An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V2) Cambridge International A Level Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.
甲部: Biology, Chemistry, and Physics Extended Theory
Answer all questions. Show all working and use appropriate units. A calculator and Periodic Table are permitted.
12 题目 · 120 分
题目 1 · structured
10 分
A young bean seedling is placed horizontally in a dark box. (a) Define the term gravitropism. [2] (b) Describe and explain how auxin controls the growth response in the shoot of this horizontal bean seedling. [5] (c) State how the response of a root to gravity differs from that of a shoot, and explain the survival advantage of this response to the plant. [3]
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解题
(a) Gravitropism is a growth response in which parts of a plant grow towards or away from gravity (the gravitational stimulus). (b) Auxin is synthesized in the shoot tip and diffuses downwards. Due to gravity, auxin accumulates in higher concentrations on the lower side of the horizontally positioned shoot. In shoots, higher auxin concentration stimulates cell elongation. Consequently, cells on the lower side elongate faster than those on the upper side, causing the shoot to bend upwards against gravity (negative gravitropism). (c) Unlike the shoot, the root shows positive gravitropism, meaning it grows downwards towards the direction of gravity. The survival advantages of this are: it anchors the plant firmly in the soil, and it allows the roots to grow downwards to reach and absorb water and dissolved mineral ions.
评分标准
(a) A growth response [1] to gravity / gravitational stimulus [1]. (b) Any five from: auxin produced in shoot tip [1]; auxin diffuses/moves down the shoot [1]; auxin accumulates on lower side of shoot due to gravity [1]; high concentration of auxin stimulates cell elongation in shoots [1]; lower side grows/elongates faster than upper side [1]; shoot bends upwards / shows negative gravitropism [1]. (c) Root grows downwards / is positively gravitropic [1]; allows root to anchor plant [1]; allows root to reach water or mineral ions [1].
题目 2 · structured
10 分
(a) Ethane \(\text{C}_2\text{H}_6\) and ethene \(\text{C}_2\text{H}_4\) are hydrocarbons. (i) State what is meant by the term hydrocarbon. [1] (ii) Describe a chemical test to distinguish between ethane and ethene, including the expected results for both compounds. [3] (b) Ethene can be polymerized to form poly(ethene). (i) Name the type of polymerization reaction that occurs. [1] (ii) State the structural difference between the monomer ethene and the repeat unit of poly(ethene). [2] (c) Ethanol can be produced by the hydration of ethene. State the temperature, pressure, and catalyst required for this industrial process. [3]
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解题
(a)(i) A hydrocarbon is a compound containing carbon and hydrogen atoms only. (ii) Add aqueous bromine (or bromine water) to both hydrocarbons. Ethane will show no change and the solution remains orange/brown. Ethene will react, causing the bromine water to decolourise (turn from orange/brown to colourless). (b)(i) Addition polymerization. (ii) Ethene contains a carbon-carbon double bond (\(\text{C}=\text{C}\)), whereas the repeat unit of poly(ethene) contains a carbon-carbon single bond (\(\text{C}-\text{C}\)) with single bonds extending outwards on each side of the carbon atoms to connect to adjacent units. (c) The reaction requires a temperature of 300 \(^{\circ}\text{C}\), a pressure of 60 atm (or 6000 kPa), and a phosphoric acid (\(\text{H}_3\text{PO}_4\)) catalyst.
评分标准
(a)(i) Compound containing carbon and hydrogen ONLY [1 mark - reject if other elements mentioned]. (ii) Add bromine water / aqueous bromine [1 mark]; ethane remains orange/brown/no change [1 mark]; ethene decolourises / turns colourless [1 mark - reject 'clear']. (b)(i) Addition (polymerization) [1 mark]. (ii) Ethene has a double carbon-carbon bond / \(\text{C}=\text{C}\) [1 mark]; repeat unit has a single carbon-carbon bond / \(\text{C}-\text{C}\) (with open-ended continuation bonds) [1 mark]. (c) Temperature of 300 \(^{\circ}\text{C}\) [1 mark]; pressure of 60 atm / 6000 kPa [1 mark]; phosphoric acid / \(\text{H}_3\text{PO}_4\) catalyst [1 mark].
