2025 CCEA AS-Level Life and Health Sciences 0008 Practice Paper with Answers

(a) A healthy adult's resting pulse rate typically lies in the range 60-80 beats per minute.
(b) Blood pressure is measured using a sphygmomanometer.

(a) 1 mark: 60-80 bpm stated (accept values within this range/close synonyms e.g. "60 to 80"). (b) 1 mark: sphygmomanometer named; [2]

(a) Heart valves function to prevent the backflow of blood, ensuring blood flows in one direction only through the heart's chambers and into the major vessels.
(b) The bicuspid (mitral) valve lies between the left atrium and left ventricle; the aortic valve, between the left ventricle and the aorta, is also on the left side.

(a) 1 mark: prevents backflow of blood/ensures one-way flow. (b) 1 mark: bicuspid/mitral valve OR aortic valve correctly named; [2]

(a) Blood plasma is composed mainly of water, in which are dissolved substances such as plasma proteins (e.g. albumin), glucose, mineral ions, hormones and waste products such as urea.
(b) Plasma's main function is to act as the transport medium for these dissolved substances around the body, and it also helps maintain blood volume and pressure.

(a) 1 mark: any two valid components named (water plus one dissolved substance, or two dissolved substances). (b) 1 mark: valid function stated (transport of nutrients/hormones/waste, or maintenance of blood volume); [2]

In cystic fibrosis, abnormally thick, sticky mucus builds up in the airways and alveoli, obstructing airflow and impairing gas exchange, as well as increasing the risk of infection. In emphysema, the walls of the alveoli break down and merge together, reducing the total surface area available for gas exchange and so reducing the efficiency of oxygen uptake.

1 mark: cystic fibrosis effect correctly described (thick mucus build-up obstructing airways/alveoli); 1 mark: emphysema effect correctly described (alveoli wall breakdown reducing surface area for gas exchange); [2]

(a) Basal metabolic rate (BMR) is the rate at which the body expends energy at complete rest in order to maintain essential life-sustaining functions (such as breathing, circulation and cell maintenance).
(b) BMR can be affected by factors such as age, sex, muscle mass or the level of thyroxine in the body.

(a) 1 mark: BMR correctly defined as energy expenditure at rest to maintain basic body functions. (b) 1 mark: any one valid factor named (age/sex/muscle mass/genetics/thyroxine level); [2]

Aerobic respiration requires oxygen and yields a large amount of ATP per glucose molecule, producing carbon dioxide and water as end products. Anaerobic respiration in humans takes place without oxygen, yields a much smaller amount of ATP per glucose molecule, and produces lactate (lactic acid) rather than carbon dioxide and water.

1 mark each for any two valid, distinct differences, e.g.: requires oxygen vs does not require oxygen; much greater ATP yield in aerobic respiration; end product lactate (anaerobic) vs carbon dioxide and water (aerobic); [2]

(a) Blood glucose concentration is regulated mainly by the hormones insulin and glucagon, both secreted by the pancreas.
(b) Insulin is released when blood glucose concentration rises too high; it stimulates cells (e.g. liver and muscle cells) to take up glucose and convert it to glycogen, lowering blood glucose back towards normal.

(a) 1 mark: insulin and glucagon both correctly named. (b) 1 mark: insulin correctly identified as being released when blood glucose is too high; [2]

(a) The normal range for blood pH is 7.35-7.45.
(b) Blood pH is monitored using arterial blood gas (ABG) analysis, in which a sample of arterial blood is tested using a blood gas analyser.

(a) 1 mark: 7.35-7.45 stated. (b) 1 mark: valid monitoring method named (arterial blood gas analysis/blood gas analyser); [2]

(a) Starchy carbohydrate foods should make up the largest proportion of a balanced diet.
(b) Sources of starchy carbohydrate include bread, rice, potatoes and pasta.

(a) 1 mark: carbohydrates/starchy foods correctly named as the largest food group. (b) 1 mark: any valid carbohydrate food source given; [2]

(a) Normal blood cholesterol level is monitored to lie in the range 4.0-6.5 mmol/L.
(b) Persistently high blood cholesterol increases the long-term risk of atherosclerosis (fatty deposits narrowing the arteries), which in turn increases the risk of coronary heart disease and stroke.

(a) 1 mark: 4.0-6.5 mmol/L stated. (b) 1 mark: valid long-term health effect given (atherosclerosis/CHD/stroke risk increased); [2]

Individual Q's diet closely matches UK dietary guidelines on every measure shown: 6 portions of fruit and vegetables (above the 5-a-day guideline), 32 g of fibre (above the 30 g guideline), only 6% of energy from added sugar and 10% from saturated fat (both within recommended limits). This diet is likely to support good long-term health.

Individual P's diet, by contrast, falls well short of the guidelines on every measure: only 1 portion of fruit and vegetables (far below 5-a-day), low fibre intake (12 g, well below the 30 g guideline), and both added sugar (16%) and saturated fat (17%) intake exceeding the recommended upper limits. This pattern of low fibre and fruit/vegetable intake combined with high sugar and saturated fat intake increases Individual P's long-term risk of obesity, type 2 diabetes, and cardiovascular disease (e.g. through raised blood cholesterol and atherosclerosis).

1 mark: Individual Q correctly evaluated as meeting/exceeding guidelines on all measures quoted with values; 1 mark: Individual P correctly evaluated as falling short of guidelines, with at least two values quoted (e.g. 1 portion fruit/veg, 12 g fibre); 1 mark: added sugar and/or saturated fat values for P correctly identified as exceeding recommended limits; 1 mark: valid long-term health risk linked to Individual P's diet (obesity/type 2 diabetes/CHD/atherosclerosis); [4]

The data show that energy intake rises steadily from 2600 kcal/day in Month 1 to 3000 kcal/day in Months 3 and 4, while energy expenditure falls steadily from 2500 kcal/day to 2350 kcal/day over the same period. This means the daily energy surplus (intake minus expenditure) increases every month: 100 kcal/day in Month 1, 350 kcal/day in Month 2, 600 kcal/day in Month 3, and 650 kcal/day in Month 4. This growing surplus of energy taken in over energy expended accounts for the accelerating gain in body mass shown (from +0.4 kg in Month 1 to +2.0 kg in Month 4), since body mass is gained when energy intake exceeds energy expenditure. If this trend continues, the individual will keep gaining body mass and is at increasing risk of becoming overweight or obese, which raises the long-term risk of health problems such as type 2 diabetes, cardiovascular disease and joint problems.

1 mark: correct trend in energy intake identified (increasing, with values quoted); 1 mark: correct trend in energy expenditure identified (decreasing, with values quoted); 1 mark: correctly links the growing energy surplus to the accelerating body mass gain shown; 1 mark: valid long-term health consequence stated if trend continues (overweight/obesity and associated disease risk); [4]

Individual X, the trained athlete, has both a lower resting pulse rate (58 bpm compared with 78 bpm for Individual Y) and a much shorter recovery time after exercise (2.5 minutes compared with 7.0 minutes). This is because regular training increases cardiac efficiency, particularly by increasing the stroke volume of the heart (the volume of blood ejected per beat). Since a trained heart pumps more blood per beat, fewer beats per minute are needed to meet the body's resting oxygen demand, giving a lower resting pulse rate. The trained cardiovascular and respiratory systems are also more efficient at delivering oxygen to, and removing waste products (such as lactate) from, the muscles during recovery, allowing pulse rate to return to its resting value more quickly after exercise.

1 mark: correct comparison of resting pulse rates with values quoted (X lower than Y); 1 mark: correct comparison of recovery times with values quoted (X faster than Y); 1 mark: valid explanation linking training to increased stroke volume/cardiac efficiency; 1 mark: valid explanation of faster recovery in trained individual (more efficient oxygen delivery/waste removal); [4]

Individuals R and S both have blood haemoglobin concentrations within the normal adult reference range (148 g/L and 138 g/L respectively, against a range of 120-170 g/L), and both have reasonably high dietary iron intakes (16 mg/day and 14 mg/day). This shows their diets are providing enough iron to support normal levels of haemoglobin production. Individual T, however, has a very low dietary iron intake of only 4 mg/day and a haemoglobin concentration of 96 g/L, which is well below the normal reference range — this is consistent with iron-deficiency anaemia, caused by an inadequate intake of iron-rich foods (in this case, a restricted vegetarian diet low in pulses and leafy green vegetables, which are important non-meat sources of iron).

