An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V3) Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.
Paper 23 (選擇題 - Extended)
Forty multiple-choice questions. Answer all questions. Choose the correct option A, B, C or D.
40 题目 · 40 分
题目 1 · 選擇題
1 分
A student places four identical pieces of potato into four sucrose solutions of different concentrations: 0.0 mol/dm^{3}, 0.2 mol/dm^{3}, 0.4 mol/dm^{3}, and 0.8 mol/dm^{3}. After two hours, which solution will contain the potato piece that has experienced the greatest increase in turgor pressure?
A.0.0 mol/dm^{3}
B.0.2 mol/dm^{3}
C.0.4 mol/dm^{3}
D.0.8 mol/dm^{3}
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解题
The turgor pressure in potato cells increases when water enters the cells by osmosis. Water moves from a region of higher water potential (less concentrated solution) to a region of lower water potential (more concentrated solution). The 0.0 mol/dm^{3} solution (pure water) has the highest water potential, causing the maximum amount of water to enter the cells and thus creating the greatest increase in turgor pressure.
评分标准
1 mark for identifying the 0.0 mol/dm^{3} solution as the correct option.
题目 2 · 選擇題
1 分
What is the volume of carbon dioxide, measured at room temperature and pressure (r.t.p.), produced when 10.0 g of calcium carbonate (CaCO_{3}) reacts completely with excess dilute hydrochloric acid? [Relative atomic masses: C = 12, O = 16, Ca = 40; the molar volume of any gas is 24.0 dm^{3} at r.t.p.]
A.1.2 dm^{3}
B.2.4 dm^{3}
C.4.8 dm^{3}
D.24.0 dm^{3}
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解题
First, calculate the relative molecular mass (Mr) of CaCO_{3}: Mr = 40 + 12 + (3 * 16) = 100. Next, calculate the moles of CaCO_{3} reacted: moles = mass / Mr = 10.0 g / 100 g/mol = 0.10 mol. The balanced equation is CaCO_{3} + 2HCl -> CaCl_{2} + H_{2}O + CO_{2}, showing a 1:1 molar ratio between CaCO_{3} and CO_{2}. Thus, 0.10 mol of CO_{2} is produced. Finally, calculate the volume of CO_{2}: volume = moles * molar volume = 0.10 mol * 24.0 dm^{3}/mol = 2.4 dm^{3}.
评分标准
1 mark for calculating Mr and moles of CaCO3; 1 mark for calculating the correct volume of CO2 gas.
题目 3 · 選擇題
1 分
A car of mass 1200 kg accelerates uniformly from rest to a speed of 20 m/s in 8.0 s. What is the average useful power developed by the car engine to achieve this speed, assuming no energy is lost to friction?
A.15 kW
B.30 kW
C.120 kW
D.240 kW
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解题
First, calculate the change in kinetic energy (which is equal to the useful work done): Ek = 0.5 * m * v^{2} = 0.5 * 1200 kg * (20 m/s)^{2} = 600 * 400 = 240,000 J. Next, use the formula for power: Power = Work / time = 240,000 J / 8.0 s = 30,000 W = 30 kW.
评分标准
1 mark for calculating the kinetic energy / work done; 1 mark for dividing by time to find the correct power in kW.
题目 4 · 選擇題
1 分
Silicon(IV) oxide, SiO_{2}, and carbon dioxide, CO_{2}, are oxides of Group IV elements. Which statement correctly explains why silicon(IV) oxide is a solid with a high melting point, while carbon dioxide is a gas at room temperature?
A.Silicon(IV) oxide has a giant covalent structure with strong covalent bonds throughout, whereas carbon dioxide has a simple molecular structure with weak intermolecular forces.
B.The covalent bonds between silicon and oxygen atoms are much stronger than the covalent bonds between carbon and oxygen atoms.
C.Silicon(IV) oxide is ionic with strong electrostatic attractions between ions, while carbon dioxide is covalent.
D.Carbon dioxide molecules are non-polar, whereas silicon(IV) oxide molecules are highly polar.
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解题
Silicon(IV) oxide possesses a giant covalent lattice structure where each silicon atom is strongly bonded to four oxygen atoms, requiring a massive amount of thermal energy to break these bonds. Carbon dioxide, however, has a simple molecular structure with weak intermolecular forces between its non-polar molecules, which are easily overcome at low temperatures.
评分标准
1 mark for correctly comparing the giant covalent lattice of silicon dioxide with the simple molecular structure of carbon dioxide.
题目 5 · 選擇題
1 分
Two resistors, one of resistance 4.0 \Omega and one of resistance 12.0 \Omega, are connected in parallel to a 6.0 V d.c. power supply. What is the total current drawn from the power supply?
A.0.38 A
B.1.5 A
C.2.0 A
D.8.0 A
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解题
First, calculate the equivalent resistance (Rp) of the parallel combination: 1/Rp = 1/4.0 + 1/12.0 = 3/12.0 + 1/12.0 = 4/12.0 = 1/3.0. Therefore, Rp = 3.0 \Omega. Next, apply Ohm's law to find the total current: I = V / Rp = 6.0 V / 3.0 \Omega = 2.0 A.
评分标准
1 mark for calculating parallel resistance; 1 mark for calculating total current from voltage and resistance.
题目 6 · 選擇題
1 分
A student investigates the rate of photosynthesis of an aquatic plant under different conditions of light intensity and temperature. At 15°C, increasing the light intensity beyond a certain point does not increase the rate of photosynthesis. However, raising the temperature to 25°C at this high light intensity increases the rate of photosynthesis. What was limiting the rate of photosynthesis at 15°C under high light intensity?
A.Carbon dioxide concentration
B.Light intensity
C.Temperature
D.Water availability
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解题
Since raising the temperature from 15°C to 25°C caused an increase in the rate of photosynthesis when light intensity was already high, temperature was the factor preventing any further increase at the lower temperature. Therefore, temperature was the limiting factor.
评分标准
1 mark for identifying temperature as the limiting factor based on the experimental results.
题目 7 · 選擇題
1 分
The reaction between hydrogen and chlorine gas to form hydrogen chloride can be represented by the equation: H_{2}(g) + Cl_{2}(g) -> 2HCl(g). The bond energies are: H–H = 436 kJ/mol, Cl–Cl = 242 kJ/mol, and H–Cl = 431 kJ/mol. What is the overall energy change for this reaction?
A.-184 kJ/mol
B.+184 kJ/mol
C.-247 kJ/mol
D.+247 kJ/mol
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解题
Energy absorbed to break bonds: 1 * (H–H) + 1 * (Cl–Cl) = 436 + 242 = 678 kJ/mol. Energy released in making bonds: 2 * (H–Cl) = 2 * 431 = 862 kJ/mol. Overall energy change = Energy absorbed - Energy released = 678 - 862 = -184 kJ/mol.
评分标准
1 mark for calculating bond-breaking energy; 1 mark for calculating bond-making energy; 1 mark for subtracting to find the correct exothermic value.
题目 8 · 選擇題
1 分
A radioactive sample has a half-life of 4.0 hours. Initially, the count rate from the sample is 480 counts per minute. What will the count rate be after 12.0 hours, assuming background radiation is negligible?
A.60 counts per minute
B.120 counts per minute
C.160 counts per minute
D.240 counts per minute
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解题
First determine the number of half-lives that have elapsed: 12.0 hours / 4.0 hours = 3 half-lives. After 1 half-life, the count rate is 480 / 2 = 240 counts per minute. After 2 half-lives, the count rate is 240 / 2 = 120 counts per minute. After 3 half-lives, the count rate is 120 / 2 = 60 counts per minute.
评分标准
1 mark for calculating the number of half-lives; 1 mark for halving the initial count rate three times correctly.
题目 9 · 選擇題
1 分
A crane lifts a cargo box of mass \(400\text{ kg}\) vertically upwards through a height of \(15\text{ m}\) in \(12\text{ s}\). The gravitational field strength, \(g\), is \(10\text{ N/kg}\). The efficiency of the crane is \(80\%\). What is the minimum electrical power input required by the crane?
A.\(4.0\text{ kW}\)
B.\(5.0\text{ kW}\)
C.\(6.25\text{ kW}\)
D.\(7.50\text{ kW}\)
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解题
1. Calculate the useful work done on the cargo box: \(W = m \times g \times h = 400\text{ kg} \times 10\text{ N/kg} \times 15\text{ m} = 60\,000\text{ J}\). 2. Calculate the useful power output: \(P_{\text{out}} = \frac{W}{t} = \frac{60\,000\text{ J}}{12\text{ s}} = 5000\text{ W} = 5.0\text{ kW}\). 3. Use the efficiency formula to find the electrical power input: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \implies 0.80 = \frac{5.0\text{ kW}}{P_{\text{in}}} \implies P_{\text{in}} = \frac{5.0\text{ kW}}{0.80} = 6.25\text{ kW}\).
评分标准
Award 1 mark for the correct option C. Incorrect options: A (multiplied by efficiency instead of dividing), B (calculated useful power output only), D (used an incorrect efficiency factor).
题目 10 · 選擇題
1 分
A circuit contains a \(12\text{ V}\) battery of negligible internal resistance 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 \(3.0\ \Omega\) resistor.
Which row gives the current in the \(3.0\ \Omega\) resistor and the potential difference across it?
| | current in the \(3.0\ \Omega\) resistor / A | potential difference across the \(3.0\ \Omega\) resistor / V | |---|---|---| | **A** | 1.3 | 4.0 | | **B** | 2.0 | 6.0 | | **C** | 2.0 | 12 | | **D** | 4.0 | 12 |
A.A
B.B
C.C
D.D
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解题
1. Find the equivalent resistance of the two \(6.0\ \Omega\) resistors in parallel: \(\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{6.0} = \frac{2}{6.0} \implies R_p = 3.0\ \Omega\). 2. Find the total resistance of the circuit: \(R_{\text{total}} = R_p + 3.0\ \Omega = 3.0 + 3.0 = 6.0\ \Omega\). 3. Find the total current leaving the battery, which is also the current passing through the \(3.0\ \Omega\) series resistor: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\). 4. Calculate the potential difference across the \(3.0\ \Omega\) resistor: \(V = I \times R = 2.0\text{ A} \times 3.0\ \Omega = 6.0\text{ V}\).
评分标准
Award 1 mark for the correct option B. Incorrect options represent various arithmetic errors or a misunderstanding of series/parallel combination properties.
