An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V3) Cambridge International A Level Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.
卷二 選擇題 (Extended)
There are forty questions on this paper. Answer all questions. Choose the one you consider correct.
40 題目 · 40 分
題目 1 · 選擇題
1 分
A car of mass 1200 kg accelerates uniformly from rest to 20 m/s in 8.0 s. What is the average useful power developed by the engine to achieve this change in kinetic energy?
First, calculate the change in kinetic energy of the car: \(\Delta E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 1200 \text{ kg} \times (20 \text{ m/s})^2 = 240\,000 \text{ J}\). Then, calculate power: \(P = \frac{W}{t} = \frac{240\,000 \text{ J}}{8.0 \text{ s}} = 30\,000 \text{ W} = 30 \text{ kW}\).
評分準則
1 mark for the correct option C.
題目 2 · 選擇題
1 分
Three resistors of values 6.0 \(\Omega\), 6.0 \(\Omega\), and 3.0 \(\Omega\) are connected together in a circuit. What is the minimum possible combined resistance of this network?
A.1.5 \(\Omega\)
B.3.0 \(\Omega\)
C.5.0 \(\Omega\)
D.15 \(\Omega\)
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解題
The minimum possible resistance is achieved when all three resistors are connected in parallel. Using the parallel resistance formula: \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{6.0} + \frac{1}{6.0} + \frac{1}{3.0} = \frac{1}{6} + \frac{1}{6} + \frac{2}{6} = \frac{4}{6} = \frac{2}{3} \text{ } \Omega^{-1}\). Thus, \(R_p = \frac{3}{2} = 1.5 \text{ } \Omega\).
評分準則
1 mark for the correct option A.
題目 3 · 選擇題
1 分
Which sequence correctly describes the flow of blood through the structures of the human circulatory system?
D.left ventricle \(\rightarrow\) aorta \(\rightarrow\) body tissues \(\rightarrow\) vena cava \(\rightarrow\) left atrium
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解題
Deoxygenated blood leaves the right ventricle via the pulmonary artery to go to the lungs. Oxygenated blood returns from the lungs to the left atrium via the pulmonary vein.
評分準則
1 mark for the correct option A.
題目 4 · 選擇題
1 分
Which features are characteristic of wind-pollinated flowers?
B.small green petals, feathery stigmas, light and smooth pollen grains
C.large petals, stigmas inside the flower, light and smooth pollen grains
D.small green petals, stigmas inside the flower, sticky pollen grains
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解題
Wind-pollinated flowers typically have small, green, or inconspicuous petals (no need to attract insects), feathery stigmas with a large surface area to catch wind-borne pollen, and light, smooth pollen grains that are easily carried by the wind.
評分準則
1 mark for the correct option B.
題目 5 · 選擇題
1 分
An experiment is carried out to investigate the effect of pH on the rate of an enzyme-controlled reaction. Which statement correctly explains why the reaction rate decreases significantly when the pH is changed from the optimum of pH 7 to pH 2?
A.The kinetic energy of the enzyme molecules decreases, causing fewer successful collisions.
B.The enzyme molecules are completely consumed by the substrate at lower pH.
C.The shape of the enzyme's active site is altered, so the substrate can no longer fit.
D.The activation energy of the reaction is lowered, slowing down the rate of product formation.
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解題
Changes in pH away from the optimum alter the 3D structure of the enzyme's active site (denaturation), meaning the substrate can no longer bind to it. Kinetic energy decreases with temperature, not pH.
評分準則
1 mark for the correct option C.
題目 6 · 選擇題
1 分
An ion of element X has a charge of 2+. The ion has 24 electrons and a nucleon number of 58. How many protons and neutrons are in the nucleus of this ion?
A.24 protons and 34 neutrons
B.26 protons and 32 neutrons
C.26 protons and 58 neutrons
D.28 protons and 30 neutrons
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解題
The ion has a 2+ charge and 24 electrons, meaning the neutral atom has 26 electrons. Therefore, the atomic number (number of protons) is 26. Since the nucleon number is 58, the number of neutrons is 58 - 26 = 32 neutrons.
評分準則
1 mark for the correct option B.
題目 7 · 選擇題
1 分
Iron(III) oxide reacts with carbon monoxide according to the equation: \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\). What mass of iron is produced from the complete reaction of 80 g of \(\text{Fe}_2\text{O}_3\)? [Relative atomic masses, \(A_r\): \(\text{O} = 16\), \(\text{Fe} = 56\)]
A.28 g
B.56 g
C.112 g
D.160 g
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解題
The relative formula mass (\(M_r\)) of \(\text{Fe}_2\text{O}_3\) is \(2(56) + 3(16) = 160\). The number of moles of \(\text{Fe}_2\text{O}_3\) is \(80 \text{ g} / 160 \text{ g/mol} = 0.5 \text{ mol}\). According to the stoichiometry of the equation, 1 mole of \(\text{Fe}_2\text{O}_3\) produces 2 moles of \(\text{Fe}\). Thus, 0.5 moles of \(\text{Fe}_2\text{O}_3\) will produce \(0.5 \times 2 = 1.0 \text{ mol}\) of \(\text{Fe}\). The mass of 1.0 mol of iron is \(1.0 \text{ mol} \times 56 \text{ g/mol} = 56 \text{ g}\).
評分準則
1 mark for the correct option B.
題目 8 · 選擇題
1 分
Which statement about the chemical reactions of alkenes is correct?
A.They react with steam in the presence of an acid catalyst to form alkanes.
B.They undergo substitution reactions with bromine water in the presence of UV light.
C.They react with aqueous bromine, causing a colour change from orange to colourless.
D.They burn in excess oxygen to produce carbon monoxide and water.
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解題
Alkenes undergo addition reactions. When reacted with aqueous bromine (bromine water), the bromine adds across the double bond, causing the orange solution to become colourless. Alkenes react with steam to form alcohols, not alkanes. They undergo addition (not substitution) and complete combustion produces carbon dioxide, not carbon monoxide.
評分準則
1 mark for the correct option C.
題目 9 · 選擇題
1 分
An enzyme-catalysed reaction is carried out at different temperatures: 10 °C, 20 °C, 30 °C, 40 °C, and 50 °C. The rate of reaction increases from 10 °C to 40 °C, but is zero at 50 °C. Which statement explains why the rate is zero at 50 °C?
A.The enzyme and substrate molecules have too much kinetic energy to collide.
B.The active site of the enzyme has changed shape, so the substrate can no longer fit.
C.The substrate molecules have denatured and cannot bind to the enzyme.
D.The activation energy of the reaction has decreased to zero.
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解題
At high temperatures (such as 50 °C), thermal energy breaks the bonds that maintain the specific three-dimensional shape of the enzyme. This causes the active site to denature (change shape), so the substrate molecules can no longer fit.
評分準則
1 mark: Identify that the active site has denatured and changed shape, preventing the substrate from fitting.
題目 10 · 選擇題
1 分
A car of mass 1200 kg accelerates uniformly from rest to 20 m/s in 8.0 s. Calculate the average useful power developed by the car's engine during this time.
A.1.5 kW
B.3.0 kW
C.30 kW
D.240 kW
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解題
First, calculate the kinetic energy gained: \(E_k = \frac{1}{2} m v^2 = 0.5 \times 1200 \times 20^2 = 240,000\) J. Next, use the power formula: \(P = \frac{E}{t} = \frac{240,000\text{ J}}{8.0\text{ s}} = 30,000\text{ W} = 30\text{ kW}\).
評分準則
1 mark: Correct calculation of kinetic energy and average power to get 30 kW.
題目 11 · 選擇題
1 分
An isotope of chlorine is represented by the symbol \(^{37}_{17}\text{Cl}^{-}\). Which row shows the correct numbers of protons, neutrons, and electrons in this ion?
A.protons: 17, neutrons: 20, electrons: 16
B.protons: 17, neutrons: 20, electrons: 18
C.protons: 18, neutrons: 19, electrons: 17
D.protons: 17, neutrons: 37, electrons: 18
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解題
The atomic number (bottom number) is 17, so there are 17 protons. The mass number (top number) is 37, so the number of neutrons is \(37 - 17 = 20\). The ion has a 1- charge, which means it has one more electron than protons: \(17 + 1 = 18\) electrons.
A circuit contains a 6.0 V battery connected in parallel to a 12 \(\Omega\) resistor and a 4.0 \(\Omega\) resistor. What is the total current in the circuit?
A.0.38 A
B.1.5 A
C.2.0 A
D.16 A
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解題
The total resistance \(R\) of the two parallel resistors is given by \(\frac{1}{R} = \frac{1}{12} + \frac{1}{4} = \frac{4}{12} = \frac{1}{3}\), which gives \(R = 3.0\ \Omega\). Applying Ohm's law: \(I = \frac{V}{R} = \frac{6.0\text{ V}}{3.0\ \Omega} = 2.0\text{ A}\).
評分準則
1 mark: Correct calculation of equivalent resistance (3.0 ohms) and total current (2.0 A).
題目 13 · 選擇題
1 分
Which sequence shows the correct pathway of water moving through a leaf during transpiration?
Water enters the leaf through xylem vessels, passes into the mesophyll cells, evaporates from the cell walls into the intercellular air spaces, and finally diffuses out through the stomata into the surrounding atmosphere.
評分準則
1 mark: Correct order of tissues and spaces for water transport.
題目 14 · 選擇題
1 分
Calculate the volume of carbon dioxide produced at r.t.p. when 5.0 g of calcium carbonate, \(\text{CaCO}_3\), is completely reacted with excess hydrochloric acid. (Relative atomic masses: \(\text{Ca} = 40\), \(\text{C} = 12\), \(\text{O} = 16\). One mole of gas occupies \(24\text{ dm}^3\) at r.t.p.)
