Cambridge IGCSE · thinka 原创模拟试题

2023 Cambridge IGCSE Science - Combined (0653) 模拟试题及答案详解

Thinka Nov 2023 (V1) Cambridge IGCSE-Style Mock — Science - Combined (0653)

200 165 分钟2023
An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V1) Cambridge IGCSE Science - Combined (0653) paper. Not affiliated with or reproduced from Cambridge.

卷一 / 卷二: 選擇題

Answer all 40 multiple choice questions. Choose one correct option out of four (A, B, C, D).
40 题目 · 40
题目 1 · 選擇題
1
An amylase enzyme is found in human saliva. At which temperature is the rate of starch breakdown catalyzed by this enzyme the highest?
  1. A.10 °C
  2. B.37 °C
  3. C.60 °C
  4. D.90 °C
查看答案详解

解题

The human body temperature is approximately 37 °C, which is the optimum temperature for human enzymes. At higher temperatures (60 °C and 90 °C), the enzyme is denatured. At lower temperatures (10 °C), the kinetic energy of the molecules is low, resulting in a slower rate of reaction.

评分标准

Award 1 mark for the correct option B.
题目 2 · 選擇題
1
Which cell structure is present in a plant cell but absent in an animal cell?
  1. A.cell membrane
  2. B.cytoplasm
  3. C.nucleus
  4. D.cell wall
查看答案详解

解题

Both plant and animal cells contain a cell membrane, cytoplasm, and a nucleus. However, only plant cells possess a cellulose cell wall.

评分标准

Award 1 mark for the correct option D.
题目 3 · 選擇題
1
A substance changes state from solid to liquid. Describe the changes in the arrangement and movement of the particles.
  1. A.The particles change from a random arrangement to a regular lattice, and their movement stops.
  2. B.The particles change from a regular lattice to a random arrangement, and they change from vibrating about fixed positions to sliding past each other.
  3. C.The particles remain in a regular lattice but move further apart and travel at high speeds in all directions.
  4. D.The particles change from sliding past each other to vibrating about fixed positions.
查看答案详解

解题

In a solid, particles are arranged in a regular lattice and vibrate about fixed positions. When it melts to a liquid, the particles lose this regular lattice, become randomly arranged, and are able to move and slide past each other.

评分标准

Award 1 mark for the correct option B.
题目 4 · 選擇題
1
Which statement describes a chemical property of Group I alkali metals?
  1. A.They react with water to form hydrogen gas and an alkaline solution.
  2. B.They are very unreactive and do not form compounds.
  3. C.They form colored compounds and act as catalysts.
  4. D.Their reactivity decreases as you go down the group.
查看答案详解

解题

Group I alkali metals are highly reactive and react vigorously with water to produce hydrogen gas and a metal hydroxide, which forms an alkaline solution. Reactivity increases down the group, and they are not transition metals (which form colored compounds).

评分标准

Award 1 mark for the correct option A.
题目 5 · 選擇題
1
Dilute hydrochloric acid is added to a test-tube containing solid copper(II) carbonate. Which observations are made?
  1. A.A colorless gas is produced, and a blue solution is formed.
  2. B.A brown gas is produced, and a colorless solution is formed.
  3. C.No reaction takes place, and the solid remains unchanged.
  4. D.Squeaky pops are heard, and a green precipitate is formed.
查看答案详解

解题

Acids react with metal carbonates to produce a salt, carbon dioxide gas, and water. Copper(II) carbonate reacts to produce carbon dioxide (a colorless gas) and copper(II) chloride, which dissolves to form a blue solution.

评分标准

Award 1 mark for the correct option A.
题目 6 · 選擇題
1
A runner travels a distance of 1200 m in a time of 5.0 minutes. What is the average speed of the runner?
  1. A.0.25 m/s
  2. B.4.0 m/s
  3. C.240 m/s
  4. D.6000 m/s
查看答案详解

解题

First, convert minutes to seconds: 5.0 minutes = 5.0 * 60 = 300 seconds. Then, calculate average speed: Speed = Distance / Time = 1200 m / 300 s = 4.0 m/s.

评分标准

Award 1 mark for the correct option B.
题目 7 · 選擇題
1
A sound wave is described as having a high pitch and being quiet. How do the frequency and amplitude of this sound wave compare to a loud sound wave with a low pitch?
  1. A.higher frequency and larger amplitude
  2. B.higher frequency and smaller amplitude
  3. C.lower frequency and larger amplitude
  4. D.lower frequency and smaller amplitude
查看答案详解

解题

Pitch is determined by frequency; a higher pitch means a higher frequency. Loudness is determined by amplitude; a quieter sound has a smaller amplitude. Therefore, a high-pitched, quiet sound has a higher frequency and smaller amplitude compared to a low-pitched, loud sound.

评分标准

Award 1 mark for the correct option B.
题目 8 · 選擇題
1
A student wants to measure the current flowing through a lamp and the potential difference (p.d.) across it. How must the ammeter and voltmeter be connected relative to the lamp?
  1. A.ammeter in parallel, voltmeter in series
  2. B.ammeter in parallel, voltmeter in parallel
  3. C.ammeter in series, voltmeter in series
  4. D.ammeter in series, voltmeter in parallel
查看答案详解

解题

To measure current, an ammeter must be connected in series with the component. To measure potential difference, a voltmeter must be connected in parallel across the component.

评分标准

Award 1 mark for the correct option D.
题目 9 · 選擇題
1
An organism detects a change in its environment and responds to it. Which characteristic of living organisms is described by this action?
  1. A.excretion
  2. B.growth
  3. C.reproduction
  4. D.sensitivityunit testing in Python is done with unittest.
查看答案详解

解题

Sensitivity is defined as the ability to detect and respond to changes in the environment.

评分标准

Award 1 mark for the correct answer D.
题目 10 · 選擇題
1
What is the net movement of water molecules from a region of higher water potential to a region of lower water potential through a partially permeable membrane?
  1. A.active transport
  2. B.diffusion
  3. C.osmosis
  4. D.transpiration
查看答案详解

解题

Osmosis is specifically the net movement of water molecules down a water potential gradient through a partially permeable membrane.

评分标准

Award 1 mark for the correct answer C.
题目 11 · 選擇題
1
An enzyme works best at a pH of 2. In which part of the human digestive system is this enzyme most likely to be active?
  1. A.large intestine
  2. B.mouth
  3. C.small intestine
  4. D.stomach
查看答案详解

解题

The stomach contains hydrochloric acid, making its environment highly acidic (around pH 2), which is the optimum pH for stomach enzymes like pepsin.

评分标准

Award 1 mark for the correct answer D.
题目 12 · 選擇題
1
An atom of sodium has a proton number of 11 and a nucleon number of 23. How many neutrons and electrons are there in a neutral sodium atom?
  1. A.11 neutrons and 12 electrons
  2. B.12 neutrons and 11 electrons
  3. C.12 neutrons and 12 electrons
  4. D.23 neutrons and 11 electrons
查看答案详解

解题

Number of neutrons = nucleon number - proton number = 23 - 11 = 12. In a neutral atom, the number of electrons is equal to the proton number, which is 11.

评分标准

Award 1 mark for the correct answer B.
题目 13 · 選擇題
1
Copper(II) oxide reacts with hydrogen gas to form copper and water: \( CuO + H_2 \rightarrow Cu + H_2O \). Which statement about this reaction is correct?
  1. A.Copper(II) oxide is oxidised because it loses oxygen.
  2. B.Copper(II) oxide is reduced because it loses oxygen.
  3. C.Hydrogen is oxidised because it loses oxygen.
  4. D.Hydrogen is reduced because it gains oxygen.
查看答案详解

解题

Copper(II) oxide loses oxygen to form copper, which is the process of reduction. Therefore, copper(II) oxide is reduced.

评分标准

Award 1 mark for the correct answer B.
题目 14 · 選擇題
1
Why is copper used to make electrical wiring?
  1. A.It is a good conductor of electricity and has a low density.
  2. B.It is a good conductor of electricity and is ductile.
  3. C.It is a poor conductor of heat and has a high melting point.
  4. D.It is unreactive with water and has a high density.
查看答案详解

解题

Copper is used for electrical wiring because it is a very good electrical conductor and is ductile, meaning it can be easily drawn into wires.

评分标准

Award 1 mark for the correct answer B.
题目 15 · 選擇題
1
A metal block has a mass of 400 g and a volume of 50 \( \text{cm}^3 \). What is the density of the metal?
  1. A.0.125 \( \text{g/cm}^3 \)
  2. B.8.0 \( \text{g/cm}^3 \)
  3. C.450 \( \text{g/cm}^3 \)
  4. D.20000 \( \text{g/cm}^3 \)
查看答案详解

解题

Density is calculated using the formula: density = mass / volume. Here, density = 400 g / 50 \( \text{cm}^3 \) = 8.0 \( \text{g/cm}^3 \).

评分标准

Award 1 mark for the correct answer B.
题目 16 · 選擇題
1
A current of 3.0 A flows through a lamp for 4.0 minutes. How much charge passes through the lamp?
  1. A.0.75 C
  2. B.12 C
  3. C.180 C
  4. D.720 C
查看答案详解

解题

Charge (Q) is calculated using the formula: Q = I * t. Time must be in seconds, so t = 4.0 minutes = 240 seconds. Q = 3.0 A * 240 s = 720 C.

