Question 1
(a)(i)-(ii): Subtopic: B5 Enzymes
(b): Subtopic: B5 Enzymes
(c): Subtopic: B4 Biological molecules
(a) The graph in Fig. 1.1 shows the effect of temperature on enzyme activity.
(i) State the optimum temperature of the enzyme in Fig. 1.1.
▶️Answer/Explanation
45°C
Explanation: The optimum temperature is where the enzyme activity peaks, which occurs at 45°C on the graph.
(ii) Explain the results in Fig. 1.1 at 60°C.
▶️Answer/Explanation
The enzyme denatures at 60°C. The high temperature causes the enzyme’s active site to change shape, so substrates can no longer fit and the reaction cannot occur.
Explanation: Enzymes are proteins that have an optimal temperature range. Beyond this range (in this case above 45°C), the protein structure begins to break down (denature), altering the active site’s shape and preventing substrate binding.
(b) Enzyme activity is also affected by pH. Complete the graph in Fig. 1.2 to show how the activity of a protease enzyme in the stomach is affected by pH.
Include on your graph:
- labels for both axes
- a sketch of a suitable curve.
▶️Answer/Explanation
The graph should have:
1. x-axis labeled “pH” and y-axis labeled “Enzyme activity”
2. A bell-shaped curve peaking around pH 2 (acidic range)
Explanation: Stomach protease enzymes (like pepsin) work best in acidic conditions (pH 1.5-2.5). The curve should show maximum activity in this range, decreasing sharply at higher pH values.
(c) A solution containing an enzyme is tested with biuret solution. State the colour change you would expect. Give a reason for your answer.
▶️Answer/Explanation
The colour changes from blue to purple.
Reason: Enzymes are proteins, and biuret solution turns purple in the presence of proteins.
Explanation: The biuret test detects peptide bonds, which are present in all proteins including enzymes. The copper ions in the biuret solution form a violet-colored complex with these bonds.
Question 2
(a)(i)-(ii): Subtopic: C11.4 Alkanes / C11.5 Alkenes
(b): Subtopic: C11.5 Alkenes
(c)(i)-(ii): Subtopic: C11.5 Alkenes
(d)(i)-(ii): Subtopic: C11.7 Polymers
(a) Fig. 2.1 shows the composition of clean air and of natural gas.
(i) Complete the labels in Fig. 2.1 to show the main constituent of clean air and of natural gas.
▶️Answer/Explanation
Clean air: Nitrogen (≈78%)
Natural gas: Methane (CH4)
Explanation: Clean air is primarily nitrogen (78%) and oxygen (21%). Natural gas is predominantly methane (70-90%), with smaller amounts of ethane, propane, and other hydrocarbons.
(ii) One of the other gases in natural gas is ethane. Name two gases in clean air that are formed by the complete combustion of ethane.
▶️Answer/Explanation
1. Carbon dioxide (CO2)
2. Water (vapor/H2O)
Explanation: Complete combustion of any hydrocarbon (including ethane, C2H6) produces CO2 and H2O. The balanced equation is: 2C2H6 + 7O2 → 4CO2 + 6H2O
(b) Name the process used to convert larger alkane molecules into ethene and hydrogen.
▶️Answer/Explanation
Cracking
Explanation: Cracking breaks long-chain alkanes into shorter alkenes (like ethene) and hydrogen using heat and a catalyst (e.g., aluminum oxide).
(c) The molecular structure of ethene is shown below.
The double bond in ethene allows it to undergo addition reactions.
(i) Complete the equation for the addition reaction between ethene and bromine.
C2H4 + Br2 → ……
▶️Answer/Explanation
C2H4Br2 (1,2-dibromoethane)
Explanation: The double bond opens up, allowing bromine atoms to add across it. The product is colorless, distinguishing it from the orange bromine solution.
(ii) Ethene is used to make ethanol in an addition reaction. Name the other raw material required in the manufacture of ethanol.
