Question 1
(a) Gills are the gas exchange surface in fish. (Subtopic – B11.1)
Fig. 1.1 is a photograph of gills in fish.
Gills have adaptive features for gas exchange.
Use your scientific knowledge and Fig. 1.1 to suggest two of these adaptive features.
▶️Answer/Explanation
Answer:
Any two from:
1. Large surface area
2. Thin surface
3. Good blood supply
Explanation:
Gills are adapted for efficient gas exchange by having a large surface area to maximize the amount of oxygen that can be absorbed. They are also thin, which allows for a short diffusion path, and have a good blood supply to transport oxygen to the rest of the body.
(b) Table 1.1 shows the composition of some gases in inspired and expired air. (Subtopic – B11.1)
| Gas | Percentage of Oxygen | Percentage of Carbon Dioxide |
|---|---|---|
| Inspired Air | 21 | 0.004 |
| Expired Air | 16 | 4 |
(i) Use Table 1.1 to calculate the difference in percentage of oxygen between inspired and expired air.
▶️Answer/Explanation
Answer:
5%
Explanation:
The difference in oxygen percentage between inspired and expired air is calculated as:
21% (inspired) – 16% (expired) = 5%.
(ii) Explain the difference in percentage of oxygen between inspired and expired air that is shown in Table 1.1.
▶️Answer/Explanation
Answer:
Oxygen is required for aerobic respiration to release energy for metabolic processes.
Explanation:
The decrease in oxygen percentage from inspired to expired air is due to the fact that oxygen is used by the body for aerobic respiration, which releases energy for various metabolic processes. As a result, the oxygen level in the expired air is lower than in the inspired air.
(iii) State one difference in composition between inspired and expired air that is not shown in Table 1.1.
▶️Answer/Explanation
Answer:
Greater percentage of water vapor in expired air.
Explanation:
Expired air contains a higher percentage of water vapor compared to inspired air due to the moisture added to the air as it passes through the respiratory system.
(c) Red blood cells have adaptive features for the efficient transport of oxygen. (Subtopic – B9.4)
State two of these features.
▶️Answer/Explanation
Answer:
Any two from:
1. Large surface area / biconcave shape
2. Contain haemoglobin
3. No nucleus
Explanation:
Red blood cells are adapted for efficient oxygen transport by having a biconcave shape, which increases their surface area for oxygen absorption. They also contain haemoglobin, which binds to oxygen, and they lack a nucleus, allowing more space for haemoglobin.
(d) Lung cancer is a disease caused by smoking. (Subtopic – B10.1)
(i) Place ticks (✓) to show two other diseases caused by smoking.
| Disease | Tick |
|---|---|
| Chronic obstructive pulmonary disease (COPD) | |
| Coronary heart disease | |
| Kwashiorkor | |
| Marasmus | |
| Scurvy |
▶️Answer/Explanation
Answer:
COPD ticked
Coronary heart disease ticked
Explanation:
Smoking is a major cause of chronic obstructive pulmonary disease (COPD) and coronary heart disease (CHD). These diseases are directly linked to the harmful effects of tobacco smoke on the lungs and cardiovascular system.
(ii) State the name of the component in tobacco smoke that causes cancer.
▶️Answer/Explanation
Answer:
Tar
Explanation:
Tar is the component in tobacco smoke that contains carcinogenic chemicals, which can lead to the development of cancer.
(iii) Describe how the goblet cells, mucus, and ciliated cells protect the gas exchange system from some of the particles in tobacco smoke.
▶️Answer/Explanation
Answer:
Goblet cells produce mucus, which traps particles/pathogens. Cilia move mucus/particles/pathogens away from the gas exchange surface.
Explanation:
Goblet cells secrete mucus that traps harmful particles and pathogens from tobacco smoke. The cilia, which are hair-like structures on the surface of the respiratory tract, then move the mucus and trapped particles away from the lungs, helping to protect the gas exchange system.
Question 2
(a) A student investigates two liquid fuels, A and B, to find out which fuel releases most energy.
Fig. 2.1 shows the apparatus used. 1.5g of each fuel is burned completely.
