Question 1
(a)(i) Subtopic: B12 Respiration
(a)(ii) Subtopic: B12 Respiration
(a)(iii) Subtopic: B12 Respiration
(b) Subtopic: B1.1 Characteristics of living organisms
(a) Respiration releases energy. It can occur aerobically or anaerobically.
(i) State the balanced chemical equation for aerobic respiration.
▶️Answer/Explanation
The balanced chemical equation for aerobic respiration is:
C6H12O6 + 6O2 → 6CO2 + 6H2O
Explanation:
– Glucose (C6H12O6) reacts with oxygen (O2)
– Produces carbon dioxide (CO2) and water (H2O)
– The equation must be balanced with equal numbers of each atom on both sides
(ii) Name the product of anaerobic respiration in muscles.
▶️Answer/Explanation
Lactic acid
Explanation:
– In muscle cells during vigorous exercise when oxygen is limited
– Glucose is broken down without oxygen to produce lactic acid
– This causes muscle fatigue and cramping
(iii) Name the two products of anaerobic respiration in yeast.
▶️Answer/Explanation
1. Ethanol (alcohol)
2. Carbon dioxide
Explanation:
– Yeast performs alcoholic fermentation
– Glucose → ethanol + carbon dioxide + energy
– This process is used in baking (CO2 makes dough rise) and brewing (ethanol production)
(b) Respiration is one of the characteristics of living organisms.
State two other characteristics of living organisms.
▶️Answer/Explanation
Any two from:
1. Movement
2. Reproduction
3. Sensitivity (to stimuli)
4. Growth
5. Excretion
6. Nutrition
Explanation:
These are part of the seven characteristics of life often remembered by the acronym “MRS GREN”:
– Movement
– Respiration
– Sensitivity
– Growth
– Reproduction
– Excretion
– Nutrition
Question 2
(a) Subtopic: C6.1 Physical and chemical changes
(b) Subtopic: C8.3 Group VII properties
(c)(i) Subtopic: B3.1 Diffusion
(c)(ii) Subtopic: B3.1 Diffusion
(c)(iii) Subtopic: C3.2 Relative masses of atoms and molecules
(c)(iv) Subtopic: B3.1 Diffusion
(a) Ammonia, NH3, is made by the reaction between nitrogen gas and hydrogen gas in the Haber process.
Construct the symbol equation for this reaction.
▶️Answer/Explanation
N2 + 3H2 → 2NH3
Explanation:
– Nitrogen gas (N2) reacts with hydrogen gas (H2)
– Produces ammonia (NH3)
– The equation must be balanced:
– 1 nitrogen molecule + 3 hydrogen molecules → 2 ammonia molecules
– This is the industrial Haber process for ammonia synthesis
(b) Identify a substance that displaces ammonia gas from ammonium chloride.
▶️Answer/Explanation
Any named base (e.g., sodium hydroxide, calcium hydroxide, potassium hydroxide)
Explanation:
– Bases displace ammonia from ammonium salts
– Example reaction: NH4Cl + NaOH → NaCl + H2O + NH3
– The ammonia gas can be detected by its pungent smell and turning damp red litmus paper blue
(c) Ammonia gas reacts with hydrogen chloride gas to form solid ammonium chloride.
Fig. 2.1 shows apparatus a teacher uses to demonstrate this reaction.
Ammonia molecules and hydrogen chloride molecules start to diffuse away from the cotton wool plugs at the same time.
The ring of white ammonium chloride forms after 1 minute.
(i) Define the term diffusion.
▶️Answer/Explanation
The (net) movement of particles from a region of their higher concentration to a region of their lower concentration, down a concentration gradient, as a result of their random movement.
Explanation:
– Particles move randomly in all directions
– Net movement occurs from high to low concentration
– Continues until equilibrium is reached (equal concentration throughout)
– A passive process that doesn’t require energy
(ii) The glass tube is 0.9 m long. The speed of each molecule is more than 1 m/s. Suggest why it takes more than 1 minute for the white ring to form.
▶️Answer/Explanation
Collisions between molecules (which slow their progress through the tube)
Explanation:
– Although individual molecules move fast (1 m/s)
– They constantly collide with other molecules and tube walls
– This zig-zag path (Brownian motion) increases the distance traveled
– Therefore the net movement is much slower than molecular speed
(iii) Show that the relative molecular mass of ammonia, NH3, is 17.
[A; H,1; N,14]
▶️Answer/Explanation
14 (N) + (3 × 1) (H) = 17
Explanation:
– Relative atomic mass of nitrogen (N) = 14
– Relative atomic mass of hydrogen (H) = 1
– NH3 contains 1 nitrogen and 3 hydrogen atoms
– Therefore: 14 + (3 × 1) = 17
(iv) The relative molecular mass of hydrogen chloride, HCl, is 36.5.
Explain how this experiment shows that the rate of diffusion depends on molecular mass.
▶️Answer/Explanation
1. White ring forms closer to the HCl end (right of center)
2. Ammonia (lighter at 17) diffuses faster than HCl (heavier at 36.5)
3. Demonstrates Graham’s Law: rate of diffusion ∝ 1/√(molecular mass)
Explanation:
– Lighter molecules diffuse faster than heavier ones at same temperature
– NH3 travels further in same time because its molecules move faster
– The position of the white ring (where they meet) is closer to the HCl end
– This proves diffusion rate is inversely proportional to square root of molecular mass
Question 3
(a) Subtopic: P4.5.1 Electromagnetic induction
(b)(i) Subtopic: P1.1 Physical quantities and measurement techniques
(b)(ii) Subtopic: P4.2.4 Resistance
(c) Subtopic: P5.2.3 Radioactive decay
(d)(i) Subtopic: P2.3.1 Conduction
(d)(ii) Subtopic: P2.2.1 Thermal expansion of solids, liquids and gases
(a) Fig. 3.1 shows a bar magnet suspended by a spring above a coil that is connected to a voltmeter.