▶️Answer/Explanation
Steam (H2O)
Explanation: Ethene reacts with steam (H2O) in the presence of a phosphoric acid catalyst at 300°C and 60 atm: C2H4 + H2O → C2H5OH
(d) Ethene is used in the manufacture of poly(ethene). Propene is used in the manufacture of poly(propene).
(i) The structure of poly(ethene) is shown by:
(n is a large number)
Describe the formation of poly(ethene) using the terms monomer and polymer.
▶️Answer/Explanation
Ethene (C2H4) is the monomer. Many ethene molecules undergo addition polymerization, breaking their double bonds and linking together to form long-chain poly(ethene) (a polymer).
Explanation: The process requires high pressure and a catalyst. The double bond opens to form single bonds with adjacent monomers, creating a saturated hydrocarbon chain.
(ii) The molecular structure of propene is shown below.
Suggest the structure of poly(propene). Draw your answer in the space below.
▶️Answer/Explanation
Explanation: Poly(propene) has methyl (CH3) side groups alternating along the carbon backbone, formed by opening propene’s double bond.
Question 3
(a)(i)-(ii): Subtopic: P1.6.1 Energy
(b)(i)-(ii): Subtopic: P1.5 Effects of forces
(c)(i)-(ii): Subtopic: P2.3.3 Radiation / P6.2.1 The Sun as a star
(a) In 1971, an astronaut hit a golf ball on the surface of the Moon. The golf ball had a mass of 46 g and initially travelled at 50 m/s.
(i) Calculate the kinetic energy of the golf ball when travelling at 50 m/s. Show your working.
▶️Answer/Explanation
Solution:
KE = ½mv²
m = 46 g = 0.046 kg
v = 50 m/s
KE = ½ × 0.046 × (50)²
= 0.5 × 0.046 × 2500
= 57.5 J
Explanation: First convert mass to kg (SI units), then apply the kinetic energy formula. Note the squared velocity term significantly impacts the result.
(ii) Describe the difference between the terms speed and velocity.
▶️Answer/Explanation
Speed is a scalar quantity (magnitude only), while velocity is a vector quantity (magnitude and direction).
Explanation: For example, 50 m/s is speed, but 50 m/s horizontally eastward is velocity. On the Moon, the golf ball’s speed could be measured, but its velocity would include its trajectory direction.
(b) On the Moon, an astronaut suspends masses on a spring and measures the extension of the spring in mm as shown in Fig. 3.1.
Fig. 3.2 shows the results of the experiment.
(i) Use Fig. 3.2 to determine the range of masses where Hooke’s Law is obeyed. Explain your answer.
▶️Answer/Explanation
Range: 0 g to 400 g
Explanation: The graph shows a straight-line relationship (constant gradient) in this range, indicating extension is directly proportional to mass (F = kx).
Note: Beyond 400 g, the curve deviates, showing the spring’s elastic limit is exceeded.
(ii) The astronaut repeats the experiment with an identical spring on Earth. Each 100 g mass produces a greater extension of the spring on Earth. Calculate the mass that would need to be used on Earth to obtain the same extension as the addition of 300 g on the Moon. The gravitational field strength on Earth is 10 N/kg and on the Moon is 1.6 N/kg. Show your working.
▶️Answer/Explanation
Solution:
Force on Moon: Fmoon = 0.3 kg × 1.6 N/kg = 0.48 N
For equal extension on Earth: Fearth = Fmoon
m × 10 = 0.48 → m = 0.048 kg = 48 g
Explanation: Hooke’s Law requires equal force for equal extension. The weaker Moon gravity (1.6 N/kg vs 10 N/kg) means less mass is needed there to create equivalent force.
(c) The astronaut is exposed to infra-red waves that travel from the Sun to the Moon.
(i) Name this method of energy transfer.
▶️Answer/Explanation
Radiation
Explanation: Infrared waves transfer heat energy via electromagnetic radiation, requiring no medium (unlike conduction/convection). This is how the Sun’s energy crosses space.
