Question 1
(a) Subtopic: B12 Respiration
(b) Subtopic: B12 Respiration
(c)(i) Subtopic: B12 Respiration
(c)(ii) Subtopic: B12 Respiration
(c)(iii) Subtopic: B12 Respiration
(a) Complete the sentence to define the term anaerobic respiration.
Anaerobic respiration is the …… reactions in cells that break down nutrient molecules to release …… without using …… .
▶️Answer/Explanation
Answer: chemical ; energy ; oxygen
Explanation: Anaerobic respiration is defined as the chemical reactions in cells that break down glucose molecules to release energy without using oxygen. This process is less efficient than aerobic respiration but allows cells to produce energy when oxygen is scarce, such as during intense exercise.
(b) Describe one advantage of using aerobic respiration rather than anaerobic respiration.
▶️Answer/Explanation
Answer: release more energy (per glucose molecule) / does not produce lactic acid
Explanation: Aerobic respiration is more advantageous because it produces significantly more ATP (36-38 molecules per glucose) compared to anaerobic respiration (only 2 ATP molecules). Additionally, aerobic respiration completely oxidizes glucose to carbon dioxide and water, while anaerobic respiration produces lactic acid as a byproduct which can cause muscle fatigue and pain.
(c) Fig. 1.1 shows a scientist monitoring the oxygen consumption of an athlete running on a treadmill.
Oxygen consumption is the volume of oxygen used by the body per minute.
The graph in Fig. 1.2 shows the oxygen consumption of the athlete before, during and after exercise.
(i) Explain the results between 2-6 minutes.
▶️Answer/Explanation
Answer: increased breathing rate or depth ; increased respiration ; increased need for energy ; increased muscle contraction
Explanation: Between 2-6 minutes, the oxygen consumption increases sharply because: 1. The athlete’s muscles require more energy for contraction during exercise 2. The breathing rate and depth increase to supply more oxygen to the muscles 3. The heart rate increases to deliver oxygenated blood to muscles faster 4. Cellular respiration increases to meet the higher energy demands This represents the body’s response to increased metabolic demands during physical activity.
(ii) Calculate the length of time it takes for the oxygen consumption of the athlete to return to the resting rate after exercise stops.
▶️Answer/Explanation
Answer: 5 (minutes)
Explanation: From the graph, when exercise stops (at about 11 minutes), the oxygen consumption gradually decreases until it reaches the resting rate at approximately 16 minutes. The difference is 16 – 11 = 5 minutes. This is called the recovery time or oxygen debt repayment period.
(iii) Explain why oxygen consumption does not return to resting rate immediately after exercise stops.
▶️Answer/Explanation
Answer: anaerobic respiration has taken place ; needs to repay oxygen debt / to remove lactic acid
Explanation: The elevated oxygen consumption continues after exercise because: 1. The body needs to oxidize the lactic acid that accumulated during anaerobic respiration 2. Muscle cells need to replenish their oxygen stores (myoglobin) 3. The body needs to restore ATP and creatine phosphate levels 4. The body temperature and breathing/heart rates take time to return to normal This continued oxygen demand is known as “excess post-exercise oxygen consumption” (EPOC) or oxygen debt.
Question 2
(a) Subtopic: C10.2 Air quality and climate
(b)(i) Subtopic: C2.5 Simple molecules and covalent bonds
(b)(ii) Subtopic: C2.5 Simple molecules and covalent bonds
(c)(i) Subtopic: C2.4 Ions and ionic bonds
(c)(ii) Subtopic: C2.4 Ions and ionic bonds
(c)(iii) Subtopic: C3.1 Formulas
(a) Fig. 2.1 is a pie chart which shows the composition of clean air.
Complete Fig. 2.1 to show the percentages of oxygen and of nitrogen in clean air.
▶️Answer/Explanation
Answer: oxygen: 21 ; nitrogen: 78
Explanation: Clean dry air at sea level consists approximately of: – Nitrogen (N₂): 78% by volume – Oxygen (O₂): 21% by volume – Other gases (Argon, CO₂, etc.): 1% combined These percentages remain relatively constant in the troposphere. The high nitrogen content makes air relatively inert, while the oxygen percentage is optimal for respiration and combustion.
(b) Fig. 2.2 shows the electronic structure of a nitrogen atom.
(i) Draw the dot-and-cross diagram to represent the bonding in a nitrogen molecule, N₂. Include only outer shell electrons.
▶️Answer/Explanation
Answer: 3 bonding pairs ; 2 lone pairs, all else correct
Explanation: The dot-and-cross diagram for N₂ should show: – A triple bond between the two nitrogen atoms (3 shared electron pairs) – 1 lone pair of electrons on each nitrogen atom – Correct electron arrangement (2 electrons in inner shell, 5 in outer shell) The triple bond (N≡N) is one of the strongest covalent bonds, with a bond energy of 945 kJ/mol, making nitrogen gas very stable.
(ii) Explain why nitrogen molecules are much less reactive than nitrogen atoms.