To correct this, Individual T's diet should be modified to include more iron-rich, vegetarian-suitable foods, such as pulses (lentils, beans, chickpeas), leafy green vegetables (e.g. spinach), and iron-fortified cereals or bread. Eating these iron sources alongside a source of vitamin C (e.g. citrus fruit) can also improve the absorption of non-haem iron from plant sources. If dietary changes alone are insufficient, an iron supplement could also be considered.

1 mark: R and S correctly identified as having normal haemoglobin levels, with values quoted and compared to the reference range; 1 mark: T correctly identified as having haemoglobin below the normal reference range, with value quoted; 1 mark: T's low haemoglobin correctly linked to T's very low dietary iron intake (iron-deficiency anaemia); 1 mark: at least one valid, specific iron-rich vegetarian food source suggested for T's diet (e.g. pulses/leafy greens/fortified cereals); 1 mark: valid additional point (e.g. role of vitamin C in improving iron absorption, or supplementation); [5]

(a) Vital capacity = tidal volume + inspiratory reserve volume + expiratory reserve volume.
Non-smoker: 0.5 + 3.1 + 1.2 = 4.8 litres.
Smoker with emphysema: 0.4 + 1.9 + 0.7 = 3.0 litres.
(b) On the spirometer trace, the smoker's trace would show a noticeably smaller vital capacity than the non-smoker's — the overall height of the trace, from the lowest point (after maximal expiration) to the highest point (after maximal inspiration), would be smaller (3.0 litres compared with 4.8 litres). This is because emphysema causes the walls of the alveoli to break down and merge, reducing the elastic recoil of the lung tissue and the surface area available for gas exchange, so the individual cannot draw in or expel as large a volume of air as a healthy non-smoker.

(a) 1 mark: correct method (sum of the three volumes) shown for at least one individual; 1 mark: both vital capacities correct (4.8 litres and 3.0 litres) [ecf from (a) working]. (b) 1 mark: smaller/reduced vital capacity in the smoker's trace correctly identified with values compared; 1 mark: valid description of trace shape (lower overall height/flatter swing); 1 mark: correct physiological explanation (alveoli wall breakdown reducing elasticity/surface area for gas exchange); [5]

(a) Energy expended in each activity = (rate per 30 min) / 30 x (minutes performed).
Walking: (120/30) x 60 = 4 x 60 = 240 kcal.
Cycling: (240/30) x 30 = 8 x 30 = 240 kcal.
Swimming: (300/30) x 20 = 10 x 20 = 200 kcal.
Total for the three activities = 240 + 240 + 200 = 680 kcal.
(b) This 680 kcal is expended in addition to the individual's resting/sedentary energy expenditure for the rest of the day (at the background rate of 50 kcal per 30 minutes, i.e. 100 kcal/hour, applied to the remaining hours of the day). Adding this background expenditure to the 680 kcal from active exercise is likely to bring total daily energy expenditure to a level close to, or somewhat above, the stated daily energy requirement of 2200 kcal. Since body mass is gained when energy intake exceeds total energy expenditure and lost when expenditure exceeds intake, with an intake of 2200 kcal on this day the individual is likely to roughly maintain their body mass, or lose a small amount of body mass if total expenditure exceeds 2200 kcal — the exercise undertaken makes weight gain on this day unlikely.

(a) 1 mark: correct method for at least two activities shown; 1 mark: correct total of 680 kcal (ecf from working). (b) 1 mark: correctly recognises that total expenditure = active energy expended plus resting/background expenditure; 1 mark: valid comparison made between total likely expenditure and the 2200 kcal intake; 1 mark: correct conclusion drawn (mass maintained or slightly lost) with reasoning linked to energy balance principle; [5]

A diet high in saturated fat and sodium combined with low physical activity has both short-term and long-term negative effects on health. In the short term, excess sodium intake raises blood pressure and can cause fluid retention, while a high-fat, low-activity lifestyle contributes to weight gain, reduced fitness, and fatigue during everyday activity. In the long term, sustained high saturated fat intake raises blood cholesterol levels, contributing to atherosclerosis (narrowing of the arteries through fatty deposits), which increases the risk of coronary heart disease and stroke. Persistently high sodium intake contributes to chronic hypertension, further raising cardiovascular risk. Combined with low physical activity and consequent weight gain, this lifestyle also raises the long-term risk of type 2 diabetes and joint problems associated with obesity.

Modifying the diet to reduce saturated fat and sodium intake, alongside increasing regular physical activity, can substantially reduce these risks. Reducing saturated fat intake lowers blood cholesterol, slowing the progression of atherosclerosis; reducing sodium intake helps normalise blood pressure. Regular physical exercise strengthens the cardiovascular system (e.g. increasing stroke volume and lowering resting pulse rate) and the respiratory system, improves the body's ability to regulate blood glucose, and helps maintain a healthy body mass by increasing energy expenditure. Together, these changes reduce the long-term risk of coronary heart disease, hypertension, type 2 diabetes and stroke, while also improving short-term energy levels and general fitness.

Level 3 (5-6 marks): Thorough, balanced evaluation covering short-term AND long-term effects of the diet/inactivity, and a well-developed evaluation of how diet modification and exercise reduce risk, using accurate technical terminology (e.g. atherosclerosis, hypertension, stroke volume) with fluent, well-organised prose and accurate spelling/grammar. Level 2 (3-4 marks): Sound coverage of most effects and mitigation strategies with some technical terminology, but limited in balance or depth (e.g. long-term effects covered well but mitigation underdeveloped, or vice versa); generally accurate spelling/grammar. Level 1 (1-2 marks): Basic, general statements about diet/exercise and health lacking specific technical detail or balance; weak use of terminology; basic spelling/grammar. [6]

The Bohr effect describes how a rise in carbon dioxide concentration (and the resulting fall in pH/rise in acidity) around haemoglobin reduces haemoglobin's affinity for oxygen, causing it to unload oxygen more readily. Contracting muscle produces more carbon dioxide, and becomes more acidic, than resting tissue because its rate of aerobic respiration is much higher. This locally raised CO2/acidity around the muscle capillaries shifts haemoglobin's oxygen affinity down, so haemoglobin releases a greater proportion of the oxygen it is carrying exactly where the contracting muscle needs it most. This benefits oxygen delivery because it ensures oxygen is preferentially unloaded at the tissues with the greatest metabolic demand, rather than being distributed evenly regardless of need.

1 mark: correctly states that rising CO2/acidity reduces haemoglobin's affinity for oxygen (the Bohr effect); 1 mark: correctly explains that this causes greater oxygen unloading/release from haemoglobin; 1 mark: correctly links contracting muscle's higher CO2 production/acidity to this effect occurring specifically at the muscle; 1 mark: correctly explains the benefit (oxygen delivered preferentially to the tissue with greatest demand); [4]

The partial pressure of oxygen is the pressure that oxygen alone would exert if it occupied the total volume of a gas mixture on its own; it is used as a measure of the concentration or availability of oxygen, whether in a gas mixture such as air or dissolved in a liquid such as blood.

In the lungs, the partial pressure of oxygen in the alveolar air is higher than in the deoxygenated blood arriving in the pulmonary capillaries. This difference in partial pressure creates a diffusion gradient, so oxygen diffuses from the alveoli into the blood plasma and then into the red blood cells. As the partial pressure of oxygen in the blood rises at the lungs, haemoglobin's affinity for oxygen at this high partial pressure means it readily binds (loads) oxygen, becoming saturated with oxygen ready for transport to the tissues of the body.

1 mark: partial pressure of oxygen correctly defined as its individual contribution to total gas pressure/a measure of oxygen availability; 1 mark: correctly states partial pressure of oxygen is higher in alveolar air than in the blood arriving at the lungs; 1 mark: correctly explains that oxygen diffuses down this partial pressure gradient into the blood; 1 mark: correctly links the resulting high partial pressure of oxygen in the blood to increased loading of oxygen onto haemoglobin; [4]

Glycolysis takes place in the cytoplasm of the cell. During glycolysis, one molecule of glucose, a six-carbon sugar, is broken down in a series of enzyme-controlled reactions into two molecules of pyruvate, each a three-carbon compound. This process also generates reduced NAD (NADH), which carries hydrogen to the electron transport chain in aerobic respiration. Although four molecules of ATP are produced during glycolysis, two molecules of ATP are used up in the initial steps to activate glucose, so the net yield of ATP from glycolysis is 2 molecules of ATP per molecule of glucose.