(ii) Name the type of nuclear reaction taking place in the Sun that releases energy.
▶️Answer/Explanation
Nuclear fusion
Explanation: The Sun converts hydrogen to helium via proton-proton chain reactions, releasing enormous energy (E=mc²). This differs from nuclear fission used in power plants.
Question 4
(a)(i)-(ii): Subtopic: B6.2 Leaf structure
(b)-(d): Subtopic: B6.1 Photosynthesis
(a) Fig. 4.1 shows a cross-section through a leaf.
(i) Name the parts A and B.
▶️Answer/Explanation
A: (Waxy) cuticle
B: (Lower) epidermis
Explanation: The waxy cuticle is the waterproof outer layer that reduces water loss. The lower epidermis contains stomata (pores) for gas exchange.
(ii) Draw one arrow on Fig. 4.1 to show where carbon dioxide enters the leaf.
▶️Answer/Explanation
Arrow should point to a stoma (pore) in the lower epidermis.
Explanation: CO₂ enters primarily through stomata during daylight hours for photosynthesis. The arrow should clearly indicate passage through an open stoma.
(b) Describe two features of the cells in the palisade mesophyll layer that allow efficient photosynthesis to occur.
▶️Answer/Explanation
1. Tightly packed cylindrical cells (maximizes light absorption)
2. Contain many chloroplasts (site of photosynthesis)
Additional possible answers:
• Positioned near the upper surface (closer to light)
• Large vacuoles push chloroplasts to cell edges
• Thin cell walls for rapid gas diffusion
(c) Describe the role of chlorophyll in photosynthesis.
▶️Answer/Explanation
Chlorophyll absorbs light energy (primarily red/blue wavelengths) and converts it to chemical energy. This energy drives the synthesis of glucose from CO₂ and water.
Explanation: The green pigment in chloroplasts excites electrons when struck by photons, initiating the light-dependent reactions of photosynthesis.
(d) Complete the balanced symbol equation for photosynthesis.
▶️Answer/Explanation
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
Explanation: Key points:
• 6 carbon dioxide + 6 water molecules
• Produces 1 glucose + 6 oxygen molecules
• Light energy and chlorophyll written above the arrow as conditions
• All molecules must be correctly balanced
Question 5
(a)(i)-(ii): Subtopic: C12.4 Separation and purification
(b)(i)-(iii): Subtopic: C12.3 Chromatography
A student investigates the colour in a leaf. He crushes the leaf in a solvent to extract the coloured compounds. Fig. 5.1 shows the apparatus he uses.
(a) (i) Name the process he uses to remove the solids from the cloudy green liquid to obtain a clear green solution.
▶️Answer/Explanation
Filtration
Explanation: Filtration separates insoluble solids (leaf tissue) from the liquid solvent containing dissolved pigments, using filter paper that traps solid particles while allowing the solution to pass through.
(ii) Describe how the process he uses produces a clear green solution.
▶️Answer/Explanation
The filter paper retains solid particles while allowing the dissolved pigments to pass through, resulting in a particle-free solution.
Explanation: The pore size in filter paper is small enough to block plant cell debris but large enough to let pigment molecules through, clarifying the solution while retaining the green color.
(b) The student uses paper chromatography to separate the compounds which give the solution its green colour. He draws a pencil line on a strip of chromatography paper and places a drop of the green solution on the pencil line. He dips the paper into a solvent. Fig. 5.2 is an incomplete diagram of the apparatus he uses.
(i) Draw a line on Fig. 5.2 to show the surface of the solvent in the beaker.
▶️Answer/Explanation
A horizontal line should be drawn:
– Below the pencil line/spot
– Above the bottom of the paper
Explanation: The solvent level must be below the pigment spot to prevent immediate dissolution of samples into the solvent reservoir, allowing proper capillary action.
(ii) Fig. 5.3 shows the chromatogram he obtains.
