2023 CCEA AS-Level Life and Health Sciences 0008 Practice Paper with Answers

The typical resting pulse rate for a healthy adult falls within the range of 60 to 80 beats per minute.

1 mark: 60 (bpm); 1 mark: 80 (bpm) — allow the range stated as 60-80 bpm for both marks.

Blood pressure is measured using a sphygmomanometer. Two values are recorded: the systolic pressure, the pressure in the arteries when the heart contracts (ventricles contract), and the diastolic pressure, the pressure in the arteries when the heart relaxes between beats.

1 mark: sphygmomanometer; 1 mark: systolic and diastolic pressure (both needed).

Air travels from the nose/mouth down through the trachea, which branches into the two bronchi (one to each lung), which in turn branch repeatedly into progressively narrower bronchioles, finally leading to the alveoli where gas exchange occurs.

1 mark each for trachea, bronchi and bronchioles named in the correct order. Max 3 marks.

A spirometer is used to measure lung volumes, including tidal volume (the volume of air breathed in or out in a normal breath) and vital capacity (the maximum volume of air that can be forcibly exhaled after a maximum inhalation). A peak flow meter is used to measure peak expiratory flow rate, the maximum speed at which a person can exhale.

1 mark: spirometer measures lung volume(s), e.g. tidal volume/vital capacity; 1 mark: peak flow meter measures peak expiratory flow rate.

ATP (adenosine triphosphate) is the molecule that directly supplies energy for most cellular processes. The great majority of a cell's ATP is produced by aerobic respiration.

1 mark: ATP (adenosine triphosphate); 1 mark: aerobic respiration.

Aerobic respiration occurs in three main stages, in sequence: glycolysis (in the cytoplasm), followed by the Krebs cycle (in the mitochondrial matrix), followed by the electron transport chain (on the inner mitochondrial membrane).

1 mark each for glycolysis, the Krebs cycle, and the electron transport chain, named in the correct order. Max 3 marks.

Blood glucose concentration is regulated by two hormones produced by the pancreas: insulin, which lowers blood glucose concentration, and glucagon, which raises it. Acting together (antagonistically), they keep blood glucose concentration within a narrow, stable range around its normal set point.

1 mark: insulin and glucagon (both needed); 1 mark: combined effect is to maintain blood glucose within a stable/narrow range (homeostasis).

Normal blood pH is maintained within the range 7.35 to 7.45. This, like many other physiological variables, is kept within its normal range by negative feedback: a mechanism in which any change away from the normal value triggers a response that counteracts (reverses) that change, returning the variable back towards normal.

1 mark: 7.35; 1 mark: 7.45 (allow range 7.35-7.45 for both marks); 1 mark: negative feedback.

A balanced diet should include an appropriate proportion of several food groups, including carbohydrates, protein, fats, vitamins, minerals, fibre and water. Each has a distinct function: for example, carbohydrates are the body's main energy source, protein is needed for the growth and repair of tissues, and fats provide a concentrated energy store and thermal insulation.

1 mark each for any three valid food groups named (carbohydrate, protein, fat, vitamins, minerals, fibre, water); 1 mark: correct function given for one named food group. Max 3 marks (at least one mark must be for a correct function).

As part of assessing nutritional health, blood levels of several vitamins are monitored, including vitamin B, vitamin C, vitamin D and vitamin E, since deficiencies in these vitamins can have significant effects on health.

1 mark each for any three of vitamin B, C, D, E named. Max 3 marks.

Normal blood cholesterol concentration is generally considered to lie within the range 4.0 to 6.5 mmol/dm³.

1 mark: 4.0 mmol/dm³; 1 mark: 6.5 mmol/dm³ (allow range 4.0-6.5 mmol/dm³ for both marks).

(a) Tachycardia is an abnormally fast, but still regular, heart rate. Patient Q has a heart rate of 145 bpm, well above the normal resting range of 60-80 bpm, but their rhythm remains regular, so their trace is most likely to show tachycardia. (b) Ventricular fibrillation is a dangerous arrhythmia characterised by a chaotic, irregular electrical rhythm in the ventricles, rather than simply a fast rate. Patient S has an irregular, chaotic rhythm (rather than merely an elevated but regular rate), which matches the characteristic pattern of ventricular fibrillation.

(a) 1 mark: Patient Q identified; 1 mark: correct reasoning (rate far above normal range but rhythm regular). (b) 1 mark: Patient S identified; 1 mark: correct reasoning (irregular, chaotic rhythm distinguishes fibrillation from a simply fast, regular tachycardia).

Patients W (93%) and Z (91%) both have SaO2 readings within the normal range of 90-95%, whereas Patients X (84%) and Y (82%) both have readings below this normal range, showing reduced SaO2. Emphysema causes progressive damage to and breakdown of the walls of the alveoli, reducing the total surface area available for gas exchange, so less oxygen can diffuse into the blood in a given time, lowering SaO2. Pneumonia causes inflammation and a build-up of fluid within the alveoli, which increases the diffusion distance for oxygen and reduces the area of alveoli available for efficient gas exchange, similarly lowering SaO2.

1 mark: X and Y identified as below the normal range (reduced SaO2), using data; 1 mark: W and Z correctly identified as within the normal range; 1 mark: emphysema explanation (damaged alveoli reduce surface area for gas exchange); 1 mark: pneumonia explanation (fluid/inflammation in alveoli impairs gas exchange).

The data show that blood lactate concentration rises only gradually between 40% and 60% of maximum exercise intensity (1.5 to 2.0 mmol/dm³), but then rises much more sharply between 60% and 100% intensity (2.0 to 9.0 mmol/dm³). At low-to-moderate exercise intensities, the cardiovascular and respiratory systems can supply the exercising muscles with enough oxygen to meet their energy demand, so the muscles mainly respire aerobically, which does not produce lactate. As exercise intensity increases towards maximum effort, the muscles' demand for energy (and therefore oxygen) increasingly exceeds the rate at which oxygen can be supplied. To meet this shortfall, the muscles increasingly respire anaerobically, producing energy without oxygen but generating lactate (lactic acid) as a by-product. This growing reliance on anaerobic respiration at high intensities explains the sharp rise in blood lactate concentration shown in the data.

1 mark: trend described (lactate increases with intensity, rise becomes much steeper at high intensity), using data; 1 mark: at lower intensities, oxygen supply meets demand and respiration is mainly aerobic; 1 mark: at high intensities, oxygen demand exceeds supply; 1 mark: muscles increasingly respire anaerobically, producing lactate, causing the sharp rise.

