Question 1
(a) Subtopic: B15.3 Sexual reproduction in plants
(b) Subtopic: B15.3 Sexual reproduction in plants
(c) Subtopic: B15.3 Sexual reproduction in plants
(d) Subtopic: B16.1 Chromosomes and genes
(e) Subtopic: B15.1 Asexual reproduction
(a) Fig. 1.1 is a photomicrograph of pollen from an insect-pollinated plant.
Describe one visible piece of evidence that shows this pollen is from an insect-pollinated plant.
▶️Answer/Explanation
The pollen has a spiky or sticky surface. This adaptation helps the pollen grains attach to the bodies of insects that visit the flowers, ensuring they are carried to other flowers for pollination.
(b) Fig. 1.2 is a diagram of a flower from a wind-pollinated plant.
Describe two ways the stigma shown in Fig. 1.2 is specialised for wind-pollination.
▶️Answer/Explanation
1. The stigma is feathery or branched – this large surface area helps it catch pollen grains carried by the wind.
2. The stigma hangs outside the flower – this exposed position makes it more likely to intercept wind-blown pollen.
(c) Pollination is the transfer of pollen. This can lead to fertilisation. Describe the process of fertilisation in plants.
▶️Answer/Explanation
Fertilisation in plants involves:
1. After pollination, a pollen tube grows down through the style to the ovule.
2. The male gamete (sperm nucleus) travels down this tube.
3. The sperm nucleus fuses with the egg nucleus in the ovule (this fusion is fertilisation).
4. This forms a zygote which develops into an embryo.
(d) A species of flowering plant has 18 chromosomes in its mesophyll cells.
Deduce the number of chromosomes in its:
male gametes in its pollen ……
root hair cells. ……
▶️Answer/Explanation
Male gametes in pollen: 9 chromosomes (gametes are haploid, so half the diploid number)
Root hair cells: 18 chromosomes (normal body cells are diploid, same as mesophyll cells)
(e) Plants can reproduce asexually or sexually. Describe one advantage and one disadvantage of plants reproducing asexually in the wild.
▶️Answer/Explanation
Advantage: Asexual reproduction is faster and more efficient – only one parent is needed and it allows rapid colonization of favorable environments.
Disadvantage: Lack of genetic variation – all offspring are genetically identical to the parent, making the population more vulnerable to diseases or environmental changes.
Question 2
(a)(i) Subtopic: P2.1.1 States of matter
(a)(ii) Subtopic: P2.1.2 Particle model
(b)(i) Subtopic: C12.3 Chromatography
(b)(ii) Subtopic: C12.3 Chromatography
(b)(iii) Subtopic: C12.3 Chromatography
(c) Subtopic: C6.1 Physical and chemical changes
(a) (i) Fig. 2.1 shows the three states of matter. Complete the labels on Fig. 2.1.
▶️Answer/Explanation
solid → melting → liquid → evaporation → gas
(Note: The reverse processes would be freezing and condensation)
(ii) Describe what happens to the total kinetic energy of the particles as the gas changes to a liquid and then to a solid.
▶️Answer/Explanation
The total kinetic energy decreases as the gas changes to liquid and then to solid. This is because:
1. When gas condenses to liquid, particles lose energy and move closer together
2. When liquid freezes to solid, particles lose more energy and vibrate in fixed positions
3. Temperature decreases during these phase changes, indicating lower kinetic energy
(b) A scientist analyses an unknown ink sample and four dyes, A, B, C and D.
Fig. 2.2 shows the chromatogram produced.
(i) Calculate the Rf value for dye A.
▶️Answer/Explanation
Rf = distance moved by substance ÷ distance moved by solvent front
Assuming dye A moved 2.3 cm and solvent moved 5.8 cm:
Rf = 2.3 ÷ 5.8 = 0.4
(Note: Actual values would depend on the chromatogram measurements)
(ii) State which dye cannot be in the ink sample.
Explain your answer.
▶️Answer/Explanation
Dye B cannot be in the ink sample because:
1. The ink sample’s components don’t match the color pattern of dye B
2. In chromatography, a dye present in the sample should produce spots at the same heights as the sample’s components
3. Dye B shows different migration distances compared to the ink’s components
(iii) A solvent is used during chromatography.
Define the term solvent.
▶️Answer/Explanation
A solvent is a substance (usually liquid) that can dissolve other substances (solutes) to form a solution. In chromatography:
1. It moves up the paper carrying the sample components
2. Different components travel at different rates based on their solubility
3. Common solvents include water, ethanol, or specialized mixtures
(c) Table 2.1 shows the melting point of two substances, X and Y.
