An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V1) Cambridge International A Level 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 題目 · 40 分
題目 1 · 選擇題
1 分
A metal block of mass 4.0 kg is released from rest at a height of 15 m above the ground. Air resistance is negligible. The acceleration of free fall, \(g\), is 10 m/s\(^2\). What is the kinetic energy of the block when it has fallen a distance of 10 m?
A.200 J
B.400 J
C.600 J
D.4000 J
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解題
The conservation of energy states that the loss in gravitational potential energy (GPE) is equal to the gain in kinetic energy (KE) when air resistance is negligible.
\(\Delta\text{GPE} = m \cdot g \cdot \Delta h\)
where: - \(m = 4.0\text{ kg}\) - \(g = 10\text{ m/s}^2\) - \(\Delta h = 10\text{ m}\) (the distance fallen)
Therefore, the kinetic energy of the block when it has fallen 10 m is 400 J.
評分準則
1 mark for the correct calculation of kinetic energy (400 J).
題目 2 · 選擇題
1 分
An analyst performs paper chromatography to separate the pigments in a sample of food colouring. The solvent front travels a distance of 8.0 cm from the baseline. One of the dye spots travels a distance of 5.0 cm from the baseline. What is the \(R_f\) value of this dye, and what type of process is chromatography?
A.0.625, chemical process
B.0.625, physical process
C.1.60, chemical process
D.1.60, physical process
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解題
The retardation factor (\(R_f\)) is calculated as: \(R_f = \frac{\text{distance moved by substance}}{\text{distance moved by solvent front}} = \frac{5.0\text{ cm}}{8.0\text{ cm}} = 0.625\).
Chromatography is a physical process because it separates the substances in a mixture based on their physical properties (solubility in the mobile phase and retention by the stationary phase) without any chemical reaction occurring.
評分準則
1 mark for identifying the correct \(R_f\) value of 0.625 and classifying chromatography as a physical process.
題目 3 · 選擇題
1 分
A ray of monochromatic light travels through the air and enters a flat glass block at an angle of incidence of \(45^\circ\). Which row correctly describes what happens to the frequency, speed, and wavelength of the light as it enters the glass?
When light travels from a less optically dense medium (air) into a more optically dense medium (glass): 1. The frequency of the wave is determined solely by the source, so it remains constant. 2. The speed of light decreases in the denser medium. 3. Since the wave equation is \(v = f \lambda\), where \(v\) is the speed, \(f\) is the frequency, and \(\lambda\) is the wavelength, a decrease in speed \(v\) while frequency \(f\) remains constant must result in a decrease in wavelength \(\lambda\).
評分準則
1 mark for identifying that frequency remains constant, speed decreases, and wavelength decreases.
題目 4 · 選擇題
1 分
Which combination of features is characteristic of a wind-pollinated flower?
Wind-pollinated flowers do not need to attract insects, so they typically have small, dull petals. Their stigmas are long, feathery, and hang outside the flower to maximize the chance of catching wind-borne pollen. Their pollen grains are light, smooth, and produced in large numbers so they can be easily carried by air currents.
評分準則
1 mark for identifying the correct combination of floral features for wind pollination.
題目 5 · 選擇題
1 分
Two resistors of resistances \(3.0\ \Omega\) and \(6.0\ \Omega\) are connected in parallel. This combination is then connected in series with a \(4.0\ \Omega\) resistor and a \(12.0\text{ V}\) power supply of negligible internal resistance. What is the current drawn from the power supply?
A.0.92 A
B.1.2 A
C.2.0 A
D.3.0 A
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解題
First, calculate the equivalent resistance of the two parallel resistors (\(3.0\ \Omega\) and \(6.0\ \Omega\)):
1 mark for the correct calculation of the total current (2.0 A).
題目 6 · 選擇題
1 分
Which statement describes the trends in physical and chemical properties of the Group VII halogens as the group is descended from chlorine to iodine?
A.The elements become darker in colour and their reactivity increases.
B.The elements become lighter in colour and their reactivity increases.
C.The elements become darker in colour and their reactivity decreases.
D.The elements become lighter in colour and their reactivity decreases.
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解題
As you descend Group VII (chlorine \(\rightarrow\) bromine \(\rightarrow\) iodine): 1. The colour becomes darker (chlorine is a pale green gas, bromine is a red-brown liquid, iodine is a grey-black solid). 2. Reactivity decreases because the distance between the nucleus and the outer shell increases, making it harder to attract and gain an incoming electron.
評分準則
1 mark for identifying that the elements become darker and their reactivity decreases down the group.
題目 7 · 選擇題
1 分
In an ecosystem, the feeding relationships are represented by the following food web: - Grass is eaten by rabbits and grasshoppers. - Grasshoppers are eaten by frogs. - Frogs are eaten by snakes. - Rabbits are eaten by snakes.
Which organism acts as both a secondary consumer and a tertiary consumer in this food web?
Therefore, the snake acts as a secondary consumer in the first pathway and as a tertiary consumer in the second pathway.
評分準則
1 mark for identifying the snake as both the secondary and tertiary consumer.
題目 8 · 選擇題
1 分
Copper(II) sulfate is a soluble salt. It can be prepared by reacting excess insoluble copper(II) oxide with dilute sulfuric acid. Which sequence of steps is correct to obtain pure, dry crystals of copper(II) sulfate?
A.filter to remove excess copper(II) oxide \(\rightarrow\) evaporate the filtrate to dryness to obtain the crystals
B.filter to remove excess copper(II) oxide \(\rightarrow\) heat the filtrate to the point of crystallization \(\rightarrow\) leave to cool and crystallize \(\rightarrow\) filter and dry the crystals
C.evaporate the mixture to dryness \(\rightarrow\) filter the solid residue \(\rightarrow\) wash with water
D.filter to remove excess copper(II) oxide \(\rightarrow\) wash the residue with water \(\rightarrow\) dry the residue to obtain the crystals
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解題
To prepare a pure sample of a soluble salt from an insoluble base: 1. The excess insoluble base (copper(II) oxide) is removed by filtration. 2. The filtrate (copper(II) sulfate solution) is heated to concentrate it until the crystallization point is reached. 3. The hot solution is allowed to cool slowly to form large, pure crystals. 4. The crystals are then filtered from the remaining liquid, washed with a small amount of cold distilled water, and dried (e.g., between pieces of filter paper).
評分準則
1 mark for identifying the correct sequence of experimental steps for crystallization of a soluble salt.
題目 9 · 選擇題
1 分
A car travels along a straight road. It accelerates uniformly from rest to a speed of 15 m/s in 6 seconds, then travels at a constant speed of 15 m/s for 10 seconds. Finally, it decelerates uniformly to a stop in 4 seconds. What is the total distance traveled by the car?
A.150 m
B.225 m
C.300 m
D.345 m
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解題
The total distance traveled can be calculated by finding the area under the speed-time graph. Area under the acceleration phase = \(0.5 \times 6 \text{ s} \times 15 \text{ m/s} = 45 \text{ m}\). Area under the constant speed phase = \(10 \text{ s} \times 15 \text{ m/s} = 150 \text{ m}\). Area under the deceleration phase = \(0.5 \times 4 \text{ s} \times 15 \text{ m/s} = 30 \text{ m}\). Total distance = \(45 \text{ m} + 150 \text{ m} + 30 \text{ m} = 225 \text{ m}\).
評分準則
1 mark for selecting the correct total distance of 225 m.
題目 10 · 選擇題
1 分
A student sets up a circuit with three resistors. Two 6.0 \(\Omega\) resistors are connected in parallel with each other. This combination is then connected in series with a 5.0 \(\Omega\) resistor. What is the total resistance of the circuit?
A.1.5 \(\Omega\)
B.5.0 \(\Omega\)
C.8.0 \(\Omega\)
D.17.0 \(\Omega\)
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解題
First, calculate the equivalent resistance of the two 6.0 \(\Omega\) resistors in parallel: \(1/R_p = 1/6.0 + 1/6.0 = 2/6.0\), which gives \(R_p = 3.0 \ \Omega\). Next, add the series resistor: \(R_{\text{total}} = R_p + 5.0 \ \Omega = 3.0 \ \Omega + 5.0 \ \Omega = 8.0 \ \Omega\).
評分準則
1 mark for identifying the correct total resistance of 8.0 \(\Omega\).
題目 11 · 選擇題
1 分
Which statement about electromagnetic waves is correct?
