, cortex, xylem, and mesophyll.
题目 25 · 選擇題
1 分
Which row correctly contrasts the features of insect-pollinated and wind-pollinated flowers?
A.Insect-pollinated: pollen grains are light and smooth. Wind-pollinated: pollen grains are sticky or spiky.
B.Insect-pollinated: petals are large and brightly coloured. Wind-pollinated: petals are small and dull-green.
C.Insect-pollinated: stigmas are feathery and hang outside the flower. Wind-pollinated: stigmas are sticky and enclosed inside the flower.
D.Insect-pollinated: anthers are versatile and swing freely. Wind-pollinated: anthers are firmly attached and enclosed.
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解题
Insect-pollinated flowers have large, brightly coloured petals to attract insects, while wind-pollinated flowers have small, inconspicuous, dull-green petals because they do not need to attract animal pollinators. Light and smooth pollen grains, feathery stigmas, and versatile anthers are features of wind-pollinated flowers.
评分标准
1 mark for identifying the correct row contrasting the petals of insect-pollinated and wind-pollinated flowers.
题目 26 · 選擇題
1 分
Which row correctly identifies the mineral ion required by plants to make chlorophyll and the direct symptom of its deficiency?
A.mineral ion: magnesium | symptom of deficiency: yellowing of leaves
B.mineral ion: magnesium | symptom of deficiency: stunted growth with purple leaves
C.mineral ion: nitrate | symptom of deficiency: yellowing of leaves
D.mineral ion: nitrate | symptom of deficiency: stunted growth with normal green leaves
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解题
Magnesium ions are a crucial component of the chlorophyll molecule, which is responsible for the green colour of leaves. A deficiency in magnesium ions prevents the plant from synthesising enough chlorophyll, leading to a yellowing of the leaves (chlorosis).
评分标准
1 mark for selecting the correct mineral ion (magnesium) and its corresponding deficiency symptom (yellowing of leaves).
题目 27 · 選擇題
1 分
As we go down Group VII (the halogens) of the Periodic Table, how do the colour intensity, state at room temperature, and reactivity of the elements change?
A.colour intensity: increases | state at r.t.p.: solid to gas | reactivity: decreases
B.colour intensity: increases | state at r.t.p.: gas to solid | reactivity: decreases
C.colour intensity: decreases | state at r.t.p.: gas to solid | reactivity: increases
D.colour intensity: decreases | state at r.t.p.: solid to gas | reactivity: increases
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解题
Down Group VII (the halogens): the elements become darker in colour (intensity increases); the physical state at room temperature and pressure changes from gas (chlorine) to liquid (bromine) to solid (iodine); and their reactivity decreases because it becomes harder for the atom to attract and gain an incoming electron into its outer shell.
评分标准
1 mark for identifying the correct trends down Group VII: increased colour intensity, gas to solid state transition, and decreased reactivity.
题目 28 · 選擇題
1 分
An ideal battery of electromotive force (e.m.f.) 12 V is connected across a parallel combination of two resistors of resistance 4.0 \(\Omega\) and 12.0 \(\Omega\). What is the total current supplied by the battery?
A.0.75 A
B.3.0 A
C.4.0 A
D.12 A
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解题
First, calculate the total equivalent resistance of the two resistors connected in parallel: 1/R_p = 1/R_1 + 1/R_2 = 1/(4.0 \(\Omega\)) + 1/(12.0 \(\Omega\)) = 3/(12.0 \(\Omega\)) + 1/(12.0 \(\Omega\)) = 4/(12.0 \(\Omega\)), giving R_p = 3.0 \(\Omega\). Next, use Ohm's law to find the total current I: I = V/R_p = 12 V / 3.0 \(\Omega\) = 4.0 A.
评分标准
1 mark for the correct calculation of total current: - Calculating parallel resistance as 3.0 \(\Omega\) - Applying I = V/R to find 4.0 A
题目 29 · 選擇題
1 分
An electric motor lifts a load of mass 50 kg vertically through a height of 12 m in a time of 15 s. The gravitational field strength g is 10 N/kg. What is the average useful power output of the motor?
A.40 W
B.400 W
C.600 W
D.6000 W
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解题
First, calculate the work done (which is equal to the increase in gravitational potential energy of the load): E_p = m * g * h = 50 kg * 10 N/kg * 12 m = 6000 J. Next, calculate the useful power output: P = Work Done / t = 6000 J / 15 s = 400 W.
评分标准
1 mark for the correct calculation: - Identifying useful work done as 6000 J - Calculating power as 400 W
题目 30 · 選擇題
1 分
A student reacts a known mass of zinc granules with an excess of dilute hydrochloric acid. Which change to the reaction conditions increases the initial rate of reaction without changing the total volume of hydrogen gas produced?
A.using a single larger piece of zinc of the same mass
B.performing the reaction at a lower temperature
C.using the same mass of zinc in the form of a fine powder
D.using a larger volume of more dilute hydrochloric acid
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解题
Using a fine powder instead of granules increases the surface area of the zinc exposed to the acid. This increases the frequency of successful collisions between reactant particles, thereby increasing the initial rate of reaction. Since the mass of zinc (the limiting reactant) is unchanged and the acid is still in excess, the total volume of hydrogen gas produced remains the same.