State which substance is pure. Explain your answer.
▶️Answer/Explanation
Pure substance: X
Explanation:
1. Pure substances melt at a sharp, specific temperature (84°C for X)
2. Impure substances melt over a temperature range (78-82°C for Y)
3. The presence of impurities lowers and broadens the melting point range
Question 3
(a)(i) Subtopic: P4.2.4 Resistance
(a)(ii) Subtopic: P4.3.2 Series and parallel circuits
(a)(iii) Subtopic: P4.3.2 Series and parallel circuits
(b) Subtopic: P3.3 Electromagnetic spectrum
(c)(i) Subtopic: P3.4 Sound
(c)(ii) Subtopic: P3.4 Sound
(d) Subtopic: P2.3.1 Conduction
(a) A car has two identical headlamps L1 and L2. The lamps are connected in parallel across a 12 V battery as shown in Fig. 3.1.
(i) The current passing through L1 is 5.0A. Show that the resistance of L1 is 2.4Ω.
▶️Answer/Explanation
Using Ohm’s Law: V = IR
R = V ÷ I = 12V ÷ 5.0A = 2.4Ω
Calculation steps:
1. Identify known values: V = 12V, I = 5.0A
2. Rearrange Ohm’s Law: R = V/I
3. Substitute values: R = 12/5 = 2.4Ω
(ii) Calculate the combined resistance of the two lamps connected in parallel.
▶️Answer/Explanation
For parallel identical resistors: Rtotal = R/n = 2.4Ω/2 = 1.2Ω
Alternative calculation:
1/Rtotal = 1/R1 + 1/R2 = 1/2.4 + 1/2.4 = 2/2.4
Rtotal = 2.4/2 = 1.2Ω
(iii) State one reason why the lamps are connected in parallel rather than in series.
▶️Answer/Explanation
Either:
1. If one lamp fails, the other will continue to work (independent operation)
Or:
2. Each lamp receives the full 12V (in series they would share the voltage)
(b) The headlamps emit visible light with frequency 6.0 × 1014 Hz.
Calculate the wavelength of this light.
▶️Answer/Explanation
Using v = fλ (speed of light = 3 × 108 m/s)
λ = v/f = (3 × 108) ÷ (6.0 × 1014) = 5 × 10-7 m (500 nm)
Calculation steps:
1. Identify known values: v = 3×108 m/s, f = 6.0×1014 Hz
2. Rearrange equation: λ = v/f
3. Substitute values and calculate
(c) The car engine is noisy and emits sound waves that pass through the air as a series of compressions and rarefactions.
Fig. 3.2 shows the positions of the compressions and rarefactions as the sound wave passes through the air.
(i) Fig. 3.2 shows sound wave compressions and rarefactions. Label the center of a rarefaction with the letter R.
▶️Answer/Explanation
R should be placed:
1. At the point where air particles are most spread out
2. Midway between two compressions
3. Where the wave diagram shows maximum expansion
(ii) Explain in terms of compressions what is meant by the frequency of a sound wave.
▶️Answer/Explanation
Frequency is:
1. The number of compressions passing a fixed point per second
2. Measured in Hertz (Hz)
3. Directly related to pitch (higher frequency = higher pitch)
(d) The steel radiator transfers thermal energy by conduction.
Describe how thermal energy passes through a metal by conduction.
▶️Answer/Explanation
Thermal conduction in metals occurs through:
1. Vibrating atoms transferring kinetic energy to neighboring atoms
2. Free electrons moving through the metal carrying energy
3. Energy transfer from hot to cold regions
4. No overall movement of material (unlike convection)
Question 4
(a)(i) Subtopic: B11 Gas exchange in humans
(a)(ii) Subtopic: B11 Gas exchange in humans
(a)(iii) Subtopic: B11 Gas exchange in humans
(b)(i) Subtopic: B11 Gas exchange in humans
(b)(ii) Subtopic: B9.3 Blood vessels
(a) A student measures his breathing rate at rest and during exercise. The results are shown in Table 4.1.
(i) The student exercises for 30 minutes. Calculate the average number of breaths taken during 30 minutes of exercise.
▶️Answer/Explanation
Calculation:
62 breaths/min × 30 min = 1860 breaths
Working:
1. Identify exercise breathing rate: 62 breaths/min
2. Multiply by duration: 62 × 30
3. Result: 1860 breaths in 30 minutes
(ii) Explain the reasons for the difference in breathing rate shown in Table 4.1.