评分标准

Award 1 mark for the correct answer D.
题目 17 · 選擇題
1
An enzyme-catalysed reaction is investigated at different temperatures. The rate of reaction is found to be highest at \(40\text{ }^{\circ}\text{C}\), but drops to zero at \(70\text{ }^{\circ}\text{C}\). What is the explanation for the rate of reaction dropping to zero at \(70\text{ }^{\circ}\text{C}\)?
  1. A.The enzyme molecules have run out of energy.
  2. B.The enzyme molecules have been denatured.
  3. C.The substrate molecules have been completely used up.
  4. D.The activation energy of the reaction has decreased.
查看答案详解

解题

At high temperatures, such as \(70\text{ }^{\circ}\text{C}\), the shape of the enzyme's active site is permanently altered. This process is called denaturation. The substrate can no longer fit into the active site, and the reaction stops.

评分标准

Award 1 mark for selecting the option stating that the enzyme molecules have been denatured (B).
题目 18 · 選擇題
1
A green plant is kept in a sealed glass container with plenty of water. It is exposed to bright sunlight for several hours. Which gas will increase in concentration inside the container?
  1. A.carbon dioxide
  2. B.nitrogen
  3. C.noble gases
  4. D.oxygen
查看答案详解

解题

In bright sunlight, the rate of photosynthesis in the plant is greater than the rate of respiration. Photosynthesis produces oxygen gas, so the concentration of oxygen inside the sealed container will increase.

评分标准

Award 1 mark for selecting oxygen (D).
题目 19 · 選擇題
1
Which statement correctly describes the arrangement and movement of particles in a liquid?
  1. A.They are closely packed in a regular arrangement and vibrate about fixed positions.
  2. B.They are close together in an irregular arrangement and are free to move past one another.
  3. C.They are far apart in a random arrangement and move rapidly in all directions.
  4. D.They are far apart in a regular arrangement and vibrate about fixed positions.
查看答案详解

解题

In a liquid, particles are close together but not in a regular pattern, and they are free to move and slide past one another. Option A describes a solid, and option C describes a gas.

评分标准

Award 1 mark for selecting the correct description for a liquid (B).
题目 20 · 選擇題
1
A student reacts marble chips (calcium carbonate) with dilute hydrochloric acid. Which change decreases the rate of this reaction?
  1. A.decreasing the concentration of the acid
  2. B.increasing the temperature of the acid
  3. C.using powdered marble chips instead of large chips
  4. D.stirring the reaction mixture continuously
查看答案详解

解题

Decreasing the concentration of the acid means there are fewer reacting particles per unit volume, which decreases the frequency of successful collisions and thus decreases the rate of reaction.

评分标准

Award 1 mark for selecting the option that decreases the concentration of the acid (A).
题目 21 · 選擇題
1
An aqueous solution turns universal indicator paper red. What is the pH of this solution?
  1. A.2
  2. B.7
  3. C.9
  4. D.14
查看答案详解

解题

Strongly acidic solutions turn universal indicator red. Strongly acidic solutions have a very low pH, typically around 1 or 2.

评分标准

Award 1 mark for selecting pH 2 (A).
题目 22 · 選擇題
1
An object of mass \(15\text{ kg}\) is on the surface of the Earth, where the gravitational field strength \(g\) is \(10\text{ N/kg}\). What is the weight of the object?
  1. A.\(1.5\text{ N}\)
  2. B.\(15\text{ N}\)
  3. C.\(150\text{ N}\)
  4. D.\(1500\text{ N}\)
查看答案详解

解题

Weight is calculated using the formula \(W = m \times g\). Here, \(W = 15\text{ kg} \times 10\text{ N/kg} = 150\text{ N}\).

评分标准

Award 1 mark for selecting 150 N (C).
题目 23 · 選擇題
1
Which type of wave is a longitudinal wave?
  1. A.infrared wave
  2. B.radio wave
  3. C.sound wave
  4. D.ultraviolet wave
查看答案详解

解题

Sound waves are longitudinal waves because the vibrations of the particles are parallel to the direction of wave travel. Infrared, radio, and ultraviolet waves are electromagnetic waves, which are transverse.

评分标准

Award 1 mark for selecting sound wave (C).
题目 24 · 選擇題
1
A lamp is connected to a \(12\text{ V}\) power supply. If the current through the lamp is \(3.0\text{ A}\), what is the electrical resistance of the lamp?
  1. A.\(0.25\text{ }\Omega\)
  2. B.\(4.0\text{ }\Omega\)
  3. C.\(15\text{ }\Omega\)
  4. D.\(36\text{ }\Omega\)
查看答案详解

解题

Resistance is calculated using the formula \(R = \frac{V}{I}\). Here, \(R = \frac{12\text{ V}}{3.0\text{ A}} = 4.0\text{ }\Omega\).

评分标准

Award 1 mark for selecting 4.0 \(\Omega\) (B).
题目 25 · 選擇題
1
A student sets up an aquatic plant in water under a bright light. Gas bubbles are observed being released from the plant. Which gas is being produced, and which process is responsible?
  1. A.carbon dioxide by respiration
  2. B.carbon dioxide by photosynthesis
  3. C.oxygen by respiration
  4. D.oxygen by photosynthesis
查看答案详解

解题

In bright light, photosynthesis produces oxygen gas at a rate faster than respiration consumes it, so oxygen bubbles are released from the plant.

评分标准

Award 1 mark for option D.
题目 26 · 選擇題
1
An enzyme-controlled reaction is carried out at different temperatures: 10 °C, 30 °C, 50 °C, and 80 °C. At which temperature will the enzyme most likely be completely denatured and show no activity?
  1. A.10 °C
  2. B.30 °C
  3. C.50 °C
  4. D.80 °C
查看答案详解

解题

High temperatures (usually above 60 °C) denature enzymes by permanently altering the shape of their active sites. At 80 °C, the enzyme is completely denatured and shows no activity.

评分标准

Award 1 mark for option D.
题目 27 · 選擇題
1
Which change will decrease the rate of the reaction between dilute hydrochloric acid and calcium carbonate chips?
  1. A.using smaller calcium carbonate chips
  2. B.heating the hydrochloric acid to a higher temperature
  3. C.adding water to dilute the hydrochloric acid
  4. D.using a higher concentration of hydrochloric acid
查看答案详解

解题

Adding water dilutes the acid, reducing the concentration of reacting acid particles. This decreases the frequency of successful collisions, thereby decreasing the rate of reaction.

评分标准

Award 1 mark for option C.
题目 28 · 選擇題
1
A student places four different metals into separate test-tubes containing dilute hydrochloric acid. Metal W fizzes rapidly, metal X does not react at all, metal Y fizzes slowly, and metal Z reacts explosively. What is the order of reactivity of these metals, from most reactive to least reactive?
  1. A.Z -> W -> Y -> X
  2. B.X -> Y -> W -> Z
  3. C.Z -> Y -> W -> X
  4. D.X -> W -> Y -> Z
查看答案详解

解题

The more rapid the reaction with acid, the more reactive the metal is. Explosive (Z) is most reactive, followed by rapid fizzing (W), slow fizzing (Y), and no reaction (X) is the least reactive. Thus, the order is Z -> W -> Y -> X.

评分标准

Award 1 mark for option A.
题目 29 · 選擇題
1
How do the amplitude and the frequency of a sound wave change when the sound becomes louder and has a higher pitch?
  1. A.amplitude decreases, frequency decreases
  2. B.amplitude decreases, frequency increases
  3. C.amplitude increases, frequency decreases
  4. D.amplitude increases, frequency increases
查看答案详解

解题

Loudness is determined by amplitude (larger amplitude means a louder sound). Pitch is determined by frequency (higher frequency means a higher pitch). Therefore, both amplitude and frequency must increase.

评分标准

Award 1 mark for option D.
题目 30 · 選擇題
1
A student wants to determine the density of an irregular piece of rock. Which two pieces of apparatus must the student use?
  1. A.a balance and a measuring cylinder with water
  2. B.a balance and a ruler
  3. C.a measuring cylinder with water and a stop-watch
  4. D.a ruler and a stop-watch
查看答案详解

解题

To find density, the mass of the rock must be measured using a balance, and the volume of the irregular rock must be measured using the displacement of water in a measuring cylinder.

评分标准

Award 1 mark for option A.
题目 31 · 選擇題
1
A current of 3.0 A flows through a lamp for 20 seconds. How much electrical charge passes through the lamp in this time?
  1. A.0.15 C
  2. B.6.7 C
  3. C.60 C
  4. D.120 C
查看答案详解

解题

Using the formula Q = I * t, where I = 3.0 A and t = 20 s: Q = 3.0 * 20 = 60 C.

评分标准

Award 1 mark for option C.
题目 32 · 選擇題
1
What is the approximate percentage by volume of nitrogen gas in clean, dry air?
  1. A.21%
  2. B.78%
  3. C.0.9%
  4. D.0.04%
查看答案详解

解题

Clean, dry air is composed of approximately 78% nitrogen, 21% oxygen, nearly 1% argon, and 0.04% carbon dioxide.