A.0.12 dm\(^3\)
B.1.2 dm\(^3\)
C.2.4 dm\(^3\)
D.12 dm\(^3\)
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解題
Calculate \(M_r\) of \(\text{CaCO}_3\): \(40 + 12 + (3 \times 16) = 100\). Find the number of moles of \(\text{CaCO}_3\): \(\text{moles} = \frac{5.0\text{ g}}{100\text{ g/mol}} = 0.05\text{ mol}\). The mole ratio of \(\text{CaCO}_3\) to \(\text{CO}_2\) is 1:1, so 0.05 mol of \(\text{CO}_2\) is produced. Volume of gas = \(0.05\text{ mol} \times 24\text{ dm}^3\text{/mol} = 1.2\text{ dm}^3\).
評分準則
1 mark: Correct calculation of moles of calcium carbonate and volume of carbon dioxide.
題目 15 · 選擇題
1 分
A sound wave of frequency 170 Hz travels through air at a speed of 340 m/s. What is the wavelength of this sound wave, and is it audible to a healthy human ear?
A.0.50 m, not audible
B.0.50 m, audible
C.2.0 m, not audible
D.2.0 m, audible
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解題
Using the wave speed equation: \(v = f \lambda \implies \lambda = \frac{v}{f} = \frac{340\text{ m/s}}{170\text{ Hz}} = 2.0\text{ m}\). Because the typical audible range of a healthy human ear is from 20 Hz to 20,000 Hz, a frequency of 170 Hz is audible.
評分準則
1 mark: Correct calculation of wavelength (2.0 m) and identification that 170 Hz is audible.
題目 16 · 選擇題
1 分
What is the primary function of the amniotic fluid during pregnancy?
A.to protect the fetus against mechanical shocks
B.to transport oxygen and glucose to the fetus
C.to remove metabolic waste from the fetus
D.to produce progesterone to maintain the uterus lining
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解題
The main function of the amniotic fluid is to cushion the fetus and protect it from mechanical shocks and physical damage. Nutrient delivery and waste removal are primary functions of the placenta and umbilical cord.
評分準則
1 mark: Correctly identifies the protective/shock absorption role of the amniotic fluid.
題目 17 · multiple_choice
1 分
A student investigates the effect of pH on the activity of amylase. The optimum pH of amylase is pH 7. Why does the activity of amylase decrease when the pH is changed from pH 7 to pH 4?
A.The kinetic energy of the amylase molecules decreases.
B.The starch molecules are broken down by the acid before they reach the active site.
C.The shape of the active site of the amylase changes, so the starch cannot bind to it.
D.The activation energy of the reaction is reduced by the acid.
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解題
At pH values away from the optimum, the active site of the enzyme changes shape (denatures). This prevents the substrate (starch) from fitting into the active site, decreasing enzyme activity.
評分準則
1 mark for the correct option (C). Correctly identifying that changes in pH alter the shape of the active site, preventing substrate binding.
題目 18 · multiple_choice
1 分
An electric motor lifts a load of mass \( 40\text{ kg} \) vertically through a height of \( 12\text{ m} \) in a time of \( 6.0\text{ s} \). The gravitational field strength, \( g \), is \( 10\text{ N/kg} \). What is the average useful power output of the motor?
A.\( 80\text{ W} \)
B.\( 480\text{ W} \)
C.\( 800\text{ W} \)
D.\( 4800\text{ W} \)
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解題
Calculate the work done: \( W = mgh = 40\text{ kg} \times 10\text{ N/kg} \times 12\text{ m} = 4800\text{ J} \). Calculate the average power output: \( P = W/t = 4800\text{ J}/6.0\text{ s} = 800\text{ W} \).
評分準則
1 mark for the correct option (C). Method: 1 mark for showing work done = 4800 J and power = 800 W. Reject all incorrect calculations.
題目 19 · multiple_choice
1 分
A uniform metal wire has a resistance of \( 8.0\ \Omega \). Another wire is made of the same metal, has twice the length and half the cross-sectional area of the first wire. What is the resistance of the second wire?
A.\( 2.0\ \Omega \)
B.\( 8.0\ \Omega \)
C.\( 16.0\ \Omega \)
D.\( 32.0\ \Omega \)
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解題
Resistance \( R \) is directly proportional to length \( l \) and inversely proportional to cross-sectional area \( A \): \( R = \rho \frac{l}{A} \). When length is doubled and area is halved: \( R_{new} = \rho \frac{2l}{0.5A} = 4 \rho \frac{l}{A} = 4R = 4 \times 8.0\ \Omega = 32.0\ \Omega \).
評分準則
1 mark for the correct option (D). Method: Recognising that doubling the length doubles the resistance, and halving the area doubles the resistance again, resulting in a fourfold increase. Reject options with incorrect scaling.
題目 20 · multiple_choice
1 分
A solution is tested for biological molecules. A sample of the solution is heated with Benedict's reagent and turns red. Another sample of the solution is treated with biuret solution and remains blue. Which biological molecules are present in the solution?
A.both reducing sugar and protein are present
B.reducing sugar is present, but protein is absent
C.reducing sugar is absent, but protein is present
D.both reducing sugar and protein are absent
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解題
Benedict's reagent turns from blue to red in the presence of reducing sugars. Biuret solution stays blue (does not turn purple) when proteins are absent. Therefore, reducing sugar is present, but protein is absent.
評分準則
1 mark for the correct option (B). Method: Correct identification of positive Benedict's test and negative Biuret test. Reject incorrect interpretations.
題目 21 · multiple_choice
1 分
An atom of an isotope of potassium is represented by \( ^{39}_{19}\text{K} \). What is the structure of a potassium ion, \( \text{K}^+ \), formed from this atom?
A.19 protons, 20 neutrons, 18 electrons
B.19 protons, 20 neutrons, 19 electrons
C.19 protons, 39 neutrons, 18 electrons
D.20 protons, 19 neutrons, 19 electrons
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解題
The atomic number is 19, so there are 19 protons. The mass number is 39, so the number of neutrons is \( 39 - 19 = 20 \). A neutral potassium atom has 19 electrons, so a \( \text{K}^+ \) ion (which has lost one electron) has \( 19 - 1 = 18 \) electrons.
評分準則
1 mark for the correct option (A). Method: 19 protons (atomic number), 20 neutrons (mass - atomic number), 18 electrons (ion has charge +1). Reject options with incorrect subatomic counts.
題目 22 · multiple_choice
1 分
What is the volume of \( 11\text{ g} \) of carbon dioxide gas, \( \text{CO}_2 \), measured at room temperature and pressure (r.t.p.)? [Relative atomic masses: \( \text{C} = 12 \), \( \text{O} = 16 \); 1 mole of any gas occupies \( 24\text{ dm}^3 \) at r.t.p.]
A.\( 1.5\text{ dm}^3 \)
B.\( 6.0\text{ dm}^3 \)
C.\( 12\text{ dm}^3 \)
D.\( 24\text{ dm}^3 \)
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解題
Calculate the relative formula mass of \( \text{CO}_2 \): \( M_r = 12 + (16 \times 2) = 44 \). Determine the number of moles of \( \text{CO}_2 \): \( \text{moles} = 11\text{ g} / 44\text{ g/mol} = 0.25\text{ mol} \). Calculate the volume at r.t.p.: \( \text{volume} = 0.25\text{ mol} \times 24\text{ dm}^3/\text{mol} = 6.0\text{ dm}^3 \).
評分準則
1 mark for the correct option (B). Method: 1 mark for calculating correct molar mass of 44, moles of 0.25, and multiplying by 24 to get 6.0. Reject other answers.
題目 23 · multiple_choice
1 分
A radio wave has a frequency of \( 1.5 \times 10^8\text{ Hz} \). The speed of electromagnetic waves in a vacuum is \( 3.0 \times 10^8\text{ m/s} \). What is the wavelength of this radio wave?
A.\( 0.5\text{ m} \)
B.\( 2.0\text{ m} \)
C.\( 4.5 \times 10^{16}\text{ m} \)
D.\( 2.0 \times 10^{-8}\text{ m} \)
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解題
Use the wave equation: \( v = f \lambda \). Rearranging for wavelength: \( \lambda = v / f = (3.0 \times 10^8\text{ m/s}) / (1.5 \times 10^8\text{ Hz}) = 2.0\text{ m} \).
評分準則
1 mark for the correct option (B). Method: 1 mark for using the wave equation and correctly dividing speed by frequency. Reject options based on multiplication or inverted division.
題目 24 · multiple_choice
1 分
Which row correctly identifies where the hormones estrogen and progesterone are produced, and what their primary function is during the human menstrual cycle?
A.Estrogen is produced in the ovaries and stimulates the repair of the uterus lining.
B.LH is produced in the ovaries and stimulates ovulation.
C.Progesterone is produced in the pituitary gland and maintains the uterus lining.
D.FSH is produced in the uterus and stimulates follicle development.
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解題
Estrogen is produced in the ovaries (developing follicle) and is responsible for repairing and thickening the uterus lining. LH and FSH are produced in the pituitary gland, not the ovaries or uterus. Progesterone is produced in the ovaries (corpus luteum), not the pituitary gland.
評分準則
1 mark for the correct option (A). Method: Correct identification of estrogen production site and its physiological function in the uterus. Reject all incorrect hormone-site combinations.
題目 25 · 選擇題
1 分
A car of mass \(1200\text{ kg}\) accelerates uniformly from rest to a speed of \(20\text{ m/s}\) in a time of \(8.0\text{ s}\). Assuming there are no energy losses, what is the average useful power developed by the engine to accelerate the car?
A.1.5 kW
B.3.0 kW
C.30 kW
D.240 kW
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解題
First, calculate the change in kinetic energy (\(E_k\)) of the car: \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 1200\text{ kg} \times (20\text{ m/s})^2 = 600 \times 400 = 240,000\text{ J}\). Next, calculate the average power (\(P\)) by dividing the work done (which equals the kinetic energy gained) by the time taken: \(P = \frac{E_k}{t} = \frac{240,000\text{ J}}{8.0\text{ s}} = 30,000\text{ W} = 30\text{ kW}\).
評分準則
1 mark for the correct calculation of kinetic energy (\(240,000\text{ J}\)) and dividing by the time (\(8.0\text{ s}\)) to obtain \(30\text{ kW}\).