题目 11 · 選擇題
1 分
The reaction between nitrogen gas and hydrogen gas to produce ammonia is represented by the equation:
| bond | bond energy / \(\text{kJ/mol}\) | |---|---| | \(\text{N}\equiv\text{N}\) | 945 | | \(\text{H}-\text{H}\) | 436 | | \(\text{N}-\text{H}\) | 390 |
What is the overall energy change, \(\Delta H\), for this reaction?
A.\(-87\text{ kJ/mol}\)
B.\(+87\text{ kJ/mol}\)
C.\(-1083\text{ kJ/mol}\)
D.\(+1083\text{ kJ/mol}\)
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解题
1. Calculate the energy required to break reactants' bonds (endothermic process): - \(1 \times \text{N}\equiv\text{N} = 945\text{ kJ}\) - \(3 \times \text{H}-\text{H} = 3 \times 436 = 1308\text{ kJ}\) - Total energy absorbed = \(945 + 1308 = 2253\text{ kJ}\).
2. Calculate the energy released when products' bonds form (exothermic process): - \(2\text{ moles of NH}_3\) contain a total of \(6\text{ moles of N}-\text{H}\) bonds. - Total energy released = \(6 \times 390 = 2340\text{ kJ}\).
Award 1 mark for the correct option A. Option B has the wrong sign. Options C and D arise from calculating with an incorrect number of N-H bonds (3 instead of 6).
题目 12 · 選擇題
1 分
An unknown solid salt, \(Y\), is analyzed using qualitative chemical tests:
- A flame test on solid \(Y\) produces a blue-green flame. - An aqueous solution of \(Y\) is acidified with dilute nitric acid, and then aqueous barium nitrate is added. A white precipitate is formed.
What is the identity of salt \(Y\)?
A.copper(II) chloride
B.copper(II) sulfate
C.iron(II) sulfate
D.iron(II) chloride
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解题
1. The blue-green flame test indicates the presence of copper(II) ions, \(\text{Cu}^{2+}\). 2. The reaction with aqueous barium nitrate in acidic solution yields a white precipitate of barium sulfate, indicating the presence of sulfate ions, \(\text{SO}_4^{2-}\). 3. Combining these ions gives copper(II) sulfate.
评分标准
Award 1 mark for the correct option B. Incorrect options contain wrong metal cations or halide anions.
题目 13 · 選擇題
1 分
A plant leaf is kept in the dark for 24 hours, then a strip of black paper is placed over part of the leaf. The plant is subsequently exposed to bright light for 4 hours.
The leaf is then tested for the presence of starch using iodine solution.
Which row correctly describes the purpose of keeping the leaf in the dark, and the result of the starch test on the covered part of the leaf?
| | purpose of keeping in the dark | result of starch test on covered part | |---|---|---| | **A** | to de-starch the leaf | blue-black | | **B** | to de-starch the leaf | brown | | **C** | to denature leaf enzymes | blue-black | | **D** | to denature leaf enzymes | brown |
A.A
B.B
C.C
D.D
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解题
1. Keeping the leaf in the dark for 24 hours ensures that any pre-existing starch is used up or transported out of the leaf (de-starching). This ensures that any starch detected at the end of the experiment was synthesized during the light exposure period. 2. The covered part of the leaf did not receive light, preventing photosynthesis, so no starch was produced there. Therefore, the starch test on the covered part is negative, leaving the iodine solution its natural brown color.
评分标准
Award 1 mark for the correct option B. Options A, C, and D contain incorrect explanations for the dark phase or incorrect test results.
题目 14 · 選擇題
1 分
Which statement about the transition elements is correct?
A.They have low densities and low melting points compared to Group I metals.
B.They form only white or colourless compounds.
C.They can act as catalysts in industrial chemical processes.
D.They react extremely rapidly with cold water to release hydrogen gas.
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解题
A is incorrect because transition elements have high densities and high melting points compared to alkali metals (Group I). B is incorrect because transition elements form colored compounds. C is correct because transition elements (such as iron or nickel) and their compounds commonly act as catalysts in industry. D is incorrect because transition elements are relatively unreactive with cold water.
评分标准
Award 1 mark for the correct option C. Options A, B, and D describe properties opposite to those of transition elements.
题目 15 · 選擇題
1 分
Amylase is an enzyme that digests starch.
Four test-tubes are prepared as follows: - test-tube 1: starch + amylase at \(37\ ^\circ\text{C}\) - test-tube 2: starch + amylase that has been boiled at \(37\ ^\circ\text{C}\) - test-tube 3: starch + amylase at \(0\ ^\circ\text{C}\) - test-tube 4: maltose + amylase at \(37\ ^\circ\text{C}\)
In which test-tube(s) will starch be completely broken down after 30 minutes?
A.1 only
B.1 and 3 only
C.1 and 4 only
D.2 and 3 only
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解题
- In test-tube 1: Amylase is active at \(37\ ^\circ\text{C}\), so it successfully breaks down starch. - In test-tube 2: Boiled amylase has been denatured and is no longer functional, so starch remains intact. - In test-tube 3: The temperature of \(0\ ^\circ\text{C}\) inactivates the enzyme, so starch is not digested. - In test-tube 4: There is no starch present to be broken down (maltose is already a simple sugar and is not a substrate for amylase).
评分标准
Award 1 mark for the correct option A. Incorrect options incorrectly suggest that denatured, inactive, or substrate-absent mixtures can successfully digest starch.
题目 16 · 選擇題
1 分
A ray of light in air is incident on the surface of a flat glass block. The angle of incidence is \(45^\circ\). The refractive index of the glass is \(1.5\).
What is the angle of refraction of the light inside the glass block?
A.\(15^\circ\)
B.\(28^\circ\)
C.\(45^\circ\)
D.\(68^\circ\)
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解题
1. Use Snell's Law: \(n = \frac{\sin(i)}{\sin(r)}\). 2. Rearrange the formula to solve for \(\sin(r)\): \(\sin(r) = \frac{\sin(i)}{n} = \frac{\sin(45^\circ)}{1.5}\). 3. Substitute the values: \(\sin(r) = \frac{0.7071}{1.5} \approx 0.4714\). 4. Calculate the angle of refraction: \(r = \arcsin(0.4714) \approx 28^\circ\).
评分标准
Award 1 mark for the correct option B. Distractors represent common computation errors (e.g. multiplying by 1.5 instead of dividing).
题目 17 · 選擇題
1 分
A rectangular block of mass 240 g has dimensions 5.0 cm by 4.0 cm by 2.0 cm. What is the maximum pressure it can exert when resting on a horizontal table? (g = 10 N/kg)
A.1200 Pa
B.2400 Pa
C.3000 Pa
D.4800 Pa
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解题
First, calculate the weight of the block: \(W = m \times g = 0.24\text{ kg} \times 10\text{ N/kg} = 2.4\text{ N}\). The maximum pressure is exerted when the contact area is at its minimum. The minimum surface area of the block is \(4.0\text{ cm} \times 2.0\text{ cm} = 8.0\text{ cm}^2 = 8.0 \times 10^{-4}\text{ m}^2\). Maximum pressure \(P = \frac{W}{A_{\text{min}}} = \frac{2.4\text{ N}}{8.0 \times 10^{-4}\text{ m}^2} = 3000\text{ Pa}\).
评分标准
1 mark for calculating the weight of 2.4 N, identifying the minimum surface area of 8.0 cm², and calculating the maximum pressure as 3000 Pa.
题目 18 · 選擇題
1 分
4.6 g of sodium reacts completely with excess water. What is the volume of hydrogen gas produced, measured at room temperature and pressure (r.t.p.)? (Ar: Na = 23; the volume of one mole of any gas is 24 dm³ at r.t.p.)
A.1.2 dm³
B.2.4 dm³
C.4.8 dm³
D.9.6 dm³
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解题
The chemical equation is \(2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\). First, calculate the moles of sodium: \(\text{moles} = \frac{4.6\text{ g}}{23\text{ g/mol}} = 0.20\text{ mol}\). According to the equation, 2 moles of Na react to produce 1 mole of \(\text{H}_2\). Therefore, the moles of \(\text{H}_2\) produced is \(\frac{0.20}{2} = 0.10\text{ mol}\). Volume of \(\text{H}_2\) gas = \(0.10\text{ mol} \times 24\text{ dm}^3\text{/mol} = 2.4\text{ dm}^3\).
评分标准
1 mark for calculating the moles of sodium (0.20 mol), using the molar ratio (2:1) to find the moles of hydrogen (0.10 mol), and converting this to the correct volume of 2.4 dm³.
题目 19 · 選擇題
1 分
Which statement about a healthy green plant at its light compensation point is correct?
A.The rate of photosynthesis is zero.
B.The rate of oxygen uptake by respiration is equal to the rate of oxygen release by photosynthesis.
C.The plant releases more carbon dioxide than it absorbs.
D.Glucose is produced by photosynthesis faster than it is consumed by respiration.
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解题
At the light compensation point, the rate of photosynthesis is exactly equal to the rate of cellular respiration. Therefore, the volume of oxygen consumed by respiration is equal to the volume of oxygen released by photosynthesis, and the net gaseous exchange with the atmosphere is zero.
评分标准
1 mark for identifying that the rate of photosynthesis equals the rate of respiration, leading to balanced gas exchange.
题目 20 · 選擇題
1 分
Silicon(IV) oxide, SiO₂, has a giant covalent macromolecular structure. How many covalent bonds are formed by each silicon atom and each oxygen atom in this structure?
A.each silicon atom forms 2 bonds; each oxygen atom forms 4 bonds
B.each silicon atom forms 4 bonds; each oxygen atom forms 2 bonds
C.each silicon atom forms 4 bonds; each oxygen atom forms 4 bonds
D.each silicon atom forms 2 bonds; each oxygen atom forms 2 bonds
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解题
In the giant tetrahedral macromolecular structure of silicon(IV) oxide, each silicon atom is covalently bonded to four oxygen atoms, and each oxygen atom is covalently bonded to two silicon atoms. This arrangement maintains the overall 1:2 stoichiometry of the SiO₂ structure.
评分标准
1 mark for correctly identifying that each silicon atom forms 4 covalent bonds and each oxygen atom forms 2 covalent bonds.
题目 21 · 選擇題
1 分
Three resistors of resistances 3.0 ohms, 6.0 ohms, and 4.0 ohms are connected to a 12 V d.c. supply. The 3.0 ohm and 6.0 ohm resistors are connected in parallel with each other, and this parallel combination is connected in series with the 4.0 ohm resistor. What is the current in the 4.0 ohm resistor?
