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HKDSE Integrated Science Practice Paper: Matter and Energy, Life and Environment, Earth and Space
HKDSE Integrated Science – 12 questions – 100 marks – 75 minutes
A 12-question HKDSE Integrated Science practice paper covering matter and energy, life and environment, and earth and space, with full worked answers and marking points.
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Paper details HKDSE Integrated Science practice paper: 12 questions, 100 marks, 75 minutes, with worked answers on moles, specific heat, ecology, genetics and earth science.
Part 1: Matter and Energy
Section A: Chemical calculation and energy transfer
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Magnesium burns in oxygen: 2Mg + O₂ → 2MgO. 4.8 g of magnesium is burned completely. (Relative atomic masses: Mg = 24.3, O = 16.0.) (a) Calculate the number of moles of magnesium. (b) Calculate the mass of magnesium oxide formed, in grams to two decimal places. (c) Calculate the moles of oxygen needed and explain where the mass gained in (b) comes from. | (a) Moles of Mg = 4.8 ÷ 24.3 = 0.1975 mol (2 marks) (b) 2 mol Mg gives 2 mol MgO, so moles of MgO = 0.1975 mol; molar mass = 40.3 g/mol; mass = 7.96 g (3 marks) (c) Moles of O₂ = 0.0988 mol, half the moles of Mg, matching the 2:1 ratio (2 marks); the 3.16 g increase is oxygen from the air combining with magnesium, so mass is conserved (1 mark) |
| 2 | An electric kettle rated at 2000 W holds 0.5 kg of water of specific heat capacity 4200 J/(kg °C). The water is heated from 20 °C to its boiling point of 100 °C. (a) Calculate the energy absorbed by the water in joules and in kilojoules. (b) Assuming no heat loss, calculate the time needed. (c) The actual heating took 105 s; calculate the efficiency of the kettle. | (a) E = mcΔT = 0.5 × 4200 × 80 = 168000 J = 168 kJ (3 marks) (b) t = E ÷ P = 168000 ÷ 2000 = 84 s (2 marks) (c) Efficiency = 84 ÷ 105 = 80% (3 marks); the remaining 20% is lost as heat to the surrounding air and the kettle body |
| 3 | An object of mass 2.0 kg is released from the top of a smooth ramp 5.0 m high and reaches the ground at 8.0 m/s. Take g = 10 m/s². (a) Calculate its gravitational potential energy at the top. (b) Calculate its kinetic energy at the bottom. (c) Find the energy lost to friction and the energy efficiency, and explain where that energy goes. | (a) PE = mgh = 2.0 × 10 × 5.0 = 100 J (2 marks) (b) KE = ½mv² = 0.5 × 2.0 × 8.0² = 64 J (2 marks) (c) Energy lost = 100 − 64 = 36 J (2 marks); efficiency = 64 ÷ 100 = 64% (2 marks); it becomes heat and sound, warming the ramp, and total energy is conserved |
| 4 | A titration is carried out: 25.0 cm³ of 0.1 M sodium hydroxide is titrated with 0.125 M hydrochloric acid (HCl + NaOH → NaCl + H₂O). (a) Calculate the moles of sodium hydroxide. (b) Find the volume of acid needed to reach the end point. (c) How would the volume change if 0.050 M acid were used? State a suitable indicator and the colour change at the end point. | (a) n(NaOH) = 25.0 ÷ 1000 × 0.1 = 0.00250 mol (2 marks) (b) 1:1 reaction, so n(HCl) = 0.00250 mol; volume = 0.00250 ÷ 0.125 × 1000 = 20.0 cm³ (3 marks) (c) Halving the concentration doubles the volume to 50.0 cm³ (2 marks); with a strong acid and strong base the end point is near pH 7, so phenolphthalein (colourless to pale pink) or bromothymol blue (blue to green) is suitable (1 mark) |
Part 2: Life and Environment
Section B: Biological systems and data analysis
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | Producers in a grassland fix 50000 kJ per square metre per year, and about 10% of the energy passes to the next trophic level. (a) Calculate the energy available to the primary, secondary and tertiary consumers. (b) Explain where the other 90% of the energy goes. (c) Explain why food chains rarely have more than four or five trophic levels. | (a) Primary 5000 kJ, secondary 500 kJ, tertiary 50 kJ (3 marks, 1 each) (b) The other 90% is released as heat by respiration, with some lost in undigested faeces and dead remains (3 marks) (c) Only about 10% passes to the next level, so too little energy remains to support further levels; food chains therefore stop at four or five (3 marks) |
| 6 | In peas, the tall allele T is dominant to the dwarf allele t. Two plants of genotype Tt are crossed and produce 400 offspring. (a) Draw a genetic diagram and state the phenotypic ratio. (b) Calculate the expected numbers of tall and dwarf plants. (c) The observed numbers are 290 tall and 110 dwarf; calculate each percentage and explain the difference from the expected values. | (a) Tt × Tt → 1 TT : 2 Tt : 1 tt, so tall : dwarf = 3 : 1 (3 marks) (b) Tall expected = 400 × 3/4 = 300; dwarf = 400 × 1/4 = 100 (3 marks) (c) Tall = 290 ÷ 400 × 100% = 72.5%, dwarf = 27.5% (2 marks); the deviation is due to random fertilisation and a finite sample, not a wrong ratio (1 mark) |
