2022 CCEA AS-Level Life and Health Sciences 0008 模擬試題連答案詳解

(a) The (superior/inferior) vena cava.

(b) The tricuspid valve.

(c) The wall of the left ventricle is much thicker (more muscular) than that of the right ventricle. This is because the left ventricle must generate a much higher pressure to pump blood all the way around the systemic circulation (to the whole body), whereas the right ventricle only needs to pump blood the shorter distance to the lungs (pulmonary circulation), which offers much less resistance.

(d) Right atrium → tricuspid valve → right ventricle → pulmonary (semilunar) valve → pulmonary artery → (lungs, gas exchange) → pulmonary vein → left atrium → bicuspid (mitral) valve → left ventricle → aortic (semilunar) valve → aorta.

(e) The heart is described as a double pump because it consists of two separate pumping sides (the right side and the left side) that operate together but pump blood around two separate circuits: the right side pumps deoxygenated blood to the lungs (pulmonary circulation), while the left side pumps oxygenated blood to the rest of the body (systemic circulation).

(a) [1] vena cava (accept superior or inferior vena cava). (b) [1] tricuspid valve. (c) [1] correctly states left ventricle wall is thicker; [1] valid reason referencing higher pressure needed to pump blood around the (longer/higher-resistance) systemic circuit compared with the pulmonary circuit. (d) [1] right atrium → tricuspid valve → right ventricle; [1] → pulmonary valve → pulmonary artery → lungs; [1] → pulmonary vein → left atrium → bicuspid/mitral valve → left ventricle; [1] → aortic valve → aorta; award marks for structures named in the correct order, penalise once only for an incorrect order not for each individual omission thereafter. (e) [1] identifies two separate pumping sides/circuits; [1] correctly names both circuits (pulmonary and systemic) and links each to the correct side of the heart.

(a) Any two of: a very large total surface area (due to the huge number of alveoli); extremely thin walls (one cell thick), giving a short diffusion distance; a dense network of capillaries surrounding each alveolus, providing a good blood supply and maintaining a steep diffusion gradient.

(b) The concentration (partial pressure) of oxygen is higher in the alveolar air than in the deoxygenated blood arriving in the surrounding capillaries, creating a diffusion gradient. Oxygen therefore diffuses down this concentration gradient, from a region of high partial pressure (the alveolus) to a region of low partial pressure (the capillary blood), across the thin alveolar and capillary walls. As blood continuously flows through the capillary, oxygenated blood is carried away and is replaced by deoxygenated blood, maintaining the gradient.

(c) The moisture allows oxygen and carbon dioxide to dissolve before diffusing across the alveolar membrane (gases must dissolve to cross a cell membrane efficiently). Without surfactant (which reduces the surface tension of this moisture film), the surface tension of the water lining could cause the moist alveolar walls to stick together and collapse, making the lungs harder to re-inflate.

(d) The trachea is supported by C-shaped rings of cartilage, which are rigid enough to hold the airway open (preventing it from collapsing during breathing) while the incomplete part of the ring allows the adjacent oesophagus to expand when swallowing food.

(a) [1] each for any two valid structural features (large surface area; thin walls/short diffusion distance; good capillary blood supply); max [2]. (b) [1] identifies higher partial pressure/concentration of O2 in alveolus than in capillary blood; [1] correct statement that diffusion occurs down this gradient, high to low concentration; [1] correctly explains gradient is maintained by continuous blood flow removing oxygenated blood. (c) [1] correctly explains gases must dissolve in the moisture to diffuse across the membrane; [1] correctly identifies risk of alveoli collapsing/sticking together without surfactant. (d) [1] identifies C-shaped cartilage rings; [1] correct functional explanation (holds airway open while allowing oesophagus to expand).

(a) glucose → lactic acid (+ small amount of ATP released).

(b) Aerobic respiration produces far more ATP per glucose molecule than anaerobic respiration (aerobic respiration fully oxidises glucose to carbon dioxide and water, releasing much more energy).

(c) Oxygen debt is the extra volume of oxygen that the body needs, above the resting level, after exercise has stopped, in order to metabolise the lactic acid that has accumulated during anaerobic respiration. It is repaid by continued heavy/deep breathing after exercise (which is why breathing and heart rate stay elevated after stopping); the extra oxygen taken in is used to oxidise lactic acid back to pyruvate, most of which is then broken down aerobically (via the liver, converting some back to glucose/glycogen).

(d) Lactic acid is acidic and its build-up lowers the pH within muscle cells; this changed pH interferes with the muscle enzymes and with the proteins involved in muscle contraction, reducing the efficiency of muscle contraction, causing fatigue, and can trigger painful cramping.

(e) Anaerobic respiration (fermentation) also occurs in yeast (a microorganism, not human tissue) — however within the human body context, red blood cells (which lack mitochondria) respire anaerobically, also producing lactic acid as the product.

(a) [1] glucose (or 'sugar') as reactant; [1] lactic acid as product (reject 'ethanol/carbon dioxide' which are yeast fermentation products, not human). (b) [1] correctly states aerobic respiration produces (much) more ATP per glucose than anaerobic. (c) [1] correct definition referencing extra oxygen required after exercise to deal with lactic acid; [1] correctly links continued heavy breathing to repaying the debt; [1] correctly explains lactic acid is oxidised/converted (e.g. back to pyruvate/glycogen) using this extra oxygen. (d) [1] identifies lactic acid lowers pH / is acidic; [1] correctly links this to reduced enzyme/muscle protein function causing fatigue/cramp. (e) [1] identifies red blood cells (accept yeast, if correctly noted as a microorganism rather than human tissue, for partial credit); [1] correctly names lactic acid (or ethanol and carbon dioxide, if yeast given) as the product.

(a) Insulin, secreted by the beta cells of the islets of Langerhans in the pancreas; glucagon, secreted by the alpha cells of the islets of Langerhans.

(b) The rise in blood glucose concentration is detected by the beta cells of the pancreas, which respond by secreting more insulin into the blood. Insulin travels in the blood to target cells (e.g. liver and muscle cells), binding to receptors on their cell surface membranes, and causes these cells to take up more glucose from the blood. In the liver specifically, insulin stimulates the conversion of glucose into glycogen (glycogenesis) for storage, which lowers the blood glucose concentration back towards normal.

