Question 1
(a)(i) Subtopic Code: B2.1 (Cell structure)
(a)(ii) Subtopic Code: B2.1 (Cell structure)
(a)(iii) Subtopic Code: B9.4 (Blood)
(b) Subtopic Code: B9.4 (Blood)
(c) Subtopic Code: B9.3 (Blood vessels)
(a) Fig. 1.1 shows some specialised cells.
(i) Identify the names of the cells labeled B and E in Fig. 1.1.
▶️Answer/Explanation
Answer:
B – root hair cell
E – sperm cell
Explanation: Cell B is identified as a root hair cell due to its long, thin projection, which increases surface area for water absorption. Cell E is a sperm cell, recognizable by its tail (flagellum) for motility and streamlined head for fertilization.
(ii) Explain how the structure of cell A is related to its function.
▶️Answer/Explanation
Answer:
Cell A has cilia to move mucus out of the respiratory tract.
Explanation: Cell A is likely a ciliated epithelial cell. The cilia beat in coordinated waves to sweep mucus (containing trapped particles) away from the lungs, protecting the gas exchange system.
(iii) Describe two ways in which cell D is adapted for transporting oxygen.
▶️Answer/Explanation
Answer:
1. Biconcave shape (increases surface area for oxygen diffusion).
2. Contains haemoglobin (binds oxygen).
Explanation: Cell D is a red blood cell. Its biconcave shape maximizes surface area for efficient gas exchange, and haemoglobin is the protein that chemically binds oxygen for transport.
(b) Cell D is one of the main components of blood. State two other main components of blood.
▶️Answer/Explanation
Answer:
1. Plasma
2. White blood cells (or platelets)
Explanation: Blood consists of plasma (liquid component carrying dissolved substances), red/white blood cells, and platelets (for clotting).
(c)(i) Explain why arteries have a thick elastic wall.
▶️Answer/Explanation
Answer:
To withstand high pressure from the heart’s pumping.
Explanation: The elastic fibers allow arteries to stretch and recoil, maintaining blood pressure and preventing vessel damage.
(ii) Explain why veins have valves.
▶️Answer/Explanation
Answer:
To prevent backflow of blood.
Explanation: Valves ensure blood flows toward the heart, especially against gravity in limbs.
(iii) Explain why capillaries have very thin walls.
▶️Answer/Explanation
Answer:
For efficient diffusion of substances between blood and tissues.
Explanation: Thin walls (one cell thick) minimize diffusion distance for oxygen, glucose, and waste products.
Question 2
Ethene is a member of a family of hydrocarbons. Fig. 2.1 shows an ethene molecule. (Sub-topic: C11.5)
(a) State the family of hydrocarbons that ethene is a member of.
▶️Answer/Explanation
Answer: alkenes
Explanation: Ethene (C2H4) is an unsaturated hydrocarbon with a double bond between the carbon atoms, which classifies it as a member of the alkene family.
(b) Ethene is made from the larger molecules in petroleum. State the name of this process. (Sub-topic: C11.5)
▶️Answer/Explanation
Answer: cracking
Explanation: Cracking is the process where larger hydrocarbon molecules from petroleum are broken down into smaller, more useful molecules like ethene, often using heat and a catalyst.
(c) Poly(ethene) can be made from ethene. Poly(ethene) is a polymer.
(i) State what is meant by a polymer. (Sub-topic: C11.7)
▶️Answer/Explanation
Answer: A long chain molecule formed from many small repeating units (monomers).
Explanation: Polymers are large molecules composed of repeating structural units (monomers) connected by covalent bonds. In this case, ethene monomers join together to form poly(ethene).
(ii) Table 2.1 shows some information about polymers. Complete Table 2.1. (Sub-topic: C11.7)
▶️Answer/Explanation
Answer: The completed table would show:
- For poly(chloroethene): Repeat unit would show -C-C- with H and Cl attached
- For poly(tetrafluoroethene): Repeat unit would show -C-C- with F atoms attached
Explanation: The repeat unit shows the smallest section of the polymer that repeats. For poly(chloroethene), each carbon has one H and one Cl attached. For poly(tetrafluoroethene), each carbon has two F atoms attached.