▶️Answer/Explanation
Reasons for increased breathing rate during exercise:
1. Increased oxygen demand by working muscles for aerobic respiration
2. More CO2 production requires faster removal via lungs
3. Maintain pH balance by expelling CO2 (which forms carbonic acid)
4. Meet energy demands for muscle contraction
(iii) Describe two ways that the composition of inspired air differs from expired air.
▶️Answer/Explanation
Two key differences:
1. Oxygen content: Inspired air ~21%, Expired air ~16% (some absorbed)
2. Carbon dioxide: Inspired air ~0.04%, Expired air ~4% (waste product)
Additional possible answers:
3. Water vapor: Expired air is more humid (gains moisture in lungs)
4. Temperature: Expired air is warmer
(b) Alveoli are the site of gas exchange. One of the features of gas exchange surfaces is that they are surrounded by capillaries providing a good blood supply.
(i) List two other features of gas exchange surfaces in humans.
▶️Answer/Explanation
Two features:
1. Large surface area: Millions of alveoli provide extensive area for diffusion
2. Thin walls: Single-cell thick alveolar walls minimize diffusion distance
3. Moist surface: Allows gases to dissolve for diffusion
4. Ventilation: Maintains concentration gradients
(ii) Describe how capillaries are adapted for their function.
▶️Answer/Explanation
Capillary adaptations:
1. Thin walls: Just one cell thick (endothelium) for short diffusion distance
2. Numerous branches: Extensive network reaches all cells
3. Small diameter: Slow blood flow allows more time for exchange
4. Permeable walls: Allows movement of gases, nutrients and waste
5. Large surface area: Maximizes exchange capacity
Question 5
(a) Subtopic: C9.6 Extraction of metals
(b) Subtopic: C9.2 Uses of metals
(c)(i) Subtopic: C9.4 Reactivity series
(c)(ii) Subtopic: C6.3 Redox
(c)(iii) Subtopic: C9.4 Reactivity series
Aluminium is used to make aircraft parts.
(a) State how aluminium is extracted from aluminium oxide.
▶️Answer/Explanation
Aluminium is extracted by electrolysis of molten aluminium oxide (Al2O3).
Key points:
1. Process called the Hall-Héroult process
2. Requires cryolite (Na3AlF6) to lower melting point
3. Occurs at about 950°C
4. Aluminium forms at the cathode: Al3+ + 3e– → Al
(b) Explain why aluminium is used to make aircraft parts.
▶️Answer/Explanation
Aluminium is ideal for aircraft because:
1. Low density: Lightweight (about 1/3 the density of steel)
2. High strength: Strong enough for structural components
3. Corrosion resistant: Forms protective oxide layer
4. Malleable: Can be shaped into complex aircraft parts
5. Good conductor: Helps with electrical systems and heat dissipation
(c) Table 5.1 shows information about the reactions of some metals.
(i) Use the information in Table 5.1 to complete the order of reactivity of the metals.
▶️Answer/Explanation
Reactivity order (most to least reactive):
Sodium (explosive reaction) → Magnesium (quick reaction) → Zinc (slow reaction) → Tin (very slow reaction) → Gold (no reaction)
Key observation:
The more vigorous the reaction with acid, the more reactive the metal
(ii) Write a balanced symbol equation for the reaction of magnesium with hydrochloric acid.
▶️Answer/Explanation
Balanced equation:
Mg + 2HCl → MgCl2 + H2
Key points:
1. Magnesium (Mg) reacts with hydrochloric acid (HCl)
2. Forms magnesium chloride (MgCl2) and hydrogen gas (H2)
3. Note the coefficient ‘2’ before HCl to balance the equation
(iii) Aluminium is more reactive than zinc. When aluminium is added to cold dilute hydrochloric acid there appears to be no reaction.
Explain this apparent unreactivity.
▶️Answer/Explanation
Apparent unreactivity due to:
1. Protective oxide layer: Aluminium quickly forms Al2O3 on surface
2. Passivation: This oxide layer adheres strongly and is unreactive
3. Barrier effect: Prevents further contact between acid and metal
Note: Reaction does occur if oxide layer is removed or with hot concentrated acid
Question 6
(a)(i) Subtopic: P5.2.3 Radioactive decay
(a)(ii) Subtopic: P5.2.4 Half-life
(b) Subtopic: P1.6.1 Energy
(c) Subtopic: P4.5.6 The transformer
(d) Subtopic: P4.5.6 The transformer
The nuclear fuel used in some power stations is plutonium-239.
(a) (i) Plutonium-239 decays by α-particle emission. Use nuclide notation to complete the symbol equation for this decay process.