评分标准
1 mark for identifying that powder increases surface area and initial rate while keeping the final gas volume constant.
题目 31 · 選擇題
1 分
An unknown aqueous solution of salt X undergoes two tests: 1. When dilute nitric acid and aqueous silver nitrate are added, a white precipitate forms. 2. When aqueous sodium hydroxide is added, a green precipitate forms that is insoluble in excess. What is the chemical name of salt X?
A.copper(II) chloride
B.iron(II) chloride
C.iron(III) chloride
D.iron(II) sulfate
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解题
The formation of a white precipitate with acidified silver nitrate confirms the presence of chloride (Cl^-) ions. The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) (Fe^2+) ions. Therefore, the salt is iron(II) chloride.
评分标准
1 mark for identifying the correct salt name based on the cation and anion test results.
题目 32 · 選擇題
1 分
A sound wave travels from air into a water pool. Which row describes how the speed, frequency, and wavelength of the sound wave change?
Sound is a mechanical wave that travels faster in denser media like water than in air because the particles are closer together, allowing vibrations to be transmitted more rapidly. Therefore, its speed increases. The frequency of a wave is determined solely by its source and does not change when the wave enters a new medium. Since v = f * lambda, an increase in speed with a constant frequency means the wavelength must also increase.
评分标准
1 mark for identifying the correct changes: speed increases, frequency is constant, and wavelength increases.
题目 33 · 選擇題
1 分
Which substances are present in a higher concentration in the blood of the umbilical artery than in the blood of the umbilical vein?
A.carbon dioxide and urea
B.glucose and oxygen
C.carbon dioxide and oxygen
D.glucose and urea
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解题
The umbilical artery carries deoxygenated blood and metabolic waste products from the fetus to the placenta to be cleared by the mother's system. Therefore, it has higher concentrations of waste products such as carbon dioxide and urea compared to the umbilical vein, which carries oxygenated blood and nutrients back to the fetus.
评分标准
A is correct: 1 mark.
题目 34 · 選擇題
1 分
A student performs paper chromatography to identify the dyes present in a green food colouring. The solvent front travels 8.0 cm from the baseline. One of the separated spots travels a distance of 6.4 cm. What is the \(R_f\) value of this dye, and what does this indicate about its solubility in the mobile phase compared to a dye with an \(R_f\) value of 0.40?
A.\(R_f = 0.80\), and it is more soluble in the mobile phase.
B.\(R_f = 0.80\), and it is less soluble in the mobile phase.
C.\(R_f = 1.25\), and it is more soluble in the mobile phase.
D.\(R_f = 1.25\), and it is less soluble in the mobile phase.
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解题
The \(R_f\) value is calculated using the formula: \(R_f = \frac{\text{distance moved by substance}}{\text{distance moved by solvent front}} = \frac{6.4}{8.0} = 0.80\). A higher \(R_f\) value means the substance travels further up the chromatogram, indicating that it has a higher solubility in the mobile phase (solvent) than a substance with a lower \(R_f\) value (0.40).
评分标准
A is correct: 1 mark.
题目 35 · 選擇題
1 分
A block of mass 12 kg is lifted vertically through a height of 5.0 m in a time of 4.0 s. The gravitational field strength, \(g\), is 10 N/kg. What is the useful power developed in lifting the block?
A.15 W
B.150 W
C.240 W
D.600 W
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解题
First, calculate the weight of the block: \(W = m \times g = 12\text{ kg} \times 10\text{ N/kg} = 120\text{ N}\). Next, find the work done in lifting it: \(E = W \times h = 120\text{ N} \times 5.0\text{ m} = 600\text{ J}\). Finally, calculate power: \(P = \frac{E}{t} = \frac{600\text{ J}}{4.0\text{ s}} = 150\text{ W}\).
评分标准
B is correct: 1 mark.
题目 36 · 選擇題
1 分
In a food chain, only a fraction of the energy entering one trophic level is transferred to the next. Which processes explain why herbivores only transfer about 10% of the energy they consume to carnivores?
1. Energy lost as heat from respiration. 2. Energy stored in organic molecules of uneaten parts of the plants. 3. Energy lost in egested waste (faeces). 4. Energy captured from sunlight by plants.
A.1, 2 and 3
B.1 and 3 only
C.2 and 4 only
D.4 only
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解题
Energy transfer efficiency from herbivores to carnivores is limited because: (1) herbivores release energy as heat during cellular respiration, (2) not all plant parts are eaten, meaning energy in those organic molecules is not ingested, and (3) some ingested energy cannot be digested and is lost as egested waste (faeces). Statement 4 describes the capture of energy by producers, which does not explain the loss of energy specifically between the herbivore and carnivore trophic levels.
评分标准
A is correct: 1 mark.
题目 37 · 選擇題
1 分
Which row correctly describes the trends in reactivity and physical properties of the Group VII elements (halogens) as the group is descended from chlorine to iodine?