When the magnet is pulled downwards into the coil and then released, it oscillates up and down inside the coil. An alternating voltage is observed on the voltmeter.
Explain why an alternating voltage is observed.
▶️Answer/Explanation
1. Voltage is induced as the coil cuts the magnetic field/magnetic flux changes
2. The voltage reverses when the magnet changes direction
Explanation:
– According to Faraday’s Law of Electromagnetic Induction, a changing magnetic flux induces an EMF
– As the magnet moves down, it induces current in one direction
– As it moves up, the direction of motion reverses, so the induced current reverses
– This continuous change in direction produces an alternating voltage
– The frequency of alternation matches the oscillation frequency of the magnet
(b) A thin piece of iron wire has a diameter of 0.20 mm.
(i) Name the device which could accurately measure very small distances such as 0.20 mm.
▶️Answer/Explanation
Micrometer screw gauge
Explanation:
– A micrometer can measure small dimensions to 0.01 mm precision
– More precise than vernier calipers which typically measure to 0.1 mm
– Works by advancing a calibrated screw to gently clamp the object
– Has a ratchet mechanism to ensure consistent measuring pressure
(ii) The wire is 0.10 m in length and has a resistance of 0.30 Ω.
Determine the resistance of a piece of wire made from the same iron metal that is 0.10 m in length but has a diameter of 0.40 mm.
▶️Answer/Explanation
0.075 Ω
Explanation:
– Resistance is inversely proportional to cross-sectional area (R ∝ 1/A)
– Area = πr² = π(d/2)², so area increases by 4× when diameter doubles
– Original diameter = 0.20 mm → New diameter = 0.40 mm (double)
– New area = (0.40/0.20)² = 4 times original area
– Therefore new resistance = 0.30 Ω ÷ 4 = 0.075 Ω
– Length remains constant at 0.10 m
(c) The isotope iron-55 has a half-life of 2.7 years. A sample of this isotope contains 8 × 1012 atoms.
Some time later 7 × 1012 atoms have decayed.
Calculate the time needed for this number of atoms to decay.
▶️Answer/Explanation
8.1 years
Calculation:
1. Initial atoms = 8 × 1012
2. Atoms remaining = 8 × 1012 – 7 × 1012 = 1 × 1012
3. Number of half-lives:
– After 1 half-life (2.7 yrs): 4 × 1012 atoms remain
– After 2 half-lives (5.4 yrs): 2 × 1012 atoms remain
– After 3 half-lives (8.1 yrs): 1 × 1012 atoms remain
4. Total time = 3 × 2.7 = 8.1 years
(d) Fig. 3.2 shows an iron rod being heated at one end by a Bunsen burner.
Thermal energy passes through the rod by conduction.
(i) Describe the process of conduction in solid iron, using ideas about the vibration of atoms.
▶️Answer/Explanation
1. Heat energy makes atoms vibrate more vigorously at the heated end
2. These vibrating atoms collide with neighboring atoms, transferring kinetic energy
3. The energy is passed along the rod through successive atomic collisions
Additional details:
– In metals, free electrons also contribute significantly to heat conduction
– The process continues until thermal equilibrium is reached
– No net movement of atoms occurs, just increased vibration energy transfer
(ii) When heated, the iron rod expands.
Explain in terms of the motion and arrangement of the atoms why iron expands when heated.
▶️Answer/Explanation
1. Atoms vibrate with greater amplitude when heated
2. The average distance between atoms increases
3. While the atoms remain in fixed positions, their increased vibration causes the overall structure to occupy more space
Key points:
– The potential energy curve for atomic bonds is asymmetric
– Increased vibration moves atoms further apart on average than it brings them closer
– This effect occurs in all dimensions, leading to volume expansion
– Known as thermal expansion, quantified by the coefficient of linear expansion
Question 4
(a)(i) Subtopic: B8.4 Translocation and B8.3 Transpiration
(a)(ii) Subtopic: B8.4 Translocation and B8.3 Transpiration
(b)(i) Subtopic: B8.4 Translocation
(b)(ii) Subtopic: B8.1 Xylem and phloem
(a) Scientists investigate where translocation and transpiration occur in a plant stem.
The scientists test three plant stems:
- Stem A is left in its natural state
- Stem B has a ring of phloem removed
- Stem C has a ring of phloem and xylem removed
Fig. 4.1 shows the stems used.
(i) Table 4.1 is used to predict which processes occur in each stem.
Complete Table 4.1 by placing ticks (✓) in the correct boxes to predict which processes occur in each stem.
▶️Answer/Explanation
| stem A | stem B | stem C | |
|---|---|---|---|
| translocation occurs | ✓ | ||
| transpiration occurs | ✓ | ✓ |
Explanation:
– Stem A (Natural): Both processes occur normally as all vascular tissues are intact
– Stem B (Phloem removed): Only transpiration occurs because xylem is intact (phloem needed for translocation)
– Stem C (Both removed): Neither process occurs as both vascular tissues are disrupted
(ii) Compare the direction of movement of substances during translocation and transpiration.
