Question 1
(a) Some plant cells are immersed in a concentrated salt solution.
Fig. 1.1 is a photomicrograph showing the appearance of the cells. (Sub-topic – B3.2)
Complete the sentences using words or phrases from the list to explain the appearance of the cells in Fig. 1.1.
Each word or phrase may be used once, more than once or not at all.
- active transport
- a higher
- a lower
- osmosis
- plasmolysis
- the same
- turgor
The solution outside the cell has ……………………………………………………… water potential than the cells.
Water diffuses across the cell membrane by ……………………………………………………… from high water potential to low water potential.
This reduces the ……………………………………………………… pressure of the cells. The cytoplasm is pulled away from the cell wall.
This is called ……………………………………………………… .
▶️Answer/Explanation
The solution outside the cell has a lower water potential than the cells.
Water diffuses across the cell membrane by osmosis from high water potential to low water potential.
This reduces the turgor pressure of the cells.
The cytoplasm is pulled away from the cell wall.
This is called plasmolysis.
(b) Fig. 1.2 shows two specialised plant cells: a root hair cell and a palisade cell. (Sub-topic – B2.1)
(i) Name the cell structure labelled X in Fig. 1.2.
▶️Answer/Explanation
The cell structure labelled X is the chloroplast.
(ii) Explain why cell structure X is not needed in root hair cells.
▶️Answer/Explanation
Root hair cells are primarily involved in the absorption of water and mineral ions from the soil. Since they are underground and do not receive light, they do not need chloroplasts for photosynthesis.
(iii) Use Fig. 1.2 to identify two cell structures found in root hair cells but not in animal cells.
▶️Answer/Explanation
Two cell structures found in root hair cells but not in animal cells are:
- Vacuole
- Cell wall
(c) Fig. 1.3 is a photomicrograph of a type of specialised animal cell. (Sub-topic – B9.4)
(i) Name the cells shown in Fig. 1.3.
▶️Answer/Explanation
The cells shown in Fig. 1.3 are red blood cells.
(ii) Describe two ways the cells shown in Fig. 1.3 are adapted for their function.
▶️Answer/Explanation
Two adaptations of red blood cells for their function are:
- They contain haemoglobin, which binds to oxygen for transport.
- They have a biconcave shape, which increases their surface area for efficient oxygen exchange.
Question 2
Solid, liquid and gas are three states of matter.
(a) Fig. 2.1 shows the arrangements of particles in these three states. (Sub-topic – C1.1)
Complete Fig. 2.1 by writing under each box the name of the state of matter shown.
▶️Answer/Explanation
Answer: liquid, solid, gas
(b) An ice cube is left in a cup in a warm room. After a few hours, liquid water can be seen in the cup. (Sub-topic – C1.1)
State the name of the process that has occurred in the cup.
▶️Answer/Explanation
Answer: melting
(c) The freezing point of water is \(0^\circ C\). (Sub-topic – C1.1)
Describe how the movement and arrangement of water particles change when water is cooled from \(10^\circ C\) to \(-10^\circ C\).
▶️Answer/Explanation
Answer:
- Movement: The particles change from moving around each other to vibrating about fixed positions.
- Arrangement: The arrangement of particles changes from random to regular.
(d) Water, \(H_2O\), is a covalent molecule. (Sub-topic – C2.5)
Complete the dot-and-cross diagram in Fig. 2.2 to show the bonding in a water molecule.
You only need to show the outer shell electrons.
▶️Answer/Explanation
Answer:
The dot-and-cross diagram should show:
- Two hydrogen atoms each sharing one electron with the oxygen atom.
- The oxygen atom has two lone pairs of electrons.
Question 3
(a) Fig. 3.1 shows a student observing an exploding firework.
The firework produces light and sound at the same time. The student measures the time between seeing the light and hearing the sound. (Subtopic – P3.4)
(i) It takes 3.50 seconds for the student to hear the sound. Calculate the distance between the student and the firework. The speed of sound in air is 340 m/s.
▶️Answer/Explanation
Solution:
To calculate the distance, we use the formula:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]
Given:
Speed of sound = 340 m/s
Time = 3.50 s
\[ \text{Distance} = 340 \times 3.50 = 1190 \, \text{m} \]
Therefore, the distance between the student and the firework is 1190 m.