评分标准

Award 1 mark for option B.
题目 33 · 選擇題
1
Pepsin is an enzyme found in the human stomach that digests proteins and has an optimum pH of approximately 2. When the stomach contents pass into the small intestine, the pH is raised to about 8 by bile and pancreatic juices. What happens to the activity of pepsin in the small intestine?
  1. A.Its rate of reaction increases due to the higher pH.
  2. B.It continues to digest proteins at its optimum rate.
  3. C.It becomes denatured and ceases to function.
  4. D.It changes its shape to digest fats instead of proteins.
查看答案详解

解题

Pepsin is adapted to work in highly acidic conditions (pH 2). When the pH rises to 8 in the small intestine, the chemical environment changes significantly. This extreme change in pH disrupts the active site of the enzyme, denaturing it. Once denatured, the enzyme's active site changes shape permanently and can no longer bind to its substrate, causing pepsin to cease functioning.

评分标准

Award 1 mark for selecting option C.
题目 34 · 選擇題
1
A student wants to measure the potential difference across a lamp and the current flowing through it in a simple circuit. How must the voltmeter and the ammeter be connected relative to the lamp?
  1. A.The ammeter is connected in parallel, and the voltmeter is connected in series.
  2. B.The ammeter is connected in series, and the voltmeter is connected in parallel.
  3. C.Both the ammeter and the voltmeter are connected in series.
  4. D.Both the ammeter and the voltmeter are connected in parallel.
查看答案详解

解题

To measure the current passing through a lamp, the ammeter must be connected in series with it so that the same current flows through both. To measure the potential difference across the lamp, the voltmeter must be connected in parallel across the lamp's terminals.

评分标准

Award 1 mark for selecting option B.
题目 35 · 選擇題
1
The reactions of four metals, W, X, Y, and Z, are studied: Metal W reacts vigorously with cold water. Metal X does not react with cold water but reacts with steam when heated. Metal Y does not react with cold water or steam, but reacts with dilute hydrochloric acid. Metal Z does not react with dilute hydrochloric acid. What is the correct order of reactivity of these metals, from most reactive to least reactive?
  1. A.W, then X, then Y, then Z
  2. B.W, then Y, then X, then Z
  3. C.Z, then Y, then X, then W
  4. D.X, then W, then Y, then Z
查看答案详解

解题

The reactivity of metals determines the conditions under which they react with water, steam, and acids. Metal W is the most reactive because it reacts with cold water. Metal X is less reactive than W as it requires steam to react. Metal Y is less reactive than X as it does not react with water or steam but still reacts with acid. Metal Z is the least reactive as it does not react with dilute acid. Therefore, the correct order is W to X to Y to Z.

评分标准

Award 1 mark for selecting option A.
题目 36 · 選擇題
1
A solid rectangular block of an unknown metal has a mass of 160 g. Its volume is measured to be 20 cubic centimeters. What is the density of this metal?
  1. A.0.125 g/cm³
  2. B.8.0 g/cm³
  3. C.32 g/cm³
  4. D.3200 g/cm³
查看答案详解

解题

The density of a substance is calculated using the formula: Density = Mass / Volume. Substituting the given values: Density = 160 g / 20 cubic centimeters = 8.0 grams per cubic centimeter.

评分标准

Award 1 mark for selecting option B.
题目 37 · 選擇題
1
Which row shows the correct approximate percentages by volume of nitrogen, oxygen, and other gases in clean, dry air?
  1. A.Nitrogen: 21%, Oxygen: 78%, Other gases: 1%
  2. B.Nitrogen: 78%, Oxygen: 21%, Other gases: 1%
  3. C.Nitrogen: 50%, Oxygen: 50%, Other gases: less than 1%
  4. D.Nitrogen: 78%, Oxygen: 1%, Other gases: 21%
查看答案详解

解题

Clean, dry air is a mixture of gases. By volume, it consists of approximately 78% nitrogen, 21% oxygen, and 1% of other gases (primarily noble gases like argon, along with carbon dioxide and water vapor).

评分标准

Award 1 mark for selecting option B.
题目 38 · 選擇題
1
Which word equation correctly summarizes the chemical process of photosynthesis in green plants?
  1. A.carbon dioxide + water -> glucose + oxygen
  2. B.glucose + oxygen -> carbon dioxide + water
  3. C.carbon dioxide + glucose -> water + oxygen
  4. D.oxygen + water -> glucose + carbon dioxide
查看答案详解

解题

In photosynthesis, light energy absorbed by chlorophyll is used to react carbon dioxide with water to produce glucose and oxygen. The correct word equation is: carbon dioxide + water -> glucose + oxygen.

评分标准

Award 1 mark for selecting option A.
题目 39 · 選擇題
1
Which statement correctly defines the process of osmosis?
  1. A.The net movement of water molecules from a region of higher water potential to a region of lower water potential through a partially permeable membrane.
  2. B.The net movement of particles from a region of their higher concentration to a region of their lower concentration down a concentration gradient.
  3. C.The net movement of water molecules from a region of lower water potential to a region of higher water potential through a cell wall.
  4. D.The net movement of solute particles from a region of lower concentration to a region of higher concentration using energy.
查看答案详解

解题

Osmosis is a specialized type of diffusion. It is defined as the net movement of water molecules from a region of higher water potential (more dilute solution) to a region of lower water potential (more concentrated solution) through a partially permeable membrane.

评分标准

Award 1 mark for selecting option A.
题目 40 · 選擇題
1
Waves can be classified as either transverse or longitudinal. How are light waves and sound waves in air classified?
  1. A.Light wave: longitudinal; Sound wave: transverse
  2. B.Light wave: transverse; Sound wave: longitudinal
  3. C.Light wave: transverse; Sound wave: transverse
  4. D.Light wave: longitudinal; Sound wave: longitudinal
查看答案详解

解题

Light waves are electromagnetic waves, which are transverse because their oscillations are perpendicular to the direction of energy propagation. Sound waves in air are longitudinal because the vibrations of air particles are parallel to the direction of energy propagation.

评分标准

Award 1 mark for selecting option B.

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Paper 3 / Paper 4: Structured Theory

Answer all 9 structured questions covering Biology, Chemistry, and Physics evenly.
40 题目 · 40
题目 1 · 選擇題
1
The temperature of an enzyme-catalysed reaction is increased from 20 °C to 35 °C. Which statement describes the correct explanation for the resulting increase in the rate of reaction?
  1. A.The kinetic energy of the reactant molecules increases, leading to more frequent successful collisions.
  2. B.The enzyme molecules undergo a permanent shape change in their active site to fit the substrate better.
  3. C.The temperature rise decreases the activation energy of the reaction.
  4. D.The enzyme molecules lose energy and undergo denaturation.
查看答案详解

解题

An increase in temperature increases the kinetic energy of both enzyme and substrate molecules. This causes them to move faster, increasing the frequency of successful collisions between the active site and substrate molecules, which increases the rate of reaction.

评分标准

1 mark for the correct answer A.
题目 2 · 選擇題
1
Which statement describes the effect of carbon monoxide from tobacco smoke on the human body?
  1. A.It damages the cilia lining the trachea, preventing them from moving mucus.
  2. B.It causes cells in the bronchi to divide uncontrollably, forming tumours.
  3. C.It binds irreversibly to haemoglobin, reducing the transport of oxygen by red blood cells.
  4. D.It causes the walls of the alveoli to break down, reducing their surface area.
查看答案详解

解题

Carbon monoxide binds irreversibly to haemoglobin in red blood cells, forming carboxyhaemoglobin. This severely limits the amount of oxygen that the blood can transport around the body.

评分标准

1 mark for the correct answer C.
题目 3 · 選擇題
1
An element X reacts with oxygen to form an ionic compound with the formula \(X_2O\). In which group of the Periodic Table is element X located?
  1. A.Group I
  2. B.Group II
  3. C.Group VI
  4. D.Group VII
查看答案详解

解题

Oxygen forms an oxide ion with a charge of 2- (\(O^{2-}\)). For the overall ionic compound to be neutral, the two ions of element X must have a combined charge of 2+, meaning each X ion has a charge of 1+ (\(X^+\)). Elements in Group I form ions with a 1+ charge by losing their single valence electron.

评分标准

1 mark for the correct answer A.
题目 4 · 選擇題
1
The reaction between copper(II) oxide and hydrogen gas is represented by the equation shown: \(CuO + H_2 \rightarrow Cu + H_2O\). Which statement about this reaction is correct?
  1. A.Copper(II) oxide is oxidised because it gains hydrogen.
  2. B.Copper(II) oxide is the reducing agent because it loses oxygen.
  3. C.Hydrogen is oxidised because it gains oxygen.
  4. D.Hydrogen is the oxidising agent because it reduces copper(II) oxide.
查看答案详解

解题

In this reaction, hydrogen (\(H_2\)) gains oxygen to form water (\(H_2O\)), which means it is oxidised. Copper(II) oxide (\(CuO\)) loses oxygen to form copper, so it is reduced. Copper(II) oxide acts as the oxidising agent, and hydrogen acts as the reducing agent.

评分标准

1 mark for the correct answer C.
题目 5 · 選擇題
1
An unstretched spring has a length of 12.0 cm. When a load of 6.0 N is suspended from the spring, its length becomes 15.0 cm. The limit of proportionality is not exceeded. What is the spring constant of the spring?
  1. A.0.50 N/cm
  2. B.2.0 N/cm
  3. C.40 N/cm
  4. D.50 N/cm
查看答案详解

解题

First, calculate the extension (\(x\)) of the spring: \(x = 15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\). Using Hooke's Law, \(F = kx\), where \(F = 6.0\text{ N}\) and \(x = 3.0\text{ cm}\), we get \(k = F / x = 6.0\text{ N} / 3.0\text{ cm} = 2.0\text{ N/cm}\).