A.1.2 A
B.2.0 A
C.3.0 A
D.6.0 A
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解题
First, calculate the equivalent resistance of the parallel combination: \(R_p = \frac{3.0 \times 6.0}{3.0 + 6.0} = 2.0\ \Omega\). Next, calculate the total resistance of the series-parallel circuit: \(R_{\text{total}} = R_p + 4.0\ \Omega = 2.0\ \Omega + 4.0\ \Omega = 6.0\ \Omega\). The total current flowing from the supply is \(I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\). Since the 4.0 ohm resistor is in series with the parallel block, the total current of 2.0 A must flow directly through it.
评分标准
1 mark for calculating the parallel resistance as 2.0 ohms, total resistance as 6.0 ohms, and the current as 2.0 A.
题目 22 · 選擇題
1 分
Hydrogen gas reacts with chlorine gas according to the equation: H₂(g) + Cl₂(g) → 2HCl(g). The bond energies are: H-H = 436 kJ/mol, Cl-Cl = 242 kJ/mol, H-Cl = 431 kJ/mol. What is the enthalpy change, ΔH, for this reaction?
A.-184 kJ/mol
B.+184 kJ/mol
C.-247 kJ/mol
D.+247 kJ/mol
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解题
Energy absorbed to break bonds in reactants = \(1 \times (\text{H-H}) + 1 \times (\text{Cl-Cl}) = 436 + 242 = 678\text{ kJ/mol}\). Energy released when bonds are formed in products = \(2 \times (\text{H-Cl}) = 2 \times 431 = 862\text{ kJ/mol}\). Enthalpy change, \(\Delta H = \text{energy absorbed} - \text{energy released} = 678 - 862 = -184\text{ kJ/mol}\).
评分标准
1 mark for calculating the correct energy absorbed (678 kJ/mol) and energy released (862 kJ/mol) to obtain the net enthalpy change of -184 kJ/mol.
题目 23 · 選擇題
1 分
A ray of light in a transparent plastic block of refractive index 1.5 strikes the boundary with air at an angle of incidence of 40 degrees. Which statement describes what happens to the ray?
A.It undergoes total internal reflection because the angle of incidence is greater than the critical angle of 42 degrees.
B.It is refracted into the air at an angle of refraction of 25 degrees.
C.It is refracted into the air at an angle of refraction of 75 degrees.
D.It passes straight through the boundary into the air without any deviation.
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解题
First, find the critical angle \(c\): \(\sin(c) = \frac{1}{n} = \frac{1}{1.5} \approx 0.667 \rightarrow c \approx 41.8^\circ\). Since the angle of incidence (\(40^\circ\)) is less than the critical angle (\(41.8^\circ\)), refraction occurs rather than total internal reflection. Using Snell's law: \(n = \frac{\sin(r)}{\sin(i)}\), where refraction is from plastic to air. Thus, \(\sin(r) = n \times \sin(i) = 1.5 \times \sin(40^\circ) = 1.5 \times 0.6428 = 0.9642\). This gives \(r = \arcsin(0.9642) \approx 74.6^\circ\) (rounded to \(75^\circ\)).
评分标准
1 mark for comparing the angle of incidence with the critical angle to conclude refraction occurs, and applying Snell's law correctly to find the angle of refraction is 75 degrees.
题目 24 · 選擇題
1 分
A person looks up from reading a book to focus on a distant mountain. Which row correctly describes the changes that occur in the eyes?
A.ciliary muscles contract; suspensory ligaments become slack; lens becomes thick
B.ciliary muscles contract; suspensory ligaments become tight; lens becomes thin
C.ciliary muscles relax; suspensory ligaments become slack; lens becomes thick
D.ciliary muscles relax; suspensory ligaments become tight; lens becomes thin
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解题
To focus on a distant object, the ciliary muscles must relax, which increases tension on the suspensory ligaments (making them tight). This pulls the lens, causing it to become thin and less convex to focus parallel light rays onto the retina.
评分标准
1 mark for identifying the correct combination of relaxed ciliary muscles, tight suspensory ligaments, and a thin lens shape.
题目 25 · multiple_choice
1 分
A toy car of mass \(1.2\text{ kg}\) moves along a straight horizontal track. The kinetic energy of the toy car is \(15\text{ J}\). What is the momentum of the toy car?
A.\(3.0\text{ kg m/s}\)
B.\(6.0\text{ kg m/s}\)
C.\(7.2\text{ kg m/s}\)
D.\(18\text{ kg m/s}\)
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解题
First, calculate the velocity of the toy car using the formula for kinetic energy: \(E_k = \frac{1}{2} m v^2\) \(15 = \frac{1}{2} \times 1.2 \times v^2\) \(15 = 0.6 \times v^2\) \(v^2 = 25\) \(v = 5.0\text{ m/s}\)
Next, calculate the momentum using the formula: \(p = m v\) \(p = 1.2\text{ kg} \times 5.0\text{ m/s} = 6.0\text{ kg m/s}\).
评分标准
Award 1 mark for the correct answer B. Award 0 marks for incorrect options: - Option A incorrectly calculates momentum by neglecting the factor of 1/2 in the kinetic energy formula. - Option C uses incorrect relationships between the variables. - Option D is an incorrect calculation.
题目 26 · multiple_choice
1 分
A cylindrical metal wire has a resistance \(R\). The wire is stretched uniformly so that its length doubles, while its volume and resistivity remain constant. What is the resistance of the stretched wire?
A.\(0.5R\)
B.\(R\)
C.\(2R\)
D.\(4R\)
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解题
The volume of a cylinder is given by \(V = L \times A\), where \(L\) is the length and \(A\) is the cross-sectional area. Since volume is constant and the length doubles to \(2L\), the cross-sectional area must halve to \(A/2\).
The original resistance is: \(R = \rho \frac{L}{A}\)
The new resistance is: \(R' = \rho \frac{2L}{A/2} = 4 \rho \frac{L}{A} = 4R\).
评分标准
Award 1 mark for correct answer D. Incorrect options: - Option A assumes that resistance decreases when stretched. - Option B assumes resistance is unchanged. - Option C neglects the effect of the changing cross-sectional area.
题目 27 · multiple_choice
1 分
The bond energies of some covalent bonds are shown in the table.
Energy required to break bonds (reactants): \(436\text{ (H-H)} + 242\text{ (Cl-Cl)} = +678\text{ kJ/mol}\)
Energy released in making bonds (products): \(2 \times 431\text{ (H-Cl)} = 862\text{ kJ/mol}\)
Energy change, \(\Delta H = \text{energy absorbed} - \text{energy released} = 678 - 862 = -184\text{ kJ/mol}\).
评分标准
Award 1 mark for correct answer A. Incorrect options: - Option B represents an incorrect subtraction method. - Option C gives the correct magnitude but the wrong sign (endothermic instead of exothermic). - Option D is calculated using incorrect values and signs.
题目 28 · multiple_choice
1 分
Which statement explains why graphite is used as a lubricant whereas diamond is used in cutting tools?
A.Graphite has a simple molecular structure, whereas diamond has a giant covalent lattice.
B.Graphite has weak forces between layers of carbon atoms which allow the layers to slide, whereas diamond has a rigid three-dimensional structure of strong covalent bonds.
C.The covalent bonds between carbon atoms in graphite are much weaker than those in diamond.
D.Carbon atoms in graphite are arranged in a tetrahedral structure, whereas in diamond they are arranged in hexagonal layers.
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解题
In graphite, the carbon atoms are arranged in hexagonal layers held together by weak forces of attraction, allowing the layers to slide past one another easily. In contrast, diamond has a rigid three-dimensional tetrahedral lattice of strong covalent bonds throughout, making it extremely hard and ideal for cutting tools.
评分标准
Award 1 mark for correct answer B. Incorrect options: - Option A incorrectly states graphite has a simple molecular structure. - Option C is scientifically false; the individual C-C covalent bonds in graphite's layers are very strong. - Option D reverses the geometries of graphite and diamond.
题目 29 · multiple_choice
1 分
An organic compound \(Y\) reacts with aqueous bromine, turning it from orange to colourless. When \(Y\) undergoes addition polymerisation, the polymer formed has the repeating unit: \(-[\text{CH}_2-\text{CH}(\text{CH}_3)]_n-\).
What is the name of monomer \(Y\) and the general formula of the homologous series to which it belongs?
A.ethane, \(\text{C}_n\text{H}_{2n+2}\)
B.ethene, \(\text{C}_n\text{H}_{2n}\)
C.propane, \(\text{C}_n\text{H}_{2n+2}\)
D.propene, \(\text{C}_n\text{H}_{2n}\)
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解题
The decolourisation of bromine water indicates that monomer \(Y\) is unsaturated (an alkene). The repeating unit shows a three-carbon monomer skeleton: \(\text{CH}_2=\text{CH}-\text{CH}_3\), which is propene. Alkenes belong to a homologous series with the general formula \(\text{C}_n\text{H}_{2n}\).
评分标准
Award 1 mark for correct answer D. Incorrect options: - Options A and C name saturated alkanes which do not undergo addition polymerisation or decolourise bromine water. - Option B names ethene, which has only two carbons per repeating unit \(-[\text{CH}_2-\text{CH}_2]_n-\).
题目 30 · multiple_choice
1 分
The rate of photosynthesis of a terrestrial plant is measured under different conditions.
- At carbon dioxide concentration of \(0.04\%\), the rate of photosynthesis increases with light intensity and then levels off at a maximum rate of \(R_1\). - At carbon dioxide concentration of \(0.12\%\), the rate of photosynthesis levels off at a higher maximum rate of \(R_2\).
The temperature is kept constant at \(25^\circ\text{C}\) in both cases.
When the rate of photosynthesis has reached its maximum level under both carbon dioxide concentrations, what is the limiting factor at \(0.04\%\) carbon dioxide?
A.carbon dioxide concentration
B.light intensity
C.oxygen concentration
D.temperature
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解题
Since increasing the carbon dioxide concentration from \(0.04\%\) to \(0.12\%\) causes the maximum rate of photosynthesis to increase (from \(R_1\) to \(R_2\)), carbon dioxide concentration must be the factor that was limiting the rate at \(0.04\%\).
评分标准
Award 1 mark for correct answer A. Incorrect options: - Option B is incorrect because at maximum level, further increase in light intensity does not increase the rate. - Option C is not a limiting factor in photosynthesis. - Option D is kept constant and is not limiting, as shown by the increase in rate when only CO2 concentration is increased.
题目 31 · multiple_choice
1 分
An enzyme-catalysed reaction is investigated at different temperatures. The time taken for the substrate to be completely broken down is recorded.
- At \(15^\circ\text{C}\), the time taken is \(120\text{ seconds}\). - At \(35^\circ\text{C}\), the time taken is \(30\text{ seconds}\). - At \(75^\circ\text{C}\), the substrate is still not broken down after \(10\text{ minutes}\).