After the meal is eaten, blood glucose concentration rises from 4.5 to a peak of 7.5 mmol/dm³ at 1 hour, as glucose is absorbed from the digested food into the blood. This rise stimulates the release of insulin from the pancreas, which causes cells (particularly in the liver and muscles) to take up glucose from the blood and store it as glycogen, bringing blood glucose back down to 5.0 mmol/dm³ by 2 hours. During the period of exercise (3 to 4 hours), blood glucose falls further, to 3.9 mmol/dm³, because the exercising muscles are respiring at a faster rate and using glucose more quickly than it is being replaced. This fall in blood glucose stimulates the release of glucagon from the pancreas, which causes the liver to break down its stored glycogen and release glucose into the blood, opposing the further fall and helping to restore blood glucose concentration towards its normal level.

1 mark: rise then fall in blood glucose after the meal, described using data; 1 mark: explains insulin causes glucose uptake by cells, lowering blood glucose after the meal; 1 mark: describes further fall in blood glucose during exercise (muscles using glucose faster); 1 mark: explains glucagon's role in opposing the fall (stimulates glycogen breakdown/glucose release from the liver).

Individual A's daily saturated fat intake (45 g) is well above the recommended maximum of less than 30 g, and their fibre intake (12 g) is well below the recommended minimum of more than 25 g; consistent with this, their blood cholesterol concentration (6.9 mmol/dm³) is above the normal range of 4.0-6.5 mmol/dm³, suggesting a raised risk of cardiovascular disease. Individual B, by contrast, has a saturated fat intake (18 g) within the recommended limit, a fibre intake (28 g) meeting the recommendation, and a blood cholesterol concentration (4.8 mmol/dm³) within the normal range. Overall, the data indicate that Individual B has the healthier, more balanced diet of the two, while Individual A's diet, being high in saturated fat and low in fibre, is associated with raised blood cholesterol and therefore an increased risk of cardiovascular disease.

1 mark: Individual A's fat intake above and fibre intake below the recommended values, using data; 1 mark: Individual A's cholesterol above the normal range, linked to raised cardiovascular disease risk; 1 mark: Individual B's values shown to be within/meeting all three recommended ranges, using data; 1 mark: valid overall conclusion that Individual B has the healthier diet, with justification.

The data show that resting tidal volume is identical for both men (500 cm³), but vital capacity (4.8 vs 3.4 dm³) and peak flow (600 vs 420 dm³/min) are both considerably lower in the smoker than the non-smoker. Long-term smoking is associated with emphysema, in which chemicals in tobacco smoke damage and break down the walls of the alveoli, reducing their elastic recoil and merging many small alveoli into fewer, larger air spaces with a smaller total surface area; this reduces the volume of air that can be forcibly exhaled, lowering vital capacity. Smoking is also associated with (chronic) bronchitis, in which the airways become inflamed and produce excess mucus, narrowing the airways and increasing resistance to airflow; this reduces the maximum rate at which air can be forced out, lowering peak flow. Together, these smoking-related changes explain why the smoker's vital capacity and peak flow are both considerably lower than the non-smoker's, even though resting tidal volume (which does not require forced or maximal effort) is unaffected.

1 mark: tidal volume similar but vital capacity and peak flow markedly lower in the smoker, using data; 1 mark: reference to emphysema (alveoli damage/reduced elastic recoil/surface area); 1 mark: reference to bronchitis/airway narrowing/inflammation and mucus; 1 mark: correct overall link between these conditions and the reduced vital capacity/peak flow values shown.

The graph should have time (s) on the x-axis and lung volume (dm³) on the y-axis, each with a suitable linear scale using at least half of the available grid, and all seven points plotted accurately and joined with straight ruled line segments between consecutive points, reflecting the way a real spirometer trace is read. (a) The three peaks occur at t = 2 s (2.5 dm³), t = 6 s (2.9 dm³) and t = 10 s (3.4 dm³); the greatest of these is at t = 10 s, with a peak lung volume of approximately 3.4 dm³, so the deepest breath occurred at this time. (b) Comparing the three breaths, each peak is higher than the last (2.5, then 2.9, then 3.4 dm³), while the baseline (2.0 dm³) and the four-second interval between successive peaks stay the same throughout; this shows that breathing became progressively deeper (tidal volume increased with each breath) while breathing rate (frequency) remained roughly constant over the 12 seconds shown.

Graph (3 marks): 1 mark for a suitable scale on each axis (using at least half the available grid) with axes correctly labelled with quantity and unit; 1 mark for all seven points plotted accurately (±½ small square); 1 mark for points joined with straight ruled line segments (not a smooth curve). (a) 1 mark: time identified as approximately 10 s (own figure, consistent with candidate's graph); 1 mark: peak volume stated as approximately 3.4 dm³ (accept range 3.2-3.6 dm³). (b) 1 mark: correctly describes breathing becoming progressively deeper (increasing peak/tidal volume) while the interval between breaths stays roughly constant.

Number of minutes in 24 hours = 24 x 60 = 1440 minutes. Total oxygen consumption = 250 cm³/min x 1440 min = 360 000 cm³. Converting to dm³ (1 dm³ = 1000 cm³): 360 000 ÷ 1000 = 360 dm³. Check by a second route: 250 cm³/min x 60 min = 15 000 cm³ per hour = 15 dm³ per hour; 15 dm³/hour x 24 hours = 360 dm³, which matches, confirming the answer.

1 mark: correct number of minutes in 24 hours calculated (1440); 1 mark: correct multiplication (250 x 1440 = 360 000 cm³); 1 mark: correct final answer with correct unit conversion, 360 dm³.

Daily energy deficit = energy expenditure − energy intake = 10 800 − 9500 = 1300 kJ. Check by a second route: energy intake (9500 kJ) plus the deficit (1300 kJ) = 10 800 kJ, which matches the given expenditure, confirming the answer. Because this person's energy expenditure (10 800 kJ) exceeds their energy intake (9500 kJ) every day, their body must break down stored energy reserves, such as fat, to make up the daily 1300 kJ shortfall. If this pattern continued, the person would therefore be expected to lose body mass over time.

1 mark: correct deficit calculated (10 800 − 9500 = 1300 kJ); 1 mark: correctly states the person would be expected to lose body mass; 1 mark: correct reason (expenditure exceeds intake, so the body draws on/breaks down stored energy reserves).