題目 26 · 選擇題
1 分
A circuit consists of a \(12\text{ V}\) d.c. power supply connected in series with a \(3.0\ \Omega\) resistor and a parallel combination of two \(6.0\ \Omega\) resistors. What is the current flowing through the \(3.0\ \Omega\) resistor?
A.1.0 A
B.2.0 A
C.3.0 A
D.4.0 A
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解題
1. Calculate the equivalent resistance of the two \(6.0\ \Omega\) resistors in parallel: \(R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{6.0 \times 6.0}{6.0 + 6.0} = 3.0\ \Omega\). 2. Calculate the total resistance of the series circuit: \(R_{\text{total}} = 3.0\ \Omega + R_p = 3.0\ \Omega + 3.0\ \Omega = 6.0\ \Omega\). 3. Find the total current leaving the power supply using Ohm's law: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\). Since the \(3.0\ \Omega\) resistor is in series with the parallel combination, the total current of \(2.0\text{ A}\) flows directly through it.
評分準則
1 mark for calculating the total resistance as \(6.0\ \Omega\) and using Ohm's Law to calculate the current of \(2.0\text{ A}\).
題目 27 · 選擇題
1 分
An experiment investigates the rate of reaction of an enzyme-catalysed reaction at different temperatures: \(10^\circ\text{C}\), \(20^\circ\text{C}\), \(30^\circ\text{C}\), \(40^\circ\text{C}\), and \(60^\circ\text{C}\). Which statement correctly explains the difference in the rate of reaction at \(60^\circ\text{C}\) compared to \(30^\circ\text{C}\)?
A.The rate is higher at \(60^\circ\text{C}\) because the molecules have more kinetic energy and collide more frequently.
B.The rate is lower at \(60^\circ\text{C}\) because the enzyme molecules have been denatured, changing the shape of the active site.
C.The rate is the same because enzyme active sites are unaffected by temperature changes.
D.The rate is lower at \(60^\circ\text{C}\) because the substrate molecules have been denatured, changing their active site.
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解題
At \(60^\circ\text{C}\), which is well above the optimum temperature for most enzymes, the high temperature disrupts the bonds holding the enzyme's three-dimensional shape. This denatures the enzyme, permanently altering the shape of its active site so that the substrate can no longer fit. Therefore, the rate of reaction decreases significantly.
評分準則
1 mark for identifying that high temperature denatures enzyme molecules and changes the active site shape.
題目 28 · 選擇題
1 分
A student tests a food sample with three reagents and records the results: - Iodine solution: blue-black colour - Benedict's solution (heated): remains blue - Biuret reagent: purple colour
Which nutrients are present in the food sample?
A.reducing sugar and starch only
B.protein and starch only
C.protein and reducing sugar only
D.protein, reducing sugar and starch
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解題
- Iodine solution turns blue-black in the presence of starch (starch is present). - Benedict's solution remains blue if reducing sugars are absent (reducing sugars are absent). - Biuret reagent turns purple/violet in the presence of protein (protein is present). Therefore, the food sample contains protein and starch only.
評分準則
1 mark for identifying that blue-black iodine indicates starch, blue Benedict's indicates no reducing sugar, and purple Biuret indicates protein.
題目 29 · 選擇題
1 分
An ion of element \(X\) is represented by the symbol \(\text{}^{37}_{17}X^-\). How many protons, neutrons, and electrons are present in this ion?
A.17 protons, 20 neutrons, and 17 electrons
B.17 protons, 20 neutrons, and 18 electrons
C.17 protons, 37 neutrons, and 18 electrons
D.18 protons, 20 neutrons, and 17 electrons
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解題
- The atomic number (proton number) is the lower number, which is 17. So, there are 17 protons. - The mass number (nucleon number) is the upper number, 37. The number of neutrons is \(\text{mass number} - \text{proton number} = 37 - 17 = 20\). - The negative charge of \(-1\) indicates that the ion has one more electron than protons. Therefore, the number of electrons is \(17 + 1 = 18\).
評分準則
1 mark for correctly determining protons (17), neutrons (20), and electrons (18).
題目 30 · 選擇題
1 分
Paper chromatography is carried out on three water-soluble dyes using water as the solvent. The chromatogram is developed and the \(R_f\) values are calculated: - Dye P: \(R_f = 0.25\) - Dye Q: \(R_f = 0.60\) - Dye R: \(R_f = 0.85\)
Which statement about these dyes is correct?
A.Dye R is the least soluble in water and travels the furthest from the baseline.
B.Dye P is the most soluble in water and travels the shortest distance from the baseline.
C.Dye R is the most soluble in water and travels the furthest from the baseline.
D.Dye P is the least soluble in water and travels the furthest from the baseline.
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解題
The retardation factor (\(R_f\)) is the ratio of the distance travelled by the solute to the distance travelled by the solvent front. A higher \(R_f\) value means the dye has travelled further up the chromatogram. Since water is the mobile phase (solvent), the substance that is most soluble in water will travel the furthest, giving it the highest \(R_f\) value. Therefore, Dye R (\(R_f = 0.85\)) is the most soluble in water and travels the furthest.
評分準則
1 mark for linking high \(R_f\) value to high solubility in the solvent and travelling the furthest distance.
題目 31 · 選擇題
1 分
A hydrocarbon compound has the molecular formula \(\text{C}_4\text{H}_8\). Which statement correctly describes the saturation of this hydrocarbon and its reaction with aqueous bromine?
A.It is a saturated hydrocarbon and will decolourise aqueous bromine.
B.It is an unsaturated hydrocarbon and will decolourise aqueous bromine.
C.It is a saturated hydrocarbon and will not decolourise aqueous bromine.
D.It is an unsaturated hydrocarbon and will not decolourise aqueous bromine.
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解題
\(\text{C}_4\text{H}_8\) fits the general formula of alkenes, \(\text{C}_n\text{H}_{2n}\). Alkenes are unsaturated hydrocarbons because they contain a carbon-carbon double bond (\(\text{C}=\text{C}\)). Unsaturated hydrocarbons react rapidly with bromine water (aqueous bromine) in an addition reaction, causing the orange bromine water to become colourless (decolourise).
評分準則
1 mark for identifying \(\text{C}_4\text{H}_8\) as an unsaturated alkene and knowing that unsaturated alkenes decolourise aqueous bromine.
題目 32 · 選擇題
1 分
A ray of light travels from air into a rectangular glass block with a refractive index of 1.50. The angle of incidence in air is \(45^\circ\). What is the angle of refraction in the glass block, to the nearest degree?
A.28°
B.30°
C.42°
D.45°
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解題
Using Snell's Law: \(n = \frac{\sin i}{\sin r}\) where \(n = 1.50\) and \(i = 45^\circ\). Rearranging to find \(\sin r\): \(\sin r = \frac{\sin i}{n} = \frac{\sin 45^\circ}{1.50}\) Since \(\sin 45^\circ \approx 0.707\): \(\sin r = \frac{0.707}{1.50} \approx 0.471\) Calculating the inverse sine: \(r = \arcsin(0.471) \approx 28.1^\circ\). To the nearest degree, the angle of refraction is \(28^\circ\).
評分準則
1 mark for using Snell's Law formula, substituting the given values correctly, and calculating the angle of refraction to be \(28^\circ\).
題目 33 · 選擇題
1 分
A spring has an unstretched length of 12.0 cm. When a load of 4.0 N is suspended from it, the length of the spring becomes 16.0 cm. Assuming the limit of proportionality is not exceeded, what is the length of the spring when a load of 10.0 N is suspended from it?
A.20.0 cm
B.22.0 cm
C.30.0 cm
D.40.0 cm
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解題
First, find the extension caused by the 4.0 N load: \(16.0\text{ cm} - 12.0\text{ cm} = 4.0\text{ cm}\). According to Hooke's law, extension is directly proportional to load below the limit of proportionality. The extension per unit load is \(4.0\text{ cm} / 4.0\text{ N} = 1.0\text{ cm/N}\). For a load of 10.0 N, the extension will be \(10.0\text{ N} \times 1.0\text{ cm/N} = 10.0\text{ cm}\). The new length is the unstretched length plus the extension: \(12.0\text{ cm} + 10.0\text{ cm} = 22.0\text{ cm}\).
評分準則
1 mark for correct calculation of the extension and new length.
題目 34 · 選擇題
1 分
Two resistors, with resistances of 4.0 \(\Omega\) and 12.0 \(\Omega\), are connected in parallel across a 6.0 V d.c. power supply. What is the total current drawn from the supply?
A.0.38 A
B.1.5 A
C.2.0 A
D.8.0 A
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解題
For parallel resistors, the combined resistance \(R\) is given by \(1/R = 1/R_1 + 1/R_2 = 1/4.0 + 1/12.0 = 3/12.0 + 1/12.0 = 4/12.0 = 1/3.0\). Therefore, \(R = 3.0\ \Omega\). Using Ohm's law, the total current \(I\) is \(I = V/R = 6.0\text{ V} / 3.0\ \Omega = 2.0\text{ A}\).
評分準則
1 mark for the correct calculation of the combined resistance and total current.
題目 35 · 選擇題
1 分
Which row correctly matches a blood vessel with the direction of blood flow and the oxygenation status of the blood it carries?
A.pulmonary artery | from heart to lungs | deoxygenated
B.pulmonary vein | from heart to lungs | oxygenated
C.vena cava | from body cells to heart | oxygenated
D.aorta | from lungs to heart | oxygenated
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解題
The pulmonary artery carries deoxygenated blood away from the right ventricle of the heart to the lungs. The pulmonary vein carries oxygenated blood from the lungs back to the left atrium of the heart. The vena cava carries deoxygenated blood from the body to the right atrium. The aorta carries oxygenated blood from the left ventricle to the body.
評分準則
1 mark for identifying the correct blood vessel, flow direction, and oxygenation status.
題目 36 · 選擇題
1 分
Which statement explains why the rate of an enzyme-catalysed reaction decreases rapidly at temperatures above the optimum temperature?
A.The kinetic energy of the substrate molecules decreases as temperature rises.
B.The enzyme molecules are denatured, changing the shape of the active site so that the substrate no longer fits.
C.The substrate molecules are denatured and can no longer bind to the active site.