Table 2.1 shows the student’s results.
| Fuel | Temperature of water at start /°C | Temperature of water at end /°C | Temperature change /°C |
|---|---|---|---|
| A | 16 | 25 | 9 |
| B | 16 | 34 | 18 |
(i) Describe how the results show which fuel releases the most energy. (Sub-topic – C5.1)
▶️Answer/Explanation
Fuel B releases more energy because it causes a greater temperature change in the water.
Explanation: The results show that fuel B releases more energy than fuel A because the temperature change for fuel B (18°C) is greater than that for fuel A (9°C). Since the same mass of each fuel was burned and the same amount of water was used, the greater temperature change indicates that fuel B released more energy during combustion.
(ii) Fig. 2.2 is the equation representing the complete combustion of ethanol. (Sub-topic – C5.1)
This reaction is exothermic. Place a tick (✓) in the box next to the correct explanation of an exothermic reaction.
- More energy is given out by bond breaking than is taken in by bond making.
- More energy is given out by bond making than is taken in by bond breaking.
- More bonds are broken than are made.
- More energy is taken in by bond breaking than is given out by bond making.
▶️Answer/Explanation
More energy is given out by bond making than is taken in by bond breaking.
(b) Fig. 2.3 shows the energy level diagram for an exothermic reaction. (Sub-topic – C5.1)
(i) Use Fig. 2.3 to calculate the energy given out in the reaction.
▶️Answer/Explanation
The energy given out in the reaction can be calculated by finding the difference in energy between the reactants and the products. From the diagram, the energy of the reactants is 150 kJ, and the energy of the products is 50 kJ. Therefore, the energy given out is:
Energy given out = Energy of reactants – Energy of products = 150 kJ – 50 kJ = 100 kJ
100 kJ
(ii) Use Fig. 2.3 to calculate the activation energy for the reaction.
▶️Answer/Explanation
The activation energy is the energy required to reach the transition state from the reactants. From the diagram, the energy of the transition state is 200 kJ, and the energy of the reactants is 150 kJ. Therefore, the activation energy is:
Activation energy = Energy of transition state – Energy of reactants = 200 kJ – 150 kJ = 50 kJ
50 kJ
(c) Ethanol can be made by: (Sub-topic – C11.6)
- the catalytic addition of steam to ethene
- fermentation.
(i) Construct the balanced symbol equation for the addition of steam to ethene to make ethanol, C2H5OH.
▶️Answer/Explanation
C2H4 + H2O → C2H5OH
(ii) Describe how ethanol is made by fermentation.
▶️Answer/Explanation
Ethanol is produced by the fermentation of sugars (e.g., glucose) using yeast in anaerobic conditions, resulting in the production of ethanol and carbon dioxide.
Explanation:
Ethanol is made by fermentation using yeast, which converts sugars (such as glucose) into ethanol and carbon dioxide in the absence of oxygen (anaerobic conditions). The process can be represented by the following equation:
C6H12O6 → 2C2H5OH + 2CO2
Question 3
(a) Fig. 3.1 shows a man in a canoe on a lake. The combined mass of the man and the canoe is 120 kg.
(i) The canoe moves at a speed of 4.0 m/s. Calculate the kinetic energy of the man and the canoe. (Sub-topic – P1.6.1)
▶️Answer/Explanation
Answer: 960 J
Explanation: The kinetic energy (KE) is calculated using the formula:
\[ KE = \frac{1}{2}mv^2 \] where \( m = 120 \, \text{kg} \) and \( v = 4.0 \, \text{m/s} \).
\[ KE = \frac{1}{2} \times 120 \times (4.0)^2 = \frac{1}{2} \times 120 \times 16 = 960 \, \text{J} \]
(ii) The canoe takes 5.0 s to slow down to a speed of 0.5 m/s. Calculate the constant deceleration of the canoe. (Sub-topic – P1.2)
▶️Answer/Explanation
Answer: 0.7 m/s²
Explanation: The deceleration (a) is calculated using the formula:
\[ a = \frac{\Delta v}{\Delta t} \] where \( \Delta v = 4.0 – 0.5 = 3.5 \, \text{m/s} \) and \( \Delta t = 5.0 \, \text{s} \).
\[ a = \frac{3.5}{5.0} = 0.7 \, \text{m/s}^2 \]
(iii) On Fig. 3.2, draw a speed–time graph to show the canoe’s deceleration. (Sub-topic – P1.2)
▶️Answer/Explanation
Answer: The graph should show a straight line starting at 4.0 m/s at time 0 s and decreasing linearly to 0.5 m/s at time 5.0 s.
