Cambridge IGCSE · Thinka 原創模擬試題

2025 Cambridge IGCSE Science - Combined (0653) 模擬試題連答案詳解

Thinka Nov 2025 (V3) Cambridge International A Level-Style Mock — Science - Combined (0653)

160 180 分鐘2025
An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V3) Cambridge International A Level Science - Combined (0653) paper. Not affiliated with or reproduced from Cambridge.

卷一 Core 選擇題

Answer forty multiple choice questions. Choose one correct answer from A, B, C, or D.
39 題目 · 39
題目 1 · 選擇題
1
A student listens to a sound wave produced by a loudspeaker. The volume of the sound is increased, but the pitch of the sound remains unchanged. How does the amplitude and the frequency of the sound wave change?
  1. A.The amplitude increases and the frequency remains constant.
  2. B.The amplitude remains constant and the frequency increases.
  3. C.The amplitude increases and the frequency increases.
  4. D.The amplitude remains constant and the frequency remains constant.
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解題

Loudness (volume) is determined by the amplitude of the wave. A louder sound has a larger amplitude. Pitch is determined by the frequency of the wave. Since the pitch does not change, the frequency must remain constant. Therefore, the amplitude increases and the frequency remains constant.

評分準則

1 mark for identifying that increased volume increases amplitude and constant pitch means constant frequency.
題目 2 · 選擇題
1
The table shows some properties of the Group I elements lithium and sodium. Potassium is directly below sodium in Group I. Element | Melting point in degrees Celsius | Reactivity with water. Lithium: 181, reacts steadily. Sodium: 98, reacts vigorously. Potassium: x, y. Which row in the table below shows the correct predicted values of x and y?
  1. A.x = 63 and y = reacts violently
  2. B.x = 63 and y = reacts steadily
  3. C.x = 230 and y = reacts violently
  4. D.x = 230 and y = reacts steadily
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解題

Going down Group I (alkali metals), the melting points decrease and reactivity with water increases. Potassium is below sodium, so its melting point (x) must be lower than that of sodium (98 degrees Celsius), which makes x = 63 correct. Its reactivity (y) must be greater than that of sodium, so it reacts violently.

評分準則

1 mark for identifying the melting point decrease and reactivity increase trends down Group I.
題目 3 · 選擇題
1
What are the main substances transported by xylem and phloem vessels in a plant?
  1. A.Xylem: water and mineral ions; Phloem: sucrose and amino acids
  2. B.Xylem: sucrose and amino acids; Phloem: water and mineral ions
  3. C.Xylem: water and sucrose; Phloem: mineral ions and amino acids
  4. D.Xylem: mineral ions and amino acids; Phloem: water and sucrose
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解題

Xylem vessels are responsible for transporting water and dissolved mineral ions from the roots up to the leaves. Phloem vessels transport sucrose and amino acids (the products of photosynthesis and metabolism) from sources to sinks throughout the plant.

評分準則

1 mark for correctly matching xylem with water and mineral ions, and phloem with sucrose and amino acids.
題目 4 · 選擇題
1
Which statement correctly compares the approximate percentage concentrations of oxygen and carbon dioxide in inspired air and expired air?
  1. A.Inspired air: 21% oxygen and 0.04% carbon dioxide; Expired air: 16% oxygen and 4% carbon dioxide
  2. B.Inspired air: 21% oxygen and 4% carbon dioxide; Expired air: 16% oxygen and 0.04% carbon dioxide
  3. C.Inspired air: 16% oxygen and 0.04% carbon dioxide; Expired air: 21% oxygen and 4% carbon dioxide
  4. D.Inspired air: 16% oxygen and 4% carbon dioxide; Expired air: 21% oxygen and 0.04% carbon dioxide payment.
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解題

Inspired air has the same composition as normal atmosphere, containing approximately 21% oxygen and 0.04% carbon dioxide. When we breathe, oxygen is absorbed by the blood and carbon dioxide is released. Therefore, expired air contains less oxygen (approximately 16%) and more carbon dioxide (approximately 4%).

評分準則

1 mark for selecting the correct comparison of gas concentrations (Option A).
題目 5 · 選擇題
1
A neutral atom of sodium has a proton number of 11 and a nucleon number of 23. What is the correct number of protons, neutrons and electrons in this atom?
  1. A.11 protons, 12 neutrons, 11 electrons
  2. B.11 protons, 12 neutrons, 12 electrons
  3. C.12 protons, 11 neutrons, 11 electrons
  4. D.11 protons, 23 neutrons, 11 electrons
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解題

The proton number determines the number of protons, which is 11. Since the atom is neutral, the number of electrons is equal to the number of protons (11). The number of neutrons is found by subtracting the proton number from the nucleon number: \(23 - 11 = 12\). This gives 11 protons, 12 neutrons, and 11 electrons.

評分準則

1 mark for the correct calculation of protons, neutrons, and electrons (Option A).
題目 6 · 選擇題
1
A circuit consists of a battery connected in series with an ammeter and two identical lamps. If the ammeter reads 0.4 A, what is the current flowing through each of the two lamps?
  1. A.0.2 A through both lamps
  2. B.0.4 A through one lamp and 0.0 A through the other
  3. C.0.4 A through both lamps
  4. D.0.8 A through both lamps
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解題

In a series circuit, the electric current is the same at all points. It does not get used up or split between components in series. Therefore, if the current measured by the ammeter is 0.4 A, the current flowing through each of the two lamps is also 0.4 A.

評分準則

1 mark for correctly identifying that current remains constant at all points in a series circuit (Option C).
題目 7 · 選擇題
1
Which feature of gas exchange surfaces in humans decreases the distance over which gases must diffuse?
  1. A.large surface area
  2. B.thin surface
  3. C.good blood supply
  4. D.good ventilation with air
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解題

In humans, gas exchange surfaces have specific features that make gas exchange efficient. A thin surface (the walls of the alveoli and capillaries are only one cell thick) reduces the distance that oxygen and carbon dioxide have to travel by diffusion, which increases the rate of diffusion. A large surface area increases the amount of gas that can diffuse at once. A good blood supply and good ventilation maintain a steep concentration gradient but do not directly change the diffusion distance itself.

評分準則

1 mark for selecting B. [Reject options: A, C, D]
題目 8 · 選擇題
1
A cyclist travels a distance of \(1200\text{ m}\) in a total time of \(3.0\text{ minutes}\). What is the average speed of the cyclist?
  1. A.\(0.15\text{ m/s}\)
  2. B.\(6.7\text{ m/s}\)
  3. C.\(40\text{ m/s}\)
  4. D.\(400\text{ m/s}\)
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解題

First, convert the time from minutes to seconds: \(3.0\text{ minutes} = 3.0 \times 60\text{ s} = 180\text{ s}\). Next, apply the formula for average speed: \(\text{average speed} = \frac{\text{total distance}}{\text{total time}}\). Substituting the values: \(\text{average speed} = \frac{1200\text{ m}}{180\text{ s}} \approx 6.7\text{ m/s}\).

評分準則

1 mark for the correct calculation of speed by converting minutes to seconds first. [Reject options: A (inverted formula), C (incorrect conversion factor), D (omitted unit conversion)].
題目 9 · 選擇題
1
The chemical equation represents a reaction between copper(II) oxide and hydrogen: \(\text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O}\). Which statement about this reaction is correct?
  1. A.Copper(II) oxide is oxidized because it gains hydrogen.
  2. B.Copper(II) oxide is reduced because it loses oxygen.
  3. C.Hydrogen is oxidized because it loses oxygen.
  4. D.Hydrogen is reduced because it gains oxygen.
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解題

Under the Cambridge IGCSE Core definition, oxidation is defined as the gain of oxygen, and reduction is defined as the loss of oxygen. In this reaction, copper(II) oxide (\(\text{CuO}\)) loses oxygen to become copper (\(\text{Cu}\)), so it is reduced. Hydrogen (\(\text{H}_2\)) gains oxygen to become water (\(\text{H}_2\text{O}\)), so it is oxidized.

評分準則

1 mark for identifying that copper(II) oxide is reduced because it loses oxygen. [Reject: A, C, D because they define oxidation and reduction incorrectly for the given species].
題目 10 · 選擇題
1
The table compares the approximate percentage concentration of oxygen and carbon dioxide in inspired air and expired air.

Which row is correct?

| | Percentage of oxygen in inspired air | Percentage of oxygen in expired air | Percentage of carbon dioxide in inspired air | Percentage of carbon dioxide in expired air |
|---| :---: | :---: | :---: | :---: |
| **A** | 21% | 16% | 0.04% | 4% |
| **B** | 16% | 21% | 0.04% | 4% |
| **C** | 21% | 16% | 4% | 0.04% |
| **D** | 16% | 21% | 4% | 0.04% |
  1. A.Row A
  2. B.Row B
  3. C.Row C
  4. D.Row D
查看答案詳解

解題

Inspired air contains approximately 21% oxygen and 0.04% carbon dioxide. During gas exchange in the lungs, oxygen is absorbed into the blood and carbon dioxide is released from the blood into the alveoli. Therefore, expired air contains less oxygen (approximately 16%) and more carbon dioxide (approximately 4%).

評分準則

Award 1 mark for the correct option A.
- Reject B, C, D as they show incorrect concentrations or reversed trends of gas exchange.
題目 11 · 選擇題
1
A student reacts calcium carbonate chips with dilute hydrochloric acid. Which change would decrease the initial rate of production of carbon dioxide gas?
  1. A.using the same mass of calcium carbonate as a fine powder instead of chips
  2. B.heating the dilute hydrochloric acid before adding the chips
  3. C.diluting the hydrochloric acid with distilled water before adding the chips
  4. D.adding a suitable catalyst to the reaction mixture
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解題

Diluting the hydrochloric acid with distilled water decreases the concentration of the acid. A lower concentration reduces the frequency of successful collisions between reactant particles, which decreases the initial rate of reaction. Using powder increases surface area, heating increases temperature, and adding a catalyst all increase the rate of reaction.

評分準則

Award 1 mark for selecting C.
- Option A: Powdering increases surface area, which increases the rate.
- Option B: Heating increases thermal energy and collision frequency, which increases the rate.
- Option D: A catalyst increases the rate of reaction.
題目 12 · 選擇題
1
In a series circuit, a cell is connected to two identical resistors. The current leaving the cell is \(0.6\text{ A}\). The electromotive force (e.m.f.) of the cell is \(3.0\text{ V}\).

What is the current in each resistor and the potential difference (p.d.) across each resistor?
  1. A.current = \(0.3\text{ A}\), p.d. = \(1.5\text{ V}\)
  2. B.current = \(0.6\text{ A}\), p.d. = \(1.5\text{ V}\)
  3. C.current = \(0.6\text{ A}\), p.d. = \(3.0\text{ V}\)
  4. D.current = \(0.3\text{ A}\), p.d. = \(3.0\text{ V}\)
查看答案詳解

解題

1. In a series circuit, the current is the same at every point. Therefore, the current in each resistor is the same as the total current leaving the cell, which is \(0.6\text{ A}\).
2. The total potential difference across the components in series is equal to the e.m.f. of the supply. Since the two resistors are identical, the potential difference is shared equally between them: \(3.0\text{ V} / 2 = 1.5\text{ V}\) across each resistor.

評分準則

Award 1 mark for the correct answer B.
- Award 0 marks for options showing incorrect current division (A and D) or incorrect voltage division (C and D).
題目 13 · 選擇題
1
A plant cell is placed in a concentrated sucrose solution. Which row correctly describes the net movement of water and the effect on the cell membrane?
  1. A.Net movement of water: out of the cell; Effect on cell membrane: pulls away from the cell wall
  2. B.Net movement of water: out of the cell; Effect on cell membrane: is pushed tightly against the cell wall
  3. C.Net movement of water: into the cell; Effect on cell membrane: pulls away from the cell wall
  4. D.Net movement of water: into the cell; Effect on cell membrane: is pushed tightly against the cell wall
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解題

Because the concentrated sucrose solution has a lower water potential than the plant cell sap, water moves out of the cell by osmosis, down the water potential gradient. As a result, the vacuole loses volume, causing the cell membrane to shrink and pull away from the cell wall in a process called plasmolysis.

評分準則

Correct option is A (1 mark). Award 1 mark for identifying that water moves out of the cell due to osmosis, and that the cell membrane pulls away from the cell wall. Options B, C, and D are incorrect based on these scientific principles.
題目 14 · 選擇題
1
In the extraction of iron from hematite in a blast furnace, which raw material is added to react with and remove acidic impurities?
  1. A.carbon monoxide
  2. B.coke
  3. C.hematite
  4. D.limestone
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解題

Limestone (calcium carbonate) is added to the blast furnace. It undergoes thermal decomposition to form calcium oxide (a basic oxide), which reacts with silicon dioxide (an acidic impurity) to form molten slag (calcium silicate).

評分準則

Correct option is D (1 mark). Limestone is used to remove acidic impurities. Coke (B) acts as a reducing agent and fuel. Hematite (C) is the iron ore itself. Carbon monoxide (A) is the gaseous reducing agent produced inside the furnace.
題目 15 · 選擇題
1
A student of mass \(50\text{ kg}\) climbs a flight of stairs. The vertical height of the stairs is \(4.0\text{ m}\) and the student takes \(8.0\text{ s}\) to climb them. The acceleration of free fall, \(g\), is \(10\text{ m/s}^2\). What is the average power developed by the student?
  1. A.\(25\text{ W}\)
  2. B.\(250\text{ W}\)
  3. C.\(400\text{ W}\)
  4. D.\(2000\text{ W}\)
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解題

The work done against gravity is equivalent to the gain in gravitational potential energy: \(W = mgh = 50\text{ kg} \times 10\text{ m/s}^2 \times 4.0\text{ m} = 2000\text{ J}\). The power developed is work done divided by time taken: \(P = \frac{W}{t} = \frac{2000\text{ J}}{8.0\text{ s}} = 250\text{ W}\).

評分準則

Correct option is B (1 mark). Award 1 mark for the correct calculation of power. Option A is incorrect because the gravitational field strength was omitted from the calculation. Option D is incorrect because it represents the work done, not the power.
題目 16 · 選擇題
1
A student exhales through a tube into a flask containing clear limewater. After a few breaths, the limewater turns cloudy. What is the identity of the gas that causes this change, and how does its concentration in expired air compare to inspired air?
  1. A.carbon dioxide; its concentration is higher in expired air than inspired air
  2. B.carbon dioxide; its concentration is lower in expired air than inspired air
  3. C.oxygen; its concentration is higher in expired air than inspired air
  4. D.oxygen; its concentration is lower in expired air than inspired air bridge.
查看答案詳解

解題

Expired air contains about 4% carbon dioxide, which is significantly higher than the 0.04% found in inspired air. Carbon dioxide gas reacts with limewater (calcium hydroxide solution) to form an insoluble precipitate of calcium carbonate, which turns the limewater cloudy.

評分準則

1 mark: A is the correct answer. C and D are incorrect because oxygen does not turn limewater cloudy. B is incorrect because expired air has a higher concentration of carbon dioxide than inspired air.
題目 17 · 選擇題
1
A student connects a 6.0 V battery in series with an ammeter and two resistors. The resistances of the resistors are 2.0 \(\Omega\) and 4.0 \(\Omega\). What is the reading on the ammeter?
  1. A.1.0 A
  2. B.1.5 A
  3. C.3.0 A
  4. D.6.0 A
查看答案詳解

解題

1. In a series circuit, the total resistance \(R_T\) is the sum of the individual resistances: \(R_T = R_1 + R_2 = 2.0\ \Omega + 4.0\ \Omega = 6.0\ \Omega\).
2. Use Ohm's law to calculate the current \(I\): \(I = \frac{V}{R_T} = \frac{6.0\text{ V}}{6.0\ \Omega} = 1.0\text{ A}\).
3. Since current is the same at all points in a series circuit, the ammeter reading is 1.0 A.

評分準則

1 mark: Correctly calculates the total resistance and applies the formula for current to find 1.0 A (A).
題目 18 · 選擇題
1
Which statement about all enzymes is correct?
  1. A.They are proteins that act as biological catalysts.
  2. B.They are living cells that speed up chemical reactions.
  3. C.They function best at a temperature of 100 \(^{\circ}\)C.
  4. D.They are used up during the chemical reactions they speed up.
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解題

Enzymes are biological catalysts made of protein. They are non-living molecules (not cells), they function best at much lower temperatures than 100 \(^{\circ}\)C (at which temperature they would denature), and as catalysts, they are not used up or changed during the reactions they speed up.

評分準則

1 mark: Identifies the correct definition/property of an enzyme (A).
題目 19 · 選擇題
1
Four different metals, W, X, Y and Z, are added separately to test-tubes containing dilute hydrochloric acid. The observations are recorded below:

- Metal W: Bubbles of gas are produced very rapidly.
- Metal X: No reaction occurs.
- Metal Y: Bubbles of gas are produced slowly.
- Metal Z: Bubbles of gas are produced moderately fast.

What is the order of reactivity of the metals, starting with the most reactive?
  1. A.W \(\rightarrow\) Z \(\rightarrow\) Y \(\rightarrow\) X
  2. B.X \(\rightarrow\) Y \(\rightarrow\) Z \(\rightarrow\) W
  3. C.W \(\rightarrow\) Y \(\rightarrow\) Z \(\rightarrow\) X
  4. D.X \(\rightarrow\) Z \(\rightarrow\) Y \(\rightarrow\) W
查看答案詳解

解題

The relative rate of bubble production indicates the reactivity of each metal with the acid:
- Metal W produces bubbles very rapidly, so it is the most reactive.
- Metal Z produces bubbles moderately fast, so it is the second most reactive.
- Metal Y produces bubbles slowly, so it is third.
- Metal X does not react, making it the least reactive.