1 mark: correctly states glucose (six-carbon) is broken down into two molecules of pyruvate (three-carbon); 1 mark: correctly states/implies that NADH (reduced NAD) is also produced; 1 mark: correct net ATP yield of 2 ATP per glucose molecule stated; [3]

During glycolysis and the Krebs cycle, only a small number of ATP molecules are produced directly, by substrate-level phosphorylation. However, both stages also produce large numbers of reduced NAD (NADH) and reduced FAD (FADH2) molecules, which carry high-energy electrons to the electron transport chain located on the inner mitochondrial membrane. As electrons pass along the chain of carriers, energy released is used to actively pump hydrogen ions across the inner mitochondrial membrane, building up a steep hydrogen ion concentration (electrochemical) gradient. The flow of hydrogen ions back across the membrane, through the enzyme ATP synthase, drives the synthesis of a very large number of ATP molecules — this process, called oxidative phosphorylation (or chemiosmosis), generates far more ATP than the direct, substrate-level phosphorylation occurring in glycolysis and the Krebs cycle, which is why the electron transport chain accounts for by far the greatest proportion of ATP produced overall.

1 mark: correctly states glycolysis/Krebs cycle produce reduced NAD/FAD which carry electrons/hydrogen to the electron transport chain; 1 mark: correctly describes hydrogen ions being pumped across the inner mitochondrial membrane to create a gradient; 1 mark: correctly explains that the flow of hydrogen ions through ATP synthase (chemiosmosis/oxidative phosphorylation) generates ATP; 1 mark: correctly contrasts this with the much smaller, direct (substrate-level) ATP yield of glycolysis/Krebs cycle; [4]

Thyroxine is secreted by the thyroid gland, and its general role is to regulate the body's basal metabolic rate, i.e. the rate at which cells throughout the body release and use energy.

The secretion of thyroxine is controlled by a negative feedback mechanism. When thyroxine levels are low, the hypothalamus secretes thyrotropin-releasing hormone (TRH), which stimulates the pituitary gland to secrete thyroid-stimulating hormone (TSH). TSH in turn stimulates the thyroid gland to secrete thyroxine. As blood thyroxine levels rise, this rise inhibits (via negative feedback) further secretion of TRH from the hypothalamus and TSH from the pituitary gland, reducing further thyroxine release and so keeping thyroxine concentration within a normal range.

1 mark: thyroid gland correctly named as the source of thyroxine; 1 mark: correct general role given (regulation of basal metabolic rate/energy release in cells); 1 mark: correct description of the hypothalamus-pituitary-thyroid pathway (TRH stimulates TSH release, TSH stimulates thyroxine release); 1 mark: correct explanation of negative feedback (rising thyroxine inhibits further TRH/TSH release); [4]

Vitamin D's main role in the body is to regulate the absorption of calcium and phosphate from the gut into the bloodstream, and it is required for the normal mineralisation (hardening) of bone and teeth. Vitamin D can be synthesised in the skin on exposure to sunlight, and is also obtained from dietary sources such as oily fish and eggs.

A deficiency of vitamin D reduces calcium absorption and bone mineralisation, leading to rickets in children (characterised by soft, weakened and deformed bones) or osteomalacia (softening of the bones) in adults.

1 mark: correct role stated (regulates calcium/phosphate absorption from the gut); 1 mark: correctly links this to bone mineralisation/hardening; 1 mark: valid deficiency effect stated (rickets in children/osteomalacia in adults); [3]

(a) In a batch process, a fixed quantity of raw materials is processed together as one batch; the process runs to completion, and the plant is then cleaned and reset before starting a new batch. In a continuous process, raw materials are fed in and products are removed continuously, without the process being stopped between production runs.
(b) The Haber process (manufacture of ammonia) is operated as a continuous process. This is because ammonia is in very high, constant industrial demand (e.g. for fertiliser manufacture), so running the process continuously avoids the lost production time and cost associated with repeatedly stopping and restarting a batch process, making large-scale continuous production more economical.

(a) 1 mark: correct distinction drawn (batch = discrete quantity processed then reset; continuous = ongoing feed and removal without stopping). (b) 1 mark: valid continuous industrial process named (e.g. Haber process/Contact process); 1 mark: valid reason given (high/constant demand avoiding shutdown costs); [3]

(a) A capital cost is the initial, one-off cost of purchasing or constructing the plant, machinery and equipment required to manufacture the chemical.
(b) A direct cost is one incurred directly in making the product, such as the cost of raw materials, labour, or energy/fuel used in the process. An indirect cost is one not directly tied to a specific unit of production but still necessary to run the business, such as plant maintenance, insurance, or waste disposal/administration costs.

(a) 1 mark: capital cost correctly defined as the one-off cost of plant/equipment. (b) 1 mark: valid direct cost example given (raw materials/labour/energy); 1 mark: valid indirect cost example given (maintenance/insurance/administration/waste disposal); [3]

Industrial-scale production differs from laboratory-scale production in several ways. Industrial processes are often run continuously rather than as batches, to maximise output for high, constant demand. Industrial production prioritises cost-effectiveness and profitability — for example, unreacted raw materials are often recycled back into the process to reduce waste and cost, whereas a laboratory procedure prioritises accuracy and purity of results over minimising cost. Because of the much larger quantities of chemicals and energy involved, industrial production also requires far greater consideration of safety, waste management and compliance with environmental regulations than a laboratory-scale reaction. Finally, industrial processes typically require substantial capital investment in specialised plant and machinery, and make use of catalysts and carefully optimised temperature and pressure conditions to maximise the rate and yield of production on a large scale.

1 mark each for any three valid, distinct differences, e.g.: continuous vs batch operation; recycling of unreacted materials to reduce cost/waste vs prioritising purity in the lab; greater emphasis on safety/waste management/environmental regulation at industrial scale; need for capital investment in specialised plant/use of catalysts and optimised conditions; [3]

A manufacturer must set the selling price of a chemical high enough to recover its total production costs — capital costs (plant and equipment), direct costs (e.g. raw materials, labour, energy) and indirect costs (e.g. maintenance, insurance, waste disposal) — while also making a sufficient profit to remain a viable business. If any of these costs rise, for example due to an increase in the price of raw materials or energy, the manufacturer will generally need to increase the selling price to maintain profitability. However, the manufacturer cannot simply raise the price indefinitely, since the price must also remain competitive with other suppliers of the same or a similar chemical, and must reflect what customers are willing to pay.

1 mark: correctly explains that selling price must cover total production costs (capital/direct/indirect) plus profit margin; 1 mark: correctly explains that rising costs generally lead to a higher selling price, constrained by market competition/demand; [2]

Moles of HCl used \( = \text{concentration} \times \text{volume} = 0.100 \times \dfrac{22.50}{1000} = 0.00225 \) mol.
From the equation, NaOH reacts with HCl in a 1:1 mole ratio, so moles of NaOH = moles of HCl = \( 0.00225 \) mol.
Concentration of NaOH \( = \dfrac{\text{moles}}{\text{volume}} = \dfrac{0.00225}{25.0/1000} = \dfrac{0.00225}{0.0250} = 0.0900\,\text{mol}\,\text{dm}^{-3} \).

1 mark: correct moles of HCl calculated \( (0.00225\,\text{mol}) \); 1 mark: correct 1:1 mole ratio applied to obtain moles of NaOH (ecf); 1 mark: correct final concentration of \( 0.0900\,\text{mol}\,\text{dm}^{-3} \) (ecf); [3]

Relative formula mass of Na₂CO₃ \( = (2 \times 23) + 12 + (3 \times 16) = 46 + 12 + 48 = 106 \).
Moles of Na₂CO₃ in \( 250\,\text{cm}^3 \) stock solution \( = \dfrac{\text{mass}}{M_r} = \dfrac{0.530}{106} = 0.00500 \) mol.
Concentration of stock Na₂CO₃ solution \( = \dfrac{0.00500}{0.250} = 0.0200\,\text{mol}\,\text{dm}^{-3} \).
Moles of Na₂CO₃ in the \( 25.0\,\text{cm}^3 \) sample \( = 0.0200 \times \dfrac{25.0}{1000} = 0.000500 \) mol.
From the equation, 1 mol Na₂CO₃ reacts with 2 mol HCl, so moles of HCl \( = 2 \times 0.000500 = 0.00100 \) mol.
Concentration of HCl \( = \dfrac{0.00100}{24.00/1000} = \dfrac{0.00100}{0.02400} = 0.0417\,\text{mol}\,\text{dm}^{-3} \) (3 s.f.).