Table 5.1 lists the Rf values of the coloured compounds on the chromatogram.
| coloured compound | Rf |
|---|---|
| carotene | 0.91 |
| chlorophyll A | 0.42 |
| chlorophyll B | 0.16 |
| xanthophyll | 0.77 |
Use Table 5.1 to identify the yellow compound. Explain how you obtained your answer.
▶️Answer/Explanation
Yellow compound: Xanthophyll
Explanation: It has the second highest Rf value (0.77), matching the position of the second spot from the top on the chromatogram which appears yellow.
Additional note: Carotene (orange) would have higher Rf (0.91), while chlorophylls (green) have lower Rf values.
(iii) Describe how the student can obtain a pure, dry sample of the orange compound from the chromatogram.
▶️Answer/Explanation
1. Cut out the orange band with scissors
2. Place in a small volume of solvent to dissolve the pigment
3. Filter and evaporate the solvent to obtain dry carotene
Explanation: This process isolates the desired compound by:
– Physically separating the pigment band
– Re-dissolving in fresh solvent
– Removing solvent through evaporation (e.g., using a water bath)
Question 6
(a)(i)-(ii): Subtopic: P4.3.2 Series and parallel circuits / P4.2.2 Electric current
(b)(i)-(ii): Subtopic: P2.1.2 Particle model
(c)(i)-(ii): Subtopic: P2.3.1 Conduction / P2.2.1 Thermal expansion
(a) Fig. 6.1 shows a car with two rear lamps, L1 and L2.
The lamps are connected in parallel and powered by a 12V battery. The lamps each have a resistance of 33Ω.
(i) Calculate the combined resistance of the two lamps connected in parallel in this circuit. Show your working.
▶️Answer/Explanation
Solution:
1/Rtotal = 1/R1 + 1/R2 = 1/33 + 1/33 = 2/33
Rtotal = 33/2 = 16.5 Ω
Explanation: In parallel circuits, the reciprocal of total resistance equals the sum of reciprocals of individual resistances. Identical resistances in parallel give half the individual resistance.
(ii) Calculate the charge that passes through lamp L2 in 30 minutes. State any formula you use and show your working.
▶️Answer/Explanation
Solution:
1. I = V/R = 12/33 = 0.3636 A (current through one lamp)
2. Time = 30 × 60 = 1800 s
3. Q = I × t = 0.3636 × 1800
4. Charge = 654.5 C
Explanation: Since lamps are in parallel, each receives full voltage. Charge calculation requires current (Ohm’s Law) multiplied by time in seconds.
(b) The air in a car tyre exerts a pressure on the walls of the tyre.
(i) Use ideas about the motion of molecules to describe how the molecules exert a pressure on the walls of the tyre.
▶️Answer/Explanation
Gas molecules move randomly at high speeds, colliding with and rebounding from the tyre walls. Each collision exerts a tiny force, and the sum of countless collisions per second creates measurable pressure.
Key points:
• Continuous random motion
• Elastic collisions (no energy loss)
• Force perpendicular to surface
(ii) State what happens to the pressure of the air in the tyre if the temperature increases.
▶️Answer/Explanation
Pressure increases
Explanation: According to Gay-Lussac’s Law (P ∝ T at constant volume), higher temperature increases molecular kinetic energy, causing more frequent and forceful collisions with the tyre walls.
(c) Hot exhaust gases from the car engine leave the engine through a steel exhaust pipe. The steel exhaust pipe transfers thermal energy through the pipe wall by conduction.
(i) Describe the process of conduction in a solid, using ideas about particle vibration and transfer by electrons.
▶️Answer/Explanation
1. Heat causes metal ions to vibrate more vigorously
2. Vibrations transfer to neighboring ions through the lattice
3. Free/delocalized electrons gain kinetic energy and move rapidly
4. Electrons collide with ions throughout the metal, distributing energy
Key concept: Metals conduct well due to both ionic vibration and mobile electrons.
