Question 1
Subtopic: B15.4 Sexual reproduction in humans
(a) Fig. 1.1 is a diagram of the female reproductive system.
Identify the letter from Fig. 1.1 that represents:
where female gametes are released ……
where fertilisation occurs ……
where implantation occurs ……
where meiosis occurs. ……
▶️Answer/Explanation
Answers:
– Female gametes are released: E (ovary)
– Fertilisation occurs: A (fallopian tube/oviduct)
– Implantation occurs: D (uterus)
– Meiosis occurs: E (ovary)
Explanation:
In the female reproductive system:
1. Gametes are released from the ovaries (E)
2. Fertilization typically occurs in the fallopian tubes (A)
3. Implantation happens in the uterus (D)
4. Meiosis occurs in the ovaries (E) during oogenesis
(b) Female gametes in humans are called egg cells.
State the name of the male gamete in humans.
▶️Answer/Explanation
Answer: sperm
Explanation:
The male gamete in humans is called sperm. Gametes are reproductive cells, with females producing egg cells and males producing sperm cells.
(c) Table 1.1 compares some features of male and female gametes.
Complete Table 1.1.
| male gamete | female gamete | |
|---|---|---|
| relative size | ||
| number released at one time | usually one | |
| motility |
▶️Answer/Explanation
Answers:
| male gamete | female gamete | |
|---|---|---|
| relative size | small | large |
| number released at one time | millions/many | usually one |
| motility | motile | non-motile |
Explanation:
Male and female gametes differ significantly:
1. Size: Sperm are much smaller than egg cells
2. Quantity: Males produce millions of sperm per ejaculation, while females typically release one egg per menstrual cycle
3. Motility: Sperm are mobile with flagella for movement, while egg cells are stationary
(d) Complete the sentences to describe the adaptive features of egg cells.
One of the adaptive features of egg cells is that it has …… stores.
The egg cell also has a …… coating that changes after fertilisation.
▶️Answer/Explanation
Answers:
– energy
– jelly
Explanation:
Egg cells have special adaptations:
1. Energy stores (in the form of yolk) to support early embryonic development
2. A protective jelly coating (zona pellucida) that changes after fertilization to prevent polyspermy (multiple sperm fertilizing the same egg)
Question 2
(a)(i) Subtopic: C11.3 Fuels
(a)(ii) Subtopic: C11.3 Fuels
(b) Subtopic: C11.3 Fuels
(c) Subtopic: C11.1 Formulas and terminology
(d) Subtopic: C6.1 Physical and chemical changes
(e) Subtopic: C6.1 Physical and chemical changes
Petroleum is a fossil fuel. It can be separated into useful fractions by fractional distillation.
Fig. 2.1 shows a diagram of a fractionating column.
(a)(i) Explain why it is possible to separate the substances in petroleum by fractional distillation.
▶️Answer/Explanation
The different fractions in petroleum have different boiling points. During fractional distillation, the mixture is heated, and each fraction vaporizes at its specific boiling point. As the vapors rise through the column, they cool and condense at different heights according to their boiling points, allowing for separation.
(ii) On Fig. 2.1, write the letter X in the coolest part of the fractionating column.
▶️Answer/Explanation
The coolest part of the fractionating column is at the top, where the temperature is lowest. The letter X should be placed in the top section of the column, above all the fraction collection points.
(b) Table 2.1 shows the uses of some of the fractions. Complete Table 2.1.
| Fraction | Use |
|---|---|
| refinery gas | bottled gas for heating |
| gasoline | |
| naphtha | feedstock for making chemicals |
| diesel oil | |
| bitumen |
▶️Answer/Explanation
| Fraction | Use |
|---|---|
| gasoline | fuel in cars (petrol) |
| diesel oil | fuel in diesel engines |
| bitumen | road surfaces and roofing |
(c) Refinery gas contains propane, C3H8. Draw a diagram to show the structure of propane.
▶️Answer/Explanation
H H H | | | H-C-C-C-H | | | H H H
Propane has a straight chain structure with three carbon atoms single-bonded to each other, and each carbon atom is fully saturated with hydrogen atoms.