题目 3 · structured
10 分
A student sets up a circuit with a 12.0 V d.c. power supply, an ammeter, and two resistors connected in parallel. Resistor A has a resistance of 4.0 \(\Omega\) and Resistor B has a resistance of 12.0 \(\Omega\). (a) Calculate the combined resistance of the two resistors connected in parallel. [2] (b) Calculate the total current measured by the ammeter in the main circuit. State the unit. [3] (c) Calculate the electrical power dissipated by the 4.0 \(\Omega\) resistor. [2] (d) The student now disconnects the resistors and reconnects the same two resistors in series with the same 12.0 V supply. (i) State and explain how the current in this series circuit compares to the current in the parallel circuit. [2] (ii) Calculate the total current in this series circuit. [1]
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解题
(a) For parallel resistors, \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{4.0} + \frac{1}{12.0} = \frac{3}{12.0} + \frac{1}{12.0} = \frac{4}{12.0}\). Therefore, \(R_p = \frac{12.0}{4} = 3.0\\ \Omega\). (b) Using Ohm's law: \(I = \frac{V}{R_p} = \frac{12.0\text{ V}}{3.0\ \Omega} = 4.0\text{ A}\). The unit of current is Amperes (A). (c) Power \(P = \frac{V^2}{R}\). Since the resistors are in parallel, the potential difference across the 4.0 \(\Omega\) resistor is 12.0 V. Thus, \(P = \frac{12.0^2}{4.0} = \frac{144}{4.0} = 36.0\text{ W}\). (d)(i) The current in the series circuit is less than in the parallel circuit because connecting resistors in series increases the total resistance (to 16.0 \(\Omega\)), and current is inversely proportional to resistance for a constant voltage. (ii) Total resistance in series \(R_s = 4.0\ \Omega + 12.0\ \Omega = 16.0\ \Omega\). Current \(I = \frac{V}{R_s} = \frac{12.0\text{ V}}{16.0\ \Omega} = 0.75\text{ A}\).
评分标准
(a) Formula used: \(R_p = \frac{R_1 R_2}{R_1 + R_2}\) or \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\) [1 mark]; Correct answer: 3.0 \(\Omega\) [1 mark]. (b) Use of \(I = \frac{V}{R}\) [1 mark]; Correct calculation: 4.0 [1 mark]; Unit: A / Ampere / Amps [1 mark]. (c) Correct power formula e.g. \(P = \frac{V^2}{R}\) or calculates branch current \(I = 3.0\text{ A}\) and uses \(P = I^2 R\) [1 mark]; Correct calculation: 36.0 W / 36.0 J/s [1 mark]. (d)(i) Current is less [1 mark]; because total resistance is higher in series / resistance is 16 \(\Omega\) compared to 3 \(\Omega\) in parallel [1 mark]. (ii) 0.75 A / 0.75 [1 mark].
题目 4 · Structured
10 分
A student accidentally touches a hot beaker in a laboratory and quickly pulls their hand away. This is a reflex action.
(a) Describe the pathway of the nerve impulses that result in this reflex action. [4]
(b) Explain how a nerve impulse is transmitted across a synapse from one neurone to the next. [3]
(c) The student's eyes also respond to the bright light in the laboratory. Describe the changes that occur in the pupil and iris when moving from a dimly lit room into a brightly lit laboratory, and explain the purpose of this response. [3]
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解题
(a) The process begins when the high temperature (stimulus) is detected by temperature receptors in the skin. This triggers an electrical impulse that travels along a sensory neurone to the central nervous system (spinal cord). Inside the spinal cord, the impulse is passed across a synapse to a relay neurone, and then across another synapse to a motor neurone. The motor neurone transmits the impulse to the effector, which is the muscle in the arm. The muscle contracts, causing the hand to pull away.
(b) When a nerve impulse reaches the end of the first (presynaptic) neurone, it causes vesicles to release chemical messenger molecules called neurotransmitters. These molecules diffuse across the narrow gap (synaptic cleft) between the neurones. They bind to specific receptor molecules on the membrane of the second (postsynaptic) neurone. This binding triggers a new electrical impulse in the second neurone.