(c) Negative feedback is a control mechanism in which a change away from the normal (set-point) value triggers a response that counteracts (reverses) that change, returning the system to its normal level. For blood glucose, a rise above normal triggers increased insulin release, which lowers glucose back towards normal; conversely a fall below normal triggers glucagon release, which raises it back towards normal — in each case the response opposes/corrects the original change.

(d) Persistently high blood glucose (hyperglycaemia) — symptoms could include excessive thirst, excessive urination, tiredness/fatigue (as in untreated diabetes mellitus).

(a) [1] insulin from beta cells; [1] glucagon from alpha cells (both cell type and hormone name required for each mark; accept 'islets of Langerhans' without full cell-type name for partial credit only if cell type omitted, but full mark requires beta/alpha specified). (b) [1] rise detected by pancreas/beta cells, more insulin secreted; [1] insulin travels in blood to target/liver cells and binds to receptors; [1] increases uptake of glucose by cells; [1] correctly identifies conversion of glucose to glycogen (glycogenesis) in the liver. (c) [1] correct general definition of negative feedback (change triggers a response that counteracts/reverses it); [1] correctly applies this using blood glucose (rise → insulin → glucose falls, or equivalent). (d) [1] any valid symptom of hyperglycaemia (e.g. excess thirst, excess urination, tiredness); reject symptoms of hypoglycaemia.

(a) A balanced diet is one that provides all of the nutrients (carbohydrate, protein, fat, vitamins, minerals, fibre and water) that the body needs, in the correct proportions and in sufficient (but not excessive) quantities, to meet an individual's energy and nutritional requirements.

(b) The active adult has a greater energy requirement because: (i) skeletal muscles do significantly more mechanical work during activity, requiring more ATP/energy from respiration; and (ii) a higher overall metabolic rate is sustained both during and for some time after exercise (e.g. increased heart rate and breathing rate, raised body temperature) which uses additional energy.

(c) (i) Regular aerobic exercise strengthens the heart muscle (myocardium), increasing stroke volume, so the heart can pump more blood per beat and does not need to beat as fast at rest (lower resting heart rate), making the heart more efficient. (ii) Regular exercise helps maintain healthy, elastic blood vessels and can help reduce blood pressure and improve blood lipid profile (e.g. raising HDL, lowering LDL cholesterol), reducing the build-up of atheroma and so lowering the risk of atherosclerosis, coronary heart disease and stroke.

(d) A chronically low-fibre diet is associated with an increased risk of constipation and of bowel disorders (e.g. diverticular disease, and an increased risk of bowel/colorectal cancer).

(e) Example: iron deficiency, leading to anaemia (reduced haemoglobin/red blood cell production, causing tiredness and breathlessness). (Other valid pairs, e.g. vitamin D deficiency → rickets/weak bones, would also be credited.)

(a) [1] provides all necessary nutrients; [1] in correct/appropriate proportions or quantities for the individual's needs. (b) [1] each for any two valid physiological reasons (e.g. more muscular work/ATP demand; sustained raised metabolic rate/heart & breathing rate); max [2]. (c) [1] first benefit named (e.g. stronger heart/increased stroke volume); [1] correct explanation (lower resting heart rate/more efficient pumping); [1] second benefit named (e.g. improved blood vessel health/lipid profile); [1] correct explanation (reduced atheroma/blood pressure, lower CHD/stroke risk); award full credit for two clearly distinct, correctly explained benefits. (d) [1] any valid risk (constipation, diverticular disease, bowel cancer risk). (e) [1] correct micronutrient with a correctly linked health consequence (e.g. iron → anaemia; vitamin D → rickets; vitamin C → scurvy); both parts required for the mark.

(a) Vital capacity is the maximum volume of air that can be forcibly expelled from the lungs after the deepest possible inspiration (i.e. TV + IRV + ERV). \( VC = TV+IRV+ERV = 0.50+3.0+1.1 = 4.6\ \text{dm}^3 \)

(b) \( TLC = VC+RV = 4.6+1.2 = 5.8\ \text{dm}^3 \)

(c) Minute ventilation = tidal volume × breathing rate \( = 0.50 \times 14 = 7.0\ \text{dm}^3\text{min}^{-1} \)

(d) Cardiac output is the volume of blood pumped by (one ventricle of) the heart in one minute; \( CO = HR \times SV \). At rest: \( CO = 72 \times 70 = 5040\ \text{cm}^3\text{min}^{-1} = 5.04\ \text{dm}^3\text{min}^{-1} \)

(e) During exercise: \( CO = 150 \times 100 = 15\,000\ \text{cm}^3\text{min}^{-1} = 15.0\ \text{dm}^3\text{min}^{-1} \). Percentage increase \( = \dfrac{15.0-5.04}{5.04}\times100 = 197.6\% \) (approximately a threefold increase).

Final answer: resting CO = 5.04 dm^3/min; exercise CO = 15.0 dm^3/min, an increase of approximately 198%.

(a) [1] correct definition (max volume expelled after deepest inspiration, or TV+IRV+ERV stated); [1] correct substitution shown; [1] \( VC=4.6\ \text{dm}^3 \). (b) [1] correct method (VC+RV); [1] \( TLC=5.8\ \text{dm}^3 \) (ECF from (a)). (c) [1] correct formula (TV × breathing rate); [1] correct substitution; [1] \( 7.0\ \text{dm}^3\text{min}^{-1} \) with correct unit. (d) [1] correct definition (volume of blood pumped by the heart per minute); [1] correct substitution into \( CO=HR\times SV \) with correct unit conversion (cm^3 to dm^3); [1] \( CO=5.04\ \text{dm}^3\text{min}^{-1} \). (e) [1] correct substitution for exercise CO; [1] \( CO_{exercise}=15.0\ \text{dm}^3\text{min}^{-1} \); [1] correct percentage-increase method; [1] \( \approx198\% \) (accept 195–200%, ECF).