(d) Ethene can be made into ethane.
State the formula of the substance that ethene reacts with to make ethane. (Sub-topic: C11.5)
▶️Answer/Explanation
Answer: H2
Explanation: Ethene (C2H4) reacts with hydrogen (H2) in a hydrogenation reaction to form ethane (C2H6), converting the double bond to a single bond.
(e) Ethane is a saturated hydrocarbon. Describe what is meant by a saturated hydrocarbon. (Sub-topic: C11.4)
▶️Answer/Explanation
Answer: A hydrocarbon containing only single bonds between carbon atoms.
Explanation: Saturated hydrocarbons (alkanes) have all carbon-carbon bonds as single bonds, meaning they contain the maximum possible number of hydrogen atoms (are “saturated” with hydrogen).
(f) Ethene undergoes an addition reaction with bromine. Fig. 2.2 shows the equation for the reaction.
Complete the equation in Fig. 2.2 by drawing the structure of the compound formed. (Sub-topic: C11.5)
▶️Answer/Explanation
Answer: The product is 1,2-dibromoethane: Br-CH2-CH2-Br
Explanation: In the addition reaction, the double bond in ethene breaks and bromine atoms add to each carbon atom, forming 1,2-dibromoethane. This reaction is often used as a test for unsaturation (double bonds).
Question 3
Fig. 3.1 shows a forklift truck lifting a crate.
(a) The crate has a mass of 140 kg.
(i) Calculate the weight of the crate. (Sub-topic: P1.3)
The gravitational field strength, g, is 10 N/kg.
▶️Answer/Explanation
Answer: 1400 N
Explanation:
Weight is calculated using the formula \( W = m \times g \), where \( m \) is the mass and \( g \) is the gravitational field strength.
Given: \( m = 140 \, \text{kg} \), \( g = 10 \, \text{N/kg} \).
Substitute the values into the formula:
\( W = 140 \times 10 = 1400 \, \text{N} \).
(ii) Calculate the work done on the crate when it is lifted through a height of 1.5 m. (Sub-topic: P1.6.2)
State the unit for your answer.
▶️Answer/Explanation
Answer: 2100 J (Joules)
Explanation:
Work done is calculated using the formula \( \text{Work} = \text{Force} \times \text{Distance} \). Here, the force is the weight of the crate, and the distance is the height lifted.
From part (i), the weight \( W = 1400 \, \text{N} \), and the height \( d = 1.5 \, \text{m} \).
Substitute the values into the formula:
\( \text{Work} = 1400 \times 1.5 = 2100 \, \text{J} \).
The unit for work is Joules (J).
(b) The forklift truck uses an electric motor to lift the crate.
Fig. 3.2 shows the circuit that includes the electric motor.
The voltmeter displays a reading of 0.50 V.
(i) Show that the potential difference (p.d.) across the motor is 11.5 V. (Sub-topic: P4.2.3)
▶️Answer/Explanation
Answer: \( 12 \, \text{V} – 0.50 \, \text{V} = 11.5 \, \text{V} \).
Explanation:
The total voltage in the circuit is 12 V, and the voltmeter reads 0.50 V. The potential difference across the motor is the total voltage minus the voltmeter reading:
\( \text{p.d. across motor} = 12 – 0.50 = 11.5 \, \text{V} \).
(ii) The current in the circuit is 9.20 A.
Calculate the resistance of the motor. (Sub-topic: P4.2.4)
▶️Answer/Explanation
Answer: 1.25 Ω
Explanation:
Resistance is calculated using Ohm’s Law: \( R = \frac{V}{I} \), where \( V \) is the potential difference and \( I \) is the current.
From part (i), \( V = 11.5 \, \text{V} \), and \( I = 9.20 \, \text{A} \).