▶️Answer/Explanation
Answer: (molecule is unreactive because of) strong (covalent) bonding between atoms / triple bond ; (atom is reactive because) atom has high tendency to gain electrons / has incomplete electron shell / does not have noble gas structure
Explanation: Nitrogen molecules (N₂) are unreactive because: 1. The triple covalent bond requires substantial energy to break (945 kJ/mol) 2. The molecule has achieved stable noble gas configuration Whereas nitrogen atoms: 1. Have 5 valence electrons and seek to gain 3 more to complete their octet 2. Are highly electronegative (3.04 on Pauling scale) 3. Readily form compounds to achieve stability (e.g., ammonia, nitrates)
(c) At high temperature nitrogen reacts with magnesium to form magnesium nitride.
(i) Explain why magnesium nitride is an ionic compound, but nitrogen atoms are covalently bonded in nitrogen molecules.
▶️Answer/Explanation
Answer: ionic bonds between metallic and non-metallic elements ; covalent bonds between non-metallic elements / nitrogen is a non-metal (so covalently bonded)
Explanation: Magnesium nitride (Mg₃N₂) is ionic because: 1. Magnesium is a metal that loses electrons to form Mg²⁺ cations 2. Nitrogen is a non-metal that gains electrons to form N³⁻ anions 3. The electrostatic attraction between oppositely charged ions forms ionic bonds Whereas N₂ molecules are covalent because: 1. Both nitrogen atoms are non-metals 2. They share electrons to achieve stable octets 3. The electronegativity difference is zero (pure covalent)
(ii) The melting point of magnesium nitride is very high. Explain why ionic compounds have high melting points.
▶️Answer/Explanation
Answer: attractive force between oppositely charged ions / strong force / bonds between ions / many forces / bonds between ions ; more energy required to overcome force / break bond
Explanation: Ionic compounds like magnesium nitride have high melting points because: 1. Strong electrostatic forces exist between oppositely charged ions in a 3D lattice 2. These ionic bonds require substantial energy (typically 600-3000°C) to overcome 3. The lattice energy is high due to the small size and high charge of Mg²⁺ and N³⁻ 4. All ions must be separated simultaneously to melt the compound
(iii) Magnesium nitride contains magnesium ions, Mg²⁺ and nitride ions, N³⁻. Deduce the formula of magnesium nitride. Explain your answer.
▶️Answer/Explanation
Answer: Mg₃N₂ ; idea of balanced charges
Explanation: The formula is determined by balancing charges: 1. Each Mg²⁺ has +2 charge 2. Each N³⁻ has -3 charge 3. Lowest common multiple of charges is 6 (2×3) 4. Therefore need 3 Mg²⁺ (total +6) to balance 2 N³⁻ (total -6) 5. Resulting formula: Mg₃N₂ This maintains electrical neutrality in the crystal lattice.
Question 3
(a)(i) Subtopic: P2.3.2 Convection
(a)(ii) Subtopic: P2.3.2 Convection
(a)(iii) Subtopic: P2.3.1 Conduction
(b)(i) Subtopic: P4.3.2 Series and parallel circuits
(b)(ii) Subtopic: P4.3.2 Series and parallel circuits
(c)(i) Subtopic: P2.2.2 Melting, boiling and evaporation
(c)(ii) Subtopic: P2.1.2 Particle model
(a) Fig. 3.1 shows a potato being baked in the oven of an electric cooker.
The potato has a metal skewer (a long metal pin) pushed through it.
(i) The air in the oven is heated by the heating element. On Fig. 3.1 draw an arrow to show how the heated air moves inside the oven.
▶️Answer/Explanation
Answer: arrow showing heated air rising
Explanation: Heated air moves upward because: 1. When air is heated, it expands and becomes less dense 2. The warmer, less dense air rises due to buoyancy forces 3. This creates a convection current where hot air rises and cooler air sinks 4. The arrow should point upward from the heating element toward the potato
(ii) Name the main method of thermal energy transfer in the heated air.
▶️Answer/Explanation
Answer: convection
Explanation: Convection is the primary heat transfer method because: 1. It involves the bulk movement of heated air molecules 2. The circulating air currents carry thermal energy throughout the oven 3. This is more efficient than conduction in gases where molecules are far apart 4. Radiation also occurs but convection dominates the air heating process
(iii) The metal skewer transfers heat to the inside of the potato by conduction. Describe the process of conduction in a solid, using ideas about particle vibration.
▶️Answer/Explanation
Answer: particles vibrate more / gain energy ; this vibration is passed through metal
Detailed Explanation: Conduction in metals occurs through: 1. Free electrons near the heat source gain kinetic energy 2. These electrons move rapidly through the metal lattice 3. They collide with metal ions and other electrons, transferring energy 4. Metal ions vibrate more vigorously at hotter regions 5. These vibrations are passed to adjacent ions through the lattice 6. No overall movement of material occurs, just energy transfer Metals are excellent conductors due to their delocalized electrons.
(b) The cooker has four electric hotplates, A, B, C and D. These are each connected in parallel with each other across the mains voltage supply.
Each hotplate is operated by a separate switch. Each hotplate has a resistance of 60 Ω.
Fig. 3.2 shows the circuit. The hotplates are represented by resistor symbols.