An indicative full-mark response would include:

Detection and coordination: the hypothalamus contains thermoreceptors that monitor the temperature of the blood flowing through it, and also receives nerve impulses from thermoreceptors in the skin; it acts as the body's thermoregulatory centre, comparing detected temperature with the set-point (normal core temperature, about 37°C) and coordinating an appropriate response via the nervous system.

Response to a rise in temperature: vasodilation — arterioles supplying the skin capillaries widen, increasing blood flow near the skin surface, so more heat is lost from the blood to the environment by radiation and convection (the skin appears flushed). Sweat glands increase their secretion of sweat onto the skin surface; the latent heat needed to evaporate this sweat is taken from the skin, cooling the body. Hair erector (arrector pili) muscles relax, so hairs lie flat, reducing the layer of insulating air trapped near the skin. Behavioural/metabolic responses may also reduce heat production (e.g. reduced physical activity).

Response to a fall in temperature: vasoconstriction — arterioles supplying the skin capillaries narrow, diverting blood flow away from the skin surface (to deeper vessels), reducing heat loss by radiation and convection (skin looks pale). Sweat gland activity decreases, reducing evaporative heat loss. Hair erector muscles contract, raising hairs to trap a thicker layer of insulating air next to the skin ('goose bumps'). Shivering — the hypothalamus stimulates rapid, involuntary contraction and relaxation of skeletal muscles; this increased muscular respiration generates extra metabolic heat, warming the body.

Overall, both responses are examples of negative feedback: a deviation of core temperature away from the normal set-point (in either direction) triggers a coordinated physiological response (via the hypothalamus acting on the skin and muscles) that acts to return the temperature back towards normal.

Levels-of-response (QWC) mark scheme, out of 12 marks.

Level 3 (9–12 marks): Detailed, accurate, well-organised account covering the role of the hypothalamus as coordinator/thermoreceptor site, AND correctly describes both responses (rise: vasodilation, sweating, hair erectors relax; fall: vasoconstriction, reduced sweating, hair erectors contract, shivering) with correct mechanistic explanation of how each response causes heat loss/gain (e.g. evaporation of sweat, insulating air layer, respiration generating heat in shivering). Correct specialist terminology throughout (vasodilation, vasoconstriction, arrector pili, negative feedback); coherent structure; accurate spelling, punctuation and grammar.

Level 2 (5–8 marks): Good coverage of both directions of response, but with some omissions (e.g. missing hair erector muscle role, or incomplete explanation of one mechanism) or a small number of inaccuracies; hypothalamus role stated but not fully explained; generally clear communication with occasional lapses.

Level 1 (1–4 marks): Limited or one-sided account (e.g. describes only the response to heat, or only names mechanisms without explaining how they cause heat loss/gain); weak terminology or organisation.

0 marks: No creditable response.

Full marks at Level 3 require substantive, accurate coverage of BOTH directions of temperature response, not just one.

(a) When [A] doubles (0.10→0.20), the rate doubles (2.0→4.0 ×10^-3); when [A] doubles again (0.20→0.40), the rate doubles again (4.0→8.0 ×10^-3). Since rate is directly proportional to [A]^1, the reaction is first order with respect to A.

(b) Doubling [B] causes the rate to quadruple (×4 = ×2^2), so the reaction is second order with respect to B.

(c) Overall rate equation: \( \text{rate}=k[A][B]^2 \). Overall order \( = 1+2 = 3 \) (third order overall).

(d) Using [A] = 0.10 mol dm^-3, [B] = 0.20 mol dm^-3, rate = 2.0×10^-3 mol dm^-3 s^-1: \( k = \dfrac{\text{rate}}{[A][B]^2} = \dfrac{2.0\times10^{-3}}{(0.10)(0.20)^2} = \dfrac{2.0\times10^{-3}}{4.0\times10^{-3}} = 0.500\ \text{mol}^{-2}\ \text{dm}^6\ \text{s}^{-1} \)

Final answer: rate \( = k[A][B]^2 \), \( k = 0.500\ \text{mol}^{-2}\ \text{dm}^6\ \text{s}^{-1} \).

(a) [1] correctly identifies rate doubles when [A] doubles (using at least two rows of data); [1] correctly identifies same doubling relationship holds for the second data pair too; [1] correct conclusion: first order with respect to A. (b) [1] correctly identifies quadrupling of rate when [B] doubles; [1] correct conclusion: second order with respect to B. (c) [1] correct rate equation \( \text{rate}=k[A][B]^2 \) (ECF from (a)/(b)); [1] correct overall order (3), consistent with their equation. (d) [1] correct rearrangement for k; [1] correct substitution of values; [1] \( k=0.500\ \text{mol}^{-2}\text{dm}^6\text{s}^{-1} \); [1] correct unit derived from the rate equation (accept unit shown as a separate derivation).

(a) \( K_c = \dfrac{[NH_3]^2}{[N_2][H_2]^3} \)

(b) \( K_c = \dfrac{(0.60)^2}{(0.40)(1.20)^3} = \dfrac{0.360}{0.40\times1.728} = \dfrac{0.360}{0.6912} = 0.521\ \text{mol}^{-2}\ \text{dm}^6 \)

(c) Increasing the pressure shifts the position of equilibrium towards the side with fewer moles of gas, to partially oppose (reduce) the increase in pressure. There are 4 moles of gas on the left (1 N2 + 3 H2) but only 2 moles of gas on the right (2 NH3), so increasing the pressure shifts equilibrium to the right, towards the products, increasing the yield of ammonia.

(d) Although a lower temperature would favour a higher equilibrium yield (since the forward reaction is exothermic, and lowering temperature shifts equilibrium towards the exothermic/forward side, by Le Chatelier's principle), a lower temperature would also make the rate of reaction (both forward and reverse) much slower, meaning it would take impractically long to reach equilibrium. A compromise temperature (around 450°C) is therefore used industrially to achieve a reasonably fast rate of reaction while still obtaining an economically acceptable (if not maximal) equilibrium yield.

Final answer: \( K_c = 0.521\ \text{mol}^{-2}\ \text{dm}^6 \).