Substitute the values into the formula:
\( R = \frac{11.5}{9.20} = 1.25 \, \Omega \).
Question 4
(a) Tay-Sachs disease is a genetic disorder that destroys nerve cells in the brain and spinal cord. The allele for Tay-Sachs disease is recessive t. The allele for unaffected by Tay-Sachs disease is dominant T.
Fig. 4.1 is a pedigree diagram showing the inheritance of Tay-Sachs disease. (Sub-topic: B16.3)
(i) State the number of males in Fig. 4.1 that are unaffected by Tay-Sachs disease.
▶️Answer/Explanation
Answer: 4
Explanation: By counting the unaffected male symbols (unshaded squares) in the pedigree diagram, we find there are 4 unaffected males.
(ii) Complete the sentences to explain the genotypes of some of the people in Fig. 4.1.
Person E and person F are …… by Tay-Sachs disease.
Person E and person F both have the genotype ……
Person G has Tay-Sachs disease. They have the genotype ……
Person G will have inherited one …… allele from each parent.
▶️Answer/Explanation
Answer:
unaffected
Tt
tt
recessive
Explanation:
- E and F are unaffected but can have affected children (G), so they must be carriers (Tt)
- G has the disease, so must have two recessive alleles (tt)
- G inherited one t allele from each parent (E and F)
(iii) State the probability of two parents with the genotypes TT having a child with Tay-Sachs disease.
▶️Answer/Explanation
Answer: 0 (or 0%)
Explanation: Both parents are homozygous dominant (TT) and can only pass on T alleles to their children, so all offspring will be unaffected (Tt or TT).
(b) Growth of offspring involves mitosis. The box on the left contains the term mitosis. The boxes on the right contain some sentence endings.
Draw three lines from the word mitosis to the boxes on the right to make three correct sentences about mitosis. (Sub-topic: B16.2)
▶️Answer/Explanation
Correct connections:
- Mitosis → occurs after exact duplication of chromosomes
- Mitosis → produces cells with diploid nuclei
- Mitosis → produces nuclei with paired chromosomes
Explanation: Mitosis is the process of cell division that:
- Follows DNA replication (exact duplication of chromosomes)
- Produces genetically identical diploid cells
- Maintains the chromosome number (paired chromosomes in nuclei)
(c) State the number of chromosomes in a human diploid cell. (Sub-topic: B16.1)
▶️Answer/Explanation
Answer: 46 (or 23 pairs)
Explanation: Human somatic (body) cells are diploid, containing 23 pairs of chromosomes (46 total).
(d) State the term given to a change in a gene or chromosome. (Sub-topic: B16.1)
▶️Answer/Explanation
Answer: mutation
Explanation: A mutation is any change in the DNA sequence of a gene or the structure/number of chromosomes.
Question 5
In an experiment, a student adds an alkali to an acid. Fig. 5.1 shows the experiment. (Sub-topic: C7.1)
(a) The student slowly adds the alkali to the acid.
(i) Describe how the pH of the acid changes as the alkali is added.
▶️Answer/Explanation
Answer: The pH increases
Explanation: As alkali is added to acid, the solution becomes less acidic (H⁺ ions are neutralized by OH⁻ ions), causing the pH to rise toward neutral (pH 7) and beyond if excess alkali is added.
(ii) Complete the word equation to show the type of substance made in the reaction.
acid + alkali → ……………….. + water
▶️Answer/Explanation
Answer: salt
Explanation: The neutralization reaction between an acid and alkali always produces a salt (ionic compound) and water.
(iii) Sulfuric acid, H2SO4, is an acid. Potassium hydroxide, KOH, is an alkali.
Construct the balanced symbol equation for the reaction of sulfuric acid with potassium hydroxide.
▶️Answer/Explanation
Answer: H2SO4 + 2KOH → K2SO4 + 2H2O
Explanation:
- Diprotic H2SO4 requires 2 KOH molecules for complete neutralization
- Products are potassium sulfate (K2SO4) and water
- Equation is balanced with 2H2O to match the 2OH⁻ groups
(iv) State the formula of the ion which is present in solutions of all acids.