(c) When entering a brightly lit room, circular muscles in the iris contract while radial muscles relax. This causes the pupil to constrict (become smaller). The purpose of this response is to limit the amount of light entering the eye, thereby protecting the retina from damage by overexposure to bright light.
评分标准
(a) Max 4 marks: - Receptor detects stimulus (heat / high temperature) [1] - Electrical impulse travels along sensory neurone to spinal cord / CNS [1] - Passed to relay neurone (via synapse) [1] - Passed to motor neurone [1] - Impulse reaches effector / muscle which contracts [1]
(b) Max 3 marks: - Impulse triggers release of neurotransmitters (from vesicles) [1] - Neurotransmitters diffuse across the synaptic cleft / gap [1] - Neurotransmitters bind to receptors on the postsynaptic membrane / second neurone [1] - Generates new electrical impulse [1]
(c) Max 3 marks: - Circular muscles of the iris contract [1] - Radial muscles of the iris relax [1] - Pupil constricts / gets smaller [1] - Purpose: protects the retina / prevents too much light entering [1]
题目 5 · Structured
10 分
Decane, \(\text{C}_{10}\text{H}_{22}\), is a long-chain alkane that can be cracked to produce ethene, \(\text{C}_2\text{H}_4\), and other useful hydrocarbons.
(a) (i) State two conditions required for industrial cracking. [2] (ii) Write a balanced chemical equation for the cracking of decane to produce ethene and one other hydrocarbon product. [1]
(b) Ethene can react with steam to form ethanol. (i) State the name of this type of reaction and the catalyst used. [2] (ii) Draw the displayed formula of ethanol, showing all atoms and bonds. [1]
(c) Ethene can undergo polymerisation to form poly(ethene). (i) Describe how ethene molecules form a polymer, including the term used for ethene in this process. [2] (ii) Draw the structure of the repeating unit of poly(ethene). [2]
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解题
(a) (i) Industrial cracking requires high temperatures (typically around \(600-700^\circ\text{C}\)) and a catalyst (such as alumina, silica, or zeolites). (ii) The balanced equation is: \(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_2\text{H}_4 + \text{C}_8\text{H}_{18}\).
(b) (i) This is an addition reaction (specifically, hydration). The catalyst used is phosphoric acid (\(\text{H}_3\text{PO}_4\)). (ii) The displayed formula of ethanol has two carbon atoms single-bonded to each other, with one carbon attached to three hydrogens, and the other attached to two hydrogens and an -OH group: H H | | H - C - C - O - H | | H H
(c) (i) Ethene acts as a monomer. Under high pressure and temperature, the double bonds (\(\text{C}=\text{C}\)) in many ethene molecules break open, allowing them to link together to form a long chain (polymer) called poly(ethene). (ii) The repeating unit of poly(ethene) is drawn with two carbon atoms single-bonded to each other, with four single-bonded hydrogen atoms, and single bonds extending through brackets to represent continuation: [ - C(H_2) - C(H_2) - ]_n
评分标准
(a)(i) - High temperature (accept range \(500-800^\circ\text{C}\) or 'heat') [1] - Catalyst (accept silica, alumina, zeolites) [1] (a)(ii) - Correct equation: \(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_2\text{H}_4 + \text{C}_8\text{H}_{18}\) [1] (b)(i) - Addition / hydration [1] - Phosphoric acid / \(\text{H}_3\text{PO}_4\) catalyst [1] (b)(ii) - Correct displayed formula showing all atoms and all bonds (C-H, C-C, C-O, O-H must be explicitly drawn) [1] (c)(i) - Identifies ethene as a monomer [1] - Explains that the double bonds break / open up to allow many molecules to join / link together into a long chain [1] (c)(ii) - Two carbon atoms single-bonded with single bonds extending out of brackets / extension lines [1] - Four hydrogen atoms correctly bonded to the carbons [1]
题目 6 · Structured
10 分
A student sets up an electrical circuit containing a \(12.0\text{ V}\) battery with negligible internal resistance, connected to three resistors. Resistor \(R_1 = 4.0\ \Omega\) is connected in series with a parallel combination of two resistors: \(R_2 = 6.0\ \Omega\) and \(R_3 = 12.0\ \Omega\).