(a) [1] correct powers matching the equation coefficients; [1] correct expression overall \( [NH_3]^2/([N_2][H_2]^3) \). (b) [1] correct substitution; [1] \( K_c=0.521 \) (accept 0.51–0.53); [1] correct unit \( \text{mol}^{-2}\text{dm}^6 \) derived correctly (allow ECF on unit if expression in (a) differs but is dimensionally consistent). (c) [1] correctly states equilibrium shifts to the side with fewer gas moles / towards products; [1] correctly identifies 4 mol gas on left vs 2 mol gas on right; [1] correct conclusion that yield of ammonia increases. (d) [1] identifies that lower temperature would give higher yield but a much slower rate of reaction; [1] correctly concludes a compromise temperature balances acceptable rate against acceptable yield (economic reasoning).

(a) Mean bond energy is the average energy required to break one mole of a given type of bond in the gaseous state, averaged over many different molecules/compounds containing that bond (since the exact value varies slightly depending on the molecule).

(b) Bonds broken: 4×(C–H) + 2×(O=O) = \( 4(413)+2(498) = 1652+996 = 2648\ \text{kJ mol}^{-1} \)

(c) Bonds formed: 2×(C=O) + 4×(O–H) = \( 2(805)+4(464) = 1610+1856 = 3466\ \text{kJ mol}^{-1} \)

(d) \( \Delta H = \text{(energy to break bonds)} - \text{(energy released forming bonds)} = 2648-3466 = -818\ \text{kJ mol}^{-1} \). Since \( \Delta H \) is negative, the reaction is exothermic.

(e) Mean bond energies are average values taken across many different compounds, whereas the actual bond energies in a specific molecule (e.g. the C–H bonds in methane specifically, or the O–H bonds in gaseous water as actually formed) may differ from this average; using averaged, rather than exact, bond energy values for these particular bonds/molecules is the main source of the discrepancy from the true (experimental) enthalpy value.

(f) Comparing bond-energy-derived enthalpies of combustion allows engineers/industry to estimate and compare the energy output (and hence efficiency/cost) of different fuels before committing to experimental testing, informing fuel choice for heating or power generation.

Final answer: \( \Delta H \approx -818\ \text{kJ mol}^{-1} \) (exothermic).

(a) [1] average energy to break one mole of the bond (in the gaseous state); [1] correctly notes this is averaged over different molecules/compounds. (b) [1] correct identification of bonds broken (4 C–H, 2 O=O); [1] correct substitution; [1] \( 2648\ \text{kJ mol}^{-1} \). (c) [1] correct identification of bonds formed (2 C=O, 4 O–H); [1] correct substitution; [1] \( 3466\ \text{kJ mol}^{-1} \). (d) [1] correct method (broken minus formed); [1] \( \Delta H=-818\ \text{kJ mol}^{-1} \) (ECF from (b)/(c)); [1] correctly states exothermic (negative \( \Delta H \)). (e) [1] correctly identifies mean/average bond energies do not exactly match the true bond energies in this specific molecule; [1] for a clearly and correctly reasoned statement (allow reference to it being an estimate rather than an exact experimental quantity). (f) [1] any valid reason (e.g. allows quick comparison/estimation of fuel energy output without needing to burn each fuel experimentally, informing fuel selection).

(a) Hess's law states that the total enthalpy change for a reaction is independent of the route taken from reactants to products, provided the initial and final conditions are the same.

(b) \( C(s)+2H_2(g)\rightarrow CH_4(g) \), \( \Delta H_f^{\ominus}=? \). Cycle: C(s) + 2H2(g) at the top, connected by the direct (formation) arrow to CH4(g); both C(s)+2H2(g) and CH4(g) are then connected by separate downward arrows to the same common products, CO2(g) + 2H2O(l), at the base of the cycle — the left-hand route via direct combustion of C(s) and 2H2(g) (\( \Delta H_c[C]+2\Delta H_c[H_2] \)), and the right-hand route via formation then combustion of CH4(g) (\( \Delta H_f[CH_4]+\Delta H_c[CH_4] \)).

(c) By Hess's law, both routes from C(s)+2H2(g) down to CO2(g)+2H2O(l) must give the same total enthalpy change: \( \Delta H_c[C]+2\Delta H_c[H_2] = \Delta H_f[CH_4]+\Delta H_c[CH_4] \), so \( \Delta H_f[CH_4] = \Delta H_c[C]+2\Delta H_c[H_2]-\Delta H_c[CH_4] \).

(d) \( \Delta H_f[CH_4] = (-394)+2(-286)-(-890) = -394-572+890 = -76\ \text{kJ mol}^{-1} \)

(e) Because the direct formation reaction cannot be carried out cleanly in the laboratory, Hess's law provides the only practical way to obtain this enthalpy value; industries handling fuel gases rely on combustion enthalpies (which they CAN measure directly, precisely and routinely, e.g. using a bomb calorimeter, and which are also directly relevant to a fuel's energy value/quality) to indirectly calculate formation enthalpies and other otherwise-inaccessible thermochemical data needed for process design, energy balance calculations and safety assessments.

Final answer: \( \Delta H_f^{\ominus}[CH_4(g)] = -76\ \text{kJ mol}^{-1} \) (close to the accepted literature value of about \( -74.8\ \text{kJ mol}^{-1} \)).

(a) [1] total enthalpy change independent of route; [1] provided initial and final states/conditions are the same. (b) [1] correct formation equation with state symbols; [1] correct cycle structure showing C(s)+2H2(g) and CH4(g) as two starting points; [1] both routes correctly directed to the common products CO2(g)+2H2O(l); [1] all four enthalpy arrows correctly labelled (\( \Delta H_f \), \( \Delta H_c[C] \), \( \Delta H_c[H_2] \), \( \Delta H_c[CH_4] \)). (c) [1] correct equating of the two routes; [1] correct rearrangement for \( \Delta H_f[CH_4] \). (d) [1] correct substitution; [1] correct arithmetic; [1] \( \Delta H_f=-76\ \text{kJ mol}^{-1} \) (ECF from (c)). (e) [1] identifies formation reaction cannot be done directly/cleanly; [1] identifies combustion data can be measured directly and routinely (e.g. bomb calorimetry, fuel quality testing); [1] correctly links this to practical importance for process design/energy calculations in industry.