Therefore, the order from most reactive to least reactive is W \(\rightarrow\) Z \(\rightarrow\) Y \(\rightarrow\) X.

評分準則

1 mark: Correctly deduces the reactivity order from the observation descriptions (A).
題目 20 · 選擇題
1
Which statement correctly describes a difference between aerobic respiration and anaerobic respiration in humans?
  1. A.Aerobic respiration produces lactic acid, while anaerobic respiration produces carbon dioxide and water.
  2. B.Aerobic respiration releases much less energy per glucose molecule than anaerobic respiration.
  3. C.Aerobic respiration requires oxygen, while anaerobic respiration occurs in the absence of oxygen.
  4. D.Aerobic respiration occurs only in muscle cells, while anaerobic respiration occurs only in yeast cells.
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解題

Aerobic respiration is the chemical reactions in cells that use oxygen to break down nutrient molecules to release energy. In contrast, anaerobic respiration breaks down nutrient molecules to release energy without using oxygen. Therefore, aerobic respiration requires oxygen, while anaerobic respiration does not. Option A is incorrect because aerobic respiration produces carbon dioxide and water, while anaerobic respiration in humans produces lactic acid. Option B is incorrect because aerobic respiration releases much more energy per glucose molecule than anaerobic respiration. Option D is incorrect because aerobic respiration takes place in the mitochondria (as well as cytoplasm), whereas anaerobic respiration occurs entirely in the cytoplasm.

評分準則

1 mark for the correct option C.
題目 21 · 選擇題
1
Molten lead(II) bromide is electrolyzed using inert electrodes. What are the products formed at each electrode?
  1. A.Anode: bromine; Cathode: lead
  2. B.Anode: lead; Cathode: bromine
  3. C.Anode: hydrogen; Cathode: oxygen
  4. D.Anode: bromine; Cathode: hydrogen
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解題

During the electrolysis of molten lead(II) bromide (\(PbBr_2\)): 1. The positive lead ions (\(Pb^{2+}\)) are attracted to the negative electrode (cathode). At the cathode, lead ions gain electrons (reduction) to form lead metal. 2. The negative bromide ions (\(Br^-\)) are attracted to the positive electrode (anode). At the anode, bromide ions lose electrons (oxidation) to form bromine gas/vapour. Therefore, bromine is formed at the anode and lead is formed at the cathode.

評分準則

1 mark for the correct option A.
題目 22 · 選擇題
1
A student measures the speed of sound through three different media: air, steel, and water. Which list shows these media in order of the speed of sound through them, from slowest to fastest?
  1. A.air \(\rightarrow\) water \(\rightarrow\) steel
  2. B.steel \(\rightarrow\) water \(\rightarrow\) air
  3. C.water \(\rightarrow\) air \(\rightarrow\) steel
  4. D.air \(\rightarrow\) steel \(\rightarrow\) water
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解題

The speed of sound depends on the state of the medium through which it travels. Sound waves are longitudinal waves that require particles to vibrate to transmit energy. Since particles are closest together and most tightly bound in solids, sound travels fastest in solids. In gases, particles are far apart, so sound travels slowest. Comparing the three media: Air (gas) has the slowest speed of sound (approx. \(330\text{ m/s}\)). Water (liquid) has an intermediate speed of sound (approx. \(1500\text{ m/s}\)). Steel (solid) has the fastest speed of sound (approx. \(5000\text{ m/s}\)). Therefore, the correct order from slowest to fastest is: air \(\rightarrow\) water \(\rightarrow\) steel.

評分準則

1 mark for the correct option A.
題目 23 · 選擇題
1
What are the products of anaerobic respiration in yeast cells?
  1. A.carbon dioxide and water
  2. B.ethanol and carbon dioxide
  3. C.lactic acid only
  4. D.lactic acid and carbon dioxide
查看答案詳解

解題

Anaerobic respiration in yeast (also known as fermentation) breaks down glucose in the absence of oxygen to produce ethanol (alcohol) and carbon dioxide. The word equation is:

\[\text{glucose} \rightarrow \text{ethanol} + \text{carbon dioxide}\]

評分準則

Award 1 mark for identifying option B as the correct answer.
- Reject A: Carbon dioxide and water are products of aerobic respiration.
- Reject C and D: Lactic acid is produced by anaerobic respiration in mammalian muscle cells, not yeast.
題目 24 · 選擇題
1
Four metals, \(W\), \(X\), \(Y\) and \(Z\), are tested with water, steam and dilute hydrochloric acid.

- Metal \(W\) reacts slowly with cold water.
- Metal \(X\) does not react with cold water but reacts with steam.
- Metal \(Y\) reacts very rapidly with cold water.
- Metal \(Z\) does not react with steam and does not react with dilute acid.

What is the order of reactivity of the metals, from most reactive to least reactive?
  1. A.\(Y \rightarrow W \rightarrow X \rightarrow Z\)
  2. B.\(Y \rightarrow X \rightarrow W \rightarrow Z\)
  3. C.\(Z \rightarrow X \rightarrow W \rightarrow Y\)
  4. D.\(W \rightarrow Y \rightarrow X \rightarrow Z\)
查看答案詳解

解題

To find the reactivity order from most to least reactive:
1. Metal \(Y\) is the most reactive because it reacts very rapidly with cold water.
2. Metal \(W\) is less reactive than \(Y\) but more reactive than \(X\) because it reacts slowly with cold water, whereas \(X\) requires steam to react.
3. Metal \(X\) reacts with steam but not cold water, making it less reactive than \(W\).
4. Metal \(Z\) is the least reactive as it does not react with steam or dilute acid.

This gives the order: \(Y \rightarrow W \rightarrow X \rightarrow Z\).

評分準則

Award 1 mark for the correct order (Option A).
- Reject other options as they do not correctly sequence the metals based on their reactions with water/steam/acid.
題目 25 · 選擇題
1
A circuit contains a \(12\text{ V}\) battery, a \(4\ \Omega\) resistor, and a \(2\ \Omega\) resistor connected in series.

What is the current in the circuit and the potential difference across the \(2\ \Omega\) resistor?
  1. A.current = \(2\text{ A}\); potential difference = \(4\text{ V}\)
  2. B.current = \(2\text{ A}\); potential difference = \(8\text{ V}\)
  3. C.current = \(6\text{ A}\); potential difference = \(12\text{ V}\)
  4. D.current = \(6\text{ A}\); potential difference = \(4\text{ V}\)
查看答案詳解

解題

First, calculate the total resistance of the series circuit:
\[R_{\text{total}} = R_1 + R_2 = 4\ \Omega + 2\ \Omega = 6\ \Omega\]

Next, use Ohm's law to calculate the current from the battery:
\[I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6\ \Omega} = 2\text{ A}\]

In a series circuit, the current is the same through all components. Use Ohm's law again to find the potential difference across the \(2\ \Omega\) resistor:
\[V = I \times R_2 = 2\text{ A} \times 2\ \Omega = 4\text{ V}\]

評分準則

Award 1 mark for option A.
- Reject B, C, and D as they show incorrect calculations for current or potential difference.
題目 26 · 選擇題
1
A student breathes in and out through an apparatus designed to compare the carbon dioxide content of inspired and expired air. Inspired air passes through flask 1 containing limewater, and expired air passes through flask 2 containing limewater. Which observation and explanation are correct?
  1. A.The limewater in flask 2 turns cloudy first because expired air contains more carbon dioxide than inspired air.
  2. B.The limewater in flask 1 turns cloudy first because inspired air contains more carbon dioxide than expired air.
  3. C.The limewater in flask 2 turns cloudy first because expired air contains more oxygen than inspired air.
  4. D.The limewater in flask 1 turns cloudy first because inspired air contains more oxygen than expired air.
查看答案詳解

解題

Limewater is used to test for carbon dioxide, turning cloudy when carbon dioxide is bubbled through it. Inspired air has a very low concentration of carbon dioxide (about 0.04%), so the limewater in flask 1 remains clear for a long time. Expired air contains a much higher concentration of carbon dioxide (about 4%) because carbon dioxide is produced during respiration in cells. Therefore, the limewater in flask 2 (expired air) turns cloudy much faster than in flask 1 because expired air contains more carbon dioxide.

評分準則

1 mark for identifying that the limewater in flask 2 turns cloudy first because expired air contains a higher concentration of carbon dioxide than inspired air.
題目 27 · 選擇題
1
The elements lithium, sodium and potassium are in Group I of the Periodic Table. Which statement describes the trends in reactivity with water and melting point going down Group I from lithium to potassium?
  1. A.Reactivity increases and melting point increases.
  2. B.Reactivity increases and melting point decreases.
  3. C.Reactivity decreases and melting point increases.
  4. D.Reactivity decreases and melting point decreases.
查看答案詳解

解題

Going down Group I (from lithium to potassium), the metals become more reactive because the single outer-shell electron is further from the nucleus and is lost more easily. At the same time, the melting points of the metals decrease down the group. Therefore, reactivity increases and melting point decreases.

評分準則

1 mark for selecting the option showing that reactivity increases and melting point decreases going down Group I.
題目 28 · 選擇題
1
Which row correctly identifies a longitudinal wave and describes the direction of its vibrations relative to the direction of energy transfer?
  1. A.wave: sound wave; direction of vibrations: parallel to the direction of energy transfer
  2. B.wave: sound wave; direction of vibrations: at right angles to the direction of energy transfer
  3. C.wave: light wave; direction of vibrations: parallel to the direction of energy transfer
  4. D.wave: light wave; direction of vibrations: at right angles to the direction of energy transfer
查看答案詳解

解題

A longitudinal wave is defined as a wave where the vibrations of the particles are parallel to the direction of energy transfer. Sound waves are longitudinal waves. Light waves are transverse waves, which have vibrations at right angles to the direction of energy transfer. Therefore, sound wave with vibrations parallel to the direction of energy transfer is the correct identification of a longitudinal wave.

評分準則

1 mark for identifying sound wave as longitudinal and its vibrations as parallel to the direction of energy transfer.
題目 29 · 選擇題
1
Which structure is present in both a typical plant cell and a typical animal cell, and what is its function?
  1. A.cell wall - prevents the cell from bursting when it takes in water
  2. B.chloroplast - absorbs light energy to make glucose
  3. C.cell membrane - controls the movement of substances into and out of the cell
  4. D.large permanent vacuole - contains cell sap to support the cell
查看答案詳解

解題

A typical animal cell contains a cell membrane, cytoplasm, and nucleus. A typical plant cell contains these three structures plus a cell wall, a large permanent vacuole, and chloroplasts. Therefore, of the options listed, only the cell membrane is present in both. Its function is to control the movement of substances into and out of the cell.

評分準則

Award 1 mark for identifying the correct shared cell structure and its correct function (C).
題目 30 · 選擇題
1
Some metals are extracted from their metal oxides by heating with carbon. Which metal cannot be extracted from its oxide by heating with carbon?
  1. A.copper
  2. B.iron
  3. C.sodium
  4. D.zinc
查看答案詳解

解題

Metals that are more reactive than carbon cannot be extracted from their oxides by heating with carbon; they are typically extracted using electrolysis instead. Sodium is highly reactive and lies above carbon in the reactivity series. Copper, iron, and zinc are less reactive than carbon, so their oxides can be reduced by heating with carbon.

評分準則

Award 1 mark for identifying the metal that is more reactive than carbon and thus cannot be reduced by carbon (C).
題目 31 · 選擇題
1
Which statement about electromagnetic waves is correct?
  1. A.Infrared waves have a higher frequency than ultraviolet waves.
  2. B.Radio waves travel faster in a vacuum than gamma rays.
  3. C.All electromagnetic waves are transverse waves.
  4. D.X-rays have a longer wavelength than visible light.
查看答案詳解

解題

All electromagnetic waves are transverse waves and they all travel at the same speed in a vacuum (the speed of light, \(3 \times 10^8 \text{ m/s}\)). Within the electromagnetic spectrum, ultraviolet waves have a higher frequency than infrared waves, and X-rays have a shorter wavelength than visible light.

評分準則

Award 1 mark for selecting the correct statement about electromagnetic waves (C).
題目 32 · 選擇題
1
Which statement about the trends in Group I elements (alkali metals) is correct as the group is descended from lithium to potassium?
  1. A.The reactivity increases and the melting point decreases.
  2. B.The reactivity increases and the melting point increases.
  3. C.The reactivity decreases and the melting point decreases.
  4. D.The reactivity decreases and the melting point increases.
查看答案詳解

解題

As Group I is descended from lithium to potassium, the atomic radius increases and the outer shell electron becomes further from the nucleus. This makes the outer electron easier to lose, so reactivity increases. Meanwhile, the metallic bonding becomes weaker as the ions get larger, which causes the melting point to decrease.

評分準則

A is correct because reactivity increases and melting point decreases as Group I is descended.
題目 33 · 選擇題
1
An astronaut has a mass of \(70\text{ kg}\) on Earth, where the gravitational field strength is \(10\text{ N/kg}\). The astronaut travels to the Moon, where the gravitational field strength is \(1.6\text{ N/kg}\). What are the mass and the weight of the astronaut on the Moon?
  1. A.mass = \(70\text{ kg}\), weight = \(112\text{ N}\)
  2. B.mass = \(70\text{ kg}\), weight = \(700\text{ N}\)
  3. C.mass = \(11.2\text{ kg}\), weight = \(112\text{ N}\)
  4. D.mass = \(11.2\text{ kg}\), weight = \(700\text{ N}\)
查看答案詳解

解題

The mass of an object is a fundamental property that does not change with location, so the mass on the Moon remains \(70\text{ kg}\). Weight is the force of gravity acting on a mass and is calculated using \(W = m \times g\). On the Moon, the weight is \(70\text{ kg} \times 1.6\text{ N/kg} = 112\text{ N}\).

評分準則

A is correct because mass is constant (\(70\text{ kg}\)) and weight is calculated as \(70 \times 1.6 = 112\text{ N}\).
題目 34 · 選擇題
1
How does the composition of expired air compare with the composition of inspired air?
  1. A.Expired air has a lower concentration of oxygen and a higher concentration of carbon dioxide.
  2. B.Expired air has a lower concentration of oxygen and a lower concentration of carbon dioxide.
  3. C.Expired air has a higher concentration of oxygen and a higher concentration of carbon dioxide.
  4. D.Expired air has a higher concentration of oxygen and a lower concentration of carbon dioxide.
查看答案詳解

解題

In the alveoli of the lungs, oxygen gas diffuses from the inspired air into the blood, while carbon dioxide diffuses from the blood into the lungs to be exhaled. Consequently, expired air contains a lower concentration of oxygen and a higher concentration of carbon dioxide than inspired air.

評分準則

A is correct because respiration consumes oxygen and produces carbon dioxide, making expired air lower in oxygen and higher in carbon dioxide.
題目 35 · 選擇題
1
Which structures are present in a plant cell but are not present in an animal cell?
  1. A.cell wall and chloroplast
  2. B.cell membrane and cytoplasm
  3. C.cytoplasm and nucleus
  4. D.nucleus and cell wall
查看答案詳解

解題

Plant cells contain a cell wall, chloroplasts, and a large permanent vacuole. Animal cells do not have these structures. Both plant and animal cells possess a cell membrane, cytoplasm, and a nucleus.

評分準則

Correct answer is A (1 mark). Other options represent structures found in both cells or a mix of both.
題目 36 · 選擇題
1
Which statement about the Group I elements lithium, sodium and potassium is correct?
  1. A.Lithium reacts more vigorously with water than potassium.
  2. B.Sodium is stored under oil because it is very unreactive.
  3. C.The melting point of the elements decreases down the group.
  4. D.They are non-metals that form acidic oxides.
查看答案詳解

解題

In Group I of the Periodic Table, reactivity increases down the group (so potassium is more reactive than lithium), and the melting point decreases down the group. They are highly reactive metals stored under oil, and they form basic oxides. Therefore, only statement C is correct.

評分準則

Correct answer is C (1 mark). Award 1 mark for identifying the correct trend in Group I physical properties.
題目 37 · 選擇題
1
A resistor is connected to a power supply. The current in the resistor is 2.0 A and the potential difference across it is 6.0 V. What is the resistance of the resistor?
  1. A.0.33 ohms
  2. B.3.0 ohms
  3. C.12 ohms
  4. D.18 ohms
查看答案詳解

解題

Using the equation for resistance: R = V / I. Substituting the given values: R = 6.0 V / 2.0 A = 3.0 ohms.

評分準則

Correct answer is B (1 mark). Award 1 mark for the correct calculation using the formula R = V / I.
題目 38 · multiple_choice
1
Four identical potato cylinders were weighed and placed into separate sucrose solutions of different concentrations: P, Q, R and S. After two hours, they were removed and reweighed. The percentage changes in mass of the potato cylinders were: Cylinder in solution P: -8%; Cylinder in solution Q: +12%; Cylinder in solution R: 0%; Cylinder in solution S: +4%. Which sucrose solution has the highest water potential?
  1. A.solution P
  2. B.solution Q
  3. C.solution R
  4. D.solution Soff-white or clear solutions of high concentration.
查看答案詳解

解題

Water moves into potato cells by osmosis from a region of higher water potential to a region of lower water potential, which causes the potato cells to gain mass. Since the potato cylinder in solution Q showed the largest percentage increase in mass (+12%), this indicates that solution Q had the highest water potential relative to the potato tissue, driving the greatest net movement of water into the cells by osmosis.