Explanation: The graph represents a constant deceleration, so the line should be straight with a negative slope.
(b) The canoe exerts a pressure of 0.5 kPa on the surface of the water. Calculate the area of the canoe in contact with the surface of the water. The gravitational field strength, \( g \), is 10 N/kg.
(Sub-topic – P1.7)
▶️Answer/Explanation
Answer: 2.4 m²
Explanation: The pressure (P) is given by:
\[ P = \frac{F}{A} \] where \( F = mg = 120 \times 10 = 1200 \, \text{N} \) and \( P = 0.5 \, \text{kPa} = 500 \, \text{Pa} \).
\[ A = \frac{F}{P} = \frac{1200}{500} = 2.4 \, \text{m}^2 \]
(c) Fig. 3.3 shows water waves on the surface of the lake. (Sub-topic – P3.1)
(i) On Fig. 3.3, draw a double-headed arrow to show the wavelength of the wave.
▶️Answer/Explanation
Answer: The arrow should span from one crest to the next crest or one trough to the next trough.
Explanation: The wavelength is the distance between two consecutive points in phase, such as crest to crest or trough to trough.
(ii) Use the words below to complete the sentences about waves. You can use each word once, more than once or not at all.
compression, energy, force, longitudinal, matter, perpendicular, parallel, rarefaction, transverse
Waves transfer …… without transferring …… . A water wave is an example of a …… wave. In a water wave, the oscillations are …… to the direction of the wave.
▶️Answer/Explanation
Answer: Waves transfer energy without transferring matter. A water wave is an example of a transverse wave. In a water wave, the oscillations are perpendicular to the direction of the wave.
Explanation: Waves transfer energy through a medium without permanently displacing the medium itself. Water waves are transverse waves, meaning the oscillations are perpendicular to the direction of wave propagation.
Question 4
(a) A student investigates the effect of light intensity on the rate of photosynthesis in an aquatic plant. The plant is placed in a beaker of water containing an excess of carbon dioxide. A lamp is placed 10 cm away from the beaker of water. The student counts the number of oxygen bubbles produced by the aquatic plant in one minute. The lamp is then moved increasing distances away from the beaker to decrease the light intensity. The number of oxygen bubbles produced is directly proportional to the rate of photosynthesis. Table 4.1 shows the results. (Sub-topic – B6.1)
Table 4.1
| Distance of lamp from the aquatic plant /cm | Number of oxygen bubbles produced per minute |
|---|---|
| 10 | 37 |
| 20 | 37 |
| 30 | 36 |
| 40 | 32 |
| 50 | 25 |
| 60 | 15 |
| 70 | 6 |
| 80 | 1 |
(i) Use Table 4.1 to describe the effect of light intensity on the rate of photosynthesis. Include data in your answer.
▶️Answer/Explanation
As the distance of the lamp from the aquatic plant increases, the light intensity decreases, and the rate of photosynthesis decreases. At 10 cm, the plant produces 37 bubbles per minute, which remains almost the same at 20 cm (37 bubbles) and 30 cm (36 bubbles). However, as the distance increases further, the number of bubbles decreases significantly: 32 bubbles at 40 cm, 25 bubbles at 50 cm, 15 bubbles at 60 cm, 6 bubbles at 70 cm, and only 1 bubble at 80 cm. This shows that light intensity is a limiting factor for photosynthesis, and as it decreases, the rate of photosynthesis also decreases.
(ii) State why an excess of carbon dioxide is provided for the aquatic plant during this investigation.
▶️Answer/Explanation
An excess of carbon dioxide is provided to ensure that carbon dioxide is not a limiting factor in the rate of photosynthesis. This allows the student to isolate the effect of light intensity on the rate of photosynthesis without interference from other variables.
(b) Complete the sentences to explain how a lack of magnesium affects plant growth. (Sub-topic – B6.1)
▶️Answer/Explanation
Magnesium is required for the synthesis of chlorophyll. This substance transfers light energy into chemical energy for the synthesis of carbohydrates. A deficiency of magnesium ions causes the leaves to turn yellow.