Which statement explains these results?
A.Raising the temperature from \(15^\circ\text{C}\) to \(35^\circ\text{C}\) denatures the enzyme, increasing its rate of reaction.
B.At \(35^\circ\text{C}\), the kinetic energy of the molecules is higher than at \(15^\circ\text{C}\), resulting in more frequent successful collisions.
C.At \(75^\circ\text{C}\), the enzyme and substrate molecules move too quickly to bind to each other.
D.Lowering the temperature to \(15^\circ\text{C}\) permanently alters the active site shape of the enzyme.
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解题
An increase in temperature from \(15^\circ\text{C}\) to \(35^\circ\text{C}\) increases the kinetic energy of the enzyme and substrate molecules. This leads to more rapid movement and more frequent successful collisions between active sites and substrate molecules, reducing the time taken for the reaction. At \(75^\circ\text{C}\), the enzyme is denatured, meaning its active site has changed shape permanently and can no longer bind the substrate.
评分标准
Award 1 mark for correct answer B. Incorrect options: - Option A incorrectly states that the enzyme is denatured at low temperatures. - Option C is incorrect because at \(75^\circ\text{C}\) the enzyme is denatured, preventing binding entirely. - Option D incorrectly identifies the temperature where the reaction stops as the optimum temperature.
题目 32 · multiple_choice
1 分
A ray of light passes from glass into air. The angle of incidence in the glass is \(30.0^\circ\). The refractive index of the glass is \(1.50\).
What is the angle of refraction in air?
A.19.5\(^\circ\)
B.30.0\(^\circ\)
C.45.0\(^\circ\)
D.48.6\(^\circ\)
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解题
By definition, the refractive index \(n\) is given by: \(n = \frac{\sin(i_{\text{air}})}{\sin(r_{\text{glass}})}\)
Rearranging to find the angle of refraction in air: \(\sin(i_{\text{air}}) = n \times \sin(r_{\text{glass}})\) \(\sin(i_{\text{air}}) = 1.50 \times \sin(30.0^\circ) = 1.50 \times 0.500 = 0.750\)
Therefore, the angle of refraction in air is: \(i_{\text{air}} = \arcsin(0.750) = 48.6^\circ\).
评分标准
Award 1 mark for correct answer D. Incorrect options: - Option A incorrectly applies the formula as \(\sin r = \frac{\sin 30^\circ}{1.5}\). - Option B is the same as the angle of incidence. - Option C is a common miscalculation.
题目 33 · 選擇題
1 分
The table describes the movement of a substance across a cell membrane.
$$\begin{array}{|c|c|c|} \hline \text{direction of movement} & \text{requires energy from respiration} & \text{requires carrier proteins} \\ \hline \text{against a concentration gradient} & \text{yes} & \text{yes} \\ \hline \end{array}$$
Which process is described by this row?
A.active transport
B.diffusion
C.osmosis
D.translocation
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解题
Active transport is the movement of particles through a cell membrane from a region of lower concentration to a region of higher concentration (against a concentration gradient), using energy from respiration and requiring carrier proteins.
评分标准
[1 mark] for selecting active transport as the process that moves substances against a concentration gradient using energy and carrier proteins.
题目 34 · 選擇題
1 分
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes.
Which ionic half-equation represents the reaction occurring at the anode (positive electrode)?
At the anode (positive electrode), negative ions (anions) are attracted. In concentrated aqueous sodium chloride, chloride ions (\(\text{Cl}^-\)) are preferentially discharged over hydroxide ions (\(\text{OH}^-\)) because they are in high concentration. The half-equation is \(2\text{Cl}^-(\text{aq}) \rightarrow \text{Cl}_2(\text{g}) + 2\text{e}^-\), representing oxidation.
评分标准
[1 mark] for selecting the correct oxidation half-equation for chloride ions at the anode.
题目 35 · 選擇題
1 分
A trolley of mass \(3.0\text{ kg}\) travelling at a velocity of \(4.0\text{ m/s}\) collides with a stationary trolley of mass \(1.0\text{ kg}\).
After the collision, the two trolleys stick together and move with a common velocity \(v\).
What is the value of \(v\)?
A.\(1.0\text{ m/s}\)
B.\(1.3\text{ m/s}\)
C.\(3.0\text{ m/s}\)
D.\(4.0\text{ m/s}\)
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解题
By the principle of conservation of momentum, total momentum before collision equals total momentum after collision.
\(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\)
\(3.0 \times 4.0 + 1.0 \times 0 = (3.0 + 1.0) v\)
\(12.0 = 4.0 v\)
\(v = 3.0\text{ m/s}\)
评分标准
[1 mark] for correctly calculating the initial total momentum, total combined mass, and resolving to find the final velocity of \(3.0\text{ m/s}\).
题目 36 · 選擇題
1 分
The graph shows the rate of photosynthesis of a plant at different light intensities under two different environmental conditions, X and Y.
- Condition X: \(0.04\%\text{ CO}_2\) concentration at \(20^\circ\text{C}\) - Condition Y: \(0.12\%\text{ CO}_2\) concentration at \(20^\circ\text{C}\)
At high light intensity, the rate of photosynthesis under condition X reaches a maximum plateau that is lower than under condition Y.
Which factor is limiting the rate of photosynthesis at high light intensity under condition X?
A.carbon dioxide concentration
B.light intensity
C.temperature
D.chlorophyll concentration
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解题
At high light intensity, increasing the light intensity further does not increase the rate of photosynthesis under condition X. This means light intensity is no longer the limiting factor. Since the rate increases when the carbon dioxide concentration is increased to \(0.12\%\) (condition Y) at the same temperature, carbon dioxide concentration must be the limiting factor under condition X.
评分标准
[1 mark] for identifying carbon dioxide concentration as the limiting factor based on the comparison of the two plateaus.
题目 37 · 選擇題
1 分
A mixture of \(40\text{ cm}^3\) of carbon monoxide and \(40\text{ cm}^3\) of oxygen is ignited to react according to the equation:
What is the total volume of gas remaining after the reaction is complete, measured at room temperature and pressure?
A.\(40\text{ cm}^3\)
B.\(60\text{ cm}^3\)
C.\(80\text{ cm}^3\)
D.\(100\text{ cm}^3\)
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解题
According to the balanced equation, 2 volumes of \(\text{CO}\) react with 1 volume of \(\text{O}_2\) to produce 2 volumes of \(\text{CO}_2\). Thus, \(40\text{ cm}^3\) of \(\text{CO}\) reacts completely with \(20\text{ cm}^3\) of \(\text{O}_2\), leaving \(20\text{ cm}^3\) of unreacted \(\text{O}_2\). The reaction produces \(40\text{ cm}^3\) of \(\text{CO}_2\). The total volume of gas remaining is \(20\text{ cm}^3\text{ (unused } \text{O}_2) + 40\text{ cm}^3\text{ (produced } \text{CO}_2) = 60\text{ cm}^3\).
评分标准
[1 mark] for correctly determining the limiting reactant, the volume of leftover reactant, and the volume of gaseous product, leading to a total volume of \(60\text{ cm}^3\).
题目 38 · 選擇題
1 分
A potential divider circuit consists of a light-dependent resistor (LDR) and a fixed resistor of resistance \(5.0\text{ k}\Omega\) connected in series across a steady \(6.0\text{ V}\) d.c. power supply.
A voltmeter is connected across the fixed resistor.
What happens to the resistance of the LDR and the reading on the voltmeter when the light level falling on the LDR increases?
When the light level increases, the resistance of the LDR decreases. Since the LDR and the fixed resistor are in series, a decrease in the LDR's resistance decreases the total resistance of the circuit, which increases the current flowing through it. Consequently, the potential difference across the fixed resistor (which is measured by the voltmeter) increases, in accordance with \(V = IR\).
评分标准
[1 mark] for identifying that increased light level decreases LDR resistance and increases the voltmeter reading across the fixed resistor.
题目 39 · 選擇題
1 分
Which ester is formed when propanoic acid reacts with ethanol in the presence of an acid catalyst?
A.ethyl propanoate
B.propyl ethanoate
C.methyl propanoate
D.ethyl ethanoate
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解题
Esters are named by taking the alkyl group from the alcohol first, followed by the carboxylate group from the carboxylic acid. Here, ethanol provides the ethyl group and propanoic acid provides the propanoate group. Therefore, the ester formed is ethyl propanoate.
评分标准
[1 mark] for identifying the correct name of the ester formed from ethanol and propanoic acid.
题目 40 · 選擇題
1 分
A ray of light travels from air into a transparent plastic block. The angle of incidence in air is \(50^\circ\) and the angle of refraction in the plastic is \(31^\circ\).
What is the refractive index of the plastic?
A.0.67
B.1.5
C.1.6
D.2.2
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解题
The refractive index \(n\) is given by Snell's Law: \(n = \frac{\sin i}{\sin r}\).
Twelve structured theory questions. Answer all questions. Show all working and use appropriate units.
12 题目 · 120 分
题目 1 · 結構題
10 分
A delivery drone of mass 15 kg starts from rest and climbs vertically. Fig. 1.1 shows the speed-time graph for the first 30 seconds of its journey.
- From 0 to 10 seconds, the drone accelerates uniformly from 0 to 20 m/s. - From 10 to 25 seconds, it travels at a constant speed of 20 m/s. - From 25 to 30 seconds, it decelerates uniformly to rest.
(a) Define acceleration. [1]
(b) Using the information above, calculate the acceleration of the drone during the first 10 seconds. [2]
(c) Calculate the total distance travelled by the drone during the 30 seconds. [3]
(d) (i) Calculate the kinetic energy of the drone when it is travelling at its constant speed. [2]
(ii) At the end of the journey, the drone has climbed to a vertical height of 80 m. Calculate the gain in gravitational potential energy of the drone. (g = 9.8 N/kg) [2]
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解题
(a) Acceleration is the rate of change of velocity.
(a) rate of change of velocity / change in velocity per unit time [1] (Reject: 'speed' instead of 'velocity')
(b) correct formula / substitution: \(20 / 10\) [1]; correct value and unit: \(2.0\text{ m/s}^2\) [1]
(c) method to find any individual area or split shape [1]; sum of all three areas shown: \(100 + 300 + 50\) [1]; final answer: \(450\text{ m}\) [1]
(d) (i) correct substitution: \(0.5 \times 15 \times 20^2\) [1]; final answer: \(3000\text{ J}\) [1] (ii) correct substitution: \(15 \times 9.8 \times 80\) [1]; final answer: \(11760\text{ J}\) [1]
题目 2 · 結構題
10 分
A student sets up a circuit to investigate a model heating system. The circuit is powered by a 12 V d.c. power supply and consists of a 4.0 \(\Omega\) resistor connected in series with a parallel combination of two identical 12 \(\Omega\) resistors.