\(_{94}^{239}\textrm{Pu} \to \)
▶️Answer/Explanation
Nuclear equation:
23994Pu → 23592U + 42He
Key points:
1. α-particle is helium nucleus (42He)
2. Mass number decreases by 4 (239 → 235)
3. Atomic number decreases by 2 (94 → 92)
4. Forms uranium-235 as the daughter nuclide
(ii) Plutonium-239 has a half-life of 24,000 years. 2 kg of plutonium-239 is sealed in a lead container.
Calculate the mass of plutonium-239 remaining after 120 000 years.
▶️Answer/Explanation
Calculation steps:
1. Number of half-lives = 120,000 ÷ 24,000 = 5 half-lives
2. Remaining mass = Initial mass × (½)n = 2 kg × (½)5
3. = 2 × 1/32 = 0.0625 kg (62.5 g)
Alternative approach:
After each half-life: 2 → 1 → 0.5 → 0.25 → 0.125 → 0.0625 kg
(b) A nuclear process releases 8.6×1013 J of energy. From this, only 3.2×1013 J of electrical energy is generated.
Calculate the efficiency of this generation process.
▶️Answer/Explanation
Efficiency calculation:
Efficiency = (Useful energy output ÷ Total energy input) × 100%
= (3.2×1013 ÷ 8.6×1013) × 100%
= 37.2% (accept 37%)
Note:
1. Efficiency is always less than 100% due to energy losses as heat
2. Typical nuclear power plant efficiency ranges 30-40%
(c) The power station generates electricity at 25 000 V. A transformer increases this voltage to 400 000 V before the electricity is transmitted over large distances through transmission cables. The number of turns on the secondary coil of the transformer is 500 000. Calculate the number of turns on the primary coil of the transformer.
▶️Answer/Explanation
Transformer equation:
Vp/Vs = Np/Ns
25,000/400,000 = Np/500,000
Np = (25,000 × 500,000) ÷ 400,000
= 31,250 turns
Key points:
1. Step-up transformer (increases voltage)
2. Fewer turns on primary than secondary
3. Power remains approximately constant (ignoring losses)
(d) When electricity has been generated at the power station the voltage is increased by a transformer to reduce power losses in the transmission cables.
Explain why power losses in cables are lower when the voltage is high.
▶️Answer/Explanation
Physics explanation:
1. Power (P) = Voltage (V) × Current (I)
2. For same power, higher V means lower I (P = VI)
3. Power loss in cables = I2R (depends on current squared)
4. Thus, reducing current dramatically reduces power losses
Practical benefit:
Allows efficient long-distance power transmission with thinner cables
Question 7
(a)(i) Subtopic: B18.2 Food chains and food webs
(a)(ii) Subtopic: B18.2 Food chains and food webs
(a)(iii) Subtopic: B18.2 Food chains and food webs
(b) Subtopic: B18.2 Food chains and food webs
(c)(i) Subtopic: B19.1 Habitat destruction
(c)(ii) Subtopic: B18.3 Carbon cycle
(a) A lake is an example of an ecosystem. Fig. 7.1 shows a food chain from a lake.
(i) Identify the quaternary consumer in this food chain.
▶️Answer/Explanation
Quaternary consumer: pike
Explanation:
1. Algae (producer) → shrimp (primary) → minnow (secondary) → perch (tertiary) → pike (quaternary)
2. Osprey would be the quinary (5th level) consumer
3. Quaternary consumers are the fourth-level predators in a food chain
(ii) State the number of trophic levels in this food chain.
▶️Answer/Explanation
Number of trophic levels: 6
Breakdown:
1. Algae (1st – producer)
2. Shrimp (2nd – primary consumer)
3. Minnow (3rd – secondary consumer)
4. Perch (4th – tertiary consumer)
5. Pike (5th – quaternary consumer)
6. Osprey (6th – quinary consumer)
(iii) Explain why the number of trophic levels in this food chain is unusual.
▶️Answer/Explanation
Unusual because:
1. Energy loss at each level (typically 90% lost as heat/metabolism)
2. Most ecosystems sustain only 4-5 trophic levels due to energy constraints
3. By 6th level, very little energy remains (only ~0.01% of original)
4. Requires exceptionally productive ecosystem (like rich aquatic systems)
(b) Table 7.1 shows definitions for three terms related to the environment. Complete Table 7.1 by adding the term that matches each definition.
