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Secondary 3 Combined Science Semestral Assessment 2 (End of Year) Paper 4

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Questions

TuitionGoWhere Practice Paper – Combined Science Secondary 3

TuitionGoWhere Secondary School (AI)

Subject: Combined Science (Physics/Chemistry/Biology)
Level: Secondary 3
Paper: SA2 – Version 4
Duration: 1 hour 30 minutes
Total Marks: 70

Name: _________________________
Class: _________________________
Date: _________________________

Instructions to Candidates

This paper consists of three sections: Section A, Section B, and Section C.
Answer all questions.
Write your answers in the spaces provided.
Show all working for calculation questions. Marks are awarded for method.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may use a calculator.

Section A: Physical Sciences – Physics (25 marks)

Answer all questions in this section.

1. A student investigates the principle of conservation of energy using a pendulum. The pendulum bob is released from point P and swings to point Q.

(a) State the principle of conservation of energy.

______________________________________________________________________________ [1]

(b) At point P, the bob has 2.5 J of gravitational potential energy and 0 J of kinetic energy. At the lowest point of the swing, the bob has 0.4 J of gravitational potential energy. Calculate the kinetic energy of the bob at the lowest point.

______________________________________________________________________________ [2]

2. A force of 15 N is applied to a box of mass 6.0 kg resting on a rough horizontal surface. The box moves with constant velocity.

(a) Explain why the box moves with constant velocity even though a force is applied.

______________________________________________________________________________ [2]

(b) Calculate the weight of the box. (Take g = 10 N/kg)

______________________________________________________________________________ [1]

3. Figure 3.1 shows a rectangular block measuring 0.20 m × 0.10 m × 0.05 m resting on a table. The block has a mass of 4.0 kg.

(a) Calculate the maximum pressure the block can exert on the table.

______________________________________________________________________________ [3]

(b) State one way the block can be positioned to exert the minimum pressure on the table.

______________________________________________________________________________ [1]

4. A student heats 500 g of water using an electric heater rated at 50 W for 10 minutes. The initial temperature of the water is 25 °C. The specific heat capacity of water is 4200 J/(kg·°C).

(a) Calculate the total energy supplied by the heater.

______________________________________________________________________________ [2]

(b) Calculate the final temperature of the water, assuming no heat loss to the surroundings.

______________________________________________________________________________ [3]

5. A ray of light travels from air into a glass block. The angle of incidence is 40° and the angle of refraction is 25°.

(a) State what is meant by refraction of light.

______________________________________________________________________________ [1]

(b) Explain why the light ray bends towards the normal as it enters the glass block.

______________________________________________________________________________ [2]

6. A student sets up a circuit with a 12 V battery connected to two resistors in series. Resistor R₁ has a resistance of 4 Ω and resistor R₂ has a resistance of 8 Ω.

(a) Calculate the total resistance in the circuit.

______________________________________________________________________________ [1]

(b) Calculate the current flowing through the circuit.

______________________________________________________________________________ [2]

(c) Calculate the potential difference across R₂.

______________________________________________________________________________ [2]

Section B: Physical Sciences – Chemistry (25 marks)

Answer all questions in this section.

7. Zinc metal can be extracted from zinc sulfide (ZnS) in a two-step process.

(a) In Step 1, zinc sulfide is roasted in air to form zinc oxide and sulfur dioxide. Write a balanced chemical equation for this reaction. Include state symbols.

______________________________________________________________________________ [2]

(b) In Step 2, zinc oxide is reduced by heating with carbon. Write a balanced chemical equation for this reaction. Include state symbols.

______________________________________________________________________________ [2]

(c) Calculate the mass of zinc that can be extracted from 9.7 g of zinc sulfide. (Relative atomic masses: Zn = 65, S = 32, O = 16)

______________________________________________________________________________ [3]

8. A student adds solid calcium carbonate to dilute hydrochloric acid in a conical flask. The flask is connected to a gas syringe.

(a) Write a balanced chemical equation for the reaction. Include state symbols.

______________________________________________________________________________ [2]

(b) Write the ionic equation for this reaction.