(d) Refinery gas also contains butane, C4H10. Butane burns in oxygen to make carbon dioxide and water. Construct the balanced symbol equation for this reaction.
▶️Answer/Explanation
2C4H10 + 13O2 → 8CO2 + 10H2O
To balance the equation:
1. Balance carbon atoms: 4 carbons in butane → 4 CO2 per butane molecule
2. Balance hydrogen atoms: 10 hydrogens → 5 H2O per butane molecule
3. Balance oxygen atoms: Requires 13/2 O2 per butane molecule
4. Multiply all coefficients by 2 to eliminate fractions
(e) Burning butane is a chemical change. Describe the difference between a chemical change and a physical change.
▶️Answer/Explanation
A chemical change results in the formation of new substances with different chemical properties (like burning butane producing CO2 and H2O), and the change is usually irreversible. A physical change only alters the physical state or appearance of a substance without changing its chemical composition (like melting ice), and the change is typically reversible.
Question 3
(a)(i) Subtopic: P1.4 Density
(a)(ii) Subtopic: P1.4 Density
(b)(i) Subtopic: P1.5.1 Effects of forces
(b)(ii) Subtopic: P1.6.4 Power
(b)(iii) Subtopic: P4.2.5 Electrical energy and electrical power
(a) Fig. 3.1 shows a piece of graphite with an irregular shape.
(i) Describe a method to determine the volume of the piece of graphite.
▶️Answer/Explanation
Method 1: Submerge the graphite in a graduated cylinder filled with water and measure the volume of water displaced.
Method 2: Measure the mass of graphite and divide by its known density (from reference tables) using V = m/d.
(ii) The piece of graphite has a mass of 33 g and a volume of 15 cm³. Calculate the density of the piece of graphite.
▶️Answer/Explanation
Solution:
Density = mass/volume = 33 g / 15 cm³ = 2.2 g/cm³
(b) Graphite can be used as a lubricant in machines with moving parts such as an electric drill.
(i) Describe, in terms of forces and energy transfers, how lubricants increase the efficiency of a machine.
▶️Answer/Explanation
Explanation:
1. Lubricants reduce friction between moving parts
2. Less heat energy is generated as waste
3. More input energy is converted to useful kinetic energy
4. This increases efficiency as less energy is wasted
(ii) An electric drill transfers 1200 J of electrical energy to 900 J of useful kinetic energy. Calculate the efficiency of the electric drill.
▶️Answer/Explanation
Calculation:
Efficiency = (Useful energy output / Total energy input) × 100%
= (900 J / 1200 J) × 100% = 75%
(iii) The electric motor in the drill has a current of 25 A when using an 18 V battery. Calculate the power output of the motor.
▶️Answer/Explanation
Calculation:
Power = Current × Voltage (P = IV)
= 25 A × 18 V = 450 W
Question 4
(a) Subtopic: B6.1 Photosynthesis
(b)(i) Subtopic: B5 Enzymes
(b)(ii) Subtopic: B5 Enzymes
(c)(i) Subtopic: B8.4 Translocation
(c)(ii) Subtopic: B4 Biological molecules
(d) Subtopic: B6.1 Photosynthesis
A student investigates the effect of temperature on the rate of photosynthesis.
Fig. 4.1 shows the apparatus they use.
The student counts the number of oxygen bubbles produced in one minute. He repeats this investigation, changing the temperature of the water each time. The number of oxygen bubbles produced per minute is equivalent to the rate of photosynthesis.
Table 4.1 shows the results.
| temperature/°C | number of bubbles produced per minute |
|---|---|
| 0 | 0 |
| 5 | 4 |
| 10 | 8 |
| 15 | 13 |
| 20 | 16 |
| 25 | 18 |
| 30 | 19 |
| 35 | 8 |
| 40 | 0 |
(a) State the temperature from Table 4.1 where the rate of photosynthesis is the highest.
▶️Answer/Explanation
Answer: 30
Explanation: The highest number of bubbles (19) is produced at 30°C, indicating the peak rate of photosynthesis at this temperature.