(ii) Suggest an appropriate measuring instrument the student uses to measure the time it takes to hear the sound.
▶️Answer/Explanation
Answer:
The student can use a stopwatch to measure the time between seeing the light and hearing the sound.
(iii) Explain why this method cannot be used to measure the speed of light.
▶️Answer/Explanation
Explanation:
The speed of light is extremely high (approximately \(3 \times 10^8 \, \text{m/s}\)), so the time taken for light to travel even a short distance is too small to measure accurately with a stopwatch. The time difference between seeing the light and hearing the sound is primarily due to the much slower speed of sound, not the speed of light.
(b) Fig. 3.2 shows a ray of light being refracted as it passes from air into glass. Calculate the refractive index of the glass block. State the formula you use and show your working. Give your answer to two significant figures. (Subtopic – P3.2.2)
▶️Answer/Explanation
Solution:
The refractive index (\(n\)) of the glass can be calculated using Snell’s Law:
\[ n = \frac{\sin i}{\sin r} \]
Where:
\(i\) = angle of incidence in air = 42°
\(r\) = angle of refraction in glass = 29°
\[ n = \frac{\sin 42°}{\sin 29°} \]
Using a calculator:
\[ \sin 42° \approx 0.6691 \]
\[ \sin 29° \approx 0.4848 \]
\[ n = \frac{0.6691}{0.4848} \approx 1.38 \]
Therefore, the refractive index of the glass is 1.4 (to two significant figures).
(c) Fig. 3.3 shows an accurate diagram of a ray of light passing into an optical fibre. (Subtopic – P3.2.2)
(i) Explain why the ray does not change direction at point X on Fig. 3.3.
▶️Answer/Explanation
Explanation:
At point X, the ray of light is traveling along the normal (perpendicular) to the boundary between the two media. When light travels along the normal, it does not change direction because the angle of incidence is 0°, and thus, there is no refraction.
(ii) State the full name of the type of reflection that occurs at point Y on Fig. 3.3.
▶️Answer/Explanation
Answer:
The type of reflection that occurs at point Y is total internal reflection.
(iii) State one use for optical fibres.
▶️Answer/Explanation
Answer:
One use for optical fibres is in communication, such as transmitting data over long distances in telecommunications.
Question 4
(a) Anaerobic respiration in yeast produces carbon dioxide.
A scientist investigates the effect of using different types of sugar on the volume of carbon dioxide produced by yeast in one minute.
The results are shown in Fig. 4.1. (Sub-topic – B12.1)
(i) Compare the results for the different types of sugar. Include comparative data from Fig. 4.1 in your answer.
▶️Answer/Explanation
The results show that glucose produces the most carbon dioxide (approximately 25 cm³), followed by sucrose and lactose (both around 15 cm³), and fructose produces the least (approximately 10 cm³). This indicates that glucose is the most effective sugar for yeast fermentation, while fructose is the least effective.
(ii) State one other product of anaerobic respiration in yeast.
▶️Answer/Explanation
Alcohol (ethanol) is another product of anaerobic respiration in yeast.
(iii) State the word equation for the anaerobic respiration in muscles.
▶️Answer/Explanation
The word equation for anaerobic respiration in muscles is:
glucose → lactic acid
(b) Carbon dioxide produced by yeast causes bread to rise. The temperature used during bread-making is carefully controlled. At very high temperatures, no carbon dioxide is produced. Explain why. Use ideas about enzymes in your answer. (Sub-topic – B5.1)
▶️Answer/Explanation
At very high temperatures, enzymes in yeast denature. Denaturation changes the shape of the enzyme’s active site, making it unable to bind with the substrate. As a result, the enzyme can no longer catalyze the reaction, and carbon dioxide is not produced. This is why temperature control is crucial in bread-making to ensure the enzymes remain active and carbon dioxide is produced to make the bread rise.
Question 5
Ammonia, NH3, is made in factories by the Haber process. Fig. 5.1 shows how ammonia is made.
Nitrogen gas and hydrogen gas are the starting materials.
(a) Describe the Haber process. (Subtopic: C5.1)
You should include:
- the sources of nitrogen gas and hydrogen gas
- the conditions used
- what the ammonia is used for.