(i) Show that the combined resistance of hotplates A and B connected in parallel is 30Ω.
▶️Answer/Explanation
Answer: use of correct formula ; evidence of correct working
Calculation: For parallel resistors: 1/Rtotal = 1/R1 + 1/R2 Given RA = RB = 60Ω 1/Rtotal = 1/60 + 1/60 = 2/60 = 1/30 Therefore Rtotal = 30Ω
This shows that identical resistors in parallel halve the total resistance.
(ii) State the combined resistance of hotplates C and D connected in parallel.
▶️Answer/Explanation
Answer: 30 Ω
Explanation: Since all hotplates have identical resistance (60Ω): 1. The parallel combination of any two will follow the same calculation 2. C and D in parallel will have the same combined resistance as A and B 3. The resistance is halved because current has two equal paths to flow through 4. This maintains consistency with the parallel resistance formula
(c) A saucepan containing water is placed on a hotplate.
When the temperature of the water reaches the boiling point of water, the water boils.
(i) State the boiling point of water.
▶️Answer/Explanation
Answer: 100 (°C)
Note: This is at standard atmospheric pressure (1 atm or 101.3 kPa). The boiling point varies with pressure – it decreases at higher altitudes where pressure is lower.
(ii) When the water boils, the liquid water turns into steam. Describe the differences between liquid water and steam in terms of:
- the forces and distances between the molecules
- the motion of the molecules.
▶️Answer/Explanation
Answer: stronger forces of attraction between water molecules in liquid ; water molecules are closer together in liquid ; water molecules have greater freedom of movement / collide less frequently in steam
Detailed Comparison:
| Property | Liquid Water | Steam |
|---|---|---|
| Intermolecular forces | Strong hydrogen bonds | Negligible forces |
| Molecular spacing | Close together | Far apart |
| Molecular motion | Vibrate and slide past neighbors | Move freely at high speed |
| Density | High (~1 g/cm³) | Low (~0.0006 g/cm³) |
| Volume | Fixed | Expands to fill container |
Question 4
(a) Subtopic: B13.1 Coordination and response
(b) Subtopic: B13.1 Coordination and response
(c) Subtopic: B13.1 Coordination and response
Fig. 4.1 shows a section through the eye.
(a) Draw an X on Fig. 4.1 to show the position of the blind spot.
▶️Answer/Explanation
Answer: centre of the letter X drawn in correct position
Explanation: The blind spot is located: – Where the optic nerve leaves the eye (optic disc) – Approximately 15-20° from the fovea (central vision point) – Contains no photoreceptors (rods or cones) – Normally we don’t notice it because the brain “fills in” the missing information
(b) Table 4.1 shows some of the parts of the eye, the letters of the parts in Fig. 4.1 and their functions.
Complete Table 4.1. One row has been done for you.
▶️Answer/Explanation
Answer:
| Part of the eye | Letter in Fig. 4.1 | Function |
|---|---|---|
| cornea | D | refracts light entering the eye |
| iris | C | controls how much light enters, the eye / pupil |
| lens | B | focuses light (on retina) |
| optic nerve | H | carries impulse to brain |
| retina | A | contains light receptor cells |
Key Functions: 1. Cornea – Primary refractive surface (fixed focus) 2. Iris – Controls pupil size (like a camera aperture) 3. Lens – Provides adjustable focus (accommodation) 4. Optic nerve – Transmits visual information 5. Retina – Contains rods (low light) and cones (color)
(c) A person changes their focus from a near object to a distant object. Describe the changes that take place in the following parts of the eye.
- ciliary muscle
- suspensory ligaments
- lens
▶️Answer/Explanation
Answer: ciliary muscle relaxes ; suspensory ligaments tighten / stretch ; lens becomes thinner / is stretched
Detailed Process: When focusing on distant objects: 1. Ciliary muscles relax (circular muscles around lens) 2. Suspensory ligaments become taut (like little strings) 3. Lens becomes thinner and flatter (reduced convexity) 4. This decreases refractive power (less bending of light) 5. Light rays from distant objects (nearly parallel) focus precisely on retina
| Structure | Near Vision | Distant Vision |
|---|---|---|
| Ciliary muscle | Contracted | Relaxed |
| Suspensory ligaments | Loose | Taut |
| Lens shape | Thick/convex | Thin/flat |
| Refractive power | Strong | Weak |
Question 5
(a) Subtopic: C10.1 Water
(b) Subtopic: C5.1 Exothermic and endothermic reactions
(c)(i) Subtopic: C7.1 The characteristic properties of acids and bases
(c)(ii) Subtopic: C7.1 The characteristic properties of acids and bases
(d) Subtopic: C7.2 Oxides
(e) Subtopic: C9.6 Extraction of metals
(f) Subtopic: C9.6 Extraction of metals
(a) Limestone consists mainly of calcium carbonate. Explain why farmers use limestone to improve soil.