A.Radio waves have a higher frequency than ultraviolet waves.
B.All electromagnetic waves travel at the same speed in a vacuum.
C.X-rays have a longer wavelength than infrared waves.
D.Sound waves are a type of electromagnetic wave.
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解題
All electromagnetic waves travel at the same high speed (approximately \(3.0 \times 10^8 \text{ m/s}\)) in a vacuum. Radio waves have lower frequencies than ultraviolet, X-rays have shorter wavelengths than infrared, and sound waves are longitudinal mechanical waves, not electromagnetic waves.
評分準則
1 mark for identifying that all electromagnetic waves travel at the same speed in a vacuum.
題目 12 · 選擇題
1 分
A solid compound X is dissolved in water to make a solution. When aqueous sodium hydroxide is added to the solution, a light blue precipitate is formed. When dilute nitric acid and aqueous silver nitrate are added to another portion of the solution, a white precipitate is formed. What is the chemical name of compound X?
A.copper(II) chloride
B.copper(II) sulfate
C.iron(II) chloride
D.iron(III) sulfate
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解題
The light blue precipitate with aqueous sodium hydroxide indicates the presence of copper(II) ions, \(\text{Cu}^{2+}\). The white precipitate formed with aqueous silver nitrate in the presence of dilute nitric acid indicates the presence of chloride ions, \(\text{Cl}^{-}\). Therefore, the compound is copper(II) chloride.
評分準則
1 mark for identifying the compound as copper(II) chloride.
題目 13 · 選擇題
1 分
Which statement about the Group I alkali metals is correct as you go down the group?
A.The melting points of the elements increase.
B.The densities of the elements decrease.
C.The reactivity with water increases.
D.The atoms lose electrons less easily.
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解題
As you go down Group I, the outermost electron is further from the nucleus and more shielded by inner electron shells, making it easier to lose. Therefore, reactivity with water increases down the group. Melting points decrease and densities generally increase.
評分準則
1 mark for identifying that reactivity with water increases down the group.
題目 14 · 選擇題
1 分
Which method is most suitable for preparing a pure, dry sample of the soluble salt copper(II) sulfate?
A.Reacting copper metal with dilute sulfuric acid, then evaporating the solution.
B.Reacting copper(II) oxide with dilute sulfuric acid, filtering off excess oxide, then crystallising the filtrate.
C.Mixing aqueous copper(II) chloride with dilute sulfuric acid and filtering the precipitate.
D.Reacting copper(II) carbonate with dilute hydrochloric acid, then evaporating to dryness.
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解題
Copper metal does not react with dilute acids, so option A is incorrect. Copper(II) oxide is an insoluble base that reacts with dilute sulfuric acid to form copper(II) sulfate. Adding excess copper(II) oxide ensures all acid is neutralised, and the unreacted oxide can be filtered off. Crystallising the filtrate yields pure copper(II) sulfate crystals. Option C produces no precipitate, and option D produces copper(II) chloride.
評分準則
1 mark for selecting the correct preparation method using copper(II) oxide and dilute sulfuric acid, followed by filtration and crystallisation.
題目 15 · 選擇題
1 分
Which combination of features is characteristic of a typical wind-pollinated flower?
C.Small, green petals; nectar guides present; sticky, heavy pollen.
D.Large, scented petals; long filaments inside the flower; nectar-producing glands.
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解題
Wind-pollinated flowers do not need to attract insects, so they have small, dull petals. They have feathery stigmas that hang outside the flower to capture drifting pollen, and they produce light, smooth pollen grains that are easily carried by the wind.
評分準則
1 mark for selecting the correct set of wind-pollination characteristics.
題目 16 · 選擇題
1 分
A student examines a cell under a light microscope. Which combination of features proves that the cell being observed is a plant cell and not an animal cell?
A.cell wall and cytoplasm
B.cell wall and chloroplasts
C.cell membrane and large permanent vacuole
D.cell wall and nucleus
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解題
Plant cells possess a cell wall and chloroplasts, which are completely absent in animal cells. Both plant and animal cells contain a nucleus, cytoplasm, and cell membrane, so those features alone cannot differentiate them.
評分準則
1 mark for identifying the correct combination of plant-specific cell structures.
題目 17 · 選擇題
1 分
An object of mass \(500\text{ g}\) is moving with a speed of \(2.0\text{ m/s}\). It accelerates to a speed of \(6.0\text{ m/s}\). What is the increase in the kinetic energy of the object?
A.\(4.0\text{ J}\)
B.\(8.0\text{ J}\)
C.\(16.0\text{ J}\)
D.\(32.0\text{ J}\)
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解題
First, convert the mass of the object from grams to kilograms: \(500\text{ g} = 0.5\text{ kg}\). Next, use the formula for kinetic energy, \(E_k = \frac{1}{2}mv^2\). The initial kinetic energy is \(\frac{1}{2} \times 0.5\text{ kg} \times (2.0\text{ m/s})^2 = 1.0\text{ J}\). The final kinetic energy is \(\frac{1}{2} \times 0.5\text{ kg} \times (6.0\text{ m/s})^2 = 9.0\text{ J}\). The increase in kinetic energy is \(9.0\text{ J} - 1.0\text{ J} = 8.0\text{ J}\).
評分準則
1 mark for the correct calculation of kinetic energy increase (8.0 J).
題目 18 · 選擇題
1 分
A student carries out chemical tests on an aqueous solution of salt X. Adding aqueous sodium hydroxide results in a green precipitate that is insoluble in excess. Adding dilute nitric acid followed by aqueous silver nitrate results in a white precipitate. What is the chemical name of salt X?
A.copper(II) chloride
B.iron(II) chloride
C.iron(II) sulfate
D.iron(III) chloride
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解題
The formation of a green precipitate with sodium hydroxide that is insoluble in excess indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\). The formation of a white precipitate with silver nitrate in the presence of nitric acid indicates the presence of chloride ions, \(\text{Cl}^{-}\). Therefore, the salt is iron(II) chloride.
評分準則
1 mark for identifying both the iron(II) cation and the chloride anion to determine iron(II) chloride.
題目 19 · 選擇題
1 分
Radio waves travel through a vacuum at a speed of \(3.0 \times 10^8\text{ m/s}\). A radio station broadcasts at a frequency of \(100\text{ MHz}\). What is the wavelength of the radio waves?
A.\(0.33\text{ m}\)
B.\(3.0\text{ m}\)
C.\(30\text{ m}\)
D.\(300\text{ m}\)
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解題
Use the wave equation \(v = f \lambda\). Rearranging the formula gives \(\lambda = v / f\). Note that \(100\text{ MHz} = 100 \times 10^6\text{ Hz} = 1.0 \times 10^8\text{ Hz}\). Thus, \(\lambda = (3.0 \times 10^8\text{ m/s}) / (1.0 \times 10^8\text{ Hz}) = 3.0\text{ m}\).
評分準則
1 mark for the correct conversion of frequency and application of the wave equation to obtain 3.0 m.
題目 20 · 選擇題
1 分
Which row correctly describes the features of a wind-pollinated flower?
A.Petals: large and brightly colored; Anthers: fixed inside the flower; Pollen: heavy and sticky
B.Petals: small and dull-colored; Anthers: hanging outside the flower; Pollen: light and smooth
C.Petals: large and brightly colored; Anthers: hanging outside the flower; Pollen: light and smooth
D.Petals: small and dull-colored; Anthers: fixed inside the flower; Pollen: heavy and sticky
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解題
Wind-pollinated flowers do not need to attract insects, so their petals are small and dull. Their anthers hang outside the flower to allow the wind to blow the pollen away easily. The pollen grains are light and smooth so they can easily float in the wind.
評分準則
1 mark for selecting the correct combination of petals, anthers, and pollen features for wind-pollination.
題目 21 · 選擇題
1 分
Three resistors, each with a resistance of \(6.0\ \Omega\), are connected in parallel. What is the total combined resistance of this network?
A.\(2.0\ \Omega\)
B.\(6.0\ \Omega\)
C.\(18.0\ \Omega\)
D.\(0.50\ \Omega\)
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解題
For resistors in parallel, the reciprocal of the total resistance is the sum of the reciprocals of the individual resistances: \(1/R_{\text{total}} = 1/R_1 + 1/R_2 + 1/R_3 = 1/6.0 + 1/6.0 + 1/6.0 = 3/6.0 = 1/2.0\). Therefore, \(R_{\text{total}} = 2.0\ \Omega\).
