Question 1
(a) (i) Subtopic: B6.1 Photosynthesis
(ii) Subtopic: B6.1 Photosynthesis
(b) Subtopic: B5 Enzymes
(c) (i) Subtopic: B6.1 Photosynthesis
(ii) Subtopic: B8.4 Translocation
A student investigates the effect of temperature on the rate of photosynthesis in elodea (an aquatic plant).
Fig. 1.1 shows the apparatus used.
The student counts the number of bubbles of gas released in one minute. They repeat this with water at different temperatures.
Fig. 1.2 is a graph of their results.
(a) (i) Name the gas released by the elodea.
▶️Answer/Explanation
Oxygen
(ii) Describe the results shown in Fig. 1.2.
Include data from Fig. 1.2 in your answer.
▶️Answer/Explanation
The number of bubbles released (rate of photosynthesis) increases with temperature up to a peak at 45°C (71 bubbles per minute), then decreases at higher temperatures.
(b) Photosynthesis is an enzyme-controlled reaction. Explain the results between 55–65°C.
▶️Answer/Explanation
At high temperatures (55-65°C), the enzymes involved in photosynthesis begin to denature. This changes the shape of their active sites, so substrates can no longer fit properly. As a result, the rate of the reaction (photosynthesis) decreases, leading to fewer bubbles being produced.
(c) Most photosynthesis occurs in the leaves.
Some of the carbohydrates produced are stored in the roots.
(i) Name a carbohydrate stored in the roots of plants.
▶️Answer/Explanation
Starch
(ii) Describe how carbohydrates are transported from the leaves to the roots.
▶️Answer/Explanation
Carbohydrates are transported as sucrose through the phloem in a process called translocation. The sucrose is actively loaded into the phloem at the source (leaves) and unloaded at the sink (roots).
Question 2
(a) Subtopic: C2.2 Atomic structure and the Periodic Table
(b) (i) Subtopic: C2.3 Isotopes
(ii) Subtopic: C2.3 Isotopes
(c) (i) Subtopic: C2.4 Ions and ionic bonds
(ii) Subtopic: C2.5 Simple molecules and covalent bonds
(a) Table 2.1 shows some information about the structure of atoms. Complete Table 2.1.
| particle | charge | relative mass |
|---|---|---|
| electron | …… | …… |
| neutron | …… | 1 |
| proton | +1 | …… |
▶️Answer/Explanation
| particle | charge | relative mass |
|---|---|---|
| electron | -1 | 0.0005 (or negligible) |
| neutron | 0 | 1 |
| proton | +1 | 1 |
(b) There are two isotopes of bromine. One isotope is called bromine-79 and the other is called bromine-81.
(i) Table 2.2 shows some information about one atom of each isotope of bromine. Complete Table 2.2.
▶️Answer/Explanation
(ii) The two isotopes of bromine have the same chemical properties. Explain why.
▶️Answer/Explanation
Isotopes have the same chemical properties because they have the same number of electrons in their outer shell (both have seven valence electrons). Chemical properties are determined by electron configuration, not by the number of neutrons.
(c) Sodium is a metal. Bromine is a non-metal.
Sodium reacts with bromine to form sodium bromide.
Sodium bromide is an ionic compound.
(i) Describe how metallic elements and non-metallic elements form ionic bonds.
▶️Answer/Explanation
Metallic elements lose electrons to form positive ions (cations), while non-metallic elements gain electrons to form negative ions (anions). The oppositely charged ions are then strongly attracted to each other by electrostatic forces, forming an ionic bond. For example, sodium (a metal) loses one electron to become Na⁺, and bromine (a non-metal) gains one electron to become Br⁻, forming NaBr.
(ii) Explain why bromine, Br₂, has a low melting point.
▶️Answer/Explanation
Bromine has a low melting point because it exists as simple covalent molecules (Br₂). The molecules are held together by weak intermolecular forces (London dispersion forces), which require little energy to overcome. The strong covalent bonds within each Br₂ molecule don’t break during melting – only the weak forces between molecules are overcome.