A diet high in saturated fat tends to raise the concentration of cholesterol in the blood. Over time, excess cholesterol can be deposited within the walls of arteries, building up as fatty plaques known as atheroma. This build-up narrows the internal diameter (lumen) of the affected arteries, restricting the flow of blood through them. If this narrowing occurs in the coronary arteries, which supply blood to the heart muscle itself, it reduces the oxygen supply to the heart; if a coronary artery becomes completely blocked, for example by a blood clot forming at the site of an atheroma, this can starve part of the heart muscle of oxygen, causing a heart attack (myocardial infarction). A lack of regular physical exercise compounds this risk: without regular exercise, the heart muscle is not strengthened, so it tends to be less efficient (lower stroke volume, higher resting heart rate), placing more strain on the cardiovascular system for a given level of activity. A sedentary lifestyle is also commonly associated with weight gain and obesity, which further raises blood pressure and adds to the workload of the heart. Together, a high intake of saturated fat and a lack of exercise substantially increase the risk of atheroma formation, high blood pressure and reduced cardiovascular fitness, all of which raise the overall risk of cardiovascular disease.

Excellent (5-6 marks): at least five relevant, accurate points made using appropriate specialist terms (e.g. cholesterol, atheroma, coronary artery, myocardial infarction, stroke volume), clearly and coherently organised with accurate spelling, punctuation and grammar. Good (3-4 marks): at least three relevant points made, with reasonable use of specialist terms and generally clear expression. Basic (1-2 marks): at least one relevant point made; answer may be simplistic, poorly organised or contain frequent errors of expression. Unworthy of credit (0 marks): no relevant content. Indicative content: saturated fat raises blood cholesterol; excess cholesterol is deposited in artery walls as atheroma; this narrows the artery lumen, restricting blood flow; narrowing/blockage of coronary arteries reduces oxygen supply to heart muscle, which can cause a heart attack; lack of exercise means the heart muscle is not strengthened (lower stroke volume/less efficient); lack of exercise is associated with weight gain/obesity, raising blood pressure; combined effect substantially increases overall cardiovascular disease risk.

Tobacco smoke contains tar and other harmful chemicals that irritate the lining of the airways, causing inflammation and stimulating the goblet cells to produce excess mucus. Smoke also damages and paralyses the cilia that normally line the airways and sweep mucus, along with trapped dust and pathogens, up and out of the respiratory system. With the cilia damaged, this mucus builds up in the airways instead of being cleared, causing frequent coughing, increased susceptibility to infection, and narrowing of the airways; this combination of symptoms is characteristic of (chronic) bronchitis. Separately, chemicals in tobacco smoke also damage the delicate walls of the alveoli over time, breaking down the elastic fibres that support them; damaged alveoli lose their elastic recoil and can merge together, forming fewer, larger air spaces with a much smaller total surface area than healthy lung tissue. This condition is called emphysema. Both bronchitis and emphysema reduce the efficiency of gas exchange: narrowed, mucus-filled airways in bronchitis restrict the flow of air into and out of the alveoli, while the reduced surface area in emphysema directly reduces the rate at which oxygen can diffuse into the blood and carbon dioxide can diffuse out. As a result, people with these smoking-related conditions typically experience breathlessness and reduced oxygen saturation of their blood, especially during exertion.

Excellent (5-6 marks): at least five relevant, accurate points made using appropriate specialist terms (e.g. cilia, mucus, alveoli, elastic recoil, surface area, gas exchange), clearly and coherently organised with accurate spelling, punctuation and grammar. Good (3-4 marks): at least three relevant points made, with reasonable use of specialist terms. Basic (1-2 marks): at least one relevant point made; answer may be simplistic or poorly organised. Unworthy of credit (0 marks): no relevant content. Indicative content: tar/chemicals in smoke irritate airways, increasing mucus production; smoke damages/paralyses cilia, so mucus is not cleared; leads to airway narrowing, coughing and infection (bronchitis); chemicals also damage alveoli walls, reducing elastic recoil; alveoli merge into fewer, larger spaces with reduced surface area (emphysema); both conditions reduce the rate/efficiency of gas exchange, causing breathlessness and reduced blood oxygen saturation.

Relative formula mass (Mr) is defined as the sum of the relative atomic masses of all the atoms present in the formula of a substance. For CaCO3: Mr = Ar(Ca) + Ar(C) + 3 x Ar(O) = 40 + 12 + (3 x 16) = 40 + 12 + 48 = 100.

1 mark: correct definition (sum of relative atomic masses of all atoms in the formula); 1 mark: correct method shown (40 + 12 + 3x16); 1 mark: correct final answer, 100.

A mole is defined as the amount of a substance that contains 6.02 x 10^23 particles (atoms, molecules or ions, as appropriate) of that substance — this number is known as the Avogadro constant. Number of moles = mass ÷ relative atomic mass = 6.0 ÷ 24 = 0.25 mol.

1 mark: correct definition (amount of substance containing 6.02x10^23 particles/Avogadro's constant); 1 mark: correct method (mass ÷ Ar, i.e. 6.0 ÷ 24); 1 mark: correct final answer, 0.25 mol.

A standard solution is a solution whose concentration is accurately known. To measure out an accurate, fixed volume of a standard solution, such as exactly 25.0 cm³, for a titration, a (volumetric/graduated) pipette is used, together with a pipette filler.

1 mark: correct definition (solution of accurately known concentration); 1 mark: (volumetric/graduated) pipette named.

Phenolphthalein indicator is colourless in acidic solution and pink (magenta) in alkaline solution. If phenolphthalein is added to an alkali (giving a pink solution) and acid is then added gradually, the solution changes colour from pink to colourless at the end point, as the solution becomes neutral/acidic.

1 mark: colourless in acid; 1 mark: pink/magenta in alkali; 1 mark: correctly describes the colour change from pink to colourless as acid is added to the alkali.

An exothermic reaction is a reaction in which energy is transferred from the reacting chemicals to the surroundings, usually as heat, so that the temperature of the surroundings increases. Common examples of exothermic reactions include combustion (the burning of a fuel) and neutralisation (an acid reacting with a base).

1 mark: correct definition (energy transferred to surroundings, causing surroundings to increase in temperature); 1 mark: valid example (e.g. combustion, neutralisation); 1 mark: example correctly justified/described as releasing energy.

Standard enthalpy changes are measured under standard conditions, defined as a pressure of 100 kPa and a temperature of 298 K. Standard enthalpy change is represented using the symbol \( \Delta H^{\theta} \).