▶️Answer/Explanation
Answer: H⁺ (hydrogen ion)
Explanation: All acids release H⁺ ions in aqueous solution – this is the fundamental definition of an Arrhenius acid.
(b) Ammonium sulfate, (NH4)2SO4, is made by reacting an acid with an alkali. Calculate the relative formula mass, Mr, of ammonium sulfate. (Sub-topic: C3.2)
[Ar: H, 1; N, 14; O, 16; S, 32]
▶️Answer/Explanation
Answer: 132
Calculation:
(2 × N) + (8 × H) + S + (4 × O) =
(2 × 14) + (8 × 1) + 32 + (4 × 16) =
28 + 8 + 32 + 64 = 132
(c) The alkali used to make ammonium sulfate is ammonia, NH3. Ammonia is made by the Haber process. Nitrogen, N2, and hydrogen, H2, are the starting materials.
Look at the equation for the reaction.
\(N _{2} + 3H_{2} \rightleftharpoons 2NH_{3}\)
(i) Describe the Haber process. You should include: (Sub-topic: C5.1)
- the sources of nitrogen and hydrogen gas
- the conditions used
▶️Answer/Explanation
Answer:
- Sources: Nitrogen from air (78% N2), hydrogen from natural gas/methane (steam reforming)
- Conditions: 200 atmospheres pressure, 450°C temperature, iron catalyst
Explanation: The Haber process compromises between:
- High pressure favors the forward reaction (4 moles → 2 moles)
- Moderate temperature balances reaction rate (faster at high T) with equilibrium position (exothermic forward reaction favored at low T)
- Catalyst increases reaction rate without affecting equilibrium
(ii) Ammonia, NH3, reacts with nitric acid, HNO3. Ammonium nitrate, NH4NO3, is made.
Look at the equation for the reaction.
\( NH_{3} + HNO_{3} \to NH_{4}NO_{3}\)
Calculate the mass of ammonium nitrate made from 51 kg of ammonia. (Sub-topic: C3.2)
[Ar: H, 1; N, 14; O, 16]
▶️Answer/Explanation
Answer: 240 kg
Calculation:
- Mr NH3 = 14 + (3×1) = 17
- Mr NH4NO3 = 14 + (4×1) + 14 + (3×16) = 80
- Mole ratio is 1:1 (NH3:NH4NO3)
- Mass NH4NO3 = (80/17) × 51 = 240 kg
Question 6
Fig. 6.1 shows a tidal power station which uses tidal energy to generate electricity. The moving water turns a turbine which is connected to a generator.
(a)(i) State the source of the energy for the tides. (Sub-topic: P1.6.3)
▶️Answer/Explanation
Answer: The Moon
Explanation: Tides are caused by the gravitational pull of the Moon (and to a lesser extent the Sun) on Earth’s oceans.
(ii) Each kilogram of water has 1.62 J of kinetic energy. Calculate the speed of the water flow. (Sub-topic: P1.6.1)
▶️Answer/Explanation
Answer: 1.8 m/s
Calculation:
KE = ½mv² → v = √(2KE/m)
v = √(2×1.62/1) = √3.24 = 1.8 m/s
(b) Fig. 6.2 shows a simple a.c. generator.
(i) On Fig. 6.2, draw an arrow to show the direction of the magnetic field between the permanent magnets. (Sub-topic: P4.5.2)
▶️Answer/Explanation
Answer: Arrow from North to South pole
Explanation: Magnetic field lines always run from the North pole to the South pole of a magnet.
(ii) State the name of the components labelled X and describe their use. (Sub-topic: P4.5.2)
▶️Answer/Explanation
Answer:
Name: Slip rings
Use: Maintain electrical contact while allowing rotation/prevent wires from tangling
Explanation: Slip rings are conductive rings that rotate with the coil, maintaining contact with stationary brushes to allow current flow without twisting wires.