(a) Calculate the total equivalent resistance of the parallel combination of \(R_2\) and \(R_3\). [2]
(b) Calculate the total resistance of the entire circuit. [1]
(c) Show that the total current leaving the battery is \(1.5\text{ A}\). [2]
(d) Calculate the potential difference across the parallel combination of \(R_2\) and \(R_3\). [2]
(e) Calculate the electrical energy transferred in resistor \(R_1\) in a time of \(5.0\text{ minutes}\). [3]
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解题
(a) To find the equivalent parallel resistance (\(R_p\)) of \(R_2\) and \(R_3\): \(\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{6.0} + \frac{1}{12.0} = \frac{2}{12.0} + \frac{1}{12.0} = \frac{3}{12.0} = \frac{1}{4.0}\) Therefore, \(R_p = 4.0\ \Omega\).
(b) The total resistance of the circuit (\(R_{total}\)) is the series combination of \(R_1\) and the parallel combination (\(R_p\)): \(R_{total} = R_1 + R_p = 4.0\ \Omega + 4.0\ \Omega = 8.0\ \Omega\).
(c) Using Ohm's Law (\(I = \frac{V}{R}\)): \(I = \frac{12.0\text{ V}}{8.0\ \Omega} = 1.5\text{ A}\).
(d) The potential difference (\(V_p\)) across the parallel network is given by: \(V_p = I \times R_p = 1.5\text{ A} \times 4.0\ \Omega = 6.0\text{ V}\).
(e) First, convert time to seconds: \(t = 5.0\text{ minutes} \times 60\text{ s/min} = 300\text{ s}\). Next, calculate the power (\(P_1\)) in resistor \(R_1\) using \(P = I^2 R\) (since the total current of \(1.5\text{ A}\) flows through \(R_1\)): \(P_1 = I^2 \times R_1 = (1.5)^2 \times 4.0 = 2.25 \times 4.0 = 9.0\text{ W}\). Finally, calculate the energy (\(E\)): \(E = P \times t = 9.0\text{ W} \times 300\text{ s} = 2700\text{ J}\) (or \(2.7\text{ kJ}\)).
(c) - Formula used: \(I = \frac{V}{R}\) [1] - Correct substitution and calculation showing \(1.5\text{ A}\) [1]
(d) - Formula or method showing p.d. across parallel portion, e.g., \(V_p = I \times R_p\) or potential divider equation [1] - Correct value: \(6.0\text{ V}\) (with unit) [1]
(e) - Correct conversion of time: \(5.0\text{ min} = 300\text{ s}\) [1] - Correct formula for energy, e.g., \(E = I^2 R t\) or \(E = P t\) with substitution [1] - Correct final value and unit: \(2700\text{ J}\) or \(2.7\text{ kJ}\) [1]
题目 7 · Structured
10 分
Amylase is an enzyme that catalyses the breakdown of starch into maltose.
(a) Define the term catalyst. [2]
(b) Describe the lock-and-key hypothesis of enzyme action. [3]
(c) An experiment was carried out to study the effect of temperature on the rate of amylase activity. The time taken for starch to be completely digested at different temperatures was recorded: - At \(20^\circ\text{C}\): \(120\text{ s}\) - At \(30^\circ\text{C}\): \(60\text{ s}\) - At \(40^\circ\text{C}\): \(30\text{ s}\) - At \(50^\circ\text{C}\): No digestion detected after \(600\text{ s}\)
(i) Calculate the rate of starch digestion at \(40^\circ\text{C}\) using the formula: \text{rate} = \frac{1000}{\text{time taken}}
Show your working and state the unit. [2]
(ii) Explain the results obtained at \(50^\circ\text{C}\) in terms of enzyme structure and kinetic theory. [3]
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解题
(a) A catalyst is defined as a substance that increases the rate of a chemical reaction and is not changed or consumed by the reaction itself.