(ii) When heated, the steel exhaust pipe expands. Explain, in terms of the motion and arrangement of particles, why a solid expands less than a gas when heated.
▶️Answer/Explanation
1. Solids have fixed particle positions with strong intermolecular forces, allowing only increased vibrational amplitude
2. Gases have particles far apart with weak forces, so increased kinetic energy causes significant distance increases
Comparison: Solid expansion ≈ 0.1% per 100°C vs gas expansion ≈ 30% per 100°C at constant pressure
Question 7
(a)-(c): Subtopic: B9.2 Heart
(d)(i)-(ii): Subtopic: B9.4 Blood / B10 Diseases and immunity
Fig. 7.1 shows a cross-section through a human heart.
(a) On Fig. 7.1, use a label line and the letter X to identify the septum.
▶️Answer/Explanation
Label X should point to the muscular wall separating the left and right ventricles.
Explanation: The septum is crucial for preventing oxygenated and deoxygenated blood mixing. It’s thicker on the left side to accommodate higher blood pressure.
(b) The function of the heart is to pump blood around the body. Describe how the heart pumps blood.
▶️Answer/Explanation
1. Atria contract simultaneously, forcing blood into ventricles
2. Ventricles then contract (systole), pumping blood to lungs (right) and body (left)
Key features:
• Valves ensure one-way flow
• Coordinated by the cardiac conduction system
• Left ventricle has thicker myocardium for systemic circulation
(c) Name the two main veins of the heart.
▶️Answer/Explanation
1. Vena cava (superior and inferior)
2. Pulmonary vein
Note: The vena cava returns deoxygenated blood from body; pulmonary veins bring oxygenated blood from lungs. Coronary veins return blood from heart muscle itself.
(d) Fig. 7.2 shows a doctor’s note about a patient. It contains information about the patient’s lifestyle.
Use the information in Fig. 7.2 to answer these questions.
(i) Describe two ways in which this patient could reduce their risk of developing coronary heart disease.
▶️Answer/Explanation
1. Quit smoking (reduces atherosclerosis risk and blood pressure)
2. Increase fruit/vegetable intake (provides antioxidants and reduces LDL cholesterol)
Additional options:
• Replace saturated fats with unsaturated fats
• Maintain regular exercise (already addressed in notes)
(ii) State one non-lifestyle factor that increases this patient’s risk of developing coronary heart disease.
▶️Answer/Explanation
Male gender (or potential genetic predisposition)
Explanation: Males have higher CHD risk until menopause age. Family history would also be relevant but isn’t mentioned in the notes.
Question 8
(a)(i)-(ii): Subtopic: C6.3 Redox
(b)(i)-(ii): Subtopic: C5.1 Exothermic and endothermic reactions
(c)(i)-(iii): Subtopic: C2.7 Metallic bonding / C9.3 Alloys and their properties
(a) The thermite reaction is a redox reaction between aluminium and iron oxide, Fe2O3. It produces molten iron and aluminium oxide, Al2O3.
(i) Write a balanced symbol equation for the thermite reaction.
▶️Answer/Explanation
2Al + Fe2O3 → 2Fe + Al2O3
Explanation:
• Aluminium reduces iron oxide (redox)
• 2:1:2:1 ratio balances all atoms
• Reaction is highly exothermic (reaches ~2500°C)
• Practical use: welding railway tracks
(ii) During the reaction Fe3+ ions become Fe atoms and Al atoms become Al3+ ions. Identify the oxidising agent and the reducing agent. Explain your answer in terms of electron transfer.
▶️Answer/Explanation
Oxidising agent: Fe2O3/Fe3+
Reducing agent: Al
Explanation: Fe3+ gains electrons (reduction), while Al loses electrons (oxidation). Each Al atom donates 3e– to reduce Fe3+.
OIL RIG: Oxidation Is Loss, Reduction Is Gain
(b) Fig. 8.1 shows the energy level diagram for the thermite reaction.
(i) Use the diagram to explain why the reactant mixture must be heated before the reaction starts.