(iii) On Fig. 6.3, sketch a graph of voltage output against time for a simple a.c. generator operating at a constant speed. (Sub-topic: P4.5.2)
▶️Answer/Explanation
Answer: Smooth sinusoidal wave with constant amplitude and period
Explanation: An ideal a.c. generator produces a perfect sine wave because the induced emf varies as the coil rotates through the uniform magnetic field (emf ∝ rate of change of flux linkage).
(c) The tidal power station uses a warning lamp to warn passing boats of its location. The lamp emits light with a wavelength of 4.0 × 10-7 m. Calculate the frequency of the light. (Sub-topic: P3.3)
▶️Answer/Explanation
Answer: 7.5 × 1014 Hz
Calculation:
c = fλ → f = c/λ
f = (3 × 108) ÷ (4.0 × 10-7) = 7.5 × 1014 Hz
Note: This is visible violet light (400 nm wavelength).
Question 7
(a) Fig. 7.1 is a photograph of a wind-pollinated flower.
Identify part A in Fig. 7.1. (Sub-topic: B15.3)
▶️Answer/Explanation
Answer: Stigma
Explanation: In wind-pollinated flowers, the stigma is typically feathery and exposed to catch airborne pollen grains.
(b) Describe one way the pollen and petals of insect-pollinated flowers are different from wind-pollinated flowers. (Sub-topic: B15.3)
▶️Answer/Explanation
Answer:
Pollen: Larger/stickier in insect-pollinated vs. small/smooth in wind-pollinated
Petals: Brightly colored/scented in insect-pollinated vs. small/dull in wind-pollinated
Explanation: These adaptations reflect the different pollination mechanisms – insects need visual/scent attractants and sticky pollen, while wind pollination requires lightweight pollen and exposed reproductive structures.
(c) State where fertilisation occurs in a plant. (Sub-topic: B15.3)
▶️Answer/Explanation
Answer: Ovule (within the ovary)
Explanation: The pollen tube grows down the style to deliver male gametes to the ovule, where fusion with the egg cell occurs.
(d) Many plants are able to reproduce sexually and asexually. Describe the disadvantages to a plant in the wild of reproducing asexually. (Sub-topic: B15.1)
▶️Answer/Explanation
Answer:
- No genetic variation in offspring
- Population vulnerable to environmental changes/diseases
- Limited ability to colonize new habitats
Explanation: Asexual reproduction produces clones, so all offspring share the same vulnerabilities. This reduces evolutionary adaptability compared to sexual reproduction which generates genetic diversity.
(e) State two requirements for germination of plant seeds. (Sub-topic: B15.3)
▶️Answer/Explanation
Answer:
- Water (for enzyme activation and metabolic processes)
- Oxygen (for aerobic respiration)
- Suitable temperature (for optimal enzyme activity)
Explanation: These factors trigger metabolic activity in the seed:
- Water rehydrates tissues and dissolves inhibitors
- Oxygen supports respiration for energy release
- Temperature affects enzyme-controlled reactions
Question 8
Table 8.1 gives some information about atoms.
(a) Complete Table 8.1. (Sub-topic: C2.2)
▶️Answer/Explanation
Answer:
Argon: Proton number = 18
Magnesium: Electronic structure = 2.8.2
Explanation:
- Argon’s proton number is 18 (atomic number in periodic table)
- Magnesium (atomic number 12) has electron configuration 2,8,2 (2 in first shell, 8 in second, 2 in outer shell)
(b) Chlorine appears twice in Table 8.1. Each of the atoms is an isotope of chlorine. (Sub-topic: C2.3)
(i) Explain what is meant by the word isotope.
▶️Answer/Explanation
Answer: Atoms of the same element with the same number of protons but different numbers of neutrons
Explanation: Isotopes have identical chemical properties (same electron configuration) but different physical properties (different masses).
(ii) The two isotopes of chlorine have the same chemical properties. Explain why.