1 mark: 100 kPa; 1 mark: 298 K; 1 mark: correct symbol, \( \Delta H^{\theta} \) (accept ΔH° or ΔH with a degree/theta symbol).

The rate of a reaction is defined as the change in concentration of a reactant (or product) per unit time. Activation energy is defined as the minimum amount of energy that colliding particles must possess in order to react — that is, the minimum energy needed to start breaking the bonds in the reactants so that a reaction can occur.

1 mark: rate of reaction correctly defined (change in concentration of reactant/product per unit time); 1 mark and 1 mark: activation energy correctly defined as the minimum energy particles must have to react (1 mark for 'minimum energy', 1 mark for correct reference to colliding particles/needed for reaction to occur).

In a batch process, a fixed quantity of raw materials is added to the reaction vessel, the reaction is allowed to proceed (often for a set period), and then the plant must be emptied and cleaned before the next batch can begin; production therefore happens in discrete, separate batches. In a continuous process, by contrast, raw materials are fed into the plant continuously and products are continuously removed, so the plant runs without stopping between batches. An advantage of a continuous process is that there is no downtime lost emptying, cleaning and reloading the plant between batches, so production can be more efficient and lower in cost per unit produced than an equivalent batch process (particularly where large quantities of a single product are required).

1 mark: correct description of a batch process (add raw materials, react, then empty/clean before next batch); 1 mark: correct description of a continuous process (raw materials fed in and products removed continuously, without stopping); 1 mark: valid advantage of a continuous process, e.g. no downtime for reloading/cleaning, more efficient/lower cost per unit produced.

A known volume (mass) of water is measured, using a measuring cylinder, into a suitable calorimeter, such as a copper can, and its initial temperature is recorded using a thermometer. A spirit burner containing the liquid fuel is weighed on a balance before use. The spirit burner is placed beneath the can and lit, heating the water; the water is stirred throughout to ensure even heating, and heating continues until the temperature has risen by a suitable amount (for example 20-30°C). The flame is then extinguished and the maximum temperature reached by the water is recorded, allowing the temperature change, ΔT, to be calculated. The spirit burner is reweighed, and the mass of fuel burned is found by subtracting the final mass from the initial mass. The heat energy transferred to the water is then calculated using \( Q = mc\Delta T \), where m is the mass of water heated, c is the specific heat capacity of water, and ΔT is the temperature change of the water. Finally, this value of Q is converted to an enthalpy change per mole of fuel by dividing by the number of moles of fuel burned (calculated from the mass of fuel burned and its relative formula mass), giving the enthalpy change of combustion. To improve accuracy, a lid and draught shield can be used to reduce heat losses to the surroundings, since heat losses mean the experimental value obtained is usually lower in magnitude than the true (data-book) enthalpy change of combustion.

Excellent (5-6 marks): at least five relevant, accurate steps described in a logical sequence, using appropriate specialist terms (e.g. calorimeter, specific heat capacity, ΔT, moles), clearly and coherently organised with accurate spelling, punctuation and grammar. Good (3-4 marks): at least three relevant steps described, with reasonable use of specialist terms. Basic (1-2 marks): at least one relevant step described; answer may be simplistic or poorly organised. Unworthy of credit (0 marks): no relevant content. Indicative content: measure a known mass/volume of water into a calorimeter and record its initial temperature; weigh the spirit burner (with fuel) before burning; heat/stir the water using the burning fuel until a suitable temperature rise occurs; extinguish the flame and record the maximum temperature reached (find ΔT); reweigh the spirit burner to find the mass of fuel burned; calculate heat released using Q = mcΔT; divide by moles of fuel burned to find enthalpy change of combustion (per mole); reference to reducing heat loss (e.g. lid/draught shield) and/or explaining why the experimental value is usually lower than the true value.

(a) Mass of water, m = 50.0 cm³ x 1.00 g/cm³ = 50.0 g. Temperature change, \( \Delta T = 47.5 - 21.0 = 26.5 \text{ degreesC} \). \( Q = mc\Delta T = 50.0 \times 4.2 \times 26.5 = 5565 \text{ J} \), which to 3 significant figures is 5570 J (5.57 kJ). (b) Heat released per gram of ethanol = total heat energy ÷ mass of ethanol burned = 5565 ÷ 0.65 = 8561.5... J/g, which to 3 significant figures is 8560 J/g. Check by a second route: 8560 J/g x 0.65 g = 5564 J, which is consistent (within rounding) with the 5565 J calculated in part (a), confirming the answer.

(a) 1 mark: mass of water correctly identified as 50.0 g; 1 mark: correct substitution into Q = mcΔT (50.0 x 4.2 x 26.5); 1 mark: correct final answer to 3 s.f., 5570 J (or 5.57 kJ). (b) 1 mark: correct method (Q from (a) ÷ mass of ethanol, i.e. 5565 ÷ 0.65); 1 mark: correct final answer to 3 s.f., 8560 J/g (allow ecf from candidate's own answer to (a)).

Energy must be put in to break the bonds in the reactants: breaking 1 mol H-H bonds and 1 mol Cl-Cl bonds requires \( 436 + 242 = 678 \text{ kJ/mol} \). Energy is released when the new bonds in the products form: forming 2 mol H-Cl bonds releases \( 2 \times 431 = 862 \text{ kJ/mol} \). Enthalpy change = energy to break bonds − energy released forming bonds \( = 678 - 862 = -184 \text{ kJ/mol} \). Check by a second route: energy released forming bonds (862) minus energy needed to break bonds (678) = 184 kJ/mol released overall, confirming a magnitude of 184 kJ/mol; since more energy is released forming the new bonds than is used breaking the old ones, the overall enthalpy change is negative, so the reaction is exothermic.

1 mark: energy to break reactant bonds correctly calculated (436 + 242 = 678 kJ/mol); 1 mark: energy released forming product bonds correctly calculated (2 x 431 = 862 kJ/mol); 1 mark: correct method (bonds broken − bonds formed); 1 mark: correct final answer, -184 kJ/mol; 1 mark: correctly identifies the reaction as exothermic, because ΔH is negative.

Moles of Mg = mass ÷ Ar = 4.8 ÷ 24 = 0.2 mol. From the balanced equation, 1 mol Mg produces 1 mol MgCl2, so moles of MgCl2 produced = 0.2 mol. Relative formula mass of MgCl2 = 24 + (2 x 35.5) = 24 + 71 = 95. Mass of MgCl2 = moles x Mr = 0.2 x 95 = 19.0 g. Check by a second route: 19.0 g ÷ 95 = 0.2 mol MgCl2, matching the moles calculated from the Mg used, confirming the answer.