(a) Draw a circuit diagram for this arrangement, including a voltmeter connected to measure the potential difference across the parallel combination. [3]
(b) Calculate the total equivalent resistance of the circuit. [3]
(c) Calculate the current drawn from the 12 V power supply. [2]
(d) Calculate the power dissipated in the 4.0 \(\Omega\) resistor. [2]
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解题
(a) The diagram must show the 12 V d.c. source, the 4.0 \(\Omega\) resistor in series, and then splitting into two parallel branches each containing one 12 \(\Omega\) resistor. The voltmeter must be connected in parallel across the parallel combination.
(b) First, calculate the resistance of the parallel combination (\(R_p\)): \(\frac{1}{R_p} = \frac{1}{12} + \frac{1}{12} = \frac{2}{12}\) \(R_p = 6.0\ \Omega\) Now, add the series resistor: \(R_{\text{total}} = R_s + R_p = 4.0 + 6.0 = 10.0\ \Omega\).
(d) Power in the 4.0 \(\Omega\) resistor: \(P = I^2 R = (1.2)^2 \times 4.0 = 1.44 \times 4.0 = 5.76\text{ W}\).
评分标准
(a) Correct symbols for power supply, resistors, and voltmeter [1]; 4.0 \(\Omega\) resistor in series with parallel combination of two 12 \(\Omega\) resistors [1]; Voltmeter connected in parallel across the two 12 \(\Omega\) resistors [1]
(b) Correct formula for parallel resistance used [1]; \(R_p = 6.0\ \Omega\) [1]; Total resistance = \(10.0\ \Omega\) [1]
(c) Correct formula \(I = V/R\) used [1]; \(1.2\text{ A}\) [1]
(d) Correct formula \(P = I^2 R\) (or \(P = V^2/R\) with correct voltage calculated) used [1]; \(5.76\text{ W}\) (accept \(5.8\text{ W}\)) [1]
题目 3 · 結構題
10 分
Photosynthesis is the process by which plants manufacture carbohydrates.
(a) Write the balanced chemical equation for photosynthesis. [2]
(b) Fig. 3.1 shows a diagram of a cross-section of a typical plant leaf (not shown here).
(i) Identify the tissue layer that is the primary site of photosynthesis and explain how its cells are adapted for this function. [2]
(ii) Describe the role of guard cells in gas exchange during the daytime. [2]
(c) A student investigates the rate of photosynthesis in Elodea (water weed). They measure the volume of oxygen gas produced per minute at different light intensities.
(i) Explain why measuring the volume of oxygen produced per minute is a suitable method to estimate the rate of photosynthesis. [2]
(ii) At very high light intensities, the rate of oxygen production ceases to increase. Suggest two limiting factors, other than light intensity, that could prevent the rate from increasing. [2]
(b) (i) Palisade mesophyll is the primary site. Cells are packed with chloroplasts and arranged vertically near the upper surface to capture maximum light. (ii) In daylight, guard cells become turgid and open the stomata, allowing carbon dioxide to diffuse into the leaf for photosynthesis, and oxygen to diffuse out.
(c) (i) Oxygen is a product of photosynthesis. The volume of oxygen produced per unit time is directly proportional to the rate of the photosynthetic reaction. (ii) Limiting factors include carbon dioxide concentration and temperature.
评分标准
(a) Correct reactants and products [1]; correct balancing [1]
(b) (i) Palisade mesophyll [1]; contains many chloroplasts / packed tightly together near upper surface [1] (ii) Guard cells open stomata when turgid [1]; allowing diffusion of carbon dioxide in / oxygen out [1]
(c) (i) Oxygen is a product of photosynthesis [1]; rate of oxygen production reflects rate of reaction [1] (ii) Temperature [1]; Carbon dioxide concentration [1]
题目 4 · 結構題
10 分
The human digestive system breaks down large, insoluble food molecules into small, soluble ones that can be absorbed.
(a) Contrast mechanical digestion and chemical digestion. [2]
(b) (i) State the function of the stomach in chemical digestion, including the name of the enzyme class involved and the nutrient it digests. [2]
(ii) Explain the importance of hydrochloric acid in the stomach. [2]
(c) (i) Bile is secreted into the small intestine. Explain how bile aids in the digestion of fats. [2]
(ii) Name the product(s) of fat digestion and the enzyme responsible for this process. [2]
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解题
(a) Mechanical digestion breaks down food into smaller pieces without chemical change, whereas chemical digestion breaks down large, insoluble molecules into small, soluble molecules using enzymes.
(b) (i) The stomach performs chemical digestion of proteins using the enzyme protease (pepsin). (ii) Hydrochloric acid kills bacteria/pathogens in food and provides the optimum acidic pH (around pH 2) for protease to function.
(c) (i) Bile emulsifies fats, breaking large droplets into smaller droplets. This increases the surface area for lipase to act upon. (ii) Fats are digested by lipase to produce fatty acids and glycerol.
评分标准
(a) Mechanical digestion does not change chemical nature of food / only physical breakdown [1]; chemical digestion uses enzymes to break chemical bonds [1]
(b) (i) Protease / pepsin [1]; digests proteins [1] (ii) Kills pathogens/bacteria [1]; provides acidic/optimum pH for stomach enzymes [1]
(c) (i) Emulsifies fats [1]; increases surface area for lipase action [1] (ii) Fatty acids and glycerol [1]; digested by lipase [1]
题目 5 · 結構題
10 分
The reaction between nitrogen gas and hydrogen gas to produce ammonia is shown by the equation:
(a) Explain, in terms of bond breaking and bond making, why this reaction is exothermic. [3]
(b) Draw an energy level diagram for this reaction. Your diagram should label: - the reactants and products - the activation energy, \(E_a\) - the overall energy change, \(\Delta H\). [3]
(c) Use the bond energies in Table 5.1 to calculate the overall energy change, \(\Delta H\), for the reaction.
Table 5.1 | Bond | Bond energy / kJ/mol | | :--- | :--- | | N\(\equiv\)N | 945 | | H-H | 436 | | N-H | 391 |
Show your working. [4]
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解题
(a) Bond breaking is endothermic (requires energy) and bond making is exothermic (releases energy). This reaction is exothermic because more energy is released when making six N-H bonds than is absorbed when breaking one N\(\equiv\)N bond and three H-H bonds.
(b) The diagram should show: - Horizontal line for reactants higher than products. - A curve going up to a peak (activation energy) and then down to the products line. - Arrow representing \(E_a\) going from reactants to peak. - Arrow representing \(\Delta H\) going downwards from reactants to products.
(c) Energy required to break bonds: - \(1 \times \text{N}\equiv\text{N} = 945\text{ kJ/mol}\) - \(3 \times \text{H-H} = 3 \times 436 = 1308\text{ kJ/mol}\) Total energy in = \(945 + 1308 = 2253\text{ kJ/mol}\)
Energy released in making bonds: - \(6 \times \text{N-H} = 6 \times 391 = 2346\text{ kJ/mol}\) Total energy out = \(2346\text{ kJ/mol}\)
(a) Bond breaking is endothermic and bond making is exothermic [1]; energy released making bonds is greater than energy required to break bonds [1]; quantitative reference to the bonds in reactants and products [1]
(b) Reactants line higher than products line [1]; activation energy \(E_a\) correctly indicated [1]; enthalpy change \(\Delta H\) correctly indicated pointing downwards [1]
(c) Calculation of energy to break reactant bonds: \(2253\text{ kJ}\) [1]; calculation of energy released making product bonds: \(2346\text{ kJ}\) [1]; subtraction of values: \(2253 - 2346\) [1]; correct value: \(-93\text{ kJ/mol}\) (accept \(93\text{ kJ/mol}\) if exothermic nature is explicitly stated) [1]
题目 6 · 結構題
10 分
Ethene, \(\text{C}_2\text{H}_4\), is an unsaturated hydrocarbon obtained by cracking long-chain alkanes.
(a) (i) State what is meant by the term unsaturated hydrocarbon. [2]
(ii) Write the word equation for the cracking of decane, \(\text{C}_{10}\text{H}_{22}\), to produce octane and ethene. [1]
(b) Describe a chemical test to distinguish between ethane (a saturated alkane) and ethene (an unsaturated alkene). State the observations for both substances.
observation with ethane: ...................................................
observation with ethene: ................................................... [3]
(c) Ethene undergoes addition polymerisation to form poly(ethene).
(i) Define polymerisation. [1]
(ii) Draw the structure of the poly(ethene) polymer showing two repeating units. [3]
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解题
(a) (i) A hydrocarbon is a compound containing hydrogen and carbon only. Unsaturated means it contains at least one carbon-carbon double bond (\(\text{C}=\text{C}\)). (ii) decane \(\rightarrow\) octane + ethene
(b) Test: Add aqueous bromine (bromine water). Observation with ethane: Remains orange/brown/yellow (no reaction). Observation with ethene: Turns colourless (decolourises).
(c) (i) Polymerisation is the process in which many small molecules (monomers) join together to form a large molecule (polymer). (ii) The structure showing two repeating units is: \(\text{H}\ \ \ \text{H}\ \ \ \text{H}\ \ \ \text{H}\) \(|\ \ \ \ |\ \ \ \ |\ \ \ \ |\) \(\text{- C - C - C - C -}\) \(|\ \ \ \ |\ \ \ \ |\ \ \ \ |\) \(\text{H}\ \ \ \text{H}\ \ \ \text{H}\ \ \ \text{H}\) With continuation bonds at each end of the carbon chain.
评分标准
(a) (i) Contains hydrogen and carbon only [1]; contains a double carbon-carbon bond [1] (ii) decane \(\rightarrow\) octane + ethene [1]
(b) Test: aqueous bromine / bromine water [1]; Ethane: remains orange / yellow / brown / no change [1]; Ethene: decolourises / turns colourless [1] (Reject: 'turns clear' for ethene)
(c) (i) Monomers joining to form a long chain/polymer [1] (ii) Four carbon atoms linked by single bonds in a row [1]; eight hydrogen atoms single-bonded to carbon atoms [1]; continuation bonds at both ends of the carbon chain [1]
题目 7 · 結構題
10 分
When a person accidentally touches a hot stove, they quickly pull their hand away. This is a reflex action.