D.The activation energy of the reaction is increased by the higher temperature.
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解題
At high temperatures above the optimum, the active site of the enzyme molecule undergoes denaturation. The high thermal energy breaks bonds holding the tertiary structure of the protein together, permanently altering the shape of the active site so that it is no longer complementary to the substrate.
評分準則
1 mark for identifying enzyme denaturation and its effect on the active site.
題目 37 · 選擇題
1 分
What is the volume of carbon dioxide gas, measured at room temperature and pressure (r.t.p.), produced when 5.0 g of calcium carbonate is completely decomposed by heating? [Relative atomic masses: Ca = 40, C = 12, O = 16. Molar volume of a gas at r.t.p. = 24 dm]
A.1.2 dm
B.2.4 dm
C.12 dm
D.24 dm
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解題
The equation for the thermal decomposition is: \(CaCO_3 \rightarrow CaO + CO_2\). Relative formula mass of \(CaCO_3 = 40 + 12 + (3 \times 16) = 100\). Number of moles of \(CaCO_3 = 5.0\text{ g} / 100\text{ g/mol} = 0.05\text{ mol}\). Since the molar ratio of \(CaCO_3\) to \(CO_2\) is 1:1, 0.05 mol of \(CO_2\) gas is produced. Volume of \(CO_2 = 0.05\text{ mol} \times 24\text{ dm}^3\text{/mol} = 1.2\text{ dm}^3\).
評分準則
1 mark for correct calculation of moles and final gas volume.
題目 38 · 選擇題
1 分
Ethane and ethene are hydrocarbons. Which reagent can be used to distinguish between them, and what is the observation when this reagent is added to ethene?
A.aqueous bromine | turns from orange-brown to colourless
D.acidified potassium manganate(VII) | turns from colourless to purple
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解題
Aqueous bromine (bromine water) is used to test for unsaturation (double bonds). Alkenes like ethene readily undergo an addition reaction with bromine, decolourising the orange-brown solution to colourless. Alkanes like ethane do not react with bromine water in the absence of UV light, so the solution remains orange-brown.
評分準則
1 mark for selecting the correct reagent and its positive result with the alkene.
題目 39 · 選擇題
1 分
Light travels from air into a rectangular glass block. Which properties of the light change as it enters the glass?
A.frequency and wavelength only
B.speed and wavelength only
C.speed and frequency only
D.speed, frequency and wavelength
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解題
When light passes from a less dense medium (air) to a more dense medium (glass), its speed decreases and its wavelength decreases proportionally. The frequency of the wave depends only on the source and remains constant across the boundary.
評分準則
1 mark for identifying that speed and wavelength change, while frequency remains constant.
題目 40 · 選擇題
1 分
A student tests an aqueous solution of an unknown compound X. Addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. Addition of dilute nitric acid followed by aqueous barium nitrate produces a white precipitate. What is the identity of compound X?
A.chromium(III) sulfate
B.iron(II) sulfate
C.iron(III) sulfate
D.iron(II) chloride
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解題
A green precipitate with aqueous sodium hydroxide that is insoluble in excess confirms the presence of iron(II) ions (\(Fe^{2+}\)). Chromium(III) also produces a green precipitate but it is soluble in excess sodium hydroxide to give a green solution. A white precipitate with dilute nitric acid and aqueous barium nitrate confirms the presence of sulfate ions (\(SO_4^{2-}\)). Therefore, the compound is iron(II) sulfate.
評分準則
1 mark for correctly matching the chemical test results to the iron(II) and sulfate ions.
Paper 4 Theory (Extended)
Answer all questions. Show your working and use appropriate units.
12 題目 · 120 分
題目 1 · Structured
10 分
A cyclist of total mass 80 kg accelerates from rest along a horizontal road. The forward thrust force is 120 N and the air resistance is 40 N.
(a)(i) Calculate the initial acceleration of the cyclist. [2]
(a)(ii) After some time, the cyclist reaches a constant speed of 12 m/s. State the size of the air resistance force at this speed. Explain your answer. [2]
(b) The cyclist then travels up a hill of vertical height 15 m. The total mass is still 80 kg.
(b)(i) Calculate the change in gravitational potential energy of the cyclist. (Use \(g = 9.8\text{ m/s}^2\)). [2]
(b)(ii) The cyclist does 14 000 J of work to climb the hill. Calculate the efficiency of the climb. [2]
(c) State the energy transfer that occurs in the cyclist's muscles to provide this work. [2]
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解題
(a)(i) Net force = Forward force - Resistance force = 120 N - 40 N = 80 N. Acceleration = Net force / mass = 80 N / 80 kg = 1.0 m/s^2. (a)(ii) Air resistance = 120 N. At constant speed, acceleration is zero, which means the forces are balanced and net force is zero. (b)(i) Change in gravitational potential energy = m * g * h = 80 kg * 9.8 m/s^2 * 15 m = 11760 J. (b)(ii) Efficiency = (useful energy output / total energy input) * 100% = (11760 J / 14000 J) * 100% = 84%. (c) Chemical energy (stored in muscles/glucose) is converted into gravitational potential energy (and kinetic energy/thermal energy).
評分準則
(a)(i) - Formula: F = ma or a = F/m [1] - Calculation: (120 - 40) / 80 = 1.0 m/s^2 [1] (a)(ii) - 120 N [1] - Explanation: Constant speed means zero acceleration / balanced forces [1] (b)(i) - Formula: GPE = mgh [1] - Calculation: 80 * 9.8 * 15 = 11760 J (allow 12000 J if g = 10 m/s^2 used) [1] (b)(ii) - Formula: Efficiency = (Useful energy / Total energy) * 100 [1] - Calculation: (11760 / 14000) * 100 = 84% (allow 85.7% or 86% if 12000 J was used) [1] (c) - Chemical energy [1] -> gravitational potential energy / kinetic energy / thermal energy [1]
題目 2 · Structured
10 分
A student sets up a circuit with a 12 V d.c. power supply and two resistors connected in parallel. Resistor R1 has a resistance of 6.0 \(\Omega\), and resistor R2 has a resistance of 3.0 \(\Omega\).
(a)(i) Calculate the combined resistance of the two resistors in parallel. [2]
(a)(ii) Calculate the total current in the circuit. [2]
(a)(iii) Calculate the electrical power dissipated by resistor R1. [2]
(b) A bar magnet is pushed into a solenoid connected to a sensitive center-zero galvanometer.
(b)(i) Describe and explain what is observed on the galvanometer as the magnet enters the solenoid. [2]
(b)(ii) State two ways to increase the magnitude of the deflection on the galvanometer. [2]
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解題
(a)(i) 1 / Rp = 1 / R1 + 1 / R2 = 1 / 6.0 + 1 / 3.0 = 1/6 + 2/6 = 3/6 = 1/2. Therefore, Rp = 2.0 \(\Omega\). (a)(ii) Total current I = V / Rp = 12 V / 2.0 \(\Omega\) = 6.0 A. (a)(iii) The voltage across R1 is 12 V (parallel circuit). Power P = V^2 / R1 = 12^2 / 6.0 = 144 / 6.0 = 24 W. (b)(i) The needle of the galvanometer deflects momentarily to one side and then returns to zero. This is because the moving magnet cuts through the magnetic field lines of the coil, inducing an electromotive force (e.m.f.) and thus a current. (b)(ii) Increase the speed of the magnet; use a stronger magnet; increase the number of turns on the solenoid coil.
評分準則
(a)(i) - Formula: 1/R = 1/R1 + 1/R2 [1] - Calculation: 2.0 \(\Omega\) [1] (a)(ii) - Formula: I = V/R [1] - Calculation: 12 / 2 = 6.0 A (accept ecf from (a)(i)) [1] (a)(iii) - Formula: P = V^2 / R or P = I^2 * R or P = V * I [1] - Calculation: 12^2 / 6 = 24 W [1] (b)(i) - Observation: Momentary deflection of the needle / needle moves and returns to zero [1] - Explanation: A changing magnetic field / magnetic field lines being cut induces an e.m.f./current [1] (b)(ii) - Any two from: move the magnet faster, use a stronger magnet, more turns in the coil, use a coil with a soft iron core [2]
題目 3 · Structured
10 分
An offshore oil exploration vessel uses ultrasound to map the seabed.
(a)(i) Define the term 'ultrasound'. [1]
(a)(ii) The speed of sound in seawater is 1500 m/s. The ship emits a pulse of ultrasound and detects the echo from the seabed 0.80 seconds later. Calculate the depth of the water. [2]
(b) A ray of red light enters a glass block from air. The angle of incidence is 45° and the angle of refraction is 28°.
(b)(i) Calculate the refractive index of the glass block. [2]
(b)(ii) Describe how the wavelength and frequency of the light wave change, if at all, as it enters the glass block. [2]
(c) Infrared radiation and ultraviolet radiation are both parts of the electromagnetic spectrum.
(c)(i) State one hazard of exposure to ultraviolet radiation. [1]
(c)(ii) State one common use of infrared radiation. [1]
(c)(iii) Compare the speeds of infrared and ultraviolet radiation in a vacuum. [1]
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解題
(a)(i) Ultrasound is sound with a frequency higher than 20 000 Hz (the upper limit of human hearing). (a)(ii) Distance travelled by pulse = speed * time = 1500 m/s * 0.80 s = 1200 m. Since the sound travels to the seabed and back, the depth is half the total distance: 1200 m / 2 = 600 m. (b)(i) Refractive index n = sin(i) / sin(r) = sin(45°) / sin(28°) = 0.7071 / 0.4695 = 1.51. (b)(ii) The wavelength decreases; the frequency remains unchanged. (c)(i) Risk of skin cancer, sunburn, or eye damage. (c)(ii) Remote controls, thermal imaging, optical fibers, or infrared heaters. (c)(iii) They both travel at the same speed (the speed of light, 3 * 10^8 m/s) in a vacuum.
