Secondary 3 Combined Science Semestral Assessment 2 (End of Year) Paper 1

TuitionGoWhere Practice Paper – Combined Science Secondary 3

SA2 – Version 1: Answer Key and Marking Scheme

Total Marks: 60

Section A: Physics – Energy, Forces & Pressure [20 marks]

Question 1: Conservation of Energy [6 marks]

(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 a complete statement including both "cannot be created or destroyed" and "converted/transferred" or "total energy remains constant". Do not award if only one part is stated.

(b) Calculate the gravitational potential energy of the ball at the point of release. [2]

Answer: GPE = mgh GPE = 0.5 × 10 × 2.0 GPE = 10 J

Marking notes: [1] for correct formula/substitution; [1] for correct answer with unit (10 J). Accept 10 J only.

(c) Assuming no energy is lost to air resistance, calculate the speed of the ball just before it hits the ground. [2]

Answer: By conservation of energy: GPE lost = KE gained mgh = ½mv² 10 = ½ × 0.5 × v² 10 = 0.25v² v² = 40 v = √40 = 6.32 m/s (or 6.3 m/s)

Marking notes: [1] for equating GPE to KE or correct substitution into ½mv²; [1] for correct answer with unit (6.32 m/s or 6.3 m/s). Accept answers rounded to 2 or 3 significant figures.

(d) Explain why this speed is lower than the value calculated in part (c). [1]

Answer: Some energy is lost/transferred as heat (and sound) due to air resistance / friction with air molecules. Not all gravitational potential energy is converted to kinetic energy.

Marking notes: [1] for mentioning energy loss due to air resistance/friction. Accept "work done against air resistance" or "energy dissipated as heat".

Question 2: Pressure in Liquids [5 marks]

(a) State the relationship between the depth of a liquid and the pressure exerted by the liquid. [1]

Answer: Pressure increases with depth / Pressure is directly proportional to depth (for a liquid of constant density).

Marking notes: [1] for stating that pressure increases with depth. Accept "the deeper the point, the greater the pressure".

(b) Explain why the water jet from hole C travels the furthest horizontal distance. [2]

Answer: Hole C is at the greatest depth, so the water pressure at C is the highest. The higher pressure causes water to exit hole C with the greatest speed/velocity, so it travels the furthest horizontal distance before hitting the ground.

Marking notes: [1] for linking greater depth to higher pressure; [1] for linking higher pressure to greater exit speed and therefore greater horizontal distance.

(c) Calculate the pressure exerted by the water at hole C. [2]

Answer: P = ρgh P = 1000 × 10 × 0.80 P = 8000 Pa (or 8.0 × 10³ Pa)

Marking notes: [1] for correct formula/substitution; [1] for correct answer with unit (8000 Pa or 8.0 kPa). Accept 8000 N/m².

Question 3: Moments and Equilibrium [5 marks]

(a) State the Principle of Moments. [1]

Answer: For an object in equilibrium, the sum of clockwise moments about a pivot equals the sum of anticlockwise moments about the same pivot.

Marking notes: [1] for a complete statement. Accept "total clockwise moments = total anticlockwise moments" provided "about the same pivot" is implied or stated.

(b) Calculate the value of the unknown weight W. [3]

Answer: Taking moments about the pivot (50 cm mark): Clockwise moment = W × (80 – 50) = W × 30 Anticlockwise moment = 2.0 × (50 – 20) = 2.0 × 30 = 60 N cm

By Principle of Moments: W × 30 = 60 W = 60 ÷ 30 = 2.0 N

Marking notes: [1] for correct distances from pivot (30 cm each); [1] for correct moment equation; [1] for correct answer with unit (2.0 N). Note: the weight of the rule (1.2 N) acts at the pivot (50 cm mark) so its moment is zero and does not affect the calculation.

(c) State one assumption made about the metre rule in this calculation. [1]

Answer: The metre rule is uniform / The weight of the rule acts at its centre (50 cm mark) / The pivot is at the centre of gravity of the rule.

Marking notes: [1] for any valid assumption. Accept "the rule is uniform" or "the centre of gravity is at the 50 cm mark".

Question 4: Work, Energy and Power [4 marks]

(a) Calculate the work done by the worker in lifting the load. [2]

Answer: Work done = Force × distance (in direction of force) Work done = 400 × 6.0 Work done = 2400 J

Marking notes: [1] for correct formula/substitution; [1] for correct answer with unit (2400 J). Accept 2.4 kJ.