▶️Answer/Explanation
1. Transpiration: Unidirectional (only upwards from roots to leaves)
2. Translocation: Bidirectional (can move both upwards and downwards in the plant)
Detailed comparison:
– Transpiration stream: Water and minerals move upwards through xylem from roots → leaves
– Translocation: Organic compounds (sugars, amino acids) move through phloem:
– Downwards: From leaves (source) to roots/storage organs (sink)
– Upwards: From storage organs to growing tips/new leaves
– Transpiration is passive (physical process), while translocation is active (requires energy)
(b) Xylem and phloem are specialised to transport substances including water around the plant.
(i) Name two other substances moved through the plant during translocation.
▶️Answer/Explanation
1. Sucrose
2. Amino acids
Additional substances:
– Other sugars (glucose, fructose)
– Hormones (auxins, cytokinins)
– Some mineral ions
– Note: These are transported in the phloem sap along with water
(ii) Describe one other function of xylem.
▶️Answer/Explanation
Mechanical support (provides structural strength to the plant)
Detailed explanation:
– Xylem vessels contain lignin (a tough, waterproof substance) in their cell walls
– This lignification makes xylem cells rigid and strong after they die
– Provides:
– Stem rigidity to keep plant upright
– Resistance to bending forces (wind, weight of branches)
– Structural framework for overall plant architecture
– Secondary function: Storage of water and nutrients
Question 5
(a)(i) Subtopic: C5.1 Exothermic and endothermic reactions
(a)(ii) Subtopic: C6.1 Physical and chemical changes
(a)(iii) Subtopic: C6.1 Physical and chemical changes
(b)(i) Subtopic: C2.5 Simple molecules and covalent bonds
(b)(ii) Subtopic: C2.5 Simple molecules and covalent bonds
(b)(iii) Subtopic: C2.5 Simple molecules and covalent bonds
(a) Fig. 5.1 shows what happens when a teacher ignites a mixture of hydrogen and air.
(i) A student concludes that the reaction between hydrogen and oxygen is exothermic.
Suggest the observation that leads him to this conclusion.
▶️Answer/Explanation
Flash/bang/explosion (or production of a flame)
Explanation:
– The rapid release of energy as heat and light indicates an exothermic reaction
– The explosive nature shows large amounts of thermal energy being released quickly
– The equation: 2H₂ + O₂ → 2H₂O + energy
– The energy released comes from forming new H-O bonds being stronger than the broken H-H and O=O bonds
(ii) The student mixes the drops of the colourless liquid that form inside the gas jar with anhydrous copper sulfate.
Describe the colour change he observes if this liquid is water.
▶️Answer/Explanation
From white to blue
Explanation:
– Anhydrous copper sulfate is a white powder
– It forms blue hydrated copper sulfate crystals (CuSO₄·5H₂O) when water is added
– This is a standard chemical test for water
– The colour change is immediate and obvious
(iii) Describe how the teacher shows that the reaction between anhydrous copper sulfate and water is reversible.
▶️Answer/Explanation
1. Heat the blue hydrated copper sulfate
2. It turns back to white anhydrous copper sulfate
Detailed process:
– First step: White CuSO₄ + water → Blue CuSO₄·5H₂O (hydration)
– Reversal: Blue CuSO₄·5H₂O + heat → White CuSO₄ + water vapor (dehydration)
– This demonstrates the reversibility: CuSO₄ + 5H₂O ⇌ CuSO₄·5H₂O
– The water of crystallization can be removed and added back
(b) Fig. 5.2 shows some molecules involved in the reaction between hydrogen and oxygen to make water.
(i) Identify the bonds which break in this reaction.
▶️Answer/Explanation
H-H bonds (in hydrogen molecules) and O=O bonds (in oxygen molecules)
Explanation:
– Reactants are H₂ and O₂ molecules
– For the reaction to occur, these covalent bonds must first break:
– The single bond in H₂ (H-H)
– The double bond in O₂ (O=O)
– This bond breaking requires energy (endothermic process)
(ii) Identify the bonds which form in this reaction.
▶️Answer/Explanation
O-H bonds (in water molecules)
Explanation:
– New covalent bonds form between oxygen and hydrogen atoms
– Each water molecule (H₂O) contains two O-H single bonds
– The bond formation releases energy (exothermic process)
– The energy released in bond formation exceeds the energy needed for bond breaking
(iii) State the type of bond in the H₂ molecule.
▶️Answer/Explanation
Covalent bond
Explanation:
– Formed by sharing of electrons between two hydrogen atoms
– Each hydrogen contributes 1 electron to form a shared pair
– A single covalent bond (one shared pair of electrons)
– Typical bond length: 74 pm
– Bond energy: 436 kJ/mol
(c) Fig. 5.3 is an energy level diagram for the reaction between hydrogen and oxygen.
(i) Explain what is meant by an exothermic reaction.
▶️Answer/Explanation
A reaction that releases thermal energy to the surroundings
Key characteristics:
– The products have less chemical energy than the reactants
– The difference in energy is released as heat
– The surroundings get warmer
– Represented by a negative ΔH (enthalpy change)
(ii) Explain how the energy level diagram shows that the reaction is exothermic.
▶️Answer/Explanation
The products are at a lower energy level than the reactants
Diagram interpretation:
– Reactants start at higher energy level
– Products finish at lower energy level
– The drop in energy level represents energy released
– The greater stability of products (H₂O) compared to reactants (H₂ + O₂)
(iii) Describe what is meant by the term activation energy.
▶️Answer/Explanation
The minimum energy required for a reaction to occur
Detailed explanation:
– Energy needed to break initial bonds
– Required to reach the transition state
– Even exothermic reactions require activation energy
– On the diagram: vertical distance from reactants to the peak
– In this reaction: needed to break H-H and O=O bonds before new O-H bonds can form
(iv) Label the activation energy on Fig. 5.3.