評分準則
(a)(i) - Sound of frequency greater than 20 000 Hz / 20 kHz [1] (a)(ii) - Distance = speed * time / use of distance = 1500 * 0.80 = 1200 m [1] - Depth = 1200 / 2 = 600 m [1] (b)(i) - Formula: n = sin(i) / sin(r) [1] - Calculation: sin(45) / sin(28) = 1.51 (accept 1.5) [1] (b)(ii) - Wavelength decreases [1] - Frequency remains unchanged/same [1] (c)(i) - Skin cancer / sunburn / cataracts / cell damage [1] (c)(ii) - Remote controls / thermal imaging / night vision / optical fibers [1] (c)(iii) - They travel at the same speed / speed of light [1]
題目 4 · Structured
10 分
Amylase is an enzyme that catalyzes the breakdown of starch into maltose.
(a)(i) State the chemical elements present in all enzyme molecules. [2]
(a)(ii) Use the lock-and-key hypothesis to explain why amylase can break down starch but cannot break down proteins. [3]
(b) An investigation was carried out to study the effect of temperature on the rate of starch breakdown by amylase.
(b)(i) Explain why the rate of reaction increases as the temperature is raised from 10 °C to 40 °C. [3]
(b)(ii) Explain why the rate of reaction drops rapidly to zero at temperatures above 60 °C. [2]
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解題
(a)(i) Enzymes are proteins, so they contain Carbon, Hydrogen, Oxygen, and Nitrogen. (a)(ii) Amylase has an active site with a specific shape. This shape is complementary only to the starch substrate (the 'key' fits the 'lock'). Proteins have a different shape and cannot fit into the active site of amylase, so no reaction occurs. (b)(i) As temperature increases, the kinetic energy of the enzyme and substrate molecules increases. This causes them to move faster, leading to a higher rate of successful collisions. Consequently, more enzyme-substrate complexes are formed per unit time. (b)(ii) At high temperatures (above 60 °C), the high thermal energy breaks the bonds holding the enzyme's 3D shape together. This changes the shape of the active site (denaturation), so the starch substrate can no longer fit into it.
評分準則
(a)(i) - Carbon, hydrogen, oxygen (any two for [1]) - Nitrogen [1] (a)(ii) - Enzyme/amylase has an active site with a specific shape [1] - Active site shape is complementary to starch / starch fits into the active site [1] - Protein shape is not complementary / does not fit [1] (b)(i) - Molecules gain kinetic energy [1] - Molecules move faster / collide more frequently [1] - More successful collisions / more enzyme-substrate complexes formed per unit time [1] (b)(ii) - High temperature denatures the enzyme [1] - Active site changes shape so substrate can no longer fit [1]
題目 5 · Structured
10 分
The human circulatory system is described as a double circulation.
(a)(i) Explain what is meant by the term 'double circulation'. [2]
(a)(ii) State one advantage of a double circulation compared to a single circulation. [1]
(b) In a vertical section through a human heart:
(b)(i) Identify the chamber of the heart that pumps blood to the lungs and state the name of the blood vessel connected to this chamber. [2]
(b)(ii) Explain why the wall of the left ventricle is much thicker than the wall of the right ventricle. [2]
(c) Coronary heart disease (CHD) is a major cause of death worldwide.
(c)(i) Describe how blockages in the coronary arteries can lead to a heart attack. [2]
(c)(ii) State one dietary change that can help prevent CHD. [1]
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解題
(a)(i) Double circulation means that for every complete circuit of the body, blood passes through the heart twice. It involves two separate circuits: the pulmonary circulation (to the lungs) and the systemic circulation (to the rest of the body). (a)(ii) It maintains high blood pressure to the body tissues, allowing faster delivery of oxygen and nutrients. (b)(i) Chamber: Right ventricle. Blood vessel: Pulmonary artery. (b)(ii) The left ventricle must pump blood to the entire body, which requires a much higher pressure and force. The right ventricle only pumps blood to the lungs, which are nearby and delicate, requiring much lower pressure. (c)(i) Blockages restrict or stop blood flow through the coronary arteries, reducing the supply of oxygen and glucose to the heart muscle cells. This prevents aerobic respiration, leading to cell death and a heart attack. (c)(ii) Reduce intake of saturated fats/cholesterol, or increase fiber/intake of fresh fruit and vegetables.
評分準則
(a)(i) - Blood passes through the heart twice for each complete circuit of the body [1] - Mention of two circuits: pulmonary (to lungs) and systemic (to body) [1] (a)(ii) - Maintains high blood pressure / faster delivery of oxygen / more efficient transport [1] (b)(i) - Chamber: Right ventricle [1] - Vessel: Pulmonary artery [1] (b)(ii) - Left ventricle pumps blood all around the body / further distance [1] - Needs to generate higher pressure / force [1] (c)(i) - Less blood/oxygen/glucose reaching heart muscle [1] - Heart muscle cells cannot respire (aerobically) / cells die [1] (c)(ii) - Reduce intake of saturated fats / cholesterol / salt OR increase fiber intake [1]
題目 6 · Structured
10 分
Carbohydrates, proteins, and lipids (fats and oils) are important biological molecules.
(a)(i) State the smaller basic units (monomers) that make up a protein. [1]
(a)(ii) Describe the chemical test used to confirm the presence of proteins in a sample of food. Include the initial and positive result colors. [3]
(b) Describe how you would test a food sample for the presence of reducing sugars (such as glucose). Describe the expected color changes for a positive result. [3]
(c) Lipids are high-energy molecules.
(c)(i) State the components that make up a lipid molecule. [2]
(c)(ii) State the name of the chemical test used to show the presence of fats and oils. [1]
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解題
(a)(i) Amino acids. (a)(ii) Add Biuret reagent (or sodium hydroxide and copper sulfate solutions) to the food sample. The initial color of the reagent is blue, and it turns purple/violet if protein is present. (b) Add Benedict's solution to the food sample, then heat the mixture in a water bath (about 80 °C) for a few minutes. If reducing sugars are present, the solution changes color from blue to green, yellow, orange, or brick-red (depending on the concentration). (c)(i) Three fatty acids and one glycerol molecule. (c)(ii) Ethanol emulsion test.
評分準則
(a)(i) - Amino acids [1] (a)(ii) - Add Biuret reagent / Biuret solution (or copper sulfate + sodium hydroxide) [1] - Initial color: blue [1] - Positive result: purple / violet / lilac [1] (b) - Add Benedict's reagent / solution [1] - Heat / place in hot water bath [1] - Color change from blue to green / yellow / orange / brick-red [1] (c)(i) - Glycerol [1] - Fatty acids (accept 3 fatty acids) [1] (c)(ii) - Ethanol emulsion test [1]
題目 7 · Structured
10 分
Hydrocarbon X contains 85.7% carbon and 14.3% hydrogen by mass.
(a)(i) Determine the empirical formula of hydrocarbon X. Show your working. [Relative atomic masses: \(A_r(C) = 12\), \(A_r(H) = 1\)] [3]
(a)(ii) The relative molecular mass (\(M_r\)) of X is 56. Deduce the molecular formula of X. [1]
(b) Calcium carbonate (\(\text{CaCO}_3\)) decomposes when heated to form calcium oxide (\(\text{CaO}\)) and carbon dioxide (\(\text{CO}_2\)) according to the equation:
(b)(i) Calculate the mass of calcium oxide produced when 25.0 g of calcium carbonate is completely decomposed. [Relative atomic masses: \(A_r(\text{Ca}) = 40\), \(A_r(\text{C}) = 12\), \(A_r(\text{O}) = 16\)] [3]
(b)(ii) Calculate the volume of carbon dioxide gas, in \(\text{dm}^3\), produced at room temperature and pressure (r.t.p.) from this reaction. [One mole of any gas occupies \(24\text{ dm}^3\) at r.t.p.] [3]
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解題
(a)(i) For Carbon: 85.7 / 12 = 7.14 mol. For Hydrogen: 14.3 / 1 = 14.3 mol. Divide by the smallest value (7.14): Carbon = 7.14 / 7.14 = 1. Hydrogen = 14.3 / 7.14 = 2. Empirical formula is \(\text{CH}_2\). (a)(ii) Empirical formula mass of \(\text{CH}_2\) = 12 + (2 * 1) = 14. Factor = \(M_r\) / empirical mass = 56 / 14 = 4. Molecular formula is \(\text{C}_4\text{H}_8\). (b)(i) \(M_r\) of \(\text{CaCO}_3\) = 40 + 12 + (3 * 16) = 100. \(M_r\) of \(\text{CaO}\) = 40 + 16 = 56. Moles of \(\text{CaCO}_3\) = mass / \(M_r\) = 25.0 / 100 = 0.25 mol. From the equation, 1 mole of \(\text{CaCO}_3\) produces 1 mole of \(\text{CaO}\). So, moles of \(\text{CaO}\) produced = 0.25 mol. Mass of \(\text{CaO}\) = 0.25 mol * 56 g/mol = 14.0 g. (b)(ii) From the equation, 1 mole of \(\text{CaCO}_3\) produces 1 mole of \(\text{CO}_2\). Moles of \(\text{CO}_2\) produced = 0.25 mol. Volume of \(\text{CO}_2\) = moles * 24 \(\text{dm}^3\)/mol = 0.25 * 24 = 6.0 \(\text{dm}^3\).
評分準則
(a)(i) - Moles of C = 85.7 / 12 = 7.14 and Moles of H = 14.3 / 1 = 14.3 [1] - Divide by smallest to get ratio: C = 1, H = 2 [1] - Empirical formula: \(\text{CH}_2\) [1] (a)(ii) - \(\text{C}_4\text{H}_8\) [1] (b)(i) - Moles of \(\text{CaCO}_3\) = 25.0 / 100 = 0.25 mol [1] - Mole ratio is 1:1, so moles of \(\text{CaO}\) = 0.25 mol [1] - Mass of \(\text{CaO}\) = 0.25 * 56 = 14.0 g [1] (b)(ii) - Moles of \(\text{CO}_2\) = 0.25 mol [1] - Formula: Volume = moles * 24 [1] - Calculation: 0.25 * 24 = 6.0 \(\text{dm}^3\) (accept ecf) [1]
題目 8 · Structured
10 分
A mixture contains sand, salt (sodium chloride), and water.