Question 3
(a) Subtopic: P1.6.1 Energy
(b) Subtopic: P1.7 Pressure
(c) Subtopic: P3.4 Sound
(d) (i)-(iii) Subtopic: P3.4 Sound
(a) An elephant of mass 3800 kg is moving at 0.4 m/s.
Calculate the kinetic energy of the elephant.
▶️Answer/Explanation
Answer: 300 J
Explanation:
To calculate kinetic energy, we use the formula:
\[ KE = \frac{1}{2}mv^2 \] where \( m = 3800 \) kg and \( v = 0.4 \) m/s.
Substituting the values:
\[ KE = \frac{1}{2} \times 3800 \times (0.4)^2 = \frac{1}{2} \times 3800 \times 0.16 = 300 \, \text{J} \]
(b) The elephant stands with all four feet on the ground. The area of each foot is 0.06 m². The gravitational field strength is 10 N/kg.
Calculate the pressure exerted by the elephant on the ground.
▶️Answer/Explanation
Answer: 160,000 N/m²
Explanation:
First, calculate the elephant’s weight (force):
\[ \text{Weight} = \text{mass} \times \text{gravitational field strength} = 3800 \times 10 = 38,000 \, \text{N} \] Next, calculate the total area of the elephant’s feet:
\[ \text{Total area} = 4 \times 0.06 = 0.24 \, \text{m}^2 \] Finally, calculate the pressure:
\[ \text{Pressure} = \frac{\text{Force}}{\text{Area}} = \frac{38,000}{0.24} = 160,000 \, \text{N/m}^2 \]
(c) Infrasound is a very low frequency sound wave which is below the lowest frequency that a human is able to hear. Elephants communicate with each other using infrasound. Suggest a possible frequency for infrasound.
Explain your answer.
▶️Answer/Explanation
Answer: 15 Hz (any value below 20 Hz is acceptable)
Explanation:
Humans can typically hear sounds in the range of 20 Hz to 20,000 Hz. Infrasound refers to sound waves with frequencies below 20 Hz, which are inaudible to humans. Elephants use these low-frequency sounds for long-distance communication.
(d) Fig. 3.1 represents the infrasound wave travelling through the air as a series of compressions and rarefactions.
(i) On Fig. 3.1 label one compression with the letter C.
(ii) On Fig. 3.1 use a double headed arrow (⇄) to indicate one wavelength.
(iii) Describe the difference between a compression and a rarefaction in terms of particles in air.
▶️Answer/Explanation
Answer:
(i) The compression (C) should be labeled where air particles are closest together.
(ii) The wavelength (⇄) should span from one compression to the next or one rarefaction to the next.
(iii) In a compression, air particles are closer together, creating a region of higher pressure. In a rarefaction, air particles are farther apart, creating a region of lower pressure.
Explanation:
Sound waves are longitudinal waves consisting of compressions (high-pressure regions) and rarefactions (low-pressure regions). The wavelength is the distance between two consecutive compressions or rarefactions.
Question 4
(a) (i)-(ii) Subtopic: B16.3 Monohybrid inheritance
(b) (i)-(ii) Subtopic: B16.3 Monohybrid inheritance
(c) Subtopic: B17.1 Variation
(d) Subtopic: B11 Gas exchange in humans
(a) Cystic fibrosis is an inherited disease.
- The allele for developing cystic fibrosis is recessive, b.
- The allele for not developing cystic fibrosis is dominant, B.
People with a heterozygous genotype are described as carriers of the disease. They can pass the allele to their offspring but do not show the symptoms of cystic fibrosis.
Fig. 4.1 is a pedigree diagram showing the inheritance of cystic fibrosis. Each person is represented by a letter.
(i) State the total number of people that are carriers of cystic fibrosis in Fig. 4.1.
▶️Answer/Explanation
Answer: 7
Explanation: Carriers have the genotype Bb. Count all individuals in the pedigree diagram with this genotype across all three generations.
(ii) Identify the letter of one person who has cystic fibrosis.