▶️Answer/Explanation
Answer: reduces acidity
Detailed Explanation: Farmers use limestone (CaCO₃) because: 1. It neutralizes acidic soils by reacting with hydrogen ions: CaCO₃ + 2H⁺ → Ca²⁺ + CO₂ + H₂O 2. Maintains optimal pH (6.0-7.0) for most crops 3. Provides essential calcium for plant cell walls 4. Improves soil structure and microbial activity 5. Enhances nutrient availability (e.g., phosphorus)
(b) Calcium carbonate is heated to produce calcium oxide. The equation for this reaction is shown.
CaCO₃ → CaO + CO₂
Complete the following sentence using words from the list, Each word may be used once, more than once or not at all.
chemical nuclear oxidation reduction thermal
The reaction is endothermic because …… energy is taken in by the reactant to produce products with greater …… energy.
▶️Answer/Explanation
Answer: thermal ; chemical
Explanation: 1. Thermal energy (heat) is required to break the CaCO₃ bonds (ΔH = +178 kJ/mol) 2. The products have greater chemical potential energy in their bonds 3. This is a decomposition reaction requiring continuous heat input 4. The reverse reaction (CaO + CO₂ → CaCO₃) is exothermic
(c) (i) Complete the general equation for the reaction between an acid and a base.
▶️Answer/Explanation
Answer: → salt + water
Example: HCl + NaOH → NaCl + H₂O
This neutralization reaction always produces a salt (ionic compound) and water.
(ii) Explain how protons (H⁺) are involved in the reaction between an acid and a base.
▶️Answer/Explanation
Answer: protons / H⁺ transferred from acid to base
Mechanism: 1. Acids donate H⁺ ions (Brønsted-Lowry definition) 2. Bases accept H⁺ ions 3. In water: H⁺ + OH⁻ → H₂O 4. The reaction is essentially proton transfer
(d) An alkali is a base dissolved in water. A student investigates the reactions of four oxides with acids and alkalis. Table 5.1 shows his results.
He correctly classifies sulfur dioxide as an acidic oxide. Classify the other oxides by completing Table 5.1.
▶️Answer/Explanation
Answer:
| Oxide | Type of Oxide |
|---|---|
| aluminium oxide | amphoteric |
| copper oxide | basic |
| nitrous oxide | neutral |
| sulfur dioxide | acidic oxide |
Key Characteristics: 1. Amphoteric: Reacts with both acids and bases (Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O) 2. Basic: Metal oxides that react with acids only (CuO + H₂SO₄ → CuSO₄ + H₂O) 3. Neutral: No reaction with acids or bases (N₂O) 4. Acidic: Non-metal oxides that react with bases (SO₂ + 2NaOH → Na₂SO₃ + H₂O)
(e) Copper metal can be extracted from copper oxide, CuO, by heating with carbon. Carbon dioxide is also formed. Write the balanced equation for this reaction. Include all of the state symbols.
▶️Answer/Explanation
Answer: 2CuO(s) + C(s) → 2Cu(s) + CO₂(g)
Explanation: 1. Redox reaction where carbon reduces copper oxide 2. Copper changes from +2 to 0 oxidation state 3. Carbon changes from 0 to +4 oxidation state 4. Typical extraction method for less reactive metals (below carbon in reactivity series)
(f) Explain why calcium cannot be extracted from calcium oxide by heating with carbon.
▶️Answer/Explanation
Answer: calcium more reactive than carbon
Detailed Explanation: 1. Calcium is above carbon in the reactivity series 2. Carbon cannot reduce calcium oxide (CaO + C → no reaction) 3. Calcium has stronger tendency to form positive ions than carbon 4. Requires electrolysis for extraction (more energy intensive method) 5. Carbon can only reduce metals below it in the series (e.g., Zn, Fe, Cu)
Question 6
(a) Subtopic: P4.1 Simple phenomena of magnetism
(b)(i) Subtopic: C2.3 Isotopes
(b)(ii) Subtopic: P5.2.3 Radioactive decay
(c) Subtopic: P4.2.4 Resistance
(d) Subtopic: P1.7 Pressure
(a) An iron magnet picks up two iron nails as shown in Fig. 6.1.
Explain why the nails do not hang vertically.
▶️Answer/Explanation
Answer: ref. to induced magnetism (in) nails ; two nail heads / north poles / like poles, will repel each other
Detailed Explanation: 1. The magnet induces temporary magnetism in the iron nails 2. Each nail becomes a magnet with: – North pole at end nearest the magnet’s south pole – South pole at end nearest the magnet’s north pole 3. The two adjacent nail heads become like poles (both north or both south) 4. Like magnetic poles repel each other (N-N or S-S repulsion) 5. This causes the nails to splay outward rather than hang straight down
(b) An isotope of iron has a nuclide notation 6026Fe and decays by beta particle emission to an isotope of cobalt.
(i) State what is meant by the term isotope.
▶️Answer/Explanation
Answer: atoms having same atomic number / proton number and different mass number / neutron number
Key Characteristics: 1. Same element (same number of protons) 2. Different numbers of neutrons 3. Same chemical properties (same electron configuration) 4. Different physical properties (density, radioactivity, etc.) 5. Example: 56Fe, 57Fe, and 58Fe are all stable iron isotopes
(ii) Use nuclide notation to complete the symbol equation for this β-decay process.