(c) Nitrate ions are required for the synthesis of amino acids. State the name of the class of large molecules made from amino acids. (Sub-topic – B4.1)
▶️Answer/Explanation
The class of large molecules made from amino acids is proteins.
Question 5
This question is about metals.
(a) Potassium is a metal in Group I of the Periodic Table. (Sub-topic – C8.2)
Fig. 5.1 shows the electronic structure of three elements.
(i) State which diagram A, B or C, shows the electronic structure of a Group I metal.
▶️Answer/Explanation
Answer: B
Explanation: Group I metals have one electron in their outermost shell. Diagram B shows an element with one electron in the outermost shell, which is characteristic of Group I metals.
(ii) A student wants to confirm that a compound contains potassium. The student uses a flame test. State what the student observes if the compound contains potassium.
▶️Answer/Explanation
Answer: Lilac flame
Explanation: Potassium ions produce a lilac (purple) flame in a flame test, which is a characteristic test for potassium.
(b) Iron is a transition element. (Sub-topic – C8.4)
Iron(II) sulfate contains iron(II) ions, Fe2+. Sodium hydroxide solution is used to test for iron(II) ions. The iron(II) ions react with OH– ions from the sodium hydroxide solution. A precipitate of iron(II) hydroxide, Fe(OH)2, is made.
(i) State the colour of the precipitate of iron(II) hydroxide.
▶️Answer/Explanation
Answer: Green
Explanation: Iron(II) hydroxide, Fe(OH)2, forms a green precipitate when iron(II) ions react with sodium hydroxide.
(ii) Construct the balanced ionic equation for the formation of Fe(OH)2. Include state symbols.
▶️Answer/Explanation
Answer: Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s)
Explanation: The balanced ionic equation shows the reaction between iron(II) ions and hydroxide ions to form solid iron(II) hydroxide.
(c) Magnesium reacts with oxygen to make magnesium oxide. (Sub-topic – C2.4)
(i) Fig. 5.2 shows the electronic structure of a magnesium atom. The proton number (atomic number) of magnesium is 12.
Fig. 5.2
Draw a diagram to show the electronic structure of an oxygen atom. The proton number (atomic number) of oxygen is 8.
▶️Answer/Explanation
Answer: Oxygen has 8 electrons, arranged as 2 in the first shell and 6 in the second shell.
Explanation: The electronic structure of oxygen is 2,6, with 2 electrons in the first shell and 6 electrons in the second shell.
(ii) When magnesium reacts with oxygen, magnesium ions and oxide ions are made. Fig. 5.3 shows the electronic structure of an oxide ion.
Fig. 5.3
Draw a diagram to show the electronic structure of a magnesium ion.
▶️Answer/Explanation
Answer: A magnesium ion (Mg2+) has 10 electrons, arranged as 2 in the first shell and 8 in the second shell.
Explanation: Magnesium loses 2 electrons to form a Mg2+ ion, resulting in an electronic structure of 2,8.
(iii) Explain why magnesium oxide has a high melting point.
▶️Answer/Explanation
Answer: Magnesium oxide has a high melting point due to the strong electrostatic attraction between the positive magnesium ions (Mg2+) and the negative oxide ions (O2-).
Explanation: The ionic bonds in magnesium oxide are very strong, requiring a large amount of energy to break, which results in a high melting point.
(d) Potassium oxide is also an ionic compound. (Sub-topic – C2.4)
Potassium ions, K+, combine with oxide ions, O2-, to form potassium oxide. Determine the formula of potassium oxide.
▶️Answer/Explanation
Answer: K2O
Explanation: Potassium oxide is formed by the combination of two potassium ions (K+) with one oxide ion (O2-), resulting in the formula K2O.
Question 6
(a) Fig. 6.1 shows a marble staircase made up of 17 steps. (Sub-topic – P1.4)
Fig. 6.2 shows the dimensions of one of the marble steps which has a mass of 72 kg.
(i) Calculate the density of the marble step.
\[ \text{density} = \frac{\text{kg}}{\text{m}^3} \]
(ii) On a hot, sunny day the marble step expands.
Suggest what happens to the density of the marble step when it expands.