(b) According to the lock-and-key model, each enzyme has an active site with a specific, highly complementary 3D shape to its substrate. The substrate (the 'key') fits precisely into the active site (the 'lock') of the enzyme to form an enzyme-substrate complex, allowing the reaction to occur and products to be released.
(c)(i) Using the formula: \text{rate} = \frac{1000}{\text{time taken}} = \frac{1000}{30} = 33.3\text{ s}^{-1} (or arbitrary units).
(c)(ii) At \(50^\circ\text{C}\), the high kinetic energy causes the enzyme molecule to vibrate excessively, breaking weak bonds that maintain its tertiary structure. The enzyme becomes denatured, which permanently alters the specific shape of the active site. As a result, the starch substrate can no longer fit into the active site, and no reaction takes place.
评分标准
(a) 1 mark: increases rate of chemical reaction. 1 mark: is not changed / consumed by the reaction.
(b) 1 mark: active site has a specific / complementary shape to the substrate. 1 mark: substrate fits into active site. 1 mark: forms enzyme-substrate complex / reaction occurs.
(c)(i) 1 mark: correct substitution (1000 / 30). 1 mark: correct answer (33.3) with correct unit (\text{s}^{-1} or arbitrary units).
(c)(ii) 1 mark: enzyme is denatured (at high temperature). 1 mark: shape of active site is changed / lost. 1 mark: substrate can no longer fit / bind to the active site / no enzyme-substrate complexes can form.
题目 8 · Structured
10 分
A student investigates the reaction between solid calcium carbonate and dilute hydrochloric acid.
(a) Balance the chemical equation for this reaction: \text{CaCO}_3(\text{s}) + \text{\dots\dots}\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) [1]
(b) The student reacts \(5.00\text{ g}\) of pure calcium carbonate with excess dilute hydrochloric acid.
(i) Calculate the number of moles of calcium carbonate reacted. [Relative atomic masses: \(A_r(\text{Ca}) = 40.0\), \(A_r(\text{C}) = 12.0\), \(A_r(\text{O}) = 16.0\)] [2]
(ii) Determine the volume of carbon dioxide gas, in \text{dm}^3\, produced at room temperature and pressure (r.t.p.). (One mole of any gas occupies \(24.0\text{ dm}^3\) at r.t.p.) [2]
(c) In a separate titration, \(25.0\text{ cm}^3\) of a sodium hydroxide solution of unknown concentration is completely neutralised by \(20.0\text{ cm}^3\) of \(0.100\text{ mol/dm}^3\) hydrochloric acid: \text{NaOH}(\text{aq}) + \text{HCl}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l})
(i) Calculate the number of moles of \text{HCl} used. [2]
(ii) Calculate the concentration, in \text{mol/dm}^3\, of the sodium hydroxide solution. [3]
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解题
(a) The balanced equation requires a coefficient of 2 in front of \text{HCl}: \text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})
(b)(i) Relative formula mass of \text{CaCO}_3: M_r = 40.0 + 12.0 + (3 \times 16.0) = 100.0 Moles of \text{CaCO}_3 = \frac{\text{mass}}{M_r} = \frac{5.00}{100.0} = 0.0500\text{ mol}
(b)(ii) According to the balanced equation, \(1\text{ mol}\) of \text{CaCO}_3 produces \(1\text{ mol}\) of \text{CO}_2. Therefore, moles of \text{CO}_2 = 0.0500\text{ mol}. \text{Volume of } \text{CO}_2 = 0.0500\text{ mol} \times 24.0\text{ dm}^3\text{/mol} = 1.20\text{ dm}^3
(c)(i) \text{Moles of } \text{HCl} = \text{concentration} \times \text{volume (in dm}^3) = 0.100\text{ mol/dm}^3 \times \frac{20.0}{1000}\text{ dm}^3 = 0.00200\text{ mol}
(c)(ii) According to the equation, the reacting ratio of \text{NaOH} : \text{HCl} is \(1:1\). Therefore, moles of \text{NaOH} = 0.00200\text{ mol}. \text{Concentration of } \text{NaOH} = \frac{\text{moles}}{\text{volume (in dm}^3)} = \frac{0.00200}{0.0250\text{ dm}^3} = 0.0800\text{ mol/dm}^3
评分标准
(a) 1 mark: correct coefficient of 2.