▶️Answer/Explanation
Answer: 6026Fe → 6027Co + 0-1e
Beta Decay Process: 1. A neutron converts to a proton + electron (β⁻ particle) 2. Atomic number increases by 1 (Fe 26 → Co 27) 3. Mass number remains same (60) 4. The emitted electron is the beta particle (0-1e) 5. Typical half-life for 60Fe is 2.6 million years
(c) An iron wire of length 0.50 m has a cross sectional area of 4.0 × 10-5 m2 and a resistance of 1.21 × 10-3 Ω.
Calculate the resistance of an iron wire of length 0.25 m that has a cross sectional area of 8.0 × 10-5 m2.
▶️Answer/Explanation
Answer: 3.0 × 10-4 Ω
Calculation Steps: Using R = ρL/A (where ρ is resistivity, constant for iron): 1. Original wire: R₁ = ρ(0.50)/(4.0×10-5) = 1.21×10-3 Ω 2. New wire has: – Half the length (0.25 m → decreases resistance) – Double the area (8.0×10-5 m² → decreases resistance) 3. Resistance changes by factor of (½)/(2) = ¼ 4. New resistance R₂ = ¼ × 1.21×10-3 = 3.025×10-4 Ω 5. Round to 2 significant figures: 3.0 × 10-4 Ω
(d) A block of iron is on a bench. The surface of the block of iron in contact with the bench has an area of 144 cm2. The mass of the block of iron is 13.6 kg. Calculate the pressure exerted by the block of iron on the bench in N/cm2. gravitational field strength = 10 N / kg
▶️Answer/Explanation
Answer: 0.94 N/cm2
Calculation: 1. Weight = mass × g = 13.6 kg × 10 N/kg = 136 N 2. Contact area = 144 cm2 3. Pressure = Force/Area = 136 N / 144 cm2 4. = 0.944… N/cm2 5. Round to 2 significant figures: 0.94 N/cm2 Key Points: – Pressure is force per unit area – 1 N/cm2 = 10,000 Pa (SI units) – Iron’s density (7.87 g/cm3) means this block has volume ≈ 1727 cm3
Question 7
(a)(i) Subtopic: B5 Enzymes
(a)(ii) Subtopic: B7.3 Digestion
(a)(iii) Subtopic: B5 Enzymes
(b)(i) Subtopic: B7.2 Digestive system
(b)(ii) Subtopic: B7.3 Digestion
(a) Lipase is a digestive enzyme. A student investigates the action of lipase on fats present in milk.
He uses an indicator that turns pink in alkaline solutions and colourless in acidic solutions.
- He adds a few drops of indicator to a test-tube containing milk. The indicator remains colourless.
- He adds sodium carbonate solution to the milk. The indicator now turns pink.
- He then adds lipase to the mixture.
- He times how long it takes the indicator to go colourless.
Fig. 7.1 shows the apparatus.
(i) Suggest why sodium carbonate is added to the mixture.
(ii) It takes 240 seconds for the mixture to turn colourless and become acidic. Explain why the mixture becomes acidic.
(iii) The investigation is repeated at a temperature of 80 °C. Explain the effect this would have on the result.
▶️Answer/Explanation
(i) Why sodium carbonate is added:
To make the solution alkaline / so the solution changes color when acid forms
(ii) Why mixture becomes acidic:
Lipase breaks down fats → fatty acids + glycerol
Fatty acids lower pH (acidic conditions)
(iii) Effect at 80°C:
1. Enzyme denatures (loses 3D structure)
2. Active site changes shape
3. No enzyme-substrate complexes form
4. Reaction stops (remains pink)
(b) Fig. 7.2 shows the alimentary canal and associated organs.
(i) Use the letters in Fig. 7.2 to identify:
- two structures where amylase is secreted
- a structure with very acidic conditions.
(ii) State two reasons why acidic conditions are needed in parts of the digestive system.
▶️Answer/Explanation
(i) Amylase secretion sites:
1. Salivary glands (mouth)
2. Pancreas (secretes into small intestine)
Acidic structure:
Stomach (pH 1-2 due to HCl)
(ii) Why acidic conditions are needed:
1. Activates pepsin enzyme
2. Kills pathogens
3. Provides optimal pH for protein digestion
Question 8
(a) Subtopic: C11.3 Fuels
(b)(i) Subtopic: C6.3 Redox
(b)(ii) Subtopic: C10.2 Air quality and climate
(b)(iii) Subtopic: C6.2 Rate of reaction
(c)(i) Subtopic: C11.4 Alkanes
(c)(ii) Subtopic: C11.4 Alkanes
A petrol engine in a car uses a mixture of air and gasoline. Gasoline is a mixture of hydrocarbons.
(a) State the products of complete combustion of a hydrocarbon.
▶️Answer/Explanation
Answer: carbon dioxide ; water
Complete Combustion Reaction:
CxHy + (x + y/4)O2 → xCO2 + (y/2)H2O
Key Points: 1. Requires sufficient oxygen supply 2. Example for octane (C8H18):
2C8H18 + 25O2 → 16CO2 + 18H2O 3. Energy released as heat (exothermic) 4. Clean combustion produces only CO2 and H2O
(b) When a petrol engine burns hydrocarbons, harmful products carbon monoxide and nitrogen monoxide form. Cars have catalytic converters which change harmful gases into less harmful gases.