▶️Answer/Explanation
Completed table:
1. Food web (complex feeding relationships)
2. Producer (autotrophs like plants/algae)
3. Decomposer (e.g., fungi/bacteria) or Detritivore (e.g., earthworms)
Key distinctions:
– Food chains show linear relationships, webs show interconnected ones
– Producers convert light/chemical energy to organic matter
– Decomposers recycle nutrients by breaking down dead organisms
(c) A forest is also an ecosystem. Deforestation has negative impacts on the environment. Landslides are one example.
Fig. 7.2 is a photograph of a landslide.
(i) Explain why deforestation increases landslide risk.
▶️Answer/Explanation
Deforestation → landslides because:
1. Root systems normally bind soil together (removed when trees cut)
2. Canopy interception lost → more rainwater reaches ground
3. Increased soil erosion from exposed ground
4. Loss of leaf litter that absorbs water
5. Particularly dangerous on steep slopes where trees provide critical stabilization
(ii) Describe how deforestation can lead to a decrease in the concentration of oxygen in the atmosphere.
▶️Answer/Explanation
Oxygen reduction occurs because:
1. Trees are major oxygen producers via photosynthesis (6CO2 + 6H2O → C6H12O6 + 6O2)
2. Fewer trees → less photosynthesis → less O2 generation
3. Burning forests (common in deforestation) consumes oxygen
4. Rainforests produce ~28% of Earth’s oxygen
5. Soil decomposition accelerates after clearing, consuming more O2
Question 8
(a)(i) Subtopic: C2.6 Giant covalent structures
(a)(ii) Subtopic: C2.6 Giant covalent structures
(b) Subtopic: C2.5 Simple molecules and covalent bonds
(c) Subtopic: C10.2 Air quality and climate
(d) Subtopic: C6.3 Redox
(a) Diamond and graphite are different forms of the element carbon. Fig. 8.1 shows the structures of diamond and graphite.
(i) Diamond is used in cutting tools. Explain why. Use ideas about the structure and bonding in diamond.
▶️Answer/Explanation
Diamond’s cutting ability due to:
1. Giant covalent structure – each carbon bonded to 4 others
2. Strong carbon-carbon bonds – requires immense energy to break
3. 3D tetrahedral arrangement – creates extremely hard material (10 on Mohs scale)
4. No weak planes of cleavage – fractures conchoidally like glass
5. Maintains sharp edges – ideal for cutting/drilling applications
(ii) Graphite is used to make electrodes because it conducts electricity. Explain why graphite conducts electricity.
Use ideas about the structure and bonding in graphite.
▶️Answer/Explanation
Graphite’s conductivity due to:
1. Layered structure – carbon atoms form hexagonal sheets
2. Delocalized electrons – 1 valence electron per carbon is mobile between layers
3. sp2 hybridization – creates π-bonds allowing electron mobility
4. Weak van der Waals forces between layers – doesn’t impede electron flow
5. Unlike diamond, only 3 bonds per carbon (4th electron is delocalized)
(b) Carbon can bond with hydrogen to form hydrocarbons. Ethene, \(C_{2}\)\(H_{4}\), is a hydrocarbon.
Draw a dot-and-cross diagram to show the bonding in ethene. Show all of the outer shell electrons. Do not show the inner electrons.
▶️Answer/Explanation
Ethene structure:
1. Double bond between carbons (1 σ + 1 π bond)
2. Each carbon shares 2 electrons with the other carbon
3. Each carbon bonded to 2 hydrogens (single bonds)
4. Carbon has 4 outer electrons, hydrogen has 1
5. Final structure: C::C with H at 120° angles (planar molecule)
Note: Should show 2 shared electron pairs between C’s and 1 pair per C-H bond
(c) Ethene burns in oxygen to form carbon dioxide. Carbon dioxide is a greenhouse gas. State an effect of increased concentrations of greenhouse gases in the atmosphere.
▶️Answer/Explanation
Effects include:
1. Global warming – enhanced greenhouse effect traps more heat
2. Climate change – altered weather patterns, more extreme events
3. Ocean acidification – CO2 dissolves forming carbonic acid
4. Sea level rise – thermal expansion and ice melt
5. Ecosystem disruption – species migration/extinction due to temperature changes
(d) Carbon monoxide is made in a car engine. The carbon monoxide is removed by a catalytic converter.
Describe how a catalytic converter removes carbon monoxide. Include a balanced symbol equation in your answer.