評分準則
1 mark for the correct calculation of parallel resistance.
題目 22 · 選擇題
1 分
What are the trends in color intensity and reactivity of the Group VII elements (halogens) as the group is descended from chlorine to iodine?
As Group VII is descended, the color intensity of the halogens increases (chlorine is a pale green gas, bromine is a red-brown liquid, and iodine is a grey-black solid). However, their chemical reactivity decreases because the atomic radius increases, making it harder to attract and gain an incoming electron.
評分準則
1 mark for correctly identifying that color intensity increases and reactivity decreases down the group.
題目 23 · 選擇題
1 分
The following food chain is found in an ecosystem: Grass -> Grasshopper -> Frog -> Snake. Which statement is correct?
A.The frog is a primary consumer because it feeds on grasshoppers.
B.The grasshopper occupies the first trophic level.
C.Energy is lost, primarily as heat, at each transfer between trophic levels.
D.The population size of snakes must be larger than that of the grasshoppers.
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解題
Energy is lost at each trophic level transfer due to processes like respiration (releasing heat), excretion, and uneaten parts of organisms. This limits the efficiency of energy transfer along the food chain.
評分準則
1 mark for identifying energy loss as heat as the correct biological principle.
題目 24 · 選擇題
1 分
Which method is most suitable for preparing a pure, dry sample of copper(II) sulfate crystals?
A.Reacting copper metal with dilute sulfuric acid, then evaporating the mixture to dryness.
B.Reacting excess copper(II) oxide with dilute sulfuric acid, filtering the mixture, heating the filtrate to saturation, and allowing it to crystallize.
C.Reacting copper(II) carbonate with aqueous sodium sulfate, then filtering to collect the precipitate.
D.Titrating aqueous copper(II) hydroxide with dilute sulfuric acid using a methyl orange indicator.
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解題
Copper is an unreactive metal and does not react with dilute acids. Copper(II) oxide is an insoluble base that reacts with dilute sulfuric acid to form soluble copper(II) sulfate. Excess oxide is added to ensure all acid is neutralized, which is then filtered off. Heating the filtrate to saturation and leaving it to cool allows pure copper(II) sulfate crystals to form.
評分準則
1 mark for identifying the correct reaction pathway and practical steps (excess insoluble base, filtration, and crystallization) for copper(II) sulfate preparation.
題目 25 · 選擇題
1 分
A block of mass 2.0 kg is pulled 4.0 m up a frictionless slope by a constant force of 15 N parallel to the slope. The vertical height gained is 2.0 m. Take \(g = 10\text{ m/s}^2\). What is the increase in kinetic energy of the block?
A.20 J
B.40 J
C.60 J
D.80 J
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解題
Work done by the pulling force is \(W = F \times d = 15\text{ N} \times 4.0\text{ m} = 60\text{ J}\). The increase in gravitational potential energy is \(\Delta E_p = mgh = 2.0\text{ kg} \times 10\text{ m/s}^2 \times 2.0\text{ m} = 40\text{ J}\). Since the slope is frictionless, the work done by the force goes into increasing both potential and kinetic energy: \(W = \Delta E_p + \Delta E_k\). Therefore, the increase in kinetic energy is \(\Delta E_k = 60\text{ J} - 40\text{ J} = 20\text{ J}\).
評分準則
1 mark for the correct calculation of kinetic energy change, matching option A.
題目 26 · 選擇題
1 分
A student performs tests on an unknown solution X.
- Adding dilute nitric acid followed by aqueous barium nitrate produces no precipitate. - Adding dilute nitric acid followed by aqueous silver nitrate produces a cream precipitate. - Adding aqueous sodium hydroxide produces a green precipitate that is insoluble in excess.
What is the identity of solution X?
A.Iron(II) bromide
B.Iron(III) bromide
C.Iron(II) chloride
D.Copper(II) bromide
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解題
The lack of precipitate with barium nitrate shows that sulfate ions are absent. The cream precipitate with nitric acid and silver nitrate shows that bromide ions are present. The green precipitate with sodium hydroxide that is insoluble in excess shows that iron(II) ions are present. Therefore, solution X is iron(II) bromide.
評分準則
1 mark for identifying the correct cation and anion from the test results, matching option A.
題目 27 · 選擇題
1 分
The speed of light in a vacuum is \(3.0 \times 10^8\text{ m/s}\). A ray of light of frequency \(5.0 \times 10^{14}\text{ Hz}\) enters a glass block. The refractive index of the glass is 1.5. What is the speed and wavelength of this light wave inside the glass?
The speed of light in glass is \(v = \frac{c}{n} = \frac{3.0 \times 10^8\text{ m/s}}{1.5} = 2.0 \times 10^8\text{ m/s}\). The frequency \(f\) of the light wave does not change when it enters a new medium. Using the wave equation \(v = f\lambda\), the wavelength inside the glass is \(\lambda = \frac{v}{f} = \frac{2.0 \times 10^8\text{ m/s}}{5.0 \times 10^{14}\text{ Hz}} = 4.0 \times 10^{-7}\text{ m}\).
評分準則
1 mark for the correct speed and wavelength calculation inside the medium, matching option A.
題目 28 · 選擇題
1 分
Which statement correctly compares human sperm and egg cells?
A.Sperm cells are produced in millions, are self-motile using a flagellum, and are very small; egg cells are produced in limited numbers, move passively, and are much larger.
B.Sperm cells are produced in limited numbers, are self-motile using a flagellum, and are large; egg cells are produced in millions, move passively, and are small.
C.Sperm cells are produced in millions, move passively, and are small; egg cells are produced in limited numbers, are self-motile, and are large.
D.Sperm cells are produced in limited numbers, move passively, and are large; egg cells are produced in millions, are self-motile, and are small.
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解題
Sperm cells are small, produced in huge numbers (millions), and have a tail (flagellum) for active movement. Egg cells are much larger (containing energy stores), produced in limited numbers (usually one per month in humans), and cannot move on their own (they are moved passively along the oviduct).
評分準則
1 mark for the correct comparison of sperm and egg cells, matching option A.
題目 29 · 選擇題
1 分
Two resistors of resistance \(3\ \Omega\) and \(6\ \Omega\) are connected in parallel. This combination is connected in series with a \(2\ \Omega\) resistor and a \(12\text{ V}\) power supply. What is the current flowing through the \(6\ \Omega\) resistor?
A.1.0 A
B.1.5 A
C.2.0 A
D.3.0 A
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解題
First, find the equivalent resistance of the parallel combination: \(R_p = \frac{3 \times 6}{3 + 6} = 2\ \Omega\). The total resistance of the circuit is \(R_{\text{total}} = R_p + 2\ \Omega = 2\ \Omega + 2\ \Omega = 4\ \Omega\). The total current from the power supply is \(I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{4\ \Omega} = 3\text{ A}\). The potential difference across the parallel combination is \(V_p = I_{\text{total}} \times R_p = 3\text{ A} \times 2\ \Omega = 6\text{ V}\). The current through the \(6\ \Omega\) resistor is \(I_{6\Omega} = \frac{V_p}{6\ \Omega} = \frac{6\text{ V}}{6\ \Omega} = 1.0\text{ A}\).
評分準則
1 mark for the correct calculation of current through the specific resistor, matching option A.
題目 30 · 選擇題
1 分
The table shows some properties of Group VII elements (halogens).
| Element | State at room temperature | Color | Reactivity | |---|---|---|---| | Chlorine | Gas | Pale green | Highly reactive | | Bromine | Liquid | Red-brown | Moderately reactive | | Iodine | Solid | Grey | Mildly reactive |
Based on these trends, which statement is correct?
A.Chlorine can displace iodine from an aqueous solution of potassium iodide, resulting in a brown color.
B.Iodine is a stronger oxidizing agent than bromine and chlorine because it is a solid.
C.Down the group, the elements become more reactive as their atomic radius decreases.
D.Adding aqueous bromine to aqueous sodium chloride produces a pale green solution of chlorine gas.
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解題
Reactivity of Group VII elements decreases down the group. Therefore, chlorine is more reactive than iodine and can displace iodide ions from potassium iodide solution: \(\text{Cl}_2(\text{aq}) + 2\text{KI}(\text{aq}) \rightarrow 2\text{KCl}(\text{aq}) + \text{I}_2(\text{aq})\). The liberated iodine turns the solution brown. Bromine is less reactive than chlorine, so it cannot displace chloride from sodium chloride.
