Secondary 3 Combined Science Scientific Inquiry Quiz

Secondary 3 Combined Science Quiz - Scientific Inquiry (Answer Key)

Total Marks: 40

Section A: Multiple Choice Questions (10 marks)

1. Answer: B [1]

Explanation: A hypothesis is a testable prediction based on observations and prior knowledge. It is not a proven fact (that would be a theory/law), nor is it a procedure or conclusion.

2. Answer: B [1]

Explanation: The independent variable is the one deliberately changed by the experimenter. Here, temperature is being changed to see its effect on enzyme activity (the dependent variable).

3. Answer: B [1]

Working: Mean = (38.2 + 38.5 + 38.1) / 3 = 114.8 / 3 = 38.266... ≈ 38.3 s (to 1 decimal place)

4. Answer: C [1]

Explanation: Control variables are kept constant to ensure a fair test. Temperature must be controlled because it also affects enzyme activity and photosynthesis rate. Light intensity is the independent variable; rate of photosynthesis and bubble count are dependent variables.

5. Answer: A [1]

Working: Mass of MgO formed = Final mass - Initial mass = 25.58 g - 25.34 g = 0.24 g

6. Answer: A [1]

Explanation: A directly proportional relationship (y ∝ x) gives a straight line graph passing through the origin (0,0). The equation is y = kx.

7. Answer: C [1]

Explanation: A line of best fit should show the general trend with roughly equal scatter of points above and below the line. It does not need to pass through any specific points.

8. Answer: B [1]

Explanation: Increasing concentration increases the number of particles per unit volume, leading to more frequent collisions. It does not increase the speed of particles (that would require temperature increase). This is a common misconception.

9. Answer: C [1]

Explanation: A pipette (especially a volumetric pipette) is designed to deliver a fixed, accurate volume (e.g., 25.0 cm³). Measuring cylinders and beakers have larger graduations and greater uncertainty.

10. Answer: B [1]

Explanation: The variable deliberately changed is the independent variable. The dependent variable is measured; control variables are kept constant.

Section B: Structured Questions (20 marks)

11. (a) Independent variable: Surface area of calcium carbonate chips (or size of chips) [1]

Marking note: Must mention surface area or particle size.

(b) Dependent variable: Volume of carbon dioxide gas produced (or rate of reaction) [1] Marking note: Accept "volume of gas" or "time taken to produce certain volume" or "rate of reaction".

(c) Two control variables (any two): [2]

• Mass of calcium carbonate
• Volume of hydrochloric acid
• Concentration of hydrochloric acid
• Temperature of reactants
• Pressure (if relevant) Marking note: 1 mark each. Must be variables that could affect the rate if not controlled.

(d) Advantage of gas syringe: [1]

• No gas dissolves in water (unlike water displacement method)
• Direct volume reading without need to level water
• Can measure larger volumes
• Less chance of gas leakage Marking note: Accept any valid advantage.

12. (a) Temperature rise = 48.5°C - 22.0°C = 26.5°C [1]

(b) Energy transferred to water = m × c × ΔT [1] = 20 g × 4.2 J/g°C × 26.5°C [1] = 2226 J [1] Marking note: 1 mark for correct substitution, 1 mark for correct answer with unit.

(c) Mass of peanut burnt = 0.85 g - 0.32 g = 0.53 g [1] Energy content = Energy transferred / Mass burnt = 2226 J / 0.53 g = 4200 J/g (or 4.20 kJ/g) [1] Marking note: 1 mark for mass burnt, 1 mark for correct calculation. Accept 4200 J/g or 4.2 kJ/g.

(d) Two reasons for lower experimental value: [2]

• Heat loss to surroundings (air, boiling tube, thermometer)
• Incomplete combustion of peanut (not all energy released)
• Some energy used to heat the boiling tube/thermometer, not just water
• Evaporation of water during heating Marking note: 1 mark each. Must relate to energy loss or incomplete transfer.

13. (a) Graph plotting: [3]

• Axes labeled with units (Load/N, Extension/cm) [1]
• Appropriate scale using >50% of grid, points plotted correctly [1]
• Line of best fit drawn (straight line through origin, ignoring anomaly at 3 N) [1] Marking note: The point at 3 N (7.6 cm) is anomalous; line should pass near (0,0), (1,2.5), (2,5.0), (4,10.0), (5,12.5).

(b) Extension at 3.5 N = 8.75 cm (accept 8.7–8.8 cm from graph) [1] Marking note: Read from candidate's line of best fit.

(c) Relationship: Extension is directly proportional to load (Hooke's Law obeyed). [1] Marking note: Must mention "directly proportional" or "linear through origin".

(d) Possible reason for anomaly at 3 N: [1]

• Parallax error reading the scale
• Spring not at equilibrium when reading taken
• Load not added gently (oscillations)
• Zero error not accounted for Marking note: Accept any valid experimental error.

14. (a) pH 7 [1]

Explanation: At pH 7, the iodine test shows yellow-brown (no starch present), meaning all starch has been digested by amylase. At other pH values, blue-black indicates starch remains undigested. [1] Marking note: 1 mark for correct pH, 1 mark for explanation linking colour to starch digestion.

(b) At pH 3 and pH 9, the enzyme is denatured / active site shape changes [1] Extreme pH alters the ionic bonds holding the enzyme's tertiary structure, changing the active site shape so substrate cannot bind effectively. [1] Marking note: 1 mark for denaturation/active site change, 1 mark for explanation of mechanism.

(c) Control variable (any one): [1]

• Temperature
• Concentration of amylase
• Concentration of starch
• Volume of solutions
• Time of reaction (5 minutes) Marking note: Must be a variable that could affect enzyme activity.