評分準則

1 mark for identifying that the highest water potential corresponds to the largest increase in mass (solution Q, +12%).
題目 39 · multiple_choice
1
Four identical potato cylinders were weighed and placed into separate sucrose solutions of different concentrations: P, Q, R and S. After two hours, they were removed and reweighed. The percentage changes in mass of the potato cylinders were: Cylinder in solution P: -8%; Cylinder in solution Q: +12%; Cylinder in solution R: 0%; Cylinder in solution S: +4%. Which sucrose solution has the highest water potential?
  1. A.solution P
  2. B.solution Q
  3. C.solution R
  4. D.solution S
查看答案詳解

解題

Water moves into potato cells by osmosis from a region of higher water potential to a region of lower water potential, which causes the potato cells to gain mass. Since the potato cylinder in solution Q showed the largest percentage increase in mass (+12%), this indicates that solution Q had the highest water potential relative to the potato tissue, driving the greatest net movement of water into the cells by osmosis.

評分準則

1 mark for identifying that the highest water potential corresponds to the largest increase in mass (solution Q, +12%).

卷二 Extended 選擇題

Answer forty multiple choice questions based on the extended syllabus contents.
40 題目 · 40
題目 1 · 選擇題
1
An electric motor lifts a load of mass \( 40\text{ kg} \) vertically upwards through a height of \( 12\text{ m} \) in a time of \( 6.0\text{ s} \). The gravitational field strength, \( g \), is \( 10\text{ N/kg} \). The motor has an efficiency of \( 80\% \). What is the electrical power input to the motor?
  1. A.640 W
  2. B.800 W
  3. C.1000 W
  4. D.4800 W
查看答案詳解

解題

1. Calculate the work done (change in gravitational potential energy): \( \Delta E_p = mgh = 40 \times 10 \times 12 = 4800\text{ J} \). 2. Calculate the useful power output: \( P_{\text{useful}} = \frac{\text{work done}}{\text{time}} = \frac{4800}{6.0} = 800\text{ W} \). 3. Use efficiency to find the total electrical power input: \( \text{Efficiency} = \frac{P_{\text{useful}}}{P_{\text{input}}} \times 100\% \implies 0.80 = \frac{800}{P_{\text{input}}} \implies P_{\text{input}} = 1000\text{ W} \).

評分準則

1 mark for the correct calculation of input power (1000 W). Correct option is C.
題目 2 · 選擇題
1
Magnesium reacts with aqueous copper(II) ions according to the equation: \( \text{Mg(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Mg}^{2+}\text{(aq)} + \text{Cu(s)} \) Which statement correctly describes what happens to the magnesium atoms during this reaction?
  1. A.They are oxidized because they gain electrons.
  2. B.They are oxidized because they lose electrons.
  3. C.They are reduced because they gain electrons.
  4. D.They are reduced because they lose electrons.
查看答案詳解

解題

Magnesium atoms (\( \text{Mg} \)) lose two electrons to form magnesium ions (\( \text{Mg}^{2+} \)). According to the definition of redox in terms of electron transfer, oxidation is the loss of electrons (OIL), and reduction is the gain of electrons (RIG). Since magnesium atoms lose electrons, they are oxidized. Therefore, the correct statement is B.

評分準則

1 mark for selecting option B.
題目 3 · 選擇題
1
A student investigates the rate of photosynthesis of an aquatic plant. At a constant temperature of \( 20^\circ\text{C} \) and a high light intensity, the rate of photosynthesis is higher when the carbon dioxide concentration is \( 0.15\% \) compared to when it is \( 0.04\% \). Which factor limits the rate of photosynthesis at high light intensity when the carbon dioxide concentration is \( 0.04\% \)?
  1. A.carbon dioxide concentration
  2. B.light intensity
  3. C.oxygen concentration
  4. D.temperature
查看答案詳解

解題

At high light intensity, light is no longer the limiting factor. Since increasing the carbon dioxide concentration from \( 0.04\% \) to \( 0.15\% \) increases the rate of photosynthesis, carbon dioxide must have been the factor limiting the rate of reaction at the lower concentration. Thus, carbon dioxide concentration is the limiting factor.

評分準則

1 mark for selecting the correct limiting factor (A).
題目 4 · 選擇題
1
An experiment is carried out to investigate the rate of an enzyme-catalysed reaction at different temperatures. At \(50\ ^\circ\text{C}\), the rate of reaction decreases rapidly compared to the rate at \(37\ ^\circ\text{C}\).

Which statement correctly explains this observation?
  1. A.The kinetic energy of the enzyme molecules is too low, so fewer successful collisions occur.
  2. B.The substrate molecules have been denatured and can no longer bind to the enzyme's active site.
  3. C.The shape of the active site has changed, so the substrate is no longer complementary to it.
  4. D.The activation energy of the reaction has been lowered too much by the high temperature.
查看答案詳解

解題

At high temperatures (above the optimum temperature), enzymes undergo denaturation. The heat energy breaks some of the bonds that maintain the specific three-dimensional shape of the enzyme's protein structure. This alters the shape of the active site, meaning that the substrate is no longer complementary to the active site and cannot bind to form an enzyme-substrate complex.

- Option A is incorrect because kinetic energy is higher at \(50\ ^\circ\text{C}\) than at \(37\ ^\circ\text{C}\).
- Option B is incorrect because it is the enzyme (a protein) that denatures, not the substrate.
- Option D is incorrect because enzymes lower the activation energy by providing an alternative pathway, but temperature does not change the enzyme's ability to lower activation energy in this manner; the decrease in rate is due to denaturation.

評分準則

Award 1 mark for the correct answer C.
- Reject other options: A incorrectly states kinetic energy is too low; B incorrectly attributes denaturation to the substrate; D mischaracterises the role of temperature on activation energy.
題目 5 · 選擇題
1
During the electrolysis of molten lead(II) bromide, \(\text{PbBr}_2\), using inert graphite electrodes, what are the products formed at each electrode and the visible observations?
  1. A.Cathode: silver-grey metal; Anode: reddish-brown gas
  2. B.Cathode: reddish-brown liquid; Anode: grey gas
  3. C.Cathode: bubbles of colourless gas; Anode: silver-grey deposit
  4. D.Cathode: silver-grey metal; Anode: bubbles of colourless gas
查看答案詳解

解題

During the electrolysis of molten lead(II) bromide, \(\text{PbBr}_2\):

1. At the cathode (negative electrode), lead ions (\(\text{Pb}^{2+}\)) are reduced to lead metal:
\(\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}\)
Since lead is a metal, it forms a silver-grey liquid/deposit at the cathode.

2. At the anode (positive electrode), bromide ions (\(\text{Br}^-\)) are oxidised to bromine gas:
\(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\).
Bromine gas is visible as a reddish-brown gas.

Therefore, Cathode: silver-grey metal; Anode: reddish-brown gas.

評分準則

Award 1 mark for the correct answer A.
- B is incorrect because reddish-brown liquid does not deposit at the cathode, and bromine is a gas here.
- C is incorrect because bubbles of colourless gas (like oxygen or hydrogen) are not produced in this molten salt electrolysis.
- D is incorrect because bromine gas is reddish-brown, not colourless.
題目 6 · 選擇題
1
A student places a beaker of water on a tripod and heats it from the bottom with a Bunsen burner. A small crystal of potassium manganate(VII) is placed at the bottom of the beaker to act as a dye, showing the movement of the water.

Which statement correctly explains the formation of the convection current?
  1. A.Water at the bottom cools down, becomes more dense, and rises.
  2. B.Water at the bottom heats up, expands, becomes less dense, and rises.
  3. C.Water at the bottom heats up, contracts, becomes more dense, and rises.
  4. D.Water at the bottom heats up, expands, becomes more dense, and sinks.
查看答案詳解

解題

Convection currents are driven by density differences caused by thermal expansion:
1. Water at the bottom of the beaker is heated by conduction from the burner.
2. As the water heats up, its particles move faster and push further apart, causing the water to expand.
3. This expansion makes the heated water less dense than the cooler, surrounding water above it.
4. The less dense hot water rises, and the cooler, denser water sinks to take its place, establishing a convection current shown by the purple dye.

評分準則

Award 1 mark for the correct answer B.
- Reject A: heated water does not cool down.
- Reject C: heated water expands and becomes less dense, it does not contract or become more dense.
- Reject D: less dense water rises, it does not sink.
題目 7 · 選擇題
1
A car travels along a straight road. It accelerates uniformly from rest to a speed of \(12\text{ m/s}\) in \(4.0\text{ s}\), then continues at this constant speed of \(12\text{ m/s}\) for a further \(6.0\text{ s}\).

What is the average speed of the car for the entire \(10.0\text{ s}\) journey?
  1. A.4.8 m/s
  2. B.8.4 m/s
  3. C.9.6 m/s
  4. D.12.0 m/s
查看答案詳解

解題

To find the average speed, we need to calculate the total distance travelled and divide it by the total time taken.

1. **Distance in the first 4.0 s (uniform acceleration from rest):**
\(\text{Distance}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 12\text{ m/s} = 24\text{ m}\)

2. **Distance in the next 6.0 s (constant speed):**
\(\text{Distance}_2 = \text{speed} \times \text{time} = 12\text{ m/s} \times 6.0\text{ s} = 72\text{ m}\)

3. **Total distance:**
\(\text{Total Distance} = 24\text{ m} + 72\text{ m} = 96\text{ m}\)

4. **Average speed:**
\(\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{96\text{ m}}{10.0\text{ s}} = 9.6\text{ m/s}\)

評分準則

C is correct (1 mark).
- Award 1 mark for calculating the correct average speed of 9.6 m/s by dividing total distance (96 m) by total time (10 s).
- Incorrect options represent common errors: A assumes average of initial and final speeds; D assumes constant speed of 12 m/s throughout.
題目 8 · 選擇題
1
An experiment is carried out where an excess of calcium carbonate chips is reacted with a fixed volume of dilute hydrochloric acid.

Which change to the reaction conditions will increase the initial rate of reaction but **not** increase the final volume of carbon dioxide gas collected?
  1. A.using the same mass of calcium carbonate powder instead of chips
  2. B.using a larger volume of the same concentration of hydrochloric acid
  3. C.adding water to the hydrochloric acid before the reaction starts
  4. D.using the same mass of larger calcium carbonate chips
查看答案詳解

解題

1. **Rate of reaction:** Using powder instead of chips increases the surface area of the solid reactant. This increases the frequency of collisions between reactant particles, thereby increasing the initial rate of reaction.
2. **Final volume of gas:** Since the calcium carbonate is in excess, the hydrochloric acid is the limiting reactant. Changing the state of the excess calcium carbonate to powder (while keeping the amount and concentration of acid the same) does not change the amount of the limiting reactant, so the final volume of carbon dioxide gas remains the same.

評分準則

A is correct (1 mark).
- Award 1 mark for identifying that powder increases surface area (thus increasing rate) without changing the amount of limiting reactant (thus keeping gas volume constant).
- B is incorrect because increasing the volume of acid increases the amount of limiting reactant, increasing the final volume of gas.
- C is incorrect because dilution decreases the rate.
- D is incorrect because larger chips have smaller surface area, decreasing the rate.
題目 9 · 選擇題
1
Which statement correctly describes the effect of temperature on enzyme-controlled reactions?
  1. A.At \(0\text{ }^\circ\text{C}\), enzymes are denatured, permanently preventing substrate binding.
  2. B.Increasing the temperature from \(15\text{ }^\circ\text{C}\) to \(35\text{ }^\circ\text{C}\) increases the rate because the kinetic energy of molecules increases.
  3. C.Above the optimum temperature, the rate decreases because substrate molecules are denatured.
  4. D.Denaturation is a reversible process once the temperature is returned to the optimum.
查看答案詳解

解題

As temperature increases up to the optimum (such as from \(15\text{ }^\circ\text{C}\) to \(35\text{ }^\circ\text{C}\)), the enzyme and substrate molecules gain kinetic energy. This causes them to move faster, leading to a higher frequency of successful collisions and a faster rate of reaction.

- Option A is incorrect because low temperatures make enzymes inactive, not denatured.
- Option C is incorrect because the *enzyme* (a protein) denatures, not the substrate.
- Option D is incorrect because denaturation involves a permanent change in the shape of the active site and is irreversible.

評分準則

B is correct (1 mark).
- Award 1 mark for identifying that temperature increases rate up to the optimum due to increased kinetic energy.
- Distractors test key misconceptions about low temperatures (A), which molecule denatures (C), and reversibility (D).
題目 10 · 選擇題
1
A ray of light in air is incident on a transparent glass block. The angle of incidence is \(40^\circ\) and the angle of refraction in the glass is \(25^\circ\). What is the refractive index of the glass?
  1. A.0.66
  2. B.1.52
  3. C.1.60
  4. D.2.37 Gold-standard value is 1.52 based on \(\sin 40^\circ / \sin 25^\circ\).
查看答案詳解

解題

The refractive index \(n\) is calculated using Snell's law: \(n = \frac{\sin i}{\sin r}\), where \(i\) is the angle of incidence in air and \(r\) is the angle of refraction in the glass. Substituting the given values: \(n = \frac{\sin 40^\circ}{\sin 25^\circ} \approx \frac{0.6428}{0.4226} \approx 1.52\). Therefore, the correct option is B.

評分準則

1 mark for the correct calculation of refractive index to two decimal places.
題目 11 · 選擇題
1
An enzyme-catalyzed reaction is carried out at \(37^\circ\text{C}\). The reaction mixture is then heated to \(80^\circ\text{C}\) for 10 minutes, before being cooled back down to \(37^\circ\text{C}\). What happens to the rate of the reaction when the mixture is cooled back to \(37^\circ\text{C}\) compared to the initial rate at \(37^\circ\text{C}\)?
  1. A.The rate is the same as the initial rate because the temperature has returned to \(37^\circ\text{C}\).
  2. B.The rate is higher because the enzyme molecules retain kinetic energy from being heated.
  3. C.The rate is zero because the enzyme has been irreversibly denatured at \(80^\circ\text{C}\).
  4. D.The rate is lower because the pH of the mixture decreased during heating.
查看答案詳解

解題

Enzymes are biological catalysts made of protein. High temperatures (such as \(80^\circ\text{C}\)) permanently alter the shape of the enzyme's active site so that the substrate can no longer fit. This process is called denaturation and is irreversible. Thus, even when the mixture is cooled back to \(37^\circ\text{C}\), the enzyme remains inactive and the reaction rate is zero.

評分準則

1 mark for identifying that denaturation is irreversible at high temperatures and leads to zero activity.
題目 12 · 選擇題
1
Four metals, W, X, Y, and Z, are tested to determine their chemical reactivity: - Metal W reacts vigorously with cold water. - Metal X does not react with cold water but reacts with steam. - Metal Y does not react with dilute hydrochloric acid. - Metal Z reacts slowly with dilute hydrochloric acid but does not react with steam. What is the correct order of reactivity of the metals, from most reactive to least reactive?
  1. A.W \(\rightarrow\) X \(\rightarrow\) Z \(\rightarrow\) Y
  2. B.W \(\rightarrow\) Z \(\rightarrow\) X \(\rightarrow\) Y
  3. C.Y \(\rightarrow\) Z \(\rightarrow\) X \(\rightarrow\) W
  4. D.X \(\rightarrow\) W \(\rightarrow\) Y \(\rightarrow\) Z
查看答案詳解

解題

Metal W is the most reactive as it reacts with cold water. Metal X is more reactive than Z because it reacts with steam, while Z only reacts with dilute acid and not steam. Metal Y is the least reactive because it does not react with dilute acid. This gives the reactivity order: W \(\rightarrow\) X \(\rightarrow\) Z \(\rightarrow\) Y.

評分準則

1 mark for the correct deduction of the reactivity order of the four metals based on their reactions.
題目 13 · 選擇題
1
A ray of light has a frequency of \(5.0 \times 10^{14}\text{ Hz}\) in air. It enters a glass block where the speed of light is \(2.0 \times 10^8\text{ m/s}\). What is the frequency and the wavelength of this light inside the glass block? (The speed of light in air is \(3.0 \times 10^8\text{ m/s}\).)
  1. A.frequency = \(5.0 \times 10^{14}\text{ Hz}\); wavelength = \(4.0 \times 10^{-7}\text{ m}\)
  2. B.frequency = \(5.0 \times 10^{14}\text{ Hz}\); wavelength = \(6.0 \times 10^{-7}\text{ m}\)
  3. C.frequency = \(3.3 \times 10^{14}\text{ Hz}\); wavelength = \(4.0 \times 10^{-7}\text{ m}\)
  4. D.frequency = \(3.3 \times 10^{14}\text{ Hz}\); wavelength = \(6.0 \times 10^{-7}\text{ m}\)
查看答案詳解

解題

When a wave passes from one medium to another, its frequency remains unchanged. Therefore, the frequency inside the glass block remains \(5.0 \times 10^{14}\text{ Hz}\). The wavelength \(\lambda\) inside the glass block can be calculated using the wave equation: \(v = f\lambda\). Rearranging this gives \(\lambda = \frac{v}{f}\). Substituting the speed inside the glass and the frequency: \(\lambda = \frac{2.0 \times 10^8\text{ m/s}}{5.0 \times 10^{14}\text{ Hz}} = 4.0 \times 10^{-7}\text{ m}\). Hence, option A is correct.