评分标准

1 mark for the correct answer B.
题目 6 · 選擇題
1
A radio station broadcasts a signal with a frequency of 100 MHz. The speed of electromagnetic waves in air is \(3.0 \times 10^8\text{ m/s}\). What is the wavelength of the radio waves?
  1. A.3.0 m
  2. B.30 m
  3. C.3.0 km
  4. D.3.0 mm
查看答案详解

解题

Use the wave equation: \(v = f \lambda\), which can be rearranged to find the wavelength: \(\lambda = v / f\). Convert the frequency to hertz: \(f = 100\text{ MHz} = 100 \times 10^6\text{ Hz} = 1.0 \times 10^8\text{ Hz}\). Calculate the wavelength: \(\lambda = (3.0 \times 10^8\text{ m/s}) / (1.0 \times 10^8\text{ Hz}) = 3.0\text{ m}\).

评分标准

1 mark for the correct answer A.
题目 7 · 選擇題
1
A circuit contains two resistors connected in parallel across a 12 V battery. The resistances of the two resistors are \(3.0\\ \Omega\) and \(6.0\\ \Omega\). What is the total current drawn from the battery?
  1. A.1.3 A
  2. B.4.0 A
  3. C.6.0 A
  4. D.18 A
查看答案详解

解题

First, find the combined parallel resistance (\(R_p\)) of the circuit: \(1/R_p = 1/3.0 + 1/6.0 = 2/6.0 + 1/6.0 = 3/6.0\), so \(R_p = 2.0\\ \Omega\). Next, calculate the total current (\(I\)) using Ohm's Law: \(I = V / R_p = 12\text{ V} / 2.0\\ \Omega = 6.0\text{ A}\).

评分标准

1 mark for the correct answer C.
题目 8 · 選擇題
1
Which metal cannot be extracted from its oxide by heating with carbon?
  1. A.copper
  2. B.iron
  3. C.zinc
  4. D.magnesium
查看答案详解

解题

Magnesium is higher than carbon in the reactivity series. Therefore, carbon is not reactive enough to reduce magnesium oxide to magnesium metal. Magnesium must be extracted by electrolysis of its molten ore instead.

评分标准

1 mark for the correct answer D.
题目 9 · 選擇題
1
A student measures the rate of an enzyme-catalysed reaction at different temperatures. Which statement correctly explains why the rate of reaction is higher at \(40\ ^\circ\text{C}\) than at \(20\ ^\circ\text{C}\) before the optimum temperature is reached?
  1. A.At \(40\ ^\circ\text{C}\), the enzyme molecules have denatured and can bind to more substrates.
  2. B.At \(40\ ^\circ\text{C}\), the kinetic energy of both the enzyme and substrate molecules is higher, leading to more frequent successful collisions.
  3. C.At \(20\ ^\circ\text{C}\), the activation energy of the reaction is higher than at \(40\ ^\circ\text{C}\).
  4. D.At \(20\ ^\circ\text{C}\), the active site of the enzyme has changed shape irreversibly.
查看答案详解

解题

Up to the optimum temperature, an increase in temperature increases the kinetic energy of both the enzyme and substrate molecules. This increases their speed, leading to more frequent successful collisions between the substrates and the active sites, thereby raising the rate of reaction.

评分标准

Award 1 mark for the correct option B. Reject options suggesting denaturation at \(40\ ^\circ\text{C}\) or change in activation energy with temperature.
题目 10 · 選擇題
1
An electric motor lifts a load of \(150\text{ N}\) vertically through a height of \(8.0\text{ m}\) in \(4.0\text{ s}\). The total electrical energy supplied to the motor is \(1600\text{ J}\). What is the efficiency of the motor system in lifting this load?
  1. A.\(25\%\)
  2. B.\(30\%\)
  3. C.\(75\%\)
  4. D.\(133\%\)
查看答案详解

解题

First, calculate the useful work done (output energy): \(W = F \times d = 150\text{ N} \times 8.0\text{ m} = 1200\text{ J}\). The total energy input is \(1600\text{ J}\). The efficiency is \(\frac{\text{useful output energy}}{\text{total input energy}} \times 100\% = \frac{1200}{1600} \times 100\% = 75\%\).

评分标准

Award 1 mark for the correct efficiency calculation leading to option C.
题目 11 · 選擇題
1
Magnesium reacts with excess hydrochloric acid according to the equation shown:
$$\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}$$
A sample of \(0.12\text{ g}\) of magnesium ribbon is fully reacted with excess acid. What is the maximum volume of hydrogen gas produced, measured at room temperature and pressure (r.t.p.)?
[Relative atomic mass: \(A_r(\text{Mg}) = 24\). The volume of one mole of any gas at r.t.p. is \(24\text{ dm}^3\).]
  1. A.\(12\text{ cm}^3\)
  2. B.\(120\text{ cm}^3\)
  3. C.\(240\text{ cm}^3\)
  4. D.\(1200\text{ cm}^3\)
查看答案详解

解题

1. Calculate the number of moles of \(\text{Mg}\): \(\text{moles of Mg} = \frac{0.12\text{ g}}{24\text{ g/mol}} = 0.005\text{ mol}\).
2. According to the balanced equation, \(1\text{ mol}\) of \(\text{Mg}\) produces \(1\text{ mol}\) of \(\text{H}_2\). Therefore, \(0.005\text{ mol}\) of \(\text{H}_2\) is produced.
3. Calculate the volume of \(\text{H}_2\): \(\text{volume} = 0.005\text{ mol} \times 24\text{ dm}^3\text{/mol} = 0.12\text{ dm}^3\).
4. Convert to \(\text{cm}^3\): \(0.12\text{ dm}^3 \times 1000 = 120\text{ cm}^3\).

评分标准

Award 1 mark for the correct molar calculation and unit conversion to option B.
题目 12 · 選擇題
1
A circuit contains a \(12\text{ V}\) battery connected to three resistors. Two \(6.0\ \Omega\) resistors are connected in parallel with each other, and this combination is connected in series with a single \(3.0\ \Omega\) resistor. What is the total current flowing from the battery?
  1. A.\(0.80\text{ A}\)
  2. B.\(1.3\text{ A}\)
  3. C.\(2.0\text{ A}\)
  4. D.\(4.0\text{ A}\)
查看答案详解

解题

1. Calculate the equivalent resistance of the two parallel \(6.0\ \Omega\) resistors: \(R_p = \frac{6.0 \times 6.0}{6.0 + 6.0} = 3.0\ \Omega\).
2. Calculate the total resistance of the circuit: \(R_{\text{total}} = R_p + 3.0\ \Omega = 3.0 + 3.0 = 6.0\ \Omega\).
3. Calculate the total current using Ohm's Law: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\).

评分标准

Award 1 mark for the correct application of parallel and series resistance rules to obtain the current in option C.
题目 13 · 選擇題
1
Plant cells are placed in a highly concentrated salt solution. Which description of the water potential gradient and the net movement of water is correct?
  1. A.Water potential is higher inside the cells than outside; net movement of water is into the cells.
  2. B.Water potential is higher inside the cells than outside; net movement of water is out of the cells.
  3. C.Water potential is higher outside the cells than inside; net movement of water is into the cells.
  4. D.Water potential is higher outside the cells than inside; net movement of water is out of the cells.
查看答案详解

解题

The concentrated salt solution has a lower water potential than the cytoplasm inside the plant cells, meaning the water potential is higher inside the cells than outside. Water moves down the water potential gradient from a region of higher water potential to a region of lower water potential, resulting in a net movement of water out of the cells.

评分标准

Award 1 mark for the correct combination of water potential description and net water movement direction (Option B).
题目 14 · 選擇題
1
A wave of frequency \(250\text{ Hz}\) travels from deep water to shallow water. In deep water, its speed is \(2.0\text{ m/s}\). In shallow water, its speed decreases to \(1.5\text{ m/s}\). What is the frequency and wavelength of the wave in shallow water?
  1. A.frequency = \(188\text{ Hz}\), wavelength = \(6.0\text{ mm}\)
  2. B.frequency = \(250\text{ Hz}\), wavelength = \(6.0\text{ mm}\)
  3. C.frequency = \(250\text{ Hz}\), wavelength = \(8.0\text{ mm}\)
  4. D.frequency = \(333\text{ Hz}\), wavelength = \(8.0\text{ mm}\)
查看答案详解

解题

1. The frequency of a wave is determined by its source and does not change when transitioning between different mediums. Therefore, the frequency in shallow water remains \(250\text{ Hz}\).
2. Wavelength is calculated using the formula \(\lambda = \frac{v}{f}\). In shallow water: \(\lambda = \frac{1.5\text{ m/s}}{250\text{ Hz}} = 0.0060\text{ m} = 6.0\text{ mm}\).

评分标准

Award 1 mark for identifying that frequency remains constant and calculating the correct wavelength (Option B).
题目 15 · 選擇題
1
Ethene (\(\text{C}_2\text{H}_4\)) contains a double covalent bond between the two carbon atoms and single covalent bonds between carbon and hydrogen atoms. What is the total number of shared electrons in one molecule of ethene?
  1. A.6
  2. B.8
  3. C.10
  4. D.12
查看答案详解

解题

Ethene has 4 single \(\text{C-H}\) bonds and 1 double \(\text{C=C}\) bond.
Each single bond contains 2 shared electrons: \(4 \times 2 = 8\) electrons.
The double bond contains 4 shared electrons: \(1 \times 4 = 4\) electrons.
Total shared electrons = \(8 + 4 = 12\) electrons.