▶️Answer/Explanation
Answer: R or V (as examples)
Explanation: Individuals with cystic fibrosis must have the genotype bb. Identify any individual in the diagram with this genotype.
(b) Generation 3 had offspring of their own.
The boxes on the left of Fig. 4.2 show the genotypes of the genetic cross.
The boxes on the right of Fig. 4.2 show the genotypes of the offspring.
(i) On Fig. 4.2, draw one line from each genetic cross to its offspring genotypes.
▶️Answer/Explanation
Answer:
cross 1: Bb × Bb → BB, Bb, bb
cross 2: BB × Bb → BB, Bb
cross 3: Bb × bb → Bb, bb
cross 4: bb × bb → bb
Explanation:
1. Bb × Bb produces 25% BB, 50% Bb, 25% bb (Mendelian ratio 1:2:1)
2. BB × Bb produces 50% BB and 50% Bb
3. Bb × bb produces 50% Bb and 50% bb
4. bb × bb produces 100% bb (pure breeding recessive)
(ii) Name the type of breeding shown in genetic cross 4 in Fig. 4.2.
▶️Answer/Explanation
Answer: pure breeding
Explanation: When two recessive homozygous individuals (bb) are crossed, they will always produce offspring with the same recessive genotype (bb), showing pure breeding for the recessive trait.
(c) Explain why cystic fibrosis is an example of discontinuous variation.
▶️Answer/Explanation
Answer:
– Limited number of phenotypes (either has CF or doesn’t)
– Caused by genes alone (not influenced by environment)
Explanation: Discontinuous variation shows distinct categories with no intermediates. CF is determined by a single gene with clear dominant/recessive alleles, resulting in either having the disease or not, with no spectrum in between.
(d) One of the symptoms of cystic fibrosis is the excess production of thick sticky mucus in the airways of the lungs.
Cilia find it difficult to remove the excess mucus.
Explain the effects of this on the gas exchange system.
▶️Answer/Explanation
Answer:
– Alveoli get blocked by mucus
– Reduces surface area for gas exchange
– Pathogens/bacteria not removed, increasing infection risk
– Causes breathing difficulties/coughing/wheezing
Explanation: The thick mucus physically blocks airways and alveoli, preventing efficient oxygen and carbon dioxide exchange. The impaired cilia function allows pathogens to accumulate, leading to frequent lung infections, which further damage lung tissue over time.
Question 5
(a) (i)-(ii) Subtopic: C4.1 Electrolysis
(b) (i)-(iii) Subtopic: C4.1 Electrolysis
(c) Subtopic: C6.3 Redox
Fig. 5.1 shows the electrolysis of dilute sulfuric acid.
(a) Hydrogen gas, H2, is made at the cathode.
(i) State the name of the gas made at the anode.
▶️Answer/Explanation
Oxygen
Explanation: During the electrolysis of dilute sulfuric acid, water is decomposed. At the anode, hydroxide ions (OH–) are discharged to form oxygen gas and water.
(ii) Write the ionic half-equation for the formation of hydrogen gas.
▶️Answer/Explanation
2H+ + 2e– → H2
Explanation: At the cathode, hydrogen ions (H+) gain electrons (reduction) to form hydrogen gas. The equation shows two hydrogen ions each gaining one electron to form one hydrogen molecule.
(b) Hydrogen gas is also made by the electrolysis of concentrated aqueous sodium chloride.
Chlorine gas is made at the anode in this process.
The ionic half-equation for the reaction is shown:
2Cl– – 2e– → Cl2
(i) State if this reaction is oxidation or reduction. Explain your answer.
▶️Answer/Explanation
Oxidation because electrons are lost (from chloride ions).
Explanation: Oxidation is defined as the loss of electrons. In this half-equation, chloride ions (Cl–) lose electrons to form chlorine gas (Cl2), which is therefore an oxidation process.
(ii) Describe the test for chlorine gas and its positive result.
▶️Answer/Explanation
Test: Damp litmus paper
Result: The paper is bleached (turns white)
Explanation: Chlorine gas is a powerful bleaching agent. When damp litmus paper is exposed to chlorine, the colored compounds in the paper are oxidized, causing the paper to lose its color.