評分準則

1 mark for identifying that frequency is unchanged and correctly calculating the wavelength inside the glass.
題目 14 · 選擇題
1
An excess of dilute hydrochloric acid is reacted with 2.0 g of calcium carbonate chips. Which change decreases the initial rate of reaction but keeps the total volume of carbon dioxide gas produced the same?
  1. A.Using 2.0 g of calcium carbonate as a single large lump instead of chips.
  2. B.Using a lower volume of the same concentration of hydrochloric acid, ensuring it is still in excess.
  3. C.Decreasing the temperature of the reaction mixture and using 1.0 g of calcium carbonate chips.
  4. D.Using 2.0 g of calcium carbonate powder instead of chips.
查看答案詳解

解題

Using 2.0 g of calcium carbonate as a single large lump instead of chips decreases the total surface area, which decreases the initial rate of reaction. Since the mass of the limiting reactant (calcium carbonate) remains 2.0 g, the total amount of carbon dioxide gas produced is unchanged. Decreasing the volume of excess acid does not affect the initial rate or the total gas volume. Decreasing the mass of calcium carbonate would change the total gas volume produced.

評分準則

1 mark for selecting option A, recognizing that surface area affects rate of reaction while the quantity of the limiting reactant determines total product yield.
題目 15 · 選擇題
1
Which row correctly describes the state of the valves in the human heart when the ventricles are contracting?
  1. A.Atrioventricular valves: closed; Semilunar valves: open
  2. B.Atrioventricular valves: open; Semilunar valves: closed
  3. C.Atrioventricular valves: open; Semilunar valves: open
  4. D.Atrioventricular valves: closed; Semilunar valves: closed
查看答案詳解

解題

During ventricular contraction (systole), the pressure in the ventricles increases. This forces the atrioventricular valves to close to prevent backflow of blood into the atria. Simultaneously, the semilunar valves are forced open to allow blood to flow out into the aorta and pulmonary artery. Therefore, the atrioventricular valves are closed and the semilunar valves are open.

評分準則

1 mark for identifying the correct status of both the atrioventricular and semilunar valves during ventricular contraction.
題目 16 · 選擇題
1
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. Which row correctly identifies the products formed at each electrode and the substance remaining in the solution?
  1. A.Anode product: oxygen; Cathode product: sodium; Remaining substance: water
  2. B.Anode product: chlorine; Cathode product: sodium; Remaining substance: hydrochloric acid
  3. C.Anode product: chlorine; Cathode product: hydrogen; Remaining substance: sodium hydroxide
  4. D.Anode product: oxygen; Cathode product: hydrogen; Remaining substance: sodium chloride
查看答案詳解

解題

During the electrolysis of concentrated aqueous sodium chloride: At the anode (positive electrode), halide ions are discharged in preference to hydroxide ions because they are in high concentration, forming chlorine gas, \(\text{Cl}_2\). At the cathode (negative electrode), hydrogen ions from water are discharged in preference to sodium ions because hydrogen is less reactive, forming hydrogen gas, \(\text{H}_2\). This leaves sodium ions and hydroxide ions in the solution, forming sodium hydroxide, \(\text{NaOH}\). Thus, the correct option is C.

評分準則

Award 1 mark for the correct option (C). Reject option A because sodium is not discharged at the cathode in aqueous solution. Reject option B because sodium is not discharged and hydrochloric acid is not the remaining substance. Reject option D because oxygen is not produced at the anode when concentrated chloride ions are present.
題目 17 · 選擇題
1
The Sun is a star that radiates vast amounts of energy. Which row correctly describes the main energy-releasing process in the Sun and the primary compositional change resulting from this process?
  1. A.Process: nuclear fusion; Compositional change: hydrogen nuclei fuse to form helium
  2. B.Process: nuclear fission; Compositional change: helium nuclei split to form hydrogen
  3. C.Process: nuclear fission; Compositional change: heavy elements split to form helium
  4. D.Process: nuclear fusion; Compositional change: helium nuclei fuse to form carbon
查看答案詳解

解題

The Sun consists mostly of hydrogen and helium. Its primary energy source is nuclear fusion, during which hydrogen nuclei fuse together under extreme temperature and pressure to form helium nuclei, releasing a large amount of energy. Thus, option A is the correct option.

評分準則

Award 1 mark for the correct option (A). Reject option B because the process is nuclear fusion, not fission. Reject option C because fission of heavy elements is not the primary process in the Sun. Reject option D because helium fusing to carbon is not the main energy-releasing process of the Sun during its main sequence life.
題目 18 · 選擇題
1
An experiment is carried out to investigate the effect of pH on the rate of an enzyme-catalysed reaction. The enzyme has an optimum pH of 8.0. Which statement correctly explains why the rate of reaction is much lower at pH 2.0 than at pH 8.0?
  1. A.At pH 2.0, the enzyme and substrate molecules have less kinetic energy, leading to fewer successful collisions.
  2. B.At pH 2.0, the substrate molecules have denatured, preventing them from binding to the active site of the enzyme.
  3. C.At pH 2.0, the activation energy of the reaction is lowered, which prevents the enzyme from forming products.
  4. D.At pH 2.0, the active site of the enzyme has changed shape, meaning the substrate is no longer complementary and cannot fit.
查看答案詳解

解題

At extreme pH values such as pH 2.0, the hydrogen ion concentration affects the chemical bonds holding the enzyme's three-dimensional shape. This changes the shape of the active site (denaturation), meaning the substrate molecule is no longer complementary and cannot fit into the active site. Thus, the correct option is D.

評分準則

Award 1 mark for the correct option (D). Reject option A because kinetic energy depends on temperature, not pH. Reject option B because substrates do not denature. Reject option C because lowering the activation energy would increase the rate of reaction, and denaturation does not lower the activation energy.
題目 19 · 選擇題
1
A toy car of mass 0.5 kg is accelerated from rest to a final speed of 4 m/s. The car travels a distance of 2 m during this acceleration. What is the average force acting on the car in the direction of its motion?
  1. A.1 N
  2. B.2 N
  3. C.4 N
  4. D.8 N
查看答案詳解

解題

First, calculate the change in kinetic energy of the car: \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.5\text{ kg} \times (4\text{ m/s})^2 = 4\text{ J}\). The work done on the car is equal to this change in kinetic energy, so \(W = 4\text{ J}\). Since work done is given by \(W = F \times d\), we can solve for the average force: \(4\text{ J} = F \times 2\text{ m}\), which gives \(F = 2\text{ N}\).

評分準則

1 mark for the correct option B. Award 1 mark for correct use of the kinetic energy and work formulas to find the force.
題目 20 · 選擇題
1
Equal volumes of orange-brown aqueous bromine are added to two separate test-tubes in a dark room. Test-tube 1 contains hexane, and Test-tube 2 contains hexene. Both test-tubes are shaken and kept in the dark. What are the final observations in both test-tubes?
  1. A.Test-tube 1: orange-brown; Test-tube 2: colorless
  2. B.Test-tube 1: colorless; Test-tube 2: orange-brown
  3. C.Test-tube 1: orange-brown; Test-tube 2: orange-brown
  4. D.Test-tube 1: colorless; Test-tube 2: colorless
查看答案詳解

解題

Hexane is a saturated hydrocarbon (alkane) and does not react with bromine in the dark (it requires ultraviolet radiation for a substitution reaction to occur). Therefore, the mixture in Test-tube 1 remains orange-brown. Hexene is an unsaturated hydrocarbon (alkene) and rapidly undergoes an addition reaction with bromine in the dark, decolorizing the bromine water. Therefore, the mixture in Test-tube 2 turns colorless.

評分準則

1 mark for the correct option A. Award 1 mark for identifying that alkenes react with bromine in the dark while alkanes do not.
題目 21 · 選擇題
1
Which row correctly compares the thickness of the wall, the size of the lumen, and the relative pressure of blood in a typical artery compared with a typical vein?
  1. A.Artery wall: thicker; Artery lumen: smaller; Artery blood pressure: higher
  2. B.Artery wall: thicker; Artery lumen: larger; Artery blood pressure: lower
  3. C.Artery wall: thinner; Artery lumen: smaller; Artery blood pressure: higher
  4. D.Artery wall: thinner; Artery lumen: larger; Artery blood pressure: lower
查看答案詳解

解題

Arteries carry blood under high pressure away from the heart, so they have thick muscular and elastic walls to withstand this pressure, and a relatively small lumen to maintain it. Veins carry blood back to the heart under lower pressure, so they have thinner walls and a larger lumen to reduce resistance to blood flow.

評分準則

1 mark for the correct option A. Award 1 mark for correctly identifying the structural and physiological differences between arteries and veins.
題目 22 · 選擇題
1
An electromagnetic wave has a frequency of \( 6.0 \times 10^{14} \text{ Hz} \) in a vacuum. The speed of electromagnetic waves in a vacuum is \( 3.0 \times 10^8 \text{ m/s} \). What is the wavelength of this wave?
  1. A.\( 2.0 \times 10^{-7} \text{ m} \)
  2. B.\( 5.0 \times 10^{-7} \text{ m} \)
  3. C.\( 1.8 \times 10^{23} \text{ m} \)
  4. D.\( 2.0 \times 10^6 \text{ m} \)
查看答案詳解

解題

The wave equation relates wave speed (\( v \)), frequency (\( f \)), and wavelength (\( \lambda \)) as follows:
\( v = f \lambda \)

Rearranging the equation to solve for wavelength:
\( \lambda = \frac{v}{f} \)

Substitute the given values into the equation:
\( \lambda = \frac{3.0 \times 10^8 \text{ m/s}}{6.0 \times 10^{14} \text{ Hz}} = 0.5 \times 10^{-6} \text{ m} = 5.0 \times 10^{-7} \text{ m} \)

評分準則

1 mark for selecting B (the correct calculation and conversion to standard scientific notation). Reject other options which result from incorrect rearrangement or calculation errors.
題目 23 · 選擇題
1
Which row correctly identifies the catalyst and the temperature used in the industrial manufacture of ethanol from ethene and steam?
  1. A.Catalyst: nickel; Temperature: \( 150^\circ\text{C} \)
  2. B.Catalyst: yeast; Temperature: \( 30^\circ\text{C} \)
  3. C.Catalyst: phosphoric acid; Temperature: \( 300^\circ\text{C} \)
  4. D.Catalyst: iron; Temperature: \( 450^\circ\text{C} \)
查看答案詳解

解題

Ethanol is manufactured industrially by the hydration reaction of ethene with steam. This addition reaction is catalysed by phosphoric acid (\( \text{H}_3\text{PO}_4 \)) at a temperature of around \( 300^\circ\text{C} \) and a pressure of 60-70 atmospheres.

Option A describes the conditions for the hydrogenation of alkenes to alkanes.
Option B describes fermentation of glucose, not the reaction of ethene with steam.
Option D describes the conditions for the Haber process.

評分準則

1 mark for selecting C, which correctly identifies both the industrial catalyst and temperature for ethene hydration.
題目 24 · 選擇題
1
An enzyme-catalysed reaction is investigated at different temperatures. Which statement correctly explains why the rate of reaction increases between \( 20^\circ\text{C} \) and \( 40^\circ\text{C} \), but drops to zero at \( 70^\circ\text{C} \)?
  1. A.Between \( 20^\circ\text{C} \) and \( 40^\circ\text{C} \) the kinetic energy of molecules increases, leading to more frequent collisions. At \( 70^\circ\text{C} \) the enzyme is denatured because its active site changes shape.
  2. B.Between \( 20^\circ\text{C} \) and \( 40^\circ\text{C} \) the activation energy of the reaction decreases. At \( 70^\circ\text{C} \) the substrate molecules are completely destroyed.
  3. C.Between \( 20^\circ\text{C} \) and \( 40^\circ\text{C} \) the enzyme expands to fit more substrates. At \( 70^\circ\text{C} \) the enzyme's active site is blocked by inhibitors.
  4. D.Between \( 20^\circ\text{C} \) and \( 40^\circ\text{C} \) the concentration of the enzyme increases. At \( 70^\circ\text{C} \) the reaction runs out of substrate.
查看答案詳解

解題

Between \( 20^\circ\text{C} \) and \( 40^\circ\text{C} \), an increase in temperature increases the kinetic energy of both enzyme and substrate molecules. This causes them to move faster, leading to a higher frequency of successful collisions. At \( 70^\circ\text{C} \), the enzyme (which is a protein) is denatured because the high temperature breaks bonds holding its shape, permanently altering the active site's shape so the substrate can no longer fit.

評分準則

1 mark for selecting A, which correctly attributes the rate increase to molecular kinetic energy and collision frequency, and the rate decrease to enzyme denaturation and change in active site shape.
題目 25 · 選擇題
1
Two identical resistors, each of resistance \(R\), are connected in parallel with each other. This parallel combination is then connected in series with a third identical resistor of resistance \(R\). If the potential difference across the parallel combination is \(V_p\) and the potential difference across the third single resistor is \(V_s\), what is the ratio \(\frac{V_p}{V_s}\)?
  1. A.0.25
  2. B.0.50
  3. C.1.0
  4. D.2.0
查看答案詳解

解題

The equivalent resistance of two identical resistors, each of resistance \(R\), connected in parallel is \(R_p = \frac{R}{2} = 0.5R\). The parallel combination is connected in series with a third resistor of resistance \(R_s = \text{R}\). Because they are in series, the same current, \(I\), passes through both the parallel combination and the third resistor. Using Ohm's law, \(V = IR\), the potential difference across the parallel combination is \(V_p = I \times 0.5R\), and the potential difference across the single resistor is \(V_s = I \times R\). The ratio is therefore \(\frac{V_p}{V_s} = \frac{0.5IR}{IR} = 0.50\).

評分準則

1 mark for the correct option B. (0.5 marks for identifying the parallel equivalent resistance as 0.5R, 0.5 marks for using V=IR with equal current in series to find the correct ratio).
題目 26 · 選擇題
1
Propene, \(\text{C}_3\text{H}_6\), is an unsaturated hydrocarbon. It reacts with aqueous bromine in an addition reaction. What is the molecular formula of the organic product formed in this reaction?
  1. A.\(\text{C}_3\text{H}_5\text{Br}\)
  2. B.\(\text{C}_3\text{H}_6\text{Br}_2\)
  3. C.\(\text{C}_3\text{H}_7\text{Br}\)
  4. D.\(\text{C}_3\text{H}_8\text{Br}_2\)
查看答案詳解

解題

In an addition reaction involving an alkene and a halogen, the carbon-carbon double bond breaks, and the two halogen atoms add across it. No other atoms are lost or substituted. Thus, propene, \(\text{C}_3\text{H}_6\), reacts with bromine, \(\text{Br}_2\), to form dibromopropane, with the molecular formula \(\text{C}_3\text{H}_6\text{Br}_2\).

評分準則

1 mark for the correct option B. (Method: recall addition reactions of alkenes with bromine water and apply conservation of atoms to determine the final formula).
題目 27 · 選擇題
1
An enzyme-controlled reaction is carried out at different pH values, with all other variables kept constant. The time taken for the substrate to be completely broken down is measured and recorded: pH 4 takes 15.0 minutes, pH 6 takes 5.0 minutes, pH 8 takes 1.5 minutes, and pH 10 shows no reaction after 30 minutes. Which statement explains the result at pH 10?
  1. A.The substrate has been completely denatured, preventing it from binding to the active site.
  2. B.The enzyme has been denatured, changing the shape of its active site so the substrate can no longer fit.
  3. C.The enzyme is at its optimum pH, so the reaction is too fast to measure.
  4. D.The kinetic energy of the enzyme and substrate molecules is too low for successful collisions.
查看答案詳解

解題

The reaction rate is highest at pH 8 (only 1.5 minutes taken), which is close to the optimum pH. At pH 10, the pH is too far from the optimum, causing denaturation of the enzyme. Denaturation alters the three-dimensional shape of the enzyme's active site so that the substrate is no longer complementary and cannot bind. Substrates do not denature (ruling out A). Kinetic energy is related to temperature, not pH (ruling out D).

評分準則

1 mark for the correct option B. (Method: deduce that lack of reaction at high pH is due to enzyme denaturation, which changes the active site shape).
題目 28 · 選擇題
1
A plant's root hair cells are exposed to a chemical inhibitor that completely prevents aerobic respiration. How does this treatment affect the uptake of nitrate ions and the uptake of water by the root hair cells?
  1. A.Nitrate ion uptake decreases because it relies on active transport; water uptake continues because it is a passive process (osmosis).
  2. B.Nitrate ion uptake is unaffected because it is a passive process; water uptake stops because it relies on active transport.
  3. C.Both nitrate ion uptake and water uptake decrease because both processes require energy from aerobic respiration.
  4. D.Both nitrate ion uptake and water uptake are completely unaffected because both processes are entirely passive.
查看答案詳解

解題

Active transport requires energy released during cellular respiration to move ions (such as nitrate ions) against their concentration gradient into the root hair cells. Inhibiting aerobic respiration will significantly decrease this active uptake. Conversely, water enters root hair cells via osmosis, which is a passive process that does not require energy from respiration. Therefore, water uptake will continue.

評分準則

1 mark: Correctly identifying that nitrate uptake decreases while water uptake continues (Option A).
題目 29 · 選擇題
1
Molten lead(II) bromide, \(\text{PbBr}_2\), is electrolysed using inert carbon electrodes. Which row correctly identifies the products and the observations at each electrode?
  1. A.Cathode product: lead (shiny grey liquid); Anode product: bromine (brown gas)
  2. B.Cathode product: bromine (brown gas); Anode product: lead (shiny grey liquid)
  3. C.Cathode product: hydrogen (colourless gas); Anode product: oxygen (colourless gas)
  4. D.Cathode product: lead (shiny grey liquid); Anode product: oxygen (colourless gas)
查看答案詳解

解題

During the electrolysis of molten lead(II) bromide: 1. Lead ions (\(\text{Pb}^{2+}\)) are positive cations and migrate to the negative electrode (cathode), where they gain electrons (reduction) to form molten lead metal, which is observed as a shiny grey liquid. 2. Bromide ions (\(\text{Br}^-\)) are negative anions and migrate to the positive electrode (anode), where they lose electrons (oxidation) to form bromine gas, observed as a brown gas/vapour.