(a)(i) Describe how pure sand and dry salt can be obtained from this mixture. [3]
(b) Paper chromatography is used to separate and identify a mixture of food colorings.
(b)(i) Explain how the Rf value of a dye is calculated. [2]
(b)(ii) Explain why the starting line in paper chromatography must be drawn in pencil rather than ink. [1]
(c) A student is given a colorless solution containing a single salt.
(c)(i) The student adds dilute sodium hydroxide solution to a sample of the solution. A green precipitate is formed, which is insoluble in excess sodium hydroxide. Identify the cation present in the solution. [1]
(c)(ii) The student adds dilute nitric acid followed by silver nitrate solution to another sample of the solution. A cream-colored precipitate is formed. Identify the anion present in the solution. [1]
(c)(iii) State the name of the gas that turns damp red litmus paper blue, and describe a test to confirm the presence of this gas. [2]
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解題
(a)(i) Filter the mixture. The sand remains on the filter paper as residue. Wash the sand with distilled water and dry it. Heat the remaining filtrate (salt solution) in an evaporating basin to evaporate the water, leaving behind dry salt. (b)(i) The Rf value is calculated by dividing the distance travelled by the dye by the distance travelled by the solvent front. (b)(ii) Ink is soluble in the solvent and would separate/run, interfering with the chromatogram. Pencil (graphite) is insoluble and will not run. (c)(i) Iron(II) ion (\(\text{Fe}^{2+}\)). (c)(ii) Bromide ion (\(\text{Br}^-\)). (c)(iii) Ammonia gas. Test: Hold a glass rod dipped in concentrated hydrochloric acid near the gas; dense white fumes of ammonium chloride are produced (or describe the pungent smell/litmus paper test).
評分準則
(a)(i) - Filtration to separate sand as residue [1] - Wash and dry sand [1] - Evaporation of filtrate to obtain salt crystals/dry salt [1] (b)(i) - Formula: distance moved by substance / distance moved by solvent front [1] - Mention of measuring from the baseline [1] (b)(ii) - Pencil lead/graphite is insoluble / ink is soluble and would spread/run [1] (c)(i) - Iron(II) / \(\text{Fe}^{2+}\) [1] (c)(ii) - Bromide / \(\text{Br}^-\) [1] (c)(iii) - Ammonia [1] - Test: turns damp red litmus paper blue / produces dense white fumes with HCl [1]
題目 9 · Structured
10 分
A student studies the mammalian circulatory system. (a) Describe what is meant by a double circulation and state one physiological advantage of this type of circulation. (4 marks) (b) (i) Identify the chamber of the human heart that pumps blood to the lungs, and state the name of the blood vessel carrying this blood. (2 marks) (ii) Explain why the muscle wall of the left ventricle is thicker than that of the right ventricle. (2 marks) (c) Describe two differences in structure between an artery and a vein. (2 marks)
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解題
(a) In a double circulation, blood passes through the heart twice for each complete circuit of the body: once through the pulmonary circulation (to the lungs) and once through the systemic circulation (to the rest of the body). The advantage is that blood can be pumped to the body tissues at a higher pressure, which increases the rate of oxygen and glucose delivery to active tissues for aerobic respiration. (b)(i) The right ventricle pumps deoxygenated blood to the lungs. The blood vessel carrying this blood is the pulmonary artery. (b)(ii) The left ventricle has a thicker muscle wall because it must generate enough force/pressure to pump blood all around the body (systemic circulation) against greater resistance. The right ventricle only pumps blood to the lungs, which are nearby and require lower pressure to prevent tissue damage. (c) Two structural differences: 1. Arteries have thicker walls of muscle and elastic tissue than veins. 2. Arteries have a narrower lumen than veins. 3. Veins contain valves to prevent the backflow of blood, whereas arteries do not have valves (except semilunar valves at the exit of the heart).
評分準則
Part (a): [Total 4 marks] - Blood passes through the heart twice for one complete circuit [1 mark]. - One circuit goes to the lungs and the other goes to the rest of the body [1 mark]. - Physiological advantage: allows blood to be pumped at a higher pressure to the body tissues [1 mark]. - Result: faster delivery of oxygen/glucose OR more efficient cellular respiration [1 mark]. Part (b)(i): [Total 2 marks] - Right ventricle [1 mark]. - Pulmonary artery [1 mark]. Part (b)(ii): [Total 2 marks] - Left ventricle pumps blood to the rest of the body (systemic circulation) [1 mark]. - Requires higher pressure to overcome greater resistance/longer distance [1 mark]. Part (c): [Total 2 marks] - Any two of: Arteries have thicker muscular/elastic walls than veins [1 mark]; Arteries have narrower lumens than veins [1 mark]; Veins have valves, arteries do not [1 mark]. (Maximum 2 marks)
題目 10 · Structured
10 分
A student sets up a series circuit containing a 12 V d.c. power supply, a resistor of resistance 6.0 ohms, and a filament lamp. (a) (i) Calculate the total resistance of the circuit when the lamp has a resistance of 4.0 ohms. (1 mark) (ii) Calculate the current in the circuit. (2 marks) (b) The lamp of resistance 4.0 ohms and the resistor of resistance 6.0 ohms are now connected in parallel with each other across the 12 V power supply. (i) Show that the combined resistance of this parallel combination is 2.4 ohms. (2 marks) (ii) Calculate the total power supplied by the 12 V power supply to this parallel circuit. (2 marks) (c) Explain, in terms of the particles in the resistor, how the electric current causes a rise in the temperature of the resistor. (3 marks)
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解題
(a)(i) In a series circuit, total resistance is the sum of the individual resistances: \(R_T = R_1 + R_2 = 6.0\text{ }\Omega + 4.0\text{ }\Omega = 10.0\text{ }\Omega\). (a)(ii) Using Ohm's law: \(I = V / R_T = 12\text{ V} / 10.0\text{ }\Omega = 1.2\text{ A}\). (b)(i) For a parallel combination, \(1/R_p = 1/R_1 + 1/R_2 = 1/6.0 + 1/4.0 = 2/12 + 3/12 = 5/12\). Thus, \(R_p = 12 / 5 = 2.4\text{ }\Omega\). (b)(ii) Power can be calculated using \(P = V^2 / R = 12^2 / 2.4 = 144 / 2.4 = 60\text{ W}\) (or alternatively, total current \(I_t = V / R_p = 12 / 2.4 = 5.0\text{ A}\), so \(P = V \times I_t = 12 \times 5.0 = 60\text{ W}\)). (c) An electric current consists of a flow of free (delocalised) electrons. As these electrons drift through the metal structure of the resistor, they collide with the positive metal ions in the lattice. These collisions transfer kinetic energy from the electrons to the lattice ions. This causes the ions to vibrate with greater amplitude, which represents an increase in the internal thermal energy of the resistor, resulting in a temperature rise.
評分準則
Part (a)(i): [Total 1 mark] - \(10.0\text{ }\Omega\) (allow 10) [1 mark]. Part (a)(ii): [Total 2 marks] - Formula: \(I = V / R\) [1 mark]. - Correct calculation: \(1.2\text{ A}\) [1 mark]. Part (b)(i): [Total 2 marks] - Formula: \(1/R = 1/R_1 + 1/R_2\) or \(R = (R_1 \times R_2) / (R_1 + R_2)\) [1 mark]. - Correct substitution showing calculation leading to \(2.4\text{ }\Omega\) [1 mark]. Part (b)(ii): [Total 2 marks] - Formula: \(P = V^2 / R\) or \(P = V \times I\) [1 mark]. - Correct calculation: \(60\text{ W}\) [1 mark]. Part (c): [Total 3 marks] - Current is the flow of free/delocalised electrons [1 mark]. - Electrons collide with positive ions in the lattice [1 mark]. - Energy is transferred, causing the ions to vibrate more / with greater amplitude (increasing thermal energy) [1 mark].
題目 11 · Structured
10 分
A student reacts 2.40 g of magnesium ribbon with excess dilute hydrochloric acid according to the equation: \(\text{Mg (s)} + 2\text{HCl (aq)} \rightarrow \text{MgCl}_2\text{ (aq)} + \text{H}_2\text{ (g)}\). (a) Calculate the number of moles in 2.40 g of magnesium. [Relative atomic mass: Mg = 24.0] (2 marks) (b) Calculate the minimum mass of hydrochloric acid (\(\text{HCl}\)) needed to react completely with 2.40 g of magnesium. [Relative formula mass: \(\text{HCl}\) = 36.5] (3 marks) (c) Calculate the volume of hydrogen gas, in \(\text{dm}^3\), produced at room temperature and pressure (r.t.p.). [Molar volume of a gas at r.t.p. is \(24.0\text{ dm}^3/\text{mol}\)] (2 marks) (d) The student collected \(2.16\text{ dm}^3\) of hydrogen gas. Calculate the percentage yield of hydrogen gas. (3 marks)
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解題
(a) \(\text{Moles of Mg} = \text{mass} / \text{Ar} = 2.40\text{ g} / 24.0 = 0.10\text{ mol}\). (b) According to the balanced equation, 1 mole of \(\text{Mg}\) reacts with 2 moles of \(\text{HCl}\). Therefore, \(\text{moles of HCl required} = 2 \times 0.10\text{ mol} = 0.20\text{ mol}\). \(\text{Mass of HCl} = \text{moles} \times \text{Mr} = 0.20\text{ mol} \times 36.5 = 7.30\text{ g}\). (c) According to the equation, 1 mole of \(\text{Mg}\) produces 1 mole of \(\text{H}_2\). Therefore, \(0.10\text{ mol}\) of \(\text{Mg}\) produces \(0.10\text{ mol}\) of \(\text{H}_2\). \(\text{Volume of H}_2 = \text{moles} \times 24.0\text{ dm}^3/\text{mol} = 0.10 \times 24.0 = 2.40\text{ dm}^3\). (d) \(\text{Percentage yield} = (\text{actual yield} / \text{theoretical yield}) \times 100\% = (2.16 / 2.40) \times 100\% = 90.0\%\).