▶️Answer/Explanation
The Haber process is used to synthesize ammonia (NH₃) from nitrogen gas (N₂) and hydrogen gas (H₂).
- Sources of nitrogen and hydrogen: Nitrogen gas is obtained from the air, which contains about 78% nitrogen. Hydrogen gas is typically obtained from natural gas (methane, CH₄) through a process called steam reforming.
- Conditions used: The reaction is carried out at high pressure (around 200 atmospheres) and moderate temperature (around 450°C). An iron catalyst is used to speed up the reaction.
- Uses of ammonia: Ammonia is primarily used to produce fertilizers, such as ammonium nitrate and urea. It is also used in the production of nitric acid, cleaning agents, and refrigerants.
(b) A factory making ammonia wants to make 680 kg of ammonia.
The balanced symbol equation for the reaction is shown:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]
(i) Calculate the mass of hydrogen, \( H_2 \), that is needed to make 680 kg of ammonia, \( NH_3 \). (Subtopic: C3.3)
[\(A_r\); H, 1; N, 14]
▶️Answer/Explanation
To calculate the mass of hydrogen required:
- First, calculate the molar mass of ammonia (NH₃): \[ \text{Molar mass of NH₃} = 14 + (3 \times 1) = 17 \, \text{g/mol} \]
- Calculate the number of moles of ammonia in 680 kg: \[ \text{Moles of NH₃} = \frac{680,000 \, \text{g}}{17 \, \text{g/mol}} = 40,000 \, \text{mol} \]
- From the balanced equation, 2 moles of NH₃ require 3 moles of H₂. Therefore, the moles of H₂ required are: \[ \text{Moles of H₂} = \frac{3}{2} \times 40,000 = 60,000 \, \text{mol} \]
- Calculate the mass of hydrogen required: \[ \text{Mass of H₂} = 60,000 \, \text{mol} \times 2 \, \text{g/mol} = 120,000 \, \text{g} = 120 \, \text{kg} \]
Therefore, the mass of hydrogen required is 120 kg.
(ii) State the chemical test and its positive result for hydrogen gas. (Subtopic: C12.5)
▶️Answer/Explanation
The chemical test for hydrogen gas involves bringing a lighted splint near the gas. If hydrogen is present, it will produce a “pop” sound as it reacts with oxygen in the air to form water.
Positive result: A “pop” sound is heard.
(iii) 560 kg of nitrogen gas, \( N_2 \), are needed in the factory. (Subtopic: C3.3)
Calculate the volume of nitrogen gas needed. The molar gas volume is 24 dm³ at room temperature and pressure (r.t.p.).
▶️Answer/Explanation
To calculate the volume of nitrogen gas:
- Calculate the molar mass of nitrogen gas (N₂): \[ \text{Molar mass of N₂} = 2 \times 14 = 28 \, \text{g/mol} \]
- Calculate the number of moles of nitrogen in 560 kg: \[ \text{Moles of N₂} = \frac{560,000 \, \text{g}}{28 \, \text{g/mol}} = 20,000 \, \text{mol} \]
- Calculate the volume of nitrogen gas at r.t.p.: \[ \text{Volume of N₂} = 20,000 \, \text{mol} \times 24 \, \text{dm³/mol} = 480,000 \, \text{dm³} \]
Therefore, the volume of nitrogen gas required is 480,000 dm³.
Question 6
(a) Metals are good conductors of thermal energy. (Subtopics: P1.5.1)
Describe the two mechanisms of energy transfer that make metals good thermal conductors.
▶️Answer/Explanation
Answer:
1. Conduction by free electrons: Metals have free electrons that can move throughout the lattice. When a metal is heated, these free electrons gain kinetic energy and move faster, transferring energy through the metal.
2. Conduction by lattice vibrations: The metal lattice consists of closely packed ions. When one part of the metal is heated, the ions vibrate more vigorously and transfer energy to neighboring ions through collisions.
(b) A student investigates how the surface colour of an object affects how fast the object loses thermal energy. (Subtopics: P2.3.3)
Fig. 6.1 shows the equipment used.