______________________________________________________________________________ [2]

(c) The student uses 2.0 g of calcium carbonate. Calculate the volume of carbon dioxide gas produced at room temperature and pressure. (Molar volume of gas at r.t.p. = 24 dm³/mol; Relative atomic masses: Ca = 40, C = 12, O = 16)

______________________________________________________________________________ [3]

9. A student performs paper chromatography to analyse three ink samples (A, B, and C) against three known dyes (X, Y, and Z). The chromatogram obtained is shown below.

Solvent front
  |
  |    * (X)    * (Y)    * (Z)
  |    
  |    *        *        *
  |    *        *        
  |    *                 *
  |
  |____A________B________C____
  Origin

(a) State which dye(s) are present in ink sample A.

______________________________________________________________________________ [1]

(b) Ink sample B contains two dyes. Identify these dyes.

______________________________________________________________________________ [1]

(c) Calculate the Rf value of dye Y, given that dye Y travelled 4.8 cm and the solvent front travelled 8.0 cm from the origin.

______________________________________________________________________________ [2]

10. A student investigates the rate of reaction between magnesium ribbon and excess dilute sulfuric acid. The volume of hydrogen gas produced is measured over time.

(a) State one observation the student would make during the reaction.

______________________________________________________________________________ [1]

(b) Explain, using collision theory, why the rate of reaction decreases as the reaction proceeds.

______________________________________________________________________________ [2]

(c) Suggest one change the student could make to increase the initial rate of reaction, other than changing the concentration of the acid.

______________________________________________________________________________ [1]

Section C: Physical Sciences – Biology (20 marks)

Answer all questions in this section.

11. Figure 11.1 shows a diagram of the human heart.

(a) Name the blood vessels labelled P and Q.

P: _________________________
Q: _________________________ [2]

(b) Describe the pathway of blood from the lungs to the aorta.

______________________________________________________________________________ [4]

12. A student sets up an experiment using potato strips placed in different concentrations of sucrose solution. After 30 minutes, the student measures the change in length of each potato strip. The results are shown in Table 12.1.

Table 12.1

Concentration of sucrose solution (mol/dm³)	Change in length (mm)
0.0	+4.0
0.2	+1.5
0.4	-1.0
0.6	-3.5
0.8	-5.0

(a) Explain why the potato strip in 0.0 mol/dm³ sucrose solution increased in length.

______________________________________________________________________________ [3]

(b) Describe the state of the potato cells in the 0.8 mol/dm³ sucrose solution.

______________________________________________________________________________ [2]

(c) Using the data in Table 12.1, estimate the concentration of sucrose solution that has the same water potential as the potato cells. Explain your answer.

______________________________________________________________________________ [2]

13. A student investigates the effect of light intensity on the rate of photosynthesis in an aquatic plant. The student counts the number of oxygen bubbles produced per minute at different distances of a lamp from the plant.

(a) Suggest how the student could vary the light intensity in this experiment.

______________________________________________________________________________ [1]

(b) State the purpose of placing a glass heat shield between the lamp and the plant.

______________________________________________________________________________ [1]

(c) The student's results are shown in Table 13.1.

Table 13.1

Distance of lamp (cm)	Number of bubbles per minute
10	42
20	28
30	18
40	12
50	8

Plot a graph of the results on the grid below. Label both axes with appropriate scales and units.

[Graph grid provided – 4 marks]

(d) Describe the relationship between the distance of the lamp and the rate of photosynthesis shown by the data.

______________________________________________________________________________ [1]

14. A student breathes in and out several times. The composition of inhaled air and exhaled air is compared.

(a) State one difference between the composition of inhaled air and exhaled air.

______________________________________________________________________________ [1]

(b) Explain why exhaled air contains less oxygen than inhaled air.

______________________________________________________________________________ [2]

END OF PAPER

Answers

TuitionGoWhere Practice Paper – Combined Science Secondary 3

SA2 Version 4 – Answer Key and Marking Scheme

Total Marks: 70

Section A: Physical Sciences – Physics (25 marks)

1. (a) State the principle of conservation of energy. [1]

Answer: Energy cannot be created or destroyed. It can only be converted/transferred from one form to another. The total energy in a closed/isolated system remains constant.

Marking notes:

Award [1] for stating both that energy cannot be created or destroyed AND that it is converted/transferred, OR that total energy remains constant.
Accept: "Energy is neither created nor destroyed, only changed from one form to another."