A.reactivity decreases, melting point increases, colour gets darker
B.reactivity increases, melting point decreases, colour gets lighter
C.reactivity decreases, melting point decreases, colour gets darker
D.reactivity increases, melting point increases, colour gets lighter
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解题
As you go down Group VII, the reactivity of the halogens decreases because the outer shell is further from the nucleus, making it harder to attract an electron. The melting and boiling points increase (going from gaseous chlorine to liquid bromine and solid iodine) due to stronger intermolecular forces. The colour of the elements also becomes progressively darker (chlorine is pale green, bromine is red-brown, and iodine is grey/black).
评分标准
A is correct: 1 mark.
题目 38 · 選擇題
1 分
A student is given an unknown solid, W. When dilute hydrochloric acid is added to W, a gas is evolved that turns limewater cloudy. When W is dissolved in water and treated with aqueous sodium hydroxide, a green precipitate is formed. What is the chemical formula of W?
A.\(\text{FeCO}_3\)
B.\(\text{FeSO}_4\)
C.\(\text{CuCO}_3\)
D.\(\text{CuSO}_4\)
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解题
The reaction of W with dilute hydrochloric acid produces carbon dioxide gas (which turns limewater cloudy), confirming the presence of carbonate ions (\(\text{CO}_3^{2-}\)). The formation of a green precipitate when dissolved W is treated with aqueous sodium hydroxide indicates the presence of iron(II) ions (\(\text{Fe}^{2+}\)). Therefore, the compound is iron(II) carbonate, \(\text{FeCO}_3\).
评分标准
A is correct: 1 mark.
题目 39 · 選擇題
1 分
An electromagnetic wave has a frequency of \(6.0 \times 10^{14}\text{ Hz}\) and travels at a speed of \(3.0 \times 10^8\text{ m/s}\) in a vacuum. What is the wavelength of this wave, and to which part of the electromagnetic spectrum does it belong?
A.\(5.0 \times 10^{-7}\text{ m}\), visible light
B.\(5.0 \times 10^{-7}\text{ m}\), infrared
C.\(2.0 \times 10^6\text{ m}\), radio waves
D.\(2.0 \times 10^6\text{ m}\), ultraviolet
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解题
The wavelength is calculated using the wave equation: \(\lambda = \frac{v}{f} = \frac{3.0 \times 10^8\text{ m/s}}{6.0 \times 10^{14}\text{ Hz}} = 5.0 \times 10^{-7}\text{ m}\). This wavelength corresponds to 500 nm, which lies within the range of visible light (approximately 400 nm to 700 nm).
评分标准
A is correct: 1 mark.
题目 40 · 選擇題
1 分
A circuit contains a 6.0 V battery connected to two resistors in parallel. One resistor has a resistance of \(3.0\ \Omega\) and the other has a resistance of \(6.0\ \Omega\). What is the total current drawn from the battery?
A.3.0 A
B.2.0 A
C.9.0 A
D.0.67 A
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解题
Method 1: Find individual currents. The potential difference across each parallel resistor is 6.0 V. \(I_1 = \frac{V}{R_1} = \frac{6.0\text{ V}}{3.0\ \Omega} = 2.0\text{ A}\) \(I_2 = \frac{V}{R_2} = \frac{6.0\text{ V}}{6.0\ \Omega} = 1.0\text{ A}\) Total current \(I = I_1 + I_2 = 2.0\text{ A} + 1.0\text{ A} = 3.0\text{ A}\).
Answer all structured questions on the space provided on the question paper.
9 题目 · 81 分
题目 1 · structured-theory
9 分
Answer all parts of this question.
(a) (i) Define the term *fertilisation* as it applies to human reproduction. [1]
(ii) State the exact part of the female reproductive system where fertilisation typically occurs. [1]
(iii) Describe what happens to the number of chromosomes during this process. [1]
(b) Wind-pollinated grass flowers have structural adaptations to ensure successful pollination. These include long, feathery stigmas and pendulous anthers that hang outside the petals.
Explain how the structure of these stigmas and anthers are adapted for wind pollination. [3]
(c) Some plants can reproduce both sexually and asexually.
State three differences between the offspring produced by asexual reproduction and those produced by sexual reproduction in plants. [3]
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解题
(a) (i) Fertilisation is the fusion of the nuclei of a male gamete (sperm) and a female gamete (egg / ovum) to form a zygote. (ii) Oviduct / fallopian tube. (iii) The haploid nuclei (each containing 23 chromosomes) fuse to form a diploid nucleus (containing 46 chromosomes) / the chromosome number doubles from haploid to diploid.
(b) Long, feathery stigmas provide a large surface area to easily catch/trap pollen grains floating in the wind. Pendulous anthers hang outside the flower where they are exposed to wind currents, allowing pollen grains to be easily dislodged and blown away.
(c) 1. Asexual offspring are genetically identical to the parent (clones), while sexual offspring show genetic variation. 2. Asexual reproduction involves only one parent, whereas sexual reproduction involves two parents. 3. Asexual reproduction does not involve gametes or fertilisation, while sexual reproduction requires the production of gametes and their subsequent fusion.