| 7 | An athlete has a resting heart rate of 72 beats per minute and a stroke volume of 70 cm³; during strenuous exercise the values are 120 beats per minute and 110 cm³. (a) Calculate the cardiac output at rest and during exercise, in cm³/min and L/min. (b) Calculate the percentage increase in cardiac output during exercise, to one decimal place. (c) Give two physiological reasons why stroke volume rises during exercise. | (a) At rest: 72 × 70 = 5040 cm³/min = 5.04 L/min; during exercise: 120 × 110 = 13200 cm³/min = 13.2 L/min (4 marks) (b) Increase = (13200 − 5040) ÷ 5040 × 100% = 161.9% (3 marks) (c) Sympathetic stimulation increases the force of ventricular contraction; greater venous return from the muscle pump fills the ventricles more, raising stroke volume (2 marks) |
| 8 | The rate of photosynthesis of an aquatic plant is measured by counting oxygen bubbles per minute at light intensities of 20, 40, 60 and 80 units, giving 12, 24, 30 and 30 bubbles per minute respectively. (a) Calculate the percentage increase in rate when intensity rises from 20 to 40 units. (b) Describe the trend beyond 60 units and explain it. (c) Suggest two changes that could raise the curve further, naming the limiting factor each addresses. | (a) Increase = (24 − 12) ÷ 12 × 100% = 100% (3 marks) (b) Beyond 60 units the rate stays at 30 bubbles per minute, a plateau (2 marks); light is no longer limiting, so carbon dioxide concentration or temperature is now the limiting factor (2 marks) (c) Add sodium hydrogencarbonate to raise dissolved CO₂, or raise the water temperature towards the optimum; both lift the plateau (2 marks) |
Part 3: Earth and Space
Section C: Earth systems and applied science
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A seismograph records a 20 s difference between the arrival of P waves and S waves. P waves travel at 6.0 km/s and S waves at 4.0 km/s in that crust. (a) Using distance divided by speed, calculate the distance from the earthquake to the station. (b) Calculate the travel time of each wave. (c) Explain why S waves cannot be used to investigate the state of the outer core. | (a) With distance d: d/4.0 − d/6.0 = 20, so d = 20 × 24.0 ÷ 2 = 240 km (4 marks) (b) P wave time = 240 ÷ 6.0 = 40 s; S wave time = 240 ÷ 4.0 = 60 s (2 marks) (c) S waves are transverse and cannot travel through liquids; the liquid outer core blocks them and creates a shadow zone, which is the evidence that the outer core is liquid (2 marks) |
| 10 | On a day when the air temperature is 25 °C, the actual water vapour content is 13.8 g/m³. The saturation vapour content is 23.0 g/m³ at 25 °C and 12.8 g/m³ at 15 °C. (a) Calculate the relative humidity. (b) If the air cools to 15 °C with no change in vapour content, calculate the mass of water condensed per cubic metre. (c) Explain why dew forms in the early morning and state what the dew point means. | (a) Relative humidity = 13.8 ÷ 23.0 × 100% = 60% (3 marks) (b) Condensed water = 13.8 − 12.8 = 1.0 g/m³ (3 marks) (c) At night the ground cools by radiation and the air near it falls below the dew point, so water vapour condenses on grass as dew (1 mark); the dew point is the temperature at which the air becomes saturated and condensation begins (1 mark) |
| 11 | A solar panel of area 2.0 m² works under solar irradiance of 1000 W/m² for 5 hours with a conversion efficiency of 18%. (a) Calculate the total incident solar energy in joules and in kWh. (b) Calculate the electrical energy actually produced in kWh. (c) If the efficiency fell to 15%, how many hours of sunlight would be needed to produce the same 1.8 kWh? | (a) Incident energy = 1000 × 2.0 × 5 × 3600 = 3.600e+07 J = 10.0 kWh (4 marks) (b) Electrical energy = 10.0 × 18% = 1.8 kWh (2 marks) (c) At 15% the output is 1000 × 2.0 × 0.15 ÷ 1000 = 0.3 kWh per hour, so 1.8 kWh needs 1.8 ÷ 0.3 = 6.0 hours (2 marks) |
| 12 | A student measures the mass of an aluminium block as 54.0 g, then lowers it into a measuring cylinder containing 30.0 cm³ of water; the level rises to 50.0 cm³. Each reading has an uncertainty of ±0.5 cm³. (a) Calculate the volume and density of the block. (b) Calculate the percentage uncertainty in volume and hence the maximum percentage error in density. (c) Would a 500 cm³ cylinder give a larger or smaller error? Explain. | (a) Volume = 50.0 − 30.0 = 20.0 cm³; density = 54.0 ÷ 20.0 = 2.70 g/cm³ (3 marks) (b) Each reading is ±0.5, so volume carries ±1.0 cm³, i.e. 5%; density has the same percentage error, about ±5% (3 marks) (c) A 500 cm³ cylinder has coarser graduations, so the same absolute reading error is a larger percentage of a smaller volume change; a narrow, finely graduated cylinder is better (2 marks) |
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