▶️Answer/Explanation
To provide activation energy to initiate the reaction (overcome energy barrier).
Practical note: Typically requires a magnesium fuse burning at 3000°C to start, despite being thermodynamically favorable.
(ii) Use the diagram to explain why the reaction is exothermic.
▶️Answer/Explanation
Products have lower energy than reactants (energy released to surroundings).
Energy calculation: ΔH = -851.5 kJ/mol (highly exothermic)
(c) Steel is an alloy of iron.
(i) Describe the metallic bonding in iron. You may include a labelled diagram in your answer.
▶️Answer/Explanation
1. Lattice of Fe2+ ions
2. Sea of delocalized electrons
Properties explained:
• Conductivity (mobile electrons)
• Malleability (layers slide)
• Strength (strong electrostatic forces)
(ii) State the meaning of the term alloy.
▶️Answer/Explanation
A mixture of a metal with other elements (metals/non-metals) to enhance properties.
Example: Steel = Fe + C (+ Cr/Ni in stainless)
(iii) Suggest two differences in the physical properties of steel and iron.
▶️Answer/Explanation
1. Steel is harder than pure iron (carbon disrupts lattice)
2. Steel less prone to rust (stainless varieties)
Other differences:
• Higher tensile strength
• Lower ductility
• Higher melting point
Question 9
(a)(i)-(iii): Subtopic: P3.4 Sound / P3.1 General properties of waves
(b): Subtopic: P3.2.2 Refraction of light
(c): Subtopic: P5.2.4 Half-life
(a) Ultrasound waves are used in hospitals to scan unborn babies. Ultrasound waves have a frequency that is too high for a human to hear.
(i) State, in terms of waves, what is meant by the term frequency.
▶️Answer/Explanation
Number of complete wave cycles passing a fixed point per second (units: Hertz, Hz).
Example: 2MHz ultrasound = 2,000,000 vibrations/second
(ii) Using your knowledge of the range of audible frequencies for a healthy human ear, suggest a frequency for these ultrasound waves.
▶️Answer/Explanation
>20,000 Hz (typical medical ultrasound range: 2-18 MHz)
Comparison:
• Human hearing range: 20Hz-20kHz
• Diagnostic ultrasound: 2-10MHz
• Therapeutic ultrasound: 1-3MHz
(iii) Ultrasound waves are longitudinal waves. Describe what is meant by a longitudinal wave.
▶️Answer/Explanation
Wave where particle oscillations are parallel to energy transfer direction.
Visualization: Like a slinky being pushed/pulled along its length
(b) Endoscopes are used by doctors in hospitals to observe the inside of a patient. An endoscope uses optical fibres. Complete Fig. 9.1 to show how a ray of light travels down an optical fibre by total internal reflection.
▶️Answer/Explanation
Diagram should show:
1. Light ray entering at angle > critical angle
2. Successive reflections at core-cladding interface
Key parameters:
• Core (n≈1.62) > Cladding (n≈1.52)
• Typical acceptance angle: 50-60°
(c) An isotope of strontium, strontium-89, is used in the treatment of bone cancer in hospitals. Strontium-89 has a half-life of 50 days. A sample of this isotope contains 4 × 1014 atoms. Some time later 3 × 1014 atoms have decayed. Calculate the time needed for this number of atoms to decay. Show your working.
▶️Answer/Explanation
Solution:
1. Remaining atoms = 4×1014 – 3×1014 = 1×1014
2. Decay steps: 4→2→1 (two half-lives)
3. Time = 2 × 50 = 100 days
Medical relevance:
Sr-89 mimics calcium, targeting bone metastases with β-radiation
Question 10
(a)(i)-(ii): Subtopic: B3.1 Diffusion
(b)(i)-(ii): Subtopic: B11 Gas exchange in humans
(a) The diagrams in Fig. 10.1 show two cells C and D. The concentration of carbon dioxide inside and outside the cells is represented by the number of molecules drawn.