(b) Photosynthesis is an enzyme-controlled reaction.
(i) Explain the results in Table 4.1 between 5°C and 15°C.
▶️Answer/Explanation
Answer: As temperature increases from 5°C to 15°C, enzyme activity increases because molecules have more kinetic energy, leading to more frequent successful collisions between enzymes and substrates. This increases the rate of photosynthesis, shown by the increasing bubble count.
Explanation: The increase in bubble production reflects higher enzyme activity. Enzymes work faster at warmer temperatures (up to their optimum) because substrates and enzymes move faster and collide more frequently with sufficient energy for reactions.
(ii) State the temperature from Table 4.1 when all the enzymes involved in photosynthesis are completely denatured.
▶️Answer/Explanation
Answer: 40
Explanation: At 40°C, no bubbles are produced, indicating all photosynthetic enzymes have denatured (lost their 3D structure and function permanently).
(c) The carbohydrate glucose is also a product of photosynthesis.
Glucose is converted to different substances for transport and storage in a plant.
(i) Describe how carbohydrates are transported in a plant.
▶️Answer/Explanation
Answer: Carbohydrates are transported as sucrose through the phloem tissue via translocation, from sources (like leaves) to sinks (like roots or fruits).
Explanation: Glucose is converted to sucrose for transport because sucrose is more stable and soluble. The phloem uses active transport and hydrostatic pressure to move sugars throughout the plant.
(ii) State the larger molecule made from glucose that is used for storage in a plant.
▶️Answer/Explanation
Answer: starch
Explanation: Starch is an insoluble polysaccharide formed from many glucose molecules, ideal for storage in amyloplasts (e.g., in potatoes or seeds).
(d) Chlorophyll is also necessary for photosynthesis. State the energy transfer that chlorophyll is responsible for.
▶️Answer/Explanation
Answer: light → chemical
Explanation: Chlorophyll absorbs light energy (primarily red and blue wavelengths) and converts it to chemical energy stored in glucose molecules during photosynthesis.
Question 5
(a) Subtopic: C4.1 Electrolysis
(b) Subtopic: C4.1 Electrolysis
(c)(i) Subtopic: C4.1 Electrolysis
(c)(ii) Subtopic: C4.1 Electrolysis
(d) Subtopic: C6.3 Redox
This question is about electrolysis.
(a) The list shows the particles found in aqueous copper(II) sulfate.
Cu2+
H+
H2O
SO42-
OH–
State the formula of one particle attracted to the cathode during electrolysis. Choose from the list.
▶️Answer/Explanation
Cu2+ or H+
Explanation: The cathode is the negative electrode in electrolysis. During electrolysis of aqueous copper(II) sulfate, positively charged ions (cations) are attracted to the cathode. From the list, both Cu2+ (copper ions) and H+ (hydrogen ions) are positively charged and would be attracted to the cathode.
(b) Aqueous copper(II) sulfate conducts electricity. Explain why.
▶️Answer/Explanation
Because it contains ions that can move.
Explanation: Aqueous copper(II) sulfate solution contains free-moving ions (Cu2+ and SO42-). These ions are charge carriers that can move through the solution when a potential difference is applied, allowing electric current to flow.
(c) Fig. 5.1 shows the apparatus used for the electrolysis of aqueous copper(II) sulfate.
(i) State the name given to the positive electrode.
▶️Answer/Explanation
Anode
Explanation: In electrolysis, the positive electrode is always called the anode. This is where oxidation occurs and negative ions (anions) are attracted.
(ii) The purification (refining) of copper uses electrolysis. Describe how impure copper is purified by electrolysis. Include ionic half-equations in your answer.
▶️Answer/Explanation
1. Use impure copper as the anode and pure copper as the cathode
2. Use copper(II) sulfate solution as the electrolyte
3. When current passes, copper from the impure anode dissolves into solution: Cu → Cu2+ + 2e–
4. Copper ions in solution gain electrons at the cathode and deposit as pure copper: Cu2+ + 2e– → Cu
5. Impurities either fall to the bottom as anode sludge or remain in solution
Explanation: This process allows for high-purity copper to be obtained (99.99% pure). The key reactions are the oxidation at the anode where copper atoms lose electrons, and the reduction at the cathode where copper ions gain electrons.