评分标准
(a) (i) Fusion of gamete nuclei (sperm and egg) [1] (ii) Oviduct [1] (reject: ovary / uterus) (iii) Haploid to diploid / number of chromosomes doubles / fuses to form 46 chromosomes [1]
(b) Any three from: - Feathery stigmas provide large surface area [1] - To trap/catch pollen from the air [1] - Hanging/pendulous anthers are exposed to wind [1] - To easily release/blow away pollen grains [1]
(c) Any three from: - Asexual offspring are genetically identical / clones AND sexual offspring show genetic variation [1] - Asexual involves one parent AND sexual involves two parents [1] - Asexual does not use gametes/fertilisation AND sexual does [1] - Sexual involves meiosis AND asexual involves mitosis [1]
题目 2 · structured-theory
9 分
Answer all parts of this question.
(a) An electric crane is used on a construction site to lift a concrete block of mass 150 kg vertically upwards. The block is lifted through a height of 8.0 m in a time of 12 s at a constant speed.
The gravitational field strength, \(g\), is 10 N/kg.
(i) Calculate the increase in gravitational potential energy (\(\Delta\text{GPE}\)) of the concrete block. Show your working. [2]
(ii) Calculate the useful power developed by the crane. State the unit of your answer. [3]
(b) The motor of the crane is actually supplied with 1500 W of electrical power.
(i) Explain why the electrical power supplied to the motor is greater than the useful power output calculated in (a)(ii). [2]
(b) (i) Some energy is wasted / dissipated as thermal energy (heat) or sound to the surroundings because of friction in the motor, gears, or moving parts of the crane. (ii) \(\text{Efficiency} = \frac{\text{Useful Power Output}}{\text{Total Power Input}} \times 100\% = \frac{1000}{1500} \times 100\% = 66.7\%\) (or 67%)
评分标准
(a) (i) - Formula: \(\Delta\text{GPE} = mgh\) or substitution of \(150 \times 10 \times 8.0\) [1] - Correct answer: 12000 (J) or 12 (kJ) [1]
(ii) - Formula: \(\text{Power} = \text{Work} / \text{time}\) or substitution of \(12000 / 12\) [1] - Correct answer: 1000 [1] - Unit: W / Watts / J/s [1]
(b) (i) - Energy is lost/wasted/dissipated [1] - As heat/thermal energy / sound due to friction in bearings/gears/motor [1]
(a) Potassium and sodium are Group I alkali metals.
(i) Describe the trend in reactivity of the alkali metals as the group is descended. [1]
(ii) Explain this trend in terms of atomic structure. [3]
(b) Chlorine is a Group VII halogen.
(i) Write a balanced chemical equation, including state symbols, for the reaction between aqueous potassium bromide and chlorine gas. [3]
(ii) Explain why a reaction does not occur when aqueous potassium chloride is mixed with iodine solution. [2]
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解题
(a) (i) Reactivity increases down the group. (ii) As Group I is descended, the number of electron shells increases (the atoms get larger). The single outer shell electron is further away from the positive nucleus, resulting in a weaker electrostatic attraction between the nucleus and the outer electron. Therefore, the outer electron is lost more easily during a reaction.
(b) (i) \(\text{Cl}_2\text{(g)} + 2\text{KBr(aq)} \rightarrow 2\text{KCl(aq)} + \text{Br}_2\text{(aq)}\) (ii) Chlorine is more reactive than iodine (or iodine is less reactive than chlorine). A less reactive halogen cannot displace a more reactive halogen from its compound.
评分标准
(a) (i) Reactivity increases down the group [1] (ii) - Atoms get larger / more shells / outer electron further from nucleus [1] - Weaker attraction between nucleus and outer electron [1] - Outer electron lost more easily [1]
(ii) - Iodine is less reactive than chlorine [1] - Less reactive halogen cannot displace a more reactive halogen / no displacement occurs [1]
题目 4 · structured-theory
9 分
Answer all parts of this question.
(a) A circuit contains a 12 V d.c. power supply connected to two resistors in parallel. Resistor \(R_1\) has a resistance of \(4.0\ \Omega\) and resistor \(R_2\) has a resistance of \(6.0\ \Omega\).
(i) Calculate the combined resistance of the two resistors connected in parallel. Show your working. [2]
(ii) Calculate the total current leaving the power supply. [2]
(iii) Calculate the current flowing through resistor \(R_1\). [2]
(b) A third resistor, \(R_3\), is now connected in series with the parallel combination.
State and explain the effect of this modification on:
(ii) \(I = \frac{V}{R} = \frac{12}{2.4} = 5.0\text{ A}\)
(iii) In parallel, the potential difference across each branch is equal to the supply voltage (12 V). \(I_1 = \frac{V}{R_1} = \frac{12}{4.0} = 3.0\text{ A}\)
(b) (i) The total resistance increases because adding a resistor in series adds directly to the resistance of the parallel combination (\(R_{\text{total}} = R_{12} + R_3\)). (ii) The total current decreases because the total resistance of the circuit has increased while the supply voltage remains constant.