(a) State three characteristics of a reflex action. [3]
(b) Describe the pathway of the nerve impulse in this reflex arc, naming the three types of neurones involved. [4]
(c) A reflex action is an example of nervous coordination. Compare nervous control and endocrine (hormonal) control in terms of: - the speed of transmission - the duration of the effect. [3]
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解题
(a) A reflex action is: rapid (fast), automatic (involuntary), and protective (does not involve conscious thought).
(b) 1. Receptors in the skin detect heat (stimulus) and generate a nerve impulse. 2. Sensory neurone transmits the impulse to the spinal cord (CNS). 3. Relay neurone in the spinal cord passes the impulse from the sensory neurone to the motor neurone. 4. Motor neurone transmits the impulse to the effector (muscle in the arm), causing it to contract and pull the hand away.
(c) - Speed of transmission: Nervous control is very fast (electrical impulses along neurones), while endocrine control is slower (chemical hormones transported in blood). - Duration of the effect: Nervous control has a short-lived/immediate effect, whereas endocrine control generally has a longer-lasting effect.
评分标准
(a) Any three from: rapid/fast [1]; automatic/involuntary [1]; protective/prevents harm [1]; does not involve conscious brain control [1]
(b) Receptor detects stimulus and sensory neurone carries impulse to CNS/spinal cord [1]; relay neurone in spinal cord connects to motor neurone [1]; motor neurone carries impulse to effector/muscle [1]; muscle contracts to cause response [1]
(c) Speed of transmission: nervous is faster / endocrine is slower [1]; Duration of effect: nervous is shorter-lived / endocrine is longer-lasting [1]; Mode of transmission: nervous uses electrical impulses / endocrine uses chemical hormones [1]
题目 8 · 結構題
10 分
A radioactive isotope of bismuth, bismuth-210 (\(^{210}_{83}\text{Bi}\)), decays by emitting a beta (\(\beta\)) particle to form an isotope of polonium (Po).
(a) (i) Write the decay equation for the beta decay of bismuth-210. [3]
(ii) Describe the nature of a beta (\(\beta\)) particle. [1]
(b) Table 8.1 shows how the activity of a sample of bismuth-210 changes over time.
(ii) Use the data in Table 8.1 to determine the half-life of bismuth-210. Explain how you obtained your answer. [2]
(c) Bismuth-210 can also emit gamma (\(\gamma\)) radiation. Compare the penetrating power and ionising ability of beta particles with gamma radiation. [3]
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解题
(a) (i) In beta decay, a neutron changes into a proton and an electron. The nucleon number of the daughter nucleus remains 210, and its proton number increases by 1 to 84. \(^{210}_{83}\text{Bi} \rightarrow ^{210}_{84}\text{Po} + \ ^0_{-1}\beta\) (or \(e\)). (ii) A beta particle is a fast-moving electron emitted from the nucleus.
(b) (i) Half-life is the time taken for the activity or number of radioactive nuclei of a sample to decrease to half of its initial value. (ii) The half-life is 5 days. From Table 8.1, the activity halves from 800 cpm to 400 cpm in the interval from day 0 to day 5 (or from 400 to 200 cpm from day 5 to day 10, which also takes 5 days).
(c) - Penetrating power: Gamma radiation is highly penetrating (requires thick lead to stop), whereas beta particles are moderately penetrating (stopped by a few mm of aluminium). - Ionising ability: Beta particles are moderately ionising, while gamma rays are weakly ionising.
评分标准
(a) (i) Polonium symbol Po with correct mass number 210 and proton number 84 [1]; beta particle symbol \(^{0}_{-1}\beta\) or \(^{0}_{-1}e\) [1]; correctly balanced equation [1] (ii) Fast-moving electron [1]
(b) (i) Time taken for half of the radioactive nuclei to decay / activity to halve [1] (ii) 5 days [1]; explanation showing calculation/interval of halving (e.g., activity goes from 800 to 400 in 5 days) [1]
(c) Gamma is more penetrating than beta [1]; beta is more ionising than gamma [1]; correct material comparison (e.g., beta stopped by aluminium, gamma requires thick lead/concrete) [1]
题目 9 · 結構題
10 分
A student investigates the acceleration of a small electric car. The car starts from rest and accelerates uniformly to a speed of \(6.0\text{ m/s}\) in \(4.0\text{ s}\). It then travels at a constant speed of \(6.0\text{ m/s}\) for \(8.0\text{ s}\).
(a) (i) Calculate the acceleration of the car during the first \(4.0\text{ s}\).
(ii) Calculate the total distance travelled by the car during the \(12.0\text{ s}\) journey.
(b) The car has a mass of \(1.2\text{ kg}\).
(i) Calculate the kinetic energy of the car when it is travelling at its constant speed.
(ii) The electric motor of the car has an efficiency of \(75\%\). Calculate the electrical energy supplied to the motor to produce a useful kinetic energy of \(21.6\text{ J}\).
(ii) Efficiency = \(\frac{\text{Useful energy output}}{\text{Total energy input}} \times 100\%\) \(75\% = \frac{21.6\text{ J}}{\text{Total energy input}} \times 100\%\) \(\text{Total energy input} = \frac{21.6\text{ J}}{0.75} = 28.8\text{ J}\).
评分标准
(a)(i) [2 marks] - 1 mark for formula \(a = \frac{\Delta v}{t}\) or working \(\frac{6.0}{4.0}\) - 1 mark for correct value with unit \(1.5\text{ m/s}^2\)
(a)(ii) [3 marks] - 1 mark for calculating the area of the acceleration triangle (\(12\text{ m}\)) - 1 mark for calculating the area of the constant speed rectangle (\(48\text{ m}\)) - 1 mark for total distance of \(60\text{ m}\)
(b)(i) [2 marks] - 1 mark for formula \(E_k = \frac{1}{2}mv^2\) or working \(\frac{1}{2} \times 1.2 \times 6.0^2\) - 1 mark for correct answer \(21.6\text{ J}\)
(b)(ii) [3 marks] - 1 mark for efficiency formula or rearangement \(\text{Total energy} = \frac{\text{Useful energy}}{\text{efficiency}}\) - 1 mark for correct substitution \(\frac{21.6}{0.75}\) - 1 mark for correct calculation \(28.8\text{ J}\)
题目 10 · 結構題
10 分
A hydrocarbon fuel, propane (\(\text{C}_3\text{H}_8\)), undergoes complete combustion.
(a) Propane is an alkane.
(i) Write a balanced chemical equation for the complete combustion of propane.
(ii) Draw the structural formula of propane showing all atoms and bonds.
(b) Propene (\(\text{C}_3\text{H}_6\)) is an alkene that can be formed by cracking propane.
(i) State the conditions required for industrial catalytic cracking.
(ii) Describe a test to distinguish between propane and propene. State the observations for each.
(c) Propene can polymerise to form poly(propene).
(i) State the type of polymerisation reaction that occurs.
(ii) Structural formula of propane: H H H | | | H-C - C - C-H | | | H H H
(b) (i) Conditions for catalytic cracking: High temperature (approximately \(500\text{ }^\circ\text{C}\) to \(700\text{ }^\circ\text{C}\)) and a catalyst (such as alumina / silica / zeolite).
(ii) Test: Add bromine water / aqueous bromine. Propene (alkene) turns the orange/brown bromine water colourless (decolourises it). Propane (alkane) does not react, and the mixture remains orange/brown.
(c) (i) Addition polymerisation.
(ii) Repeat unit of poly(propene): H H | | -[C - C]- | | H CH3
评分标准
(a)(i) [3 marks] - 1 mark for correct formulas of reactants (\(\text{C}_3\text{H}_8\) and \(\text{O}_2\)) - 1 mark for correct formulas of products (\(\text{CO}_2\) and \(\text{H}_2\text{O}\)) - 1 mark for correct balancing coefficients (5, 3, 4)
(a)(ii) [1 mark] - 1 mark for fully drawn structure of propane with single covalent bonds and correct valencies (3 carbons with 8 hydrogens).
(b)(i) [2 marks] - 1 mark for mentioning high temperature (accept range \(500\text{ }^\circ\text{C}\) to \(700\text{ }^\circ\text{C}\)) - 1 mark for catalyst (accept alumina, silica, zeolite)
(b)(ii) [2 marks] - 1 mark for testing with bromine water - 1 mark for stating correct results: propene decolourises / turns colourless and propane remains orange/brown
(c)(i) [1 mark] - 1 mark for addition polymerisation
(c)(ii) [1 mark] - 1 mark for repeat unit with single C-C bond in backbone, continuation bonds extending outside the brackets, and correct side-groups (H, H, H, \(\text{CH}_3\))
题目 11 · 結構題
10 分
A student accidentally touches a hot object and pulls their hand away rapidly. This is a reflex action.
(a) (i) Describe the pathway of the nerve impulse from the receptor in the skin to the effector muscle. Include the three types of neurones involved.
(ii) State two features of a reflex action that distinguish it from a voluntary action.
(b) The response to a dangerous situation also involves a hormonal response.
(i) Name the hormone released in a 'fight or flight' situation and the gland that secretes it.
(ii) State two physiological changes caused by this hormone that prepare the body for action.
(iii) State the method of transport of this hormone to its target organs.
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解题
(a) (i) The stimulus (heat) is detected by receptors in the skin. An electrical impulse is generated and travels along the sensory neurone to the central nervous system (spinal cord). Inside the spinal cord, the impulse crosses a synapse to a relay neurone, and then across another synapse to a motor neurone. The impulse travels along the motor neurone to the effector (the muscle), which contracts to move the hand.
(ii) Features of a reflex action: 1. It is involuntary / automatic / does not involve conscious thought from the brain. 2. It is extremely rapid / fast.
(b) (i) The hormone is adrenaline, secreted by the adrenal glands.
(ii) Physiological changes include: increased heart rate, increased breathing rate, dilation of pupils, and conversion of glycogen to glucose in the liver.
(iii) Transported in the blood plasma.
评分标准
(a)(i) [3 marks] - 1 mark for pathway from receptor via sensory neurone to CNS / spinal cord. - 1 mark for relay neurone inside the CNS / crossing synapse. - 1 mark for pathway via motor neurone to the effector muscle.
(a)(ii) [2 marks] - 1 mark for stating it is involuntary / does not require conscious thought. - 1 mark for stating it is rapid / fast.
(b)(i) [2 marks] - 1 mark for adrenaline. - 1 mark for adrenal gland(s).
(b)(ii) [2 marks] - Any two from: increased heart rate, increased breathing rate, dilated pupils, glycogen breakdown to glucose.
(b)(iii) [1 mark] - 1 mark for transport in blood (plasma).