She uses two identical aluminium cans, one of which has been painted black and the other white. She fills both cans with an equal volume of hot water at the same temperature and records the temperature of the water every minute for 60 minutes. Fig. 6.2 shows her results.
(i) Use the information in Fig. 6.2 to state the temperature of the room.
▶️Answer/Explanation
Answer:
The temperature of the room is 20°C. This is the temperature at which both cans stabilize after losing heat to the surroundings.
(ii) State which line in Fig. 6.2, P or Q, is for the can painted black. Explain your answer in terms of energy transfer by radiation.
▶️Answer/Explanation
Answer:
Line P is for the can painted black. Black surfaces are better emitters of thermal radiation compared to white surfaces. Therefore, the black can loses heat faster, causing its temperature to drop more quickly than the white can.
(c) The student reheats the water in one of the cans using an electric immersion heater. (Subtopic: P1.6.3)
The heater has a power rating of 1.5 kW and is switched on for 120 seconds.
(i) Calculate the amount of energy used by the electric immersion heater.
▶️Answer/Explanation
Answer:
Energy = Power × Time
Power = 1.5 kW = 1500 W
Time = 120 s
Energy = 1500 W × 120 s = 180,000 J (or 180 kJ).
(ii) State the amount of electrical work done by the heater during this process.
▶️Answer/Explanation
Answer:
The amount of electrical work done by the heater is 180,000 J (or 180 kJ). This is equal to the energy used by the heater.
(d) During the experiment, the student spills some water on the table. (Subtopic: P2.2.2)
The water evaporates. State two ways to increase the rate of evaporation.
▶️Answer/Explanation
Answer:
1. Increase the temperature: Higher temperatures provide more energy to water molecules, increasing the rate of evaporation.
2. Increase the surface area: Spreading the water over a larger surface area allows more molecules to escape into the air, increasing the rate of evaporation.
Question 7
(a) State where chromosomes are found in cells. [Subtopic: B16.1]
▶️Answer/Explanation
Chromosomes are found in the nucleus of cells.
(b) State the female and male sex chromosomes in humans. [Subtopic: B16.1]
▶️Answer/Explanation
Female: XX, Male: XY
(c) State the number of chromosomes in each human body cell. [Subtopic: B16.1]
▶️Answer/Explanation
46 (or 23 pairs)
(d) Table 7.1 shows the number of chromosomes found in each body cell of four different organisms. [Subtopic: B16.1]
| Organism | Number of Chromosomes |
|---|---|
| Goldfish | 94 |
| Potato | 48 |
| Pea | 14 |
| Fruit fly | 8 |
(i) State the number of chromosomes in each body cell of a goldfish.
▶️Answer/Explanation
94
(ii) State the number of chromosomes in one gamete of a fruit fly.
▶️Answer/Explanation
4
(e) A mutation is a change in a chromosome. State one factor that increases the rate of mutation in cells. [Subtopic: B16.1]
▶️Answer/Explanation
Ionising radiation or chemicals.
(f) State two roles of mitosis in the body. [Subtopic: B16.2]
▶️Answer/Explanation
1. Growth
2. Repair of damaged tissues
(g) The list shows some statements about meiosis. Place ticks (✓) in the boxes next to all the correct statements. [Subtopic: B16.2]
| Statement | Correct? |
|---|---|
| Meiosis produces genetically identical cells. | |
| Meiosis produces sperm in humans. | |
| Meiosis is a type of cell division. | |
| Meiosis results in the production of diploid cells. | |
| Meiosis only occurs when there is a mutation in cells. |
▶️Answer/Explanation
Meiosis produces sperm in humans ticked ;
Meiosis is a type of cell division ;
Question 8
A student is investigating indigestion tablets.
Indigestion tablets neutralise excess acid in the stomach. The student adds one tablet to 50 cm3 of dilute hydrochloric acid. He measures the time taken for the tablet to completely react.
The student repeats the experiment using different concentrations of hydrochloric acid. The temperature of the acid is always 25°C.
Fig. 8.1 shows the apparatus he uses.
Fig. 8.2 shows a graph of the student’s results.
(a) Look at Fig. 8.2. State how long it takes for the tablet to fully react when the student uses hydrochloric acid with a concentration of 1.0 mol/dm3. (Subtopic – C6.2)
▶️Answer/Explanation
Answer: 50 seconds
From the graph, the time taken for the tablet to fully react when the concentration of hydrochloric acid is 1.0 mol/dm3 is 50 seconds.