Fig. 8.1 shows the position of the catalytic converter in a car.
(i) In a catalytic converter carbon monoxide is oxidised. State the product of this oxidation.
▶️Answer/Explanation
Answer: carbon dioxide
Oxidation Reaction:
2CO + O2 → 2CO2
Catalyst Details: 1. Platinum/palladium/rhodium catalysts 2. Honeycomb structure maximizes surface area 3. Operating temperature: 400-600°C 4. Removes 90% of CO emissions
(ii) In a catalytic converter nitrogen monoxide (NO) is reduced to nitrogen. State one harmful effect of nitrogen oxides on the environment.
▶️Answer/Explanation
Answer: acid rain / respiratory disease / smog formation / ozone depletion
Environmental Impacts: 1. Acid Rain: NOx + H2O → HNO3 (nitric acid) 2. Respiratory Issues: Lung irritation, asthma exacerbation 3. Photochemical Smog: NOx + VOCs + sunlight → ozone + PANs 4. Eutrophication: Excess nitrogen in ecosystems
(iii) A catalyst works by reducing activation energy in a reaction. Explain what is meant by the activation energy.
▶️Answer/Explanation
Answer: minimum energy for particles/reactants to react
Detailed Explanation: 1. Energy barrier between reactants and products 2. Represented as peak in reaction profile diagrams 3. Catalysts provide alternative reaction pathway 4. Typical values: 50-100 kJ/mol for many reactions 5. Does not affect thermodynamics (ΔH) of reaction
(c) Octane and butane are in the same homologous series.
(i) Describe what is meant by homologous series.
▶️Answer/Explanation
Answer: same general formula ; similar chemical properties ; gradual change in physical properties
Characteristics of Homologous Series: 1. Alkanes: CnH2n+2 (e.g., methane, ethane, propane) 2. Each member differs by CH2 unit 3. Similar chemical reactivity 4. Gradual change in: – Boiling points – Viscosity – Flammability 5. Same functional group
(ii) Complete Fig. 8.2 to show the structure of a butane molecule.
▶️Answer/Explanation
Answer:
H H H H
| | | |
H-C-C-C-C-H
| | | |
H H H H
Butane Properties: 1. Molecular formula: C4H10 2. Straight-chain alkane 3. Boiling point: -1°C to 1°C 4. Used in lighters and camping gas 5. Isomer: methylpropane (branched chain)
Question 9
(a) Subtopic: P1.6.1 Energy
(b) Subtopic: P1.2 Motion
(c)(i) Subtopic: P1.6.2 Work
(c)(ii) Subtopic: P1.6.2 Work
(a) An aircraft has a mass of 400,000 kg. Calculate the kinetic energy of the aircraft when travelling at 50 m/s.
▶️Answer/Explanation
Answer: 500,000 kJ
Calculation:
KE = ½mv²
= ½ × 400,000 kg × (50 m/s)²
= ½ × 400,000 × 2,500
= 500,000,000 J = 500,000 kJ
Key Points: 1. Kinetic energy depends on mass and velocity squared 2. 1 kJ = 1,000 J 3. At this speed (180 km/h), the aircraft would be taking off 4. For comparison: 500,000 kJ ≈ energy from burning 12 kg of jet fuel
(b) The pilot says the velocity is 50 m/s. The co-pilot says the speed is 50 m/s. State the difference between velocity and speed.
▶️Answer/Explanation
Answer: velocity has direction but speed does not
Detailed Comparison:
| Property | Speed | Velocity |
|---|---|---|
| Definition | Distance/time | Displacement/time |
| Direction | Not specified | Specified |
| Type | Scalar quantity | Vector quantity |
| Example | 50 m/s | 50 m/s NE |
(c) Fig. 9.1 shows an aircraft passenger pulling her suitcase.
The passenger pulls the suitcase with a horizontal force of 15 N for 150 m.
(i) State the formula that relates force, work done and distance moved.
▶️Answer/Explanation
Answer: work done = force × distance (moved in direction of force)
Formula Details: 1. W = F × d × cosθ (θ = angle between force and displacement) 2. For horizontal pulling (θ=0°), cosθ=1 3. Unit: joule (J) = newton × meter 4. Only the force component in the displacement direction does work
(ii) Calculate the work done on the suitcase by the passenger. State the unit of your answer.
▶️Answer/Explanation
Answer: 2,250 J
Calculation:
Work = Force × Distance
= 15 N × 150 m
= 2,250 J
Question 10
(a)(i) Subtopic: B8.3 Transpiration
(a)(ii) Subtopic: B8.3 Transpiration
(a)(iii) Subtopic: B8.3 Transpiration
(b) Subtopic: B8.1 Xylem and phloem
(c)(i) Subtopic: B16.1 Chromosomes and genes
(c)(ii) Subtopic: B16.2 Cell division
(d) Subtopic: B16.2 Cell division
(e) Subtopic: B16.2 Cell division
(a) A student investigates the effect of temperature on the rate of transpiration. The apparatus in Fig. 10.1 is used to measure the rate of water uptake by a plant shoot. This is approximately equal to the rate of transpiration.