题目 12 · 結構題
10 分
A student investigates a circuit containing a resistor and a lamp connected in series to a DC power supply.
(a) A \(12\text{ V}\) power supply is connected to a series combination of a \(4.0\text{ }\Omega\) resistor and a lamp. The current in the circuit is \(1.5\text{ A}\).
(i) Calculate the potential difference across the \(4.0\text{ }\Omega\) resistor.
(ii) Show that the resistance of the lamp is \(4.0\text{ }\Omega\).
(iii) Calculate the power dissipated by the lamp.
(b) The circuit is now modified so that another identical \(4.0\text{ }\Omega\) lamp is connected in parallel with the first lamp. The series resistor of \(4.0\text{ }\Omega\) remains in series with this parallel combination.
(i) Calculate the combined resistance of the two lamps in parallel.
(ii) Determine the total resistance of the modified circuit.
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解题
(a) (i) \(V_{\text{resistor}} = I \times R = 1.5\text{ A} \times 4.0\text{ }\Omega = 6.0\text{ V}\).
(ii) Total resistance of the circuit: \(R_{\text{total}} = \frac{V}{I} = \frac{12\text{ V}}{1.5\text{ A}} = 8.0\text{ }\Omega\). Since the components are in series, \(R_{\text{total}} = R_{\text{resistor}} + R_{\text{lamp}}\). \(R_{\text{lamp}} = R_{\text{total}} - R_{\text{resistor}} = 8.0\text{ }\Omega - 4.0\text{ }\Omega = 4.0\text{ }\Omega\).
(iii) Power of the lamp: \(P = I^2 R_{\text{lamp}} = (1.5\text{ A})^2 \times 4.0\text{ }\Omega = 2.25 \times 4.0 = 9.0\text{ W}\) (or \(P = V_{\text{lamp}} \times I = (12 - 6)\text{ V} \times 1.5\text{ A} = 9.0\text{ W}\)).
(ii) Total resistance is the series resistor plus the combined parallel resistance: \(R_{\text{total}} = R_{\text{resistor}} + R_p = 4.0\text{ }\Omega + 2.0\text{ }\Omega = 6.0\text{ }\Omega\).
评分标准
(a)(i) [2 marks] - 1 mark for formula \(V = IR\) or substitution \(1.5 \times 4.0\) - 1 mark for correct value with unit \(6.0\text{ V}\)
(a)(ii) [2 marks] - 1 mark for calculating total resistance \(R_{\text{total}} = \frac{12}{1.5} = 8.0\text{ }\Omega\) - 1 mark for subtracting \(4.0\text{ }\Omega\) series resistor to yield \(4.0\text{ }\Omega\) lamp resistance
(a)(iii) [2 marks] - 1 mark for power formula \(P = I^2R\) or \(P = VI\) with substitution - 1 mark for correct calculation \(9.0\text{ W}\)
(b)(i) [2 marks] - 1 mark for formula \(R_p = \frac{R_1 R_2}{R_1 + R_2}\) or parallel calculation step - 1 mark for correct value \(2.0\text{ }\Omega\)
(b)(ii) [2 marks] - 1 mark for adding the parallel resistance and series resistor - 1 mark for correct final resistance \(6.0\text{ }\Omega\)
Paper 63 (Alternative to Practical)
Six alternative to practical questions. Answer all questions. Write your answers in the spaces provided.
6 题目 · 60 分
题目 1 · Practical / Experimental Question
10 分
A student investigates osmosis using cylinders cut from a fresh potato.
The student: - Cuts five potato cylinders of equal initial length (\(40.0\text{ mm}\)). - Measures the initial mass of each cylinder. - Places each cylinder in a test-tube containing a different concentration of sucrose solution (\(0.0, 0.2, 0.4, 0.6, \text{ and } 0.8\text{ mol/dm}^3\)) for 60 minutes. - Removes the cylinders, blots them dry, and measures their final mass.
(a) Fig. 1.1 shows the balance readings for the initial and final mass of the potato cylinder placed in the \(0.4\text{ mol/dm}^3\) solution.
(i) On a grid, plot a graph of the percentage change in mass (vertical axis) against sucrose concentration. Start both axes from (0,0) but allow negative values on the vertical axis. [3] (ii) Draw the best-fit straight line. [1]
(c) Use your graph to estimate the concentration of sucrose inside the potato cells. Explain your answer. [2]
(d) Explain why it is important to blot the potato cylinders dry before weighing them. [1]
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解题
(a)(i) Initial mass = \(2.45\text{ g}\) Final mass = \(2.18\text{ g}\)
(a)(ii) Change in mass = \(2.18 - 2.45 = -0.27\text{ g}\) Percentage change = \(\frac{-0.27}{2.45} \times 100 \approx -11.0\%\)
(b)(i) Axes correctly labelled with units: Sucrose concentration on x-axis (\(\text{mol/dm}^3\)) and Percentage mass change on y-axis (\%) Scales must be linear and cover at least half of the grid. All 5 points plotted correctly. (b)(ii) Single straight line of best fit drawn with a ruler, showing even distribution of points.
(c) The estimated concentration is the point where the line crosses the x-axis (percentage change in mass is \(0\)), which is approximately \(0.48\text{ mol/dm}^3\) (accept \(0.46 - 0.50\text{ mol/dm}^3\)). At this concentration, there is no net movement of water into or out of the potato cells because the water potential inside the cells is equal to the water potential of the surrounding sucrose solution.
(d) To remove excess liquid / surface water from the outside of the potato cylinder, which would otherwise add to the mass and make the measurement inaccurate.
评分标准
Total 10 marks: - (a)(i) [1 mark] for both masses correct (2.45 g and 2.18 g). - (a)(ii) [2 marks] for correct working shown and percentage change calculation (-11.0%). Allow ECF from (a)(i). - (b)(i) [3 marks] as follows: * 1 mark for correctly labelled axes with units. * 1 mark for suitable linear scales covering at least half of the grid. * 1 mark for correct plotting of all 5 points. - (b)(ii) [1 mark] for drawing a single, straight line of best fit using a ruler. - (c) [2 marks] as follows: * 1 mark for reading the correct concentration where percentage change is 0 (allow 0.46 to 0.50 mol/dm^3 or according to student's line of best fit). * 1 mark for explaining that at this point there is no net osmosis / water potential is equal. - (d) [1 mark] for explaining that surface liquid adds to the measured mass / ensures only the internal mass change is measured.
题目 2 · Practical / Experimental Question
10 分
Plan an investigation to determine how pH affects the activity of the enzyme amylase when breaking down starch.
You are provided with: - Amylase enzyme solution - Starch solution - Iodine solution - Buffer solutions of different pH values (\(pH\ 4.0, 5.0, 6.0, 7.0, 8.0, 9.0\)) - Common laboratory apparatus (test-tubes, pipettes, water-baths, spotting tiles, stopwatches)
Do not carry out this investigation.
Include in your plan: - the apparatus needed - a brief description of the method, including how you will determine when starch has been completely broken down - the variables you will control and how you will control them - how you will process your results to draw a conclusion. [10]
Method: 1. Label test-tubes for each pH buffer solution. 2. Add a fixed volume (e.g., \(2\text{ cm}^3\)) of starch solution and a fixed volume (e.g., \(1\text{ cm}^3\)) of amylase solution into separate test-tubes containing the respective pH buffer. 3. Place the test-tubes in a water-bath at a constant temperature (e.g., \(37^\circ\text{C}\)) for 5 minutes to equilibrate. 4. Put one drop of iodine solution into each well of a spotting tile. 5. Mix the starch and amylase solutions together, start the stopwatch immediately. 6. Every 30 seconds, take a sample of the mixture using a pipette and add it to a well containing iodine solution on the spotting tile. 7. Record the time taken for the iodine to remain yellow-brown (indicating all starch is broken down / end-point reached). 8. Repeat the procedure for all other pH buffer solutions.
Control variables: - Temperature: kept constant by using a water-bath (and monitored with a thermometer). - Concentration and volume of starch and amylase: kept constant by using the same stock solutions and measuring volumes accurately with pipettes.
Processing results and conclusion: - Calculate the rate of reaction as \(\frac{1}{\text{time taken}}\). - Plot a graph of rate of reaction against pH. The pH with the highest rate of reaction is the optimum pH for amylase activity.
评分标准
Total 10 marks (maximum 1 mark per point): - 1 mark for mentioning spotting tile, pipettes/syringes, and stopwatch. - 1 mark for mixing amylase, starch, and buffer solution of a specified pH. - 1 mark for using iodine solution in wells of a spotting tile to test for starch. - 1 mark for taking samples at regular time intervals (e.g., every 30 seconds) and starting stopwatch immediately upon mixing. - 1 mark for stating that the end-point is reached when the iodine solution remains yellow/brown (no longer turns blue-black). - 1 mark for repeating the entire process for at least 3 different pH values. - 1 mark for keeping temperature constant using a water-bath. - 1 mark for keeping the volumes/concentrations of starch and amylase solutions constant. - 1 mark for calculating rate as 1 / time, or plotting a graph of time taken/rate against pH. - 1 mark for stating that the lowest time (or highest rate) indicates the optimum pH for the enzyme.
题目 3 · Practical / Experimental Question
10 分
A student investigates the temperature change when different metals react with dilute hydrochloric acid.
The student: 1. Places \(25\text{ cm}^3\) of dilute hydrochloric acid into a polystyrene cup. 2. Measures the initial temperature of the acid. 3. Adds \(1.0\text{ g}\) of zinc powder, stirs, and records the highest temperature reached. 4. Repeats the steps using \(1.0\text{ g}\) of magnesium powder and \(1.0\text{ g}\) of iron powder.