(d) Look at this ionic half-equation:
Al3+ + 3e– → Al
State if this reaction is an example of oxidation or reduction. Explain your answer.
▶️Answer/Explanation
Reduction
Explanation: This is reduction because the aluminum ion (Al3+) is gaining electrons (3e–). Reduction is defined as the gain of electrons, while oxidation is the loss of electrons. In this case, the aluminum ion is being reduced to aluminum metal.
Question 6
(a) Subtopic: P1.2 Motion
(b) Subtopic: P1.2 Motion
(c) Subtopic: P1.6.1 Energy
(d)(i) Subtopic: P2.2.2 Melting, boiling and evaporation
(d)(ii) Subtopic: P2.2.2 Melting, boiling and evaporation
Fig. 6.1 shows a cheetah. Cheetahs are the fastest land animal and have a top speed of 30 m/s.
(a) State the difference between speed and velocity.
▶️Answer/Explanation
Answer: Velocity has direction while speed does not.
Explanation: Speed is a scalar quantity that only refers to how fast an object is moving (magnitude only), while velocity is a vector quantity that includes both speed and direction of motion.
(b) Fig. 6.2 shows a speed-time graph for a cheetah’s journey.
Describe the motion of the cheetah shown by the graph.
▶️Answer/Explanation
Answer: The cheetah:
1. Initially accelerates (increasing speed)
2. Then moves at constant speed (zero acceleration)
3. Finally decelerates (decreasing speed)
Explanation: The graph shows three distinct phases: an upward sloping line (positive acceleration), a horizontal line (constant speed, zero acceleration), and a downward sloping line (negative acceleration or deceleration).
(c) The mass of the cheetah is 42 kg. Calculate the kinetic energy of the cheetah when it is running at its maximum speed of 30 m/s.
▶️Answer/Explanation
Answer: 18,900 J
Explanation:
Using the kinetic energy formula: KE = ½mv²
Where:
m = 42 kg
v = 30 m/s
KE = 0.5 × 42 × (30)²
KE = 0.5 × 42 × 900
KE = 18,900 J
(d) A cheetah drinks water from a puddle. Over time, the water in the puddle evaporates. Evaporation and boiling both turn liquid water into a gas.
(i) State one difference between evaporation and boiling.
▶️Answer/Explanation
Answer: Evaporation occurs only at the surface of the liquid at any temperature, while boiling occurs throughout the liquid at a specific temperature (boiling point).
Alternative answer: Evaporation is a slow process that happens below the boiling point, while boiling is rapid and occurs at the boiling point.
(ii) State two ways to increase the rate of evaporation from the puddle.
▶️Answer/Explanation
Answer: Any two from:
1. Increase the temperature of the water
2. Increase the surface area of the puddle
3. Increase air movement (wind/draught) over the puddle
4. Reduce humidity in the surrounding air
Explanation: These factors all increase the rate at which water molecules can escape from the liquid surface into the air.
Question 7
(a)(i) Subtopic: B13.2 Hormones
(a)(ii) Subtopic: B13.2 Hormones
(b)(i) Subtopic: B13.1 Coordination and response
(b)(ii) Subtopic: B13.1 Coordination and response
(c)(i) Subtopic: B13.3 Homeostasis
(c)(ii) Subtopic: B13.3 Homeostasis
(c)(iii) Subtopic: B13.3 Homeostasis
(d) Subtopic: B13.1 Coordination and response
(a) The effect of an injection of adrenaline on pulse rate is recorded. The adrenaline is injected at 1 minute.
Fig. 7.1 shows a graph of the results.
(i) Identify in Fig. 7.1 the pulse rate before the adrenaline injection.
▶️Answer/Explanation
65 beats per minute
(ii) Describe the immediate effect of the adrenaline injection on pulse rate shown in Fig. 7.1. Use data from the graph to support your answer.