(iii) The total volume of chlorine gas produced at 25°C in an electrolysis experiment is 4.8 cm3.
Calculate the number of moles of chlorine gas in 4.8 cm3.
The molar gas volume is 24 dm3.
▶️Answer/Explanation
Number of moles = 2.0 × 10-4 or 0.0002
Calculation:
1. Convert cm3 to dm3: 4.8 cm3 = 0.0048 dm3
2. Number of moles = Volume ÷ Molar volume = 0.0048 ÷ 24 = 0.0002 mol
Alternatively: 4.8 ÷ (24 × 1000) = 0.0002 mol
(c) Chlorine reacts with aqueous sodium iodide, NaI.
State the formulae of the products made in this reaction.
▶️Answer/Explanation
NaCl and I2
Explanation: This is a displacement reaction where the more reactive chlorine displaces iodine from sodium iodide. The products are sodium chloride (NaCl) and iodine (I2). The balanced equation is: Cl2 + 2NaI → 2NaCl + I2
Question 6
(a) Subtopic: P2.3.1 Conduction
(b) Subtopic: P2.2.2 Melting, boiling and evaporation
(c) Subtopic: P2.2.1 Thermal expansion of solids, liquids and gases
(d) Subtopic: P4.2.4 Resistance
(e) Subtopic: P4.3.2 Series and parallel circuits
(f) (i)-(ii) Subtopic: P4.5.2 The a.c. generator
(a) Describe how thermal energy passes through copper by conduction.
▶️Answer/Explanation
Thermal energy is transferred through copper via the vibration of atoms. These vibrations are passed from one atom to neighboring atoms. Additionally, delocalized electrons in the metal carry energy through the material, contributing to efficient heat transfer.
(b) Copper boils at 2562°C. Describe two differences between boiling and evaporation.
▶️Answer/Explanation
1. Boiling occurs at a specific temperature (boiling point) throughout the liquid, while evaporation can occur at any temperature and only at the liquid’s surface.
2. During boiling, bubbles of vapor form within the liquid, whereas evaporation involves molecules escaping from the surface without bubble formation.
(c) Equal volumes of air, copper and water are heated from 10°C to 90°C. State which of these materials will expand:
▶️Answer/Explanation
most – air
least – copper
(d) A copper wire of length 0.5 m has a resistance of 0.02 Ω. Determine the resistance of another copper wire of length 0.25 m that has twice the cross-sectional area.
▶️Answer/Explanation
Using the formula R = ρL/A:
Original wire: 0.02 = ρ(0.5)/A
New wire: R = ρ(0.25)/2A = (ρ(0.5)/A)/4 = 0.02/4 = 0.005 Ω
resistance = 0.005 Ω
(e) Two wires are connected in parallel. One wire has a resistance of 0.40 Ω. The other wire has a resistance of 0.60 Ω.
Calculate the combined resistance of the two wires connected together in parallel.
▶️Answer/Explanation
Using the parallel resistance formula: 1/Rtotal = 1/R1 + 1/R2
1/Rtotal = 1/0.40 + 1/0.60 = 2.5 + 1.666… = 4.166…
Rtotal = 1/4.166… ≈ 0.24 Ω
resistance = 0.24 Ω
(f) Copper wire is used in the coil of a generator. Fig. 6.1 shows a simple a.c. generator.
(i) On Fig. 6.1 label the coil with the letter C.
(ii) An electromotive force (e.m.f.) is induced in the rotating coil. State two factors that would increase the magnitude of the induced e.m.f.
▶️Answer/Explanation
(i) [The coil should be labeled with a “C” in the diagram]
(ii) 1. Increasing the speed of rotation of the coil
2. Increasing the strength of the magnetic field
Question 7
(a) (i)-(ii) Subtopic: B15.4 Sexual reproduction in humans
(b) (i)-(iii) Subtopic: B15.4 Sexual reproduction in humans
(a) Fig. 7.1 is a diagram of a fetus inside a uterus.