(a) Fig. 3.1 shows the thermometer readings for the reaction with zinc powder. - Initial temperature reading: \(19.5^\circ\text{C}\) - Highest temperature reading: \(31.0^\circ\text{C}\)
(i) Record the temperatures to the nearest \(0.5^\circ\text{C}\). Initial temperature = .................... \(^\circ\text{C}\) Highest temperature = .................... \(^\circ\text{C}\) [2]
(ii) Calculate the temperature rise (\(\Delta T\)) for the reaction with zinc. \(\Delta T = \) .................... \(^\circ\text{C}\) [1]
(b) The temperature rises for magnesium and iron are shown below: - Magnesium: \(\Delta T = 18.5^\circ\text{C}\) - Iron: \(\Delta T = 4.0^\circ\text{C}\)
(i) Construct a results table showing the metal used, the initial temperature, the highest temperature, and the temperature rise (\(\Delta T\)) for all three metals. Assume the initial temperature of the acid was the same (\(19.5^\circ\text{C}\)) for all experiments. [3]
(ii) Use the results to list the three metals in order of their reactivity, starting with the most reactive. [1]
(c) State why a polystyrene cup is used instead of a glass beaker in this experiment. [1]
(d) Explain how a lid on the polystyrene cup would affect the accuracy of the temperature changes measured. [1]
(e) State one safety precaution the student should take when handling dilute hydrochloric acid. [1]
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解题
(a)(i) Initial temperature = \(19.5^\circ\text{C}\) Highest temperature = \(31.0^\circ\text{C}\)
(a)(ii) \(\Delta T = 31.0 - 19.5 = 11.5^\circ\text{C}\)
(b)(i) Table with correct headings and units:
| Metal | Initial temperature / \(^\circ\text{C}\) | Highest temperature / \(^\circ\text{C}\) | Temperature rise \(\Delta T\) / \(^\circ\text{C}\) | |---|---|---|---| | Magnesium | 19.5 | 38.0 | 18.5 | | Zinc | 19.5 | 31.0 | 11.5 | | Iron | 19.5 | 23.5 | 4.0 |
(b)(ii) Magnesium, Zinc, Iron.
(c) Polystyrene is a good thermal insulator, which reduces heat loss to the surroundings, making the measured temperature rise more accurate.
(d) It would reduce heat loss from the surface of the liquid to the air, resulting in a higher and more accurate measured temperature rise.
Total 10 marks: - (a)(i) [2 marks] 1 mark for each correct temperature reading (initial = 19.5 °C, highest = 31.0 °C). - (a)(ii) [1 mark] for correct calculation of temperature rise (11.5 °C). Allow ECF from (a)(i). - (b)(i) [3 marks] as follows: * 1 mark for drawing a table with correct headings and units. * 1 mark for correctly entering the zinc data. * 1 mark for correctly calculating the highest temperatures for magnesium (38.0 °C) and iron (23.5 °C) and entering them. - (b)(ii) [1 mark] for correct reactivity order: Magnesium, Zinc, Iron. - (c) [1 mark] for stating that polystyrene is an insulator / reduces heat loss. - (d) [1 mark] for stating that a lid reduces heat loss to the air / increases the measured temperature rise. - (e) [1 mark] for stating a valid safety precaution (e.g., wearing safety goggles / gloves).
题目 4 · Practical / Experimental Question
10 分
A student is given a solid sample of salt **W** to analyze. The student dissolves **W** in distilled water to make solution **W**.
The student performs five tests on solution **W**. Table 4.1 shows the tests and the student's observations.
Table 4.1 - Test 1: Add a few drops of aqueous sodium hydroxide -> Observation: Green precipitate forms - Test 2: Add excess aqueous sodium hydroxide -> Observation: Precipitate is insoluble in excess - Test 3: Add a few drops of aqueous ammonia -> Observation: Green precipitate forms - Test 4: Add excess aqueous ammonia -> Observation: Precipitate is insoluble in excess - Test 5: Add dilute hydrochloric acid followed by a few drops of aqueous barium chloride -> Observation: White precipitate forms
(a) Identify the cation present in solution **W**. [1]
(b) Identify the anion present in solution **W**. [1]
(c) Deduce the chemical formula of salt **W**. [1]
(d) Write the ionic equation for the reaction that occurs in Test 5. Include state symbols. [2]
(e) In Test 5, the student added dilute hydrochloric acid before adding barium chloride solution. (i) Explain why the acid is added first. [1] (ii) State why dilute sulfuric acid cannot be used to acidify the mixture in Test 5. [1]
(f) The student performs a flame test on solid **W**. (i) State how a flame test is carried out. [2] (ii) State why a blue Bunsen burner flame is used instead of a yellow flame. [1]
(e)(i) To react with and remove any carbonate impurities (which would otherwise produce a false positive white precipitate of barium carbonate). (e)(ii) Dilute sulfuric acid contains sulfate ions (\(\text{SO}_4^{2-}\)) which would react with barium chloride to form a white precipitate of barium sulfate, invalidating the test.
(f)(i) Dip a clean nichrome/platinum wire into concentrated hydrochloric acid, then dip it into the solid sample, and place the wire into the hot/blue Bunsen burner flame. (f)(ii) A blue flame is non-luminous, so it does not mask the flame colour of the metal ion, unlike a yellow flame which is bright and yellow.
评分标准
Total 10 marks: - (a) [1 mark] for iron(II) / Fe^2+. - (b) [1 mark] for sulfate / SO4^2-. - (c) [1 mark] for FeSO4. Allow ECF from (a) and (b). - (d) [2 marks] as follows: * 1 mark for correct formulas of reactants and products (Ba^2+ + SO4^2- -> BaSO4). * 1 mark for correct state symbols ((aq) for ions, (s) for BaSO4). - (e)(i) [1 mark] for explaining that it removes carbonate ions / prevents barium carbonate precipitate. - (e)(ii) [1 mark] for stating that sulfuric acid contains sulfate ions / would form a precipitate with barium ions. - (f)(i) [2 marks] as follows: * 1 mark for using a clean wire dipped in acid. * 1 mark for placing it into the Bunsen flame. - (f)(ii) [1 mark] for stating that the blue flame is non-luminous / doesn't mask the flame colour.
题目 5 · Practical / Experimental Question
10 分
A student determines the density of a small, irregularly shaped pebble and a regular cuboid block of wood.
(a) Fig. 5.1 shows a balance measuring the mass of the pebble. - Balance reading: \(42.6\text{ g}\)
State the mass \(m\) of the pebble. \(m = \) .................... g [1]
(b) To determine the volume of the pebble, the student uses the displacement of water in a measuring cylinder. Fig. 5.2 shows the measuring cylinder containing water before and after the pebble is immersed. - Initial water level: \(54.0\text{ cm}^3\) - Final water level: \(71.5\text{ cm}^3\)
(i) State the initial and final volume of water. Initial volume \(V_1 = \) .................... \(\text{cm}^3\) Final volume \(V_2 = \) .................... \(\text{cm}^3\) [2]
(ii) Calculate the volume \(V_p\) of the pebble. \(V_p = \) .................... \(\text{cm}^3\) [1]
(c) Calculate the density \(\rho\) of the pebble. State the unit of your answer. Density \(\rho = \) .................... [2]
(d) State how the student should read the scale on the measuring cylinder to avoid a parallax error. [1]
(e) The student measures the dimensions of the rectangular wood block with a ruler: - Length = \(5.0\text{ cm}\) - Width = \(4.0\text{ cm}\) - Height = \(2.5\text{ cm}\)
(i) Calculate the volume of the wood block. Volume = .................... \(\text{cm}^3\) [1]
(ii) The student attempts to measure the volume of this wood block using the displacement of water method as in part (b). Suggest why this method is not suitable for wood. [2]
(e)(ii) Wood floats on water / does not submerge completely, so the displaced volume cannot be measured easily without pushing it down. Also, wood absorbs water, which decreases the measured volume change / makes the mass calculation inaccurate.
评分标准
Total 10 marks: - (a) [1 mark] for 42.6 g. - (b)(i) [2 marks] for each reading correct (V1 = 54.0 cm^3, V2 = 71.5 cm^3). - (b)(ii) [1 mark] for volume = 17.5 cm^3. Allow ECF from (b)(i). - (c) [2 marks] as follows: * 1 mark for correct calculation (2.43 g/cm^3) to 2 or 3 sig figs. * 1 mark for correct unit (g/cm^3). - (d) [1 mark] for reading at eye level / perpendicular to the meniscus / at the bottom of the meniscus. - (e)(i) [1 mark] for 50 cm^3. - (e)(ii) [2 marks] as follows: * 1 mark for stating that wood floats (requires a sinker to submerge). * 1 mark for stating that wood absorbs water (which affects volume accuracy).
题目 6 · Practical / Experimental Question
10 分
A student investigates how the length of a resistance wire affects its resistance.
(a) Draw a circuit diagram to show how the student should connect the apparatus. The circuit must contain: - a power supply (cell) - an ammeter to measure the current in the test wire - a voltmeter connected to measure the potential difference across a variable length of the test wire - a switch - a sliding contact (such as a crocodile clip) to vary the length of the test wire connected in the circuit. [3]
(b) The student connects the circuit and measures the potential difference \(V\) and current \(I\) for different lengths \(L\) of the wire.
(i) Calculate the resistance of the \(80.0\text{ cm}\) length of wire. Record your answer to three significant figures in Table 6.1. [1]
(ii) State the relationship between the length \(L\) of the wire and its resistance \(R\) shown by the results in Table 6.1. [1]
(c) In this experiment, the current is kept constant at \(0.38\text{ A}\). (i) Explain why the student should open the switch between readings. [1] (ii) Suggest how a heating effect in the wire would affect its resistance. [1]
(d) The student repeats the experiment using a wire of the same material and length, but with a larger cross-sectional area. (i) State how the resistance of this thicker wire compares with the original wire. [1] (ii) Predict the current in the thicker wire if the potential difference is kept at \(1.30\text{ V}\) (as it was for \(40.0\text{ cm}\ of the original wire). Explain your answer. [2]
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解题
(a) Diagram should show: - Cell, switch, ammeter, and test wire connected in series. - Voltmeter connected in parallel across the portion of the test wire between the fixed end and the sliding contact. - Correct symbols used for cell, switch, ammeter, voltmeter, and resistor/wire.
(b)(ii) Resistance is directly proportional to length (or as length increases, resistance increases linearly).
(c)(i) To prevent the wire from heating up (which changes its resistance and is a safety hazard). (c)(ii) Heating the wire would increase its resistance.
(d)(i) Its resistance would be lower. (d)(ii) The current would be higher because a lower resistance allows more current to flow at the same potential difference (\(I = \frac{V}{R}\)).
评分标准
Total 10 marks: - (a) [3 marks] as follows: * 1 mark for correct series connection of cell, switch, ammeter, and test wire. * 1 mark for connecting the voltmeter in parallel across the variable section of the test wire. * 1 mark for all correct circuit symbols. - (b)(i) [1 mark] for 6.84 \Omega. - (b)(ii) [1 mark] for stating that resistance is directly proportional to length / as length increases, resistance increases. - (c)(i) [1 mark] for explaining that it prevents the wire from heating up. - (c)(ii) [1 mark] for stating that heating increases resistance. - (d)(i) [1 mark] for stating that resistance is lower. - (d)(ii) [2 marks] as follows: * 1 mark for predicting higher current. * 1 mark for explaining that lower resistance leads to higher current (or referencing Ohm's law, I = V/R).
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