评分标准
(a) (i) - Formula used: \(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\) or \(\frac{R_1R_2}{R_1+R_2}\) [1] - Correct answer: 2.4 (\(\Omega\)) [1]
(ii) - Calculation: \(I = 12 / 2.4\) (allow ecf from (a)(i)) [1] - Correct answer: 5.0 (A) [1]
(b) (i) - Statement: total resistance increases [1] - Explanation: resistance in series adds up / total resistance becomes \(2.4 + R_3\) [1]
(ii) - Statement: total current decreases (since resistance increases and voltage is constant) [1]
题目 5 · structured-theory
9 分
Answer all parts of this question.
(a) Describe the pathway and mechanism of water transport in a flowering plant from the soil, through the roots and stem, and out into the atmosphere. [4]
(b) A student sets up a potometer to measure the rate of transpiration of a leafy shoot under different conditions.
State and explain how the rate of transpiration changes when:
(i) the humidity of the surrounding air increases [2]
(ii) the temperature of the surrounding air increases [3]
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解题
(a) Water is absorbed from the soil into the root hair cells by osmosis (down a water potential gradient). It then moves across the root cortex into the xylem vessels. Water is pulled upwards through the xylem of the stem and into the leaves in a continuous column due to the transpiration pull (tension created by water loss at the leaves). Finally, water evaporates from the surfaces of the spongy mesophyll cells into the air spaces, and water vapour diffuses out of the leaf through the stomata.
(b) (i) Transpiration rate decreases. This is because higher humidity increases the concentration of water vapour outside the leaf, which reduces the water vapour concentration gradient between the inside and the outside of the leaf. (ii) Transpiration rate increases. Higher temperatures increase the kinetic energy of water molecules, leading to faster evaporation of water from the mesophyll cell walls. It also lowers the relative humidity of the air outside, increasing the concentration gradient.
评分标准
(a) Any four from: - Water enters root hair cells by osmosis [1] - Down a water potential gradient [1] - Water travels into xylem vessels [1] - Pulled up stem by transpiration pull / tension / continuous column of water [1] - Evaporation from mesophyll cell surfaces into air spaces [1] - Water vapour diffuses out through stomata [1]
(ii) - Statement: Transpiration rate increases [1] - Explanation: Water molecules gain kinetic energy, increasing rate of evaporation [1] - Higher temperature reduces external relative humidity / increases concentration gradient [1]
题目 6 · structured-theory
9 分
Answer all parts of this question.
(a) Magnesium sulfate is a soluble salt. It can be prepared by reacting dilute sulfuric acid with insoluble magnesium carbonate.
(i) Write a balanced chemical equation for this reaction. State symbols are not required. [2]
(ii) Describe the steps involved in preparing a pure, dry sample of magnesium sulfate crystals starting from dilute sulfuric acid and solid magnesium carbonate. [5]
(b) State the test and observation used to confirm the presence of sulfate ions in the resulting solution. [2]
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解题
(a) (i) \(\text{MgCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{MgSO}_4 + \text{H}_2\text{O} + \text{CO}_2\) (ii) 1. Add excess magnesium carbonate to a measured volume of warm dilute sulfuric acid to ensure all acid is neutralised. 2. Filter the mixture to remove the unreacted/excess solid magnesium carbonate. 3. Heat the filtrate (magnesium sulfate solution) to evaporate some water until the saturated point/crystallization point is reached. 4. Allow the saturated solution to cool slowly so that crystals form. 5. Filter off the crystals, wash them with a small amount of cold distilled water, and dry them on filter paper or in a warm oven.
(b) Add dilute nitric acid (or hydrochloric acid) to the solution, followed by aqueous barium nitrate (or barium chloride). A white precipitate of barium sulfate forms if sulfate ions are present.
(ii) - Step 1: Add excess magnesium carbonate to (warm) sulfuric acid [1] - Step 2: Filter to remove excess magnesium carbonate [1] - Step 3: Heat/evaporate filtrate to crystallization point / saturation point [1] - Step 4: Allow saturated solution to cool (to form crystals) [1] - Step 5: Filter crystals, wash with cold distilled water AND dry with filter paper / in warm oven [1]
(b) - Test: Add dilute nitric/hydrochloric acid, then add aqueous barium nitrate/chloride [1] - Observation: White precipitate [1]
题目 7 · structured-theory
9 分
Answer all parts of this question.
(a) Electromagnetic waves travel through a vacuum at a constant speed of \(3.0 \times 10^8\text{ m/s}\).
(i) State one feature that is common to all electromagnetic waves, other than their speed in a vacuum. [1]
(ii) Microwaves are used for satellite communications.
Calculate the wavelength of a microwave signal with a frequency of \(1.5 \times 10^9\text{ Hz}\). Show your working. [2]
(b) Light travels from air into a rectangular glass block.
(i) Describe what happens to the speed, frequency, and wavelength of the light as it enters the glass block from air. [3]
(ii) Explain the process of refraction (bending of light) in terms of the wave speed at the boundary. [3]
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解题
(a) (i) They are all transverse waves / can travel through a vacuum / transfer energy without transferring matter. (ii) \(\lambda = \frac{v}{f} = \frac{3.0 \times 10^8\text{ m/s}}{1.5 \times 10^9\text{ Hz}} = 0.20\text{ m}\) (or 20 cm)
(b) (i) - Speed: decreases. - Frequency: remains constant. - Wavelength: decreases. (ii) Glass is a denser medium than air, so light travels slower in glass. When a wavefront of light hits the boundary at an angle, one side of the wavefront enters the glass and slows down before the other side. This difference in speed across the wavefront causes the wavefront to change direction and bend towards the normal.