1 mark: correct \( M_r \) of Na₂CO₃ \( (106) \) and moles in stock solution \( (0.00500\,\text{mol}) \); 1 mark: correct moles of Na₂CO₃ in the \( 25.0\,\text{cm}^3 \) sample \( (0.000500\,\text{mol}) \) and correct 1:2 mole ratio applied to get moles of HCl \( (0.00100\,\text{mol}) \) (ecf); 1 mark: correct final concentration of HCl \( = 0.0417\,\text{mol}\,\text{dm}^{-3} \), accept \( 0.0416\text{-}0.0417 \) (ecf); [3]

(a) The rough titre of \( 24.60\,\text{cm}^3 \) should be excluded, since it was obtained as a preliminary run (added quickly, without careful dropwise addition near the end point) and so is not sufficiently accurate. The titre of \( 22.85\,\text{cm}^3 \) should also be excluded, since it is not concordant with \( 22.35\,\text{cm}^3 \) and \( 22.30\,\text{cm}^3 \) (which agree with each other to within \( 0.10\,\text{cm}^3 \)); \( 22.85\,\text{cm}^3 \) differs from these by roughly \( 0.5\,\text{cm}^3 \), suggesting it is an anomalous, less reliable result.
(b) Using only the two concordant, accurate titres: mean \( = \dfrac{22.35 + 22.30}{2} = \dfrac{44.65}{2} = 22.325\,\text{cm}^3 \), which rounds to \( 22.33\,\text{cm}^3 \).

(a) 1 mark: \( 24.60\,\text{cm}^3 \) correctly excluded as the rough titre; 1 mark: \( 22.85\,\text{cm}^3 \) correctly excluded as non-concordant/anomalous. (b) 1 mark: correct mean of \( 22.33\,\text{cm}^3 \) calculated from the two concordant titres only (ecf); [3]

Moles of Mg \( = \dfrac{\text{mass}}{A_r} = \dfrac{2.4}{24} = 0.10 \) mol.
From the equation, Mg reacts with H₂SO₄ in a 1:1 mole ratio to form MgSO₄ in a 1:1 mole ratio, so moles of MgSO₄ formed \( = 0.10 \) mol.
Mass of MgSO₄ \( = \text{moles} \times M_r = 0.10 \times 120 = 12.0\,\text{g} \).

1 mark: correct moles of Mg calculated \( (0.10\,\text{mol}) \) and 1:1 ratio correctly applied to give moles of MgSO₄; 1 mark: correct final mass of \( 12.0\,\text{g} \) (ecf); [2]

Relative formula mass \( = \dfrac{\text{mass}}{\text{moles}} = \dfrac{8.0}{0.20} = 40 \).

1 mark: correct relative formula mass of \( 40 \), with working shown; [1]

To prepare the standard solution: accurately weigh the required mass of anhydrous sodium carbonate on a balance. Dissolve this fully in a small volume of distilled water in a beaker, stirring with a glass rod. Transfer this solution quantitatively into a \( 250\,\text{cm}^3 \) volumetric flask using a funnel, rinsing the beaker and funnel thoroughly with distilled water to ensure no solid or solution is left behind, and adding these washings to the flask. Make the solution up to the \( 250\,\text{cm}^3 \) graduation mark with distilled water, adding the last few drops with a dropper so the bottom of the meniscus sits exactly on the mark, then stopper the flask and invert it several times to mix thoroughly.

To carry out the titration: rinse a burette with a small amount of the hydrochloric acid (to avoid dilution), then fill it above the zero mark and run it down to remove air bubbles from the tap, recording the initial (accurate) burette reading. Using a pipette and pipette filler, transfer exactly \( 25.0\,\text{cm}^3 \) of the standard sodium carbonate solution into a clean conical flask, and add a few drops of methyl orange indicator. Place the flask on a white tile under the burette, and add the hydrochloric acid from the burette to the flask, swirling continuously; as the end point approaches (indicated by a temporary colour change), add the acid dropwise, using a wash bottle to rinse any splashes down into the flask, until the indicator just changes colour from yellow to orange (the end point) and record the final burette reading.

To obtain a reliable mean titre: repeat the titration at least twice more, discarding the first (rough) titre, until two or more concordant titres (agreeing within \( 0.10\,\text{cm}^3 \) of each other) are obtained, then calculate the mean of only these concordant titres to use in further calculations.

Bullet 1 - preparing the standard solution (up to 3 marks): 1 mark: accurately weighing the solid on a balance; 1 mark: dissolving and quantitatively transferring to the volumetric flask (rinsing beaker/funnel); 1 mark: making up to the graduation mark with distilled water (using a dropper for final additions) and inverting to mix. Bullet 2 - carrying out the titration (up to 3 marks): 1 mark: rinsing/filling the burette with acid and recording the initial reading; 1 mark: pipetting \( 25.0\,\text{cm}^3 \) of standard solution into a conical flask and adding indicator; 1 mark: adding acid dropwise near the end point and correctly identifying the colour change end point, recording the final reading. Bullet 3 - obtaining a reliable mean titre (up to 2 marks): 1 mark: repeating until concordant titres (within \( 0.10\,\text{cm}^3 \)) are obtained; 1 mark: correctly excluding the rough titre and averaging only the concordant titres; [8]

(a) \( Q = mc\Delta T = 200 \times 4.18 \times (56.8 - 20.0) = 200 \times 4.18 \times 36.8 = 30\,764.8\,\text{J} \), i.e. approximately \( 30\,800\,\text{J} \) \( (30.8\,\text{kJ}) \).
(b) Moles of ethanol \( = \dfrac{\text{mass}}{M_r} = \dfrac{1.15}{46} = 0.0250 \) mol.
(c) Enthalpy change of combustion \( = -\dfrac{\text{heat released (kJ)}}{\text{moles of ethanol}} = -\dfrac{30.7648}{0.0250} = -1230.6\,\text{kJ}\,\text{mol}^{-1} \), which is \( -1230\,\text{kJ}\,\text{mol}^{-1} \) to 3 significant figures.
(d) This experimental value \( (-1230\,\text{kJ}\,\text{mol}^{-1}) \) is less exothermic than the accepted value \( (-1367\,\text{kJ}\,\text{mol}^{-1}) \) because heat is lost to the surroundings (e.g. to the air and the calorimeter apparatus itself) rather than all of it being transferred to the water, so the temperature rise recorded, and hence the calculated enthalpy change, underestimates the true heat released per mole.

(a) 1 mark: correct substitution into \( Q=mc\Delta T \); 1 mark: correct value of \( 30\,700\text{-}30\,800\,\text{J} \) \( (30.7\text{-}30.8\,\text{kJ}) \). (b) 1 mark: correct moles of ethanol \( (0.0250\,\text{mol}) \). (c) 1 mark: correct method (heat in kJ / moles) shown (ecf); 1 mark: correct final answer of \( -1230\,\text{kJ}\,\text{mol}^{-1} \) (accept \( -1220 \) to \( -1231 \)), with negative sign (ecf). (d) 1 mark: valid reason given (heat loss to surroundings/calorimeter, incomplete combustion, or evaporation of ethanol); [6]

(a) Total mass of solution \( = 50.0 + 50.0 = 100.0\,\text{g} \). \( Q = mc\Delta T = 100.0 \times 4.18 \times (26.2 - 19.5) = 100.0 \times 4.18 \times 6.7 = 2800.6\,\text{J} \), i.e. \( 2800\,\text{J} \) \( (2.80\,\text{kJ}) \).
(b) Moles of HCl \( = 1.00 \times \dfrac{50.0}{1000} = 0.0500 \) mol; moles of NaOH \( = 1.00 \times \dfrac{50.0}{1000} = 0.0500 \) mol. Since HCl and NaOH react in a 1:1 ratio and equal moles are present, all of both are neutralised, forming \( 0.0500 \) mol of water.
(c) Enthalpy change of neutralisation \( = -\dfrac{\text{heat released (kJ)}}{\text{moles of water formed}} = -\dfrac{2.8006}{0.0500} = -56.0\,\text{kJ}\,\text{mol}^{-1} \).
(d) The reaction occurring is fundamentally the same in every case: a strong acid and a strong base are essentially completely dissociated into their ions in aqueous solution, so the reaction that releases the heat is always simply H⁺(aq) + OH⁻(aq) → H₂O(l), regardless of which specific strong acid and strong base are used. Since it is this same ionic reaction that releases the enthalpy of neutralisation, the value obtained per mole of water formed is approximately the same in each case.