(b)(i) 1 mark: correct calculation of \(M_r(\text{CaCO}_3) = 100\). 1 mark: correct moles calculation: \(0.05\) or \(0.0500\text{ mol}\).
(b)(ii) 1 mark: state/use mole ratio \(1:1\) so moles of \(\text{CO}_2 = 0.0500\). 1 mark: correct volume calculation: \(1.2\) or \(1.20\text{ dm}^3\).
(c)(i) 1 mark: correct volume conversion to \(\text{dm}^3\) (e.g., \(0.020\)). 1 mark: correct moles calculation: \(0.002\) or \(0.00200\text{ mol}\).
(c)(ii) 1 mark: state/use mole ratio \(1:1\) so moles of \(\text{NaOH} = 0.00200\). 1 mark: correct formula / substitution: \(0.00200 / 0.025\). 1 mark: correct concentration value: \(0.08\) or \(0.0800\text{ mol/dm}^3\).
题目 9 · Structured
10 分
A circuit consists of a battery of electromotive force (e.m.f.) \(12.0\text{ V}\) connected in series with a switch, and two resistors, \(R_1 = 4.0\ \Omega\) and \(R_2 = 12.0\ \Omega\), connected in parallel with each other.
(a) Calculate the combined (equivalent) resistance of the parallel combination of \(R_1\) and \(R_2\). [3]
(b) State the potential difference across the resistor \(R_1\). [1]
(c) Calculate the current in:
(i) resistor \(R_1\) [1]
(ii) the main circuit. [2]
(d) Calculate the electrical energy delivered by the battery to the circuit in a time of \(5.0\text{ minutes}\). State the unit. [3]
(c)(ii) The total current in the main circuit is equal to the sum of the currents in the individual parallel branches, or can be found using the total combined resistance: \text{Method 1: } I_{\text{total}} = \frac{V}{R_p} = \frac{12.0}{3.0} = 4.0\text{ A} \text{Method 2: } I_2 = \frac{12.0}{12.0} = 1.0\text{ A} \Rightarrow I_{\text{total}} = 3.0 + 1.0 = 4.0\text{ A}
(d) First, convert time to seconds: \(t = 5.0\text{ minutes} = 5.0 \times 60 = 300\text{ s}\). Now, calculate the electrical energy delivered: \(E = I \times V \times t\) (or \(E = P \times t = I^2 R t\)) \(E = 4.0\text{ A} \times 12.0\text{ V} \times 300\text{ s} = 14400\text{ J}\) (or \(14.4\text{ kJ}\)).
评分标准
(a) 1 mark: correct formula \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\). 1 mark: correct substitution: \(\frac{1}{4} + \frac{1}{12}\). 1 mark: correct final answer: \(3\) or \(3.0\ \Omega\).
(b) 1 mark: \(12\) or \(12.0\text{ V}\).
(c)(i) 1 mark: \(3\) or \(3.0\text{ A}\) (allow ecf from (b)).
(c)(ii) 1 mark: evidence of adding branch currents or using \(I = \frac{V}{R_{\text{total}}}\). 1 mark: correct final answer: \(4\) or \(4.0\text{ A}\).
(d) 1 mark: correct time conversion: \(300\text{ s}\). 1 mark: correct formula/working: \(4.0 \times 12.0 \times 300\) (allow ecf from (c)(ii)). 1 mark: correct final answer with unit: \(14400\text{ J}\) / \(14.4\text{ kJ}\) / \(1.44 \times 10^4\text{ J}\) (reject without unit or with incorrect unit).