(b) The student does the experiment again. He makes only one change. He uses dilute hydrochloric acid at a temperature of 35°C instead of 25°C. Sketch a line on Fig. 8.2 to predict the results at 35°C. (Subtopic – C6.2)
▶️Answer/Explanation
Answer: The line should be the same shape as the original but always below it, indicating faster reaction times at 35°C.
(c) The student’s results show that it takes less time for indigestion tablets to react when the acid is more concentrated. Explain why reactions are faster when reactants are more concentrated. Explain your answer in terms of collisions between particles. (Subtopic – C6.2)
▶️Answer/Explanation
Answer: Higher concentration means more particles per unit volume, leading to more frequent collisions and a faster reaction rate.
(d) Indigestion tablets contain calcium carbonate, CaCO3. Look at the symbol equation for the reaction of calcium carbonate with dilute hydrochloric acid. The equation is not balanced.
Balance the equation. (Subtopic – C3.1)
▶️Answer/Explanation
Answer: CaCO3 + 2HCl → CaCl2 + CO2 + H2O
The unbalanced equation is:
CaCO3 + HCl → CaCl2 + CO2 + H2O
The balanced equation is:
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
(e) In one experiment the student measures the temperature of the acid before he adds the tablet. He also measures the temperature after all the tablet has reacted. The temperature decreases. State the name for this type of energy transfer. (Subtopic – C5.1)
▶️Answer/Explanation
Answer: Endothermic
(f) Dilute hydrochloric acid is an acid. Define an acid in terms of proton transfer. (Subtopic – C7.1)
▶️Answer/Explanation
Answer: An acid is a proton donor.
Question 9
(a) Name the process which releases energy from uranium in nuclear power stations. [Subtopic: P5.1]
Name the process which releases energy from uranium in nuclear power stations.
▶️Answer/Explanation
Answer: Nuclear fission
(b) Uranium-238 is unstable and decays to produce an isotope of thorium. [Subtopic: P5.2.3]
Use the correct nuclide notation to complete the symbol equation for this decay process.
▶️Answer/Explanation
Answer:
(c) The isotope of thorium produced is also unstable and decays releasing more ionising radiation. [Subtopic: P5.2.4]
Fig. 9.1 shows how the activity of a sample of thorium-234 varies over time.
Use Fig. 9.1 to calculate the half-life of thorium-234.
▶️Answer/Explanation
Answer: The half-life of thorium-234 is 25 days.
(d) Fig. 9.2 shows radioactive emissions passing between two oppositely charged plates. [Subtopic: P5.2.2]
(i) An electric field exists between the charged plates.
Describe what is meant by an electric field.
▶️Answer/Explanation
Answer: An electric field is a region in which an electric charge experiences a force.
(ii) Complete Fig. 9.2 to show the paths of an \(\alpha\)-particle, a \(\beta\)-particle and a \(\gamma\)-ray as they pass through the electric field. [Subtopic: P5.2.2]
Complete Fig. 9.2 to show the paths of an \(\alpha\)-particle, a \(\beta\)-particle and a \(\gamma\)-ray as they pass through the electric field.
▶️Answer/Explanation
Answer: – The \(\alpha\)-particle will be deflected towards the negative plate due to its positive charge. – The \(\beta\)-particle will be deflected towards the positive plate due to its negative charge. – The \(\gamma\)-ray will not be deflected as it has no charge.
Question 10
(a) Fig. 10.1 shows a plant’s response to light. [Subtopic: B13.1]
(i) Name the tropic response shown in Fig. 10.1.
▶️Answer/Explanation
Answer: Phototropism
Explanation: Phototropism is the growth response of a plant toward or away from light. In Fig. 10.1, the plant is bending toward the light source, which is a classic example of positive phototropism.
(ii) Explain the response to light seen in Fig. 10.1. Use ideas about auxin in your answer.
▶️Answer/Explanation
Answer: Auxin is produced in the tip of the plant and moves to the shaded side, promoting cell elongation on that side, causing the plant to bend toward the light.