(a) 1 mark: correct total mass \( (100.0\,\text{g}) \) used with correct substitution into \( Q=mc\Delta T \); 1 mark: correct value of \( 2800\,\text{J} \) \( (2.80\,\text{kJ}) \). (b) 1 mark: correct moles of water formed \( (0.0500\,\text{mol}) \). (c) 1 mark: correct method (heat in kJ / moles of water) shown (ecf); 1 mark: correct final answer of \( -56.0\,\text{kJ}\,\text{mol}^{-1} \) (ecf), with negative sign. (d) 1 mark: correctly explains that the same ionic reaction, H⁺(aq) + OH⁻(aq) → H₂O(l), occurs regardless of the specific strong acid/base pair; [6]

(a) Hess's law states that the total enthalpy change for a chemical reaction is the same regardless of the route taken from reactants to products, provided the starting and finishing conditions are the same.
(b) In the enthalpy cycle, the direct route is C(s) + 2H₂(g) → CH₄(g) (enthalpy change \( \Delta H \), to be found). The indirect route goes via combustion: C(s) + 2H₂(g) + 3O₂(g) → [combustion products: CO₂(g) + 2H₂O(l)] → CH₄(g) + 2O₂(g), where the first arrow has enthalpy change \( [\Delta H_c(\text{C}) + 2\Delta H_c(\text{H}_2)] \) and the second arrow (reverse combustion of methane) has enthalpy change \( -\Delta H_c(\text{CH}_4) \).
By Hess's law:
\( \Delta H = [\Delta H_c(\text{C}) + 2 \times \Delta H_c(\text{H}_2)] - \Delta H_c(\text{CH}_4) \)
\( = [(-394) + 2 \times (-286)] - (-890) \)
\( = [-394 - 572] - (-890) \)
\( = -966 + 890 \)
\( = -76\,\text{kJ}\,\text{mol}^{-1} \).
(c) This calculated value \( (-76\,\text{kJ}\,\text{mol}^{-1}) \) is close to, but may differ slightly from, the accepted data book value (approximately \( -75\,\text{kJ}\,\text{mol}^{-1} \)) because the enthalpies of combustion used in the calculation are themselves experimentally measured (and rounded) quantities, each carrying a small degree of experimental uncertainty, which propagates into the final calculated answer.

(a) 1 mark: Hess's law correctly stated (total enthalpy change independent of route, same start/end conditions). (b) 1 mark: correct enthalpy cycle/method shown \( (\Delta H = \) sum of \( \Delta H_c \) of reactants \( - \Delta H_c \) of product\( ) \); 1 mark: correct substitution of values; 1 mark: correct intermediate value of \( -966\,\text{kJ}\,\text{mol}^{-1} \) (or equivalent) shown; 1 mark: correct final answer of \( -76\,\text{kJ}\,\text{mol}^{-1} \) (ecf) with correct sign. (c) 1 mark: valid reason given (combustion data are themselves experimentally determined/rounded values carrying uncertainty); [6]

(a) Average (mean) bond enthalpy is the average amount of energy required to break one mole of a particular covalent bond, taken (averaged) across a range of different gaseous compounds containing that bond.
(b) Energy is absorbed to break bonds in the reactants, and released when bonds are formed in the products.
Bonds broken (reactants): 1 × H-H + 1 × Cl-Cl \( = 436 + 243 = 679\,\text{kJ}\,\text{mol}^{-1} \).
Bonds formed (products): 2 × H-Cl \( = 2 \times 432 = 864\,\text{kJ}\,\text{mol}^{-1} \).
Enthalpy change \( = \) energy to break bonds \( - \) energy released forming bonds \( = 679 - 864 = -185\,\text{kJ}\,\text{mol}^{-1} \).
(c) Bond enthalpy calculations use mean bond enthalpies, which are average values calculated across many different compounds containing that type of bond, rather than the exact bond energy for that specific bond in that specific molecule; the true bond energies in H₂, Cl₂ and HCl may differ slightly from these average values, so the calculated enthalpy change is an approximation and may differ from the experimentally measured value.

(a) 1 mark: average bond enthalpy correctly defined (average energy to break one mole of a given bond, averaged over many compounds, gaseous state). (b) 1 mark: correct bonds broken identified and summed \( (679\,\text{kJ}\,\text{mol}^{-1}) \); 1 mark: correct bonds formed identified and summed \( (864\,\text{kJ}\,\text{mol}^{-1}) \); 1 mark: correct method (bonds broken - bonds formed) applied (ecf); 1 mark: correct final answer of \( -185\,\text{kJ}\,\text{mol}^{-1} \) (ecf) with correct sign. (c) 1 mark: valid reason given (mean/average bond enthalpies are not exact for the specific molecules involved); [6]

When choosing a site for a new large-scale continuous chemical plant, a company must weigh up several economic and environmental factors. Economically, the site should ideally be close to sources of raw materials and to good transport links (road, rail or port), to minimise the cost of transporting materials and products. Access to a reliable energy supply and to a suitable workforce are also important, as is the cost of the land itself, which is often lower away from city centres. Environmentally, the company must consider how waste products and emissions from the plant will be managed and disposed of safely, in line with environmental regulations (such as those covering waste management set out by the relevant environment agency); this includes assessing the risk to nearby watercourses, air quality and local communities, and may require costly abatement or treatment equipment to reduce pollution.

Using a solid (heterogeneous) catalyst in the process brings clear economic benefit. A catalyst increases the rate of the reaction without being permanently used up, allowing the desired yield to be achieved more quickly, or allowing the reaction to be run effectively at a lower temperature and/or pressure than would otherwise be needed; since heating and pressurising large industrial reactors is very energy-intensive and costly, this significantly reduces energy costs, improving the overall economics of the process. A further practical benefit of a solid catalyst used with gaseous or liquid reactants is that it is easily separated from the product stream (since it exists in a different phase), allowing it to be recovered and reused repeatedly, further reducing costs. However, this benefit must be weighed against the risk and cost of catalyst poisoning, where impurities in the feedstock bind to and block the catalyst's active sites, gradually reducing its effectiveness and requiring periodic replacement or regeneration.

Level 3 (5-6 marks): Thorough, balanced evaluation of BOTH site-selection factors (economic AND environmental, with specific detail) AND the economic benefit/limitations of a solid catalyst, using accurate technical terminology and fluent, well-organised prose with accurate spelling/grammar. Level 2 (3-4 marks): Sound coverage of most points but limited in balance or depth (e.g. site factors covered well but catalyst discussion underdeveloped, or vice versa; or economic factors covered but environmental factors weak); generally accurate spelling/grammar. Level 1 (1-2 marks): Basic, general statements with limited technical detail or balance across the two parts of the question; weak use of specialist terminology; basic spelling/grammar. [6]

Increasing the temperature of a reaction mixture increases the rate of reaction. This can be explained using the Maxwell-Boltzmann distribution: increasing temperature increases the average kinetic energy of the particles, which shifts the distribution curve to the right and flattens it, increasing the proportion of particles possessing kinetic energy equal to or greater than the activation energy required for a reaction to occur. According to collision theory, particles must collide with energy at least equal to the activation energy, and with the correct orientation, for a reaction to take place; at higher temperature, both the proportion of particles with sufficient energy and the overall frequency of collisions between particles increase, so the frequency of successful collisions increases, increasing the rate of reaction.

1 mark: rate of reaction correctly stated to increase with temperature; 1 mark: correctly describes the Maxwell-Boltzmann distribution shifting/increasing the proportion of particles with energy ≥ activation energy at higher temperature; 1 mark: correctly links this (and/or increased collision frequency) to an increased frequency of successful collisions, increasing rate; [3]

(a) Catalyst poisoning occurs when impurity molecules present in the reaction mixture bind strongly (often irreversibly) to the active sites on the surface of a solid catalyst, permanently blocking those sites from further use.
(b) A solid catalyst works by chemisorption: reactant molecules form weak chemical bonds with atoms at specific active sites on the catalyst's surface, which weakens the bonds within the reactant molecules and lowers the activation energy needed for the reaction to proceed. When a catalyst becomes poisoned, impurity molecules instead chemisorb onto these active sites, occupying them and preventing reactant molecules from adsorbing there. As the number of available active sites falls, fewer reactant molecules can be catalysed at any one time, so the overall rate of the catalysed reaction decreases.