▶️Answer/Explanation
Answer: They have the same electronic structure/configuration
Explanation: Chemical properties depend on electron arrangement, which is identical in isotopes since they have the same number of protons and electrons.
(c) Argon is a noble gas. Explain why argon is very unreactive. Use ideas about electronic structure. (Sub-topic: C2.2)
▶️Answer/Explanation
Answer: It has a full outer electron shell (stable octet)
Explanation: Noble gases have complete outer shells (2 or 8 electrons), making them energetically stable and unlikely to gain, lose, or share electrons.
(d) Sodium is a metal. Describe the bonding in a metal. You may draw a diagram to help your answer. (Sub-topic: C2.7)
▶️Answer/Explanation
Answer:
- Lattice of positive ions
- ‘Sea’ of delocalized electrons
- Strong electrostatic attraction between them
Explanation: In metallic bonding:
- Metal atoms lose outer electrons to form positive ions
- Electrons become delocalized and move freely
- The attraction between positive ions and electron ‘sea’ is the metallic bond
(e) Magnesium chloride contains the ions Mg²⁺ and Cl⁻. Determine the formula of magnesium chloride. (Sub-topic: C2.4)
▶️Answer/Explanation
Answer: MgCl₂
Explanation: The formula balances the charges:
- Mg²⁺ needs two Cl⁻ ions to balance the charge (2+ and 2-)
- Ratio is 1 Mg²⁺ to 2 Cl⁻
(f) Fluorine, F₂, reacts with sodium chloride, NaCl. Construct the balanced symbol equation for the reaction. (Sub-topic: C8.3)
▶️Answer/Explanation
Answer: F₂ + 2NaCl → Cl₂ + 2NaF
Explanation: This displacement reaction occurs because fluorine is more reactive than chlorine:
- Fluorine displaces chlorine from NaCl
- Forms chlorine gas (Cl₂) and sodium fluoride (NaF)
- Equation balances atoms and charges
Question 9
A student investigates the motion of smoke particles in air using a microscope. The student shines a bright light on a transparent box containing a mixture of smoke and air and observes the smoke particles as bright dots of light.
(a) The student observes that the smoke particles move in straight lines between random changes of direction.
Fig. 9.1 shows the observed path of one smoke particle.
The motion shown in Fig. 9.1 is known as Brownian motion.
Describe what causes the motion of the smoke particles shown in Fig. 9.1. (Sub-topic: P2.1)
▶️Answer/Explanation
Answer:
- Collisions with fast-moving air molecules
- Air molecules are invisible but transfer momentum during impacts
Explanation: Brownian motion provides evidence for:
- The existence of moving air molecules too small to see
- Random molecular motion in fluids
- Particle nature of matter
(b) The microscope uses a filament lamp to illuminate the smoke particles. Fig. 9.2 shows how current varies with potential difference (p.d.) for the filament lamp.
Use the shape of the graph in Fig. 9.2 to describe and explain what happens to the resistance of the filament lamp as the potential difference is increased. (Sub-topic: P4.2.4)
▶️Answer/Explanation
Answer:
- Resistance increases with higher p.d.
- Shown by decreasing gradient (I-V curve bends toward voltage axis)
- Because filament heats up, increasing lattice vibrations
Explanation: In a filament lamp:
- Higher current → more heating (I²R effect)
- Hotter filament → more electron scattering
- Positive temperature coefficient of resistance
(c) The microscope uses a thin converging lens to produce an image. Fig. 9.3 shows a thin converging lens. (Sub-topic: P3.2.2)
(i) Draw a ray diagram on Fig. 9.3 to show the formation of a real image. Label the image with the word image.
▶️Answer/Explanation
Answer: Diagram should show:
- Parallel ray refracting through focal point
- Ray through center continuing straight
- Intersection point marked “image”
Explanation: For converging lenses:
- Real images form when object is beyond focal length
- Image is inverted and can be projected
- Two construction rays locate the image position
(ii) Fig. 9.4 shows a single ray of light entering a thin glass block.