Explanation: When light is unevenly distributed, auxin (a plant hormone) accumulates on the shaded side of the plant. This causes the cells on the shaded side to elongate more than the cells on the illuminated side, resulting in the plant bending toward the light source.
(b) Adrenaline is a hormone released in the human body during stressful situations. [Subtopic: B13.2]
(i) Suggest the name of the target organ for this hormonal response.
▶️Answer/Explanation
Answer: Liver
Explanation: Adrenaline targets the liver to increase blood glucose concentration by stimulating the breakdown of glycogen into glucose.
(ii) State two other effects of adrenaline on the human body.
▶️Answer/Explanation
Answer: 1. Increased heart rate
2. Widened pupils
Explanation: Adrenaline prepares the body for a “fight or flight” response by increasing heart rate to pump more blood and widening pupils to improve vision in low light conditions.
(iii) Name one other hormone that increases the blood glucose concentration.
▶️Answer/Explanation
Answer: Glucagon
Explanation: Glucagon is another hormone that increases blood glucose levels by stimulating the liver to convert stored glycogen into glucose.
(c) Describe two ways that the transmission of information by hormonal control is different from nervous control. [Subtopic: B13.1]
▶️Answer/Explanation
Answer: 1. Hormonal control is slower than nervous control.
2. Hormonal control involves chemical substances (hormones) transmitted via the bloodstream, while nervous control involves electrical impulses transmitted via neurons.
Explanation: 1. Hormonal responses take longer to initiate and have longer-lasting effects compared to the rapid, short-lived responses of the nervous system.
2. Hormones are released into the bloodstream and travel to target organs, whereas nerve impulses are transmitted quickly along neurons to specific target cells.
Question 11
(a) Table 11.1 shows some information about particles found in an atom. Complete Table 11.1. [Subtopic: B2.1]
Table 11.1
| Particle | Relative Mass | Relative Charge |
|---|---|---|
| Electron | \( \frac{1}{1840} \) | |
| Neutron | 0 | |
| Proton | 1 |
▶️Answer/Explanation
Answer:
Electron: Relative Charge = -1
Neutron: Relative Mass = 1
Proton: Relative Charge = +1
(b) The diagrams in Fig. 11.1 each show the nucleus of a different atom. [Subtopic: B2.1]
(i) State which atom has a proton number (atomic number) of 3.
(ii) State which atom has a nucleon number (mass number) of 6.
(iii) State which two atoms are isotopes of the same element.
▶️Answer/Explanation
Answer:
(i) Atom C has a proton number of 3.
(ii) Atom C has a nucleon number of 6.
(iii) Atoms D and E are isotopes of the same element.
Question 12
Fig. 12.1 shows a cyclist.
(a) The cyclist starts from rest and accelerates with constant acceleration. The cyclist reaches 12 m/s after 20 seconds. He then continues at this constant speed for 15 seconds. (Sub-topic – P1.2)
(i) On Fig. 12.2, plot a speed–time graph for the cyclist.
▶️Answer/Explanation
Solution:
The speed-time graph for the cyclist can be plotted as follows:
1. From 0 to 20 seconds, the cyclist accelerates uniformly from 0 m/s to 12 m/s. This is a straight line with a positive slope.
2. From 20 to 35 seconds, the cyclist maintains a constant speed of 12 m/s. This is a horizontal line.
The graph will have time (in seconds) on the x-axis and speed (in m/s) on the y-axis.
(ii) Calculate the acceleration of the cyclist during the first 20 seconds. State the unit for your answer. (Sub-topic – P1.2)
▶️Answer/Explanation
Solution:
The acceleration \( a \) can be calculated using the formula:
\[ a = \frac{\Delta v}{\Delta t} \]
Where:
\(\Delta v = 12 \, \text{m/s} – 0 \, \text{m/s} = 12 \, \text{m/s}\)
\(\Delta t = 20 \, \text{s}\)
\[ a = \frac{12 \, \text{m/s}}{20 \, \text{s}} = 0.6 \, \text{m/s}^2 \]
The acceleration of the cyclist during the first 20 seconds is \( 0.6 \, \text{m/s}^2 \).