(a) 1 mark: catalyst poisoning correctly described as impurities binding to/blocking active sites on the catalyst surface. (b) 1 mark: correctly describes chemisorption of reactants at active sites lowering activation energy; 1 mark: correctly explains that poison molecules occupy/block active sites, reducing sites available to reactants and so reducing rate; [3]

Two further factors that affect the rate of a chemical reaction are the concentration of reactants in solution (a higher concentration increases collision frequency and so increases rate) and the pressure of gaseous reactants (a higher pressure increases the frequency of collisions between gas particles and so increases rate).

1 mark: concentration (of reactants) correctly stated; 1 mark: pressure (of gaseous reactants) correctly stated; [2]

(a) A reversible reaction is one in which the products formed can react together to re-form the original reactants, so the reaction is able to proceed in both the forward direction and the reverse direction.
(b) Dynamic equilibrium is the state reached in a closed system (at constant temperature) in which the rate of the forward reaction is exactly equal to the rate of the reverse reaction.
(c) At dynamic equilibrium, because the forward and reverse reactions are occurring at exactly the same rate, reactants are being converted into products at precisely the same rate as products are being converted back into reactants. The two opposing changes therefore cancel out, so although both reactions are continuously occurring, there is no overall (net) change in the concentrations of reactants and products over time.

(a) 1 mark: reversible reaction correctly defined (products can re-form reactants; reaction proceeds both ways). (b) 1 mark: dynamic equilibrium correctly defined (rate of forward reaction = rate of reverse reaction in a closed system). (c) 1 mark: correctly explains that equal forward/reverse rates mean no net change in concentration despite ongoing reaction; [3]

(a) The reactant side of the equation has 1 + 3 = 4 moles of gas, while the product side has only 2 moles of gas. According to Le Chatelier's principle, increasing the pressure on a gaseous equilibrium shifts the position of equilibrium towards the side with the fewer number of gas molecules, in order to partially oppose (reduce) the increase in pressure. Here, that is the product side (ammonia), so increasing pressure shifts the equilibrium to the right, increasing the yield of ammonia.
(b) The forward reaction (N2 + 3H2 → 2NH3) is exothermic \( (\Delta H = -92\,\text{kJ}\,\text{mol}^{-1}) \), so the reverse reaction is endothermic. By Le Chatelier's principle, increasing the temperature shifts the position of equilibrium in the endothermic direction, to partially oppose the increase in temperature — here, that is the reverse (leftward) direction, so increasing temperature shifts equilibrium away from ammonia, decreasing its yield.

(a) 1 mark: correctly identifies fewer moles of gas on the product (ammonia) side; 1 mark: correctly concludes equilibrium shifts right/yield of ammonia increases with increased pressure, with reasoning (Le Chatelier's principle/opposing the change). (b) 1 mark: correctly concludes equilibrium shifts left/yield of ammonia decreases with increased temperature, correctly linked to the exothermic forward reaction; [3]

(a) A: Aluminium is a metal. B: Reinforced concrete is a composite (steel reinforcing bars embedded in a concrete/ceramic matrix). C: Porcelain is a ceramic. D: Polyethylene is a polymer. E: Window glass is a glass.
(b) For example: aluminium (metal) is malleable and a good electrical/thermal conductor, due to its metallic bonding and sea of delocalised electrons; porcelain (ceramic) is hard, brittle and heat-resistant, due to its strong ionic/covalent bonding in a rigid lattice; polyethylene (polymer) is flexible and of low density, due to its long-chain molecular structure held together by weak intermolecular forces.
(c) In a composite material, two or more distinct materials are combined physically, without dissolving into or chemically bonding with each other to form a new single substance. Each constituent therefore keeps its own individual properties within the composite structure. In reinforced concrete, the steel reinforcing bars provide high tensile strength (resisting stretching forces), while the surrounding concrete provides high compressive strength (resisting crushing forces); combined, the composite has both properties, which neither steel alone nor concrete alone would provide as effectively for the same application.

(a) 1 mark per two correct categorisations, up to 3 marks for all five correct (metal/composite/ceramic/polymer/glass). (b) 1 mark each for any two valid material-property justifications linked to the correct category (up to 2 marks). (c) 1 mark: correctly explains that constituents remain physically distinct/retain individual properties within the composite structure; [6]

A: Shape-memory alloy - (v) returns to a pre-set shape when heated above its transition temperature (e.g. Nitinol used in stents/orthodontic wires).
B: Piezoelectric material - (iii) generates a voltage when mechanically stressed/deformed (and, conversely, deforms when a voltage is applied).
C: Quantum-tunnelling composite - (ii) is normally an electrical insulator, but becomes conductive when compressed, due to electrons tunnelling between conductive particles forced closer together.
D: Thermochromatic material - (vi) changes colour in response to a change in temperature.
E: Photochromatic material - (iv) changes colour (typically darkens) in response to a change in light intensity, such as exposure to UV light.
F: Electroluminescent material - (i) emits light when an electric voltage/current is applied across it.

1 mark for each correct match (A-(v), B-(iii), C-(ii), D-(vi), E-(iv), F-(i)); [6]

(a) A simple Bohr model diagram must correctly show the nucleus (containing the protons and neutrons) at the centre, with the electrons arranged in shells (energy levels) at set distances around it.
(b) In the structure of a metal, atoms are arranged in a regular, closely-packed lattice. Each metal atom loses its outer-shell (valence) electrons, which become delocalised — free to move throughout the whole structure rather than staying attached to a particular atom. This leaves behind a lattice of positively charged metal ions, held together by their electrostatic attraction to the surrounding 'sea' of delocalised electrons (metallic bonding). Because these delocalised electrons are not bound to individual atoms, they are free to drift through the metal's structure when a voltage is applied, carrying electric current — this mobility of delocalised electrons is what gives metals their high electrical conductivity.

(a) 1 mark: nucleus (protons/neutrons) and electrons in shells both correctly stated. (b) 1 mark: correctly describes the regular lattice of positive metal ions; 1 mark: correctly describes delocalised/free electrons ('sea of electrons') surrounding the ions; 1 mark: correctly links the mobility of delocalised electrons to high electrical conductivity; [4]

In a crystalline material, the particles (atoms, ions or molecules) are arranged in a highly regular, repeating, ordered pattern that extends throughout the structure in three dimensions — quartz and metals such as iron are examples of crystalline materials. In an amorphous material, by contrast, the particles are arranged randomly, with no long-range repeating order, although there may be some short-range order — glass is a typical example of an amorphous material.

This structural difference affects physical properties: a crystalline material typically has a sharp, well-defined melting point, because the regular lattice structure requires a specific amount of thermal energy to break down uniformly at a particular temperature. An amorphous material, lacking this regular structure, instead softens gradually over a range of temperatures as it is heated, rather than melting sharply at a single temperature.

1 mark: crystalline structure correctly described as regular/repeating/ordered, with a valid named example; 1 mark: amorphous structure correctly described as random/disordered, with a valid named example (e.g. glass); 1 mark: valid physical property difference identified (sharp melting point vs gradual softening); 1 mark: correctly explains why this difference arises from the structural difference; [4]

In a thermoplastic polymer, individual polymer chains are held together only by relatively weak intermolecular forces (such as van der Waals forces), with no covalent bonds linking one chain to another. On heating, these weak forces are overcome, allowing the chains to slide past one another, so the material softens and can be melted and reshaped (and later re-melted/recycled).

In a thermosetting polymer, the polymer chains are joined to one another by strong covalent cross-links, forming a single, rigid three-dimensional network structure. Because these are strong covalent bonds rather than weak intermolecular forces, the chains cannot slide past one another when heated. As a result, a thermosetting polymer does not soften or melt on heating; instead, at high enough temperatures it chars or decomposes, and it cannot be reshaped or recycled by re-melting.

1 mark: thermoplastic structure correctly described (chains held by weak intermolecular forces only, no cross-links); 1 mark: thermosetting structure correctly described (chains joined by strong covalent cross-links, rigid network); 1 mark: correct behaviour on heating stated for both (thermoplastic softens/melts and can be reshaped; thermosetting chars/decomposes and cannot be reshaped); [3]

In glass-fibre-reinforced plastic, thin, strong, stiff glass fibres are embedded within a surrounding polymer resin matrix. When the composite is placed under mechanical (e.g. tensile) stress, the stiff glass fibres bear the majority of the load, since they are much stronger and stiffer than the surrounding resin. The polymer matrix serves to bind the fibres together, transferring load between adjacent fibres so the material acts as a single, cohesive structure, and also protects the fibres from surface damage and moisture. As a result, GRP combines the high tensile strength and stiffness contributed by the glass fibres with the toughness, lower brittleness and mouldability contributed by the polymer matrix — giving an overall composite material with better mechanical performance for many applications than either the glass fibres or the polymer resin would have on their own.