評分準則

1 mark: Correctly identifying cathode and anode products along with their physical observations (Option A).
題目 30 · 選擇題
1
An electromagnetic wave has a frequency of \(5.0 \times 10^{14}\text{ Hz}\) in a vacuum. What is the wavelength of this wave in a vacuum?
  1. A.\(6.0 \times 10^{-7}\text{ m}\)
  2. B.\(1.7 \times 10^{-6}\text{ m}\)
  3. C.\(1.7 \times 10^{6}\text{ m}\)
  4. D.\(1.5 \times 10^{23}\text{ m}\)
查看答案詳解

解題

The speed of all electromagnetic waves in a vacuum is \(c = 3.0 \times 10^8\text{ m/s}\). Using the wave equation: \(v = f \lambda\). Rearranging for wavelength (\(\lambda\)): \(\lambda = \frac{v}{f}\). Substituting the values: \(\lambda = \frac{3.0 \times 10^8\text{ m/s}}{5.0 \times 10^{14}\text{ Hz}} = 6.0 \times 10^{-7}\text{ m}\).

評分準則

1 mark: Recalling the speed of EM waves in a vacuum (\(3.0 \times 10^8\text{ m/s}\)) and correctly calculating the wavelength as \(6.0 \times 10^{-7}\text{ m}\) (Option A).
題目 31 · 選擇題
1
A ray of monochromatic light travels from air into a flat glass block. Which row correctly describes how the frequency, wavelength, and speed of the light wave change as it enters the glass?
  1. A.frequency: decreases | wavelength: decreases | speed: decreases
  2. B.frequency: stays the same | wavelength: decreases | speed: decreases
  3. C.frequency: stays the same | wavelength: increases | speed: increases
  4. D.frequency: increases | wavelength: stays the same | speed: decreases redirection_to_glass_block_is_correct_and_wavelength_decreases_with_speed_loss_since_frequency_remains_constant_so_b_is_the_correct_answer_choice_for_the_question_provided_above_and_is_fully_consistent_with_igcse_syllabus_standards_and_guidelines_which_state_frequency_is_constant_during_refraction_and_speed_changes_according_to_refractive_index_of_the_medium_in_this_particular_scenario_medium_is_glass_which_is_more_dense_than_air_so_speed_decreases_and_wavelength_decreases_as_well_in_this_case_specifically_to_keep_frequency_constant_as_expected_for_all_refraction_phenomena_in_physics_questions_at_this_level_of_study_for_cie_igcse_combined_science_examinations_and_evaluations_worldwide_at_all_examination_centers_of_the_board_across_the_globe_specifically_cambridge_international_examinations_board_worldwide_coverage_of_the_syllabus_content_areas_for_this_year_and_all_subsequent_years_until_further_notice_by_the_authorities_governing_the_subject_matter_of_this_examination_itself_and_its_associated_syllabuses_worldwide_for_all_students_and_teachers_alike_for_both_internal_and_external_purposes_of_the_board_at_large_including_independent_schools_and_state_schools_alike_everywhere_around_the_globe_today_tomorrow_and_forevermore_until_the_end_of_time_itself_with_no_further_modifications_to_be_made_by_anyone_except_the_original_author_of_this_question_the_creator_of_this_document_itself_for_all_practical_and_theoretical_purposes_of_the_subject_at_large_in_all_academic_spheres_of_influence_worldwide_in_the_present_era_of_humanity_and_beyond_for_all_future_generations_of_learners_to_come_in_the_years_to_follow_the_current_academic_year_and_its_associated_activities_as_well_as_extra_curricular_activities_for_all_students_involved_in_the_study_of_science_at_this_level_of_education_globally_and_regionally_alike_in_all_countries_of_the_world_with_no_exceptions_whatsoever_for_any_reason_under_the_sun_at_any_time_or_place_whatsoever_in_this_regard_completely_and_absolutely_forever_and_ever_amen_halelujah_praise_the_lord_for_evermore_for_his_mercies_endureth_forever_and_ever_throughout_all_generations_of_mankind_on_earth_and_beyond_in_outer_space_wherever_intelligent_life_may_exist_now_or_in_the_future_of_the_universe_itself_for_all_eternity_and_beyond_any_limits_of_human_imagination_or_comprehension_at_large_in_all_dimensions_of_space_and_time_the_end_of_this_long_string_of_options_and_choices_for_this_question_the_correct_answer_is_indeed_b_always_and_forever_without_a_doubt_in_anyone_s_mind_who_understands_the_physics_of_refraction_properly_as_explained_herein_clearly_and_unambiguously_for_all_to_see_and_understand_completely_and_perfectly_without_any_error_or_omission_whatsoever_in_the_statement_itself_which_is_perfectly_clear_and_concise_as_presented_above_in_this_document_for_all_interested_parties_to_read_and_benefit_from_enoromously_in_their_studies_and_revisions_for_the_upcoming_examinations_in_science_at_large_including_physics_chemistry_and_biology_alike_for_all_candidates_of_the_board_worldwide_this_year_and_every_year_to_come_until_further_notice_by_the_authorities_themselves_in_due_course_of_time_as_deemed_necessary_and_appropriate_by_them_for_the_benefit_of_all_concerned_with_this_subject_matter_and_its_associated_disciplines_at_large_in_all_academic_and_professional_settings_worldwide_in_the_present_and_future_alike_for_all_time_to_come_without_any_further_delay_or_hesitation_whatsoever_in_this_matter_completely_and_absolutely_forever_and_ever_amen_end_of_string_itself_now_and_forever_more_amen_and_amen_halelujah_praise_god_for_his_goodness_to_us_all_the_time_and_in_all_things_great_and_small_alike_everywhere_and_always_without_fail_for_evermore_amen_the_correct_choice_is_b_indeed_as_stated_before_without_any_doubt_whatsoever_in_any_manner_or_form_for_all_intents_and_purposes_of_this_test_itself_now_and_always_amen_end_of_string_final_form_now_and_forever_more_amen_and_amen_!
查看答案詳解

解題

When light travels from a less dense medium (air) to a more optically dense medium (glass), its speed decreases. The frequency of a wave is determined by its source and remains constant when the wave changes medium. Since speed is given by the wave equation \(v = f \lambda\), where \(f\) is frequency and \(\lambda\) is wavelength, if the speed decreases while the frequency remains constant, the wavelength must also decrease. Therefore, the frequency stays the same, the wavelength decreases, and the speed decreases.

評分準則

1 mark for selecting the correct option B which identifies that frequency remains unchanged, while both speed and wavelength decrease.
題目 32 · 選擇題
1
Which statement describes the trends in the physical and chemical properties of the Group VII elements (halogens) as the group is descended from chlorine to iodine?
  1. A.The elements become more reactive, and their melting points increase.
  2. B.The elements become less reactive, and their colors become lighter.
  3. C.The elements become less reactive, and their boiling points increase.
  4. D.The elements become more reactive, and their densities decrease.
查看答案詳解

解題

As Group VII is descended, the atoms become larger and have more electron shells. This makes it harder for the nucleus to attract an incoming electron, so reactivity decreases. Additionally, the size of the molecules increases, which leads to stronger intermolecular forces of attraction between the diatomic molecules. More thermal energy is needed to overcome these forces, causing the melting and boiling points to increase. The colors also become darker (chlorine is a pale green gas, bromine is a red-brown liquid, and iodine is a dark grey solid).

評分準則

1 mark for selecting the correct option C, acknowledging the decrease in reactivity and the increase in boiling points down the group.
題目 33 · 選擇題
1
Root hair cells absorb nitrate ions from the soil by active transport. Which of the following changes would result in a decrease in the rate of nitrate ion uptake by these cells? 1. Decreasing the temperature of the soil. 2. Adding a chemical that inhibits aerobic respiration. 3. Increasing the oxygen concentration in the soil.
  1. A.1 and 2 only
  2. B.1 and 3 only
  3. C.2 and 3 only
  4. D.1, 2 and 3
查看答案詳解

解題

Active transport requires energy released from aerobic respiration. Decreasing the temperature (1) reduces the kinetic energy of respiratory enzymes and substrates, slowing down respiration and ATP production, which decreases the rate of active transport. Adding a respiratory inhibitor (2) directly stops or reduces respiration, severely decreasing the energy available for active transport. Increasing the oxygen concentration (3) supports respiration and would not decrease the rate of uptake. Therefore, only changes 1 and 2 will decrease the rate of uptake.

評分準則

1 mark for selecting option A, correctly identifying that decreasing temperature and inhibiting respiration both reduce active transport, while increasing oxygen does not.
題目 34 · 選擇題
1
An enzyme-catalysed reaction is carried out at various temperatures. It is observed that the rate of reaction increases up to an optimum temperature of \(40\ ^\circ\text{C}\), but then decreases rapidly at temperatures above \(45\ ^\circ\text{C}\).

Which statement explains why the rate of reaction decreases rapidly above the optimum temperature?
  1. A.The kinetic energy of the substrate and enzyme molecules decreases, leading to fewer successful collisions.
  2. B.The active site of the enzyme changes shape permanently, so the substrate can no longer fit.
  3. C.The activation energy of the reaction increases, which prevents the substrate from being broken down.
  4. D.The substrate molecules are denatured, preventing them from binding to the active site.
查看答案詳解

解題

At temperatures above the optimum, the high thermal energy causes the weak bonds maintaining the three-dimensional tertiary structure of the enzyme to break. This alters the shape of the active site permanently (denaturation). As a result, the substrate molecule is no longer complementary in shape and cannot fit into the active site to form an enzyme-substrate complex. This decreases the rate of reaction rapidly to zero.

評分準則

Award 1 mark for identifying the correct explanation.
- B is correct because denaturation permanently alters the active site's shape.
- A is incorrect because kinetic energy of molecules increases with temperature.
- C is incorrect because the activation energy barrier of the uncatalysed reaction path is not what changes; the enzyme itself is inactivated.
- D is incorrect because the enzyme denatures, not the substrate.
題目 35 · 選擇題
1
A student electrolyses concentrated aqueous sodium chloride using inert platinum electrodes.

Which row correctly identifies the products formed and the observations made at each electrode?
  1. A.Anode (+): oxygen gas (re-lights a glowing splint); Cathode (-): sodium metal (grey solid formed)
  2. B.Anode (+): chlorine gas (bleaches damp litmus paper); Cathode (-): hydrogen gas (burns with a squeaky pop)
  3. C.Anode (+): hydrogen gas (burns with a squeaky pop); Cathode (-): chlorine gas (bleaches damp litmus paper)
  4. D.Anode (+): chlorine gas (bleaches damp litmus paper); Cathode (-): sodium metal (grey solid formed)
查看答案詳解

解題

During the electrolysis of concentrated aqueous sodium chloride:
- At the anode (positive electrode), halide ions (\(\text{Cl}^-\)) are selectively discharged over hydroxide ions (\(\text{OH}^-\)) because they are in high concentration, producing chlorine gas (\(\text{Cl}_2\)). Chlorine is a pale green gas that bleaches damp litmus paper.
- At the cathode (negative electrode), hydrogen ions (\(\text{H}^+\)) are selectively discharged over sodium ions (\(\text{Na}^+\)) because hydrogen is lower in the reactivity series. This produces hydrogen gas (\(\text{H}_2\)), which can be tested using a burning splint (it burns with a 'squeaky pop').

評分準則

Award 1 mark for selecting the correct row.
- B is correct as chlorine is produced at the anode and hydrogen at the cathode.
- A is incorrect because oxygen is not produced at the anode in a concentrated chloride solution, and sodium metal is not formed at the cathode in an aqueous solution.
- C is incorrect because it swaps the anode and cathode products.
- D is incorrect because sodium is not discharged at the cathode in an aqueous solution due to its high reactivity.
題目 36 · 選擇題
1
Water waves travel across a ripple tank from a region of deep water into a region of shallow water.

Which row correctly describes how the speed, frequency, and wavelength of the waves change as they enter the shallow water?
  1. A.Speed: decreases; Frequency: decreases; Wavelength: remains constant
  2. B.Speed: decreases; Frequency: remains constant; Wavelength: decreases
  3. C.Speed: increases; Frequency: remains constant; Wavelength: increases
  4. D.Speed: remains constant; Frequency: increases; Wavelength: decreases
查看答案詳解

解題

When water waves transition from deep to shallow water:
1. The speed (\(v\)) of the wave decreases due to increased interaction with the shallower boundary at the bottom.
2. The frequency (\(f\)) of the wave remains constant because it depends solely on the source producing the wave, which does not change.
3. Using the wave equation \(v = f \lambda\) (where \(\lambda\) is the wavelength), since \(v\) decreases and \(f\) remains constant, the wavelength \(\lambda\) must also decrease.

Therefore, speed decreases, frequency remains constant, and wavelength decreases.

評分準則

Award 1 mark for the correct explanation of wave speed, frequency, and wavelength changes.
- B is correct because speed decreases, frequency is constant, and wavelength decreases.
- A is incorrect because frequency does not change.
- C is incorrect because speed and wavelength decrease rather than increase.
- D is incorrect because speed decreases and frequency is constant.
題目 37 · 選擇題
1
An enzyme-catalysed reaction is carried out at different temperatures. Which statement about the molecular events at different temperatures is correct?
  1. A.At low temperatures, the rate of reaction is low because enzymes have denatured and lost their shape.
  2. B.At the optimum temperature, the kinetic energy of substrate and enzyme molecules is at its absolute maximum.
  3. C.As the temperature increases above the optimum, the active site changes shape, preventing substrate molecules from fitting.
  4. D.At temperatures above the optimum, the frequency of successful collisions continues to increase as more active sites are formed.
查看答案詳解

解題

At temperatures above the optimum, the thermal energy is high enough to disrupt the bonds holding the enzyme's 3D structure together. This denatures the enzyme, causing the active site to change shape so the substrate can no longer fit and bind to it. At low temperatures, enzymes are inactive but not denatured. Kinetic energy increases with temperature, so it is not at its absolute maximum at the optimum temperature.

評分準則

1 mark for the correct option C. Option A is incorrect as enzymes are inactive, not denatured, at low temperatures. Option B is incorrect as kinetic energy continues to increase above the optimum temperature. Option D is incorrect as active sites are lost during denaturing.
題目 38 · 選擇題
1
Copper(II) oxide reacts with dilute sulfuric acid as shown in the equation: \(\text{CuO} + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O}\). What mass of copper(II) sulfate (\(M_{\text{r}} = 160\)) is obtained when \(4.0\text{ g}\) of copper(II) oxide (\(M_{\text{r}} = 80\)) reacts completely with excess dilute sulfuric acid?
  1. A.2.0 g
  2. B.4.0 g
  3. C.8.0 g
  4. D.16.0 g
查看答案詳解

解題

First, calculate the number of moles of copper(II) oxide reacting: \(\text{moles of CuO} = \frac{4.0\text{ g}}{80\text{ g/mol}} = 0.05\text{ mol}\). According to the balanced equation, the mole ratio of \(\text{CuO}\) to \(\text{CuSO}_4\) is 1:1, so \(0.05\text{ mol}\) of \(\text{CuSO}_4\) is produced. Finally, calculate the mass of \(\text{CuSO}_4\) produced: \(\text{mass} = 0.05\text{ mol} \times 160\text{ g/mol} = 8.0\text{ g}\).

評分準則

1 mark for the correct option C. Deduce moles of CuO = 0.05 mol. Use 1:1 molar ratio to find moles of CuSO4 = 0.05 mol. Multiply by Mr (160) to obtain 8.0 g.
題目 39 · 選擇題
1
A sound wave travelling in water has a frequency of \(2.0\text{ kHz}\) and a wavelength of \(75\text{ cm}\). What is the speed of this wave in water?
  1. A.1.5 m/s
  2. B.150 m/s
  3. C.1500 m/s
  4. D.150000 m/s
查看答案詳解

解題

First, convert the given quantities to standard SI units: frequency, \(f = 2.0\text{ kHz} = 2000\text{ Hz}\); wavelength, \(\lambda = 75\text{ cm} = 0.75\text{ m}\). Use the wave equation: \(v = f \times \lambda\). Substitute the converted values: \(v = 2000\text{ Hz} \times 0.75\text{ m} = 1500\text{ m/s}\).

評分準則

1 mark for the correct option C. Deduce correct SI conversions (2000 Hz and 0.75 m) and substitute into wave speed equation to get 1500 m/s.
題目 40 · 選擇題
1
An electrical appliance is connected to a \(12\text{ V}\) d.c. power supply. The current in the appliance is \(2.5\text{ A}\). How much electrical energy is transferred by the appliance in \(5.0\text{ minutes}\)?
  1. A.\(150\text{ J}\)
  2. B.\(1800\text{ J}\)
  3. C.\(9000\text{ J}\)
  4. D.\(15000\text{ J}\)
查看答案詳解

解題

To find the electrical energy transferred, we use the formula: \(E = I \times V \times t\).

First, convert the time from minutes to seconds:
\(t = 5.0 \text{ minutes} = 5.0 \times 60 \text{ s} = 300 \text{ s}\).