(iii) Describe how to calculate the distance travelled by the cyclist using the speed–time graph. (Sub-topic – P1.2)
▶️Answer/Explanation
Solution:
The distance travelled by the cyclist can be calculated by finding the area under the speed-time graph.
1. For the first 20 seconds (acceleration phase), the area is a triangle:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \, \text{s} \times 12 \, \text{m/s} = 120 \, \text{m} \]
2. For the next 15 seconds (constant speed phase), the area is a rectangle:
\[ \text{Area} = \text{base} \times \text{height} = 15 \, \text{s} \times 12 \, \text{m/s} = 180 \, \text{m} \]
The total distance travelled is \( 120 \, \text{m} + 180 \, \text{m} = 300 \, \text{m} \).
(b) State one difference and one similarity between speed and velocity. (Sub-topic – P1.2)
▶️Answer/Explanation
Solution:
Difference:
Speed is a scalar quantity, meaning it only has magnitude (e.g., 12 m/s). Velocity is a vector quantity, meaning it has both magnitude and direction (e.g., 12 m/s north).
Similarity:
Both speed and velocity measure the rate of change of distance with respect to time, and they share the same units (m/s).
(c) Fig. 12.3 shows the forces acting on the cyclist while he is travelling at constant speed.
(i) State the size of force R on Fig. 12.3.
▶️Answer/Explanation
460 (N) ;
(ii) Suggest the cause of force R on Fig. 12.3.
▶️Answer/Explanation
air resistance / friction / drag ;
Question 13
The electrolysis of concentrated aqueous sodium chloride, NaCl, produces two useful gases.
Fig. 13.1 shows the apparatus used.
(a) State the formulae of all the ions present in concentrated aqueous sodium chloride. (Sub-topic – C4.1)
▶️Answer/Explanation
Solution:
The ions present in concentrated aqueous sodium chloride are:
1. Sodium ions: \( \text{Na}^+ \)
2. Chloride ions: \( \text{Cl}^- \)
3. Hydrogen ions: \( \text{H}^+ \) (from water)
4. Hydroxide ions: \( \text{OH}^- \) (from water)
The formulae of the ions are: \( \text{Na}^+ \), \( \text{Cl}^- \), \( \text{H}^+ \), and \( \text{OH}^- \).
(b) State the name of the gas that forms at each electrode. (Sub-topic – C4.1)
▶️Answer/Explanation
Solution:
The gases formed at each electrode are:
1. At the anode: Chlorine gas (\( \text{Cl}_2 \))
2. At the cathode: Hydrogen gas (\( \text{H}_2 \))
Chlorine gas is produced at the anode, and hydrogen gas is produced at the cathode.
(c) State the name of the solution remaining after electrolysis. Explain why the solution is alkaline. (Sub-topic – C4.1)
▶️Answer/Explanation
Solution:
The solution remaining after electrolysis is sodium hydroxide (\( \text{NaOH} \)).
The solution is alkaline because hydroxide ions (\( \text{OH}^- \)) remain in the solution after hydrogen ions (\( \text{H}^+ \)) are discharged at the cathode. The accumulation of \( \text{OH}^- \) ions makes the solution alkaline.
(d) Lead is extracted from molten lead(II) bromide, \( \text{PbBr}_2 \), by electrolysis. The ionic half-equation is shown:
\[ \text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb} \]
Explain, in terms of electrons, if lead ions are oxidised or reduced in this reaction. (Sub-topic – C4.1)
▶️Answer/Explanation
Solution:
In the given half-equation:
\[ \text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb} \]
Lead ions (\( \text{Pb}^{2+} \)) gain electrons (\( \text{e}^- \)).
Reduction is the gain of electrons, so lead ions are reduced in this reaction.
(e) Aluminium is extracted from aluminium oxide, \( \text{Al}_2\text{O}_3 \), by electrolysis. Construct the ionic half-equation for the formation of aluminium at the cathode. Use \( \text{e}^- \) to represent an electron. (Sub-topic – C4.1)
▶️Answer/Explanation
Solution:
The ionic half-equation for the formation of aluminium at the cathode is:
\[ \text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al} \]
In this reaction, aluminium ions (\( \text{Al}^{3+} \)) gain three electrons (\( \text{e}^- \)) to form aluminium atoms (\( \text{Al} \)).