The rate of water uptake is determined by measuring the distance travelled by the air bubble in a set time.
(i) At 30°C the air bubble travels 8 mm in 2 minutes. Calculate the rate of water uptake.
▶️Answer/Explanation
Answer: 4 mm/min
Calculation:
Rate = Distance/Time = 8 mm / 2 min = 4 mm/min
Experimental Notes: 1. The bubble moves because water is absorbed by roots and transpired by leaves 2. This potometer measures the rate of water uptake, which approximates transpiration rate 3. Standard conditions: 30°C, moderate humidity, light breeze
(ii) The student repeats the investigation at different temperatures. Sketch a line on Fig 10.2 to show the expected results.
▶️Answer/Explanation
Graph Characteristics: 1. Curve rises steeply from 10-30°C 2. Peaks around 30-35°C 3. Declines sharply above 40°C
Explanation:
| Temperature Range | Effect on Transpiration |
|---|---|
| 10-30°C | Increased kinetic energy → faster evaporation |
| 30-35°C | Optimum enzyme activity → maximum stomatal opening |
| >40°C | Stomata close → reduced transpiration |
(iii) Suggest a reason for the shape of your graph in (a)(ii).
▶️Answer/Explanation
Key Factors: 1. Low temperatures: Slow molecular motion → limited evaporation 2. Optimum range: – Increased water molecule kinetic energy – Maximum stomatal aperture 3. High temperatures: – Stomatal closure to prevent dehydration – Possible enzyme denaturation
(b) Describe the mechanism that moves water up the xylem.
▶️Answer/Explanation
Cohesion-Tension Theory: 1. Transpiration pull: Water evaporates from leaf mesophyll cells 2. Cohesion: Hydrogen bonds between water molecules 3. Adhesion: Water molecules stick to xylem walls 4. Capillary action: Narrow xylem vessels enhance water rise 5. Root pressure: Additional push from root osmosis
Process Flow: Evaporation → Negative pressure → Continuous water column → Water uptake
(c) Fig. 10.3 shows a photograph of a plant cell undergoing cell division by mitosis.
(i) State where chromosomes are found in a cell.
▶️Answer/Explanation
Answer: in the nucleus
Chromosome Details: 1. Composed of DNA and histone proteins 2. Visible only during cell division 3. Human cells: 46 chromosomes (23 pairs) 4. Plant cells: Varies by species (e.g., wheat has 42)
(ii) Complete the sentence about the chromosomes in cell A:
“The chromosomes on the left-hand side of the cell are genetically …… to the chromosomes on the right-hand side.”
▶️Answer/Explanation
Answer: identical
Mitosis Significance: 1. Produces genetically identical daughter cells 2. Maintains chromosome number (2n → 2n) 3. Ensures equal distribution of genetic material 4. Critical for growth and tissue repair
(d) One of the roles of mitosis is growth. State two other roles of mitosis.
▶️Answer/Explanation
Key Roles: 1. Asexual reproduction: – Binary fission in bacteria – Plant cuttings – Starfish regeneration 2. Tissue repair: – Skin cell replacement – Liver regeneration – Blood cell production
Cell Cycle Facts: – Interphase: 90% of cycle – Mitosis: 10% (Prophase, Metaphase, Anaphase, Telophase)
(e) State how the number of chromosomes in a cell produced by meiosis is different to the number of chromosomes in a cell produced by mitosis.
▶️Answer/Explanation
Answer: half the number of chromosomes (haploid vs diploid)
Comparison Table:
| Feature | Mitosis | Meiosis |
|---|---|---|
| Chromosome number | 2n → 2n | 2n → n |
| Genetic variation | None | Crossing over + independent assortment |
| Number of divisions | 1 | 2 |
| Resulting cells | 2 identical | 4 unique |
Question 11
(a)(i) Subtopic: C8.3 Group VII properties
(a)(ii) Subtopic: C3.2 Relative masses of atoms and molecules
(b)(i) Subtopic: C1.2 Diffusion
(b)(ii) Subtopic: C8.3 Group VII properties
(a) Table 11.1 contains data about some elements.
(i) Predict the color and physical state of fluorine at room temperature. Explain your answers.
▶️Answer/Explanation
Answer: pale yellow / lighter than chlorine ; gas ; use of trend to predict property
Halogen Trends:
| Element | State (RT) | Color | Explanation |
|---|---|---|---|
| Fluorine (F₂) | Gas | Pale yellow | Lightest halogen, weak intermolecular forces |
| Chlorine (Cl₂) | Gas | Greenish-yellow | Trend: Color darkens down group |
| Bromine (Br₂) | Liquid | Red-brown | Increased molecular mass |
| Iodine (I₂) | Solid | Dark grey | Strongest London forces |
(ii) Explain why the relative molecular mass of fluorine is 38. [\(A_{r}\) : F,19]
▶️Answer/Explanation
Answer: fluorine is diatomic / formula is F₂
Calculation: 1. Atomic mass of fluorine (Aᵣ) = 19 2. F₂ molecule contains 2 atoms: 19 × 2 = 38 3. All halogens exist as diatomic molecules (F₂, Cl₂, Br₂, I₂) 4. This explains why bromine (Br₂) has Mᵣ = 160 (80 × 2)
(b) Fig. 11.1 shows a gas jar filled with nitrogen over a gas jar filled with bromine gas. The gases are separated by a glass plate.