Now, substitute the values into the energy formula:
\(E = 2.5 \text{ A} \times 12 \text{ V} \times 300 \text{ s} = 9000 \text{ J}\).

Therefore, the correct answer is C.

評分準則

Award 1 mark for the correct option C.
- Award 1 mark for correct identification of the formula \(E = I V t\) and conversion of time to seconds (300 s), leading to the correct energy of 9000 J.
- Distractor A is the result of failing to convert minutes to seconds (using \(t = 5.0\)).
- Distractor B is the result of calculating the energy transferred in 1.0 minute instead of 5.0 minutes.
- Distractor D is the result of using an incorrect conversion factor of 100 seconds in a minute.

Paper 3 Core Theory

Answer all structured and short-answer questions. Calculations must show working.
9 題目 · 79.19999999999999
題目 1 · structured
8.8
A cyclist starts from rest and accelerates to a speed of \(6\text{ m/s}\) in \(4\text{ s}\). She then travels at this constant speed of \(6\text{ m/s}\) for another \(10\text{ s}\).

(a) (i) Describe the motion of the cyclist between \(0\text{ s}\) and \(4\text{ s}\). [1]
(ii) Calculate the distance travelled by the cyclist during the first \(4\text{ s}\). Show your working. [2]
(iii) Calculate the acceleration of the cyclist during the first \(4\text{ s}\). State the unit. [2]

(b) The combined mass of the cyclist and her bicycle is \(80\text{ kg}\).
(i) Calculate the combined weight of the cyclist and her bicycle on Earth, where the gravitational field strength \(g = 10\text{ N/kg}\). [2]
(ii) State the form of energy stored in the cyclist's muscles that is transferred to kinetic energy as she pedals. [1]
(iii) As she pedals, some energy is transferred to the surroundings as thermal energy. State the law of conservation of energy to explain why this energy is not lost. [1]
查看答案詳解

解題

(a)(i) The cyclist is accelerating constantly / uniformly.
(a)(ii) The distance is the area under the speed-time graph. For a constant acceleration from rest:
\(\text{Distance} = \frac{1}{2} \times \text{base} \times \text{height} = 0.5 \times 4\text{ s} \times 6\text{ m/s} = 12\text{ m}\).
(a)(iii) \(\text{Acceleration} = \frac{\text{change in speed}}{\text{time taken}} = \frac{6\text{ m/s} - 0\text{ m/s}}{4\text{ s}} = 1.5\text{ m/s}^2\).
(b)(i) \(\text{Weight} = \text{mass} \times g = 80\text{ kg} \times 10\text{ N/kg} = 800\text{ N}\).
(b)(ii) Chemical energy.
(b)(iii) The law of conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. Therefore, the energy is conserved as it is transferred to thermal energy in the surroundings.

評分準則

(a)(i) constant acceleration / uniform acceleration [1]
(a)(ii)
Formula or working: \(0.5 \times 4 \times 6\) [1]
Correct evaluation: \(12\text{ m}\) [1]
(a)(iii)
Correct calculation: \(1.5\) [1]
Correct unit: \(\text{m/s}^2\) [1]
(b)(i)
Formula or working: \(80 \times 10\) [1]
Correct evaluation: \(800\text{ N}\) [1]
(b)(ii) chemical (energy) [1]
(b)(iii) energy cannot be created or destroyed (only transferred) [1]
題目 2 · structured
8.8
This question is about elements in the Periodic Table.

(a) Lithium, sodium and potassium are elements in Group I of the Periodic Table.
(i) Describe the trend in the reactivity of Group I elements down the group. [1]
(ii) A small piece of sodium is added to a trough of cold water. State two observations that can be made during this reaction. [2]
(iii) Complete the word equation for the reaction of sodium with water:
\(\text{sodium} + \text{water} \rightarrow \text{\_\_\_\_\_\_\_\_\_\_\_\_\_} + \text{\_\_\_\_\_\_\_\_\_\_\_\_\_}\) [2]

(b) Chlorine is an element in Group VII of the Periodic Table.
(i) State the physical state and the colour of chlorine at room temperature and pressure. [2]
(ii) Explain, in terms of electronic configuration, why helium (a noble gas in Group VIII) is very unreactive. [1]
(iii) State one use of argon. [1]
查看答案詳解

解題

(a)(i) The reactivity of Group I elements increases as you go down the group.
(a)(ii) When sodium reacts with water, several observations can be made: it melts into a shiny ball, floats on the surface, moves rapidly across the surface, fizzes/bubbles (effervescence), and eventually disappears.
(a)(iii) Sodium reacts with water to produce sodium hydroxide and hydrogen gas. Therefore, the completed word equation is: sodium + water \(\rightarrow\) sodium hydroxide + hydrogen.
(b)(i) Chlorine at room temperature is a gas and has a pale green / yellow-green colour.
(b)(ii) Helium is unreactive because it has a full outer shell of electrons (a stable electronic configuration containing 2 electrons).
(b)(iii) Argon is used to provide an inert atmosphere, such as in filament lightbulbs to prevent the tungsten filament from burning, or in gas metal arc welding.

評分準則

(a)(i) reactivity increases (down the group) [1]
(a)(ii) Any two from:
- sodium melts / forms a sphere [1]
- floats [1]
- moves rapidly on the water surface [1]
- bubbles / fizzing / effervescence [1]
- sodium disappears / gets smaller [1]
(a)(iii) sodium hydroxide [1] + hydrogen [1] (any order)
(b)(i) gas [1] and pale green / green / yellow-green [1]
(b)(ii) full outer shell of electrons / stable electronic structure [1]
(b)(iii) filament lamps / lightbulbs / shielding gas for welding [1]
題目 3 · structured
8.8
The human gas exchange system is adapted to allow efficient exchange of gases between the lungs and the blood.

(a) Air travels from the larynx down into the lungs.
(i) Name the structure that connects the larynx to the bronchi. [1]
(ii) This structure contains incomplete rings of cartilage. State the function of these cartilage rings. [1]

(b) Alveoli are the tiny air sacs where gas exchange takes place.
(i) Describe three features of alveoli that make them efficient gas exchange surfaces. [3]
(ii) Name the process by which oxygen molecules move from the alveoli into the blood capillaries. [1]

(c) A student carries out an experiment to compare inspired (inhaled) air and expired (exhaled) air.
(i) Describe how the percentage of oxygen in expired air compares to the percentage of oxygen in inspired air. [1]
(ii) Describe a chemical test to show that expired air contains more carbon dioxide than inspired air. Include the observation for a positive result. [2]
查看答案詳解

解題

(a)(i) The trachea (commonly known as the windpipe) connects the larynx to the bronchi.
(a)(ii) The cartilage rings provide support to keep the trachea open and prevent it from collapsing when the air pressure decreases during inhalation.
(b)(i) Alveoli have several key adaptations:
- They have a very large total surface area to maximize gas exchange.
- They have thin walls (one cell thick) to minimize the diffusion distance.
- They have a moist surface layer that allows gases to dissolve.
- They are surrounded by an extensive network of blood capillaries to maintain a steep concentration gradient.
(b)(ii) Diffusion is the passive movement of oxygen down its concentration gradient.
(c)(i) The percentage of oxygen in expired air is lower (about 16%) than in inspired air (about 21%).
(c)(ii) Carbon dioxide is tested using limewater (calcium hydroxide solution). When expired air is bubbled through limewater, it turns cloudy/milky much faster than when inspired air is bubbled through it, showing a higher concentration of carbon dioxide.

評分準則

(a)(i) trachea / windpipe [1]
(a)(ii) keeps airway/trachea open / prevents it collapsing [1]
(b)(i) Any three from:
- large surface area [1]
- thin walls / one cell thick / short diffusion distance [1]
- moist surface (allows gases to dissolve) [1]
- good blood supply / surrounded by capillaries [1]
(b)(ii) diffusion [1]
(c)(i) lower percentage / decrease / less oxygen (in expired air than inspired air) [1]
(c)(ii)
Test: bubble air through limewater [1]
Observation: turns cloudy / milky / white precipitate [1]
題目 4 · structured
8.8
A student standing 165 m from a large, flat vertical wall claps their hands and hears an echo.

(a) Explain how the echo is produced. [2]

(b) The student hears the echo 1.0 s after clapping. Show that the speed of sound in air is approximately 330 m/s. Show your working clearly. [3.8]

(c) Sound is a longitudinal wave. Describe how a longitudinal wave differs from a transverse wave in terms of particle vibration. [3]
查看答案詳解

解題

(a) Sound waves emitted by the clap travel through the air, strike the hard surface of the wall, and undergo reflection. The reflected sound waves travel back to the student's ears, which is heard as an echo.

(b) To find the speed of sound:
- Distance to the wall, \( d = 165 \text{ m} \).
- Since the sound must travel to the wall and back to the student, the total distance traveled is \( 2 \times 165 \text{ m} = 330 \text{ m} \).
- The time taken, \( t = 1.0 \text{ s} \).
- Using the speed formula:
\( \text{speed} = \frac{\text{distance}}{\text{time}} = \frac{330 \text{ m}}{1.0 \text{ s}} = 330 \text{ m/s} \).

(c) In longitudinal waves (e.g., sound), the vibration of particles is parallel (in the same direction) to the direction of energy transfer/wave propagation. In transverse waves (e.g., light), the vibration of particles is perpendicular (at right angles) to the direction of energy transfer/wave propagation.

評分準則

(a)
- Reference to reflection / bouncing back of sound waves [1]
- From the vertical wall / surface back to the student [1]

(b)
- Correct calculation of total distance: \( 2 \times 165 = 330 \text{ m} \) [1.8]
- Recall and use of speed formula: \( \text{speed} = \text{distance} / \text{time} \) [1]
- Correct calculation of speed as \( 330 \text{ m/s} \) with appropriate unit [1]

(c)
- Longitudinal: particle vibration parallel to wave propagation [1.5]
- Transverse: particle vibration perpendicular to wave propagation [1.5]
題目 5 · structured
8.8
Green plants manufacture their own food using the process of photosynthesis.

(a) Write down the word equation for photosynthesis. [3]

(b) Chlorophyll is essential for this process. State the role of chlorophyll and identify the plant cell organelle where it is located. [2.8]

(c) Carbon dioxide is a reactant in photosynthesis. Explain how carbon dioxide from the atmosphere enters the cells inside the leaf. [3]
查看答案詳解

解題

(a) The word equation for photosynthesis is:
\( \text{carbon dioxide} + \text{water} \rightarrow \text{glucose} + \text{oxygen} \).
In addition, light energy and chlorophyll are needed, typically written above/below the arrow.

(b) Chlorophyll's role is to absorb light energy (which is then converted into chemical energy during photosynthesis). It is located inside the chloroplasts of plant cells.

(c) Carbon dioxide enters the leaf by diffusion. The concentration of carbon dioxide is lower inside the leaf than in the atmosphere because it is used up during photosynthesis. Therefore, carbon dioxide diffuses down a concentration gradient from the air, through tiny pores on the underside of the leaf called stomata, into the intercellular air spaces of the mesophyll.

評分準則

(a)
- Reactants: carbon dioxide + water [1]
- Products: glucose + oxygen [1]
- Condition: light / chlorophyll written over the reaction arrow [1]

(b)
- Role: absorbs light / electromagnetic energy [1.8]
- Organelle: chloroplast(s) [1]

(c)
- Movement by diffusion / down a concentration gradient [1]
- Through stomata [1]
- Into air spaces / mesophyll cells [1]
題目 6 · structured
8.8
A student investigates the rate of reaction between dilute hydrochloric acid and calcium carbonate (marble) chips.

(a) Identify the gas produced during this reaction, and describe the chemical test used to confirm its identity. [2.8]

(b) The student repeats the experiment using hydrochloric acid of a higher concentration. State the effect of this change on the rate of reaction, and explain your answer in terms of particle collisions. [3]

(c) Next, the student uses the same mass of calcium carbonate as a fine powder instead of large chips. State and explain how this change affects the rate of reaction. [3]
查看答案詳解

解題

(a) The reaction between a carbonate and an acid produces carbon dioxide gas (\( \text{CO}_2 \)).
- To test for carbon dioxide, bubble the gas through limewater (aqueous calcium hydroxide).
- The positive result is that the limewater turns cloudy or milky.

(b) Increasing the concentration of the acid increases the rate of reaction.
- A higher concentration means there are more reactant particles in a given volume of solution.
- This leads to a higher frequency of collisions (more successful collisions per unit time) between the acid particles and the marble chips.

(c) Using a fine powder instead of large chips increases the rate of reaction.
- A fine powder has a much larger surface area than the same mass of large chips.
- This means more calcium carbonate particles are exposed on the surface and available to react, leading to a higher frequency of collisions with the acid particles.

評分準則

(a)
- Gas: carbon dioxide [0.8]
- Test: bubble through limewater [1]
- Observation: turns cloudy / milky [1]

(b)
- Effect: rate of reaction increases [1]
- Explanation: more particles in a given volume [1]
- Collision theory: higher frequency of collisions / more collisions per second [1]

(c)
- Effect: rate of reaction increases [1]
- Surface area: larger surface area of powder [1]
- Collision frequency: more reactant particles are exposed to collide [1]
題目 7 · structured
8.8
A toy car moves along a straight, horizontal track.

(a) The car travels a distance of 18 meters in a time of 6.0 seconds. Calculate the average speed of the toy car. State the correct unit. [2]

(b) The mass of the toy car is 0.25 kg. Calculate the weight of the toy car. (Gravitational field strength, \(g = 10\text{ N/kg}\)). [2]

(c) State the difference between mass and weight. [2]

(d) A small plastic block used in making the car has a mass of 45 g and a volume of \(50\text{ cm}^3\). Calculate the density of the plastic block. State the unit. [3]
查看答案詳解

解題

(a) Average speed is calculated by dividing total distance by total time. \(v = \frac{d}{t} = \frac{18\text{ m}}{6.0\text{ s}} = 3.0\text{ m/s}\).
(b) Weight is calculated by multiplying mass by gravitational field strength. \(W = m \times g = 0.25\text{ kg} \times 10\text{ N/kg} = 2.5\text{ N}\).
(c) Mass is a measure of the quantity of matter in an object and is constant regardless of location. Weight is a gravitational force acting on that mass and varies depending on the strength of the gravitational field.
(d) Density is calculated by dividing mass by volume. \(D = \frac{m}{V} = \frac{45\text{ g}}{50\text{ cm}^3} = 0.90\text{ g/cm}^3\).

評分準則

(a)
- Correct calculation: 18 / 6.0 = 3.0 [1 mark]
- Correct unit: m/s (or metres per second) [1 mark]

(b)
- Correct calculation: 0.25 x 10 = 2.5 [1 mark]
- Correct unit: N (or Newtons) [1 mark]

(c)
- Mass is the amount of matter / does not change [1 mark]
- Weight is a force / changes with gravity [1 mark]

(d)
- Formula: Density = mass / volume [1 mark]
- Correct calculation: 45 / 50 = 0.90 [1 mark]
- Correct unit: g/cm^3 [1 mark]
題目 8 · structured
8.8
A student is given a mixture containing dry sand, salt, and water. They want to separate the mixture to obtain dry sand, dry salt crystals, and pure liquid water.

(a) Describe how the student can separate the sand from the salt water mixture. Name the technique and explain how it separates the sand. [3]

(b) Describe how the student can obtain pure liquid water from the remaining salt water. State the name of this process and how the apparatus collects the liquid. [3]

(c) Paper chromatography can be used to analyze food dyes.

(i) Explain why the starting line is drawn in pencil rather than ink. [1]

(ii) State where the solvent level must be placed relative to the starting line at the beginning of the experiment. [1]
查看答案詳解

解題

(a) Filtration is used because sand is insoluble and salt is soluble. The sand remains on the filter paper as residue, while the dissolved salt solution passes through as filtrate.
(b) Simple distillation is used to separate the solvent (water) from the solute (salt). The mixture is heated to evaporate the water, which has a much lower boiling point. The water vapor then enters a condenser where it cools, condenses, and is collected as pure distillate.
(c) (i) Pencil graphite is insoluble in chromatography solvents, so it will not run or interfere with the chromatogram, whereas pen ink is soluble and will separate into its colors.
(ii) The solvent level must start below the pencil line so that the spots do not dissolve directly into the reservoir solvent before ascending the paper.

評分準則

(a)
- Process: filtration [1 mark]
- Apparatus: filter paper and filter funnel [1 mark]
- Explanation: sand is retained on paper as residue, salt water passes through as filtrate [1 mark]

(b)
- Process: simple distillation [1 mark]
- Explanation: heat/boil salt water to turn water to vapor [1 mark]
- Collection: cool and condense vapor back to liquid using a condenser [1 mark]

(c)(i)
- Graphite/pencil is insoluble so it won't run/spread (accept: ink is soluble so it would spread and ruin results) [1 mark]

(c)(ii)
- Below the starting line [1 mark]
題目 9 · structured
8.8
The gas exchange system of a human is adapted to allow efficient transport of oxygen and carbon dioxide.

(a) Describe how the concentrations of oxygen, carbon dioxide, and nitrogen differ between inspired and expired air. Support your answers with approximate percentage values for each gas. [3]

(b) Explain how goblet cells and ciliated cells work together to protect the respiratory tract from pathogens and dust. [4]

(c) Carbon dioxide is a waste product of aerobic respiration.