题目 10 · Structured
10 分
(a) State what is meant by the term gravitropism. [2] (b) A student sets up an experiment where a plant shoot is illuminated from one side only. Explain how auxin causes the shoot to bend towards the light. [4] (c) Chemical coordination also occurs in animals. (i) Name the hormone released in a 'fight or flight' situation and describe one of its physiological effects. [2] (ii) Compare hormonal coordination with nervous coordination by stating two differences in how they function. [2]
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解题
(a) Gravitropism is a growth response in which a plant grows towards or away from gravity, where gravity acts as the stimulus. (b) Auxin is produced in the shoot tip and diffuses downwards. When the shoot is illuminated from one side, the auxin moves/accumulates on the shaded side of the shoot. The higher concentration of auxin on the shaded side causes the cells on this side to elongate more than the cells on the illuminated side, resulting in the shoot bending towards the light. (c) (i) The hormone is adrenaline. Its physiological effects include increased heart rate, increased breathing rate, dilated pupils, or increased blood glucose concentration. (ii) Hormonal coordination uses chemical messengers (hormones) transported in the blood, which has a slower transmission speed but longer-lasting effects. In contrast, nervous coordination uses electrical impulses transmitted along neurones, which is much faster but has short-lived effects.
评分标准
Part (a): Response in which a plant grows towards/away from gravity [1] Gravity acts as the stimulus [1]. Part (b): Auxin is produced in the shoot tip [1] Auxin diffuses down the shoot [1] Auxin accumulates/concentrates on the shaded side [1] Causes cell elongation on the shaded side (causing bending) [1]. Part (c)(i): Adrenaline [1] Increases heart rate / breathing rate / dilates pupils / increases blood glucose [1]. Part (c)(ii) (Any two differences, 1 mark each): Hormonal transmission is slower than nervous transmission (or vice versa) [1] Hormonal effects last longer than nervous effects (or vice versa) [1] Hormonal signals are chemical (hormones) whereas nervous signals are electrical (impulses) [1] Hormones travel through the bloodstream whereas nerve impulses travel along neurones [1].
题目 11 · Structured
10 分
(a) Write a balanced chemical equation for the complete combustion of propane, \(\text{C}_3\text{H}_8\). [2] (b) Calculate the volume of carbon dioxide, \(\text{CO}_2\), measured at room temperature and pressure (r.t.p.), produced when \(8.8\text{ g}\) of propane is completely combusted. [Relative atomic masses, \(A_r\): \(\text{H} = 1.0\), \(\text{C} = 12.0\). Molar volume of a gas at r.t.p. is \(24\text{ dm}^3/\text{mol}\).] [4] (c) If the actual volume of carbon dioxide gas collected in the experiment was \(11.5\text{ dm}^3\) at r.t.p., calculate the percentage yield of carbon dioxide. Give your answer to 3 significant figures. [2] (d) State the test for carbon dioxide gas and the positive result observed. [2]
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解题
Part (a): The balanced chemical equation for the complete combustion of propane is \(\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\). Part (b): 1. Calculate the relative formula mass (\(M_r\)) of propane: \(M_r(\text{C}_3\text{H}_8) = (3 \times 12.0) + (8 \times 1.0) = 44.0\text{ g/mol}\). 2. Calculate the moles of propane reacting: \(n(\text{C}_3\text{H}_8) = \frac{8.8\text{ g}}{44.0\text{ g/mol}} = 0.20\text{ mol}\). 3. From the equation, 1 mole of \(\text{C}_3\text{H}_8\) produces 3 moles of \(\text{CO}_2\): \(n(\text{CO}_2) = 0.20 \times 3 = 0.60\text{ mol}\). 4. Calculate the volume of \(\text{CO}_2\) at r.t.p.: \(V = 0.60\text{ mol} \times 24\text{ dm}^3/\text{mol} = 14.4\text{ dm}^3\). Part (c): \(\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\% = \frac{11.5}{14.4} \times 100\% = 79.861\% \approx 79.9\%\). Part (d): The test for carbon dioxide is bubbling the gas through limewater. The positive result is that the limewater turns cloudy (or milky).