1. (b) Calculate the kinetic energy of the bob at the lowest point. [2]

Working:

Total energy at P = GPE at P + KE at P = 2.5 J + 0 J = 2.5 J
By conservation of energy, total energy at lowest point = 2.5 J
Total energy at lowest point = GPE at lowest point + KE at lowest point
2.5 J = 0.4 J + KE
KE = 2.5 J – 0.4 J = 2.1 J

Answer: 2.1 J

Marking notes:

Award [1] for stating or showing that total energy is conserved (total energy = 2.5 J).
Award [1] for correct calculation and answer with unit.
Accept 2.1 J with correct working.

2. (a) Explain why the box moves with constant velocity even though a force is applied. [2]

Answer: The applied force is balanced by an equal and opposite frictional force acting on the box. Since the resultant/net force on the box is zero, according to Newton's First Law, the box continues moving at constant velocity.

Marking notes:

Award [1] for stating that friction opposes the applied force / forces are balanced.
Award [1] for linking to zero resultant force and constant velocity / Newton's First Law.

2. (b) Calculate the weight of the box. [1]

Working: Weight = mass × gravitational field strength = 6.0 kg × 10 N/kg = 60 N

Answer: 60 N

Marking notes:

Award [1] for correct answer with unit.

3. (a) Calculate the maximum pressure the block can exert on the table. [3]

Working:

Weight of block = mg = 4.0 kg × 10 N/kg = 40 N
Force on table = 40 N
Maximum pressure occurs when area of contact is smallest.
Smallest area = 0.10 m × 0.05 m = 0.0050 m²
Pressure = Force / Area = 40 N / 0.0050 m² = 8000 Pa (or 8.0 × 10³ Pa)

Answer: 8000 Pa (or 8.0 kPa)

Marking notes:

Award [1] for calculating weight/force (40 N).
Award [1] for identifying the smallest area (0.0050 m²).
Award [1] for correct pressure calculation with unit.

3. (b) State one way the block can be positioned to exert the minimum pressure on the table. [1]

Answer: Place the block so that the largest face (0.20 m × 0.10 m) is in contact with the table.

Marking notes:

Award [1] for stating the largest area face should be in contact.
Accept: "Place the 0.20 m × 0.10 m face on the table."

4. (a) Calculate the total energy supplied by the heater. [2]

Working:

Time = 10 minutes = 10 × 60 = 600 s
Energy = Power × Time = 50 W × 600 s = 30,000 J (or 30 kJ)

Answer: 30,000 J (or 30 kJ)

Marking notes:

Award [1] for correct conversion of time to seconds.
Award [1] for correct calculation with unit.

4. (b) Calculate the final temperature of the water, assuming no heat loss. [3]

Working:

Energy supplied = 30,000 J
Mass of water = 500 g = 0.50 kg
E = mcΔθ
30,000 = 0.50 × 4200 × Δθ
Δθ = 30,000 / (0.50 × 4200) = 30,000 / 2100 = 14.3 °C (to 1 d.p.)
Final temperature = 25 °C + 14.3 °C = 39.3 °C

Answer: 39.3 °C (accept 39 °C or 39.3 °C)

Marking notes:

Award [1] for correct substitution into E = mcΔθ.
Award [1] for correct calculation of temperature rise (14.3 °C).
Award [1] for adding to initial temperature with correct final answer and unit.

5. (a) State what is meant by refraction of light. [1]

Answer: Refraction is the bending/changing direction of light as it passes from one medium to another of different optical density.

Marking notes:

Award [1] for stating bending/change in direction AND change in medium/optical density.

5. (b) Explain why the light ray bends towards the normal as it enters the glass block. [2]

Answer: Glass is optically denser than air. When light enters the glass, its speed decreases. This causes the light ray to bend towards the normal.

Marking notes:

Award [1] for stating that glass is optically denser / light slows down in glass.
Award [1] for linking decreased speed to bending towards the normal.

6. (a) Calculate the total resistance in the circuit. [1]

Working: R_total = R₁ + R₂ = 4 Ω + 8 Ω = 12 Ω

Answer: 12 Ω

Marking notes:

Award [1] for correct answer with unit.

6. (b) Calculate the current flowing through the circuit. [2]

Working: I = V / R = 12 V / 12 Ω = 1.0 A

Answer: 1.0 A

Marking notes:

Award [1] for correct formula/substitution.
Award [1] for correct answer with unit.