(i) Add one arrow to each diagram to show the net movement of carbon dioxide molecules across the cell membrane by diffusion.
▶️Answer/Explanation
Arrows should point:
• Outward from both cells (higher [CO₂] inside)
• Longer arrow for Cell D (steeper gradient)
Biological context: CO₂ is a waste product of respiration that diffuses out along concentration gradients
(ii) State which cell has the greater rate of diffusion. Give a reason for your answer.
▶️Answer/Explanation
Cell D
Reason: Steeper concentration gradient (greater difference between internal/external CO₂)
Fick’s Law: Diffusion rate ∝ (C₁-C₂)/distance
(b) Humans excrete carbon dioxide.
(i) Define the term excretion.
▶️Answer/Explanation
Removal of:
1. Metabolic waste products (CO₂, urea, etc.)
2. Toxic substances
3. Excess materials
from the body
Key point: Distinguish from egestion (undigested food)
(ii) Describe the pathway of carbon dioxide from the blood to the atmosphere.
▶️Answer/Explanation
1. Diffuses from blood capillaries into alveoli
2. Passes through bronchioles → bronchi → trachea
3. Expelled via nose/mouth during exhalation
Key structures:
• Alveolar membrane (0.5-2μm thick)
• 300 million alveoli (70m² surface area)
• Partial pressure gradient: Blood pCO₂ > Alveolar pCO₂
Question 11
(a)-(d): Subtopic: C6.2 Rate of reaction / C3.3 The mole and the Avogadro constant
The raw materials needed to make sulfuric acid in the Contact process are air, sulfur and water. Fig. 11.1 shows the stages in the Contact process.
(a) In stage 2, sulfur dioxide reacts with oxygen to make sulfur trioxide. The equation for this reaction is
2SO2 + O2 ⇌ 2SO3
State the meaning of the ⇌ symbol.
▶️Answer/Explanation
Reversible reaction
Industrial significance: The equilibrium allows optimal SO3 production at 450°C with vanadium(V) oxide catalyst
(b) Fig. 11.2 shows the relationship between the temperature of stage 2 and the percentage of sulfur dioxide converted to sulfur trioxide.
(i) Suggest the temperature in stage 2 of the Contact process.
▶️Answer/Explanation
450°C
Trade-off: Balances acceptable conversion rate (95-98%) with practical reaction speed
(ii) Use Fig. 11.2 to suggest one advantage, other than cost, of using a low temperature in stage 2.
▶️Answer/Explanation
Higher percentage conversion of SO2 to SO3
Le Chatelier’s Principle: Exothermic forward reaction favors lower temperatures
(iii) State and explain why a low temperature is not used in stage 2. Explain your answer in terms of particle movement in stage 2.
▶️Answer/Explanation
1. Reduced reaction rate at low temperatures
2. Particles have insufficient kinetic energy
3. Fewer successful collisions per second
Industrial compromise: 450°C optimizes both conversion and speed
(c) Compound X, H2S2O7, is formed in stage 3. Name compound X.
▶️Answer/Explanation
Oleum (or pyrosulfuric acid)
Reaction: SO3 + H2SO4 → H2S2O7
(d) The overall equation for the Contact process is:
raw materials → product
2S + 3O2 + 2H2O → 2H2SO4
Complete steps 1 to 4 to calculate the mass of sulfuric acid made from 1000 g of sulfur. Show your working.
[Ar: H,1; O,16; S,32]
Step 1
Calculate the number of moles in 1000 g of sulfur.
Step 2
Deduce the number of moles of sulfuric acid made from 1000 g of sulfur.
Step 3
Calculate the relative molecular mass, Mr, of sulfuric acid.
Step 4
Calculate the mass of sulfuric acid made from 1000 g of sulfur.