評分準則
Part (a): [Total 2 marks] - Formula: \(\text{moles} = \text{mass} / \text{Ar}\) [1 mark]. - Correct calculation: \(0.10\text{ mol}\) [1 mark]. Part (b): [Total 3 marks] - Correct mole ratio: \(\text{moles of HCl} = 2 \times \text{moles of Mg} = 0.20\text{ mol}\) [1 mark]. - Formula: \(\text{mass} = \text{moles} \times \text{Mr}\) [1 mark]. - Correct calculation: \(7.30\text{ g}\) [1 mark]. Part (c): [Total 2 marks] - Deduction that \(0.10\text{ mol}\) of \(\text{H}_2\) is produced [1 mark]. - Correct calculation: \(2.40\text{ dm}^3\) [1 mark]. Part (d): [Total 3 marks] - Formula for percentage yield: \((\text{actual} / \text{theoretical}) \times 100\%\) [1 mark]. - Correct substitution: \(2.16 / 2.40\) [1 mark]. - Correct calculation: \(90.0\%\) [1 mark].
題目 12 · Structured
10 分
An electromagnetic wave has a frequency of \(6.0 \times 10^{14}\text{ Hz}\) in a vacuum. (a) (i) State the speed of electromagnetic waves in a vacuum. (1 mark) (ii) Calculate the wavelength of this wave in a vacuum. State the formula used and show your working. (3 marks) (b) Identify the region of the electromagnetic spectrum to which this wave belongs. (1 mark) (c) The wave enters a block of transparent glass. The speed of the wave in the glass is \(2.0 \times 10^8\text{ m/s}\). (i) State the frequency of the wave inside the glass. (1 mark) (ii) Calculate the refractive index of the glass. (2 marks) (d) State two differences between electromagnetic waves and sound waves. (2 marks)
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解題
(a)(i) The speed of all electromagnetic waves in a vacuum is \(3.0 \times 10^8\text{ m/s}\). (a)(ii) Using the wave equation: \(v = f \lambda\), so \(\lambda = v / f\). Wavelength \(\lambda = (3.0 \times 10^8\text{ m/s}) / (6.0 \times 10^{14}\text{ Hz}) = 5.0 \times 10^{-7}\text{ m}\) (or \(500\text{ nm}\)). (b) A wavelength of \(5.0 \times 10^{-7}\text{ m}\) belongs to the visible light region of the electromagnetic spectrum. (c)(i) The frequency of a wave depends on the source and does not change when it enters a new medium. Therefore, the frequency in glass is \(6.0 \times 10^{14}\text{ Hz}\). (c)(ii) Refractive index \(n = \text{speed in vacuum} / \text{speed in medium} = (3.0 \times 10^8\text{ m/s}) / (2.0 \times 10^8\text{ m/s}) = 1.5\). (d) Two differences: 1. Electromagnetic waves are transverse waves, while sound waves are longitudinal waves. 2. Electromagnetic waves can travel through a vacuum, whereas sound waves require a physical medium to travel through (cannot travel through a vacuum).
評分準則
Part (a)(i): [Total 1 mark] - \(3.0 \times 10^8\text{ m/s}\) [1 mark]. Part (a)(ii): [Total 3 marks] - Formula: \(v = f \lambda\) or equivalent [1 mark]. - Substitution: \(3.0 \times 10^8 / 6.0 \times 10^{14}\) [1 mark]. - Correct calculation with unit: \(5.0 \times 10^{-7}\text{ m}\) [1 mark]. Part (b): [Total 1 mark] - Visible light (accept green/yellow light) [1 mark]. Part (c)(i): [Total 1 mark] - \(6.0 \times 10^{14}\text{ Hz}\) (frequency remains unchanged) [1 mark]. Part (c)(ii): [Total 2 marks] - Formula: \(n = c / v\) [1 mark]. - Correct calculation: \(1.5\) (no unit) [1 mark]. Part (d): [Total 2 marks] - EM waves are transverse AND sound waves are longitudinal [1 mark]. - EM waves can travel in a vacuum AND sound waves cannot / need a medium [1 mark].
Paper 6 Alternative to Practical
Answer all questions. Write your answers in the spaces provided.
6 題目 · 60 分
題目 1 · Practical Structured
10 分
A student investigates the effect of pH on the rate of starch breakdown by amylase. Starch and amylase are mixed in buffer solutions of different pH values: pH 5.0, pH 6.0, and pH 7.0.
Every 30 seconds, a sample of the mixture is removed and added to one drop of iodine solution on a white spotting tile. This is repeated until the mixture no longer changes the color of the iodine solution.
(a) Name the reagent used to test for starch and state the color change when starch is present. [2]
(b) The student counts the number of blue-black spots formed before the iodine remains yellow-brown. Each spot represents a time interval of 30 seconds. - For pH 5.0, there are 5 blue-black spots. - For pH 6.0, there are 2 blue-black spots. - For pH 7.0, there are 4 blue-black spots.
Calculate the time taken for starch to be completely digested at each pH: [3] - Time for pH 5.0 = ........ s - Time for pH 6.0 = ........ s - Time for pH 7.0 = ........ s
(c) For this investigation, identify: (i) the independent variable. [1] (ii) two variables that must be controlled to ensure a fair test. [2]
(d) Describe how the student could modify the experiment to find a more precise value for the optimum pH. [2]
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解題
(a) The reagent for starch testing is iodine solution [1]. A positive result is a change from yellow-brown/orange to blue-black [1]. (b) Multiply the number of spots by 30 seconds: - pH 5.0: 5 * 30 = 150 s [1] - pH 6.0: 2 * 30 = 60 s [1] - pH 7.0: 4 * 30 = 120 s [1] (c) (i) The independent variable is the one being changed: pH [1]. (ii) Controlled variables include temperature [1] and volume/concentration of starch or amylase [1]. (d) To find a more precise optimum pH, the student should test more pH values at smaller intervals [1] between pH 5.0 and pH 7.0 [1].
評分準則
- (a) Iodine solution [1]; blue-black [1] (reject: blue, black alone; accept: blue/black) - (b) 150 [1], 60 [1], 120 [1] - (c)(i) pH [1] - (c)(ii) Any two from: temperature, volume of starch, concentration of starch, volume of amylase, concentration of amylase [2] (1 mark each) - (d) Test smaller pH intervals / steps [1]; in the range of pH 5.0 to 7.0 [1]
題目 2 · Practical Structured
10 分
A student carries out a series of tests to identify the ions in a green salt, Salt X. Salt X contains one cation and two anions.
(a) Salt X is dissolved in distilled water to form Solution Y. (i) To a 2 \(\text{cm}^3\) sample of Solution Y, aqueous sodium hydroxide is added dropwise until in excess. A green precipitate is formed, which remains insoluble in excess. Identify the cation present. [1] (ii) To another 2 \(\text{cm}^3\) sample of Solution Y, dilute nitric acid is added, followed by aqueous silver nitrate. A cream precipitate is formed. Identify the first anion present. [1] (iii) To a third 2 \(\text{cm}^3\) sample of Solution Y, dilute hydrochloric acid is added, followed by aqueous barium chloride. A white precipitate is formed. Identify the second anion present. [1]
(b) Explain why dilute nitric acid must be added before aqueous silver nitrate in test (a)(ii). [1]
(c) Solution Y contains some undissolved green residue. Describe how the student can separate the undissolved solid to obtain a clear, green filtrate. [2]
(d) Plan an experiment to show that Salt X is a hydrated salt (contains water of crystallization). In your plan, you should: - list the apparatus required, - describe the procedure, - explain how the measurements or observations would show that Salt X is hydrated. [4]
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解題
(a) (i) Iron(II) ions (\(\text{Fe}^{2+}\)) form a green precipitate with sodium hydroxide that is insoluble in excess [1]. (ii) Bromide ions (\(\text{Br}^-\)) form a cream-colored precipitate with acidified silver nitrate [1]. (iii) Sulfate ions (\(\text{SO}_4^{2-}\)) form a white precipitate with acidified barium chloride [1]. (b) Nitric acid destroys any carbonate ions (\(\text{CO}_3^{2-}\)) that might be present. Carbonates form an insoluble white precipitate with silver ions, which would interfere with the halide test [1]. (c) Use filtration: Place filter paper in a funnel [1]. Pour the mixture through and collect the filtrate in a flask or beaker [1]. (d) Apparatus: Crucible, Bunsen burner, tripod, clay pipe triangle, balance (or test-tube, delivery tube, etc.) [1]. Procedure: Weigh the solid before heating [1]. Heat the solid strongly [1]. Observation/Measurement: Reweigh after heating to show a decrease in mass (loss of water) OR observe condensation on the cooler upper walls of the tube [1].
評分準則
- (a)(i) Iron(II) / \(\text{Fe}^{2+}\) [1] (reject: iron / Fe) - (a)(ii) Bromide / \(\text{Br}^-\)[1] - (a)(iii) Sulfate / \(\text{SO}_4^{2-}\) [1] - (b) To remove carbonate ions / prevent false positive precipitate of silver carbonate [1] - (c) Filter paper and funnel [1]; collect filtrate in beaker/flask [1] - (d) Crucible / test-tube / boiling tube and balance / Bunsen burner [1]; weigh mass before heating [1]; heat strongly [1]; reweigh to show loss of mass / observe water droplets condensing [1]
題目 3 · Practical Structured
10 分
A student investigates how the resistance of a wire depends on its length.
(a) Draw a circuit diagram showing a power source, an ammeter, a voltmeter, and a length of resistance wire connected with a sliding contact (jockey). The voltmeter must be connected to measure the potential difference across the length of wire being tested. [2]
(b) The student measures the potential difference \(V\) and current \(I\) for different lengths \(L\) of the wire. For \(L = 40.0\text{ cm}\): - The voltmeter reading is 1.80 V. - The ammeter reading is 0.40 A.