6. (c) Calculate the potential difference across R₂. [2]

Working: V = IR = 1.0 A × 8 Ω = 8.0 V

Answer: 8.0 V

Marking notes:

Award [1] for correct formula/substitution.
Award [1] for correct answer with unit.
Accept alternative method using potential divider principle.

Section B: Physical Sciences – Chemistry (25 marks)

7. (a) Write a balanced chemical equation for roasting of zinc sulfide. [2]

Answer: 2ZnS(s) + 3O₂(g) → 2ZnO(s) + 2SO₂(g)

Marking notes:

Award [1] for correct reactants and products.
Award [1] for correct balancing and state symbols.
Accept: 2ZnS(s) + 3O₂(g) → 2ZnO(s) + 2SO₂(g)

7. (b) Write a balanced chemical equation for reduction of zinc oxide. [2]

Answer: 2ZnO(s) + C(s) → 2Zn(s) + CO₂(g)
OR ZnO(s) + C(s) → Zn(s) + CO(g)

Marking notes:

Award [1] for correct reactants and products.
Award [1] for correct balancing and state symbols.
Accept either equation with CO₂ or CO as product, provided it is balanced.

7. (c) Calculate the mass of zinc extracted from 9.7 g of ZnS. [3]

Working:

Mᵣ of ZnS = 65 + 32 = 97
Moles of ZnS = 9.7 g / 97 g/mol = 0.10 mol
From equation: 2 mol ZnS → 2 mol Zn, so mole ratio is 1:1
Moles of Zn = 0.10 mol
Mass of Zn = 0.10 mol × 65 g/mol = 6.5 g

Answer: 6.5 g

Marking notes:

Award [1] for calculating Mᵣ of ZnS and moles of ZnS.
Award [1] for using correct mole ratio.
Award [1] for correct mass of Zn with unit.

8. (a) Write a balanced chemical equation for the reaction. [2]

Answer: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

Marking notes:

Award [1] for correct reactants and products.
Award [1] for correct balancing and state symbols.

8. (b) Write the ionic equation for this reaction. [2]

Answer: CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + CO₂(g) + H₂O(l)

Marking notes:

Award [1] for correct reactants (CaCO₃ and H⁺).
Award [1] for correct products, balancing, and state symbols.
Spectator ions (Cl⁻) must not be included.

8. (c) Calculate the volume of CO₂ produced. [3]

Working:

Mᵣ of CaCO₃ = 40 + 12 + (3 × 16) = 100
Moles of CaCO₃ = 2.0 g / 100 g/mol = 0.020 mol
From equation: 1 mol CaCO₃ → 1 mol CO₂
Moles of CO₂ = 0.020 mol
Volume of CO₂ = 0.020 mol × 24 dm³/mol = 0.48 dm³ (or 480 cm³)

Answer: 0.48 dm³ (or 480 cm³)

Marking notes:

Award [1] for calculating moles of CaCO₃.
Award [1] for using correct mole ratio.
Award [1] for correct volume with unit.

9. (a) State which dye(s) are present in ink sample A. [1]

Answer: Dye X only.

Marking notes:

Award [1] for correctly identifying dye X.
Accept: "X" or "Dye X."

9. (b) Identify the dyes in ink sample B. [1]

Answer: Dyes X and Y.

Marking notes:

Award [1] for correctly identifying both dyes X and Y.

9. (c) Calculate the Rf value of dye Y. [2]

Working: Rf = distance travelled by substance / distance travelled by solvent front = 4.8 cm / 8.0 cm = 0.60

Answer: 0.60 (or 0.6)

Marking notes:

Award [1] for correct formula/substitution.
Award [1] for correct answer (no units required for Rf).

10. (a) State one observation during the reaction. [1]

Answer: Effervescence / bubbles of gas produced / magnesium ribbon dissolves/disappears / magnesium ribbon becomes smaller.

Marking notes:

Award [1] for any one correct observation.

10. (b) Explain, using collision theory, why the rate decreases as the reaction proceeds. [2]

Answer: As the reaction proceeds, the concentration of sulfuric acid decreases. This means there are fewer hydrogen ions per unit volume. The frequency of effective collisions between reactant particles decreases, so the rate of reaction decreases.

Marking notes:

Award [1] for stating that concentration of acid/reactant decreases.
Award [1] for linking to decreased frequency of effective collisions.