Table 7.1 shows the functions of some parts shown in Fig. 7.1.
(i) Complete Table 7.1.
| name of part | letter in Fig. 7.1 | function |
|---|---|---|
| …… | F | carries nutrient rich blood to fetus |
| …… | …. | contains amniotic fluid |
| amniotic fluid | …. | protects baby from mechanical damage |
| …… | ….. | site of exchange between the blood of the fetus and the mother |
▶️Answer/Explanation
| name of part | letter in Fig. 7.1 | function |
|---|---|---|
| umbilical cord | F | carries nutrient rich blood to fetus |
| amnion/amniotic sac | D | contains amniotic fluid |
| amniotic fluid | E | protects baby from mechanical damage |
| placenta | A | site of exchange between the blood of the fetus and the mother |
Explanation: The umbilical cord connects the fetus to the placenta and transports nutrients. The amnion is the sac that contains the amniotic fluid, which cushions the fetus. The placenta is where exchange of materials between mother and fetus occurs.
(ii) Name one gas that is transferred from the blood of the fetus to the mother’s blood.
▶️Answer/Explanation
carbon dioxide
Explanation: Carbon dioxide is a waste product of fetal metabolism that diffuses from the fetal blood into the mother’s blood at the placenta to be excreted.
(b) Pregnancy can occur after fertilisation.
(i) Name the two types of gametes involved in fertilisation in humans.
▶️Answer/Explanation
sperm and egg (ova)
Explanation: Fertilization occurs when a male gamete (sperm) fuses with a female gamete (egg/ovum) to form a zygote.
(ii) Name the process that is involved in the production of gametes.
▶️Answer/Explanation
meiosis
Explanation: Gametes are produced through meiosis, which halves the chromosome number to create haploid cells.
(iii) Describe two differences between the nuclei in gametes and those in body cells.
▶️Answer/Explanation
1. Gametes have haploid nuclei (23 chromosomes) while body cells have diploid nuclei (46 chromosomes).
2. Gametes have unpaired chromosomes while body cells have paired homologous chromosomes.
Explanation: Gametes are haploid as they result from meiotic division, while body (somatic) cells are diploid as they result from mitotic division. This ensures the correct chromosome number is maintained when gametes fuse during fertilization.
Question 8
(a) Subtopic: C6.1 Physical and chemical changes
(b) Subtopic: C3.3 The mole and the Avogadro constant
(c) Subtopic: C6.2 Rate of reaction
(d) (i)-(ii) Subtopic: C5.1 Exothermic and endothermic reactions
Calcium carbonate, CaCO3, reacts with dilute hydrochloric acid. Calcium chloride, CaCl2, carbon dioxide and water are made.
(a) Write the balanced symbol equation for this reaction.
▶️Answer/Explanation
CaCO3 + 2HCl → CaCl2 + H2O + CO2
Solution:
1. Write the correct formulas for all reactants and products:
– Reactants: CaCO3 (calcium carbonate) + HCl (hydrochloric acid)
– Products: CaCl2 (calcium chloride) + H2O (water) + CO2 (carbon dioxide)
2. Balance the equation by ensuring equal numbers of each atom on both sides:
– Start with 1 CaCO3 + 1 HCl → 1 CaCl2 + 1 H2O + 1 CO2 (unbalanced)
– Notice we need 2 Cl atoms on the left, so change to 2 HCl
– Now the equation is balanced with:
1 Ca, 1 C, 3 O, 2 H, and 2 Cl atoms on both sides
(b) The hydrochloric acid used in the experiment is made by dissolving 0.75 moles of hydrogen chloride in 500 cm3 of water.
Calculate the concentration of the hydrochloric acid in mol/dm3.
▶️Answer/Explanation
1.5 mol/dm3
Solution:
1. Convert volume from cm3 to dm3:
500 cm3 = 500 ÷ 1000 = 0.5 dm3
2. Use the concentration formula:
Concentration (mol/dm3) = Number of moles ÷ Volume (dm3)
= 0.75 moles ÷ 0.5 dm3
= 1.5 mol/dm3
(c) The rate of this reaction can be changed by changing the concentration of the acid. Explain the effect of changing the concentration of the acid on the rate of the reaction. Use ideas about particles.