评分标准
Part (a): Correct formulas of reactants and products [1] Correctly balanced equation: \(\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\) [1]. Part (b): Calculating \(M_r\) of propane as \(44.0\) [1] Calculating moles of propane as \(0.20\text{ mol}\) [1] Using mole ratio 1:3 to get \(0.60\text{ mol}\) of \(\text{CO}_2\) [1] Volume of \(\text{CO}_2 = 14.4\text{ dm}^3\) [1]. Part (c): Correct formula for percentage yield shown (\(11.5 / 14.4\)) [1] Answer of \(79.9\%\) to 3 sig figs [1]. Part (d): Test: bubble through limewater [1] Result: turns cloudy / milky / white precipitate [1].
题目 12 · Structured
10 分
A battery of electromotive force (e.m.f.) \(12.0\text{ V}\) is connected in a circuit. Two resistors with resistances of \(6.0\ \Omega\) and \(12.0\ \Omega\) are connected in parallel. This parallel combination is then connected in series with a \(4.0\ \Omega\) resistor and the battery. (a) (i) Calculate the combined resistance of the \(6.0\ \Omega\) and \(12.0\ \Omega\) resistors connected in parallel. [2] (ii) State the total resistance of the entire circuit. [1] (iii) Calculate the current drawn from the battery. State the unit. [2] (b) Calculate the rate of thermal energy dissipation (power) in the \(4.0\ \Omega\) resistor. State the unit. [2] (c) The circuit is disconnected. A bar magnet is then pushed into a solenoid connected to a sensitive galvanometer to investigate electromagnetic induction. (i) State two ways to increase the magnitude of the induced electromotive force (e.m.f.). [2] (ii) State the effect on the galvanometer needle when the magnet is pulled out of the solenoid at the same speed. [1]
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解题
Part (a)(i): The formula for two resistors in parallel is: \(R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{6.0 \times 12.0}{6.0 + 12.0} = \frac{72.0}{18.0} = 4.0\ \Omega\). Part (a)(ii): Since the parallel combination is in series with the \(4.0\ \Omega\) resistor: \(R_{\text{total}} = R_p + R_3 = 4.0\ \Omega + 4.0\ \Omega = 8.0\ \Omega\). Part (a)(iii): Using Ohm's law: \(I = \frac{V}{R_{\text{total}}} = \frac{12.0\text{ V}}{8.0\ \Omega} = 1.5\text{ A}\). Part (b): The current through the \(4.0\ \Omega\) resistor is the total circuit current, \(I = 1.5\text{ A}\). Power \(P = I^2 \times R = (1.5\text{ A})^2 \times 4.0\ \Omega = 2.25 \times 4.0 = 9.0\text{ W}\). (Alternatively, the voltage drop across the \(4.0\ \Omega\) resistor is \(V = I \times R = 1.5\text{ A} \times 4.0\ \Omega = 6.0\text{ V}\). Power \(P = V \times I = 6.0\text{ V} \times 1.5\text{ A} = 9.0\text{ W}\).) Part (c)(i): The induced e.m.f. can be increased by: 1. Moving the magnet faster into the coil. 2. Using a stronger magnet. 3. Increasing the number of turns in the solenoid coil. Part (c)(ii): The galvanometer needle will deflect in the opposite direction (with the same magnitude of deflection since the speed is the same).
评分标准
Part (a)(i): Formula \(1/R_p = 1/R_1 + 1/R_2\) or \(R_p = (R_1 \times R_2)/(R_1 + R_2)\) [1] \(4.0\ \Omega\) [1]. Part (a)(ii): \(8.0\ \Omega\) (allow ECF from a(i)) [1]. Part (a)(iii): Formula \(I = V/R\) used [1] \(1.5\text{ A}\) (accept unit written as Amperes or A) [1]. Part (b): Correct formula used, e.g., \(P = I^2R\) or \(P = VI\) after finding \(V = 6\text{ V}\) [1] \(9.0\text{ W}\) (accept unit written as Watts or W) [1]. Part (c)(i) (Any two, 1 mark each): Move the magnet faster [1] Use a stronger magnet [1] Increase the number of turns on the solenoid coil [1]. Part (c)(ii): Deflection in the opposite direction [1].
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