(b) Calculate the power output of the worker. [2]

Answer: Power = Work done ÷ Time Power = 2400 ÷ 15 Power = 160 W

Marking notes: [1] for correct formula/substitution; [1] for correct answer with unit (160 W). Accept ecf from part (a) if work done is incorrect but power calculation is correct.

Section B: Chemistry – Reactions, Calculations & Analysis [20 marks]

Question 5: Chemical Equations and Stoichiometry [7 marks]

(a) Write a balanced chemical equation for the reaction between zinc oxide and carbon. Include state symbols. [2]

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

Marking notes: [1] for correct formulae of reactants and products; [1] for correct balancing and state symbols. Accept ZnO(s) + C(s) → Zn(s) + CO(g) only if the question context allows for carbon monoxide as a product; however, the standard extraction equation with CO₂ is expected. Deduct [1] if state symbols are missing or incorrect.

(b) Calculate the maximum mass of zinc that can be obtained. [3]

Answer: Mr of ZnO = 65 + 16 = 81 Moles of ZnO = mass ÷ Mr = 8.1 ÷ 81 = 0.10 mol

From equation: 2 mol ZnO produces 2 mol Zn So 0.10 mol ZnO produces 0.10 mol Zn

Mass of Zn = moles × Ar = 0.10 × 65 = 6.5 g

Marking notes: [1] for correct calculation of moles of ZnO; [1] for correct mole ratio (1:1 from balanced equation); [1] for correct mass of Zn with unit (6.5 g). Award ecf if mole ratio is used incorrectly but subsequent calculation is consistent.

(c) Calculate the percentage yield of zinc. [2]

Answer: Percentage yield = (actual yield ÷ theoretical yield) × 100% Percentage yield = (5.8 ÷ 6.5) × 100% Percentage yield = 89.2% (or 89%)

Marking notes: [1] for correct formula/substitution; [1] for correct answer (89.2% or 89%). Accept ecf from part (b) for theoretical yield.

Question 6: Acid-Base Reactions [6 marks]

(a) Write a balanced chemical equation for the reaction between calcium carbonate and hydrochloric acid. Include state symbols. [2]

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

Marking notes: [1] for correct formulae of reactants and products; [1] for correct balancing and state symbols. Deduct [1] if state symbols are missing or incorrect.

(b) Explain why the mass of the apparatus decreases during the first four minutes. [2]

Answer: Carbon dioxide gas is produced during the reaction. The gas escapes from the flask into the atmosphere, so the total mass of the apparatus (flask + contents) decreases. The mass stops decreasing after 4 minutes because the reaction is complete / all the calcium carbonate has reacted.

Marking notes: [1] for stating that CO₂ gas is produced and escapes; [1] for linking gas escape to mass decrease. Accept "gas is lost to the surroundings".

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

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

Marking notes: [1] for correct reactants (CaCO₃ and H⁺); [1] for correct products and balancing. Spectator ions (Cl⁻) must be omitted. Deduct [1] if spectator ions are included.

Question 7: Chromatography [7 marks]

(a) State the principle behind the separation of substances by paper chromatography. [1]

Answer: Different substances have different solubilities in the solvent and different degrees of adsorption/affinity for the stationary phase (paper), so they travel at different rates and are separated.

Marking notes: [1] for mentioning differential solubility or differential affinity/adsorption. Accept "substances have different Rf values" only if explained in terms of solubility/affinity.

(b) Which pen, P, Q, or R, contains only one dye? Explain your answer. [2]

Answer: Pen R contains only one dye. Only one spot is visible in the R column on the chromatogram, indicating that the ink is a single substance / pure dye.

Marking notes: [1] for identifying pen R; [1] for explaining that only one spot indicates one dye/substance.

(c) Identify which reference ink(s), if any, are present in pen Q. Explain your reasoning. [2]

Answer: Both Reference Ink A and Reference Ink B are present in pen Q. The two spots in the Q column match the positions (Rf values) of Reference Ink A and Reference Ink B on the chromatogram.

Marking notes: [1] for identifying both reference inks; [1] for explaining that the spots match in position/Rf value. Award [1] only if both inks are correctly identified.