1 mark: correctly identifies that glass fibres provide strength/stiffness and bear the mechanical load; 1 mark: correctly describes the role of the resin matrix (binding fibres, transferring load, protecting fibres); 1 mark: correctly explains that the composite combines properties of both constituents (strength from fibres, toughness/mouldability from matrix); [3]

Graphene consists of a single layer of carbon atoms, just one atom thick, arranged in a two-dimensional hexagonal (honeycomb) lattice. Each carbon atom is covalently bonded to three neighbouring carbon atoms within the plane of the sheet.

This structure gives graphene a very high tensile strength, since the strong covalent bonds extend throughout the entire lattice in the plane of the sheet, making it extremely difficult to break. Graphene also has a very high electrical conductivity, because each carbon atom's fourth outer electron is delocalised across the sheet, allowing charge to move freely through the structure.

1 mark: correctly describes graphene as a single (one-atom-thick) layer of carbon atoms; 1 mark: correctly describes the hexagonal/honeycomb lattice arrangement with covalent bonding between carbon atoms; 1 mark: valid physical property correctly linked to this structure (high tensile strength from covalent bonding, or high electrical conductivity from delocalised electrons); [3]

A carbon nanotube is essentially a cylindrical, tube-shaped structure formed by rolling a single sheet of graphene — a two-dimensional hexagonal lattice of covalently bonded carbon atoms — into a seamless tube, typically only a few nanometres in diameter.

This structure gives carbon nanotubes potential healthcare applications, including as vehicles for targeted drug delivery (exploiting their hollow interior and high loading capacity for drug molecules), as biosensors (for example in glucose detection biosensors), as scaffolding to support tissue regeneration, and in the selective destruction of cancer cells.

1 mark: correctly describes a carbon nanotube as a cylindrical/tube structure formed by rolling a graphene sheet; 1 mark: correctly describes the underlying hexagonal carbon lattice/covalent bonding; 1 mark: valid healthcare application named (drug delivery/biosensors/tissue scaffolding/cancer cell destruction); [3]

Silicon atoms have four electrons in their outer (valence) shell. This allows each silicon atom to form four covalent bonds with four neighbouring silicon atoms, creating a stable, regular crystal lattice structure in which (at low temperature) essentially all outer-shell electrons are held within these covalent bonds. Because there are very few free (mobile) electrons available to carry charge, pure silicon conducts electricity poorly at low temperature — much more poorly than a metal.

However, as temperature increases, a small number of electrons gain enough thermal energy to break free from their covalent bonds and become mobile charge carriers (leaving behind a positive 'hole' that can also carry charge). This gives silicon an electrical conductivity that lies between that of a good conductor (such as a metal) and a good insulator. Crucially, this conductivity can also be deliberately and precisely increased by doping the silicon with other elements, which is what makes silicon such a useful and controllable semiconductor material for electronic devices.

1 mark: correctly states silicon has four outer-shell electrons, forming four covalent bonds in a lattice structure; 1 mark: correctly explains that few free charge carriers exist at low temperature, giving conductivity between conductor and insulator; 1 mark: correctly explains that conductivity increases with temperature/doping (electrons freed from bonds become mobile charge carriers); [3]

(a) In n-type doping, silicon is doped with atoms possessing five outer-shell electrons (such as phosphorus). Four of these electrons form covalent bonds with neighbouring silicon atoms, leaving one extra, loosely-held electron per dopant atom free to act as a mobile negative charge carrier. In p-type doping, silicon is doped with atoms possessing only three outer-shell electrons (such as boron). These can only form three of the four covalent bonds expected in the silicon lattice, leaving an electron vacancy, or 'hole', which behaves as a mobile positive charge carrier as neighbouring electrons move to fill it.
(b) At a p-n junction, when the p-type side is connected to the positive terminal of a supply and the n-type side to the negative terminal (forward bias), the free electrons in the n-type region and the holes in the p-type region are both pushed towards the junction; they can cross the junction and recombine, allowing a current to flow through the diode. When the connections are reversed (reverse bias), the free electrons and holes are instead pulled away from the junction, widening the depletion region at the junction and preventing charge carriers from crossing it, so no (significant) current flows.

(a) 1 mark: n-type correctly described (five-outer-electron dopant, e.g. phosphorus, gives free electron/negative carrier); 1 mark: p-type correctly described (three-outer-electron dopant, e.g. boron, gives hole/positive carrier). (b) 1 mark: correctly explains forward bias allows carriers to cross/recombine at the junction (current flows), while reverse bias widens the depletion region and prevents current flow; [3]

Strength-to-weight ratio can be estimated by dividing tensile strength by density: mild steel = 400/7.8 ≈ 51; aluminium alloy = 310/2.7 ≈ 115; titanium alloy = 900/4.5 = 200. Titanium alloy has by far the highest value of these three, meaning it provides the greatest tensile strength for the least mass, making it the material that would give the bicycle frame the best strength-to-weight ratio.

1 mark: titanium alloy correctly identified as giving the best strength-to-weight ratio; 1 mark: valid justification with reference to the data (e.g. comparing tensile strength relative to density, or calculating/comparing an approximate ratio for at least two materials); [2]

The data show that alloying copper with another metal increases its hardness: pure copper has a hardness of only 40 HV, whereas bronze (copper alloyed with 10% tin) has a hardness of 70 HV, and brass (copper alloyed with 35% zinc) has a hardness of 80 HV — both considerably harder than pure copper, with brass (containing the larger proportion of a different metal) the hardest of the three. This occurs because the atoms of the alloying metal (zinc or tin) are a different size from copper atoms, so they distort the regular arrangement of atoms in the copper lattice. This distortion makes it more difficult for layers of atoms to slide over one another when a force is applied, increasing the hardness of the alloy compared with the pure metal.

1 mark: correctly identifies from the data that alloying increases hardness (with values quoted, e.g. 40 HV pure copper vs 70-80 HV for the alloys); 1 mark: correct explanation given (differently-sized alloying atoms distort the lattice, hindering layers of atoms sliding over each other); [2]

Silicon is the semiconductor in the table. Its electrical conductivity (approximately \( 1.6 \times 10^{-3}\,\text{S/m} \)) is many orders of magnitude lower than that of copper, a metal and good electrical conductor (approximately \( 6 \times 10^7\,\text{S/m} \)), but many orders of magnitude higher than that of glass and rubber, both good electrical insulators (with conductivities in the range of roughly \( 10^{-11} \) to \( 10^{-15}\,\text{S/m} \)). This intermediate level of conductivity, between that of a typical conductor and a typical insulator, is the defining characteristic of a semiconductor material.

1 mark: silicon correctly identified as the semiconductor; 1 mark: valid justification referencing the data (conductivity intermediate between copper and glass/rubber, by many orders of magnitude); [2]

Considering mechanical performance first, stainless steel has the highest tensile strength (600 MPa), followed by mild steel (400 MPa) and then bronze (350 MPa), so stainless steel would best withstand mechanical stress in this application.

Considering durability in a marine (seawater) environment, corrosion resistance is critical: mild steel has poor corrosion resistance and would be expected to corrode relatively quickly in seawater, potentially leading to early failure of the component and requiring frequent replacement; bronze has good corrosion resistance; stainless steel has excellent corrosion resistance, due to the formation of a passive, self-repairing chromium oxide layer on its surface, making it highly durable in a marine environment.

Considering cost, mild steel is the cheapest option (£1.0 per kg), stainless steel is intermediate (£4.5 per kg), and bronze is the most expensive (£6.0 per kg). Although mild steel is cheapest, its poor corrosion resistance makes it a poor choice overall for this application, since the component would likely need frequent, costly replacement, offsetting its low initial material cost. Bronze offers good corrosion resistance and reasonable strength, but at the highest cost of the three, and with lower tensile strength than stainless steel.

Overall, stainless steel appears to be the most suitable material for this marine engine component: it combines the highest tensile strength with excellent corrosion resistance, at a cost that, while higher than mild steel, is lower than bronze — offering the best overall balance of mechanical performance, durability and cost for a demanding, corrosive marine environment.