1 mark: moles of Mg correctly calculated (4.8 ÷ 24 = 0.2 mol); 1 mark: correct mole ratio applied from the equation (1:1, so 0.2 mol MgCl2); 1 mark: relative formula mass of MgCl2 correctly calculated (95); 1 mark: correct method for final mass (0.2 x 95); 1 mark: correct final answer, 19.0 g.

Moles of NaOH = concentration x volume (in dm³) = \( 0.100 \times \frac{25.0}{1000} = 0.00250 \text{ mol} \). From the balanced equation, NaOH reacts with HCl in a 1:1 mole ratio, so moles of HCl = 0.00250 mol. Concentration of HCl = moles ÷ volume (in dm³) = \( 0.00250 \div \frac{22.5}{1000} = 0.00250 \div 0.0225 = 0.1111... \text{ mol/dm}^3 \), which to 3 significant figures is 0.111 mol/dm³. Check by a second route: 0.111 mol/dm³ x 0.0225 dm³ = 0.0025 mol, matching the moles of HCl calculated, confirming the answer.

1 mark: moles of NaOH correctly calculated (0.100 x 0.0250 = 0.00250 mol); 1 mark: correct mole ratio applied (1:1, so 0.00250 mol HCl); 1 mark: correct formula for concentration (moles ÷ volume in dm³); 1 mark: correct substitution (0.00250 ÷ 0.0225); 1 mark: correct final answer to 3 s.f., 0.111 mol/dm³.

(a) Percentage yield = (actual yield ÷ theoretical yield) x 100 = (6.20 ÷ 8.00) x 100 = 0.775 x 100 = 77.5%. Check by a second route: 77.5% of 8.00 g = 0.775 x 8.00 = 6.20 g, matching the actual yield given, confirming the answer. (b) In practice, some of the product is often lost during the preparation, for example during filtration (some crystals remain on the filter paper) or transfer between containers, or some product may remain dissolved in the solution rather than crystallising out, all of which would reduce the yield below 100%.

(a) 1 mark: correct formula (actual yield ÷ theoretical yield x 100); 1 mark: correct substitution (6.20 ÷ 8.00 x 100); 1 mark: correct final answer, 77.5%. (b) 1 mark: valid reason, e.g. product lost during filtration/transfer, or some product remained dissolved in solution.

(a) Temperature change, \( \Delta T = 26.0 - 19.5 = 6.5 \text{ degreesC} \). (b) Mass of solution, m = 25.0 cm³ x 1.00 g/cm³ = 25.0 g. \( Q = mc\Delta T = 25.0 \times 4.2 \times 6.5 = 682.5 \text{ J} \). Check by a second route: \( 682.5 \div (25.0 \times 4.2) = 6.5 \text{ degreesC} \), matching the temperature change given, confirming the answer. (c) The reaction is exothermic, because the temperature of the solution increased during the reaction, showing that heat energy was transferred from the reacting chemicals to the surroundings (the solution itself), rather than being absorbed from them.

(a) 1 mark: correct ΔT, 6.5 °C. (b) 1 mark: correct substitution into Q = mcΔT (25.0 x 4.2 x 6.5); 1 mark: correct final answer, 682.5 J (accept 683 J or 0.6825/0.683 kJ). (c) 1 mark: exothermic, with correct reason (temperature increased, so heat was released to the surroundings).

The Haber process is typically carried out at a temperature of about 450 °C and a pressure of about 200 atmospheres, using an iron catalyst. The reversible reaction taking place is represented by the balanced symbol equation: \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \).

1 mark: temperature of approximately 450 °C; 1 mark: pressure of approximately 200 atmospheres; 1 mark: iron catalyst; 1 mark: correct balanced equation with state symbols and reversible arrow, N2(g) + 3H2(g) ⇌ 2NH3(g).

The reactant side of the equation has 1 mol of N2 plus 3 mol of H2, a total of 4 mol of gas, while the product side has 2 mol of NH3. Increasing the pressure on a gaseous equilibrium shifts the position of equilibrium towards the side with fewer moles of gas, because this reduces the total number of gas molecules present and so acts to reduce the pressure, partially opposing the increase. Since the product side (2 mol) has fewer gas molecules than the reactant side (4 mol), increasing the pressure shifts the position of equilibrium to the right, towards ammonia, increasing the equilibrium yield of ammonia.

1 mark: correctly identifies that the reactant side has more moles of gas (4 mol) than the product side (2 mol); 1 mark: states that equilibrium shifts towards the side with fewer gas molecules when pressure increases; 1 mark: correctly concludes the equilibrium shifts to the right/towards ammonia; 1 mark: correctly links this to an increased yield of ammonia.

Because the forward reaction is exothermic, lowering the temperature shifts the position of equilibrium further towards the products (ammonia), increasing the equilibrium yield. However, lowering the temperature also reduces the rate of reaction, since particles have less kinetic energy and collide less frequently and less energetically, meaning it would take much longer to reach equilibrium. A very low temperature would therefore give a high yield but only after an uneconomically long time, reducing the overall rate of ammonia production and profitability. A compromise (moderate) temperature is therefore chosen that gives a reasonably high equilibrium yield within a reasonably short time, maximising the overall rate of profitable ammonia production rather than simply maximising the equilibrium yield alone.

1 mark: lower temperature would give a higher equilibrium yield, because the forward reaction is exothermic; 1 mark: lower temperature would also give a slower rate of reaction; 1 mark: a very low temperature would therefore be uneconomical (high yield but very slow); 1 mark: compromise temperature balances a reasonably high yield against a reasonably fast rate, maximising overall economic efficiency.

On a reaction profile diagram for an exothermic reaction, energy is plotted on the vertical axis against the progress of the reaction on the horizontal axis. The curve starts at the energy level of the reactants, rises to a peak (representing the unstable, high-energy transition state through which the reactants must pass), and then falls to the energy level of the products; for an exothermic reaction, this products' energy level is lower than that of the reactants, since energy has been transferred to the surroundings overall. The activation energy is defined, and shown on the diagram, as the difference in energy between the reactants and the peak of the curve — the minimum energy the reacting particles need in order to reach the transition state and react. The overall enthalpy change, \( \Delta H \), is defined, and shown on the diagram, as the difference in energy between the reactants and the products; because the products are at a lower energy than the reactants in an exothermic reaction, this enthalpy change is negative. Adding a catalyst provides an alternative reaction pathway that has a lower activation energy; on the reaction profile diagram, this is shown as a lower peak than for the uncatalysed reaction. Crucially, the energy levels of the reactants and products themselves are unaffected by the catalyst, so the overall enthalpy change, ΔH, for the reaction remains exactly the same; only the height of the peak (the activation energy) is reduced.