▶️Answer/Explanation
Explanation:
1. Higher concentration means more acid particles per unit volume
2. This leads to more frequent collisions between acid particles and calcium carbonate particles
3. More successful collisions occur per unit time (collision theory)
4. Therefore, the rate of reaction increases with higher concentration
Detailed reasoning:
When the concentration of hydrochloric acid increases, there are more HCl molecules in the same volume of solution. These molecules move randomly and collide with the calcium carbonate surface. With more HCl molecules present, the frequency of collisions between HCl and CaCO3 increases. Since chemical reactions occur when particles collide with sufficient energy (activation energy) and proper orientation, more collisions mean more successful reactions per second, resulting in a faster overall reaction rate.
(d) The reaction between calcium carbonate and hydrochloric acid is exothermic.
(i) State the meaning of an exothermic reaction.
▶️Answer/Explanation
A reaction that releases thermal energy/heat to the surroundings
Additional information:
In exothermic reactions, the total energy of the products is less than the total energy of the reactants. The “missing” energy is released as heat. This can be measured as a temperature increase in the reaction mixture. Examples include combustion reactions and many neutralization reactions.
(ii) Fig. 8.1 shows an energy level diagram for an exothermic reaction.
State which arrow, A, B or C, shows the activation energy for the reaction.
▶️Answer/Explanation
B
Explanation:
In an energy profile diagram for an exothermic reaction:
– Arrow A would typically show the energy of reactants
– Arrow B shows the activation energy (energy barrier between reactants and transition state)
– Arrow C would show the energy change (ΔH) of the reaction
Activation energy is always the vertical distance from the reactants to the peak of the energy curve, regardless of whether the reaction is exothermic or endothermic.
Question 9
(a) (i)-(ii) Subtopic: P3.3 Electromagnetic spectrum
(b) (i)-(ii) Subtopic: P3.2.2 Refraction of light
(c) (i)-(ii) Subtopic: P5.2.5 Applications and safety precautions
▶️Answer/Explanation
Answer: 3 × 108 m/s
Explanation: All electromagnetic waves, including visible light and γ-radiation, travel at the same speed in a vacuum, which is the speed of light (3 × 108 m/s).
State the unit of your answer.
▶️Answer/Explanation
Answer: 3.75 × 1019 Hz
Explanation:
Using the wave equation: v = fλ
Where: v = speed (3 × 108 m/s), λ = wavelength (8 × 10-12 m)
f = v/λ = (3 × 108)/(8 × 10-12) = 3.75 × 1019 Hz
The unit is hertz (Hz), which is the SI unit for frequency.
(i) Fig. 9.1 shows a ray of light passing into an optical fibre.
On Fig. 9.1 continue the ray of light to show its path through the optical fibre. 
▶️Answer/Explanation
Answer: The light ray should zigzag down the fiber, reflecting off the walls at angles greater than the critical angle.
Explanation: The light ray undergoes total internal reflection, bouncing off the inner walls of the fiber at angles that maintain the reflection (greater than the critical angle for the fiber material).
(ii) Explain why total internal reflection occurs.
▶️Answer/Explanation
Answer: Total internal reflection occurs when light travels from a denser medium to a less dense medium and the angle of incidence exceeds the critical angle.
Explanation: When light tries to pass from the optically denser core of the fiber (higher refractive index) to the less dense cladding (lower refractive index) at an angle greater than the critical angle, it is completely reflected back into the core rather than being refracted out.
Small quantities of I-123 are absorbed by the thyroid gland.
I-123 emits γ-radiation which is detected outside the body.
I-123 has a half-life of 13 hours.
(i) Give two reasons why I-123 is suitable for use inside the body.