评分标准

Award 1 mark for the correct total number of shared electrons (Option D).
题目 16 · 選擇題
1
Four metals, \(P, Q, R,\) and \(S\), are tested as follows:
- Metal \(P\) reacts vigorously with cold water.
- Metal \(Q\) does not react with cold water but reacts with steam.
- Metal \(R\) does not react with steam, but its oxide is reduced by heating with carbon.
- Metal \(S\) does not react with dilute acids and is found uncombined in the Earth's crust.
What is the correct order of reactivity of these metals, from most reactive to least reactive?
  1. A.\(P \rightarrow Q \rightarrow R \rightarrow S\)
  2. B.\(P \rightarrow R \rightarrow Q \rightarrow S\)
  3. C.\(S \rightarrow R \rightarrow Q \rightarrow P\)
  4. D.\(Q \rightarrow P \rightarrow R \rightarrow S\)
查看答案详解

解题

Reacting with cold water represents the highest reactivity (Metal \(P\)). Reacting with steam but not cold water is next (Metal \(Q\)). Not reacting with steam but being extractable by carbon reduction is lower still (Metal \(R\)). Found uncombined and not reacting with acids represents the lowest reactivity (Metal \(S\)). Thus, the order is \(P \rightarrow Q \rightarrow R \rightarrow S\).

评分标准

Award 1 mark for correctly ordering the metals according to the given chemical behavior (Option A).
题目 17 · 選擇題
1
Plant cells are placed in a beaker containing a solution with a higher water potential than the cell sap. Which statement correctly describes the movement of water and the final state of the cells?
  1. A.Water moves into the cells down a water potential gradient, and the cells become turgid.
  2. B.Water moves into the cells up a water potential gradient, and the cells become plasmolysed.
  3. C.Water moves out of the cells down a water potential gradient, and the cells become plasmolysed.
  4. D.Water moves out of the cells up a water potential gradient, and the cells become turgid.
查看答案详解

解题

Water moves from an area of higher water potential to an area of lower water potential. Since the external solution has a higher water potential than the cell sap, water moves into the cells by osmosis down the water potential gradient. This influx of water causes the vacuole to swell, pressing the cytoplasm against the cell wall, making the cells turgid.

评分标准

Correct option is A (1 mark).
题目 18 · 選擇題
1
Which statement correctly describes why the rate of an enzyme-catalysed reaction decreases rapidly at temperatures above the optimum temperature?
  1. A.The substrate and enzyme molecules have too much kinetic energy, so they collide too frequently.
  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 slows down the chemical process.
  4. D.The substrate molecules are denatured by the high thermal energy.
查看答案详解

解题

At high temperatures, the thermal energy is sufficient to break the bonds holding the enzyme's three-dimensional structure together. This denatures the enzyme, causing a permanent change in the shape of its active site. Consequently, the substrate molecule can no longer fit into the active site, and the rate of reaction drops.

评分标准

Correct option is B (1 mark).
题目 19 · 選擇題
1
Dilute hydrochloric acid reacts with excess calcium carbonate. The concentration of the hydrochloric acid is doubled, while the temperature and mass of calcium carbonate are kept constant. How does this change affect the activation energy and the frequency of successful collisions?
  1. A.The activation energy decreases and the frequency of successful collisions increases.
  2. B.The activation energy remains the same and the frequency of successful collisions increases.
  3. C.The activation energy decreases and the frequency of successful collisions remains the same.
  4. D.The activation energy remains the same and the frequency of successful collisions decreases.
查看答案详解

解题

An increase in concentration increases the number of reactant particles per unit volume, which increases the frequency of collisions and thus the frequency of successful collisions. However, concentration has no effect on the activation energy; only adding a catalyst can change the activation energy.

评分标准

Correct option is B (1 mark).
题目 20 · 選擇題
1
Which metal can be extracted from its oxide by heating with carbon, but does not react with dilute hydrochloric acid?
  1. A.copper
  2. B.iron
  3. C.magnesium
  4. D.zinc
查看答案详解

解题

Copper is below carbon in the reactivity series, so copper(II) oxide can be reduced by carbon when heated. Copper is also below hydrogen in the reactivity series, which means it is unreactive towards dilute acids like hydrochloric acid.

评分标准

Correct option is A (1 mark).
题目 21 · 選擇題
1
A toy car of mass \( 500\text{ g} \) is traveling along a horizontal track at a constant speed of \( 12\text{ m/s} \). What is the kinetic energy of the car?
  1. A.\( 3.0\text{ J} \)
  2. B.\( 36\text{ J} \)
  3. C.\( 72\text{ J} \)
  4. D.\( 36\,000\text{ J} \)
查看答案详解

解题

First, convert the mass from grams to kilograms: \( m = 500\text{ g} = 0.5\text{ kg} \). Next, use the formula for kinetic energy: \( KE = \frac{1}{2} m v^2 \). Substituting the values: \( KE = 0.5 \times 0.5 \times 12^2 = 0.25 \times 144 = 36\text{ J} \).

评分标准

Correct option is B (1 mark).
题目 22 · 選擇題
1
An object is placed in front of a thin converging lens. The distance between the object and the lens is less than the focal length of the lens. What is the nature of the image formed?
  1. A.virtual, right way up and larger than the object
  2. B.virtual, upside down and smaller than the object
  3. C.real, right way up and larger than the object
  4. D.real, upside down and larger than the object
查看答案详解

解题

When an object is placed closer to a converging lens than its focal point (\( u < f \)), the lens acts as a magnifying glass. The refracted rays diverge, so they must be projected backwards to form a virtual, upright (right way up), and magnified (larger than the object) image.

评分标准

Correct option is A (1 mark).
题目 23 · 選擇題
1
Three identical resistors, each of resistance \( 6.0\ \Omega \), are connected in a circuit. Two of the resistors are connected in parallel with each other, and this combination is connected in series with the third resistor. What is the total resistance of this network?
  1. A.\( 2.0\ \Omega \)
  2. B.\( 3.0\ \Omega \)
  3. C.\( 9.0\ \Omega \)
  4. D.\( 18.0\ \Omega \)
查看答案详解

解题

First, calculate the resistance of the two parallel resistors: \( R_p = \frac{6.0 \times 6.0}{6.0 + 6.0} = 3.0\ \Omega \). Next, add the resistance of the series resistor: \( R_{\text{total}} = R_p + R_s = 3.0 + 6.0 = 9.0\ \Omega \).

评分标准

Correct option is C (1 mark).
题目 24 · 選擇題
1
Which row correctly identifies the typical features of sexual reproduction?
  1. A.genetic variation in offspring, two parents required, zygote produced
  2. B.genetic variation in offspring, one parent required, no zygote produced
  3. C.genetically identical offspring, two parents required, zygote produced
  4. D.genetically identical offspring, one parent required, no zygote produced
查看答案详解

解题

Sexual reproduction involves the fusion of two haploid gametes (two parents) to produce a diploid zygote, which results in genetic variation among the offspring.

评分标准

Correct option is A (1 mark).
题目 25 · 選擇題
1
A medical scan uses a type of electromagnetic radiation with a wavelength shorter than ultraviolet radiation but longer than gamma radiation. Which row correctly identifies this radiation and its speed in a vacuum?
  1. A.infrared, \(3.0 \times 10^8\text{ m/s}\)
  2. B.microwave, \(330\text{ m/s}\)
  3. C.X-rays, \(3.0 \times 10^8\text{ m/s}\)
  4. D.X-rays, \(330\text{ m/s}\)
查看答案详解

解题

X-rays have a wavelength between ultraviolet and gamma rays on the electromagnetic spectrum. All electromagnetic waves, including X-rays, travel at the speed of light (\(3.0 \times 10^8\text{ m/s}\)) in a vacuum.

评分标准

Award 1 mark for choosing option C because it correctly identifies the radiation as X-rays and its speed in a vacuum as \(3.0 \times 10^8\text{ m/s}\).
题目 26 · 選擇題
1
The reaction between copper(II) oxide and hydrogen gas is represented by the equation:

\[\text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O}\]

Which statement about this reaction is correct?
  1. A.\(\text{CuO}\) acts as the reducing agent because it loses oxygen.
  2. B.\(\text{CuO}\) is oxidized to \(\text{Cu}\).
  3. C.\(\text{H}_2\) acts as the reducing agent because it gains oxygen.
  4. D.\(\text{H}_2\) is reduced to \(\text{H}_2\text{O}\).
查看答案详解

解题

In this reaction, hydrogen (\(\text{H}_2\)) gains oxygen to become water, so it is oxidized. Since it causes the copper(II) oxide to be reduced, hydrogen acts as the reducing agent. Copper(II) oxide loses oxygen, so it is reduced, acting as the oxidizing agent.