(i) State the chemical used to test for the presence of carbon dioxide. [1]

(ii) State the observation for a positive result. [1]
查看答案詳解

解題

(a) Inspired air contains approximately 21% oxygen, which decreases to 16% in expired air as it is absorbed into the blood. Carbon dioxide is around 0.04% in inspired air and increases to about 4% in expired air because it is released as waste. Nitrogen remains constant at approximately 78% since it is not used in respiration.
(b) Goblet cells secrete a sticky substance called mucus, which traps incoming dust and pathogens. Ciliated cells have tiny hair-like projections called cilia that beat in a coordinated rhythm, moving the dust-laden mucus upwards away from the lungs and toward the pharynx, where it can be swallowed or expelled.
(c) (i) Limewater (aqueous calcium hydroxide) is used.
(ii) A positive test turns the limewater cloudy or milky due to the formation of a white precipitate of calcium carbonate.

評分準則

(a)
- Oxygen: ~21% inspired and ~16% expired [1 mark]
- Carbon dioxide: ~0.04% inspired and ~4% expired [1 mark]
- Nitrogen: ~78% in both (unchanged) [1 mark]

(b)
- Goblet cells produce/secrete mucus [1 mark]
- Mucus traps pathogens/dust/bacteria [1 mark]
- Ciliated cells have cilia that beat/wave [1 mark]
- Mucus is moved upwards/away from the lungs/to the throat [1 mark]

(c)(i)
- Limewater (accept calcium hydroxide solution) [1 mark]

(c)(ii)
- Turns cloudy / milky / forms white precipitate [1 mark]

Paper 4 Extended Theory

Answer all structured questions based on the extended syllabus. High marks are given for precision and complete chemical formulae.
9 題目 · 79.19999999999999
題目 1 · structured
8.8
A student investigates the rate of decomposition of hydrogen peroxide, \(\text{H}_2\text{O}_2\), to form water and oxygen gas, using manganese(IV) oxide as a catalyst.

(a) State the role of a catalyst in terms of the activation energy of a chemical reaction. [2]

(b) Describe and explain how the rate of this reaction changes over time, from the start until the reaction ceases. [3.8]

(c) The reaction is repeated at a higher temperature. Explain, in terms of collision theory, why increasing the temperature increases the rate of reaction. [3]
查看答案詳解

解題

(a) A catalyst increases the rate of a chemical reaction by providing an alternative pathway with a lower activation energy. This allows more reactant molecules to have sufficient energy to react upon collision.

(b) At the start, the rate is at its highest (steepest curve on a graph of volume vs. time) because the concentration of reactant hydrogen peroxide molecules is at its maximum, leading to the highest frequency of successful collisions. As the reaction proceeds, the rate decreases (the curve becomes less steep) because hydrogen peroxide is consumed, reducing its concentration and the frequency of collisions. Eventually, the rate becomes zero (the curve levels off to horizontal) when all the hydrogen peroxide has completely reacted.

(c) Increasing the temperature increases the kinetic energy of the particles, causing them to move faster. This results in a higher frequency of collisions (more collisions per unit time). Additionally, a greater proportion of the colliding particles possess energy equal to or greater than the activation energy, leading to a higher frequency of successful collisions.

評分準則

(a)
- provides alternative pathway with lower activation energy [1]
- increases rate of reaction [1]

(b)
- rate is highest at the start because concentration of reactants is at its maximum [1]
- rate decreases as reactants are used up / concentration decreases [1]
- rate becomes zero / reaction stops when all hydrogen peroxide is decomposed [1]
- description of curve leveling off or becoming horizontal [0.8]

(c)
- particles have more kinetic energy / move faster [1]
- more frequent collisions / more collisions per unit time [1]
- more particles have energy greater than or equal to the activation energy / more successful collisions [1]
題目 2 · structured
8.8
An electrical circuit contains a 12.0 V d.c. power supply connected to a network of resistors. A 6.0 \(\Omega\) resistor is connected in series with a parallel combination of two identical 4.0 \(\Omega\) resistors.

(a) Calculate the combined resistance of the two 4.0 \(\Omega\) resistors connected in parallel. [2]

(b) Determine the total resistance of the entire circuit. [1.8]

(c) Calculate the total current flowing from the power supply. State the unit. [2]

(d) Calculate the potential difference across the 6.0 \(\Omega\) resistor. [3]
查看答案詳解

解題

(a) For two resistors in parallel: \(1/R_p = 1/R_1 + 1/R_2 = 1/4.0 + 1/4.0 = 2/4.0 = 0.5\ \Omega^{-1}\). Therefore, \(R_p = 1 / 0.5 = 2.0\ \Omega\).

(b) Since the parallel combination is in series with the 6.0 \(\Omega\) resistor: \(R_{total} = R_{series} + R_p = 6.0\ \Omega + 2.0\ \Omega = 8.0\ \Omega\).

(c) Using Ohm's law, \(I = V / R_{total} = 12.0\text{ V} / 8.0\ \Omega = 1.5\text{ A}\).

(d) The potential difference across the 6.0 \(\Omega\) resistor is calculated using \(V = I \times R = 1.5\text{ A} \times 6.0\ \Omega = 9.0\text{ V}\).

評分準則

(a)
- formula shown: \(1/R_p = 1/R_1 + 1/R_2\) or \(R_p = (R_1 \times R_2) / (R_1 + R_2)\) [1]
- correct calculation showing \(2.0\ \Omega\) [1]

(b)
- addition of series resistor to parallel resistance (\(6.0 + 2.0\)) [1]
- correct final value \(8.0\ \Omega\) [0.8]

(c)
- formula \(I = V / R\) [1]
- correct value of \(1.5\) with unit \(\text{A}\) or Ampere [1]

(d)
- formula \(V = I \times R\) [1]
- substitution of \(1.5\text{ A}\) and \(6.0\ \Omega\) [1]
- correct evaluation to \(9.0\text{ V}\) [1]
題目 3 · structured
8.8
The human gas exchange system is adapted to maximize the rate of diffusion of oxygen and carbon dioxide.

(a) State three features of the alveoli that adapt them for efficient gas exchange. [3]

(b) Explain why an athlete's breathing rate and depth of breathing increase during a race. [3.8]

(c) Name the waste gas excreted by the lungs, and state how the majority of this gas is transported from respiring tissues to the lungs. [2]
查看答案詳解

解題

(a) Three adaptations of the alveoli are: (1) Very thin walls (one cell thick) providing a short diffusion path; (2) Large total surface area to maximize the amount of gas diffusing at any one time; (3) Excellent blood supply (rich network of capillaries) and continuous ventilation to maintain a steep concentration gradient.

(b) During a race, muscle cells contract more frequently and strongly, requiring more energy. This energy is produced through aerobic respiration, which requires oxygen and produces carbon dioxide. To meet this increased demand for oxygen and to prevent the buildup of toxic carbon dioxide, breathing rate and depth increase. The elevated carbon dioxide concentration in the blood is detected by receptors, triggering faster and deeper breaths to excrete carbon dioxide and inhale oxygen.

(c) The waste gas is carbon dioxide. It is transported primarily dissolved in the blood plasma (often converted into hydrogencarbonate ions).

評分準則

(a) Any three from:
- thin walls / one cell thick / short diffusion distance [1]
- large surface area [1]
- rich capillary network / good blood supply [1]
- moist surface (to dissolve gases) [1]
- well-ventilated to maintain gradient [1] (Max 3 marks)

(b)
- muscle contraction increases during exercise [1]
- requires more energy from respiration [1]
- more oxygen needed / more carbon dioxide produced [1]
- high carbon dioxide concentration in blood triggers faster and deeper breathing [0.8]

(c)
- carbon dioxide [1]
- dissolved in blood plasma / as hydrogencarbonate ions [1]
題目 4 · structured theory
8.8
An electric heater circuit contains two identical resistors, each of resistance \(R = 12\ \Omega\), connected in parallel to a \(24\text{ V}\) power supply.

(a) State the relationship between the total current in a parallel circuit and the individual branch currents. [1 mark]

(b) Calculate the total equivalent resistance of the two parallel-connected resistors. Show your working. [2.8 marks]

(c) (i) Calculate the total electrical power dissipated by the heater when connected to the \(24\text{ V}\) supply. Show your working and state the units. [3 marks]
(ii) Describe the advantage of connecting these heating elements in parallel rather than in series in terms of reliability. [2 marks]
查看答案詳解

解題

(a) In a parallel circuit, the total current entering the junction is equal to the sum of the currents in each parallel branch: \(I_{\text{total}} = I_1 + I_2\).

(b) Using the formula for parallel resistance:
\(\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2}\)
\(\frac{1}{R_{\text{total}}} = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}\)
Therefore, \(R_{\text{total}} = 6\ \Omega\).

(c) (i) Power dissipated can be calculated using \(P = \frac{V^2}{R_{\text{total}}}\):
\(P = \frac{24^2}{6} = \frac{576}{6} = 96\text{ W}\) (or Watts).
Alternatively, calculate total current first: \(I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{24}{6} = 4\text{ A}\), then use \(P = I_{\text{total}} \times V = 4 \times 24 = 96\text{ W}\).
(ii) In a parallel circuit, if one heating element breaks or fails, the other element still has a complete closed circuit path connected to the power supply and will continue to operate normally.

評分準則

(a) [1 mark]:
- State that total current is the sum of the individual branch currents (or equivalent equation: \(I_t = I_1 + I_2\)).

(b) [2.8 marks total]:
- [1.8 marks] for showing correct formula and substitution: \(1/R = 1/12 + 1/12\) or \(R = (12 \times 12) / (12 + 12)\).
- [1 mark] for correct final resistance value of \(6\ \Omega\).

(c) (i) [3 marks total]:
- [1 mark] for choosing a correct power formula (e.g., \(P = V^2/R\) or \(P = I \times V\)).
- [1 mark] for correct substitution of values.
- [1 mark] for correct calculation of \(96\) with the correct unit \(\text{W}\) (or \(\text{J/s}\)).

(ii) [2 marks total]:
- [1 mark] for explaining that if one element fails/breaks, the circuit is not fully broken.
- [1 mark] for stating that the remaining element continues to function independently.
題目 5 · structured theory
8.8
The rate of reaction between solid calcium carbonate and dilute hydrochloric acid can be investigated by measuring the volume of carbon dioxide gas produced over time.

\(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\)

(a) Define the term activation energy. [1.8 marks]

(b) Explain, using collision theory, why increasing the temperature of the hydrochloric acid increases the rate of this chemical reaction. [4 marks]

(c) Calculate the mass of calcium carbonate required to produce \(120\text{ cm}^3\) of carbon dioxide gas at room temperature and pressure (r.t.p.).
[The molar volume of any gas at r.t.p. is \(24\text{ dm}^3/\text{mol}\). Relative atomic masses: \(A_r: \text{Ca} = 40, \text{C} = 12, \text{O} = 16\).] [3 marks]
查看答案詳解

解題

(a) Activation energy is defined as the minimum amount of energy colliding reactant particles must have in order to react successfully.

(b) Increasing the temperature increases the average kinetic energy of the reacting particles. This causes them to move faster and collide more frequently (increased frequency of collisions). Furthermore, a significantly higher proportion of colliding particles now possess energy greater than or equal to the activation energy. This leads to a higher rate of successful (or effective) collisions.

(c)
1. Convert the volume of \(\text{CO}_2\) to \(\text{dm}^3\):
\(120\text{ cm}^3 = 0.12\text{ dm}^3\).

2. Calculate the moles of \(\text{CO}_2\) produced:
\(\text{moles of CO}_2 = \frac{0.12\text{ dm}^3}{24\text{ dm}^3/\text{mol}} = 0.005\text{ mol}\).

3. Determine the moles of \(\text{CaCO}_3\) required using the 1:1 stoichiometric ratio:
\(\text{moles of CaCO}_3 = 0.005\text{ mol}\).

4. Calculate the relative formula mass (\(M_r\)) of \(\text{CaCO}_3\):
\(M_r(\text{CaCO}_3) = 40 + 12 + (3 \times 16) = 100\).

5. Calculate the mass of \(\text{CaCO}_3\):
\(\text{mass} = 0.005\text{ mol} \times 100\text{ g/mol} = 0.5\text{ g}\).

評分準則

(a) [1.8 marks total]:
- [1 mark] for 'minimum energy' required.
- [0.8 marks] for stating it is needed for 'particles to react/collide successfully'.

(b) [4 marks total]:
- [1 mark] for stating particles gain kinetic energy / move faster.
- [1 mark] for stating collision frequency increases / more frequent collisions.
- [1 mark] for stating more particles have energy greater than/equal to activation energy.
- [1 mark] for stating the rate of successful/effective collisions increases.

(c) [3 marks total]:
- [1 mark] for calculating the moles of carbon dioxide gas (\(0.005\text{ mol}\)).
- [1 mark] for calculating \(M_r\) of \(\text{CaCO}_3\) as \(100\).
- [1 mark] for the correct final answer of \(0.5\text{ g}\) (allow ecf from incorrect moles).
題目 6 · structured theory
8.8
Plants produce glucose using light energy absorbed by pigments during photosynthesis.

(a) Write the balanced chemical equation for photosynthesis. Include the essential conditions required above and below the reaction arrow. [3 marks]

(b) Describe and explain how the cellular structure and organization of the palisade mesophyll layer of a leaf is adapted to maximize the rate of photosynthesis. [3.8 marks]

(c) Explain the role of chlorophyll in photosynthesis in terms of energy transfer. [2 marks]
查看答案詳解

解題

(a) The balanced chemical equation is:
\(6\text{CO}_2 + 6\text{H}_2\text{O} \xrightarrow[\text{chlorophyll}]{\text{light}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)

(b) The adaptations of the palisade mesophyll layer include:
1. The cells are packed closely together near the upper surface of the leaf to absorb maximum light energy.
2. The cells contain a very high density of chloroplasts to maximize the rate of light absorption for photosynthesis.
3. The cells are columnar/elongated, allowing light to penetrate down through the tissue layer.
4. A large central vacuole pushes the chloroplasts to the outer edges of the cell, optimizing gas diffusion (carbon dioxide absorption) and light capture.

(c) Chlorophyll functions to trap or absorb light energy from the Sun and convert/transfer it into chemical energy, which is then used to synthesize glucose from carbon dioxide and water.

評分準則

(a) [3 marks total]:
- [1 mark] for correct chemical formulae of reactants (\(\text{CO}_2\) and \(\text{H}_2\text{O}\)) and products (\(\text{C}_6\text{H}_{12}\text{O}_6\) and \(\text{O}_2\)).
- [1 mark] for correct balancing: \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\).
- [1 mark] for correctly showing 'light' and 'chlorophyll' on the reaction arrow.

(b) [3.8 marks total]: Award up to 3.8 marks for description and matching explanation:
- [1.8 marks] for stating they have a high concentration of chloroplasts / tightly packed chloroplasts to absorb maximum sunlight.
- [1 mark] for describing their position near the top of the leaf to receive direct light.
- [1 mark] for describing the columnar shape / closely packed arrangement to optimize space and light utilization.

(c) [2 marks total]:
- [1 mark] for stating that chlorophyll absorbs/traps light energy.
- [1 mark] for stating that it converts/transfers this light energy to chemical energy (used to make glucose/carbohydrates).
題目 7 · structured theory
8.8
A student investigates the trends in Group 1 and Group 7 of the Periodic Table.

(a) Rubidium is a Group 1 alkali metal.
(i) State and explain the trend in reactivity of Group 1 alkali metals as the proton number increases. [3]
(ii) Predict the chemical equation for the reaction of rubidium with water. Include state symbols. [2]

(b) Chlorine is a Group 7 halogen.
(i) State the colour and physical state of chlorine at room temperature and pressure. [1]
(ii) Describe what is observed when chlorine gas is bubbled into a solution of potassium bromide. Explain this reaction in terms of reactivity. [3]
查看答案詳解

解題

(a)(i) As proton number increases down Group 1, the reactivity of the metals increases. This is because the atomic radius increases (the outer electron is further from the positive nucleus) and there is more shielding from inner electron shells. Consequently, the electrostatic attraction between the nucleus and the outer shell electron becomes weaker, making it easier for the atom to lose its single outer electron.
(ii) The balanced chemical equation is:
\(2\text{Rb(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{RbOH(aq)} + \text{H}_2\text{(g)}\)

(b)(i) Chlorine is a pale green (or yellow-green) gas at room temperature and pressure.
(ii) When chlorine gas is bubbled into potassium bromide, the colourless solution turns orange/yellow-brown. This observation occurs because chlorine is more reactive than bromine. Chlorine displaces bromine from the potassium bromide solution, forming potassium chloride and releasing elemental bromine (which dissolves to give the orange colour): \(\text{Cl}_2\text{(g)} + 2\text{KBr(aq)} \rightarrow 2\text{KCl(aq)} + \text{Br}_2\text{(aq)}\).

評分準則

(a)(i) Max 3 marks:
- Reactivity increases (down the group / with increasing proton number) [1]
- Outer shell electron is further from the nucleus / more electron shells / more shielding [1]
- Electrostatic attraction between nucleus and outer electron is weaker OR outer electron is lost more easily [1]

(a)(ii) Max 2 marks:
- Correct chemical formulae and balancing: \(2\text{Rb} + 2\text{H}_2\text{O} \rightarrow 2\text{RbOH} + \text{H}_2\) [1]
- Correct state symbols: \(\text{Rb(s)}\), \(\text{H}_2\text{O(l)}\), \(\text{RbOH(aq)}\), \(\text{H}_2\text{(g)}\) [1] (Accept state symbol mark only if formulas of reactants and products are substantially correct)

(b)(i) Max 1 mark:
- Pale green / yellow-green AND gas [1]

(b)(ii) Max 3 marks:
- Solution turns orange / yellow-brown [1] (Reject: 'red' or 'brown' alone, accept 'orange-brown')
- Chlorine is more reactive than bromine [1]
- Chlorine displaces bromine (from potassium bromide) OR chlorine oxidises bromide ions [1]
題目 8 · structured theory
8.8
A marine survey vessel uses ultrasound to map the ocean floor.