▶️Answer/Explanation
The pulse rate increases sharply immediately after the adrenaline injection. From the graph, we can see it rises from 65 beats per minute to 98 beats per minute, an increase of 33 beats per minute. This shows adrenaline’s rapid effect on heart rate.
(b) (i) State one effect adrenaline has on the eye.
▶️Answer/Explanation
Adrenaline causes the pupils to dilate (widen).
(ii) Name the nerve that carries impulses from the eye to the brain.
▶️Answer/Explanation
Optic nerve
(c) A hormone decreases blood glucose concentration by causing the glucose to be stored.
(i) State the name of the hormone that decreases blood glucose concentration.
▶️Answer/Explanation
Insulin
(ii) State the name of the organ that produces this hormone.
▶️Answer/Explanation
Pancreas
(iii) State the name of the organ that stores the excess glucose.
▶️Answer/Explanation
Liver
(d) Describe two ways the actions of the hormonal system are different from the nervous system.
▶️Answer/Explanation
1. The hormonal system acts more slowly than the nervous system (seconds/minutes/hours vs milliseconds for nervous system)
2. The effects of hormones last longer than nervous impulses (hours/days vs milliseconds/seconds)
Question 8
(a) Subtopic: C1.1 Solids, liquids and gases
(b) Subtopic: C1.1 Solids, liquids and gases
(c)(i) Subtopic: C6.2 Rate of reaction
(c)(ii) Subtopic: C6.2 Rate of reaction
(c)(iii) Subtopic: C6.2 Rate of reaction
(d) Subtopic: C6.2 Rate of reaction
(e) Subtopic: C2.4 Ions and ionic bonds
Fig. 8.1 shows the arrangement of ions in magnesium metal at 25°C.
(a) Describe the changes in the arrangement and movement of magnesium ions when magnesium melts.
▶️Answer/Explanation
Changes in arrangement: The regular/ordered arrangement of ions becomes more random/irregular when magnesium melts.
Changes in movement: The ions change from vibrating about fixed positions to moving around each other when magnesium melts.
(b) Magnesium melts at 650°C and boils at 1090°C. In the box, draw the arrangement of ions in magnesium at 1800°C.
▶️Answer/Explanation
At 1800°C (above boiling point), magnesium exists as a gas. The ions should be drawn:
– Randomly arranged
– Not touching each other
– Far apart from each other
(c) Magnesium reacts with dilute hydrochloric acid. Hydrogen gas is made in the reaction. A student investigates this reaction.
Fig. 8.2 shows the apparatus he uses.
Every 10 seconds, the student measures the total volume of hydrogen gas made.
Fig. 8.3 shows the graph the student plots of his results.
(i) State the volume of gas collected after 40 seconds.
▶️Answer/Explanation
From the graph, the volume of gas collected after 40 seconds is 48 cm³.
(ii) The reaction is fastest during the first 10 seconds. Explain why.
▶️Answer/Explanation
The reaction is fastest initially because:
1. The hydrochloric acid is at its highest concentration at the start
2. The magnesium ribbon has its maximum surface area at the beginning
These factors lead to more frequent successful collisions between reactant particles.
(iii) The student repeats the experiment using the same volume of acid and mass of magnesium, but increases the temperature of the hydrochloric acid. All of the magnesium reacts with the acid.
On Fig. 8.3, sketch the shape of the graph you would expect this time.
▶️Answer/Explanation
The new graph would show:
1. A steeper initial slope (faster initial reaction rate due to higher temperature)
2. The curve would level off at the same final volume (58 cm³) as all magnesium reacts
3. The reaction would complete in less time than the original experiment
(d) The rate of the reaction can be increased by increasing the concentration of the dilute hydrochloric acid. Explain why. Use ideas about collisions between particles.
▶️Answer/Explanation
Increasing concentration increases the reaction rate because:
1. There are more HCl particles per unit volume
2. This leads to more frequent collisions between magnesium and HCl particles
3. Therefore, there are more successful collisions per unit time
The increased collision frequency directly increases the reaction rate.