评分标准

Award 1 mark for choosing option C, which correctly states that hydrogen acts as the reducing agent because it gains oxygen.
题目 27 · 選擇題
1
A small vehicle of mass \(800\text{ kg}\) is travelling at a constant speed of \(36\text{ km/h}\). What is the kinetic energy of the vehicle?
  1. A.\(40\text{ kJ}\)
  2. B.\(160\text{ kJ}\)
  3. C.\(518\text{ kJ}\)
  4. D.\(1440\text{ kJ}\)
查看答案详解

解题

First, convert speed from \(\text{km/h}\) to \(\text{m/s}\):
\[36\text{ km/h} = \frac{36 \times 1000}{3600} = 10\text{ m/s}\]

Then use the kinetic energy formula:
\[E_k = \frac{1}{2} m v^2\]
\[E_k = \frac{1}{2} \times 800\text{ kg} \times (10\text{ m/s})^2 = 400 \times 100 = 40\,000\text{ J} = 40\text{ kJ}\]

评分标准

Award 1 mark for option A, after correctly converting speed to \(10\text{ m/s}\) and using the formula to compute \(40\text{ kJ}\).
题目 28 · 選擇題
1
An enzyme is active at pH 7. When the pH is decreased to pH 2, the rate of reaction drops to zero. Which statement explains this observation?
  1. A.The acidic conditions increase the kinetic energy of the substrate molecules, preventing them from binding.
  2. B.The enzyme molecules are denatured and the active site changes shape, so the substrate no longer fits.
  3. C.The substrate molecules are denatured and can no longer fit into the unchanged active site of the enzyme.
  4. D.The low pH acts as a competitive inhibitor by permanently binding to the active site.
查看答案详解

解题

Enzymes are protein molecules that function as biological catalysts. Extreme pH values away from the optimum pH cause denaturation of the enzyme. This permanently changes the shape of the active site, meaning the substrate is no longer complementary and cannot bind to form enzyme-substrate complexes.

评分标准

Award 1 mark for choosing option B because it correctly explains that extreme pH denatures the enzyme and alters the active site shape.
题目 29 · 選擇題
1
A circuit contains a \(12\text{ V}\) battery connected in series with two resistors, \(R_1\) and \(R_2\), that are connected in parallel with each other. The resistance of \(R_1\) is \(6.0\ \Omega\) and the resistance of \(R_2\) is \(12\.0\ \Omega\). What is the total current leaving the battery?
  1. A.\(0.67\text{ A}\)
  2. B.\(2.0\text{ A}\)
  3. C.\(3.0\text{ A}\)
  4. D.\(4.0\text{ A}\)
查看答案详解

解题

First, find the combined resistance of the parallel combination, \(R_p\):
\[\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6.0} + \frac{1}{12.0} = \frac{3}{12.0}\]
\[R_p = 4.0\ \Omega\]
Since there are no other resistors in series, the total resistance is \(4.0\ \Omega\). Now calculate current using Ohm's Law:
\[I = \frac{V}{R} = \frac{12\text{ V}}{4.0\ \Omega} = 3.0\text{ A}\]

评分标准

Award 1 mark for option C, calculated by finding parallel resistance as \(4.0\ \Omega\) and then division from \(12\text{ V}\) to get \(3.0\text{ A}\).
题目 30 · 選擇題
1
Three metals, \(W\), \(X\), and \(Y\), are tested to find their place in the reactivity series.
- Metal \(W\) can be extracted from its oxide by heating with carbon.
- Metal \(X\) reacts violently with cold water to produce hydrogen.
- Metal \(Y\) does not react with dilute hydrochloric acid.

Which list shows the metals in order of decreasing reactivity (most reactive first)?
  1. A.\(W \rightarrow X \rightarrow Y\)
  2. B.\(X \rightarrow W \rightarrow Y\)
  3. C.\(X \rightarrow Y \rightarrow W\)
  4. D.\(Y \rightarrow W \rightarrow X\)
查看答案详解

解题

Metal \(X\) is highly reactive because it reacts violently with cold water. Metal \(W\) is moderately reactive because carbon is able to reduce its oxide. Metal \(Y\) is very unreactive as it does not react with dilute hydrochloric acid. This gives the decreasing order of reactivity: \(X \rightarrow W \rightarrow Y\).

评分标准

Award 1 mark for option B because it correctly orders the metals from most reactive to least reactive based on the observations.
题目 31 · 選擇題
1
Three separate food samples are treated with enzymes:
- Amylase is added to Sample 1.
- Protease is added to Sample 2.
- Lipase is added to Sample 3.

Which row correctly identifies the smaller soluble molecules produced in each sample?
  1. A.Sample 1: glucose; Sample 2: amino acids; Sample 3: fatty acids and glycerol
  2. B.Sample 1: glycerol; Sample 2: fatty acids; Sample 3: glucose
  3. C.Sample 1: maltose; Sample 2: fatty acids; Sample 3: amino acids
  4. D.Sample 1: glucose; Sample 2: fatty acids and glycerol; Sample 3: amino acids
查看答案详解

解题

Amylase breaks down starch to simpler sugars such as maltose and glucose. Protease breaks down proteins into amino acids. Lipase breaks down fats into fatty acids and glycerol.

评分标准

Award 1 mark for option A, which correctly matches all three enzyme digestive enzymes with their respective end-products.
题目 32 · 選擇題
1
A molecule of ethene has the formula \(\text{C}_2\text{H}_4\). What is the total number of shared electrons in all the covalent bonds of one molecule of ethene?
  1. A.6
  2. B.8
  3. C.12
  4. D.16
查看答案详解

解题

Ethene, \(\text{H}_2\text{C=CH}_2\), has four single carbon-hydrogen (\(\text{C–H}\)) covalent bonds and one double carbon-carbon (\(\text{C=C}\)) covalent bond. Each single covalent bond consists of 2 shared electrons (total of 8 electrons for the four \(\text{C–H}\) bonds) and the double covalent bond consists of 4 shared electrons.

Total shared electrons = \(8 + 4 = 12\).

评分标准

Award 1 mark for option C, obtained by correctly identifying 6 total covalent bonds (each sharing 2 electrons) in ethene, giving 12 shared electrons.
题目 33 · 選擇題
1
An electromagnetic wave travels through a vacuum with a frequency of \(6.0 \times 10^{14}\text{ Hz}\). What is the wavelength of this wave? (The speed of electromagnetic waves in a vacuum is \(3.0 \times 10^8\text{ m/s}\)).
  1. A.\(2.0 \times 10^{-6}\text{ m}\)
  2. B.\(5.0 \times 10^{-7}\text{ m}\)
  3. C.\(1.8 \times 10^{23}\text{ m}\)
  4. D.\(5.0 \times 10^{5}\text{ m}\)
查看答案详解

解题

Using the wave equation, \(v = f \lambda\), where \(v\) is the speed of light (\(3.0 \times 10^8\text{ m/s}\)) and \(f\) is the frequency (\(6.0 \times 10^{14}\text{ Hz}\)), we can rearrange to find wavelength: \(\lambda = \frac{v}{f} = \frac{3.0 \times 10^8}{6.0 \times 10^{14}} = 5.0 \times 10^{-7}\text{ m}\).

评分标准

Award 1 mark for selecting correct option B.
题目 34 · 選擇題
1
A student investigates the reaction between calcium carbonate chips and excess dilute hydrochloric acid at \(20\text{ }^\circ\text{C}\). Which change decreases the initial rate of this reaction?
  1. A.adding a suitable catalyst to the mixture
  2. B.grinding the calcium carbonate chips into a fine powder
  3. C.increasing the temperature of the hydrochloric acid to \(30\text{ }^\circ\text{C}\)
  4. D.adding distilled water to dilute the hydrochloric acid
查看答案详解

解题

Diluting the acid decreases the concentration of acid particles in the solution. This reduces the frequency of collisions between reactant particles, which decreases the initial rate of reaction. Grinding increases surface area, heating increases kinetic energy, and a catalyst provides an alternative pathway with lower activation energy; all of these increase the rate of reaction.

评分标准

Award 1 mark for selecting correct option D.
题目 35 · 選擇題
1
An object of mass \(4.0\text{ kg}\) moves with a constant kinetic energy of \(72\text{ J}\). What is the speed of this object?
  1. A.\(3.0\text{ m/s}\)
  2. B.\(6.0\text{ m/s}\)
  3. C.\(18\text{ m/s}\)
  4. D.\(36\text{ m/s}\)
查看答案详解

解题

The formula for kinetic energy is \(E_k = \frac{1}{2}mv^2\). Substituting the given values: \(72 = \frac{1}{2} \times 4.0 \times v^2 \Rightarrow 72 = 2.0 \times v^2 \Rightarrow v^2 = 36 \Rightarrow v = 6.0\text{ m/s}\).

评分标准

Award 1 mark for selecting correct option B.
题目 36 · 選擇題
1
The rate of an enzyme-catalyzed reaction decreases rapidly at temperatures above the optimum temperature. Which statement explains this decrease?
  1. A.The kinetic energy of the substrate molecules decreases.
  2. B.The enzyme molecules are denatured, changing the shape of the active site.
  3. C.The activation energy of the reaction is increased by the higher temperature.
  4. D.The substrate molecules are thermally decomposed by the high temperature.
查看答案详解

解题

At high temperatures, the chemical bonds holding the enzyme's three-dimensional structure together are broken. This denatures the enzyme, permanently changing the shape of its active site so that the substrate can no longer fit, resulting in a rapid decrease in the rate of reaction.

评分标准

Award 1 mark for selecting correct option B.
题目 37 · 選擇題
1
A \(6.0\text{ }\Omega\) resistor and a \(3.0\text{ }\Omega\) resistor are connected in parallel across a \(12\text{ V}\) d.c. power supply. What is the total current drawn from the power supply?
  1. A.\(1.3\text{ A}\)
  2. B.\(2.0\text{ A}\)
  3. C.\(6.0\text{ A}\)
  4. D.\(18\text{ A}\)
查看答案详解

解题

The combined resistance \(R\) of two resistors in parallel is given by \(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{3}{6.0} \Rightarrow R = 2.0\text{ }\Omega\). Using Ohm's Law, the total current is \(I = \frac{V}{R} = \frac{12\text{ V}}{2.0\text{ }\Omega} = 6.0\text{ A}\).