(d) Calculate the Rf value of the dye spot in pen P. [2]

Answer: Rf = distance moved by substance ÷ distance moved by solvent front Rf = 3.2 ÷ 8.0 Rf = 0.40 (or 0.4)

Marking notes: [1] for correct formula/substitution; [1] for correct answer (0.40 or 0.4). Rf has no units.

Question 8: Rate of Reaction [4 marks]

(a) State two variables that must be kept constant in this experiment. [2]

Answer: Any two from:

• Temperature of the acid
• Volume of hydrochloric acid
• Mass/length/surface area of magnesium ribbon
• Same apparatus setup

Marking notes: [1] for each valid controlled variable (max 2). Accept any reasonable variable that would affect the rate.

(b) Explain, using collision theory, why increasing the concentration of hydrochloric acid increases the rate of reaction. [2]

Answer: Increasing the concentration of hydrochloric acid means there are more H⁺ ions per unit volume. This increases the frequency of collisions between reactant particles (H⁺ ions and magnesium atoms). More frequent effective collisions per unit time leads to a faster rate of reaction.

Marking notes: [1] for stating that there are more particles/ions per unit volume; [1] for linking this to increased collision frequency and faster rate. Accept "more particles per unit volume" or "particles are closer together".

Section C: Biology – Transport, Respiration & Enzymes [20 marks]

Question 9: The Circulatory System [7 marks]

(a) Name the chambers labelled A and B. [2]

Answer: A: Right atrium (accept: right auricle) B: Left ventricle

Marking notes: [1] for each correct chamber. Must specify "right" for A and "left" for B. Do not accept "atrium" or "ventricle" alone.

(b) Describe the pathway of deoxygenated blood from the body through the heart to the lungs. [3]

Answer: Deoxygenated blood from the body enters the right atrium via the vena cava. The right atrium contracts, pushing blood through the tricuspid valve into the right ventricle. The right ventricle contracts, pumping blood through the pulmonary valve into the pulmonary artery, which carries it to the lungs.

Marking notes: [1] for stating blood enters right atrium from vena cava; [1] for blood passing from right atrium to right ventricle; [1] for blood leaving right ventricle via pulmonary artery to lungs. Award marks for correct sequence. Accept "vena cava" or "superior/inferior vena cava".

(c) Explain why the wall of chamber B is thicker than the wall of chamber A. [2]

Answer: Chamber B is the left ventricle, which pumps blood to the entire body (systemic circulation). This requires higher pressure to overcome the resistance of the extensive network of blood vessels. Chamber A is the right atrium, which only pumps blood to the nearby right ventricle at lower pressure. The thicker muscular wall of the left ventricle generates the greater force needed.

Marking notes: [1] for stating that the left ventricle pumps blood to the whole body / systemic circulation; [1] for linking this to the need for higher pressure and thicker muscle. Accept comparison with right atrium's function (pumping only to right ventricle).

Question 10: Respiration [5 marks]

(a) Write the word equation for aerobic respiration in humans. [1]

Answer: Glucose + Oxygen → Carbon dioxide + Water (+ energy/ATP)

Marking notes: [1] for correct word equation. "Energy" or "ATP" may be included but is not required for the mark. Do not accept chemical formulae unless the question explicitly allows them.

(b) Write the word equation for anaerobic respiration in human muscle cells. [1]

Answer: Glucose → Lactic acid (+ energy/ATP)

Marking notes: [1] for correct word equation. Note: no oxygen is involved. Do not accept "ethanol + carbon dioxide" (this is for yeast, not human muscle).

(c) Explain why the sprinter's muscles carry out anaerobic respiration during the race, even though aerobic respiration is also occurring. [2]

Answer: During intense exercise/sprinting, the oxygen supply to the muscles is insufficient to meet the high energy demand. Aerobic respiration alone cannot provide energy quickly enough, so the muscles supplement energy production with anaerobic respiration, which does not require oxygen and can release energy rapidly.

Marking notes: [1] for stating that oxygen supply is insufficient / oxygen debt occurs; [1] for explaining that anaerobic respiration provides additional/rapid energy. Accept "oxygen cannot be delivered fast enough".