評分準則
1 mark for correctly identifying the displacement reaction and color change, matching option A.
題目 31 · 選擇題
1 分
In an ecosystem, feeding relationships are as follows:
- Oak trees are eaten by caterpillars and aphids. - Caterpillars are eaten by blue tits. - Aphids are eaten by ladybirds. - Ladybirds are eaten by blue tits. - Blue tits are eaten by sparrowhawks.
At which trophic levels does the sparrowhawk feed?
A.4 and 5
B.3 and 4
C.4 only
D.3, 4 and 5
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解題
Let's trace the food chains starting from the producer (Oak tree, trophic level 1): 1. Oak tree (1) \(\rightarrow\) Caterpillar (2) \(\rightarrow\) Blue tit (3) \(\rightarrow\) Sparrowhawk (4). 2. Oak tree (1) \(\rightarrow\) Aphid (2) \(\rightarrow\) Ladybird (3) \(\rightarrow\) Blue tit (4) \(\rightarrow\) Sparrowhawk (5). In the first chain, the sparrowhawk is at trophic level 4. In the second chain, the sparrowhawk is at trophic level 5. Thus, the sparrowhawk feeds at both trophic levels 4 and 5.
評分準則
1 mark for identifying both pathways and their corresponding trophic levels, matching option A.
題目 32 · 選擇題
1 分
Which method is most suitable for preparing a pure, dry sample of the insoluble salt, barium sulfate?
A.Mix aqueous barium nitrate and aqueous sodium sulfate, filter the mixture, wash the residue with distilled water, and dry it.
B.Mix solid barium carbonate and dilute sulfuric acid, filter the mixture, evaporate the filtrate, and crystallize.
C.Titrate aqueous barium hydroxide with dilute sulfuric acid, and evaporate the resulting solution to dryness.
D.Heat barium metal with sulfur powder in a crucible, wash with water, and dry.
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解題
Barium sulfate is an insoluble salt. Insoluble salts are prepared by precipitation, which involves mixing two soluble salt solutions (aqueous barium nitrate and aqueous sodium sulfate). The solid barium sulfate precipitate is separated from the soluble sodium nitrate by filtration. The residue (barium sulfate) is then washed with distilled water to remove impurities and dried.
評分準則
1 mark for identifying the precipitation method as correct for insoluble salt preparation, matching option A.
題目 33 · 選擇題
1 分
An object of mass 4.0 kg is accelerated from rest to a speed of 6.0 m/s in 3.0 s. What is the average power developed during this time?
A.8.0 W
B.24 W
C.48 W
D.72 W
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解題
The initial kinetic energy of the object is 0 J since it starts from rest. The final kinetic energy is calculated using the formula KE = 0.5 * m * v^2. Substituting the values: KE = 0.5 * 4.0 kg * (6.0 m/s)^2 = 2.0 * 36 = 72 J. The work done on the object is equal to the change in kinetic energy, which is 72 J. Power is the rate of doing work: Power = Work Done / time = 72 J / 3.0 s = 24 W.
評分準則
1 mark for the correct option (b). Method: KE = 0.5 * m * v^2 is used to find work done (72 J), and Power = Work / time is used to find average power (24 W).
題目 34 · 選擇題
1 分
Which cell structures are present in both a root hair cell and a palisade mesophyll cell of a plant?
A.cell wall, cell membrane and chloroplasts
B.cell wall, cell membrane and vacuole
C.cell membrane, vacuole and chloroplasts
D.nucleus, cytoplasm and chloroplasts
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解題
Both root hair cells and palisade mesophyll cells are plant cells and therefore possess a cell wall, a cell membrane, cytoplasm, a nucleus, and a large central vacuole. However, root hair cells are located underground and do not contain chloroplasts as they do not perform photosynthesis. Therefore, any option containing chloroplasts is incorrect, leaving cell wall, cell membrane, and vacuole as the correct answer.
評分準則
1 mark for the correct option (b). Correctly identifies that root hair cells do not contain chloroplasts but share all other major plant cell structures with palisade mesophyll cells.
題目 35 · 選擇題
1 分
In the preparation of the soluble salt copper(II) sulfate, excess insoluble copper(II) oxide is reacted with warm dilute sulfuric acid. Why is excess copper(II) oxide added?
A.to ensure that all of the copper(II) oxide reacts completely
B.to increase the rate of the reaction
C.to ensure that all of the sulfuric acid is fully neutralised
D.to prevent the copper(II) sulfate crystals from decomposing
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解題
Excess copper(II) oxide is added to ensure that all of the sulfuric acid is completely reacted (neutralised). Since copper(II) oxide is insoluble, any unreacted excess can be easily removed by filtration, leaving a pure solution of copper(II) sulfate. If acid were left unreacted, it would contaminate the salt crystals when the water is evaporated.
評分準則
1 mark for the correct option (c). Recognises that using excess insoluble reactant ensures all of the limiting reactant (acid) is fully consumed and neutralised.
題目 36 · 選擇題
1 分
An electromagnetic wave has a frequency of 6.0 x 10^(14) Hz in a vacuum. What is its wavelength? (The speed of electromagnetic waves in a vacuum is 3.0 x 10^(8) m/s)
A.2.0 x 10^(-6) m
B.5.0 x 10^(-7) m
C.1.8 x 10^(23) m
D.2.0 x 10^(6) m
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解題
Using the wave equation: v = f * lambda, where v is the speed of light, f is the frequency, and lambda is the wavelength. Rearranging for wavelength: lambda = v / f = (3.0 x 10^8 m/s) / (6.0 x 10^14 Hz) = 0.5 x 10^-6 m = 5.0 x 10^-7 m.
評分準則
1 mark for the correct option (b). Correctly applies the wave equation v = f * lambda with proper exponent arithmetic.
題目 37 · 選擇題
1 分
Which statement correctly compares asexual and sexual reproduction?
A.Asexual reproduction produces genetically different offspring, while sexual reproduction produces genetically identical offspring.
B.Asexual reproduction involves meiosis, while sexual reproduction involves only mitosis.
C.Asexual reproduction requires only one parent, while sexual reproduction involves the fusion of haploid nuclei.
D.Asexual reproduction produces diploid gametes, while sexual reproduction produces haploid gametes.
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解題
Asexual reproduction requires only a single parent and results in genetically identical offspring (clones) via mitosis. Sexual reproduction involves the fusion of haploid gamete nuclei (fertilisation) to form a diploid zygote, resulting in genetically diverse offspring.
評分準則
1 mark for the correct option (c). Correctly identifies key differences in the number of parents and the process of gamete fusion between the two forms of reproduction.
題目 38 · 選擇題
1 分
Which statement about the fractional distillation of petroleum is correct?
A.Fractions with larger molecules condense at the top of the fractionating column where it is coolest.
B.Fractions with smaller molecules have higher boiling points and are collected at the bottom.
C.The temperature decreases up the fractionating column, so smaller molecules with lower boiling points condense near the top.
D.Petroleum is separated into different fractions based on the different solubilities of the hydrocarbons in water.
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解題
In a fractionating column, the temperature is highest at the bottom and lowest at the top. Hydrocarbons with smaller molecules have lower boiling points and do not condense easily; they rise up the column to the cooler regions at the top before they condense. Larger molecules have higher boiling points and condense near the bottom.
評分準則
1 mark for the correct option (c). Recognises the temperature gradient in the column and the relation between molecular size, boiling point, and height of condensation.
題目 39 · 選擇題
1 分
Two resistors, one of 3.0 ohms and one of 6.0 ohms, are connected in parallel. This combination is then connected in series with a 2.0 ohm resistor and a 12 V d.c. power supply. What is the current flowing from the power supply?
A.1.1 A
B.2.0 A
C.3.0 A
D.6.0 A
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解題
First, calculate the equivalent resistance of the two parallel resistors (3.0 ohms and 6.0 ohms): 1/Rp = 1/3 + 1/6 = 2/6 + 1/6 = 3/6, so Rp = 2.0 ohms. Next, calculate the total resistance of the circuit by adding the series resistor: Rtotal = Rp + 2.0 ohms = 2.0 ohms + 2.0 ohms = 4.0 ohms. Finally, use Ohm's law to find the total current: I = V / Rtotal = 12 V / 4.0 ohms = 3.0 A.