评分标准

Award 1 mark for selecting correct option C.
题目 38 · 選擇題
1
Three metal oxides, \(X\text{O}\), \(Y\text{O}\), and \(Z\text{O}\), are heated separately with carbon powder. Oxide \(X\text{O}\) reacts to form metal \(X\) and carbon dioxide. Oxide \(Y\text{O}\) does not react. Oxide \(Z\text{O}\) reacts to form metal \(Z\) and carbon dioxide. Metal \(Z\) is less reactive than metal \(X\). What is the order of reactivity of the metals, from most reactive to least reactive?
  1. A.\(X \rightarrow Z \rightarrow Y\)
  2. B.\(Y \rightarrow X \rightarrow Z\)
  3. C.\(Y \rightarrow Z \rightarrow X\)
  4. D.\(Z \rightarrow X \rightarrow Y\)
查看答案详解

解题

Oxide \(Y\text{O}\) does not react with carbon, meaning metal \(Y\) is more reactive than carbon and cannot be reduced by it. Oxides \(X\text{O}\) and \(Z\text{O}\) react with carbon, meaning metals \(X\) and \(Z\) are less reactive than carbon. Since \(Z\) is less reactive than \(X\), the overall order from most to least reactive is \(Y \rightarrow X \rightarrow Z\).

评分标准

Award 1 mark for selecting correct option B.
题目 39 · 選擇題
1
Which row correctly identifies a digestive enzyme, the organ where it is secreted, and the product of its reaction?
  1. A.amylase | salivary glands | maltose
  2. B.lipase | stomach | fatty acids and glycerol
  3. C.protease | pancreas | glucose
  4. D.amylase | ileum | amino acids
查看答案详解

解题

Amylase is secreted by salivary glands (and the pancreas) and digests starch to produce maltose. Lipase is secreted by the pancreas (not the stomach), protease produces amino acids or peptides (not glucose), and amylase does not produce amino acids.

评分标准

Award 1 mark for selecting correct option A.
题目 40 · 選擇題
1
A molecule of methane, \(\text{CH}_4\), contains covalent bonds. How many shared pairs of electrons are there in one molecule of methane?
  1. A.2
  2. B.4
  3. C.8
  4. D.10
查看答案详解

解题

Each carbon-hydrogen single bond in a methane molecule consists of one shared pair of electrons. Since carbon forms four single covalent bonds with four hydrogen atoms, there are four shared pairs of electrons in a molecule of methane.

评分标准

Award 1 mark for selecting correct option B.

Paper 5 / Paper 6: Practical / Alternative to Practical

Answer all practical questions involving experimental observations, graph drawing, data analysis, and investigation planning.
9 题目 · 81
题目 1 · structured
9
A surveyor uses a sound pulse device to measure the distance to a vertical cliff face. (a) State the type of wave that a sound wave is. [1] (b) Explain what is meant by the frequency of a wave. [1] (c) The surveyor sends a sound pulse of frequency 400 Hz. The pulse is reflected back from the cliff face. (i) Explain, in terms of air particles, how a sound wave is transmitted. [2] (ii) The sound wave has a wavelength of 0.85 m. Calculate the speed of this sound wave in air. Show your working and state the unit. [3] (iii) The surveyor hears the echo exactly 1.8 seconds after the pulse is emitted. Calculate the distance between the surveyor and the cliff face. [2]
查看答案详解

解题

(a) Sound is a longitudinal wave. (b) Frequency is defined as the number of oscillations or waves per unit time (usually per second). (c)(i) Sound travels by causing air particles to collide, forming regions of high pressure (compressions) and low pressure (rarefactions). (c)(ii) Using the wave equation: v = f * lambda = 400 Hz * 0.85 m = 340 m/s. (c)(iii) The echo takes 1.8 s to travel to the cliff and back, so the one-way travel time is 0.9 s. Distance = speed * time = 340 m/s * 0.9 s = 306 m.

评分标准

(a) longitudinal [1]. (b) number of waves per second / unit time [1]. (c)(i) particles vibrate [1]; parallel to the direction of wave energy transfer / back and forth [1]. (c)(ii) speed = frequency * wavelength [1]; 400 * 0.85 = 340 [1]; m/s [1]. (c)(iii) distance = speed * time / 2 OR time = 1.8 / 2 = 0.9 s [1]; 340 * 0.9 = 306 (m) [1].
题目 2 · structured
9
A student investigates the reaction between solid copper(II) carbonate and dilute hydrochloric acid. (a) Complete the word equation for this reaction: copper carbonate + hydrochloric acid -> [product 1] + [product 2] + [product 3] [2] (b) State the chemical test used to show that carbon dioxide gas is produced, including the positive result. [2] (c) The student performs the reaction twice. Experiment 1: uses large lumps of copper carbonate. Experiment 2: uses powdered copper carbonate of the same mass. State and explain how the rate of reaction in Experiment 2 compares to Experiment 1. Use ideas about particles in your answer. [3] (d) Copper carbonate can also be decomposed by heating it strongly. State the name of this type of chemical reaction. [2]
查看答案详解

解题

(a) When a metal carbonate reacts with an acid, it produces a salt, carbon dioxide, and water. Here, the salt is copper chloride. (b) Carbon dioxide gas is tested by bubbling it into limewater, which turns milky/cloudy. (c) Powdering a solid reactant increases its surface area. This exposes more particles to collisions per unit time, resulting in a higher frequency of successful collisions and thus a faster rate of reaction. (d) Strong heating of a single compound to break it down into simpler substances is called thermal decomposition.

评分标准

(a) copper chloride [1]; carbon dioxide + water [1]. (b) limewater [1]; turns cloudy / milky [1]. (c) rate increases / reaction is faster [1]; larger surface area of powder [1]; more frequent collisions between reacting particles [1]. (d) thermal [1]; decomposition [1].
题目 3 · structured
9
Amylase is a digestive enzyme that breaks down starch into simpler sugars. (a) Define the term catalyst. [2] (b) Amylase activity was measured at different temperatures. The enzyme activity increased up to 40 degrees C, peaked at 40 degrees C, and then dropped rapidly to zero by 60 degrees C. (i) State the optimum temperature for this amylase enzyme. [1] (ii) Describe and explain the effect on the enzyme activity as the temperature increases from 40 degrees C to 60 degrees C. Use the term denatured in your answer. [3] (c) Digestion of starch occurs in the human digestive system. (i) State the name of one organ where amylase is produced. [1] (ii) State the name of the simpler molecule produced when starch is completely digested by amylase and maltase. [2]
查看答案详解

解题

(a) A catalyst speeds up a chemical reaction without being consumed by the process. (b)(i) The optimum temperature is the temperature at which the enzyme activity is highest, which is 40 degrees C. (b)(ii) As temperature rises past the optimum, the thermal energy disrupts the bonds holding the enzyme's shape. The active site loses its specific shape (denatures) so the starch substrate cannot bind to it, causing activity to drop to zero. (c)(i) Amylase is produced in the salivary glands and the pancreas. (c)(ii) Starch is initially broken down to maltose by amylase, which is then broken down to glucose by maltase.

评分标准

(a) speeds up a reaction [1]; unchanged at the end of the reaction [1]. (b)(i) 40 degrees C [1]. (b)(ii) activity decreases / drops to zero [1]; enzyme is denatured [1]; active site changes shape so substrate no longer fits [1]. (c)(i) salivary gland / pancreas [1]. (c)(ii) glucose [2] (allow maltose [1]).
题目 4 · structured
9
A forklift truck in a warehouse is used to lift a heavy crate. (a) The forklift truck lifts a crate of mass 240 kg vertically upwards through a height of 3.0 m. (i) Calculate the weight of the crate. (The gravitational force on unit mass g is 10 N/kg). [2] (ii) Calculate the work done by the forklift truck in lifting this crate. Show your working and state the unit. [3] (b) The forklift then moves the crate horizontally at a constant speed of 1.5 m/s. (i) State whether any work is being done vertically on the crate while it moves horizontally. Explain your answer. [2] (ii) Calculate the kinetic energy of the crate while it is moving at 1.5 m/s. [2]
查看答案详解

解题

(a)(i) Weight = m * g = 240 kg * 10 N/kg = 2400 N. (a)(ii) Work done is calculated using W = F * d. Since the lifting force is equal to the weight of 2400 N, W = 2400 N * 3.0 m = 7200 J. (b)(i) Work is done only when there is a force acting in the direction of movement. Since there is no vertical movement, no vertical work is done. (b)(ii) Kinetic energy = 0.5 * m * v^2 = 0.5 * 240 kg * (1.5 m/s)^2 = 120 * 2.25 = 270 J.