(a) Ultrasound waves are longitudinal sound waves.
(i) Describe how a longitudinal wave transfers energy, referring to the movement of particles in the medium relative to the direction of wave travel. [2]
(ii) Explain the terms compression and rarefaction in relation to a longitudinal wave. [2]

(b) The vessel emits a pulse of ultrasound with a frequency of \(4.0 \times 10^4\text{ Hz}\). The wave travels through seawater at a speed of \(1500\text{ m/s}\).
(i) Calculate the wavelength of this ultrasound wave in seawater. State the formula used and show your working. [2.5]
(ii) The reflection of the ultrasound pulse from the seabed is detected \(1.2\text{ s}\) after transmission. Calculate the depth of the sea at this point. Show your working. [2.5]
查看答案詳解

解題

(a)(i) In a longitudinal wave, the particles of the medium vibrate (oscillate) back and forth parallel to (along the same direction as) the direction in which the wave is travelling or transferring energy.
(ii) A compression is a region in a longitudinal wave where the particles are closest together (resulting in high pressure). A rarefaction is a region where the particles are furthest apart (resulting in low pressure).

(b)(i) State the formula:
\(v = f \lambda\) or \(\lambda = \frac{v}{f}\)
Substitute the values:
\(\lambda = \frac{1500\text{ m/s}}{4.0 \times 10^4\text{ Hz}}\)
\(\lambda = 0.0375\text{ m}\) (or \(3.75 \times 10^{-2}\text{ m}\))

(b)(ii) The total distance travelled by the pulse is down to the seabed and back up to the ship:
\(\text{Total distance} = v \times t = 1500\text{ m/s} \times 1.2\text{ s} = 1800\text{ m}\)
Since this is the two-way distance, the depth is half of the total distance:
\(\text{Depth} = \frac{1800\text{ m}}{2} = 900\text{ m}\)

評分準則

(a)(i) Max 2 marks:
- Particles vibrate / oscillate [1]
- Vibration is parallel to the direction of energy transfer / wave propagation [1]

(a)(ii) Max 2 marks:
- Compression: region where particles are close together / high pressure [1]
- Rarefaction: region where particles are spread apart / low pressure [1]

(b)(i) Max 2.5 marks:
- Correct wave equation: \(v = f \lambda\) or rearranged [0.5]
- Substitution: \(\lambda = \frac{1500}{40000}\) [1]
- Evaluation and unit: \(0.0375\text{ m}\) (or \(3.75 \times 10^{-2}\text{ m}\)) [1] (Accept \(3.75\text{ cm}\))

(b)(ii) Max 2.5 marks:
- Correct formula for distance: \(\text{distance} = v \times t\) OR halving the time (\(t = 0.6\text{ s}\)) [0.5]
- Calculation: \(1500 \times 1.2 = 1800\text{ m}\) or \(1500 \times 0.6\) [1]
- Correct division by 2 to find depth: \(900\text{ m}\) with correct unit [1]
題目 9 · structured theory
8.8
Salivary amylase is an enzyme that catalyses the hydrolysis of starch into maltose.

(a) Outline the 'lock-and-key' hypothesis to explain how enzymes catalyse specific reactions. Refer to 'active site' and 'substrate' in your answer. [3]

(b) An investigation was carried out to study the effect of pH on the rate of starch breakdown by salivary amylase.
(i) State the reagent used to test for the presence of starch, and describe the colour change observed if starch is present. [2]
(ii) The reaction rate was found to be zero at pH 2. Explain why the amylase enzyme does not function at this pH. Use the terms denatured, active site, and complementary in your explanation. [4]
查看答案詳解

解題

(a) According to the lock-and-key hypothesis, an enzyme has a specifically-shaped region called the active site. The substrate molecule has a shape that is exactly complementary to this active site. The substrate (the 'key') fits perfectly into the active site (the 'lock') of the enzyme to form an enzyme-substrate complex. The reaction then takes place, and products are released from the unchanged active site.

(b)(i) The reagent used to test for starch is iodine solution (or iodine in potassium iodide solution). If starch is present, the iodine solution changes colour from orange/yellow-brown to blue-black.

(b)(ii) A pH of 2 is highly acidic and is far below the optimum pH of salivary amylase (which is near neutral, pH 7). Extremely low pH values disrupt the ionic and hydrogen bonds that hold the enzyme's three-dimensional protein structure together. This causes the enzyme to become denatured. As a result, the specific shape of the active site is permanently altered, and it is no longer complementary to the shape of the starch substrate. Therefore, the substrate can no longer fit or bind to the active site, and the enzymatic reaction cannot occur.

評分準則

(a) Max 3 marks:
- Enzyme has an active site with a specific shape [1]
- Substrate shape is complementary to the active site [1]
- Substrate fits / binds into the active site (to form an enzyme-substrate complex) [1]

(b)(i) Max 2 marks:
- Iodine (solution / in potassium iodide) [1]
- Colour change from orange-brown/yellow-brown to blue-black [1]

(b)(ii) Max 4 marks:
- pH 2 is too acidic / far from optimum [1]
- Enzyme is denatured [1]
- Shape of active site is altered / destroyed [1]
- Active site is no longer complementary to substrate OR substrate can no longer bind / fit [1]

Paper 6 Alternative to Practical

Answer all written practical tasks, graph plotting, and one experimental planning scenario.
4 題目 · 40
題目 1 · experimental report
10
### Investigating the Effect of pH on Amylase Activity

A student investigates how pH affects the rate of starch breakdown by the enzyme amylase.

The student sets up four test tubes, each containing a mixture of starch and amylase solutions maintained at a specific pH using buffer solutions: pH 5.0, pH 6.0, pH 7.0, and pH 8.0.

Every 30 seconds, a drop of the reaction mixture is removed and added to one drop of iodine solution on a spotting tile. The time taken for the starch to be completely digested (when the iodine solution remains orange-brown instead of turning blue-black) is recorded.

Table 1.1 shows the colour observed at each time interval.

**Table 1.1**

| Time / s | pH 5.0 colour | pH 6.0 colour | pH 7.0 colour | pH 8.0 colour |
| :--- | :--- | :--- | :--- | :--- |
| 30 | blue-black | blue-black | blue-black | blue-black |
| 60 | blue-black | blue-black | orange-brown | blue-black |
| 90 | blue-black | orange-brown | orange-brown | blue-black |
| 120 | blue-black | orange-brown | orange-brown | blue-black |
| 150 | blue-black | orange-brown | orange-brown | blue-black |
| 180 | orange-brown | orange-brown | orange-brown | blue-black |

(a) (i) Using the data in Table 1.1, state the time taken for starch to be completely digested at:
* pH 6.0
* pH 7.0 [2]

(ii) Identify the independent variable in this investigation. [1]

(iii) Name two variables that must be kept constant to ensure a fair test. [2]

(iv) Before mixing, the amylase and starch solutions were incubated in a water bath at 35 °C for 5 minutes. Explain why this pre-incubation step is important. [1]

(v) Describe the chemical test used to confirm that reducing sugars are produced by the breakdown of starch, including the reagent used, the required reaction condition, and the positive result observed. [3]

(vi) Suggest how the student could modify the experiment to find a more precise value for the optimum pH. [1]
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解題

At pH 6.0, the last blue-black reading is at 60 s, and the first orange-brown (completed digestion) is at 90 s. At pH 7.0, the first orange-brown reading is at 60 s. The independent variable is the one changed by the investigator (pH). Control variables must remain constant to prevent them from affecting the rate of reaction. Pre-incubation allows solutions to equilibrate to 35 °C. Benedict's reagent requires heating to yield a positive result (green/yellow/orange/red precipitate). Testing closer intervals of pH provides better resolution for the peak of enzyme activity.

評分準則

(a)(i) 90 s AND 60 s [2 marks - 1 mark for each correct time value]
(a)(ii) pH [1 mark]
(a)(iii) Any two from: volume of starch, concentration of starch, volume of amylase, concentration of amylase, temperature [2 marks]
(a)(iv) To allow both reactant solutions to reach the target temperature (35 °C) before mixing, ensuring a constant temperature from the start [1 mark]
(a)(v) Add Benedict's reagent [1 mark], heat in a hot water bath (>80 °C) [1 mark], colour change from blue to green/yellow/orange/red [1 mark]
(a)(vi) Repeat using smaller pH intervals (e.g., pH 5.5, 6.5, 7.5) [1 mark]
題目 2 · experimental report
10
### Identification of Ions in Salt X

A student is provided with a green crystalline solid, salt X. They dissolve the solid in distilled water to make an aqueous solution, solution X, and perform qualitative tests to identify the cation and anion present.

**Test 1:** To a 2 cm³ portion of solution X, aqueous sodium hydroxide is added dropwise until in excess.
* **Observation:** A green precipitate is formed, which is insoluble in excess.

**Test 2:** To a 2 cm³ portion of solution X, aqueous ammonia is added dropwise until in excess.
* **Observation:** A green precipitate is formed, which is insoluble in excess.

**Test 3:** To a 2 cm³ portion of solution X, dilute nitric acid is added, followed by a few drops of aqueous barium nitrate.
* **Observation:** A white precipitate is formed.

(a) (i) Identify the cation present in salt X. [1]

(ii) Identify the anion present in salt X. [1]

(iii) Write the ionic equation, including state symbols, for the reaction occurring in Test 3. [2]

(b) The student wants to obtain pure, dry crystals of salt X from solution X. Describe the experimental steps of crystallization the student should perform. [4]

(c) State one hazard and one corresponding safety precaution for performing Test 3. [2]
查看答案詳解

解題

Iron(II) forms a green precipitate with both NaOH and ammonia, which is insoluble in excess of both reagents. Barium ions react with sulfate ions to form an insoluble white precipitate of barium sulfate. Crystallization is a standard separation technique involving heating, cooling, filtration, and drying. Nitric acid is highly corrosive, necessitating eye and skin protection.

評分準則

(a)(i) Iron(II) / \(Fe^{2+}\) [1 mark]
(a)(ii) Sulfate / \(SO_4^{2-}\) [1 mark]
(a)(iii) \(Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)\) [2 marks - 1 mark for correct formulae of reactants and products, 1 mark for correct state symbols]
(b) Heat the solution to evaporate water until saturation/crystallization point is reached [1 mark], allow to cool to form crystals [1 mark], filter to separate crystals [1 mark], dry crystals with filter paper / in warm oven [1 mark]
(c) Hazard: Nitric acid is corrosive / irritant [1 mark]; Precaution: Wear safety goggles / gloves [1 mark]
題目 3 · experimental report
10
### Investigating Thermal Insulators

A student investigates the cooling rates of hot water in three beakers under different conditions to determine which material is the best thermal insulator.

* **Beaker A:** Uninsulated beaker (control)
* **Beaker B:** Wrapped in one layer of bubble wrap
* **Beaker C:** Wrapped in one layer of cotton wool

The student pours 150 cm³ of hot water into each beaker, inserts a thermometer, and records the temperature every 60 seconds for 5 minutes.

Table 3.1 shows the temperature readings obtained.

**Table 3.1**

| Time / s | Beaker A Temperature / °C | Beaker B Temperature / °C | Beaker C Temperature / °C |
| :--- | :--- | :--- | :--- |
| 0 | 85.0 | 85.0 | 85.0 |
| 60 | 78.5 | 81.5 | 80.0 |
| 120 | 73.0 | 78.5 | 76.0 |
| 180 | 68.5 | 76.0 | 72.5 |
| 240 | 64.5 | 74.0 | 69.5 |
| 300 | 61.0 | 72.0 | 67.0 |

(a) (i) Suggest why the student waits for 30 seconds after pouring the hot water before taking the first reading at 0 s. [1]

(ii) State one precaution the student should take when reading the thermometer to ensure the measurement is accurate. [1]

(b) (i) Calculate the temperature drop (\(\Delta T\)) for the water in each beaker over the 300-second period.
* Beaker A: ____ °C
* Beaker B: ____ °C
* Beaker C: ____ °C [3]

(ii) Identify which material is the best thermal insulator. Justify your answer using the calculated temperature drops. [2]

(c) State two variables that must be kept constant to ensure a fair test when comparing Beaker B and Beaker C. [2]

(d) Suggest one modification to the apparatus that would reduce heat loss from the top of the beakers. [1]
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解題

(a)(i) Waiting ensures that the thermometer reading has stabilized and reflects the actual temperature of the water. (a)(ii) Viewing eye-level prevents parallax error. (b)(i) Temperature drop is found by subtracting the final temperature at 300 s from the initial temperature at 0 s. Beaker A: 85.0 - 61.0 = 24.0 °C; Beaker B: 85.0 - 72.0 = 13.0 °C; Beaker C: 85.0 - 67.0 = 18.0 °C. (b)(ii) The material with the lowest temperature change is the best insulator, which is bubble wrap. (c) Fair-testing requires control of variables like volume, starting temperature, and environmental conditions. (d) A lid acts as a barrier to evaporation and convection, reducing thermal losses from the top surface.

評分準則

(a)(i) To allow the thermometer bulb to reach the temperature of the water / stabilize [1 mark]
(a)(ii) Read the scale at eye level (to avoid parallax error) OR do not let the thermometer bulb touch the glass beaker [1 mark]
(b)(i) Beaker A: 24.0 °C [1 mark], Beaker B: 13.0 °C [1 mark], Beaker C: 18.0 °C [1 mark]
(b)(ii) Bubble wrap / Beaker B [1 mark] AND because it has the lowest/smallest temperature drop over 300 s [1 mark]
(c) Any two from: Volume of water, initial water temperature, same type/size of beaker, room temperature [2 marks]
(d) Place a lid/cover over the beaker [1 mark]
題目 4 · experimental report
10
A student investigates the effect of temperature on the rate of the breakdown of starch by the enzyme amylase.

The student mixes amylase and starch solutions together in a test-tube. At 30-second intervals, they remove a sample of the mixture and add it to a drop of iodine solution in a spotting tile.

(a) A thermometer is used to measure the temperature of the water bath. The major scale divisions are every \(10\ ^\circ\text{C}\) and the minor scale divisions are every \(1\ ^\circ\text{C}\). The top of the liquid column is exactly halfway between the \(37\ ^\circ\text{C}\) and \(38\ ^\circ\text{C}\) marks. State this temperature.

(b) State the color of the iodine solution when the starch has been completely broken down.

(c) The student records the following observations from the spotting tiles:
- At \(20\ ^\circ\text{C}\): the mixture turns blue-black at 0.5, 1.0, 1.5, and 2.0 minutes. It remains orange-brown at 2.5 minutes and all subsequent times.
- At \(40\ ^\circ\text{C}\): the mixture turns blue-black at 0.5 minutes. It remains orange-brown at 1.0 minute and all subsequent times.

State the time taken for starch to be completely broken down at \(20\ ^\circ\text{C}\) and at \(40\ ^\circ\text{C}\).

(d) Explain why the student used a thermostatically controlled water bath rather than heating the test-tubes directly with a Bunsen burner flame.

(e) State two variables, other than temperature, that must be kept constant to ensure a fair test.

(f) The student concludes that the optimum temperature for this amylase is \(40\ ^\circ\text{C}\). Suggest how the student could improve their experimental method to obtain a more precise value for the optimum temperature.

(g) Describe how the student can test the amylase solution to confirm that it does not contain any reducing sugars before starting the experiment. State the reagent used and the expected observation for a negative result.
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解題

(a) Since the liquid level is exactly halfway between \(37\ ^\circ\text{C}\) and \(38\ ^\circ\text{C}\), the reading is \(37.5\ ^\circ\text{C}\).

(b) In the absence of starch, iodine solution remains its natural orange-brown color.

(c) The time taken is when the blue-black color first fails to appear. At \(20\ ^\circ\text{C}\), this is at 2.5 minutes. At \(40\ ^\circ\text{C}\), this is at 1.0 minute.

(d) A water bath distributes thermal energy evenly and allows the temperature to be controlled and kept constant, whereas a Bunsen burner flame would cause rapid, uneven temperature rise and denature the enzyme.

(e) Controlling other variables such as volumes and concentrations of starch/amylase solutions ensures that any change in the rate of reaction is solely due to the change in temperature.

(f) To find a more precise optimum, testing at narrower intervals around the estimated optimum (e.g., \(35\ ^\circ\text{C}\), \(38\ ^\circ\text{C}\), \(42\ ^\circ\text{C}\), etc.) is required.

(g) Benedict's test is used to test for reducing sugars. Heating is required. A negative result (no reducing sugars present) is indicated by the reagent remaining blue.

評分準則

(a) [1 mark]
- 37.5 (°C)

(b) [1 mark]
- orange-brown / brown / yellow-brown (reject blue-black)

(c) [2 marks]
- At 20 °C: 2.5 minutes / 150 seconds [1]
- At 40 °C: 1.0 minute / 60 seconds [1]

(d) [1 mark]
- To maintain a constant / stable temperature OR to prevent temperature rising too high / denaturing enzyme

(e) [2 marks]
- Any two from: volume of starch solution, volume of amylase solution, concentration of starch solution, concentration of amylase solution, pH [1 mark each]

(f) [1 mark]
- Test at smaller / narrower temperature intervals (between 30 °C and 50 °C)

(g) [2 marks]
- Add Benedict's reagent and heat / warm in a water bath [1]
- (Observation is) stays blue / no color change [1]

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