評分準則
1 mark for the correct option (c). Awarded for finding parallel equivalent resistance of 2.0 ohms, total resistance of 4.0 ohms, and correct final current calculation.
題目 40 · 選擇題
1 分
Which statement correctly describes the trends in properties of the Group VII halogens as you go down the group?
A.The elements become more reactive and their color gets lighter.
B.The elements become less reactive and their boiling point increases.
C.The elements become more reactive and their density decreases.
D.The elements become less reactive and their color gets lighter.
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解題
As you go down Group VII (the halogens): reactivity decreases (chlorine is more reactive than bromine, which is more reactive than iodine); density increases; boiling points and melting points increase (fluorine and chlorine are gases, bromine is a liquid, iodine is a solid); and the colors become darker (chlorine is pale green, bromine is red-brown, iodine is grey-black). Therefore, they become less reactive and their boiling point increases.
評分準則
1 mark for the correct option (b). Correctly recalls the periodic trends for reactivity and boiling points of halogens down the group.
Paper 41 (Theory - Extended)
Answer all structured questions on the space provided on the question paper.
9 題目 · 79.92 分
題目 1 · structured-theory
8.88 分
A toy car of mass \(0.25\text{ kg}\) is released from rest from the top of a ramp of height \(0.80\text{ m}\). It travels down the ramp and reaches a speed of \(3.2\text{ m/s}\) at the bottom. Use \(g = 9.8\text{ m/s}^2\).
(a) (i) Calculate the gravitational potential energy (\(E_p\)) of the toy car at the top of the ramp. (ii) Calculate the kinetic energy (\(E_k\)) of the car at the bottom of the ramp. (iii) State the law of conservation of energy and explain why the kinetic energy at the bottom is less than the gravitational potential energy at the top.
(b) The car then travels along a flat horizontal surface and is brought to rest by a constant braking force of \(0.40\text{ N}\). Calculate the distance the car travels before coming to rest.
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解題
Detailed Solution: (a) (i) \(E_p = mgh = 0.25\text{ kg} \times 9.8\text{ m/s}^2 \times 0.80\text{ m} = 1.96\text{ J}\). (ii) \(E_k = \frac{1}{2}mv^2 = 0.5 \times 0.25\text{ kg} \times (3.2\text{ m/s})^2 = 0.125 \times 10.24 = 1.28\text{ J}\). (iii) The law of conservation of energy states that energy cannot be created or destroyed, but can only be changed from one form to another. The kinetic energy is less than the initial gravitational potential energy because friction and air resistance act on the car, transferring some energy into thermal energy and sound energy to the surroundings. (b) Work done by the braking force = change in kinetic energy of the car. \(W = F \times d\) \(1.28\text{ J} = 0.40\text{ N} \times d\) \(d = \frac{1.28}{0.40} = 3.2\text{ m}\).
評分準則
Marking Scheme (9 Marks Total): (a) (i) [2 marks] - Formula: \(E_p = mgh\) or correct substitution: \(0.25 \times 9.8 \times 0.80\) (1 mark) - Correct answer: \(1.96\text{ J}\) (accept \(2.0\text{ J}\) if \(g=10\) is used) (1 mark)
(a) (iii) [2 marks] - Statement of conservation of energy: energy cannot be created or destroyed, only converted/transferred (1 mark) - Explanation: energy is dissipated/transferred to surroundings as thermal energy/heat/sound due to friction (1 mark)
(b) [3 marks] - Equating work done to loss of kinetic energy: \(W = F \times d\) or \(F \times d = E_k\) (1 mark) - Substitution: \(0.40 \times d = 1.28\) (or ecf from a(ii)) (1 mark) - Correct calculation of distance: \(3.2\text{ m}\) (1 mark)
題目 2 · structured-theory
8.88 分
A student investigates an unknown green solution, X, which contains transition metal ions.
(a) Describe how paper chromatography can be used to show that the green solution X is a mixture of different coloured substances.
(b) To identify the cations present, the student adds aqueous sodium hydroxide to a sample of solution X. A green precipitate is formed which is insoluble in excess sodium hydroxide. (i) Identify the cation responsible for this observation. (ii) Write the ionic equation, with state symbols, for the formation of this green precipitate.
(c) To identify an anion present in solution X, the student adds dilute nitric acid followed by aqueous barium nitrate. A white precipitate forms. (i) Identify the anion. (ii) State why the nitric acid is added before the barium nitrate.
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解題
Detailed Solution: (a) Paper chromatography is carried out by drawing a pencil line near the bottom of a strip of chromatography paper. A small spot of solution X is applied on the line. The paper is suspended in a container containing a suitable solvent (such as water or ethanol), ensuring the solvent level is below the pencil line. As the solvent rises through the paper, the components of the mixture travel at different speeds depending on their solubility and affinity to the paper. If solution X is a mixture, it will separate into two or more distinct spots at different heights. (b) (i) Iron(II) ions, \(Fe^{2+}\), form a green precipitate of iron(II) hydroxide, \(Fe(OH)_2\), with aqueous sodium hydroxide, which does not dissolve in excess NaOH. (ii) The ionic equation is: \(Fe^{2+}(aq) + 2OH^-(aq) \rightarrow Fe(OH)_2(s)\). (c) (i) Barium ions react with sulfate ions to form an insoluble white precipitate of barium sulfate, \(BaSO_4\). Therefore, the anion is sulfate, \(SO_4^{2-}\). (ii) Dilute nitric acid is added to eliminate any carbonate ions (\(CO_3^{2-}\)) present in the solution. Carbonate ions would react with barium to form barium carbonate, which is also a white precipitate and would give a false positive result for sulfate.
評分準則
Marking Scheme (9 Marks Total): (a) [3 marks] - Draw start line in pencil and place spot of sample on it (1 mark) - Place paper in solvent such that the solvent level is below the start line (1 mark) - Observe separation of the green spot into multiple spots as solvent moves up (1 mark)
(b) (ii) [2 marks] - Correct formulae of reactants and products: \(Fe^{2+} + 2OH^- \rightarrow Fe(OH)_2\) (1 mark) - Correct state symbols: \((aq)\) for ions, \((s)\) for precipitate (1 mark)
(c) (ii) [2 marks] - To react with/remove carbonate impurities/ions (1 mark) - To prevent a false positive result / prevent barium carbonate precipitate from forming (1 mark)
題目 3 · structured-theory
8.88 分
This question compares electromagnetic waves and sound waves.
(a) Contrast electromagnetic waves and sound waves in terms of: (i) their nature as transverse or longitudinal waves. (ii) their ability to travel through a vacuum.
(b) A red laser light has a wavelength of \(6.5 \times 10^{-7}\text{ m}\) in air. (i) State the speed of this light wave in air. (ii) Calculate the frequency of the red laser light.
(c) This laser light travels from air into a glass block. State and explain what happens, if anything, to: (i) the frequency of the light wave. (ii) the speed of the light wave.
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解題
Detailed Solution: (a) (i) Electromagnetic waves are transverse waves, meaning their vibrations are perpendicular to the direction of wave travel. Sound waves are longitudinal waves, meaning their vibrations are parallel to the direction of wave travel. (ii) Electromagnetic waves do not require a physical medium and can travel through a vacuum. Sound waves require a medium to propagate; they cannot travel through a vacuum. (b) (i) The speed of all electromagnetic waves in air or vacuum is approximately \(3.0 \times 10^8\text{ m/s}\). (ii) Using the wave equation: \(v = f \lambda\) \(f = \frac{v}{\lambda} = \frac{3.0 \times 10^8\text{ m/s}}{6.5 \times 10^{-7}\text{ m}} \approx 4.6 \times 10^{14}\text{ Hz}\). (c) (i) The frequency remains unchanged because the frequency of a wave is determined solely by the source, not the medium. (ii) The speed of the light wave decreases because glass has a higher optical density (or refractive index) than air, which slows down the electromagnetic wave.
評分準則
Marking Scheme (9 Marks Total): (a) (i) [2 marks] - Electromagnetic waves are transverse (1 mark) - Sound waves are longitudinal (1 mark)
(a) (ii) [2 marks] - Electromagnetic waves can travel through a vacuum (1 mark) - Sound waves cannot travel through a vacuum / require a medium (1 mark)
(c) [2 marks] - (i) Frequency remains unchanged because frequency depends on the source (1 mark) - (ii) Speed decreases because glass has a higher refractive index than air (1 mark)
題目 4 · structured-theory
8.88 分
Human reproduction involves the fusion of specialized gametes.