评分标准

(a)(i) weight = mass x g OR 240 x 10 [1]; 2400 (N) [1]. (a)(ii) work = force x distance OR 2400 x 3 [1]; 7200 [1]; Joules / J [1]. (b)(i) no (vertical work is done) [1]; because there is no vertical movement / displacement is at right angles to the gravity force [1]. (b)(ii) KE = 0.5 x m x v^2 OR 0.5 x 240 x 1.5^2 [1]; 270 (J) [1].
题目 5 · structured
9
A student investigates electrical circuits using a 6.0 V battery and two identical resistors. (a) First, the two resistors are connected in series with the 6.0 V battery. The current in the circuit is 0.15 A. (i) Calculate the total resistance of the circuit. State the unit. [3] (ii) Deduce the resistance of one of these resistors. [1] (b) The student wants to measure the potential difference across one of the resistors. (i) Name the device used to measure potential difference. [1] (ii) State how this device must be connected to the resistor. [1] (c) The student now reconnects the same two identical resistors in parallel with the 6.0 V battery. (i) State how the total resistance of this parallel circuit compares to the series circuit. [1] (ii) Explain how the reading on an ammeter connected next to the battery will change. [2]
查看答案详解

解题

(a)(i) Total resistance is found using Ohm's Law: R = V / I = 6.0 V / 0.15 A = 40 ohms. (a)(ii) In series, total resistance is the sum of individual resistances. Since they are identical, R1 = R2 = 40 / 2 = 20 ohms. (b)(i) Potential difference is measured using a voltmeter. (b)(ii) A voltmeter is always connected in parallel across the component of interest. (c)(i) Connecting resistors in parallel always reduces the total resistance below that of any individual resistor, so it is much lower than in series. (c)(ii) Since total resistance decreases, the total current drawn from the battery increases, so the ammeter reading increases.

评分标准

(a)(i) R = V / I OR 6.0 / 0.15 [1]; 40 [1]; ohms / Omega [1]. (a)(ii) 20 (ohms) [1]. (b)(i) voltmeter [1]. (b)(ii) in parallel [1]. (c)(i) lower / less [1]. (c)(ii) increases [1]; because the total resistance of the circuit is less [1].
题目 6 · structured
9
Iron is a transition element, whereas sodium is a Group I alkali metal. (a) (i) State two physical properties of iron that are typical of transition elements but are not shown by sodium. [2] (ii) State one chemical property of transition elements shown by iron compounds but not by sodium compounds. [1] (b) Iron is extracted from iron(III) oxide, Fe2O3, in a blast furnace. (i) Name the substance used to reduce iron(III) oxide in the blast furnace. [1] (ii) Write the word equation for this reduction reaction. [2] (c) Steel is an alloy of iron containing carbon. (i) State what is meant by the term alloy. [1] (ii) State why steel, rather than pure iron, is used to build bridges and buildings. [2]
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解题

(a)(i) Transition elements have high densities and high melting points, unlike Group I metals which are soft and have low densities and melting points. (a)(ii) Transition elements typically form coloured compounds and exhibit catalytic activity. (b)(i) Carbon monoxide is the main reducing agent in the blast furnace. (b)(ii) Carbon monoxide reacts with iron(III) oxide to form iron and carbon dioxide gas. (c)(i) An alloy is a mixture of a metal with other elements. (c)(ii) The different-sized atoms of carbon disrupt the regular layers of iron atoms, preventing them from sliding over each other easily, making steel stronger and harder than pure iron.

评分标准

(a)(i) high melting point [1]; high density [1] (allow hard / strong). (a)(ii) colored compounds / acts as a catalyst [1]. (b)(i) carbon monoxide [1] (allow carbon). (b)(ii) iron oxide + carbon monoxide -> iron + carbon dioxide [2] (allow 1 mark if one product or reactant is incorrect). (c)(i) mixture of a metal with other elements [1]. (c)(ii) stronger / harder [1]; less malleable / layers cannot slide [1].
题目 7 · structured
9
Pregnant women have specific nutritional and health requirements. (a) (i) State two reasons why a pregnant woman requires more iron in her diet. [2] (ii) State one food that is a rich source of iron. [1] (b) Calcium is also highly recommended during pregnancy. (i) State the function of calcium in the development of the fetus. [1] (ii) Name the deficiency disease caused by a lack of calcium or vitamin D in children. [1] (c) Nutrients are transferred from the mother's blood to the fetus across a specialized organ. (i) Name this specialized organ. [1] (ii) State the function of the amniotic sac during pregnancy. [1] (iii) State one substance, other than nutrients and oxygen, that passes from the mother to the fetus across this organ. [2]
查看答案详解

解题

(a)(i) Iron is needed to manufacture hemoglobin for the increased volume of red blood cells required by both mother and fetus. (a)(ii) Rich sources of dietary iron include red meat, liver, and dark green leafy vegetables like spinach. (b)(i) Calcium is essential for mineralizing and forming the skeleton (bones and teeth) of the fetus. (b)(ii) A deficiency of calcium or vitamin D leads to rickets, where bones are soft and deformed. (c)(i) The placenta allows exchange of materials between mother and fetus. (c)(ii) The amniotic sac contains amniotic fluid which cushions and protects the developing fetus from mechanical shock. (c)(iii) Harmful substances like alcohol, nicotine, or pathogens, as well as beneficial substances like antibodies, can cross the placenta.

评分标准

(a)(i) to make hemoglobin [1]; to supply iron to the fetus [1]. (a)(ii) red meat / liver / spinach [1]. (b)(i) bone / teeth development [1]. (b)(ii) rickets [1]. (c)(i) placenta [1]. (c)(ii) protects fetus from physical shock / damage [1]. (c)(iii) any two from: antibodies / alcohol / nicotine / carbon monoxide / pathogens [2].
题目 8 · structured
9
The Earth's atmosphere is impacted by human activities. (a) Clean, dry air is a mixture of gases. (i) State the approximate percentage of nitrogen gas in clean, dry air. [1] (ii) Name the noble gas that is present in the largest percentage in clean, dry air. [1] (b) Carbon dioxide and methane are greenhouse gases. (i) State one human activity that increases the concentration of methane in the atmosphere. [1] (ii) Explain how an increase in the concentration of greenhouse gases leads to global warming. [3] (c) Sulfur dioxide is an atmospheric pollutant. (i) State the source of sulfur dioxide in the atmosphere. [1] (ii) Describe one environmental consequence of acid rain caused by sulfur dioxide. [2]
查看答案详解

解题

(a)(i) Nitrogen makes up approximately 78% of dry air. (a)(ii) Argon is the most abundant noble gas in air, at about 0.9%. (b)(i) Activities like decomposition of waste in landfills, raising livestock (cattle), and cultivating wet rice fields release methane. (b)(ii) Short-wavelength radiation from the Sun passes through the atmosphere. The Earth's surface absorbs this and re-emits longer-wavelength infrared radiation. Greenhouse gases absorb this infrared radiation, preventing it from escaping into space, thereby heating up the atmosphere. (c)(i) Coal and oil contain sulfur impurities; when burned, the sulfur reacts with oxygen to form sulfur dioxide gas. (c)(ii) Sulfur dioxide dissolves in rainwater to form acid rain, which lowers the pH of lakes (harming aquatic life) and leaches nutrients from soils (damaging forests).

评分标准

(a)(i) 78 (%) [1]. (a)(ii) argon [1]. (b)(i) cattle farming / rice growing / landfill sites [1]. (b)(ii) infrared / thermal radiation emitted from Earth [1]; is absorbed / trapped by greenhouse gases [1]; re-radiated back towards Earth, increasing global temperature [1]. (c)(i) combustion of coal / fossil fuels containing sulfur [1]. (c)(ii) any two from: acidifies lakes / kills aquatic life / damages trees / erodes limestone buildings [2].
题目 9 · structured
9
A student investigates the electrical properties of a resistor, \(R\).

(a) (i) State the name of the instrument used to measure the current flowing through resistor \(R\).

(ii) State how a voltmeter must be connected to measure the potential difference across resistor \(R\).

(b) The current in resistor \(R\) is \(0.40\text{ A}\) when the potential difference across it is \(6.0\text{ V}\).

Calculate the resistance of resistor \(R\) and state the unit of your answer.

(c) The student now connects an identical resistor in series with resistor \(R\).

(i) State the effect this has on the total resistance of the circuit.

(ii) Explain how this change in resistance affects the current in the circuit, assuming the potential difference of the power supply remains constant.

(d) Calculate the charge that flows through the circuit when a current of \(0.40\text{ A}\) is maintained for \(2.5\text{ minutes}\). Show your working and state the unit.
查看答案详解

解题

(a) (i) An ammeter is used to measure current.
(ii) A voltmeter must be connected in parallel across the resistor.

(b) Use the resistance formula:
\[R = \frac{V}{I}\]
\[R = \frac{6.0\text{ V}}{0.40\text{ A}} = 15\ \Omega\]
The unit is the ohm (\(\Omega\)).

(c) (i) The total resistance of the circuit increases.
(ii) The current decreases because the resistance is larger, which opposes the flow of charge more for the same potential difference.

(d) Convert time to seconds:
\[t = 2.5\text{ minutes} \times 60\text{ s/minute} = 150\text{ s}\]
Calculate the charge using \(Q = I \times t\):
\[Q = 0.40\text{ A} \times 150\text{ s} = 60\text{ C}\]
The unit is the coulomb (C).

评分标准

(a) (i) ammeter [1]
(ii) in parallel [1]

(b)
- recall of formula \(R = \frac{V}{I}\) or correct substitution \(\frac{6.0}{0.40}\) [1]
- resistance value of 15 [1]
- unit of \(\Omega\) or ohm(s) [1]

(c)
(i) (total resistance) increases / doubles [1]
(ii) current decreases because resistance has increased (for the same voltage) [1]

(d)
- conversion of time to seconds (150 s) seen or implied [1]
- correct calculation of charge (60 C / coulombs) with unit [1]

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