(e) Magnesium chloride is also made in the reaction between magnesium and dilute hydrochloric acid. Magnesium chloride contains the ions Mg²⁺ and Cl⁻. Determine the formula of magnesium chloride.
▶️Answer/Explanation
The formula is MgCl₂ because:
1. Magnesium ion has 2+ charge (Mg²⁺)
2. Chloride ion has 1- charge (Cl⁻)
3. Two chloride ions are needed to balance one magnesium ion’s charge
Question 9
(a)(i) Subtopic: P4.2.4 Resistance
(a)(ii) Subtopic: P4.2.1 Electrical charge
(b)(i) Subtopic: P3.1 General properties of waves
(b)(ii) Subtopic: P3.1 General properties of waves
(b)(iii) Subtopic: P3.3 Electromagnetic spectrum
A student investigates the effect of changing light levels on the resistance of a light-dependent resistor (LDR). The student shines a torch (flashlight) on to the LDR. She then places glass slides between the LDR and the torch (flashlight) to reduce the light intensity (amount of light) reaching the LDR.
Fig. 9.1 shows the equipment she uses.
The student places more glass slides between the torch (flashlight) and the LDR and measures the resistance, in kilo-ohms (kΩ), using a resistance meter.
(a) Fig. 9.2 shows a graph of the student’s results.
(i) Use Fig. 9.2 to describe how the resistance of the LDR varies with changing light intensity.
▶️Answer/Explanation
The resistance of the LDR increases as light intensity decreases (more glass slides are added). The relationship is non-linear – the increase in resistance becomes less dramatic with each additional glass slide, showing diminishing returns in the resistance change.
(ii) The resistance meter provides a potential difference (p.d.) of 14 V across the LDR. Calculate the charge flowing through the LDR in 1 minute when 3 glass slides are used.
▶️Answer/Explanation
Solution:
1. From graph at 3 slides: R ≈ 700 kΩ = 700,000 Ω
2. Using Ohm’s Law: I = V/R = 14/700,000 = 0.00002 A (20 μA)
3. Charge (Q) = I × t = 0.00002 A × 60 s = 0.0012 C
4. Final answer: 0.0012 C
(b) The lamp emits visible light at a frequency of 5.0 × 1014 Hz.
(i) State the meaning of the word frequency.
▶️Answer/Explanation
Frequency is the number of complete wave cycles (oscillations) passing a fixed point per second, measured in Hertz (Hz).
(ii) Calculate the wavelength of this visible light.
▶️Answer/Explanation
Solution:
1. Wave equation: c = fλ (where c = speed of light ≈ 3×108 m/s)
2. Rearrange: λ = c/f = (3×108)/(5.0×1014)
3. Calculation: λ = 6.0×10-7 m (600 nm)
4. Final answer: 6.0 × 10-7 m
(iii) State one form of electromagnetic radiation that has a frequency higher than visible light.
▶️Answer/Explanation
Ultraviolet (UV) radiation / X-rays / Gamma rays
Question 10
(a)(i) Subtopic: B9.4 Blood
(a)(ii) Subtopic: B9.4 Blood
(a)(iii) Subtopic: B17.1 Variation
(a)(iv) Subtopic: B16.1 Chromosomes and genes
(b) Subtopic: B17.2 Selection
(c) Subtopic: B15.2 Sexual reproduction
(a) The blood group of some patients in hospital is recorded.
Fig. 10.1 shows a bar chart of the results.
(i) One of the blood groups in Fig. 10.1 is not labelled. State this blood group.
▶️Answer/Explanation
AB
(ii) Identify the most common blood group in Fig. 10.1.
▶️Answer/Explanation
O
(iii) Describe evidence from Fig. 10.1 that shows that this characteristic is an example of discontinuous variation.
▶️Answer/Explanation
There are a limited number of distinct phenotypes (A, B, AB, O) with no intermediate values between them.
(iv) State the cause of the variation seen in Fig. 10.1.
▶️Answer/Explanation
Different alleles/genes inherited from parents
(b) Adaptations in populations can be inherited through natural selection or selective breeding. Table 10.1 compares some features of natural selection and selective breeding.