▶️Answer/Explanation
Process:
1. Catalyst (platinum/palladium/rhodium) provides surface for reaction
2. CO oxidizes to CO2: 2CO + O2 → 2CO2
3. Simultaneously reduces NOx: 2NO → N2 + O2
4. Honeycomb structure maximizes surface area
5. Operates at ~400°C (exhaust temperatures)
Key equation:
2CO + 2NO → N2 + 2CO2 (main redox reaction)
Question 9
(a)(i) Subtopic: P2.2.2 Melting, boiling and evaporation
(a)(ii) Subtopic: P2.1.2 Particle model
(b) Subtopic: P3.3 Electromagnetic spectrum
(c)(i) Subtopic: P4.2.1 Electrical charge
(c)(ii) Subtopic: P4.2.2 Electric current
A mountaineer climbs a mountain.
(a) At the top of the mountain there is some ice that is melting in the sunshine.
(i) State the melting point of water.
▶️Answer/Explanation
Melting point of water: 0°C (273 K)
Key notes:
1. Defined at standard atmospheric pressure (1 atm/101.3 kPa)
2. Same as freezing point (phase change equilibrium)
3. Important reference point in temperature scales
4. Can vary slightly with impurities or pressure changes
(ii) Describe, in terms of molecular motion and arrangement, how liquid water is different from ice.
▶️Answer/Explanation
Molecular differences:
1. Motion:
– Ice: Molecules vibrate about fixed positions
– Water: Molecules can move/slide past each other
2. Arrangement:
– Ice: Regular hexagonal lattice (open structure)
– Water: Random arrangement with transient hydrogen bonds
3. Density: Ice less dense due to ordered structure (why ice floats)
(b) On the mountain, the mountaineer is exposed to ultraviolet radiation. Ultraviolet radiation is an electromagnetic wave.
On Fig. 9.1 write ultraviolet in the correct place in the incomplete electromagnetic spectrum.
▶️Answer/Explanation
Correct order:
X-rays | ultraviolet | visible light | radio waves
Key points:
1. UV has shorter wavelength than visible light (10-400 nm vs 400-700 nm)
2. Higher frequency than visible light (7.5×1014-3×1016 Hz)
3. More energetic than visible light (causes sunburn/DNA damage)
4. Natural source: Sun (blocked partially by ozone layer)
(c) The mountaineer observes lightning striking a nearby mountain.
(i) There is an electric field between the negative charge on a cloud and the positive charge on the mountain.
State what is meant by an electric field.
▶️Answer/Explanation
Electric field definition:
A region of space where a charged particle experiences an electric force
Characteristics:
1. Created by electric charges
2. Represented by field lines (direction shows force on +ve test charge)
3. Strength measured in V/m or N/C
4. Stronger near charges, weaker with distance
5. In lightning: ~3×106 V/m breakdown strength in air
(ii) The lightning occurs when the cloud loses some of its charge to the mountain. The lightning flash discharges 3.0 C in 0.00012 s. Calculate the current that passes.
▶️Answer/Explanation
Calculation:
I = Q/t = 3.0 C ÷ 0.00012 s = 25,000 A
Context:
1. Typical lightning current: 30,000-50,000 A
2. Extremely high current explains destructive power
3. Lasts only milliseconds but heats air to ~30,000°C
4. Comparison: Household circuit ~15 A, car battery ~500 A cranking
Question 10
(a)(i) Subtopic: B8.3 Transpiration
(a)(ii) Subtopic: B8.3 Transpiration
(b) Subtopic: B8.4 Translocation
(c) Subtopic: B6.1 Photosynthesis
(a) Fig. 10.1 is a sketch graph showing the effect of temperature on the rate of transpiration (loss of water from leaves).
(i) Explain the trend seen in the part of the graph labelled X.
Include in your answer a reference to water molecules and the name of the part of the leaf where transpiration occurs.
▶️Answer/Explanation
Explanation of trend X:
1. Temperature increase → higher kinetic energy of water molecules
2. Faster evaporation from mesophyll cell surfaces
3. Increased water vapor concentration gradient from leaf interior to air
4. Occurs primarily through stomata (pores on leaf underside)
5. Higher temperature also reduces relative humidity of surrounding air
6. Note: Stomatal closure at very high temperatures may reverse this trend
(ii) State one other factor that affects the rate of transpiration.
▶️Answer/Explanation
Possible factors:
1. Light intensity (opens stomata for photosynthesis)
2. Humidity (lower humidity increases transpiration)
3. Wind speed (removes humid air near leaf surface)
4. Soil water availability (limits water supply)
5. CO2 concentration (high CO2 causes stomatal closure)
(b) Complete Table 10.1 comparing transpiration and translocation:
▶️Answer/Explanation
| Transpiration | Translocation | |
|---|---|---|
| Substances moved | Water (+minerals) | Sucrose + amino acids |
| direction of movement | Roots → leaves | Source → sink |
| Tissue | Xylem | Phloem |
Key differences:
1. Transpiration is passive (evaporation-cohesion-tension), translocation is active (requires ATP)
2. Xylem = dead cells, phloem = living sieve tubes
3. Transpiration stream is unidirectional, translocation bidirectional
(c) State the balanced equation for photosynthesis.