(iii) Explain, in terms of particle movement, why the marble expands.
▶️Answer/Explanation
(i) Solution:
Volume of the marble step = length × width × height = 0.90 m × 0.16 m × 0.20 m = 0.0288 m³
Density = mass / volume = 72 kg / 0.0288 m³ = 2500 kg/m³
Answer: 2500 kg/m³
(ii) Solution:
When the marble step expands, its volume increases while its mass remains the same. Therefore, the density decreases.
Answer: The density decreases.
(iii) Solution:
When the marble is heated, the particles gain kinetic energy and vibrate more vigorously. This causes the particles to move further apart, leading to an increase in volume and thus expansion.
Answer: The particles move further apart due to increased kinetic energy, causing the marble to expand.
(b)(i) On a hot, sunny day the marble steps feel cold because of conduction. (Sub-topic – P2.3.1)
Describe the process of conduction in marble.
(ii) Explain why conduction causes the marble to feel cold.
▶️Answer/Explanation
(i) Solution:
Conduction in marble occurs through the vibration and collision of particles. When one part of the marble is heated, the particles in that region vibrate more and transfer their energy to neighboring particles through collisions. This process continues, transferring heat energy through the marble.
Answer: Conduction in marble involves the transfer of heat energy through the vibration and collision of particles.
(ii) Solution:
Marble feels cold because it is a good conductor of heat. When you touch the marble, heat from your hand is quickly conducted away by the marble, making your hand feel cooler. This rapid heat transfer gives the sensation of coldness.
Answer: The marble feels cold because it conducts heat away from your hand quickly, making your hand feel cooler.
Question 7 (Sub-topic – B12.1)
(a) Yeast produces carbon dioxide during anaerobic respiration.
A scientist adds 2 g of yeast to 250 cm3 of glucose solution and leaves the mixture for 10 days. Each day, at the same time, he records the volume of carbon dioxide produced in one hour. Fig. 7.1 shows the results.
(i) Use Fig. 7.1 to identify the day with the greatest rate of anaerobic respiration.
▶️Answer/Explanation
Answer: Day 3
Explanation: The graph shows that the volume of carbon dioxide produced is highest on day 3, indicating the greatest rate of anaerobic respiration on that day.
(ii) Explain why the volume of carbon dioxide decreases as shown in Fig. 7.1.
▶️Answer/Explanation
Answer: The glucose in the solution is used up over time, reducing the substrate available for anaerobic respiration, which leads to a decrease in carbon dioxide production.
Explanation: As the yeast consumes the glucose, the amount of substrate available for respiration decreases, resulting in less carbon dioxide being produced over time.
(b) The investigation is repeated with boiled yeast.
Explain with reference to enzymes why no carbon dioxide is produced.
▶️Answer/Explanation
Answer: Boiling denatures the enzymes in the yeast, preventing them from catalyzing the anaerobic respiration process.
Explanation: Enzymes are proteins that can be denatured by high temperatures, such as boiling. When the enzymes in the yeast are denatured, they lose their shape and function, making them unable to catalyze the reactions necessary for anaerobic respiration, thus no carbon dioxide is produced.
(c) Table 7.1 shows some products of different types of respiration.
Complete Table 7.1 by placing ticks (✓) to show the correct products of each type of respiration. One has been done for you.
| Type of Respiration | Carbon Dioxide | Water | Lactic Acid |
|---|---|---|---|
| Aerobic in humans | |||
| Anaerobic in humans | |||
| Anaerobic in yeast | ✓ |
▶️Answer/Explanation
Answer:
| Type of Respiration | Carbon Dioxide | Water | Lactic Acid |
|---|---|---|---|
| Aerobic in humans | ✓ | ✓ | |
| Anaerobic in humans | ✓ | ||
| Anaerobic in yeast | ✓ |
Explanation:
- Aerobic respiration in humans produces carbon dioxide and water.
- Anaerobic respiration in humans produces lactic acid.
- Anaerobic respiration in yeast produces carbon dioxide and ethanol (not shown in the table).
(d) Substances enter and leave a yeast cell by diffusion.
A student writes an incorrect definition of diffusion. Circle the two words that are not correct.