1 mark: profile correctly described as reactants rising to a peak (transition state) then falling to products; 1 mark: products shown/stated as lower in energy than reactants (consistent with an exothermic reaction); 1 mark: activation energy correctly defined/located as the energy difference between reactants and the peak; 1 mark: enthalpy change correctly defined/located as the energy difference between reactants and products (negative for exothermic); 1 mark: catalyst correctly described as lowering the peak/activation energy while leaving the reactant and product energy levels (and therefore ΔH) unchanged.

Ductility describes a material's ability to be drawn or stretched out into a wire without breaking. Malleability describes a material's ability to be hammered, rolled or pressed into a new shape, such as a thin sheet, without breaking or cracking.

1 mark: correct definition of ductility (can be drawn into a wire without breaking); 1 mark: correct definition of malleability (can be hammered/pressed into shape without breaking/cracking).

Stress is calculated using the equation \( \sigma = \dfrac{F}{A} \), where σ is the stress, F is the force applied (in newtons, N), and A is the cross-sectional area over which the force acts (in square metres, m²). Stress is measured in pascals (Pa), equivalent to N/m².

1 mark: correct equation, stress = force ÷ cross-sectional area, with force and area correctly defined; 1 mark: correct unit, pascals (Pa) or N/m².

Strain is calculated using the equation \( \varepsilon = \dfrac{\Delta l}{l_0} \), where ε is the strain, Δl is the change in length (extension), and l0 is the original length. Because strain is the ratio of one length to another, the units cancel, so strain has no unit (it is dimensionless).

1 mark: correct equation, strain = extension ÷ original length, with terms correctly defined; 1 mark: correctly states strain has no unit (dimensionless ratio).

The Young modulus of a material is a measure of its stiffness, that is, its resistance to elastic (reversible) deformation under stress: a material with a higher Young modulus is stiffer, deforming less for a given applied stress. It is calculated using the equation \( E = \dfrac{\sigma}{\varepsilon} \), the stress divided by the strain, within the material's elastic region.

1 mark: correct definition (a measure of stiffness/resistance to elastic deformation); 1 mark: correct equation, Young modulus = stress ÷ strain.

Materials studied in this course can be grouped into five general categories: metals, ceramics, glasses, polymers and composites. An example from one of these categories is steel, which is a metal (other valid examples include porcelain as a ceramic, or fibreglass as a composite).

1 mark: correctly names at least four of the five categories (metals, ceramics, glasses, polymers, composites); 1 mark: valid example of a material correctly matched to one of the named categories.

In a thermosetting polymer, the individual polymer chains are joined together by strong covalent cross-links, forming a single, rigid, interconnected network; these cross-links do not break when the material is reheated, so thermosetting polymers do not melt or soften once set, and will char or decompose rather than remould if heated strongly. In a thermoplastic, by contrast, individual polymer chains are held together only by weak intermolecular forces, with no covalent cross-links between them; when heated, these weak forces are easily overcome, allowing the chains to slide past one another, so the material softens and can be remoulded, then re-hardens on cooling.

1 mark: thermosetting polymers have strong covalent cross-links between chains (which do not break on heating); 1 mark: thermoplastics have separate chains held by weak intermolecular forces only (no cross-links), allowing chains to slide/soften on heating.

Metals have a giant metallic structure consisting of a regular lattice of positively charged metal ions. The outer-shell electrons of the metal atoms are not held by any one particular ion; instead they become delocalised, forming a 'sea' of free electrons that move throughout the whole structure, surrounding and holding together the lattice of positive ions. Because these delocalised electrons are free to move, they can carry electrical charge through the structure when a potential difference is applied, which is why metals are good conductors of electricity.

1 mark: metal structure described as a lattice of positive metal ions; 1 mark: surrounded by a 'sea'/delocalised electrons; 1 mark: these delocalised electrons are free to move and carry charge, allowing conduction of electricity.

An alloy is defined as a mixture consisting of a metal combined with one or more other elements, which are usually other metals but can include non-metals such as carbon. Bronze is an example of an alloy that is a mixture of copper and tin.

1 mark: correct definition of alloy (mixture of a metal with one or more other elements); 1 mark: bronze correctly named as the copper-tin alloy.

Annealing is a heat treatment process used to soften a metal, relieve internal stresses that have built up (for example during previous working or cooling), and make the metal more ductile and easier to shape further. It is carried out by heating the metal to a specific temperature (below its melting point) and then allowing it to cool slowly, rather than being cooled rapidly.

1 mark: correct purpose (softens the metal/relieves internal stress, making it easier to work/more ductile); 1 mark: correct process (heating to a set temperature, then cooling slowly).

A bioactive biomaterial is one that actively interacts with the surrounding living tissue, for example by stimulating a biological response such as new bone growth, allowing it to bond with the tissue. A bioinert biomaterial, in contrast, does not interact or react chemically with the surrounding tissue at all; it produces minimal biological response and is simply tolerated by the body, without bonding to or stimulating the tissue around it.

1 mark: bioactive material correctly described as interacting with/bonding to surrounding tissue (e.g. stimulating a biological response); 1 mark: bioinert material correctly described as not interacting/reacting with surrounding tissue (minimal biological response).

A smart material is defined as a material that has one or more properties which can be significantly and reversibly changed, in a controlled and useful way, by an external stimulus, such as a change in temperature, stress, light, or electric or magnetic field. An example of a smart material is a shape-memory alloy, which can be deformed and then return to a pre-set shape when heated (other valid examples include piezoelectric materials, thermochromic materials, photochromic materials, electroluminescent materials, or quantum-tunnelling composites).

1 mark: correct definition of smart material (properties change in a controlled, reversible, useful way in response to a stimulus); 1 mark: valid named example (e.g. shape-memory alloy, piezoelectric material, thermochromic material, photochromic material, electroluminescent material, quantum-tunnelling composite).

Graphene is a form of carbon consisting of a single layer, just one atom thick, of carbon atoms arranged in a repeating hexagonal (honeycomb) lattice. Carbon nanotubes, which are related structures, have a very high tensile strength combined with a very low density (they are extremely lightweight for their strength); this combination makes them useful for reinforcing composite materials, since adding carbon nanotubes can substantially increase the strength of a composite without adding much extra mass.