评分标准
(a) (i) Transverse / can travel through vacuum / transfer energy [1] (ii) - Calculation: \(\lambda = v / f\) or substitution of \((3.0 \times 10^8) / (1.5 \times 10^9)\) [1] - Correct answer: 0.20 m / 20 cm [1]
(ii) - Light travels slower in glass than in air [1] - One side of the wavefront slows down before the other (at the boundary) [1] - This difference in speed causes a change in direction / bending towards the normal [1]
题目 8 · structured-theory
9 分
Answer all parts of this question.
(a) Hexane (\(\text{C}_6\text{H}_{14}\)) is a saturated hydrocarbon.
(i) Define the term *hydrocarbon*. [1]
(ii) Draw the displayed structure of a hexane molecule showing all atoms and covalent bonds. [2]
(b) Cracking is used to break down larger, less useful alkane molecules into smaller, more useful alkenes and alkanes.
(i) Write a balanced chemical equation for the cracking of decane (\(\text{C}_{10}\text{H}_{22}\)) to produce one molecule of ethene (\(\text{C}_2\text{H}_4\)) and one other alkane molecule. [2]
(ii) Describe a chemical test to distinguish between the products of this cracking reaction (ethene and the other alkane product). State the test and the observations for both substances. [3]
(c) State one major environmental concern associated with the incomplete combustion of hydrocarbons in car engines. [1]
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解题
(a) (i) A compound containing only carbon and hydrogen atoms. (ii) Displayed formula of hexane (six carbons in a straight chain, each bonded to hydrogens to satisfy valency of 4): H H H H H H | | | | | | H-C - C - C - C - C - C-H | | | | | | H H H H H H
(b) (i) \(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_2\text{H}_4 + \text{C}_8\text{H}_{18}\) (ii) Test: Add aqueous bromine / bromine water to both substances. Observation for Ethene (alkene): The orange/brown bromine water is decolourised (turns colourless). Observation for the Alkane (octane): No change (the solution remains orange/brown).
(c) Incomplete combustion produces carbon monoxide (CO), which is a highly toxic/poisonous gas, or particulates/soot (carbon), which causes respiratory problems and smog.
评分标准
(a) (i) Compound containing carbon and hydrogen ONLY [1] (ii) - Six carbon chain with single bonds [1] - Correct number of hydrogen atoms (14) with all single C-H bonds shown [1]
(b) (i) - Correct formula for product alkane (\(\text{C}_8\text{H}_{18}\)) [1] - Fully balanced equation: \(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_2\text{H}_4 + \text{C}_8\text{H}_{18}\) [1]
(ii) - Test: Add aqueous bromine / bromine water [1] - Alkene (ethene) observation: decolourises / turns colourless [1] - Alkane observation: remains orange/brown / no change [1]
(c) Carbon monoxide produced which is toxic/poisonous OR carbon/soot produced which causes breathing difficulties/smog [1]
题目 9 · structured-theory
9 分
A toy cart of mass \(2.5\text{ kg}\) moves along a straight, horizontal track. The cart starts from rest and accelerates uniformly to a speed of \(8.0\text{ m/s}\) in a time of \(5.0\text{ s}\). It then travels at this constant speed of \(8.0\text{ m/s}\) for a further \(10\text{ s}\).
(a) (i) Calculate the acceleration of the cart during the first \(5.0\text{ s}\). Show your working and state the unit. [3]
(ii) Calculate the total distance travelled by the cart during the \(15\text{ s}\ interruption of its motion. Show your working. [3] (b) Calculate the kinetic energy of the cart when it is travelling at its constant speed of \)8.0\text{ m/s}\). Show your working. [2]
(c) State the relationship between work done, force and distance. [1]
(a) (i) - \(\text{acceleration} = \frac{\text{change in speed}}{\text{time}}\) or \(\frac{8.0}{5.0}\) [1] - \(1.6\) [1] - \(\text{m/s}^2\) or \(\text{m s}^{-2}\) [1]
(ii) - calculation of distance during acceleration phase (\(20\text{ m}\)) [1] - calculation of distance during constant speed phase (\(80\text{ m}\)) [1] - correct total distance (\(100\text{ m}\)) [1]
(b) - correct working: \(0.5 \times 2.5 \times 8.0^2\) or \(E_k = \frac{1}{2} m v^2\) [1] - \(80\text{ J}\) [1] (unit required for second mark)
Answer all questions. Show your working where appropriate and use the tables/grids provided.