An indicative full-mark response would include:

Temperature: the forward reaction is exothermic, so by Le Chatelier's principle a lower temperature favours a higher equilibrium yield of ammonia. However, a lower temperature also gives a much slower rate of reaction (fewer molecules have enough energy to overcome the activation energy), so equilibrium would take a very long time to reach, which is not economically viable. A moderate/compromise temperature of about 450°C is therefore chosen: it sacrifices some equilibrium yield (yield is lower than it would be at a lower temperature) in exchange for a much faster, more economically practical rate of production.

Pressure: there are fewer moles of gas on the product side (2 mol NH3) than on the reactant side (4 mol, N2+3H2), so by Le Chatelier's principle increasing pressure shifts the equilibrium towards the products, increasing yield; higher pressure also increases the rate of reaction (higher concentration/collision frequency of reactant molecules). However, very high pressures require thick-walled, expensive reaction vessels and pipework (capital cost) and are more hazardous (safety risk of leaks or vessel failure), and compressing gases to very high pressure is itself energy-intensive and costly to run. A compromise pressure of about 200 atmospheres is therefore used: high enough to give a reasonably good yield and rate, but not so high that the cost and safety risk of the equipment become prohibitive.

Catalyst: the iron catalyst increases the rate at which equilibrium is reached by providing an alternative reaction pathway with a lower activation energy, without itself being used up and without affecting the position of equilibrium (it does not change the equilibrium yield, only the speed at which that yield is achieved). This allows a lower temperature to be used than would otherwise be needed to achieve an economically acceptable rate, partially offsetting the yield loss caused by using a compromise (rather than very low) temperature.

Overall evaluation: the chosen conditions (450°C, 200 atm, iron catalyst) do not maximise equilibrium yield, but instead represent a carefully optimised industrial compromise that balances an economically acceptable rate of production, a reasonable (though not maximal) yield, and manageable capital, running and safety costs.

Levels-of-response (QWC) mark scheme, out of 9 marks.

Level 3 (7–9 marks): Full, accurate, well-balanced evaluation covering: correct Le Chatelier reasoning for temperature (exothermic forward reaction, lower T favours yield but slows rate); correct Le Chatelier reasoning for pressure (fewer moles of gas on product side, higher P favours yield and rate); correct economic/safety reasoning for both (temperature: rate vs yield trade-off; pressure: cost/safety of high-pressure vessels); and correct role of the catalyst (speeds up attainment of equilibrium via lower activation energy pathway, without shifting the equilibrium position). Coherent, well-organised answer with accurate scientific terminology, spelling, punctuation and grammar.

Level 2 (4–6 marks): Good coverage of most of the above points, but with one significant omission (e.g. no discussion of pressure's economic/safety trade-off, or catalyst role stated without full explanation) or some inaccuracy; reasonably clear communication.

Level 1 (1–3 marks): Limited or partial coverage (e.g. only discusses temperature OR pressure, or lists conditions without explaining the underlying compromise); weak terminology or structure.

0 marks: No creditable response.

Full marks require a genuinely balanced treatment of temperature, pressure AND the catalyst, not just one or two of these three elements.

(a) Concordant titres are those that agree with each other within 0.10 cm^3 (or within 0.2 cm^3 of the closest other reading). Titres 1 (22.45), 2 (22.35) and 3 (22.40) are all within 0.10 cm^3 of each other, so all three are concordant: mean titre \( = \dfrac{22.45+22.35+22.40}{3} = 22.40\ \text{cm}^3 \).

(b) The rough titration is carried out quickly (adding acid in large volumes/rapidly) purely to find an approximate end point, so that subsequent, more careful titrations can be run slowly near this volume; it is therefore much less precise/accurate than the following careful titrations and is excluded from the mean.

(c) \( n(HCl) = c\times V = 0.100 \times \dfrac{22.40}{1000} = 2.24\times10^{-3}\ \text{mol} \)

(d) From the equation, the mole ratio of NaOH : HCl is 1 : 1, so \( n(NaOH) = n(HCl) = 2.24\times10^{-3}\ \text{mol} \)

(e) \( c(NaOH) = \dfrac{n}{V} = \dfrac{2.24\times10^{-3}}{25.0/1000} = 0.0896\ \text{mol dm}^{-3} \)

(f) \( \text{concentration in g dm}^{-3} = c\times M_r = 0.0896\times40.0 = 3.58\ \text{g dm}^{-3} \)

(g) The meniscus should be read with the eye level with the bottom of the meniscus (at eye level with the scale), not looking down or up at it, to avoid a parallax error.

(h) Phenolphthalein changes colour sharply within the pH range produced at the equivalence point of a strong-acid–strong-base titration such as this one (HCl and NaOH are both strong), so its end point coincides accurately with the true (stoichiometric) equivalence point of this particular titration; methyl orange would also work for a strong acid–strong base titration, but phenolphthalein is a conventional, clearly-visible choice here.

(i) Small, genuinely random variations can arise from slight differences in judging the exact end-point colour change each time (a small amount of indicator error/subjectivity in judging when the colour change is complete), or from very small variations in the volume of a single drop added near the end point, which are not eliminated by care but are an inherent limitation of the technique.

Final answer: \( c(NaOH) = 0.0896\ \text{mol dm}^{-3} = 3.58\ \text{g dm}^{-3} \).