Complete Table 10.1 by placing ticks (✓) in the boxes to show the correct features.
| involves passing on of alleles to offspring | is used to improve domesticated animals | occurs over many generations | keeps the features best suited to the environment | |
|---|---|---|---|---|
| natural selection | ||||
| selective breeding |
▶️Answer/Explanation
| involves passing on of alleles to offspring | is used to improve domesticated animals | occurs over many generations | keeps the features best suited to the environment | |
|---|---|---|---|---|
| natural selection | ✓ | ✓ | ✓ | |
| selective breeding | ✓ | ✓ | ✓ |
(c) Sexual reproduction is involved in both natural selection and selective breeding. State two disadvantages of sexual reproduction to a population of species in the wild.
▶️Answer/Explanation
1. Requires more time and energy to find a mate
2. May produce offspring with unfavorable combinations of genes that reduce survival
Question 11
(a) Subtopic: C6.2 Rate of reaction
(b)(i) Subtopic: C6.2 Rate of reaction
(b)(ii) Subtopic: C6.2 Rate of reaction
(c) Subtopic: C5.1 Exothermic and endothermic reactions
(d) Subtopic: C3.3 The mole and the Avogadro constant
Sulfuric acid is made by the Contact process. Sulfur, air and water are raw materials used to make sulfuric acid.
Look at the equations for the first two stages in the Contact process.
stage 1 …… + …… → sulfur dioxide
stage 2 sulfur dioxide + oxygen ⇌ sulfur trioxide
(a) Complete the word equation for stage 1 of the Contact process.
▶️Answer/Explanation
Answer: sulfur + oxygen → sulfur dioxide
Explanation: In stage 1 of the Contact process, sulfur (S) burns in oxygen (O2) from the air to produce sulfur dioxide (SO2). This is a combustion reaction that requires heat to initiate.
(b) The conditions used for stage 2 are:
- 450°C
- atmospheric pressure
- a catalyst.
(i) State the name of the catalyst used.
▶️Answer/Explanation
Answer: vanadium(V) oxide / vanadium pentoxide (V2O5)
Explanation: Vanadium(V) oxide is the catalyst used in the Contact process. It provides an alternative reaction pathway with lower activation energy, allowing the reaction to proceed faster at the moderate temperature of 450°C.
(ii) Explain why a catalyst and a temperature of 450°C are used in stage 2 of the Contact process.
Use ideas about:
- the percentage of sulfur trioxide made
- the rate of reaction.
▶️Answer/Explanation
Explanation:
Catalyst:
• Increases the rate of reaction by providing an alternative reaction pathway with lower activation energy
• Does not affect the percentage yield of sulfur trioxide as it speeds up both forward and reverse reactions equally
450°C:
• This is a compromise temperature that gives a reasonable reaction rate while maintaining a good yield
• The reaction is exothermic, so lower temperatures would favor higher yield (Le Chatelier’s principle), but the rate would be too slow
• Higher temperatures would decrease the percentage yield but increase the rate too much, making the process uneconomical
• 450°C provides the optimal balance between rate and yield for industrial production
(c) The reaction in stage 2 of the Contact process is exothermic. Fig. 11.1 shows the energy level diagram for the reaction. Complete the labels on Fig. 11.1. Choose the labels from the list.
energy given out
energy taken in
products
reactants
reactants have less energy
▶️Answer/Explanation
Answer:
Top label: reactants
Arrow label: energy given out
Bottom label: products
Explanation: For an exothermic reaction:
1. The reactants are at higher energy level (top label)
2. Energy is released to the surroundings (arrow label)
3. The products are at lower energy level (bottom label)
The difference in energy between reactants and products is the enthalpy change (ΔH), which is negative for exothermic reactions.
(d) In an experiment, 200 g of sulfur trioxide, SO3, is made. Calculate the volume occupied by 200 g of sulfur trioxide gas. The relative molecular mass, Mr, of sulfur trioxide is 80. The volume of one mole of any gas is 24 dm3 at room temperature and pressure (r.t.p.).
Show your working.