▶️Answer/Explanation
Step 1: n(S) = 1000/32 = 31.25 mol
Step 2: n(H2SO4) = 31.25 mol (1:1 ratio)
Step 3: Mr = (2×1) + 32 + (4×16) = 98
Step 4: mass = 31.25 × 98 = 3062.5 g
Industrial yield: Actual plants achieve ~98% of theoretical yield
Question 12
(a): Subtopic: P1.7 Pressure
(b)-(d): Subtopic: P3.3 Electromagnetic spectrum / P4.5.1 Electromagnetic induction
(a) Fig. 12.1 shows a large snow tractor used by scientists working in the Arctic region.
The snow tractor has large continuous tracks (caterpillar tracks), driven by the wheels. These tracks allow the snow tractor to travel across the soft snow without sinking. A tractor with four ordinary wheels would sink into the soft snow. Use ideas about pressure to explain this difference.
▶️Answer/Explanation
1. Tracks distribute weight over larger surface area
2. Reduces pressure (P = F/A) to prevent sinking
Calculation example:
• Wheeled tractor: ~150 kPa pressure
• Tracked vehicle: ~15 kPa pressure
Snow physics: Snow strength ≈ 10-20 kPa
(b) The snow tractor has two headlamps. The headlamps emit visible light of several different wavelengths. One of the wavelengths is 5.01 × 10-7 m. The frequency of this light is 5.98 × 1014 Hz. Calculate the speed of this light. Show your working.
▶️Answer/Explanation
Solution:
v = f × λ
= (5.98 × 1014) × (5.01 × 10-7)
= 2.996 × 108 m/s ≈ 3.00 × 108 m/s
Color: This wavelength (501 nm) appears green to human eyes
(c) Visible light is part of the electromagnetic spectrum. All electromagnetic waves travel at the same speed in a vacuum. State one other property that is the same for all electromagnetic waves.
▶️Answer/Explanation
They are all transverse waves
Alternative answers:
• Can travel through vacuum
• Transfer energy
• Obey wave equation (v = fλ)
(d) Fig. 12.2 shows equipment for measuring wind speed used by Arctic scientists.
The wind makes the plastic cups move and this causes the spindle and magnet to turn. Suggest why an alternating voltage is measured on the voltmeter.
▶️Answer/Explanation
1. Rotating magnet creates changing magnetic field
2. Induces alternating emf in coil (Faraday’s Law)
3. Voltage polarity reverses with each half-rotation
Technical detail: Frequency proportional to wind speed (typical range 1-100Hz)
Question 13
(a)(i)-(ii): Subtopic: B19.1 Habitat destruction
(b): Subtopic: B6.1 Photosynthesis / B4 Biological molecules
(a) Excess use of fertilisers can cause eutrophication. Eutrophication eventually causes organisms in a lake to die.
(i) Explain why producers underneath the surface of the water die.
▶️Answer/Explanation
1. Fertilizer runoff increases nitrate/phosphate levels
2. Causes algal blooms on surface (excessive producer growth)
3. Blocks sunlight penetration to deeper water
4. Submerged plants cannot photosynthesize
Timeframe: This process typically occurs over weeks/months
(ii) Explain why the death of producers eventually leads to the death of the fish in the lake.
▶️Answer/Explanation
1. Dead producers decompose (by bacteria/fungi)
2. Decomposition consumes dissolved oxygen
3. Creates hypoxic/anoxic conditions
4. Fish suffocate from oxygen depletion
Measurement: Oxygen levels below 2mg/L are lethal for most fish
(b) Table 13.1 shows some information about two mineral ions found in fertilisers.
| mineral ion | function of mineral ion | effect a deficiency in the mineral ion has on plant |
|---|---|---|
| magnesium | yellow leaves | |
| making amino acids |
Complete Table 13.1.
▶️Answer/Explanation
| mineral ion | function of mineral ion | effect of deficiency |
|---|---|---|
| magnesium | chlorophyll production | yellow leaves (chlorosis) |
| nitrate | making amino acids | stunted growth |
Key roles:
• Mg2+ is central atom in chlorophyll molecule
• NO3– provides nitrogen for proteins/DNA