▶️Answer/Explanation
Answer:
1. Relatively short half-life (13 hours) means it doesn’t remain radioactive in the body for too long
2. γ-radiation can pass through body tissues to be detected externally
Explanation: I-123 is ideal because its radiation can be detected outside the body while minimizing patient exposure time due to its moderate half-life. Additionally, γ-radiation is less damaging than other types of radiation (like alpha or beta) as it’s less ionizing.
(ii) A sample of I-123 contains 8 × 1014 atoms.
Sometime later 6 × 1014 atoms have decayed.
Calculate the time needed for this number of atoms to decay.
▶️Answer/Explanation
Answer: 26 hours
Explanation:
Initial atoms = 8 × 1014
Atoms remaining = 8 × 1014 – 6 × 1014 = 2 × 1014
Fraction remaining = 2/8 = 1/4
This represents two half-lives (1/2 → 1/4)
Time = 2 × half-life = 2 × 13 hours = 26 hours
Question 10
(a) (i)-(iii) Subtopic: B13.1 Coordination and response
(b) Subtopic: B13.1 Coordination and response
(c) Subtopic: B6.1 Photosynthesis
(d) (i)-(ii) Subtopic: B1.1 Characteristics of living organisms
(a) Plant shoots respond to stimuli such as light.
Fig. 10.1 shows the growth response of a shoot to light.
(i) Name the response shown in Fig. 10.1.
▶️Answer/Explanation
phototropism
(ii) Draw an X on Fig. 10.1 to show the area with the greatest cell elongation.
▶️Answer/Explanation
X should be drawn on the upper side of the shoot bend (the shaded/darker side opposite the light source)
(iii) Name the hormone that controls cell elongation.
▶️Answer/Explanation
auxin
(b) Fig. 10.2 shows a plant shoot with the tip removed.
State why the shoot in Fig. 10.2 did not bend.
▶️Answer/Explanation
The shoot tip produces auxin, which is necessary for the phototropic response. Without the tip, there is no auxin production and thus no differential growth causing bending.
(c) Explain why plants need magnesium ions for photosynthesis and healthy growth. Use ideas about energy in your answer.
▶️Answer/Explanation
Magnesium ions are essential for:
- Chlorophyll production – magnesium forms the central atom in chlorophyll molecules
- Light absorption – chlorophyll captures light energy which is converted to chemical energy during photosynthesis
- Growth – the chemical energy is used to synthesize glucose, which provides energy and building materials for plant growth
Without magnesium, plants cannot produce sufficient chlorophyll, leading to reduced photosynthesis and stunted growth.
(d) Growth is one of the characteristics of living things.
(i) Complete the definition of the term growth.
Growth is a …… increase in size and dry …… by an increase in cell number or cell size or both.
▶️Answer/Explanation
Growth is a permanent increase in size and dry mass by an increase in cell number or cell size or both.
(ii) State the name of two other characteristics of living things.
▶️Answer/Explanation
Any two from:
- Movement
- Reproduction
- Sensitivity
- Excretion
- Nutrition
- Respiration
Question 11
(a) Subtopic: C11.3 Fuels
(b) (i)-(iii) Subtopic: C11.4 Alkanes
(c) (i)-(ii) Subtopic: C11.6 Alcohols
Petroleum is separated into useful chemicals by fractional distillation.
Fig. 11.1 shows a fractionating column.
(a) Table 11.1 shows the uses of some of the fractions. Complete Table 11.1.
| fraction | use |
|---|---|
| bitumen | …… |
| diesel oil | fuel in diesel engines |
| naphtha | feedstock for making chemicals |
| gasoline | …… |
| refinery gas | …… |
▶️Answer/Explanation
Answer:
bitumen – road surfaces;
gasoline – fuel in cars;
refinery gas – bottled gas for heating or cooking or camping;
(b) Butane is a hydrocarbon found in refinery gas.
(i) State two properties of butane gas.
▶️Answer/Explanation
Answer:
1. Low boiling point
2. Colorless
3. Ignites easily (any two properties of gases)
(ii) Butane is a type of hydrocarbon called an alkane. Complete Fig. 11.2 to show the structure of a butane molecule. Show all of the atoms and all of the covalent bonds.