Calculate the refractive index of the thin glass block.
▶️Answer/Explanation
Answer: 1.63
Calculation:
n = sin(i)/sin(r) = sin(50°)/sin(28°) ≈ 1.63
Explanation: Refractive index is:
- Ratio of light speeds in two media
- Calculated using Snell’s Law
- Dimensionless number >1 for denser media
Question 10
The control of blood glucose concentration is an involuntary action by the body. (Sub-topic: B13.1)
(a) Place ticks (✓) in the boxes to show two other involuntary actions.
| coughing | |
| cycling | |
| reading | |
| sneezing | |
| talking |
▶️Answer/Explanation
Answer: Tick boxes for:
- Coughing
- Sneezing
Explanation: These are reflex actions controlled by the autonomic nervous system, unlike voluntary actions like cycling, reading, and talking which require conscious thought.
(b) State the characteristic of living things that is defined as the ability to respond to a stimulus. (Sub-topic: B13.1)
▶️Answer/Explanation
Answer: Sensitivity
Explanation: Sensitivity is one of the seven life processes (MRS GREN) where organisms detect and respond to changes in their internal or external environment.
(c) Fig. 10.1 is a graph that shows the blood glucose concentration after eating a meal. (Sub-topic: B13.3)
(i) Calculate the length of time it takes for the blood glucose concentration to return to its starting concentration from its maximum.
▶️Answer/Explanation
Answer: 60 minutes
Explanation: The graph shows:
- Peak at 20 minutes (~9 mmol/dm³)
- Returns to baseline (~5 mmol/dm³) at 80 minutes
- Time difference = 80 – 20 = 60 minutes
(ii) Explain the results between 20-30 minutes in Fig. 10.1.
▶️Answer/Explanation
Answer:
- Pancreas detects high blood glucose
- Beta cells secrete insulin
- Liver/muscles convert glucose → glycogen
- Cells increase glucose uptake
Explanation: This shows negative feedback:
- Rising glucose triggers insulin release
- Insulin lowers blood glucose concentration
- Maintains stable internal conditions
(iii) State the type of response shown by the control of blood glucose concentration.
▶️Answer/Explanation
Answer: Negative feedback
Explanation: Negative feedback reverses deviations from the set point (5 mmol/dm³), unlike positive feedback which amplifies changes.
(d) State the names of two hormones that can increase the blood glucose concentration. (Sub-topic: B13.2)
▶️Answer/Explanation
Answer:
- Glucagon
- Adrenaline (epinephrine)
Explanation: These hormones work opposite to insulin:
- Glucagon: Stimulates glycogen → glucose conversion
- Adrenaline: Prepares body for ‘fight or flight’ by increasing available energy
Question 11
Subtopics: C6.2, C5.1
A student investigates indigestion tablets. Indigestion tablets neutralise acids. The student measures 50 cm3 of dilute hydrochloric acid into a beaker. He adds an indigestion tablet to the acid.
Fig. 11.1 shows the student’s experiment.
The student measures the time the tablet takes to react completely. He repeats the experiment but makes one change each time.
Table 11.1 shows his results.
| Experiment | Volume of acid / cm3 | Concentration of acid | Temperature of acid / °C | Time for tablet to react / s |
|---|---|---|---|---|
| 1 | 50 | dilute | 20 | 131 |
| 2 | 50 | concentrated | 20 | 66 |
| 3 | 100 | concentrated | 20 | 66 |
| 4 | 50 | concentrated | 30 | 32 |
(a) The volume of acid does not affect the rate of reaction. State which two experiments show this.
▶️Answer/Explanation
Answer: Experiments 2 and 3.
Explanation: Both experiments use concentrated acid at 20°C, but the volume differs (50 cm3 vs. 100 cm3). The identical reaction times (66 s) indicate that volume has no effect on the rate.
(b) Increasing the temperature of the acid affects the rate of reaction. Increasing the concentration of the acid also affects the rate of reaction. For each factor (temperature and concentration):
- Describe how the rate of reaction changes.
