S3 Integrated Science Practice Paper (12 Questions, 100 Marks, 60 Minutes)

2026-11-02

S3 Integrated Science Full Paper

S3 Integrated Science practice paper: 12 original questions in 60 minutes for 100 marks, covering density and floating, average speed, parallel and series circuits, acid-base

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Paper details　Level: S3 | Subject: Integrated Science | Time: 60 minutes | Marks: 100 | Questions: 12

Section A (Short Questions, 4 marks each, 24 marks in total)

Write the formula, substitute the numbers and give the units.

#	Question	Answer and marking notes
1	A metal block has a mass of 78.0 g and a volume of 30.0 cm3. Find its density (g/cm3) and state whether it floats or sinks in water (density 1.0 g/cm3).	Density = 78.0 / 30.0 = 2.6 g/cm3, which is greater than the density of water, so the block sinks.
2	A student runs 150 m at a steady speed in 12 s. Find the average speed in m/s and in km/h.	Average speed = 150 / 12 = 12.5 m/s; in km/h, 12.5 x 3.6 = 45 km/h.
3	A 6 ohm resistor and a 3 ohm resistor are connected in parallel across a 12 V battery. Find the equivalent resistance and the total current from the battery.	1/R = 1/6 + 1/3 = 1/2, so R = 2 ohm; I = V / R = 12 / 2 = 6 A.
4	25.0 cm3 of 0.20 M sodium hydroxide solution is exactly neutralised by 0.25 M hydrochloric acid. Find the volume of acid needed (cm3).	Moles of NaOH = 0.20 x 0.0250 = 0.0050 mol; the ratio is 1:1, so 0.0050 mol of HCl is needed; volume = 0.0050 / 0.25 = 0.020 dm3 = 20.0 cm3.
5	Photosynthesis follows 6CO2 + 6H2O -> C6H12O6 + 6O2. If 264 g of carbon dioxide reacts completely, find the mass of glucose formed (g). (Relative atomic masses: C = 12, H = 1, O = 16)	Moles of CO2 = 264 / 44 = 6.0 mol; 6 mol of CO2 gives 1 mol of glucose; molar mass = 6×12 + 12×1 + 6×16 = 180 g/mol; mass = 180 g.
6	In a food chain the producers fix 100 000 kJ of energy. If only 10% of the energy passes to the next trophic level, find the energy available at the fourth trophic level (kJ).	Second level = 100 000 x 0.1 = 10 000 kJ; third level = 1 000 kJ; fourth level = 100 kJ.

Section B (Data and Calculation Questions, 8 marks each, 24 marks in total)

Show every step; an answer without working earns at most half the marks.

#	Question	Answer and marking notes
7	A rectangular metal block measures 20 cm x 10 cm x 5 cm and has a mass of 2.0 kg. Take g = 10 N/kg. (a) Find its weight. (b) Find the maximum pressure it exerts on a table when resting flat (Pa). (c) Find the minimum pressure (Pa). (d) Explain why one object can produce different pressures.	(a) Weight = 2.0 x 10 = 20 N. (b) Smallest area = 0.05 x 0.10 = 0.005 m2, so maximum pressure = 20 / 0.005 = 4000 Pa. (c) Largest area = 0.20 x 0.10 = 0.020 m2, so minimum pressure = 20 / 0.020 = 1000 Pa. (d) The weight is unchanged; pressure is force divided by contact area, so a smaller area gives a larger pressure.
8	A 4 ohm resistor and a 6 ohm resistor are connected in series across a 6 V battery. (a) Find the total resistance. (b) Find the current. (c) Find the voltage across each resistor. (d) Find the power used by the whole circuit.	(a) Total resistance = 4 + 6 = 10 ohm. (b) I = 6 / 10 = 0.6 A. (c) Across 4 ohm: 0.6 x 4 = 2.4 V; across 6 ohm: 0.6 x 6 = 3.6 V; total 6 V. (d) P = VI = 6 x 0.6 = 3.6 W.
9	A 50.0 g mixture of sand and salt is dissolved in water, filtered, and the filtrate evaporated to dryness, leaving 18.0 g of sand. (a) Find the mass of salt in the mixture. (b) Find the percentage by mass of sand and of salt. (c) Find the mass of salt obtainable from 250 g of the same mixture. (d) Describe the role of filtration and evaporation in this separation.	(a) Salt = 50.0 – 18.0 = 32.0 g. (b) Sand 18.0 / 50.0 x 100% = 36%; salt 32.0 / 50.0 x 100% = 64%. (c) 250 x 0.64 = 160 g. (d) Filtration traps the insoluble sand on the filter paper while the salt solution passes through; evaporation removes the water so that salt crystallises out.

Section C (Extended Questions, 52 marks in total)

Calculate first, then explain in full sentences.

#	Question	Answer and marking notes
10	A lake has pH 3 and a volume of 1.0 x 10^6 L. Calcium carbonate is added to neutralise the acid: CaCO3 + 2H+ -> Ca2+ + CO2 + H2O. (a) Find the hydrogen ion concentration (mol/L). (b) Find the total number of moles of hydrogen ions in the lake. (c) Find the mass of calcium carbonate needed (kg), taking its molar mass as 100 g/mol. (d) Explain why a large excess of lime must be avoided.	(a) pH 3 means [H+] = 10^-3 mol/L. (b) 1.0 x 10^6 L x 10^-3 mol/L = 1000 mol. (c) CaCO3 reacts with H+ in the ratio 1:2, so 500 mol is needed; mass = 500 x 100 = 50 000 g = 50 kg. (d) Excess lime makes the lake alkaline, harming aquatic life, and raises the calcium ion concentration too high.
11	An electric kettle of power 2000 W heats 1.5 kg of water from 25 C to 100 C. The specific heat capacity of water is 4200 J/(kg C). (a) Find the energy absorbed by the water (J). (b) Find the time needed if no heat is lost (s). (c) The actual heating takes 280 s; find the efficiency of the kettle. (d) Find the electrical energy used (kWh) and the cost at 1.2 dollars per kWh.	(a) E = mc(dT) = 1.5 x 4200 x 75 = 472 500 J. (b) t = 472 500 / 2000 = 236 s. (c) Efficiency = 472 500 / (2000 x 280) x 100% = 84.4%. (d) Energy = 2000 x 280 = 560 000 J = 0.156 kWh; cost = 0.156 x 1.2 = 0.19 dollars.
12	A car travels at 20 m/s. The driver takes 0.8 s to react, then brakes with a uniform deceleration of 5 m/s2 until stopping. (a) Find the reaction distance. (b) Find the braking distance and the total stopping distance. (c) On a wet road the deceleration is only 2.5 m/s2; find the new braking distance and total stopping distance. (d) Explain why wet roads are dangerous and suggest two measures to reduce rear-end collisions.	(a) Reaction distance = 20 x 0.8 = 16 m. (b) Braking distance = v^2 / (2a) = 400 / 10 = 40 m; total = 56 m. (c) Wet braking distance = 400 / 5 = 80 m; total = 96 m. (d) Wet roads give less friction and slower deceleration, so the stopping distance rises sharply; keep a longer following distance, slow down and brake earlier.

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