(a) [1] correct criterion stated (within 0.10 cm^3, or equivalent); [1] correctly identifies all three titres 1–3 as concordant; [1] correct mean \( =22.40\ \text{cm}^3 \). (b) [1] valid reason (rough titre used only to locate approximate end point / less precise, added too quickly). (c) [1] correct substitution; [1] \( n(HCl)=2.24\times10^{-3}\ \text{mol} \). (d) [1] correct 1:1 ratio identified; [1] \( n(NaOH)=2.24\times10^{-3}\ \text{mol} \) (ECF). (e) [1] correct method (n/V with correct unit conversion); [1] \( c(NaOH)=0.0896\ \text{mol dm}^{-3} \) (ECF). (f) [1] correct method (×Mr); [1] \( 3.58\ \text{g dm}^{-3} \) (ECF). (g) [1] valid precaution (read at eye level with bottom of meniscus). (h) [1] correctly identifies this is a strong acid–strong base titration; [1] correctly links this to phenolphthalein's sharp colour change occurring within the steep pH-change region at the equivalence point. (i) [1] valid source identified (e.g. subjectivity/difficulty judging exact end-point colour change, or the finite size of a single drop near the end point); [1] correctly explains why this causes small, random (not systematic) variation between titres.

(a) \( A=\pi(d/2)^2 = \pi\left(\dfrac{0.60\times10^{-3}}{2}\right)^2 = 2.83\times10^{-7}\ \text{m}^2 \)

(b) Using \( \sigma=F/A \) and \( \varepsilon=x/L \), the table of stress and strain values is exactly proportional to the load and extension values respectively (e.g. at F=100.0 N, x=6.06mm: \( \sigma=100.0/(2.83\times10^{-7})=3.53\times10^{8}\ \text{Pa} \), \( \varepsilon=6.06\times10^{-3}/1.200=5.05\times10^{-3} \)); each row scales linearly with load.

(c) Points plotted accurately (±1 small square) using scales occupying at least half the grid in each direction; single straight best-fit line through the origin.

(d) The data give an exactly straight line; using the extremes: gradient \( = \dfrac{3.53\times10^{8}-0}{5.05\times10^{-3}-0} = 7.00\times10^{10}\ \text{Pa} \). The gradient of a stress–strain graph represents the Young modulus of the material.

(e) \( E = 7.00\times10^{10}\ \text{Pa} \) (70.0 GPa), which is consistent with the accepted value for aluminium.

(f) Beyond the elastic limit, the stress–strain graph would no longer be a straight line — the graph would curve, with strain increasing disproportionately faster than stress (the material undergoes permanent, plastic deformation and would not return to its original length if unloaded).

Final answer: \( E = 7.00\times10^{10}\ \text{Pa} \).

(a) [1] correct substitution; [1] \( A=2.83\times10^{-7}\ \text{m}^2 \). (b) [1] correct method shown for at least one stress value; [1] correct method shown for at least one strain value; [1] all values correctly calculated and consistent (accept a fully correct table for full credit even without separate working shown for each row). (c) [1] suitable linear scales using at least half the grid; [1] all 6 points plotted accurately (±1 small square, ECF from (b)); [1] single straight best-fit line; [1] line passes through/consistent with the origin. (d) [1] triangle/points used span at least half the line; [1] correct gradient calculation shown; [1] correctly identifies gradient as the Young modulus. (e) [1] \( E\approx7.00\times10^{10}\ \text{Pa} \) (accept 6.6–7.4 x10^10, ECF); [1] correct unit (Pa or N m^-2). (f) [1] correctly identifies the graph would no longer be a straight line/would curve past the elastic limit; [1] correctly links this to permanent/plastic deformation (material does not return to original length).

(a) (i) Reinforced concrete is a composite (concrete matrix reinforced with steel bars). (ii) Polyethylene is a polymer. (iii) Aluminium oxide (alumina) is a ceramic.

(b) Ceramics generally have very high melting points and retain their strength/rigidity well at high temperatures, making them useful for high-temperature applications (e.g. furnace linings, cutting tools); however, ceramics are typically brittle (they have very low toughness/fracture toughness), so they crack or shatter easily under sudden impact loading rather than deforming plastically.

(c) A composite combines two (or more) materials so that the resulting material can have the best properties of each constituent (e.g. the reinforcing fibres/bars provide high tensile strength/stiffness, while the matrix provides toughness, shape and protects/binds the reinforcement), giving overall mechanical properties (e.g. a combination of strength and toughness) that neither constituent could achieve alone.

Final answer: (i) composite, (ii) polymer, (iii) ceramic.

(a) [1] each for correct classification of (i), (ii), (iii); max [3]. (b) [1] valid high-temperature property (e.g. high melting point, thermal stability); [1] valid limiting property (brittleness/low impact resistance/low toughness). (c) [1] correct explanation referencing combination of beneficial properties from each constituent (e.g. strength from one component, toughness/binding from the other).

(a) Nitinol (a nickel–titanium shape-memory alloy), which responds to a change in temperature. (Other valid examples: piezoelectric materials, responding to mechanical stress; thermochromic materials, responding to temperature; photochromic materials, responding to light.)

(b) A shape-memory alloy can be deformed (bent out of shape) while relatively cool, but when it is subsequently heated above a certain transition temperature, it reverts back to ('remembers') its original, pre-set shape; this shape change is reversible and can be repeated over many cycles.

(c) Example: Nitinol stents used to hold open a narrowed blood vessel. The stent is compressed to a small diameter (cooled) so that it can be inserted into the body via a small incision/catheter; once in position at body temperature, it expands back to its pre-set, larger shape, holding the vessel open — this smart behaviour allows a minimally invasive method of insertion that would not be possible with a rigid, fixed-shape stent.

(a) [1] valid smart material named; [1] correct corresponding stimulus. (b) [1] correctly describes deformation while cool; [1] correctly describes reversion to original shape on heating above a transition temperature. (c) [1] valid application named; [1] correct explanation of why the smart (shape-recovery) behaviour is specifically useful in that application.

(a) Thermoplastic polymers consist of long-chain molecules held together only by weak intermolecular (van der Waals/secondary) forces between separate, unlinked chains. Thermosetting polymers have their chains permanently joined together by strong covalent cross-links, forming a single, rigid, three-dimensional molecular network.

(b) On heating, a thermoplastic's weak intermolecular forces between chains are easily overcome, allowing the chains to slide past one another, so the material softens/melts and can be reshaped (remoulded); on cooling, the chains simply re-solidify in the new shape. A thermosetting polymer's chains are held by strong covalent cross-links that do not break on heating (until the material chars/decomposes at very high temperature), so the rigid network structure — and hence the shape — cannot be softened or remoulded once set.