▶️Answer/Explanation
[Diagram note: Arrow pointing from reactants level to the peak of the energy curve, labeled “activation energy”]
Visual representation:
– The “hump” in the energy profile represents the activation energy barrier
– The higher the activation energy, the slower the reaction at a given temperature
– In this reaction, the activation energy is relatively high, explaining why hydrogen and oxygen can mix at room temperature without immediately reacting
Question 6
(a)(i) Subtopic: P1.2 Motion
(a)(ii) Subtopic: P1.2 Motion
(b) Subtopic: P1.5.2 Turning effect of forces
(c) Subtopic: P1.4 Density
(d) Subtopic: P2.2.2 Melting, boiling and evaporation
(a) In a cartoon, a mouse is being chased by a cat.
The mouse accelerates constantly from rest for 1 second and reaches a speed of 3 m/s and then moves at a constant speed of 3 m/s for 8 seconds.
(i) On the grid in Fig. 6.1 draw the speed-time graph to show the motion of the mouse.
▶️Answer/Explanation
[Graph description:]
1. First segment (0-1s): Straight line with positive gradient from (0,0) to (1,3)
2. Second segment (1-9s): Horizontal line at v=3 m/s from (1,3) to (9,3)
Key features:
– Acceleration phase: Linear increase in speed (constant acceleration)
– Constant speed phase: Zero acceleration (flat line)
– Axes: Time (x-axis) 0-9s, Speed (y-axis) 0-3 m/s
– Gradient: First segment gradient = acceleration (3 m/s²)
(ii) The cat accelerates constantly from rest for 9 seconds and reaches a speed of 2 m/s.
Calculate the acceleration of the cat.
▶️Answer/Explanation
0.22 m/s² (or 2/9 m/s²)
Calculation:
a = Δv/Δt = (2 m/s – 0 m/s) ÷ 9 s = 2/9 ≈ 0.22 m/s²
Explanation:
– Acceleration = change in velocity ÷ time taken
– Initial velocity (u) = 0 m/s (from rest)
– Final velocity (v) = 2 m/s
– Time (t) = 9 s
– Units must be in m/s²
(b) Fig. 6.2 shows the mouse sitting on a cube of cheese, which is on a wooden beam pivoted in the middle.
The cat sits on the other end of the beam and balances it.
The weight of the cat is 50 N and the combined weight of the mouse and cheese is 21 N.
Calculate the distance d when the beam is balanced.
▶️Answer/Explanation
8.4 cm
Calculation:
Using principle of moments (clockwise moment = anticlockwise moment):
50 × d = 21 × 20
d = (21 × 20) ÷ 50 = 420 ÷ 50 = 8.4 cm
Explanation:
– Pivot is at the center (moment arm for mouse+cheese = 20 cm)
– System is in equilibrium when moments balance
– Clockwise moment (cat) = F × d = 50 × d
– Anticlockwise moment (mouse+cheese) = 21 × 20
– Solve for d to find cat’s position
(c) Each side of the cube of cheese is 12 cm.
The weight of the cube of cheese is 20.5 N.
Calculate the density of the cube of cheese in g/cm³.
gravitational field strength = 10 N/kg
▶️Answer/Explanation
1.2 g/cm³
Calculation steps:
1. Volume: (12 cm)³ = 1728 cm³
2. Mass: Weight ÷ g = 20.5 N ÷ 10 N/kg = 2.05 kg
3. Convert mass: 2.05 kg × 1000 = 2050 g
4. Density: mass/volume = 2050 g ÷ 1728 cm³ ≈ 1.2 g/cm³
Key formulas:
– Volume of cube = side³
– Weight = mass × gravitational field strength
– Density = mass/volume
(d) Water evaporates from the cat’s bowl.
Liquid water turns into water vapour when it evaporates. Water also turns into water vapour when water boils.
State two differences between the processes of evaporation and boiling.
▶️Answer/Explanation
Any two differences:
1. Temperature: Evaporation occurs at any temperature, boiling only at boiling point
2. Location: Evaporation only at surface, boiling throughout liquid
3. Energy source: Evaporation uses liquid’s internal energy, boiling requires external heat
4. Rate: Evaporation is slow, boiling is rapid
5. Bubble formation: No bubbles in evaporation, bubbles form in boiling
Additional notes:
– Evaporation causes cooling effect (energy taken from remaining liquid)
– Boiling maintains constant temperature at boiling point
Question 7
(a) Subtopic: B7.2 Digestive system
(b) Subtopic: B7.3 Digestion
(c) Subtopic: B7.3 Digestion
(d) Subtopic: B5 Enzymes
Fig. 7.1 shows an X-ray of a molar tooth.
(a) Identify the parts labelled A, B and C in Fig. 7.1.
▶️Answer/Explanation
A: Enamel
B: Dentine
C: Pulp
Tooth structure details:
– Enamel: Hardest substance in the body (96% mineral), covers crown
– Dentine: Softer than enamel (70% mineral), forms bulk of tooth
– Pulp: Soft tissue containing nerves and blood vessels
– X-rays show enamel as brightest (most dense), dentine as gray, pulp as darkest
(b) Consuming sugary food and drinks can increase the risk of tooth decay.
Describe, in detail, the process of tooth decay.
▶️Answer/Explanation
1. Bacteria in plaque feed on sugars, producing acids
2. Acids dissolve minerals in enamel (demineralization)
3. Decay progresses into dentine, forming cavities
4. Untreated decay reaches pulp, causing pain/infection
Key stages:
– Plaque formation: Bacteria + food debris form sticky biofilm
– Acid production: Bacteria ferment sugars → lactic acid
– Enamel erosion: Acid dissolves calcium phosphate crystals
– Cavitation: Structural collapse creates holes in tooth
– pH < 5.5 initiates demineralization
(c) Describe the role of teeth in terms of mechanical digestion.