‘Diffusion is the total movement of particles from a region of their higher concentration to a region of their lower concentration up a concentration gradient, as a result of their random movement.’
▶️Answer/Explanation
Answer: “total” and “up”
Explanation: Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration down a concentration gradient, as a result of their random movement. The words “total” and “up” are incorrect in this context.
Question 8
(a) Water must be treated so that it is safe to drink. (Sub-topic – C10.1)
Draw lines to link each stage in the water treatment process to the reason why it is used.
| Stage | Reason |
|---|---|
| Filtration | Traps finer particles using sand |
| Chlorination | Kills microbes |
| Allows larger solid particles to settle out |
▶️Answer/Explanation
- Filtration – Traps finer particles using sand
- Chlorination – Kills microbes
(b) Water can be tested to identify some of the chemicals in it. (Sub-topic – C12.5)
A scientist tests a sample of water from a river with acidified aqueous silver nitrate and also with acidified aqueous barium chloride. Table 8.1 shows the results.
| Test | With acidified aqueous silver nitrate | With acidified aqueous barium chloride |
|---|---|---|
| Result | Cream precipitate | White precipitate |
The scientist thinks that the water contains both chloride and sulfate ions. State whether the scientist is correct. Explain your answer.
▶️Answer/Explanation
The scientist is not entirely correct. The cream precipitate with silver nitrate indicates the presence of bromide ions (Br⁻) or chloride ions (Cl⁻), but since bromide ions give a cream precipitate and chloride ions give a white precipitate, it is likely that bromide ions are present. The white precipitate with barium chloride confirms the presence of sulfate ions (SO₄²⁻). Therefore, the water contains sulfate ions, but the presence of chloride ions is not confirmed.
(c) Barium chloride, BaCl2, reacts with sodium sulfate, Na2SO4. (Sub-topic – C3.1)
Barium sulfate, BaSO4, and sodium chloride, NaCl, are made. Construct the balanced symbol equation for this reaction.
▶️Answer/Explanation
The balanced symbol equation for the reaction is:
BaCl2 + Na2SO4 → BaSO4 + 2NaCl
(d) Carbon dioxide dissolves in rainwater to make the water weakly acidic. (Sub-topic – C7.1)
Suggest the pH of the rainwater produced.
▶️Answer/Explanation
The pH of rainwater containing dissolved carbon dioxide is typically around 5.6, making it weakly acidic.
(e) The atoms in carbon dioxide, CO2, are bonded by sharing electrons. (Sub-topic – C2.5)
Fig. 8.1 shows some dot-and-cross diagrams.
State which diagram A, B, C or D, shows the arrangement of the outer shell electrons in carbon dioxide.
▶️Answer/Explanation
Diagram C shows the correct arrangement of the outer shell electrons in carbon dioxide, where carbon shares electrons with two oxygen atoms.
(f) Complete the following sentences about some of the problems caused by carbon dioxide. (Sub-topic – C10.2)
Choose words from the list. Each word or phrase may be used once, more than once or not at all.
| Word List |
|---|
| climate change |
| greenhouse |
| oxidation |
| noble |
| rusting |
Carbon dioxide is a …… gas. Increased concentrations of carbon dioxide in the atmosphere contribute to …… .
▶️Answer/Explanation
Carbon dioxide is a greenhouse gas. Increased concentrations of carbon dioxide in the atmosphere contribute to climate change.
Question 9
(a) Fig. 9.1 shows the equipment used by a teacher to demonstrate the properties of ionising radiation to a group of students. They are using a source which emits β-particles. (Sub-topic – P5.2.1)
The radioactive source can be moved further away from the radiation detector. The teacher measures the distance between the source and the radiation detector and records the count rate using the laptop.
Fig. 9.2 shows the results plotted as a graph.
(i) Describe the trend shown in Fig. 9.2.
(ii) Use Fig. 9.2 to explain why the teacher tells the students to stand at least 2 m away from the radioactive source for their own safety.
(iii) The teacher replaces the radioactive source with one which only emits α-particles. The source which only emits α-particles also measures a count rate of 200 per minute at a distance of 0 m. On Fig. 9.2, draw a line to show the results the teacher obtains when using the source which emits only α-particles.