(c) Increasing crystallinity generally increases stiffness (and hardness/density), because the chains are packed more closely and regularly, increasing the intermolecular forces resisting chain movement/deformation. However, increasing crystallinity generally decreases transparency (makes the polymer more opaque/translucent), because the crystalline regions scatter light differently from the surrounding amorphous regions, and this difference in refractive index at the boundaries scatters light passing through the material.

Final answer: thermoplastics = weak intermolecular forces between separate chains (remouldable); thermosets = covalent cross-links (not remouldable).

(a) [1] thermoplastic: chains held by weak intermolecular forces, not linked; [1] thermoset: chains joined by strong covalent cross-links. (b) [1] correct explanation for thermoplastic (weak forces overcome, chains slide, softens/remoulds); [1] correct explanation for thermoset (covalent cross-links do not break on heating, cannot remould). (c) [1] correctly states stiffness increases with crystallinity, with valid reasoning (closer/more regular packing, stronger intermolecular forces resisting deformation); [1] correctly states transparency decreases with crystallinity, with valid reasoning (light scattering at crystalline/amorphous boundaries).

(a) The width and thickness of the strip should each be measured at several points along its length using a micrometer or digital calipers (which have much finer resolution than a ruler), and a mean value taken; averaging several readings at different points reduces the uncertainty in A (and accounts for any small variation in the strip's dimensions along its length).

(b) The ammeter should be connected in series in the main circuit (so the full current through the sample passes through it), and the voltmeter should be connected in parallel directly across the sample only (not across the whole circuit, including any variable resistor/cell), so that it reads only the potential difference across the semiconductor strip itself.

(c) \( R = \dfrac{V}{I} = \dfrac{0.360}{0.0150} = 24.0\ \Omega \)

(d) \( \rho = \dfrac{RA}{L} = \dfrac{(24.0)(2.5\times10^{-6})}{0.020} = 3.00\times10^{-3}\ \Omega\text{m} \)

(e) The resistance of the semiconductor would be expected to decrease as temperature increases. This is because, in a semiconductor, raising the temperature provides more thermal energy to promote additional charge carriers (electrons) across the material's band gap into the conduction band, greatly increasing the number density of free charge carriers available to carry current, which outweighs any increase in lattice vibration/scattering, so overall resistance falls. This contrasts with a metal conductor, whose resistance normally increases with temperature, because a metal already has a very large, roughly constant number of free electrons, so the dominant effect of raising temperature is increased vibration of the fixed metal ions, which scatters the conduction electrons more and so increases resistance.

(f) A small current should be used (as low as practicable while still giving a measurable, accurate voltmeter reading), and/or readings should be taken quickly, to minimise resistive (\( I^2R \)) heating of the sample during the measurement, which would otherwise raise its temperature and change (lower) its resistance while the measurement was being taken.

Final answer: \( \rho = 3.00\times10^{-3}\ \Omega\text{m} \).

(a) [1] correct instrument named (micrometer/digital calipers) for measuring width/thickness; [1] correctly states repeating at several points and averaging reduces uncertainty. (b) [1] correctly states ammeter in series; [1] correctly states voltmeter in parallel, directly across the sample. (c) [1] correct substitution into \( R=V/I \); [1] \( R=24.0\ \Omega \). (d) [1] correct substitution into \( \rho=RA/L \); [1] \( \rho=3.00\times10^{-3}\ \Omega\text{m} \) (ECF from (c)); [1] correct unit (\( \Omega\text{m} \)). (e) [1] correctly states resistance decreases with increasing temperature for the semiconductor; [1] correct explanation referencing increased number of charge carriers/promotion across the band gap; [1] correct contrast with a metal (resistance increases with temperature due to increased ion vibration/electron scattering, carrier number roughly constant). (f) [1] valid precaution (use small current and/or take readings quickly); [1] correct reasoning linking this to minimising \( I^2R \) self-heating that would otherwise change the sample's resistance during measurement.

(a) \( V = \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi(5.0\times10^{-9})^3 = 5.24\times10^{-25}\ \text{m}^3 \)

(b) Surface area-to-volume ratio \( = \dfrac{4\pi r^2}{\frac43\pi r^3} = \dfrac{3}{r} \). For the nanoparticle: \( \dfrac{3}{5.0\times10^{-9}} = 6.0\times10^{8}\ \text{m}^{-1} \). For the bulk sphere (r = 1.0 cm = 1.0×10^-2 m): \( \dfrac{3}{1.0\times10^{-2}} = 3.0\times10^{2}\ \text{m}^{-1} \). Ratio: \( \dfrac{6.0\times10^{8}}{3.0\times10^{2}} = 2.0\times10^{6} \) — the nanoparticle's surface-area-to-volume ratio is two million times greater than that of the bulk sphere.

(c) Catalysis occurs at the surface of a catalyst (reactant molecules must come into contact with the catalyst's surface). Since the nanoparticle has a vastly greater surface area per unit volume (and hence per unit mass, given the density is the same as bulk), a given mass of gold nanoparticles presents far more surface area available for reactant molecules to interact with than the same mass of bulk gold, making it a much more effective (and efficient, requiring less gold) catalyst gram for gram.

(d) Other properties that can differ include: melting point (often lower for nanoparticles than bulk); optical properties/colour (nanoparticles can absorb/scatter visible light differently, e.g. gold nanoparticle colloids appear red rather than gold-coloured); electrical or magnetic properties. (Any one valid example.)

Final answer: the nanoparticle's surface-area-to-volume ratio is \( 2.0\times10^{6} \) times greater than the bulk sphere's.

(a) [1] correct substitution into \( \frac43\pi r^3 \); [1] \( V=5.24\times10^{-25}\ \text{m}^3 \). (b) [1] correct SA:V expression/method (e.g. 3/r, or separate SA and V calculated and divided); [1] correct nanoparticle SA:V \( =6.0\times10^{8}\ \text{m}^{-1} \); [1] correct bulk sphere SA:V \( =3.0\times10^{2}\ \text{m}^{-1} \); [1] correct ratio \( =2.0\times10^{6} \). (c) [1] correctly identifies catalysis occurs at the surface / greater surface area means more reactant contact; [1] correctly links this to more effective catalysis per unit mass, given the same density/less material needed. (d) [1] any valid distinct nanoscale property (melting point, optical properties/colour, electrical/magnetic behaviour).