(a) (i) Define the term fertilization in humans. (ii) State the exact structure in the female reproductive system where fertilization normally occurs.
(b) Compare male and female gametes (sperm and egg cells) in humans in terms of: - their relative size - their mobility (motility) - the relative numbers produced.
(c) Describe two structural adaptations of a human sperm cell and explain how each adaptation relates to its function.
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解題
Detailed Solution: (a) (i) Fertilization is defined as the fusion of the nucleus of a male gamete (sperm) with the nucleus of a female gamete (egg/ovum) to form a zygote (with a diploid nucleus). (ii) Fertilization normally takes place in the oviduct (or fallopian tube). (b) - Relative size: Sperm cells are microscopic and very small compared to egg cells, which are much larger because they contain food stores (cytoplasm with yolk). - Mobility: Sperm cells are highly mobile (motile) and can actively swim using their flagellum. Egg cells are non-motile and are swept along the oviduct by ciliated cells. - Relative numbers: Sperm cells are produced in massive numbers (millions), whereas egg cells are released in very limited numbers (typically only one per menstrual cycle). (c) Standard adaptations of a sperm cell: 1. Flagellum (tail): Provides motility, allowing the sperm to swim to the egg. 2. Acrosome: Contains digestive enzymes that are released to break down the jelly coat/outer protective layers of the egg cell so the sperm nucleus can enter. 3. Mitochondria-rich midpiece: Mitochondria release energy through aerobic respiration to power the flagellum.
評分準則
Marking Scheme (9 Marks Total): (a) (i) [2 marks] - Fusion of nuclei (1 mark) - of male and female gametes / sperm and egg (to form a zygote) (1 mark)
(b) [3 marks] - Size: Egg is larger / sperm is smaller (1 mark) - Mobility: Sperm is motile / egg is non-motile (1 mark) - Numbers: Sperm are produced in larger quantities / eggs are produced in smaller quantities (1 mark)
(c) [3 marks] - Any two adaptations stated with correct functional explanations (1.5 marks per pair): - Flagellum/tail (1 mark) + to swim to the egg (1/2 mark) - Midpiece containing many mitochondria (1 mark) + to provide/release energy for swimming (1/2 mark) - Acrosome (1 mark) + containing enzymes to digest/penetrate the outer layer of the egg (1/2 mark) - Max 3 marks total.
題目 5 · structured-theory
8.88 分
A student constructs a circuit containing a \(12\text{ V}\) d.c. power supply, a \(4.0\ \Omega\) fixed resistor, and a thermistor connected in series. A voltmeter is connected to measure the potential difference across the thermistor.
(a) Draw a circuit diagram for this setup using standard electrical symbols.
(b) Initially, at room temperature, the resistance of the thermistor is \(8.0\ \Omega"). (i) Calculate the total resistance of the circuit. (ii) Calculate the current flowing through the circuit. (iii) Determine the reading on the voltmeter.
(c) The thermistor is now heated using a hairdryer. State and explain what happens to the reading on the voltmeter.
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解題
Detailed Solution: (a) The circuit diagram must show: 1. A battery/DC source symbol connected in series with other components. 2. A series loop containing the fixed resistor (rectangular box) and the thermistor (rectangular box with a diagonal line starting from the bottom-left and ending with a flat line at the top-right). 3. A voltmeter (circle with 'V') connected in parallel across the thermistor only. (b) (i) In a series circuit, total resistance is the sum of the individual resistances: \(R_T = 4.0\ \Omega + 8.0\ \Omega = 12.0\ \Omega\). (ii) Using Ohm's Law for the whole circuit: \(I = \frac{V}{R_T} = \frac{12\text{ V}}{12.0\ \Omega} = 1.0\text{ A}\). (iii) The voltmeter measures the potential difference across the thermistor: \(V_{\text{thermistor}} = I \times R_{\text{thermistor}} = 1.0\text{ A} \times 8.0\ \Omega = 8.0\text{ V}\). (c) When the thermistor is heated, its resistance decreases. Because the fixed resistor and thermistor form a potential divider, a decrease in the resistance of the thermistor means it takes a smaller fraction of the total potential difference (12 V). Alternatively, the total current increases, increasing the potential difference across the fixed resistor, leaving less potential difference across the thermistor. Therefore, the reading on the voltmeter decreases.
評分準則
Marking Scheme (9 Marks Total): (a) [3 marks] - Correct battery/power supply symbol connected in series with other components (1 mark) - Correct symbols for fixed resistor and thermistor (1 mark) - Voltmeter connected in parallel across the thermistor (1 mark)
(b) (ii) [2 marks] - Formula/substitution: \(I = V / R = 12 / 12\) (1 mark) - Correct answer: \(1.0\text{ A}\) (including unit) (1 mark)
(b) (iii) [1 mark] - Correct calculation: \(8.0\text{ V}\) (allow ecf from b(i) and b(ii)) (1 mark)
(c) [2 marks] - Voltmeter reading decreases (1 mark) - Explanation: heating decreases the resistance of the thermistor, so a smaller proportion of the voltage is dropped across it / current increases, so voltage across the fixed resistor increases (1 mark)
題目 6 · structured-theory
8.88 分
The Periodic Table displays trends in physical and chemical properties of elements.
(a) Lithium, sodium, and potassium are in Group 1 of the Periodic Table. (i) State how the reactivity of Group 1 elements with water changes down the group. (ii) Write a balanced chemical equation, including state symbols, for the reaction of sodium with water.
(b) Chlorine, bromine, and iodine are in Group 7 (the halogens). (i) Describe the physical states of chlorine, bromine, and iodine at room temperature and pressure. (ii) Aqueous chlorine is added to a solution of potassium bromide. Describe the observation and write the ionic equation for the reaction that occurs.
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解題
Detailed Solution: (a) (i) Down Group 1, the outer shell electron is further from the nucleus, meaning there is more shielding and less electrostatic attraction. Thus, the outer electron is lost more easily, and reactivity increases down the group. (ii) Sodium reacts with water to form sodium hydroxide and hydrogen gas: \(2\text{Na(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)} + \text{H}_2\text{(g)}\). (b) (i) At room temperature and pressure, chlorine is a gas, bromine is a liquid, and iodine is a solid. (ii) Chlorine is more reactive than bromine because it is higher up in Group 7 and can attract electrons more readily. Therefore, chlorine displaces bromide ions from potassium bromide. The reaction produces bromine (\(\text{Br}_2\)), which turns the colourless solution orange or brown. The ionic equation is: \(\text{Cl}_2 + 2\text{Br}^- \rightarrow 2\text{Cl}^- + \text{Br}_2\).
評分準則
Marking Scheme (9 Marks Total): (a) (i) [1 mark] - Reactivity increases (down the group) (1 mark)
An ecosystem consists of several interacting populations: grass, grasshoppers, frogs, snakes, and fungi.
(a) (i) Construct a food chain containing four of these organisms. Draw arrows to show the flow of energy between the trophic levels. (ii) State which organism in your food chain is the producer and define this term.
(b) Explain why food chains rarely contain more than five trophic levels, referring to the efficiency of energy transfer.
(c) State the role of fungi in this ecosystem and explain why their activity is essential for the continuous growth of the grass.
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解題
Detailed Solution: (a) (i) The correct sequence representing energy flow is: grass \(\rightarrow\) grasshopper \(\rightarrow\) frog \(\rightarrow\) snake. The arrows must point from the organism being eaten to the consumer eating it. (ii) The producer is grass. A producer is an organism that synthesizes its own organic nutrients (food) from inorganic molecules, using energy from sunlight via photosynthesis. (b) Energy transfer between trophic levels is highly inefficient; typically only about 10% of the energy from one trophic level is passed on to the next. The remaining 90% is lost as heat, used in respiration, lost in excretion (urine/feces), or remains in uneaten parts. By the time energy reaches the fourth or fifth trophic level, the total amount of energy available is too small to support another consumer population. (c) Fungi act as decomposers in the ecosystem. They break down dead organic matter and animal wastes. This process is essential because it recycles nutrients, releasing inorganic minerals (such as nitrates and magnesium ions) back into the soil, where they can be absorbed by the grass roots to support new growth.