1 mark: graphene correctly described as a single, one-atom-thick layer of carbon atoms; 1 mark: arranged in a hexagonal lattice/structure; 1 mark: valid useful property of carbon nanotubes stated (e.g. very high tensile strength and/or low density), correctly linked to reinforcing composites without adding much mass.

(a) \( \sigma = \dfrac{F}{A} = \dfrac{60.0}{1.50 \times 10^{-6}} = 4.00 \times 10^{7} \text{ Pa} \). (b) Extension \( \Delta l = 1.20 \text{ mm} = 1.20 \times 10^{-3} \text{ m} \). \( \varepsilon = \dfrac{\Delta l}{l_0} = \dfrac{1.20 \times 10^{-3}}{2.000} = 6.00 \times 10^{-4} \) (no unit). (c) \( E = \dfrac{\sigma}{\varepsilon} = \dfrac{4.00 \times 10^{7}}{6.00 \times 10^{-4}} = 6.6\overline{6} \times 10^{10} \text{ Pa} \), which to 3 significant figures is 6.67 x 10^10 Pa. Check by a second route: \( E \times \varepsilon = 6.67 \times 10^{10} \times 6.00 \times 10^{-4} = 4.00 \times 10^{7} \text{ Pa} \), which matches the stress calculated in (a), confirming the answer.

(a) 1 mark: correct substitution (60.0 ÷ 1.50x10^-6); 1 mark: correct final answer to 3 s.f., 4.00 x 10^7 Pa. (b) 1 mark: extension correctly converted to metres (1.20x10^-3 m); 1 mark: correct final answer, 6.00 x 10^-4 (no unit). (c) 1 mark: correct method (stress ÷ strain); 1 mark: correct substitution using candidate's own values from (a) and (b) (ecf); 1 mark: correct final answer to 3 s.f., 6.67 x 10^10 Pa.

(a) \( \rho = \dfrac{m}{V} = \dfrac{47.25}{6.30} = 7.50 \text{ g/cm}^3 \). (b) Since 1 g/cm³ = 1000 kg/m³, and 1 dm³ = 10^-3 m³, then 1 g/cm³ is equivalent to 1000 kg/m³ x 10^-3 m³/dm³ = 1 kg/dm³; so a density of 7.50 g/cm³ is equivalent to 7.50 kg/dm³. (c) Ratio = alloy density ÷ other metal's density = \( 7.50 \div 2.70 = 2.777... \), which to 3 significant figures is 2.78. Check by a second route: 2.70 g/cm³ x 2.78 = 7.506 g/cm³, which is consistent (within rounding) with the alloy's density of 7.50 g/cm³, confirming the answer.

(a) 1 mark: correct substitution (47.25 ÷ 6.30); 1 mark: correct final answer to 3 s.f., 7.50 g/cm³. (b) 1 mark: correct conversion relationship identified (1 g/cm³ = 1 kg/dm³); 1 mark: correct final answer, 7.50 kg/dm³. (c) 1 mark: correct method (alloy density ÷ other metal's density); 1 mark: correct substitution (7.50 ÷ 2.70); 1 mark: correct final answer to 3 s.f., 2.78 (times).

(a) \( E_A = \dfrac{\sigma}{\varepsilon} = \dfrac{2.4 \times 10^{8}}{0.0016} = 1.50 \times 10^{11} \text{ Pa} \). (b) \( E_B = \dfrac{\sigma}{\varepsilon} = \dfrac{1.8 \times 10^{8}}{0.0030} = 6.00 \times 10^{10} \text{ Pa} \). Check by a second route for (b): \( 6.00 \times 10^{10} \times 0.0030 = 1.8 \times 10^{8} \text{ Pa} \), matching the given stress, confirming the answer. (c) Material A has the higher Young modulus (1.50 x 10^11 Pa, compared with 6.00 x 10^10 Pa for Material B); a higher Young modulus indicates greater stiffness (greater resistance to elastic deformation), so Material A is the stiffer of the two. (d) The Young modulus of Material B is lower than that of Material A, meaning Material B is less stiff: for the same applied stress, a lower Young modulus produces a greater strain (since strain = stress ÷ E). For samples of the same original length, a greater strain corresponds to a greater extension (since extension = strain x original length), so Material B would show the greater extension for the same applied stress.

(a) 1 mark: correct substitution (2.4x10^8 ÷ 0.0016); 1 mark: correct final answer, 1.50 x 10^11 Pa. (b) 1 mark: correct substitution (1.8x10^8 ÷ 0.0030); 1 mark: correct final answer, 6.00 x 10^10 Pa. (c) 1 mark: Material A identified as stiffer, with correct reason (higher Young modulus). (d) 1 mark: Material B identified as showing greater extension; 1 mark: correct reason (lower Young modulus means greater strain, and therefore greater extension, for the same stress and original length).

(a) Converting to metres: 4.0 mm = 0.0040 m and 5.0 mm = 0.0050 m. Area \( = 0.0040 \times 0.0050 = 2.0 \times 10^{-5} \text{ m}^2 \). (b) \( \sigma = \dfrac{F}{A} = \dfrac{300}{2.0 \times 10^{-5}} = 1.50 \times 10^{7} \text{ Pa} \). (c) The applied stress calculated in (b), 1.50 x 10^7 Pa, is greater than the material's yield strength of 1.2 x 10^7 Pa. Because the applied stress exceeds the yield strength, the strut is at risk of permanent (plastic) deformation under this load, rather than deforming only elastically.

(a) 1 mark: correct unit conversion (4.0 mm and 5.0 mm to metres); 1 mark: correct final answer, 2.0 x 10^-5 m². (b) 1 mark: correct substitution (300 ÷ 2.0x10^-5); 1 mark: correct final answer to 3 s.f., 1.50 x 10^7 Pa. (c) 1 mark: correctly identifies applied stress exceeds yield strength (using own value from (b), ecf); 1 mark: correctly concludes the strut is at risk of permanent deformation, with valid reasoning.