▶️Answer/Explanation
(i) The trend shown in Fig. 9.2 is that as the distance from the radioactive source increases, the count rate decreases. The decrease is most rapid at shorter distances and levels off as the distance increases further.
(ii) The teacher tells the students to stand at least 2 m away from the radioactive source because β-particles are ionising and can cause damage to cells or DNA. At a distance of 2 m, the count rate is almost zero, meaning the radiation exposure is minimal, ensuring the students’ safety.
(iii) The line for α-particles would start at 200 counts per minute at 0 m and drop sharply to almost zero at a very short distance (e.g., within a few centimeters), as α-particles have much lower penetrating power compared to β-particles.
(b) Fig. 9.3 shows the information sticker on the laptop. (Sub-topic – P4.2.5)
(i) The laptop has an efficiency of 80%. Calculate the useful power output of the laptop.
(ii) Power for the laptop comes from a 230 V supply through a device in the charger which changes the potential difference to 19.5 V. State the name of this device.
▶️Answer/Explanation
(i) The useful power output can be calculated using the formula: \[ \text{Power Output} = \text{Efficiency} \times \text{Power Input} \] Given the efficiency is 80% (or 0.8) and the power input is 65 W: \[ \text{Power Output} = 0.8 \times 65 = 52 \, \text{W} \]
(ii) The device that changes the potential difference from 230 V to 19.5 V is a step-down transformer.
Question 10
(a) Albinism is a condition that results in a lack of colour in the skin, causing a very pale appearance. The allele for albinism is recessive, a. The allele for no albinism is dominant, A. Fig. 10.1 is a pedigree diagram showing the inheritance of albinism. (Sub-topic – B16.1)
Draw two lines from person P and two lines from person Q to make four correct statements.
Person P
- has a heterozygous genotype.
- has albinism.
- has a homozygous recessive genotype.
- has the genotype AA.
- is male.
- is female without albinism.
Person Q
- has a heterozygous genotype.
- has albinism.
- has a homozygous recessive genotype.
- has the genotype AA.
- is male.
- is female without albinism.
▶️Answer/Explanation
Person P:
- has a heterozygous genotype.
- is female without albinism.
Person Q:
- has a homozygous recessive genotype.
- has albinism.
(b) Alleles are passed to offspring during sexual reproduction. The sex cell in human males is sperm. (Sub-topic – B15.2)
(i) State the name of the sex cell in human females.
(ii) State the number of chromosomes in one human sperm cell.
(iii) State the number of sex chromosomes in a body cell of a human male.
(iv) State the number of parents required for asexual reproduction.
▶️Answer/Explanation
(i) The sex cell in human females is the egg cell or ovum.
(ii) The number of chromosomes in one human sperm cell is 23.
(iii) The number of sex chromosomes in a body cell of a human male is 2.
(iv) The number of parents required for asexual reproduction is 1.
Question 11
(a) Diamond is one form of carbon. (Sub-topic – C2.6)
(i) State the name of another form of carbon.
▶️Answer/Explanation
Answer: Graphite
(ii) Diamond is used in cutting tools such as those shown in Fig. 11.1.
State why diamond is used.
▶️Answer/Explanation
Answer: Diamond is hard.
(b) Silicon dioxide, SiO2, has a similar structure to diamond. (Sub-topic – C2.6)
Fig. 11.2 shows the structure of silicon dioxide.
Describe the structure and bonding in silicon dioxide.
▶️Answer/Explanation
Answer: Silicon dioxide has a giant covalent structure where each silicon atom is covalently bonded to four oxygen atoms, and each oxygen atom is bonded to two silicon atoms. This creates a strong, rigid lattice structure.
(c) One of the isotopes of carbon is called carbon-12 and the other is called carbon-14. (Sub-topic – C2.3)
Table 11.1 shows some information about carbon-12.
Complete the table for carbon-14.
| number of protons | number of neutrons | number of electrons | |
|---|---|---|---|
| carbon-12 | 6 | 6 | 6 |
| carbon-14 |
▶️Answer/Explanation
Answer: Carbon-14 has 6 protons, 8 neutrons, and 6 electrons.
(d) Relative atomic mass, Ar, is defined in terms of a carbon atom. (Sub-topic – C3.2)
Complete the definition of relative atomic mass.