The glass plate is removed.
Fig. 11.2 shows the colour changes in the gas jars after 15 minutes and after 30 minutes.
(i) Describe the process that causes the effect in Fig. 11.2. Explain this process in terms of movement of molecules.
▶️Answer/Explanation
Answer: diffusion ; molecules move from high concentration to low concentration / random molecular motion
Detailed Process: 1. Bromine molecules (Br₂) move into nitrogen gas 2. Nitrogen molecules (N₂) move into bromine gas 3. Net movement from high → low concentration 4. Brown color spreads through both jars 5. Eventually reaches uniform concentration
Graham’s Law: Rate ∝ 1/√(molar mass) → Br₂ diffuses slower than N₂
(ii) Predict how the results of the experiment would be different if chlorine is used instead of bromine.
▶️Answer/Explanation
Answer: shorter times / faster diffusion
Scientific Reason: 1. Chlorine (Cl₂) has lower molecular mass (71) vs bromine (160) 2. Lighter molecules move faster at same temperature 3. Diffusion rate inversely proportional to √molecular mass 4. Expected time difference: √(160/71) ≈ 1.5× faster
Question 12
(a)(i) Subtopic: P3.4 Sound
(a)(ii) Subtopic: P3.1 General properties of waves
(a)(iii) Subtopic: P3.1 General properties of waves
(b) Subtopic: P3.2.1 Reflection of light
(c)(i) Subtopic: P4.5.2 The a.c. generator
(c)(ii) Subtopic: P4.5.1 Electromagnetic induction
(a) Ultrasound is very high frequency sound. A submarine uses ultrasound to determine the distance to the sea bed. Pulses of ultrasound are sent out through the water.
The ultrasound pulses reflect off the sea bed and are detected in the submarine 1.2 seconds later. Ultrasound waves travel through water at a speed of 1500 m / s.
(i) Calculate the sea bed depth below the submarine.
▶️Answer/Explanation
Answer: 900 m
Calculation: 1. Total distance = speed × time = 1500 m/s × 1.2 s = 1800 m 2. Depth = ½ × total distance (echo travels down & back) = 1800/2 = 900 m
Key Concepts: – Sonar principle – Echo timing – Depth accuracy ±1% in clear water – Commercial subs typically operate at 300-600m depth
(ii) The wavelength is 5 cm. Show that the frequency is 30,000 Hz.
▶️Answer/Explanation
Proof: v = fλ → f = v/λ = 1500 m/s ÷ 0.05 m = 30,000 Hz
Ultrasound Characteristics: 1. Frequency range: >20 kHz (human hearing limit) 2. Typical submarine sonar: 10-50 kHz 3. Higher frequency → better resolution but less penetration 4. 30 kHz balances range (5-10km) and resolution (~10cm)
(iii) Ultrasound waves travel as a series of compressions and rarefactions. Fig. 12.1 shows the positions of some compressions and rarefactions of an ultrasound wave.
On Fig. 12.1 label a compression with the letter C and a rarefaction with the letter R.
▶️Answer/Explanation
Wave Features: 1. Compression: – High pressure region – Molecules densely packed 2. Rarefaction: – Low pressure region – Molecules spread apart
Sound Wave Properties: – Longitudinal wave – Requires medium (can’t travel in vacuum) – Pressure variations typically 1-10 Pa underwater
(b) Submarines use periscopes to view ships on the surface of the sea. Fig. 12.2 shows an incomplete simple periscope.
On Fig. 12.2, draw:
- two plane mirrors in position so that a ray of light from the ship passing through the periscope will be reflected by both mirrors to the observer’s eye
- the path of this ray of light from the ship, through the periscope, to the observer’s eye.
▶️Answer/Explanation
Mirror Configuration: 1. First mirror: – 45° angle to incoming light – Reflects light downward 2. Second mirror: – 45° angle to vertical light – Reflects light to observer’s eye
Light Path Rules: – Angle of incidence = Angle of reflection – Total internal reflection maintains image clarity – Typical periscope length: 9-15m for submarines
(c) The submarine has a generator to generate electricity. Fig. 12.3 shows a simple generator.
(i) Name the parts of the generator labelled Z on Fig. 12.3.
▶️Answer/Explanation
Answer: slip rings
Generator Components: 1. Slip rings: – Maintain electrical contact with rotating coil – Allow current reversal in AC generators 2. Alternative: Commutator (for DC generators) 3. Modern subs use both AC and DC systems
(ii) Explain why a rotating-coil generator produces an alternating current.
▶️Answer/Explanation
AC Production: 1. Coil cuts magnetic flux lines → induced EMF 2. During 180° rotation: – First 90°: Current flows one direction (max at 90°) – Next 90°: Current reverses direction 3. Complete cycle produces sine wave