▶️Answer/Explanation
Break food into smaller pieces (increasing surface area for enzymes)
Detailed functions:
– Incisors: Cutting food (8 front teeth)
– Canines: Tearing food (4 pointed teeth)
– Premolars/Molars: Grinding/crushing (12 flat-surfaced teeth)
– Mastication mixes food with saliva
– Creates bolus for swallowing
– Mechanical breakdown precedes chemical digestion
(d) Chemical digestion also occurs in the mouth.
Describe the role of enzymes in the chemical digestion that occurs in the mouth.
▶️Answer/Explanation
1. Salivary amylase breaks starch into maltose
2. Lingual lipase begins fat digestion (minor role)
3. Enzymes work best at mouth’s pH (6.2-7.4)
4. Converts complex carbs to simpler sugars
Enzyme specifics:
– Amylase: Optimal pH 6.7, inactivated by stomach acid
– Substrate: Starch (amylose/amylopectin)
– Products: Maltose/maltotriose/dextrins
– Activation: Requires chloride ions (Cl⁻) as cofactor
Question 8
(a)(i) Subtopic: C8.3 Group VII properties
(a)(ii) Subtopic: C8.3 Group VII properties
(b)(i) Subtopic: C6.3 Redox
(b)(ii) Subtopic: C6.3 Redox
Fig. 8.1 shows Group VII of the Periodic Table.
(a) A student adds aqueous chlorine to colourless aqueous sodium bromide and to colourless aqueous sodium iodide as shown in Fig. 8.2.
She repeats her experiment adding aqueous bromine to aqueous sodium chloride and to aqueous sodium iodide.
Table 8.1 shows some of her observations.
(i) Complete Table 8.1.
▶️Answer/Explanation
| Halogen solutions | aqueous sodium chloride | aqueous sodium bromide | aqueous sodium iodide |
|---|---|---|---|
| aqueous chlorine (colourless) | No change/Colorless | Orange/yellow | Brown |
| aqueous bromine (orange) | Pale orange | No change/Colorless | Brown |
Explanation:
– Chlorine displaces bromide (orange) and iodide (brown) but not chloride
– Bromine displaces iodide (brown) but not chloride or bromide
– Color changes indicate halogen formation (Cl₂=colorless, Br₂=orange, I₂=brown)
– Reactivity decreases down group: Cl₂ > Br₂ > I₂
(ii) Explain the observations when aqueous bromine is added to aqueous sodium chloride and to aqueous sodium iodide.
Use ideas about the relative reactivities of the halogens in your answer.
▶️Answer/Explanation
With sodium chloride:
– No reaction (pale orange remains)
– Bromine is less reactive than chlorine (cannot displace chloride ions)
With sodium iodide:
– Solution turns brown
– Bromine is more reactive than iodine (displaces iodide ions: Br₂ + 2I⁻ → 2Br⁻ + I₂)
– Iodine forms brown solution
Reactivity trend:
Fluorine > Chlorine > Bromine > Iodine (decreasing oxidising ability)
– More reactive halogens displace less reactive ones from compounds
– Displacement reactions demonstrate this reactivity series
(b) The ionic equation for the reaction between bromine and sodium iodide is shown.
Br₂ + 2I⁻ → 2Br⁻ + I₂
This reaction does not involve oxygen.
(i) Explain in detail why this is a redox reaction.
▶️Answer/Explanation
1. Oxidation occurs: I⁻ loses electrons (2I⁻ → I₂ + 2e⁻)
2. Reduction occurs: Br₂ gains electrons (Br₂ + 2e⁻ → 2Br⁻)
3. Electron transfer: Bromine oxidizes iodide ions
4. Oxidation numbers:
– Iodine changes from -1 to 0 (oxidation)
– Bromine changes from 0 to -1 (reduction)
Redox characteristics:
– Always involves simultaneous oxidation and reduction
– Electron transfer is fundamental (OIL RIG: Oxidation Is Loss, Reduction Is Gain)
– No oxygen required – many redox reactions occur without oxygen
(ii) Identify the oxidising agent in this reaction.
▶️Answer/Explanation
Bromine (Br₂)
Explanation:
– Oxidizing agent causes oxidation of another species
– Br₂ causes I⁻ to lose electrons (oxidizes iodide)
– Br₂ itself gets reduced (gains electrons)
– Stronger oxidizing agents have greater tendency to gain electrons
– In Group VII, oxidizing ability decreases down the group
Question 9
(a)(i) Subtopic: P4.2.5 Electrical energy and electrical power
(a)(ii) Subtopic: P4.2.2 Electric current
(a)(iii) Subtopic: P4.5.5 The d.c. motor
(b) Subtopic: P1.6.1 Energy
Fig. 9.1 shows a golf cart used to carry golfers and their golf clubs around a golf course.
(a) The cart contains an electric motor powered by a 36V battery. The power rating of the motor is 3000W.
(i) Calculate the maximum current that passes through the motor.
▶️Answer/Explanation
83.3 A
Calculation:
P = IV → I = P/V = 3000W ÷ 36V ≈ 83.3A
Key concepts:
– Power (P) = Current (I) × Voltage (V)
– Units: Watts (W) = Amperes (A) × Volts (V)
– This is the theoretical maximum current under full load
– Actual current may be lower due to efficiency losses
(ii) Calculate the charge flowing through the motor when it is used at a maximum current for 5 minutes.