Choose words from the list. Each word may be used once, more than once or not at all.
average compound density element formula mass
Relative atomic mass is the …… mass of naturally occurring atoms of an …… on a scale where the 12C atom has a …… of exactly 12 units.
▶️Answer/Explanation
Answer: Relative atomic mass is the average mass of naturally occurring atoms of an element on a scale where the 12C atom has a mass of exactly 12 units.
(e) Calculate the number of moles in 0.6 g of carbon.
[Ar : C, 12;]
▶️Answer/Explanation
(moles = 0.6 ÷ 12 =) 0.05 ;
Question 12
Fig. 12.1 shows a circuit containing two resistors connected in parallel with a 9.0 V battery.
(a)(i) Calculate the total resistance of the circuit shown in Fig. 12.1. (Sub-topic – P4.3.1)
▶️Answer/Explanation
Solution:
For resistors in parallel, the total resistance \( R_T \) is given by:
\[ \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} \]
Substituting the given values:
\[ \frac{1}{R_T} = \frac{1}{6.0} + \frac{1}{12.0} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \]
Therefore, the total resistance \( R_T \) is:
\[ R_T = 4.0 \, \Omega \]
Answer: 4.0 Ω
(ii) Calculate the current passing through the 6.0 Ω resistor.
▶️Answer/Explanation
Solution:
The current \( I \) through a resistor is given by Ohm’s Law:
\[ I = \frac{V}{R} \]
For the 6.0 Ω resistor:
\[ I = \frac{9.0}{6.0} = 1.5 \, \text{A} \]
Answer: 1.5 A
(b) The 9.0 V battery is connected in series with a lamp, a variable resistor, and a switch. (Sub-topic – P4.3.2)
(i) Draw a circuit diagram showing a 9.0 V battery connected in series with a lamp, a variable resistor, and a switch.
▶️Answer/Explanation
Solution:
The circuit diagram should show the following components connected in series:
- A 9.0 V battery
- A lamp
- A variable resistor
- A switch
The components should be connected in a single loop, with the battery providing the voltage, the lamp and variable resistor in series, and the switch controlling the flow of current.
Answer: The circuit diagram should show a 9.0 V battery connected in series with a lamp, a variable resistor, and a switch.
(ii) The variable resistor is used to change the voltage across and the current in the lamp. On Fig. 12.2, sketch a graph showing the current-voltage characteristic of a filament lamp.
▶️Answer/Explanation
Solution:
The current-voltage (I-V) characteristic of a filament lamp is non-linear due to the heating effect of the filament. As the voltage increases, the filament heats up, and its resistance increases, causing the current to increase at a decreasing rate.
The graph should show:
- Current (I) on the y-axis and Voltage (V) on the x-axis.
- A curve that starts steep and gradually flattens as voltage increases.
Answer: The graph should show a non-linear I-V characteristic curve for a filament lamp, with the current increasing at a decreasing rate as voltage increases.
(c) Fig. 12.3 shows a circuit containing a thermistor and a lamp in series with an ammeter. A voltmeter is connected in parallel across the thermistor. (Sub-topic – P4.3.2)
(i) Describe what happens to the readings on the ammeter and voltmeter when the temperature of the thermistor increases.
▶️Answer/Explanation
Solution:
A thermistor is a temperature-dependent resistor. As the temperature increases:
- The resistance of the thermistor decreases.
- This causes the total resistance of the circuit to decrease.
- According to Ohm’s Law (\( V = IR \)), the current in the circuit increases, so the ammeter reading increases.
- The voltage across the thermistor decreases because the resistance of the thermistor decreases, so the voltmeter reading decreases.
Answer: The ammeter reading increases, and the voltmeter reading decreases as the temperature of the thermistor increases.
(ii) Explain why the brightness of the lamp changes as the temperature of the thermistor increases.
▶️Answer/Explanation
Solution:
As the temperature of the thermistor increases:
- The resistance of the thermistor decreases, causing the total resistance of the circuit to decrease.
- This results in an increase in the current through the circuit (ammeter reading increases).
- Since the lamp is in series with the thermistor, the increased current causes the lamp to glow brighter.
Answer: The brightness of the lamp increases because the current through the lamp increases as the temperature of the thermistor increases.