(a) \( V = 4.00\times3.00\times1.50 = 18.0\ \text{cm}^3 \) for each sample (both have the same dimensions).

(b) Alloy P: \( \rho = \dfrac{m}{V} = \dfrac{54.0\ \text{g}}{18.0\ \text{cm}^3} = 3.00\ \text{g cm}^{-3} = 3.00\times10^{3}\ \text{kg m}^{-3} \) (since 1 g cm^-3 = 1000 kg m^-3). Alloy Q: \( \rho = \dfrac{67.5}{18.0} = 3.75\ \text{g cm}^{-3} = 3.75\times10^{3}\ \text{kg m}^{-3} \)

(c) Mass of a 50.0 cm^3 component: Alloy P: \( m=\rho V = 3.00\times50.0 = 150\ \text{g} \). Alloy Q: \( m=3.75\times50.0=187.5\ \text{g} \). Neither of these masses is below 90.0 g, so — recalculating correctly using densities in the same units — as both calculated masses (150 g for P, 187.5 g for Q) exceed the 90.0 g limit for a 50.0 cm^3 component, NEITHER alloy meets the mass requirement for this particular application at this volume; alloy P is the lighter (lower-density) of the two, and would be closer to meeting a relaxed requirement, but a lower-density alloy or a smaller/hollow design would need to be considered to meet 90.0 g at 50.0 cm^3.

(d) Other relevant properties include: strength (e.g. tensile strength/yield stress) and stiffness (Young modulus), so the component does not fail or deform excessively in use; also relevant: corrosion resistance, cost, and ease of manufacture/machinability. (Any one valid property.)

Final answer: \( \rho_P=3.00\times10^{3}\ \text{kg m}^{-3} \), \( \rho_Q=3.75\times10^{3}\ \text{kg m}^{-3} \); neither alloy meets the 90.0 g / 50.0 cm^3 requirement (P would give 150 g, Q would give 187.5 g).

(a) [1] \( V=18.0\ \text{cm}^3 \) for each. (b) [1] correct method (m/V) for P; [1] \( \rho_P=3.00\times10^{3}\ \text{kg m}^{-3} \) with correct unit conversion shown; [1] correct method for Q; [1] \( \rho_Q=3.75\times10^{3}\ \text{kg m}^{-3} \). (c) [1] correct method (mass = density × 50.0 cm^3, in consistent units) for both alloys; [1] correct masses (150 g for P, 187.5 g for Q); [1] correct conclusion that neither alloy meets the 90.0 g requirement (accept any answer with correct supporting calculation, even if the initial conclusion differs, provided the arithmetic and final comparison against 90.0 g are correct). (d) [1] any valid additional material property relevant to structural component selection (strength, stiffness/Young modulus, corrosion resistance, cost, machinability).

An indicative full-mark response would include:

Titanium alloy: titanium alloys have a high strength-to-weight ratio and good toughness (they can absorb energy and deform slightly rather than shattering under sudden/impact loading), which is important given the repeated, sometimes sudden, mechanical loads a hip joint experiences during walking, running or a fall. Titanium alloys are also biocompatible (they do not react readily with body tissue and resist corrosion in the body's chemical environment) and can bond well with surrounding bone (osseointegration). A limitation is that titanium's Young modulus, though lower than steel, is still considerably higher (stiffer) than natural bone, which can lead to 'stress shielding' — the surrounding bone bears less mechanical load than normal (because the stiffer implant carries more of the load) and can gradually weaken/resorb over time. Metal surfaces can also gradually wear, producing metal debris particles.

Ceramic (alumina): ceramics such as alumina are extremely hard and have excellent wear resistance (very low friction and very slow wear when articulating against another ceramic or polyethylene surface), which is valuable for the long service life required (many years without needing replacement/revision surgery), and are chemically very inert/highly biocompatible. However, ceramics are brittle, with low toughness/fracture toughness — they have very little ability to deform before fracturing, so a ceramic component can fail suddenly and catastrophically (rather than gradually) under an unexpectedly high or sudden load (e.g. a fall), which is a serious safety concern for a load-bearing implant.

Conclusion: in practice, a combination is often used to exploit the advantages of each material class while minimising their individual weaknesses — for example, a titanium alloy stem/socket shell (for strength, toughness and good bonding with bone) combined with a ceramic (or polyethylene) bearing surface at the actual ball-and-socket articulation (for very low wear and low friction). This combined approach gives a implant that is both tough enough to withstand sudden loads without catastrophic failure and long-wearing enough at the bearing surface to last for many years, better balancing the trade-off between toughness and wear resistance than either material used alone throughout.

Levels-of-response (QWC) mark scheme, out of 9 marks.

Level 3 (7–9 marks): Balanced, detailed evaluation of BOTH material classes, correctly referencing specific relevant properties (strength/toughness and biocompatibility for the metal; wear resistance and brittleness for the ceramic; stiffness mismatch/stress shielding for the metal), and reaching a well-justified conclusion (e.g. recognising a combined/hybrid approach is generally used, with reasoning for why). Correct, precise terminology throughout (toughness, stiffness, wear resistance, biocompatibility, brittle); coherent structure; accurate spelling, punctuation and grammar.

Level 2 (4–6 marks): Reasonable coverage of both materials with correctly identified properties, but with less depth or balance (e.g. limited discussion of one material, or conclusion asserted without full justification); generally clear communication with occasional lapses.

Level 1 (1–3 marks): Limited or one-sided evaluation (e.g. discusses only one material, or lists properties without evaluative reasoning or a conclusion); weak terminology or organisation.

0 marks: No creditable response.

Full marks at Level 3 require genuine evaluation (weighing advantages against disadvantages for both material classes) and a clearly justified conclusion, not simply a list of properties.