Calculate the resistance \(R\) of the 40.0 cm length of wire using the formula: \[R = \frac{V}{I}\]
\(R\) = ........ \(\Omega\) [1]
(c) The student plots a graph of Resistance \(R\) (on the y-axis) against Length \(L\) (on the x-axis). (i) State the features of the line of best fit that would show that resistance is directly proportional to length. [2] (ii) Explain why the line might not pass exactly through the origin, even if the variables are directly proportional. [1]
(d) During the experiment, the wire can become hot. (i) State one safety precaution the student should take because of this. [1] (ii) Explain how a hot wire could affect the accuracy of the resistance measurements. [1] (iii) Describe a practical action the student can take to minimize this heating effect. [2]
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解題
(a) The circuit diagram must show: - Power supply, ammeter, and resistance wire with sliding contact in a single series loop [1]. - Voltmeter connected in parallel across the resistance wire [1]. (b) \(R = V/I = 1.80 / 0.40 = 4.5\text{ }\Omega\) [1]. (c) (i) Directly proportional means the graph must be a straight line [1] and pass through the origin \((0,0)\) [1]. (ii) Contact resistance at the terminals/connections means resistance is not zero when length is zero [1]. (d) (i) Turn off the current between readings to prevent burns / avoid touching the hot wire [1]. (ii) Resistance of a metal increases as temperature increases [1]. (iii) Keep the current low [1] and open the switch / turn off power when not taking measurements [1].
評分準則
- (a) Correct symbols for power supply/cell, ammeter in series, variable resistance wire, and sliding contact [1]; Voltmeter connected in parallel across the active part of the wire [1] - (b) 4.5 (or 4.50) [1] - (c)(i) Straight line [1]; passing through origin [1] - (c)(ii) Contact resistance / resistance of connecting leads [1] - (d)(i) Wear insulated gloves / do not touch wire / switch off when not reading [1] - (d)(ii) Resistance increases with temperature [1] - (d)(iii) Use low currents [1]; turn off switch between readings [1]
題目 4 · Practical Structured
10 分
A student tests a clear fruit extract to identify the biological molecules present.
(a) Complete Table 4.1 by filling in the missing reagent, conditions, or positive observation for each food test. [5]
Table 4.1: | Molecule tested | Reagent used | Condition (if any) | Observation for positive result | | :--- | :--- | :--- | :--- | | Reducing Sugar | **(i)** ................ | Heat in water bath | **(ii)** ................ | | Protein | Biuret reagent | Room temperature | **(iii)** ................ | | Starch | **(iv)** ................ | Room temperature | **(v)** ................ |
(b) The student also wants to test the fruit extract for the presence of lipids (fats). (i) Name the test used to identify lipids. [1] (ii) Describe the steps of the procedure for this lipid test. [2] (iii) State the positive observation for this test. [1]
(c) State one safety precaution, other than wearing safety goggles, that should be taken when performing the reducing sugar test. [1]
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解題
(a) - (i) Benedict's reagent is used to test for reducing sugars [1]. - (ii) A positive result for reducing sugar is a change to green, yellow, orange, or red precipitate [1]. - (iii) Biuret reagent changes from blue to purple/lilac/violet in the presence of protein [1]. - (iv) Iodine solution is used to test for starch [1]. - (v) Starch turns iodine blue-black [1]. (b) (i) The test is the emulsion test [1]. (ii) Dissolve the lipid in ethanol [1], then pour this liquid into water [1]. (iii) A positive result is a milky-white / cloudy emulsion [1]. (c) Use a water bath to avoid heating flammable ethanol/reagents with a naked flame, or use test-tube holders to avoid burns [1].
評分準則
- (a)(i) Benedict's [1] - (a)(ii) Red / orange / yellow / green (precipitate) [1] - (a)(iii) Purple / lilac / violet / mauve [1] (reject: blue, pink) - (a)(iv) Iodine (solution) [1] - (a)(v) Blue-black [1] (reject: blue, black alone) - (b)(i) Emulsion test [1] - (b)(ii) Add ethanol and shake/mix [1]; pour/add to water [1] - (b)(iii) Cloudy / white / milky emulsion [1] - (c) Use water bath / use test-tube holders / keep ethanol away from open flames [1]
題目 5 · Practical Structured
10 分
A student investigates the temperature change during the neutralization of dilute hydrochloric acid with aqueous sodium hydroxide.
25.0 \(\text{cm}^3\) of dilute hydrochloric acid is placed in a polystyrene cup. The student records the initial temperature of the acid, which is 21.5 °C. Different volumes of aqueous sodium hydroxide are added. After each addition, the mixture is stirred, and the maximum temperature reached is recorded.
(a) In one run, when 20.0 \(\text{cm}^3\) of sodium hydroxide is added, the thermometer shows a maximum temperature of 28.2 °C. (i) Calculate the temperature rise, \(\Delta T\), for this volume. [1]
\(\Delta T\) = ........ °C
(ii) Explain why a polystyrene cup is used instead of a glass beaker. [1]
(i) Plot a graph of Temperature rise \(\Delta T\) (y-axis) against Volume of NaOH added (x-axis) on grid paper. [3] (ii) Draw two straight lines of best fit: one through the first three points (increasing section) and one through the last four points (decreasing section). Extrapolate both lines until they intersect. [1] (iii) Use your graph to find the volume of sodium hydroxide that corresponds to the maximum temperature rise (the intersection point). [1]
Volume of NaOH = ........ \(\text{cm}^3\)
(c) (i) Explain why the temperature rises during the initial additions of sodium hydroxide. [1] (ii) Explain why the temperature decreases when volumes of sodium hydroxide greater than the intersection volume are added. [2]
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解題
(a) (i) \(\Delta T = 28.2 - 21.5 = 6.7\text{ }^\circ\text{C}\) [1]. (ii) Polystyrene is a thermal insulator [1] which reduces heat loss to the surroundings, making the temperature measurements more accurate. (b) (i) Graph plotting: - Axes labelled with units: \(\Delta T\) / °C on y-axis, Volume / \(\text{cm}^3\) on x-axis [1]. - Scales chosen such that the plotted points occupy more than half the grid [1]. - All points plotted accurately [1]. (ii) Two straight lines of best fit drawn using a ruler, intersecting cleanly [1]. (iii) Reading the x-value of the intersection point: typically around 20.6 \(\text{cm}^3\) (accept 20.0 to 21.5 \(\text{cm}^3\)) [1]. (c) (i) The reaction between \(\text{H}^+\) and \(\text{OH}^-\)\ is exothermic / releases heat [1]. (ii) Once all the acid has reacted, the reaction is complete and no more heat is produced [1]. Adding further cold sodium hydroxide solution cools down the existing mixture by dilution [1].
評分準則
- (a)(i) 6.7 [1] - (a)(ii) Thermal insulator / reduces heat loss [1] - (b)(i) Correct axes and scales [1]; correct plotting of all points to within half a small square [1]; smooth/straight lines [1] - (b)(ii) Two straight lines of best fit extrapolated to cross [1] - (b)(iii) Correct reading of volume from the intersection on their graph [1] (allow 20.0 to 21.5 cm³) - (c)(i) Exothermic reaction / heat released [1] - (c)(ii) Reaction is complete / no more heat produced [1]; excess cold solution added cools the mixture down [1]
題目 6 · Practical Structured
10 分
A student determines the focal length \(f\) of a thin convex lens. The student places an illuminated object at the 0.0 cm mark on a optical bench. A convex lens and a screen are placed on the bench.
(a) Draw a labelled diagram showing the setup of the illuminated object, lens, screen, and a metre rule to measure distances. [2]
(b) The distance between the object and the lens is \(u\). The distance between the lens and the screen when a sharp, in-focus image is formed is \(v\).
(i) For Trial 1, the lens is at the 15.0 cm mark. The screen is adjusted until a sharp image is formed at the 45.0 cm mark. Calculate: - \(u_1\) = ........ cm [1] - \(v_1\) = ........ cm [1]
Calculate the focal length \(f_1\) for Trial 1 using the formula: \[f = \frac{uv}{u+v}\]
\(f_1\) = ........ cm [1]
(ii) For Trial 2, the lens is moved to the 30.0 cm mark. The screen is adjusted until a sharp image is formed at the 45.0 cm mark. Calculate: - \(u_2\) = ........ cm - \(v_2\) = ........ cm
Calculate the focal length \(f_2\) for Trial 2 using the same formula: [1]
\(f_2\) = ........ cm
(c) State two precautions that the student should take to obtain accurate measurements of \(v\). [2]
(d) Describe how the student can check that the image on the screen is in focus. [1]
(e) State the nature of the image formed on the screen. [1]
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解題
(a) Diagram should show: - Object, lens, and screen arranged in a straight line [1]. - Labelled metre rule positioned parallel to the bench to measure positions [1]. (b) (i) - \(u_1 = 15.0\text{ cm} - 0.0\text{ cm} = 15.0\text{ cm}\) [1] - \(v_1 = 45.0\text{ cm} - 15.0\text{ cm} = 30.0\text{ cm}\) [1] - \(f_1 = \frac{15.0 \times 30.0}{15.0 + 30.0} = \frac{450}{45} = 10.0\text{ cm}\) [1] (ii) - \(u_2 = 30.0\text{ cm} - 0.0\text{ cm} = 30.0\text{ cm}\) - \(v_2 = 45.0\text{ cm} - 30.0\text{ cm} = 15.0\text{ cm}\) - \(f_2 = \frac{30.0 \times 15.0}{30.0 + 15.0} = \frac{450}{45} = 10.0\text{ cm}\) [1] (both correct values and \(f_2\)) (c) Precautions: - Use a dark room to see the image clearly [1]. - View the screen perpendicular to avoid parallax error when reading the rule [1]. (d) Move screen back and forth until the image is at its clearest / sharpest outline [1]. (e) Real (or inverted) [1].
評分準則
- (a) Object, lens, and screen in order on bench [1]; Metre rule shown parallel to bench with labels [1] - (b)(i) u1 = 15.0 cm [1]; v1 = 30.0 cm [1]; f1 = 10.0 cm [1] - (b)(ii) u2 = 30.0 cm and v2 = 15.0 cm, giving f2 = 10.0 cm [1] - (c) Any two from: perform in dark room, move screen slowly back and forth, ensure apparatus is aligned horizontally/parallel, avoid parallax error when reading rule [2] (1 mark each) - (d) Move screen slightly back and forth to find sharpest image [1] - (e) Real / inverted [1]
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