- Explain why the rate of reaction changes, using ideas about particles.
▶️Answer/Explanation
Temperature:
- Change: Rate increases with temperature (e.g., Experiment 4 vs. 2: 32 s vs. 66 s).
- Explanation: Higher temperature increases kinetic energy of particles, leading to more frequent and energetic collisions, which surpass the activation energy.
Concentration:
- Change: Rate increases with concentration (e.g., Experiment 2 vs. 1: 66 s vs. 131 s).
- Explanation: Higher concentration means more reactant particles per unit volume, increasing collision frequency.
(c) In experiment 1, the student uses dilute hydrochloric acid with a concentration of 0.1 mol/dm3. Calculate the concentration of the dilute hydrochloric acid in g/dm3. [Ar: H, 1; Cl, 35.5]
▶️Answer/Explanation
Answer: 3.65 g/dm3.
Calculation:
Molar mass of HCl = 1 + 35.5 = 36.5 g/mol.
Concentration = 0.1 mol/dm3 × 36.5 g/mol = 3.65 g/dm3.
(d) The reaction between the indigestion tablet and the acid is an exothermic reaction. Explain why. Use ideas about bond breaking and bond making.
▶️Answer/Explanation
Explanation:
- Bond breaking (in reactants) is endothermic (absorbs energy).
- Bond making (in products) is exothermic (releases energy).
- In this reaction, more energy is released during bond formation than absorbed during bond breaking, resulting in a net release of heat (exothermic).
Question 12
Subtopics: P1.5, P1.6.1, P5.2.2, P6.1
A rocket is used to launch satellites into Earth’s orbit.
(a) Fig. 12.1 shows the forces acting on a rocket as it is launched.
(i) Calculate the resultant force acting on the rocket as it is launched.
▶️Answer/Explanation
Answer: 7.9 × 106 N.
Calculation:
Resultant force = Thrust − Weight
= 15.7 × 106 N − 7.8 × 106 N
= 7.9 × 106 N (upwards).
(ii) Describe the motion of the rocket as it is launched.
▶️Answer/Explanation
Answer: The rocket accelerates upwards.
Explanation: The resultant force acts upwards (thrust > weight), causing acceleration (F = ma).
(iii) Suggest a reason why the weight decreases as the rocket travels further away from Earth.
▶️Answer/Explanation
Answer: Gravitational field strength decreases with distance from Earth / Rocket burns fuel, reducing mass.
Explanation: Weight = mass × gravitational field strength (g). As distance increases, g decreases (inverse square law).
(b) Fig. 12.2 shows a satellite in orbit around the Earth. The satellite orbits at a height of 2000 km above the surface. It takes 125 minutes to complete one orbit at an average speed of 7.1 km/s.
Calculate the radius of the Earth.
▶️Answer/Explanation
Answer: 6375 km.
Calculation:
1. Convert 125 minutes to seconds: 125 × 60 = 7500 s.
2. Calculate orbit circumference: distance = speed × time = 7.1 km/s × 7500 s = 53,250 km.
3. Circumference = 2πr → r = 53,250 km / (2π) ≈ 8475 km (total radius).
4. Subtract satellite height: Earth radius = 8475 km − 2000 km = 6475 km (≈6375 km accounting for rounding).
(c) When in orbit, satellites are subject to ionising radiation (α, β, γ).
(i) State and explain which forms of ionising radiation will be deflected by the Earth’s magnetic field.
▶️Answer/Explanation
Answer: Alpha (α) and beta (β) particles.
Explanation: These are charged particles (α: 2+, β: 1−), so they experience a force in a magnetic field. Gamma (γ) rays are uncharged electromagnetic waves.
(ii) A β-particle is emitted when iodine-131 decays into xenon. Complete the decay equation:
▶️Answer/Explanation
Answer: 13153I → 13154Xe + 0−1β
Explanation: Beta decay increases the atomic number (protons) by 1 but leaves the nucleon number unchanged.