▶️Answer/Explanation
25,000 C
Calculation:
Q = It = 83.3A × (5 × 60)s = 83.3 × 300 = 24,990C ≈ 25,000C
Explanation:
– Charge (Q) = Current (I) × Time (t)
– Time must be converted to seconds (5 mins = 300s)
– 1 Coulomb = 1 Ampere × 1 Second
– This large charge demonstrates why high-current systems need thick cables
(iii) Fig. 9.2 shows a simple d.c. electric motor.
On Fig. 9.2, label the split-ring commutator with the letter X and the coil with the letter C.
▶️Answer/Explanation
[Diagram labels:]
– X: The copper contacts that rotate with the coil (split-ring)
– C: The wire loops wound around the armature
Motor components explained:
1. Split-ring commutator: Reverses current direction every half-rotation to maintain torque
2. Coil: Becomes an electromagnet when current flows, interacting with permanent magnets
3. Function: Converts electrical energy to mechanical rotation via the motor effect
(b) A golfer hits a golf ball.
At one moment, the golf ball has 22.5J of kinetic energy. The mass of the golf ball is 50g.
Calculate the speed of the golf ball at that moment.
▶️Answer/Explanation
30 m/s
Calculation:
KE = ½mv² → v = √(2KE/m) = √(2×22.5/0.05) = √(900) = 30 m/s
Step-by-step:
1. Convert mass to kg: 50g = 0.05kg
2. Rearrange kinetic energy formula
3. Substitute values: 2 × 22.5 = 45 → 45 ÷ 0.05 = 900
4. Square root gives velocity
Real-world context:
– 30 m/s ≈ 108 km/h (67 mph) – typical drive speed for amateur golfers
– Professional golfers can reach 80 m/s (288 km/h)
Question 10
(a) Subtopic: B17.2 Selection
(b)(i) Subtopic: B16.1 Chromosomes and genes
(b)(ii) Subtopic: B17.2 Selection
(c) Subtopic: B15.1 Asexual reproduction
(a) MRSA is a strain of bacteria that is resistant to antibiotics.
Table 10.1 compares the number of cases of infection caused by MRSA bacteria in one hospital between 1998 and 2008.
Calculate the percentage increase in number of cases between 2006 and 2008.
▶️Answer/Explanation
7.7%
Calculation:
1. Increase = 167 – 155 = 12 cases
2. Percentage increase = (12/155) × 100 ≈ 7.7%
Mathematical steps:
– Percentage change = (New value – Original value)/Original value × 100
– 2006 value = 155 cases (original)
– 2008 value = 167 cases (new)
– Shows accelerating antibiotic resistance problem
(b) The resistant allele in MRSA bacteria developed due to a mutation.
(i) Define the term mutation.
▶️Answer/Explanation
A random change in the DNA sequence/genetic material
Key characteristics:
– Can occur spontaneously or be induced by mutagens
– May affect single nucleotides (point mutations) or larger segments
– Results in new alleles/variations
– Source of genetic variation for natural selection
– Can be beneficial, harmful, or neutral depending on environment
(ii) With reference to natural selection, describe how MRSA bacteria have evolved to become resistant to antibiotics.
▶️Answer/Explanation
1. Random mutation creates antibiotic resistance gene
2. Non-resistant bacteria die when exposed to antibiotics
3. Resistant bacteria survive and reproduce
4. Resistance gene spreads through population
5. Overuse of antibiotics increases selection pressure
Evolutionary mechanism:
– Demonstrates Darwin’s “survival of the fittest”
– Vertical gene transfer (binary fission) spreads resistance
– Horizontal gene transfer (plasmids) can accelerate spread
– Medical consequence: Creates “superbugs” resistant to multiple drugs
(c) Bacteria reproduce by asexual reproduction.
Describe one disadvantage to bacteria without the resistant allele of reproducing asexually.
▶️Answer/Explanation
Primary disadvantage:
No genetic variation in offspring (all clones)
Consequences:
– Population cannot adapt quickly to environmental changes
– If environment changes (e.g., antibiotics introduced), entire population may die
– Contrasts with sexual reproduction which generates diversity through recombination
– Mutation is only source of variation in asexual populations
Survival implication:
Non-resistant strains may be completely eliminated by antibiotics
Question 11
(a)(i) Subtopic: C11.5 Alkenes
(a)(ii) Subtopic: C11.4 Alkanes and C11.5 Alkenes
(b)(i) Subtopic: C3.3 The mole and the Avogadro constant
(b)(ii) Subtopic: C10.2 Air quality and climate
(c) Subtopic: C11.5 Alkenes
(d)(i) Subtopic: C11.7 Polymers
(d)(ii) Subtopic: C11.7 Polymers
(d)(iii) Subtopic: C11.7 Polymers
A homologous series is a family of compounds which have the same general formula and similar chemical properties.
Alkanes and alkenes are examples of homologous series.
Ethane, C2H6, and propane, C3H8, are alkanes.
Ethene, C2H4, and propene, C3H6, are alkenes.
(a) (i) The general formula for alkanes is CnH2n+2.
Suggest the general formula for alkenes.
▶️Answer/Explanation
CnH2n
Explanation:
– Alkenes have one C=C double bond
– Each carbon forms 4 bonds total (2 in double bond + 2 single bonds)
– For every n carbon atoms, there are 2n hydrogen atoms
– Examples: Ethene (C2H4), Propene (C3H6)
– Contrast with alkanes (CnH2n+2) which have only single bonds
(ii) Complete Fig. 11.1 to show the structures of an ethane molecule and an ethene molecule.