評分準則
Marking Scheme (9 Marks Total): (a) (i) [2 marks] - Correct order: grass, grasshopper, frog, snake (1 mark) - Correct direction of arrows pointing from food to consumer (1 mark)
(a) (ii) [2 marks] - Producer identified: grass (1 mark) - Definition: organism that makes its own organic nutrients/food using energy from light / via photosynthesis (1 mark)
(b) [3 marks] - Mentions that energy is lost at each trophic level / only ~10% of energy is transferred (1 mark) - Mentions ways energy is lost: respiration / heat / movement / excretion / uneaten parts (1 mark) - Concludes that after 4 or 5 levels, there is insufficient energy remaining to support another population (1 mark)
(c) [2 marks] - Role: Decomposers / decay organisms (1 mark) - Explanation: release mineral ions/nutrients back into the soil, which are absorbed by grass roots for growth (1 mark)
題目 8 · structured-theory
8.88 分
A skydiver of mass \(70\text{ kg}\) jumps from an aircraft. Assume the acceleration of free fall \(g = 9.8\text{ m/s}^2\).
(a) (i) Immediately after jumping, before she has accelerated significantly, air resistance is negligible. State the magnitude and direction of the initial acceleration of the skydiver. (ii) Calculate the initial downward force acting on the skydiver.
(b) As she falls, she speeds up and eventually reaches terminal velocity. (i) Describe how the air resistance and the acceleration of the skydiver change as she speeds up, before reaching terminal velocity. (ii) State the resultant force acting on the skydiver when she is falling at terminal velocity.
(c) The skydiver now opens her parachute. Explain, in terms of forces, why her speed decreases.
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解題
Detailed Solution: (a) (i) Before air resistance builds up, the only force acting on the skydiver is her weight. Therefore, her initial acceleration is the acceleration of free fall, which has a magnitude of \(9.8\text{ m/s}^2\) and is directed downwards. (ii) The initial downward force is her weight: \(W = m \times g = 70\text{ kg} \times 9.8\text{ m/s}^2 = 686\text{ N}\). (b) (i) As the skydiver's speed increases, the air resistance acting on her increases. This upward air resistance opposes her downward weight, reducing the net/resultant downward force. Since \(F = ma\), as the resultant force decreases, her acceleration decreases towards zero. (ii) At terminal velocity, the upward air resistance is equal in magnitude and opposite in direction to her downward weight. The forces are balanced, so the resultant force is \(0\text{ N}\). (c) When the parachute opens, the surface area increases significantly, causing a large and sudden increase in air resistance. This upward air resistance force becomes much larger than her downward weight. Consequently, there is an upward resultant force, which causes her to decelerate (decrease speed).
(a) (ii) [1 mark] - Correct calculation: \(686\text{ N}\) (accept \(700\text{ N}\) if \(10\) is used) (1 mark)
(b) (i) [3 marks] - Air resistance increases as speed increases (1 mark) - Resultant downward force decreases (1 mark) - Acceleration decreases (1 mark)
(b) (ii) [1 mark] - Zero / \(0\text{ N}\) (1 mark)
(c) [2 marks] - Opening the parachute increases surface area, dramatically increasing air resistance (1 mark) - Air resistance is now greater than weight, causing an upward resultant force / deceleration (1 mark)
題目 9 · structured-theory
8.88 分
An electric winch is used to pull a box of mass \(120\text{ kg}\) up a rough wooden ramp. The ramp is \(15\text{ m}\) long and the box is raised to a vertical height of \(4.5\text{ m}\).
Take \(g = 9.8\text{ m/s}^2\).
(a) Calculate the gravitational potential energy (\(\Delta E_p\)) gained by the box. [2]
(b) The winch exerts a constant force of \(550\text{ N}\) parallel to the ramp to pull the box to the top. Calculate the work done by the winch. [2]
(c) State the energy transfer that explains why the work done by the winch is greater than the gravitational potential energy gained by the box, and calculate the amount of energy lost to the surroundings in this process. [2]
(d) The winch takes \(12\text{ s}\) to pull the box to the top of the ramp. Calculate the average power supplied by the winch. [2.88]
(b) \(W = F \times d = 550 \times 15 = 8250\text{ J}\) (or \(8300\text{ J}\) to 2 significant figures).
(c) Work is done against friction between the box and the ramp, transferring mechanical energy into thermal energy of the box, ramp, and surroundings. Energy lost = Work done - \(\Delta E_p\) = \(8250 - 5292 = 2958\text{ J}\) (or \(3000\text{ J}\) to 2 significant figures).
(a) [2 marks] - 1 mark for formula \(\Delta E_p = mgh\) or substitution \(120 \times 9.8 \times 4.5\). - 1 mark for correct answer: \(5292\text{ J}\) (accept \(5300\text{ J}\); accept \(5400\text{ J}\) if \(g = 10\text{ m/s}^2\) is used).
(b) [2 marks] - 1 mark for formula \(W = F \times d\) or substitution \(550 \times 15\). - 1 mark for correct answer: \(8250\text{ J}\) (accept \(8300\text{ J}\)).
(c) [2 marks] - 1 mark for stating that energy is lost as thermal energy due to friction / work done against friction. - 1 mark for correct subtraction calculation: \(2958\text{ J}\) (allow error carried forward, e.g., \(2850\text{ J}\) if \(g = 10\text{ m/s}^2\) was used, or \(8250 - \text{part (a)}\)).
(d) [2.88 marks] - 1 mark for formula \(P = \frac{W}{t}\). - 1 mark for substitution \(\frac{8250}{12}\). - 0.88 mark for correct value: \(687.5\text{ W}\) (accept \(690\text{ W}\) or allow error carried forward from (b)) with correct unit (\(\text{W}\) or \(\text{J/s}\)).
Paper 61 (Alternative to Practical)
Answer all questions. Show your working where appropriate and use the tables/grids provided.
4 題目 · 40 分
題目 1 · practical-task
10 分
A student investigates the effect of changing the concentration of a sucrose solution on the mass of potato cylinders.
The student prepares five potato cylinders, each initially measuring exactly 5.0 cm in length and weighing approximately 4.20 g.
Each cylinder is placed in a different beaker containing sucrose solutions of concentrations: 0.0, 0.2, 0.4, 0.6, and 0.8 mol/dm\(^{3}\) respectively.
After 45 minutes, the potato cylinders are removed, blotted dry, and reweighed.
(a) Explain why the student must blot the potato cylinders dry with a paper towel before reweighing them. [2]
(b) State two variables, other than the initial length, that must be kept constant to ensure a fair test. [2]
(c) The cylinder placed in the 0.8 mol/dm\(^{3}\) sucrose solution showed a decrease in mass. Explain, in terms of water potential and osmosis, why this mass loss occurred. [2]
(d) Describe how the student can use a graph of percentage change in mass against sucrose concentration to determine the concentration of sucrose inside the potato cells. [2]
(e) State one way the reliability of this experiment could be improved. [2]
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解題
(a) Blotting the potato cylinders dry removes any excess liquid adhering to their surface. If this liquid is not removed, it will add extra mass to the cylinder when weighed, causing an inaccurate measurement of the actual change in mass of the potato tissue.
(b) Two control variables are: the temperature of the solutions and the volume of the sucrose solutions in each beaker.
(c) The water potential of the 0.8 mol/dm\(^{3}\) sucrose solution is lower than the water potential inside the potato cells. Water moves out of the potato cells by osmosis, across the partially permeable cell membranes, down the water potential gradient, resulting in a loss of mass.
(d) The student should plot a graph of the percentage change in mass (y-axis) against sucrose concentration (x-axis) and draw a line of best fit. The point where the line intersects the x-axis (where percentage change in mass is 0%) represents the concentration where there is no net movement of water, which is equivalent to the sucrose concentration inside the potato cells.
(e) The student can repeat the experiment with three potato cylinders for each concentration, calculate the mean percentage change in mass for each concentration, and identify and exclude any anomalous results.
評分準則
(a) - To remove excess surface solution / water on the outside of the potato cylinder [1] - Which would artificially increase the measured final mass / lead to inaccurate mass readings [1] (b) - Any two from: temperature of the solutions, volume of solution, duration of immersion, source of potato (same potato), diameter/surface area of cylinder [2, 1 mark each] - Reject: initial length / initial mass (already stated in question) (c) - The solution has a lower water potential than the potato cells [1] - Water moves out of the potato cells by osmosis / down a water potential gradient [1] (d) - Identify the concentration where the percentage change in mass is zero [1] - By finding where the line of best fit crosses the x-axis [1] (e) - Repeat the experiment (at least three times for each concentration) [1] - Calculate the mean / average and identify anomalies [1]
題目 2 · practical-task
10 分
A student is provided with a green crystalline solid, salt **X**.