▶️Answer/Explanation
Answer:
H H H H
| | | |
H-C-C-C-C-H
| | | |
H H H H
(iii) The alkanes are a homologous series. Describe what is meant by a homologous series.
▶️Answer/Explanation
Answer:
A family of compounds with:
1. The same general formula
2. Similar chemical properties
3. Gradual change in physical properties
4. Each member differs by CH₂ from the next
(c) Ethene, C₂H₄, is an alkene made by the cracking of large alkane molecules.
Ethene reacts with steam to make ethanol, C₂H₅OH.
C₂H₄ + H₂O → C₂H₅OH
(i) State one other method of making ethanol.
▶️Answer/Explanation
Answer:
Fermentation of sugars (using yeast)
(ii) In an experiment 5.6 g of ethene reacts with excess steam. Calculate the maximum mass of ethanol that can be made.
▶️Answer/Explanation
Answer:
1. Calculate moles of ethene: 5.6g ÷ 28g/mol = 0.2 mol
2. 1:1 mole ratio means 0.2 mol ethanol produced
3. Mass of ethanol: 0.2 mol × 46g/mol = 9.2g
Final answer: 9.2 g
Question 12
(a) Subtopic: P4.2.1 Electrical charge
(b) (i)-(ii) Subtopic: P1.2 Motion
(c) (i)-(ii) Subtopic: P2.1.2 Particle model
(a) Fig. 12.1 shows an aircraft being refuelled using a plastic pipe.
As the fuel flows through the pipe, the fuel and pipe become electrically charged.
Explain why the fuel becomes negatively charged and the pipe becomes positively charged.
▶️Answer/Explanation
When fuel flows through the plastic pipe, friction occurs between the fuel molecules and the pipe’s inner surface. This friction causes electrons to transfer from the pipe material to the fuel. Since electrons carry a negative charge, the fuel gains electrons and becomes negatively charged, while the pipe loses electrons and becomes positively charged. This is an example of static electricity generation through friction (triboelectric effect).
(b) Fig. 12.2 is the speed-time graph for the aircraft during take-off.
(i) Calculate the acceleration at 25 seconds.
▶️Answer/Explanation
Acceleration = Change in velocity ÷ Time taken
From the graph at t=25s, velocity = 50 m/s (assuming uniform acceleration from t=5s to t=45s)
Acceleration = (50 m/s – 0 m/s) ÷ (45s – 5s) = 50 ÷ 40 = 1.25 m/s²
Note: The exact values would depend on the graph’s scale.
(ii) State how the graph shows that the acceleration of the aircraft is constant between 5.0 s and 45.0 s.
▶️Answer/Explanation
The graph shows a straight line with constant slope between 5.0s and 45.0s. In a speed-time graph, constant slope indicates constant acceleration because acceleration is the gradient of the speed-time graph.
(c)(i) During the flight the pressure inside the aircraft cabin decreases but the temperature is kept constant.
Use ideas about gas molecules to describe the change in pressure in terms of the arrangement and motion of molecules.
▶️Answer/Explanation
As pressure decreases at constant temperature:
1. The gas molecules become farther apart (less dense arrangement)
2. The molecules maintain the same average speed (since temperature is constant)
3. Fewer collisions occur per unit time between molecules and the cabin walls
4. Each collision exerts the same force, but the reduced frequency of collisions results in lower pressure
(ii) The aircraft flies at a high altitude. Some water on the outside of the aircraft body turns to ice.
Describe in terms of molecular motion and arrangement how ice differs from liquid water.
▶️Answer/Explanation
In ice compared to liquid water:
1. Molecular motion: Molecules vibrate about fixed positions rather than moving freely
2. Molecular arrangement: Molecules form a regular crystalline lattice structure instead of random arrangement
3. Spacing: Molecules are slightly farther apart in ice (which is why ice is less dense than water)
4. Energy: Molecules have lower kinetic energy in ice (potential energy increases due to hydrogen bonding)