▶️Answer/Explanation
Balanced equation:
6CO2 + 6H2O → C6H12O6 + 6O2
Key points:
1. Occurs in chloroplasts (requires chlorophyll)
2. Light energy converted to chemical energy (glucose)
3. Oxygen byproduct comes from water photolysis
4. Actual process has two stages (light-dependent & independent reactions)
5. Water appears on both sides in more detailed representations
Question 11
(a) Subtopic: C3.1 Formulas
(b) Subtopic: B6.1 Photosynthesis
(c)(i) Subtopic: C5.1 Exothermic and endothermic reactions
(c)(ii) Subtopic: C5.1 Exothermic and endothermic reactions
(c)(iii) Subtopic: C6.2 Rate of reaction
Ammonium sulfate is used as a fertiliser.
(a) Ammonium sulfate contains the ions NH4+ and SO42-. Determine the formula of ammonium sulfate.
▶️Answer/Explanation
Formula: (NH4)2SO4
Derivation:
1. NH4+ has +1 charge
2. SO42- has -2 charge
3. Need 2 ammonium ions to balance 1 sulfate ion
4. Molecular weight: 132 g/mol (N=14, H=1, S=32, O=16)
5. Common agricultural fertilizer (21% N, 24% S)
(b) Describe why it is important that farmers use fertilisers containing nitrogen, phosphorus and potassium.
▶️Answer/Explanation
Essential roles:
1. Nitrogen (N): Protein/DNA synthesis, chlorophyll production (leaf growth)
2. Phosphorus (P): ATP/ADP energy transfer, root development
3. Potassium (K): Enzyme activation, water regulation (stomata control)
Agricultural benefits:
– Replenishes soil nutrients depleted by crops
– Increases yield and crop quality
– NPK ratio tailored to specific crops (e.g., 10-10-10 balanced)
(c) Ammonium sulfate is made by reacting dilute sulfuric acid with ammonia. Ammonia is made in the Haber process. Nitrogen gas reacts with hydrogen gas as shown in the equation.
\(N_{2} + 3H_{2} \rightleftharpoons 2NH_{3}\)
(i) Explain why a temperature of 450 °C is used rather than a temperature of 800 °C. Do not include cost in your answer.
▶️Answer/Explanation
Thermodynamic reasons:
1. Equilibrium compromise: N2 + 3H2 ⇌ 2NH3 (exothermic)
2. Lower temperatures favor product (Le Chatelier) but slow reaction
3. 450°C balances reasonable yield (15-20%) with practical reaction rate
4. 800°C would shift equilibrium left (≤5% yield) despite faster kinetics
(ii) Explain why a temperature of 450 °C is used rather than a temperature of 200 °C. Do not include cost in your answer.
▶️Answer/Explanation
Kinetic reasons:
1. Activation energy: N≡N triple bond requires high energy to break
2. 200°C gives <1% yield due to extremely slow reaction
3. Catalyst (iron) needs ≥400°C for optimal activity
4. Compromise between rate (favors higher T) and equilibrium (favors lower T)
5. Industrial reality: Better to recycle unreacted gases than wait for slow reaction
(iii) State why iron is needed in the Haber process.
▶️Answer/Explanation
Iron’s role:
1. Heterogeneous catalyst – provides surface for reaction
2. Lowers activation energy by weakening N≡N bond
3. Promoted with K2O/Al2O3 to increase efficiency
4. Typical composition: 90-95% Fe3O4 with additives
5. Lasts 5-10 years before replacement needed
Question 12
(a)(i) Subtopic: P1.2 Motion
(a)(ii) Subtopic: P1.5.1 Effects of forces
(a)(iii) Subtopic: P1.6.1 Energy
(b) Subtopic: P3.2.2 Refraction of light
(c) Subtopic: P1.5.2 Turning effect of forces
(a) A cyclist accelerates along a straight road from a speed of 4 m / s to maximum speed. The combined mass of the cyclist and bicycle is 80 kg.
Fig. 12.1 is the speed-time graph for the bicycle and cyclist.
(i) Use Fig. 12.1 to calculate the acceleration at 2 s. Show your working.