Level 3 (5-6 marks): Thorough, balanced evaluation considering ALL THREE factors (tensile strength, corrosion resistance, cost) with specific values/data quoted for each material, reaching a well-justified overall conclusion; accurate technical terminology and fluent, well-organised prose with accurate spelling/grammar. Level 2 (3-4 marks): Sound evaluation covering most factors and materials with some data referenced, reaching a conclusion, but with limited depth or balance in places (e.g. cost not fully integrated into the conclusion); generally accurate spelling/grammar. Level 1 (1-2 marks): Basic, general comparison of the materials with limited reference to specific data, and/or no clear justified conclusion; weak use of terminology; basic spelling/grammar. [6]

(a) \( \text{Stress} = \dfrac{\text{force}}{\text{cross-sectional area}} = \dfrac{90}{1.5 \times 10^{-6}} = 6.0 \times 10^7\,\text{Pa} \).
(b) \( \text{Strain} = \dfrac{\text{change in length}}{\text{original length}} = \dfrac{1.2 \times 10^{-3}}{2.000} = 6.0 \times 10^{-4} \) (strain has no units).
(c) \( \text{Young modulus} = \dfrac{\text{stress}}{\text{strain}} = \dfrac{6.0 \times 10^7}{6.0 \times 10^{-4}} = 1.0 \times 10^{11}\,\text{Pa} \).
(d) The Young modulus has the same units as stress, since strain is dimensionless: the pascal, Pa (equivalent to \( \text{N}\,\text{m}^{-2} \)).

(a) 1 mark: correct stress of \( 6.0 \times 10^7\,\text{Pa} \). (b) 1 mark: correct strain of \( 6.0 \times 10^{-4} \). (c) 1 mark: correct method (stress/strain) shown (ecf); 1 mark: correct final answer of \( 1.0 \times 10^{11}\,\text{Pa} \) (ecf). (d) 1 mark: pascal (Pa)/\( \text{N}\,\text{m}^{-2} \) correctly stated; [5]

(a) \( \text{Stress} = \dfrac{\text{force}}{\text{area}} = \dfrac{40}{2.0 \times 10^{-5}} = 2.0 \times 10^6\,\text{Pa} \). \( \text{Strain} = \dfrac{\text{extension}}{\text{original length}} = \dfrac{5.0 \times 10^{-3}}{0.500} = 0.010 \) (i.e. 1.0%).
(b) This kind of permanent (non-reversible) deformation is called plastic deformation (the material shows plasticity).
(c) A typical stress-strain graph for such a material starts as a straight line through the origin, showing stress directly proportional to strain (the elastic region, obeying Hooke's law), up to the elastic limit (limit of proportionality). Beyond this point, the graph becomes non-linear (curving), entering the plastic region; within the plastic region, if the applied force were removed, the material would not return to its original length, retaining a permanent extension — as observed for the sample described in this question, which must therefore correspond to a point on the graph beyond the elastic limit, within the plastic region.

(a) 1 mark: correct stress of \( 2.0 \times 10^6\,\text{Pa} \); 1 mark: correct strain of \( 0.010 \) (1.0%). (b) 1 mark: plastic deformation/plasticity correctly named. (c) 1 mark: correctly describes the initial linear/elastic region up to the elastic limit; 1 mark: correctly describes the non-linear plastic region beyond the elastic limit and correctly places the described permanent deformation within it; [5]

(a) Young modulus \( = \dfrac{\text{stress}}{\text{strain}} \).
Wire X: \( E = \dfrac{2.5 \times 10^8}{0.0020} = 1.25 \times 10^{11}\,\text{Pa} \).
Wire Y: \( E = \dfrac{1.8 \times 10^8}{0.0020} = 9.0 \times 10^{10}\,\text{Pa} \).
(b) Wire X is stiffer than wire Y, because it has the higher Young modulus \( (1.25 \times 10^{11}\,\text{Pa} \), compared with \( 9.0 \times 10^{10}\,\text{Pa} \) for wire Y) — a higher Young modulus means a material is more resistant to elastic deformation, producing a smaller strain for a given applied stress.
(c) Rearranging the Young modulus equation: \( \text{strain} = \dfrac{\text{stress}}{\text{Young modulus}} = \dfrac{4.0 \times 10^8}{2.0 \times 10^{11}} = 2.0 \times 10^{-3} \) (i.e. a strain of \( 0.0020 \)).

(a) 1 mark: correct \( E(\text{X}) = 1.25 \times 10^{11}\,\text{Pa} \); 1 mark: correct \( E(\text{Y}) = 9.0 \times 10^{10}\,\text{Pa} \). (b) 1 mark: wire X correctly identified as stiffer, with valid reason (higher Young modulus/less strain for given stress) (ecf). (c) 1 mark: correct rearrangement (strain = stress/E) shown; 1 mark: correct final answer of \( 2.0 \times 10^{-3} \) \( (0.0020) \); [5]

(a) In the linear (elastic) region, Young modulus \( = \dfrac{\text{stress}}{\text{strain}} \), and this ratio is constant. Using the data point at \( 200\,\text{MPa} \) and \( 0.00100 \) strain: \( E = \dfrac{200 \times 10^6}{0.00100} = 2.0 \times 10^{11}\,\text{Pa} \). (This is confirmed using any other point in the linear region, e.g. \( 100\,\text{MPa}/0.00050 \): \( E = \dfrac{100 \times 10^6}{0.00050} = 2.0 \times 10^{11}\,\text{Pa} \) — the same value, confirming the region is linear.)
(b) If the material remained perfectly elastic (obeying Hooke's law) beyond \( 200\,\text{MPa} \), a stress of \( 220\,\text{MPa} \) would be expected to produce a strain of \( \dfrac{220 \times 10^6}{2.0 \times 10^{11}} = 0.0011 \). However, the table shows the actual strain at \( 220\,\text{MPa} \) is \( 0.00150 \), considerably higher than this expected value — stress and strain are no longer directly proportional. This shows that the elastic limit has been exceeded somewhere between \( 200\,\text{MPa} \) and \( 220\,\text{MPa} \).
(c)(i) Tensile strength is the maximum stress that a material can withstand before it begins to fail (before necking/fracture); from the table, this corresponds to the maximum stress value recorded, \( 235\,\text{MPa} \).
(ii) The point of fracture is the point at which the material finally breaks/separates into two pieces; from the table, this corresponds to the final data point, at a (nominal) stress of \( 190\,\text{MPa} \) and a strain of \( 0.00650 \). The nominal stress at fracture is lower than the tensile strength because, beyond the maximum stress, the sample undergoes 'necking' (local narrowing of its cross-section), so the actual force needed to continue stretching it falls even though the true stress in the narrowed region continues to rise.

(a) 1 mark: correct method shown (stress/strain using a data point from the linear region); 1 mark: correct final answer of \( 2.0 \times 10^{11}\,\text{Pa} \). (b) 1 mark: correctly identifies the elastic limit is exceeded between \( 200\text{-}220\,\text{MPa} \), with valid reasoning from the data (e.g. comparing actual vs expected strain at \( 220\,\text{MPa} \), or noting strain increasing disproportionately). (c) 1 mark: tensile strength correctly defined AND correct value \( (235\,\text{MPa}) \) given; 1 mark: point of fracture correctly defined AND correct value(s) given \( (190\,\text{MPa}/0.00650 \) strain\( ) \); [5]

(a) Young modulus of bronze \( = \dfrac{\text{stress}}{\text{strain}} = \dfrac{1.0 \times 10^8}{0.00125} = 8.0 \times 10^{10}\,\text{Pa} \).
(b) Comparing the two values, steel \( (2.0 \times 10^{11}\,\text{Pa}) \) has a considerably higher Young modulus than bronze \( (8.0 \times 10^{10}\,\text{Pa}) \), so steel is the stiffer material — it deforms elastically less than bronze for the same applied stress.
(c) Although bronze is less stiff than steel, bronze has excellent resistance to corrosion in seawater, since it does not rust in the way that steel does when exposed to salt water and oxygen; steel is much more prone to corrosion in a marine environment unless specially protected or alloyed (as in stainless steel). Bronze is also known for good wear resistance and low friction. For a marine application where long-term durability and resistance to a corrosive environment are critical, these properties can outweigh the benefit of steel's greater stiffness, making bronze the more suitable choice despite being less stiff.

(a) 1 mark: correct method (stress/strain) shown; 1 mark: correct final answer of \( 8.0 \times 10^{10}\,\text{Pa} \). (b) 1 mark: steel correctly identified as stiffer, with comparison of the two Young modulus values (ecf). (c) 1 mark: valid material property advantage of bronze given (corrosion resistance in seawater/wear resistance/low friction); 1 mark: correctly explains why this property may outweigh lower stiffness for a marine application; [5]