They carry out the following tests to identify the cation and the anion present in **X**.
(a) The student dissolves a sample of solid **X** in distilled water to make solution **Y**.
(i) They add aqueous sodium hydroxide to a portion of solution **Y**. A green precipitate is formed which is insoluble in excess. Identify the transition metal cation present in salt **X**. [1]
(ii) Suggest the formula of the green precipitate formed in this reaction. [1]
(b) To another portion of solution **Y**, the student adds dilute nitric acid followed by aqueous barium nitrate. A white precipitate is formed. Identify the anion present in salt **X**. [2]
(c) To investigate if salt **X** is hydrated (contains water of crystallization), the student heats a dry sample of solid **X** in a hard-glass test-tube.
(i) State two observations that would confirm water of crystallization is being released during heating. [2]
(ii) Draw a simple labeled diagram to show how a sample of solid can be safely heated in a dry test-tube. [2]
(iii) Describe a chemical test, including the observation for a positive result, to confirm that the liquid condensed in the test-tube is water. [2]
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解題
(a) (i) The formation of a green precipitate with aqueous sodium hydroxide that is insoluble in excess identifies the presence of the iron(II) ion, \(Fe^{2+}\). (ii) The green precipitate is iron(II) hydroxide, which has the chemical formula \(Fe(OH)_2\).
(b) The addition of dilute nitric acid followed by aqueous barium nitrate forms a white precipitate of barium sulfate, confirming the presence of the sulfate ion, \(SO_4^{2-}\).
(c) (i) Two observations confirming the release of water of crystallization are: condensation or droplets of a colorless liquid forming on the cooler upper parts of the test-tube, and the green solid changing color (e.g., turning white or brown/yellow) as it becomes anhydrous. (ii) The drawing should show: a test-tube held by a holder, tilted slightly downwards or horizontally (away from the open end), containing solid **X** at the bottom, heated by a Bunsen burner flame. Labels must include: 'test-tube holder', 'Bunsen burner / flame', and 'solid X'. (iii) Add the liquid to anhydrous cobalt(II) chloride paper. A positive result is a color change from blue to pink. Alternatively, add to anhydrous copper(II) sulfate; a positive result is a color change from white to blue.
評分準則
(a) (i) Iron(II) / \(Fe^{2+}\) [1] (Reject: iron / \(Fe^{3+}\)) (ii) \(Fe(OH)_2\) [1] (b) - Sulfate / \(SO_4^{2-}\) [1] - White precipitate is barium sulfate / \(BaSO_4\) [1] (c) (i) - Droplets of water / condensation at the top / mouth of the tube [1] - Solid changes color / steam is released [1] (ii) - Clear sketch of test-tube being heated with a Bunsen burner [1] - Labels showing test-tube holder AND the tube tilted with mouth pointed away from the user [1] (iii) - Test: anhydrous cobalt(II) chloride paper [1] - Observation: turns from blue to pink [1] (Accept: anhydrous copper(II) sulfate [1] turns from white to blue [1])
題目 3 · practical-task
10 分
A student investigates the extension of a spring. The apparatus consists of a spring suspended from a clamp stand, with a meter rule positioned vertically alongside.
The initial length of the unstretched spring, \(L_0\), is measured as 2.5 cm.
(a) The student hangs a load of 1.0 N on the spring. The new length of the spring, \(L\), is 4.2 cm. Calculate the extension, \(e_1\), of the spring. [1]
(b) The student repeats the procedure for different loads, \(F\), and records the lengths. The results are shown in Table 3.1.
Calculate the extension \(e_2\) for a load of 3.0 N. [1]
(c) Describe the relationship shown between the load and the extension of the spring in Table 3.1. [1]
(d) Calculate the spring constant, \(k\), of the spring using the formula:
\[k = \frac{F}{e}\]
Use the values for \(F = 4.0\text{ N}\). Give your answer to 2 significant figures and include the appropriate unit. [3]
(e) State one precaution the student must take when reading the ruler to avoid a parallax error. [1]
(f) The student is warned not to put extremely heavy loads on the spring. Explain the scientific reason for this precaution. [1]
(g) In a second experiment, the student plots a graph of load against extension. Explain how they would use this graph to identify if the spring has exceeded its limit of proportionality. [2]
(c) The extension is directly proportional to the load (as the load increases by equal increments of 1.0 N, the extension increases by equal increments of 1.7 cm).
(d) Using \(F = 4.0\text{ N}\) and \(e = 6.8\text{ cm}\): \[k = \frac{4.0\text{ N}}{6.8\text{ cm}} \approx 0.5882\text{ N/cm}\] To 2 significant figures, \(k = 0.59\text{ N/cm}\) (or \(59\text{ N/m}\) if converted).
(e) The student must view the ruler and the pointer at eye level, perpendicularly to the scale, to prevent parallax error.
(f) Very heavy loads can permanently deform the spring (exceed the elastic limit), meaning it will not return to its original length when the load is removed.
(g) The limit of proportionality is identified where the line on the graph ceases to be a straight line passing through the origin and begins to curve.
評分準則
(a) \(1.7\text{ cm}\) [1] (b) \(5.1\text{ cm}\) [1] (c) Extension is directly proportional to load / linear relationship [1] (d) - Calculation: \(4.0 / 6.8 = 0.588...\) [1] - Value to 2 sig figs: \(0.59\) [1] - Unit: \(\text{N/cm}\) (or \(59\text{ N/m}\)) [1] (e) View the scale perpendicularly / at eye level [1] (f) To avoid permanently stretching / damaging the spring / exceeding the elastic limit [1] (g) - Look for where the line is no longer straight [1] - The curve starts to bend / the gradient changes [1]
題目 4 · practical-task
10 分
A student investigates how the resistance of a resistance wire varies with its length.
(a) Draw a circuit diagram of the circuit used to carry out this investigation. Your diagram must include: - a cell, - a switch, - an ammeter connected to measure the current in the test wire, - a voltmeter connected to measure the potential difference across the test wire, - a sliding contact (jockey) to change the length of the test wire connected in the circuit. [3]
(b) The student connects a 50.0 cm length of the test wire. The voltmeter reads \(1.8\text{ V}\) and the ammeter reads \(0.30\text{ A}\).
Calculate the resistance, \(R\), of this length of wire. State the unit in your answer. [2]
(c) During the experiment, the student is instructed to close the switch, take the readings quickly, and open the switch immediately. Explain the practical reason for this instruction. [2]
(d) State one safety hazard in this experiment and describe how to minimize the risk. [2]
(e) Suggest one way the student can ensure a reliable and consistent electrical contact between the sliding contact (jockey) and the resistance wire. [1]
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解題
(a) The circuit diagram should show a cell connected in series with a switch, an ammeter, and a length of resistance wire. The voltmeter must be connected in parallel across the variable section of the resistance wire determined by the position of the sliding contact (jockey).
(c) Leaving the switch closed for too long causes current to flow continuously, which heats up the wire. An increase in temperature increases the resistance of the metal wire, making the test unfair and inaccurate.
(d) Hazard: The wire can become very hot and cause burns if touched. Minimization: Do not touch the wire with bare skin while the switch is closed, use low currents, and turn off the power supply between readings.
(e) Ensure the wire is straight and clean/polished (e.g., using emery paper to remove any oxide layer) and apply a constant downward pressure with the jockey.
評分準則
(a) - Correct symbols for cell, switch, ammeter in series with the test wire [1] - Voltmeter connected in parallel across the length of the test wire [1] - Sliding contact / jockey represented correctly on the resistance wire [1] (b) - Calculation: \(1.8 / 0.30 = 6.0\) [1] - Unit: \(\Omega\) / ohms [1] (c) - To prevent the wire from heating up [1] - As temperature increase would increase the resistance / change the wire properties [1] (d) - Hazard: Hot wire / burn to skin [1] - Minimization: Use low current / open switch between readings / do not touch bare wire when current is flowing [1] (e) - Clean the wire with emery paper / steel wool to remove oxide layer OR apply steady/firm pressure with the jockey [1]
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