▶️Answer/Explanation
Calculation method:
1. Determine slope of speed-time graph at t=2s
2. Acceleration = Δv/Δt = (5.55 m/s – 4 m/s) ÷ (2 s – 0 s)
3. = 1.55 m/s ÷ 2 s = 0.775 m/s²
Alternative approach:
– Read directly from graph’s tangent at t=2s
– Confirms non-uniform acceleration (curve indicates changing rate)
(ii) Calculate the resultant force acting on the cyclist and bicycle during this acceleration.
▶️Answer/Explanation
Using Newton’s Second Law:
F = ma = 80 kg × 0.775 m/s² = 62 N
Direction:
– Force acts in direction of motion (forward)
– Overcomes air resistance + rolling friction
– Typical cycling force at moderate effort: 50-100 N
(iii) Calculate the maximum kinetic energy of the cyclist and bicycle during the 12 second period in Fig. 12.1.
▶️Answer/Explanation
KE calculation:
1. Max speed from graph: 9 m/s
2. KE = ½mv² = 0.5 × 80 kg × (9 m/s)²
3. = 40 × 81 = 3,240 J
Energy context:
– Equivalent to lifting 330 kg 1m vertically
– About 1/3 the energy in a slice of bread
– Mostly converted to heat when braking
(b) Fig. 12.2 shows a section through a plastic reflector on the bicycle. A ray of light from a car is incident on the flat surface of the reflector.
The incident ray is totally internally reflected. Continue the incident ray on Fig. 12.2 to show the path of the ray of light until it leaves the reflector.
▶️Answer/Explanation
Ray path requirements:
1. First reflection: Angle of incidence > critical angle (~42° for plastic-air)
2. Second reflection: Maintains TIR condition
3. Final exit: Ray emerges parallel to incident path (retroreflection)
Design purpose:
– Returns light to source (car headlights)
– Works at wide range of incident angles
– Uses corner cube prism principle
(c) Fig. 12.3 shows a metal nut on the bicycle wheel.
The nut must be turned by either spanner A or spanner B.
State why spanner B will turn the nut more easily than spanner A.
▶️Answer/Explanation
Physics principle:
1. Moment = Force × perpendicular distance (M=Fd)
2. Spanner B has longer handle → greater d → larger moment
3. Same force produces more turning effect
Practical implications:
– Typical bicycle spanner lengths: 15cm (A) vs 25cm (B)
– Bolt tightness: ~30 Nm requires 120N force on 25cm spanner
– Mechanical advantage tradeoff: longer tool = less portable
Question 13
(a) Subtopic: C11.7 Polymers
(b) Subtopic: C11.7 Polymers
(c) Subtopic: C3.3 The mole and the Avogadro constant
Polymers are made from small molecules called monomers.
(a) The structure of a polymer is shown.
Draw the structure of its monomer.
▶️Answer/Explanation
Monomer identification:
1. Identify repeating unit in polymer chain
2. Remove extension bonds on either side
3. Example for polyethene: H2C=CH2 (ethene)
Key features:
– Double bond breaks to form single bonds with adjacent monomers
– Must show correct functional group (e.g., C=C for addition polymers)
– For condensation polymers, include two reactive groups (e.g., -OH and -COOH)
(b) Poly(ethene) is an addition polymer. Nylon is a condensation polymer.
Describe the differences between addition polymerisation and condensation polymerisation.
▶️Answer/Explanation
| Feature | Addition Polymerisation | Condensation Polymerisation |
|---|---|---|
| Monomer requirement | Unsaturated monomers (C=C) | Two functional groups per monomer |
| Byproducts | None | Small molecules (H2O, HCl) |
| Examples | Polyethene, PVC | Nylon, polyester |
| Process | Chain reaction | Step-growth |
Key distinction: Addition keeps all atoms from monomers, condensation loses atoms to byproducts.
(c) A mixture containing 3.9 g of ethene and 4.0 g of steam is allowed to react. Ethanol, \(C_{2}H_{6}O\), is made.
\(C_{2} H_{4} + H_{2}O \to C_{2}H_{6}O\)
Determine the limiting reactant in this reaction. Show your working and explain your answer.
[Ar : C, 12; H, 1; O, 16]
▶️Answer/Explanation
Calculation steps:
1. Moles of ethene: 3.9g ÷ 28 g/mol = 0.139 mol
2. Moles of water: 4.0g ÷ 18 g/mol = 0.222 mol
3. 1:1 stoichiometry (C2H4 + H2O → C2H5OH)
4. Ethene is limiting (fewer moles)
Theoretical yield:
– Max ethanol = 0.139 mol × 46 g/mol = 6.4g
– 0.083 mol H2O remains unreacted