(d) State one disadvantage of anaerobic respiration compared to aerobic respiration. [1]

Answer: Any one from:

• Anaerobic respiration releases much less energy (per glucose molecule) than aerobic respiration.
• Anaerobic respiration produces lactic acid, which causes muscle fatigue/cramps/soreness.
• Lactic acid is toxic and must be removed/oxidised later (oxygen debt).

Marking notes: [1] for any valid disadvantage. Accept "produces less ATP/energy" or "causes muscle fatigue".

Question 11: Enzymes [8 marks]

(a) State the optimum temperature for amylase activity based on the results. [1]

Answer: 40 °C

Marking notes: [1] for 40 °C. This is the temperature at which the time to break down starch is shortest (30 s), indicating the fastest rate of reaction.

(b) Explain why the time taken to break down starch decreases as the temperature increases from 10 °C to 40 °C. [2]

Answer: As temperature increases, the kinetic energy of enzyme and substrate molecules increases. This leads to more frequent collisions between enzyme (amylase) and substrate (starch) molecules. More enzyme-substrate complexes are formed per unit time, so the rate of reaction increases and the time taken decreases.

Marking notes: [1] for stating that kinetic energy of molecules increases; [1] for linking this to increased collision frequency / more enzyme-substrate complexes formed. Accept reference to collision theory.

(c) Explain why no reaction occurs at 70 °C. [2]

Answer: At 70 °C, the high temperature has denatured the amylase enzyme. The enzyme's active site has lost its specific three-dimensional shape permanently, so the starch substrate can no longer fit into the active site. No enzyme-substrate complexes can form, so no reaction occurs.

Marking notes: [1] for stating that the enzyme is denatured; [1] for explaining that the active site shape is changed/lost and substrate cannot bind. Accept "active site is destroyed" or "enzyme loses its shape permanently".

(d) Using the lock-and-key hypothesis, explain how amylase breaks down starch. [3]

Answer: According to the lock-and-key hypothesis, the starch molecule (substrate) has a shape that is complementary to the active site of the amylase enzyme (like a key fitting a lock). The starch molecule fits into the active site, forming an enzyme-substrate complex. The enzyme then catalyses the breakdown of starch into maltose. The products (maltose) are released, and the enzyme remains unchanged and can be reused.

Marking notes: [1] for stating that the substrate (starch) has a complementary shape to the enzyme's active site; [1] for describing formation of enzyme-substrate complex; [1] for stating that products are released and enzyme is unchanged/reusable. Award marks for clear description of the lock-and-key model.

Question 12: Transport in Plants [6 marks]

(a) Name the tissue in the celery stalk that is stained red. [1]

Answer: Xylem

Marking notes: [1] for "xylem". Do not accept "phloem" or "vascular bundle" alone.

(b) State the function of the tissue named in part (a). [1]

Answer: Xylem transports water and dissolved mineral salts from the roots to the rest of the plant. It also provides mechanical support.

Marking notes: [1] for stating transport of water and mineral salts. Accept "transport of water" alone. "Mechanical support" is not required but may be mentioned.

(c) Name the force that is mainly responsible for the movement of the red dye up the celery stalk. [1]

Answer: Transpiration pull

Marking notes: [1] for "transpiration pull". Accept "transpiration" or "transpiration stream". Do not accept "capillary action" as the main force (though it contributes).

(d) Describe the pathway of water from the soil into the root of a plant and its subsequent transport to the leaves. [3]

Answer: Water from the soil enters the root hair cells by osmosis (from a region of higher water potential in the soil to lower water potential in the root hair cell). The water then moves through the root cortex cells and passes through the endodermis into the xylem vessels in the centre of the root. Water is transported up the stem through the xylem vessels to the leaves, where it is used in photosynthesis or lost as water vapour through the stomata (transpiration).

Marking notes: [1] for stating water enters root hair cells by osmosis; [1] for stating water moves through cortex/root tissues to xylem; [1] for stating water is transported up through xylem to leaves. Accept "root hair → cortex → endodermis → xylem → stem → leaves" as a sequence.

END OF ANSWER KEY

Marking scheme notes:

• Award marks for correct scientific reasoning even if wording differs from model answers.
• For calculation questions, award ecf (error carried forward) where a student uses an incorrect value from a previous part but applies the correct method.
• Spelling errors should not be penalised unless the word becomes ambiguous or unrecognisable.
• Where a question asks for an explanation, look for cause-and-effect reasoning, not just a statement of fact.