▶️Answer/Explanation
Answer: 60 dm3
Working:
1. Calculate moles of SO3:
moles = mass/Mr = 200g/80g/mol = 2.5 moles
2. Calculate volume at r.t.p.:
volume = moles × molar volume = 2.5 × 24 dm3 = 60 dm3
Explanation: Using the given molar mass and standard molar volume, we first determine how many moles of SO3 are present in 200g, then convert this to volume using the fact that 1 mole of any gas occupies 24 dm3 at r.t.p.
Question 12
(a) Subtopic: P1.6.2 Work
(b) Subtopic: P1.5.3 Centre of gravity
(c)(i) Subtopic: P4.5.4 Force on a current-carrying conductor
(c)(ii) Subtopic: P4.5.4 Force on a current-carrying conductor
(d)(i) Subtopic: P5.2.2 The three types of nuclear emission
(d)(ii) Subtopic: P5.2.2 The three types of nuclear emission
Fig. 12.1 shows a forklift truck lifting a crate.
(a) The forklift truck does 2750 J of work on the crate when the crate is lifted through a height of 2.2 m. The gravitational field strength, g, is 10 N/kg.
Calculate the mass of the crate.
▶️Answer/Explanation
Solution:
We use the formula for work done against gravity:
Work = Force × Distance = mass × gravitational field strength × height
Given:
Work = 2750 J
g = 10 N/kg
height = 2.2 m
Rearranging the formula: mass = Work / (g × height)
mass = 2750 / (10 × 2.2) = 2750 / 22 = 125 kg
Answer: 125 kg
(b) Fig. 12.2 shows the same forklift truck after it has lowered the crate.
Explain why the forklift truck is more stable after it has lowered the crate. Use ideas about centre of mass in your answer.
▶️Answer/Explanation
Solution:
When the crate is lowered, the overall center of mass of the forklift truck system is lowered. A lower center of mass increases stability because it reduces the tendency of the system to topple over. This is because the torque (turning effect) created by any sideways force would be smaller when the center of mass is closer to the ground.
Answer: The forklift truck is more stable because lowering the crate lowers the center of mass.
(c) The forklift truck uses an electric motor to lift the crate. Fig. 12.3 shows a simple d.c. motor.
(i) A current flows through the coil. Draw arrows on Fig. 12.3 to show the direction of the force acting on points X and Z on the coil.
▶️Answer/Explanation
Solution:
Using Fleming’s Left Hand Rule:
– For point X: Current flows towards us (out of page), magnetic field left to right → force upwards
– For point Z: Current flows away from us (into page), magnetic field left to right → force downwards
Answer:
X: Upward arrow
Z: Downward arrow
(ii) State why point Y does not experience a force.
▶️Answer/Explanation
Solution:
Point Y doesn’t experience a force because at this point, the current is flowing parallel to the magnetic field lines. According to the motor effect, a force is only produced when current flows at an angle to the magnetic field (preferably perpendicular). When current and field are parallel, no force is generated.
Answer: The current at Y is parallel to the magnetic field.
(d) A β-particle passes between the poles of a permanent magnet.
(i) Suggest why a β-particle is deflected when moving through a magnetic field.
▶️Answer/Explanation
Solution:
A β-particle is a high-energy electron, which carries a negative charge. When a charged particle moves through a magnetic field, it experiences a force perpendicular to both its direction of motion and the magnetic field direction (Lorentz force). This causes the particle to follow a curved path, resulting in deflection.
Answer: Because it is a charged particle moving through a magnetic field.
(ii) State and explain how the deflection direction of an α-particle would differ from that of the β-particle.
▶️Answer/Explanation
Solution:
An α-particle would be deflected in the opposite direction to the β-particle because:
1. α-particles are positively charged (He nuclei, +2 charge) while β-particles are negatively charged (electrons, -1 charge)
2. The direction of deflection depends on the charge of the particle (Fleming’s Left Hand Rule)
Additionally, α-particles would show less deflection because they have much greater mass (about 7000 times more massive than electrons) while carrying only twice the charge.
Answer: α-particle would deflect in the opposite direction because it’s positively charged, and the deflection would be less due to its greater mass.