15. (a) Use titrations 2, 3, and 4 [1]

Explanation: These are concordant (within 0.10 cm³ of each other). Titration 1 is a rough titration and not precise enough. [1] Marking note: 1 mark for identifying correct titres, 1 mark for explanation of concordance.

(b) Average titre = (23.85 + 23.90 + 23.80) / 3 = 71.55 / 3 = 23.85 cm³ [1] Marking note: Must use concordant titres only. Answer to 2 decimal places.

(c) Moles of HCl = concentration × volume (dm³) = 0.100 × (23.85/1000) = 0.002385 mol [1] Mole ratio HCl:NaOH = 1:1, so moles NaOH = 0.002385 mol [1] Concentration NaOH = moles / volume (dm³) = 0.002385 / (25.0/1000) = 0.0954 mol/dm³ [1] Marking note: 1 mark for moles HCl, 1 mark for concentration calculation. Accept 0.0954 mol/dm³ or 0.095 mol/dm³.

Section C: Data Analysis and Experimental Design (10 marks)

16. (a) Melting point = 70°C [1]

Explanation: The plateau on a cooling curve represents the change of state (freezing), which occurs at the melting/freezing point.

(b) During the plateau (4–7 minutes), stearic acid is changing state from liquid to solid. [1] The energy lost is latent heat of fusion, used to overcome intermolecular forces and form the solid structure, not to lower temperature. [1] Marking note: 1 mark for state change, 1 mark for latent heat explanation.

(c) Sketch should show: [2]

• Same melting point (70°C plateau) [1]
• Longer plateau (more time to solidify larger mass) [1]
• Same initial cooling rate (if same conditions) or slightly slower initial cooling Marking note: 1 mark for same plateau temperature, 1 mark for longer plateau duration.

17. Experimental Design for Osmosis Investigation [5]

Marking scheme (1 mark each for 5 key points):

1. Independent variable: Concentration of salt solution (e.g., 0.0, 0.2, 0.4, 0.6, 0.8, 1.0 mol/dm³) [1]
2. Dependent variable: Percentage change in mass of potato cylinders = (final mass - initial mass) / initial mass × 100% [1]
3. Two control variables: [1]
   • Length/dimensions of potato cylinders (use cork borer, cut to same length)
   • Temperature of solutions
   • Time immersed
   • Volume of solution
   • Same potato (or same variety)
4. Brief procedure: [1]
   • Cut potato cylinders of equal size using cork borer and scalpel
   • Measure and record initial mass of each cylinder
   • Place each cylinder in a different salt concentration
   • Leave for fixed time (e.g., 30 minutes)
   • Remove, blot dry, measure final mass
   • Calculate percentage change in mass
5. Reliability: [1]
   • Repeat each concentration at least 3 times (replicates)
   • Calculate mean percentage change
   • Identify and exclude anomalies

Teaching notes: A good experimental design must be specific about how variables are controlled and measured. Percentage change in mass is better than absolute mass change because initial masses may vary slightly. Replicates are essential for reliability.

18. (a) Graph plotting: [3]

• Axes labeled with units (Distance/cm, Bubbles per minute) [1]
• Appropriate scale, points plotted correctly [1]
• Smooth curve of best fit (decreasing curve, not straight line) [1] Marking note: Relationship is non-linear (inverse square law for light intensity), so curve is expected.

(b) As distance increases (light intensity decreases), the rate of photosynthesis decreases. [1] The relationship is non-linear: rate decreases rapidly at first, then more slowly at greater distances. [1] Marking note: 1 mark for inverse relationship, 1 mark for non-linear description.

(c) The conclusion is not valid because: [2]

• Other factors (CO₂ concentration, temperature) could become limiting at high light intensities
• The experiment only varies light intensity; it does not test whether other factors are limiting
• At the closest distances, the rate may be limited by CO₂ or enzyme saturation, not light Marking note: 1 mark for identifying other limiting factors, 1 mark for explaining why the experiment doesn't prove light is the only factor.

19. (a) Diameter = Sleeve reading + Thimble reading = 5.5 mm + 0.28 mm = 5.78 mm [1]

Marking note: Sleeve: 5.5 mm (5 mm + 0.5 mm half-scale). Thimble: 28 × 0.01 mm = 0.28 mm.

(b) Average = (5.78 + 5.80 + 5.77 + 5.79 + 5.81) / 5 = 28.95 / 5 = 5.79 mm [1]

(c) Taking multiple readings reduces random errors (e.g., slight variations in wire diameter, parallax errors, positioning errors). Averaging gives a more reliable estimate of the true diameter. [1] Marking note: Must mention random errors or variability in the wire itself.

20. (a) Electrical energy = Power × Time = V × I × t [1]

= 12.0 V × 2.50 A × (5.00 × 60) s [1] = 12.0 × 2.50 × 300 = 9000 J [1] Marking note: 1 mark for correct formula and time conversion to seconds, 1 mark for correct answer with unit.

(b) Energy = m × c × ΔT → c = Energy / (m × ΔT) [1] ΔT = 55.0 - 20.0 = 35.0°C c = 9000 J / (1.00 kg × 35.0°C) = 257 J/kg°C [1] Marking note: 1 mark for correct rearrangement and substitution, 1 mark for answer with unit.

(c) Percentage difference = |Experimental - Accepted| / Accepted × 100% [1] = |257 - 450| / 450 × 100% = 193 / 450 × 100% = 42.9% [1] Marking note: 1 mark for correct calculation.

(d) Reason for difference: [1]

• Heat loss to surroundings (not all electrical energy heats the block)
• Heat absorbed by container/thermometer/heater itself
• Inaccurate temperature measurement (thermometer not in good contact)
• Non-uniform heating of block Marking note: Accept any valid source of energy loss or measurement error.

End of Answer Key