▶️Answer/Explanation
Ethane: C-C single bond with 3 H atoms on each C
Ethene: C=C double bond with 2 H atoms on each C
Structural details:
Ethane (C2H6):
– Single covalent bond between carbons
– Each carbon has 3 hydrogens (sp3 hybridization)
– Tetrahedral geometry (109.5° bond angles)
Ethene (C2H4):
– Double covalent bond between carbons (1 σ + 1 π bond)
– Each carbon has 2 hydrogens (sp2 hybridization)
– Planar geometry (120° bond angles)
– π bond prevents rotation about C=C
(b) (i) The equation for the complete combustion of propane is shown.
C3H8 + 5O2 → 3CO2 + 4H2O
Complete steps 1 to 3 to calculate the volume of carbon dioxide when 1000 dm3 of propane is burned. All gas volumes are measured at room temperature and pressure. The volume of 1 mole of any gas is 24 \(dm_{3}\) at room temperature and pressure.
step 1
Calculate the number of moles in 1000 \(dm_{3}\) of propane.
step 2
Use your answer to step 1 and the balanced equation to calculate the number of moles
of carbon dioxide produced by burning 1000 \(dm_{3}\) of propane.
step 3
Calculate the volume of carbon dioxide produced by burning 1000 \(dm_{3}\) of propane.
▶️Answer/Explanation
Step 1: 1000 ÷ 24 = 41.67 moles propane
Step 2: 41.67 × 3 = 125 moles CO2
Step 3: 125 × 24 = 3000 dm3 CO2
Calculation breakdown:
1. Convert volume to moles (1 mole gas = 24 dm3 at RTP)
2. Use molar ratio from equation (1:3 propane:CO2)
3. Convert moles back to volume
Key concept:
– Avogadro’s Law: Equal volumes of gases contain equal numbers of molecules at same T&P
– All gas volumes measured at room temperature and pressure (RTP)
(ii) Describe the effect of increased emission of carbon dioxide on the environment.
▶️Answer/Explanation
1. Enhanced greenhouse effect: CO2 traps infrared radiation
2. Global warming: Rising average temperatures worldwide
3. Climate change: Altered weather patterns, more extreme events
4. Ocean acidification: CO2 dissolves in seawater forming carbonic acid
Additional impacts:
– Melting polar ice caps and glaciers
– Rising sea levels
– Disruption of ecosystems
– Increased frequency of heatwaves and droughts
(c) Two reactions of the alkenes ethene and propene are:
- combustion
- polymerisation
Describe one other chemical reaction of alkenes. Explain why alkenes can undergo this chemical reaction.
▶️Answer/Explanation
Addition reactions:
1. Hydrogenation: C=C + H2 → C-C (forms alkanes)
2. Halogenation: C=C + X2 → CXCX (X = Cl, Br, etc.)
3. Hydration: C=C + H2O → alcohol (with H3PO4 catalyst)
Mechanism explanation:
– C=C double bond is electron-rich (nucleophilic)
– Undergoes electrophilic addition reactions
– π bond breaks to form two new σ bonds
– Markovnikov’s rule applies for unsymmetrical alkenes
(both have) double bond / are unsaturated
(d) (i) State one difference between addition polymerisation and condensation polymerisation.
▶️Answer/Explanation
Key difference:
Addition polymerization produces no byproducts, while condensation polymerization releases small molecules (e.g., H2O, HCl)
Comparison:
Addition:
– Only monomers join together
– Typically involves C=C double bonds opening
– Example: Polyethene from ethene
Condensation:
– Monomers react with functional groups
– Byproduct molecules formed
– Example: Nylon formation produces water
(ii) Nylon is a condensation polymer made from monomer molecules A and B.
Fig. 11.2 shows a few of these monomer molecules.
Fig. 11.3 shows an incomplete section of the nylon molecule.
Complete Fig. 11.3 to show how a molecule of monomer B has chemically combined with a molecule of monomer A.
▶️Answer/Explanation
[Structure: -NH-(CH2)x-CO-]
Polymerization details:
– Amine group (-NH2) reacts with carboxyl group (-COOH)
– Forms peptide/amide bond (-CONH-)
– Water molecule eliminated
– Repeating unit has alternating monomers
– Chain grows at both ends
(iii) State the formula of the other compound that is formed during the polymerisation to make nylon.
▶️Answer/Explanation
H2O (water)
Polymerization equation:
H2N-R-NH2 + HOOC-R’-COOH → -[HN-R-NH-OC-R’-CO-]n– + 2nH2O
Alternative byproducts:
– Some nylon types produce HCl instead
– Depends on specific monomers used
– Water is most common byproduct in biological polymers too
Question 12
(a) Subtopic: P4.4 Electrical safety
(b)(i) Subtopic: P3.4 Sound
(b)(ii) Subtopic: P3.4 Sound
(b)(iii) Subtopic: P3.1 General properties of waves
(c) Subtopic: P3.2.1 Reflection of light
(d) Subtopic: P2.3.3 Radiation
A gardener cuts grass with an electric mower.
(a) Use the information in Fig. 12.1 to explain why the cut in insulation is an electrical hazard.
▶️Answer/Explanation
Hazard explanation:
1. Exposed live wires can cause electric shock if touched
2. Damp grass increases conductivity (water is a good conductor when contains ions)
3. Risk of short circuit if conductors touch
4. Potential fire hazard from sparks
Safety implications:
– 230V mains electricity can be lethal
– Current as low as 50mA can cause cardiac arrest
– Damaged insulation violates electrical safety regulations
– Proper repair requires complete rewrapping with insulating tape or cable replacement
(b) The mower is noisy. Sound waves from the lawn mower pass through the air as a series of compressions and rarefactions.