10. (c) Suggest one change to increase the initial rate of reaction. [1]

Answer: Use magnesium powder instead of ribbon (increase surface area) / Heat the acid (increase temperature) / Use a catalyst.

Marking notes:

Award [1] for any one correct suggestion.
Accept: "Use powdered magnesium" or "Warm the acid."

Section C: Physical Sciences – Biology (20 marks)

11. (a) Name the blood vessels labelled P and Q. [2]

Answer:

P: Pulmonary vein
Q: Aorta

Marking notes:

Award [1] for each correct answer.

11. (b) Describe the pathway of blood from the lungs to the aorta. [4]

Answer:

Oxygenated blood returns from the lungs to the left atrium via the pulmonary veins. [1]
The left atrium contracts, increasing pressure and pushing blood into the left ventricle. [1]
The left ventricle contracts, increasing pressure further. [1]
Blood is forced through the aortic valve/semilunar valve into the aorta and then to the rest of the body. [1]

Marking notes:

Award [1] for each distinct step.
Accept answers that mention pressure changes and valve function.

12. (a) Explain why the potato strip in 0.0 mol/dm³ sucrose solution increased in length. [3]

Answer:

The water potential of the solution (pure water) is higher than the water potential of the potato cells. [1]
Water enters the potato cells by osmosis. [1]
The cells become turgid, causing the potato strip to increase in length. [1]

Marking notes:

Award [1] for identifying water potential gradient.
Award [1] for stating osmosis.
Award [1] for linking to turgidity and increase in length.

12. (b) Describe the state of the potato cells in 0.8 mol/dm³ sucrose solution. [2]

Answer: The cells are plasmolysed. The cell membrane/cytoplasm has shrunk/pulled away from the cell wall due to water leaving the cells by osmosis.

Marking notes:

Award [1] for stating "plasmolysed."
Award [1] for describing cell membrane shrinking away from cell wall.

12. (c) Estimate the concentration of sucrose solution with the same water potential as the potato cells. [2]

Answer: Approximately 0.28–0.32 mol/dm³. This is the concentration at which there is no change in length (0 mm change), meaning no net movement of water by osmosis.

Marking notes:

Award [1] for a value between 0.28 and 0.32 mol/dm³.
Award [1] for explaining that no change in length indicates equal water potential / no net osmosis.

13. (a) Suggest how the student could vary the light intensity. [1]

Answer: Change the distance of the lamp from the plant / Use lamps of different power/wattage / Use different numbers of lamps.

Marking notes:

Award [1] for any one correct method.

13. (b) State the purpose of the glass heat shield. [1]

Answer: To prevent heat from the lamp from affecting the temperature of the plant/water / To ensure temperature is kept constant / To ensure temperature is a controlled variable.

Marking notes:

Award [1] for stating that it controls temperature or prevents heating.

13. (c) Plot a graph of the results. [4]

Graph requirements:

X-axis: Distance of lamp (cm) – linear scale from 0 to 50 [1]
Y-axis: Number of bubbles per minute – linear scale from 0 to 50 [1]
All 5 points plotted correctly (± half a small square) [1]
Smooth curve or appropriate line of best fit drawn [1]

Marking notes:

Award marks as indicated above.
Deduct [1] if axes are not labelled with quantities and units.

13. (d) Describe the relationship shown by the data. [1]

Answer: As the distance of the lamp increases, the rate of photosynthesis (number of bubbles per minute) decreases. / The rate of photosynthesis decreases with increasing distance from the light source.

Marking notes:

Award [1] for stating the inverse relationship.

14. (a) State one difference between inhaled and exhaled air. [1]

Answer: Exhaled air contains less oxygen / more carbon dioxide / more water vapour than inhaled air.

Marking notes:

Award [1] for any one correct difference.

14. (b) Explain why exhaled air contains less oxygen than inhaled air. [2]

Answer: Oxygen is used by body cells for aerobic respiration. In the lungs, oxygen diffuses from the alveoli into the blood, while carbon dioxide diffuses from the blood into the alveoli to be exhaled. Therefore, exhaled air has a lower oxygen concentration.

Marking notes:

Award [1] for stating that oxygen is used for respiration.
Award [1] for linking to gas exchange in the lungs/alveoli.

END OF ANSWER KEY