4 题目 · 40 分
题目 1 · practical-task
10 分
A student investigates the effect of light intensity on the rate of photosynthesis in the aquatic plant Cabomba. The apparatus is set up with a lamp placed at various distances from a boiling tube containing the plant in water. (a) (i) Fig. 1.1 shows a diagram of the ruler used to measure the distance of the lamp. The pointer on the lamp indicates its position on the ruler. State the distance, d, of the lamp shown in Fig. 1.1. d = .................... cm [1] (ii) The student counts the oxygen bubbles produced by the plant in one minute. Fig. 1.2 shows the bubbles observed in the test-tube during this period. Count and record the number of bubbles shown in Fig. 1.2. number of bubbles = .................... [1] (b) The student repeats the experiment for other distances. Table 1.1 shows the results. Table 1.1: Distance d / cm: [10, 15, 20, 25, 30], Bubble count in Minute 1: [42, 28, 19, (value from (a)(ii)), 8], Bubble count in Minute 2: [44, 30, 21, 13, 9], Average bubble count: [43.0, 29.0, 20.0, ...................., 8.5]. (i) Complete Table 1.1 by writing the bubble count from (a)(ii) and calculating the average bubble count at d = 25 cm. [2] (c) Plot a graph of the average bubble count (vertical axis) against the distance of the lamp d (horizontal axis) on the grid. Draw the curve of best fit. [3] (d) State the relationship between the distance of the lamp and the rate of bubble production. [1] (e) Identify one variable, other than the light intensity, that must be kept constant in this investigation, and explain how the student could keep it constant. [2]
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解题
(a) (i) Reading from Fig. 1.1 is 25.0 cm. (ii) Counting the bubbles in Fig. 1.2 gives 14 bubbles. (b) (i) Table entry for Minute 1 at 25 cm is 14. Average count is (14 + 13) / 2 = 13.5. (c) Graph axes labelled 'Average bubble count' (no unit) and 'Distance d / cm' with linear scales; all points plotted correctly; a smooth curve of best fit drawn. (d) As the distance of the lamp increases, the rate of bubble production decreases. (e) Temperature of the water; kept constant by using a glass pane between the lamp and the tube to absorb heat, or placing the tube in a constant-temperature water bath.
评分标准
1(a)(i) 25.0 (cm) [1]; 1(a)(ii) 14 [1]; 1(b)(i) 14 written in table [1], 13.5 calculated and written in table [1]; 1(c) Axes correctly labelled with units on x-axis [1], appropriate linear scale occupying at least half of the grid [1], all 5 points plotted correctly to within half a small square and smooth curve of best fit drawn [1]; 1(d) as distance increases, bubble count / rate of photosynthesis decreases [1]; 1(e) temperature / carbon dioxide concentration [1] and detail of how it is controlled (e.g. use of water bath / add a fixed amount of sodium hydrogencarbonate) [1].
题目 2 · practical-task
10 分
A student investigates the thermal decomposition of a green powder, copper(II) carbonate. The student heats a sample of copper(II) carbonate in a crucible with a lid. (a) Fig. 2.1 shows the digital balance display for the mass of the crucible, lid, and residue after heating. State the mass shown on the balance in Fig. 2.1. mass = .................... g [1] (b) The student's recorded masses are: Mass of empty crucible and lid = 18.45 g, Mass of crucible, lid, and copper(II) carbonate before heating = 21.93 g. Calculate: (i) the mass of copper(II) carbonate heated. mass of copper(II) carbonate = .................... g [1] (ii) the mass of the copper(II) oxide residue obtained. mass of residue = .................... g [1] (iii) the mass of carbon dioxide gas lost during heating. mass of carbon dioxide lost = .................... g [1] (c) (i) The theoretical mass of carbon dioxide that should be lost from this sample is 1.24 g. Calculate the percentage yield of carbon dioxide gas. Show your working. percentage yield = .................... % [2] (ii) Suggest one reason why the calculated percentage yield might be less than 100%. [1] (d) Describe a chemical test, including the observation, to confirm that the gas evolved is carbon dioxide. test .................... observation .................... [2] (e) State one safety precaution the student should take when handling the hot crucible. [1]
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解题
(a) Reading from Fig. 2.1 is 20.71 g. (b) (i) Mass heated = 21.93 g - 18.45 g = 3.48 g. (ii) Mass of residue = 20.71 g - 18.45 g = 2.26 g. (iii) Mass of CO2 lost = 21.93 g - 20.71 g = 1.22 g. (c) (i) Percentage yield = (1.22 / 1.24) * 100 = 98.387% = 98.4% (or 98%). (ii) Some reactant remained unreacted / incomplete decomposition. (d) Bubble the gas through limewater; the limewater turns cloudy / milky. (e) Use crucible tongs to handle the hot crucible / let it cool on a heatproof mat before weighing.
评分标准
2(a) 20.71 (g) [1]; 2(b)(i) 3.48 (g) [1]; 2(b)(ii) 2.26 (g) [1]; 2(b)(iii) 1.22 (g) [1]; 2(c)(i) evidence of division of actual mass by theoretical mass (1.22 / 1.24) [1], correct calculation to 2 or more significant figures (98% or 98.4%) [1]; 2(c)(ii) incomplete decomposition / some CO2 dissolved in water / unreacted copper(II) carbonate [1]; 2(d) bubble gas through limewater [1], turns cloudy / milky / white precipitate [1]; 2(e) use crucible tongs / wear heat-resistant gloves / place on heatproof mat [1].