(i) State what is meant by a compression.
▶️Answer/Explanation
A region of high pressure where air particles are closer together
Sound wave details:
– Longitudinal wave characteristic
– Particles vibrate parallel to wave direction
– Alternates with rarefactions (low pressure regions)
– Distance between compressions = wavelength
– Frequency of compressions determines pitch
(ii) Describe the wavelength of a sound wave in terms of compressions.
▶️Answer/Explanation
The distance between two consecutive compressions (or between two consecutive rarefactions)
Wave characteristics:
– Measured in meters (m)
– Inversely proportional to frequency (λ = v/f)
– Typical audible sound wavelengths range from 17mm to 17m
– Determines the pitch we perceive – shorter wavelengths = higher pitch
(iii) Sound waves are longitudinal waves.
Describe the differences between longitudinal and transverse waves.
▶️Answer/Explanation
| Characteristic | Longitudinal Waves | Transverse Waves |
|---|---|---|
| Particle motion | Parallel to wave direction | Perpendicular to wave direction |
| Examples | Sound waves, seismic P-waves | Light waves, seismic S-waves |
| Medium required | Yes (solid, liquid, gas) | Only for mechanical waves |
| Wave features | Compressions and rarefactions | Crests and troughs |
Key difference: The relationship between particle motion and wave propagation direction
(c) The gardener places mirrors in his garden to scare cats away. When a cat sees its image in the mirror it runs away.
Describe the image formed in a plane mirror by using three words or phrases from the list.
laterally inverted magnified not upside down real
same size smaller upside down virtual
▶️Answer/Explanation
1. Virtual
2. Laterally inverted
3. Same size
Plane mirror image properties:
– Virtual: Cannot be projected on a screen (light rays appear to come from behind mirror)
– Laterally inverted: Left-right reversal (like text in a mirror)
– Same size: Magnification = 1 (object and image dimensions identical)
– Additional correct properties: Upright, same distance behind mirror as object is in front
(d) Fig. 12.2 shows a heater in the garden. The heater burns butane gas.
The underside surface of the hood is shiny and light in colour.
Suggest why this is a more suitable surface than a dull and dark colour.
▶️Answer/Explanation
Thermal radiation benefits:
1. Shiny surface reflects infrared radiation downwards
2. Light color emits less thermal radiation upwards
3. Increases heating efficiency by directing heat where needed
Physics principles:
– Good reflectors are poor emitters/absorbers (Kirchhoff’s Law)
– Dark surfaces would absorb and re-radiate heat upwards
– Shiny aluminum has low emissivity (ε ≈ 0.03)
– Maximizes downward heat transfer to people below
Question 13
(a)(i) Subtopic: B13.2 Hormones
(a)(ii) Subtopic: B13.2 Hormones
(a)(iii) Subtopic: B13.3 Homeostasis
(b) Subtopic: B13.3 Homeostasis
(c) Subtopic: B13.1 Coordination and response
(a) The blood glucose concentration of a person is monitored for 180 minutes after eating a meal.
Fig. 13.1 shows the results.
(i) State the name of the hormone that causes the change between 60-120 minutes.
▶️Answer/Explanation
Insulin
Physiological role:
– Secreted by β-cells of pancreatic islets
– Promotes glucose uptake by liver/muscle/fat cells
– Stimulates glycogenesis (glucose → glycogen conversion)
– Lowers blood glucose concentration
– Peak secretion occurs 30-60 minutes after eating
(ii) Suggest one other way to cause a similar change to blood glucose concentration as shown between 60-120 minutes.
▶️Answer/Explanation
Exercise/physical activity
Mechanism:
– Muscles use glucose for energy during contraction
– Increases cellular glucose uptake independent of insulin
– Can lower blood glucose by 1-2 mmol/L during moderate exercise
– Particularly effective for reducing postprandial hyperglycemia
(iii) Describe how the liver and pancreas work together to cause the changes shown between 120-150 minutes.
▶️Answer/Explanation
1. Pancreatic α-cells detect falling blood glucose
2. Secretes glucagon hormone
3. Liver converts glycogen to glucose (glycogenolysis)
4. Gluconeogenesis may also occur (from amino acids/fats)
5. Glucose released into bloodstream maintains homeostasis
Negative feedback loop:
– Blood glucose drops below ~5 mmol/L triggers response
– Glucagon opposes insulin’s effects
– Prevents dangerous hypoglycemia
– Typically occurs 2-5 hours after eating
(b) State a term that can be used to describe the control of blood glucose concentration.
▶️Answer/Explanation
Negative feedback
Control system characteristics:
– Deviation from set point triggers corrective response
– Insulin/glucagon act as antagonistic effectors
– Maintains homeostasis within narrow range (4-7 mmol/L)
– Similar to thermostat regulation
– Involves both hormonal and neural inputs
(c) The blood glucose concentration is controlled by hormones. Some of the body’s responses are controlled by the nervous system.
Table 13.1 compares some of the features of the hormonal and nervous control systems.
Complete Table 13.1 to compare the hormonal and nervous control systems.
▶️Answer/Explanation
| Type of control system | Hormonal | Nervous |
|---|---|---|
| Method of information transfer | Chemical hormones | Electrical impulses |
| Speed of information transfer | Slow (seconds-minutes) | Fast (milliseconds) |
| Longevity of action | Long-lasting (minutes-days) | Short-lived |
Key comparisons:
– Specificity: Nervous system has precise targetting vs hormonal broadcast
– Duration: Hormones degraded slowly vs instant neurotransmitter clearance
– Amplification: Hormones can affect multiple organs simultaneously