(a) Mass = density x volume = \( 8.90 \times 45 = 400.5 \text{ g} \). (b) Mass = density x volume = \( 2.70 \times 45 = 121.5 \text{ g} \). (c) Percentage reduction \( = \dfrac{400.5 - 121.5}{400.5} \times 100 = \dfrac{279}{400.5} \times 100 = 69.66...\% \), which to 3 significant figures is 69.7%. Check by a second route: 400.5 g reduced by 69.7% is \( 400.5 \times (1 - 0.697) = 400.5 \times 0.303 \approx 121.4 \text{ g} \), consistent (within rounding) with the 121.5 g calculated in (b), confirming the answer. (d) If the application required high strength or stiffness (for example, a structural or load-bearing component), a lower-density alloy might not be suitable if it also has a lower yield strength or Young modulus than the original alloy, since it could then deform or fail under loads the original material would have withstood; mass reduction should not come at the expense of the component being unable to perform its structural role safely.

(a) 1 mark: correct final answer, 400.5 g. (b) 1 mark: correct final answer, 121.5 g. (c) 1 mark: correct method/substitution (using candidate's own values from (a) and (b), ecf); 1 mark: correct final answer to 3 s.f., 69.7%. (d) 1 mark: valid situation identified (e.g. where high strength/stiffness is required); 1 mark: valid reason given (lower-density alloy may also have lower strength/stiffness, unsuitable for that structural role).

Several material properties are relevant to choosing a casing material for a portable electronic device. Density is important because a lighter device is more convenient to carry; thermoplastic polymers generally have a much lower density than metal alloys, making them better suited to a lightweight, portable product. Electrical conductivity is also important: polymers are electrical insulators, so a polymer casing poses no risk of short-circuiting the device's internal electronic components if they were to come into contact with the casing, whereas a metal casing conducts electricity and would need an internal insulating layer to be used safely. On the other hand, metals have a much higher thermal conductivity than polymers, so a metal casing would conduct heat generated by the device's electronics away more effectively, helping to prevent the device from overheating, which is an advantage of metal over polymer in this respect. In terms of toughness, thermoplastics can be moulded (for example by injection moulding) into shapes designed to flex and absorb impact without cracking, whereas metal casings, while often strong, may dent under impact. Considering industrial factors, thermoplastics can typically be injection moulded quickly and cheaply into complex shapes suited to mass production, likely making them less expensive to manufacture in large volumes than machining or pressing an equivalent metal casing. In terms of environmental impact, many thermoplastics are derived from non-renewable crude oil and some types are difficult to recycle, whereas many metals can be recycled relatively easily, although extracting and refining metal ores also has a significant environmental impact of its own. Finally, consumer demand for lightweight, portable electronic devices generally favours materials with lower density, such as thermoplastic polymers. Overall, considering low density (portability), electrical insulation (safety) and low-cost mass production, a thermoplastic polymer is likely to be the more suitable choice for this casing overall, despite metal offering better heat dissipation and, for some metals, easier recyclability.

Excellent (5-6 marks): at least five relevant points made, covering both material properties and industrial considerations, using appropriate specialist terms (e.g. density, thermal/electrical conductivity, toughness, injection moulding, recyclability), clearly organised into a coherent evaluation with a justified conclusion, with accurate spelling, punctuation and grammar. Good (3-4 marks): at least three relevant points made, with reasonable use of specialist terms. Basic (1-2 marks): at least one relevant point made; answer may be simplistic, one-sided or poorly organised. Unworthy of credit (0 marks): no relevant content. Indicative content: polymer has lower density than metal, aiding portability; polymer is an electrical insulator, improving safety; metal has higher thermal conductivity, aiding heat dissipation; polymer can be moulded to absorb impact/toughness; polymer casings can be cheaply mass-produced by injection moulding; environmental impact of polymer (crude-oil-derived, recycling difficulty) versus metal (more readily recyclable, but extraction/refining impact); consumer demand favours lightweight devices; an overall, justified conclusion is reached.

Silicon atoms have four electrons in their outer shell. In pure (intrinsic) silicon, each atom forms four covalent bonds with neighbouring silicon atoms, using all four outer-shell electrons, so almost all electrons are held firmly within these bonds rather than being free to move. This means pure silicon has very few free (mobile) charge carriers available at room temperature, giving it an electrical conductivity that lies between that of a good conductor (such as a metal) and an insulator. Because this intermediate conductivity can be precisely controlled, for example by doping, silicon is a useful semiconductor material.

1 mark: silicon has four electrons in its outer shell; 1 mark: these electrons are held in covalent bonds within the crystal lattice, so pure silicon has very few free charge carriers at room temperature; 1 mark: this gives silicon a controllable, intermediate conductivity between a conductor and an insulator, making it useful as a semiconductor.

N-type doping involves adding a small number of atoms that have five electrons in their outer shell, one more than silicon. Four of these electrons form covalent bonds with neighbouring silicon atoms, but the fifth electron is not needed for bonding and is free to move through the lattice, acting as a mobile (negative) charge carrier. P-type doping involves adding atoms that have only three outer-shell electrons, one fewer than silicon; this leaves one covalent bond incomplete, creating a 'hole' (a vacancy where an electron is missing). Electrons from neighbouring atoms can move into this hole, which effectively causes the hole itself to move through the lattice, acting as a mobile (positive) charge carrier. In both cases, doping increases the number of mobile charge carriers compared with pure silicon, allowing current to flow more easily.

1 mark: n-type doping described (atoms with 5 outer-shell electrons provide spare/free electrons as charge carriers); 1 mark: p-type doping described (atoms with 3 outer-shell electrons create holes as charge carriers); 1 mark: both types of doping correctly linked to an increase in mobile charge carriers compared with pure silicon, allowing easier current flow.

A diode is constructed from a junction between a region of p-type silicon and a region of n-type silicon. When the diode is connected in forward bias (with the applied voltage in the direction that favours current flow), the free electrons in the n-type region and the holes in the p-type region are both pushed towards the junction; there, electrons cross into the p-type region and combine with holes (and holes effectively move into the n-type region), allowing a continuous flow of charge carriers, and so a current, across the junction. When the diode is instead connected in reverse bias (with the applied voltage in the opposite direction), the free electrons and holes are pulled away from the junction instead, widening the region around the junction that is depleted of mobile charge carriers; with very few charge carriers available near the junction, almost no current can flow. Because current can flow readily in forward bias but not in reverse bias, the diode allows current to flow through it in one direction only.

1 mark: diode correctly described as formed from a p-n junction; 1 mark: forward bias correctly explained (electrons/holes pushed towards the junction, combine, allowing current to flow); 1 mark: reverse bias correctly explained (electrons/holes pulled away from the junction, depleted region widens, little/no current flows); 1 mark: correct overall conclusion that current can therefore flow through the diode in one direction only.