题目 3 · practical-task
10 分
A student investigates the cooling and solidifying of stearic acid. A boiling tube containing liquid stearic acid is placed in a beaker of cold water, and the temperature is recorded every minute. (a) Fig. 3.1 shows the thermometer reading at time t = 0.0 min. State the temperature shown on the thermometer in Fig. 3.1. temperature = .................... °C [1] (b) Table 3.1 shows the temperature, T, at various times, t. Table 3.1: Time t / min: [0.0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0], Temperature T / °C: [...................., 79.5, 75.0, 71.0, 69.5, 69.5, 69.5, 68.0, 65.5, 63.0, 61.0]. (i) Complete Table 3.1 by inserting the temperature at t = 0.0 min from (a). [1] (ii) On the grid, plot a graph of Temperature T (vertical axis) against Time t (horizontal axis). Draw a smooth curve of best fit. [4] (iii) State the melting point of stearic acid obtained from your graph and explain how the graph shows this value. melting point = .................... °C, explanation: .................... [2] (iv) State the state of matter of the stearic acid at t = 5.0 min. [1] (v) State one precaution that the student should take to avoid a parallax error when reading the thermometer. [1]
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解题
(a) The thermometer reading is 84.5 °C. (b) (i) 84.5 inserted in the table at t = 0.0. (ii) Plotting graph: Axes labelled T / °C and t / min with correct linear scales; all points plotted correctly to within half a small square; smooth curve of best fit drawn, showing a flat plateau. (iii) Melting point = 69.5 °C; explanation: the graph is flat/horizontal, showing that temperature remains constant while the stearic acid changes state. (iv) Mixture of liquid and solid (both co-exist during freezing). (v) Look at the thermometer with the line of sight perpendicular to the thermometer tube / at eye level with the top of the liquid meniscus.
评分标准
3(a) 84.5 (°C) [1]; 3(b)(i) 84.5 written in table [1]; 3(b)(ii) axes labelled with units [1], appropriate linear scale occupying at least half of the grid [1], all 11 points plotted correctly to within half a small square [1], smooth curve of best fit drawn [1]; 3(b)(iii) 69.5 (°C) [1], flat / horizontal section / temperature remains constant during phase change [1]; 3(b)(iv) mixture of liquid and solid [1]; 3(b)(v) view the scale perpendicular to the thermometer stem / at eye level with meniscus [1].
题目 4 · practical-task
10 分
A student investigates the refraction of a ray of light as it passes from air into a semi-circular transparent plastic block. (a) Fig. 4.1 shows the path of a ray of light incident on the flat face of the block at point P. (i) On Fig. 4.1, draw the normal to the flat surface at point P. [1] (ii) Measure and record the angle of incidence, i, between the incident ray and the normal. i = .................... degrees [1] (iii) Measure and record the angle of refraction, r, between the refracted ray and the normal. r = .................... degrees [1] (b) The student repeats the experiment for other angles of incidence. Table 4.1 shows the results. Table 4.1: Angle of incidence, i / degrees: [20, 30, 40, 50, 60], Angle of refraction, r / degrees: [13, 19, ...................., 31, 35]. (i) Complete Table 4.1 by inserting your measured value of r from (a)(iii). [1] (ii) State the relationship between the angle of incidence and the angle of refraction. [1] (c) The refractive index, n, of the plastic block is given by the formula: n = sin(i) / sin(r). (i) Calculate the value of n for i = 30 degrees and r = 19 degrees. Show your working and give your answer to two significant figures. n = .................... [2] (ii) State why the refractive index, n, has no unit. [1] (d) State two sources of experimental error when tracing rays with pins or using a light box, and describe how to minimise them. [2]
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解题
(a) (i) Normal drawn as a straight dashed line perpendicular to the flat surface of the block at point P. (ii) i = 40 degrees (allow 38 to 42). (iii) r = 25 degrees (allow 24 to 26). (b) (i) 25 written in Table 4.1. (ii) As the angle of incidence increases, the angle of refraction increases. (c) (i) n = sin(30) / sin(19) = 0.500 / 0.3256 = 1.536. To 2 s.f., n = 1.5. (ii) It is a ratio of two sines, which are dimensionless numbers. (d) Source of error 1: The ray of light from the light box is thick. Minimisation: Use a very narrow slit / mark the exact center of the ray. Source of error 2: Pins used for tracing are not perfectly vertical. Minimisation: Ensure pins are vertical / look at the bases of the pins on the paper.
评分标准
4(a)(i) normal drawn perpendicular to flat face at P [1]; 4(a)(ii) 40 (degrees) (allow 38-42) [1]; 4(a)(iii) 25 (degrees) (allow 24-26) [1]; 4(b)(i) 25 (or candidate's value) inserted in table [1]; 4(b)(ii) as angle of incidence increases, angle of refraction increases [1]; 4(c)(i) calculation shown using correct sine values (0.500 / 0.326) [1], correct value to 2 s.f. (1.5) [1]; 4(c)(ii) it is a ratio of two identical units / sine has no unit [1]; 4(d) any two valid sources of error and their corresponding improvements (e.g., ray is thick - use narrow slit / pins not straight - align using pin bases / difficulty aligning block - trace outline with sharp pencil) [2].
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