GCSE Combined Science Higher Short Paper B
Covers Quantitative Chemistry, Chemical Changes, Energy and 3 more.
Questions
Question 1 [4 marks]
Quantitative Chemistry
A solution contains 0.05 moles of sodium hydroxide dissolved in 500 cm^3 of solution.
Calculate the concentration of the sodium hydroxide solution, in mol/dm^3.
Question 2 [4 marks]
Electricity
A current of 3.0 A flows through a component for 2 minutes.
Calculate the charge that flows through the component in this time. Use Q = I t.
Question 3 [5 marks]
Energy
A weightlifter raises a 120 kg barbell through a height of 2.0 m in 1.5 s.
Calculate the work done raising the barbell, and calculate the power developed by the weightlifter.
Question 4 [5 marks]
Particle Model of Matter
A bicycle tyre has a fixed volume.
More air is pumped into the tyre, increasing the number of gas particles inside it, while the temperature stays constant.
Explain, in terms of the particle model, why pumping more air into the tyre increases the pressure inside it.
Question 5 [5 marks]
Forces
A ball of mass 0.40 kg travelling at 8.0 m/s hits a wall and rebounds at 6.0 m/s in the opposite direction.
The collision with the wall lasts for 0.02 s.
Calculate the force exerted by the wall on the ball. Use force = change in momentum / time.
Question 6 [5 marks]
Particle Model of Matter
A fixed volume of gas is heated in a sealed container.
Explain, in terms of the particle model, why increasing the temperature of a gas in a sealed, fixed-volume container increases the pressure of the gas.
Question 7 [6 marks]
Chemical Changes
Concentrated sodium chloride solution (brine) is electrolysed using inert electrodes, producing hydrogen gas at the cathode, chlorine gas at the anode, and sodium hydroxide solution.
Explain, in terms of the ions present and their discharge at each electrode, why hydrogen gas (not sodium) forms at the cathode, and why chlorine gas forms at the anode.
Question 8 [6 marks]
Energy
An electric motor is used to lift a load of mass 40 kg through a height of 6.0 m. The motor is 60% efficient. Use g = 9.8 N/kg.
Calculate the useful energy transferred to the gravitational potential energy store of the load, and calculate the total input energy supplied to the motor to achieve this lift.
Model solutions
| Question 1[4 marks] | |
|---|---|
| Answer or working | Marks |
| converting 500 cm^3 to dm^3, 500 / 1000 | M1 |
| using concentration = moles / volume | M1 |
| substituting 0.05 / 0.5 | M1 |
| 0.1 mol/dm^3 | A1 |
| Question 2[4 marks] | |
|---|---|
| Answer or working | Marks |
| converting 2 minutes to seconds, 2 x 60 | M1 |
| substituting into Q = I t, 3.0 x 120 | M1 |
| 360 | A1 |
| the unit coulombs (C) | B1 |
| Final answer: 360 C | |
| Question 3[5 marks] | |
|---|---|
| Answer or working | Marks |
| using work done = GPE = m g h | M1 |
| substituting 120 x 9.8 x 2.0 | M1 |
| 2352 J | A1 |
| power = work done / time, 2352 / 1.5 | M1 |
| 1568 W | A1 |
| Final answer: 2352 J of work done; 1568 W of power | |
| Question 4[5 marks] | |
|---|---|
| Answer or working | Marks |
| there being more gas particles inside the same fixed volume | B1 |
| a higher density of particles inside the tyre | B1 |
| particles colliding with the walls of the tyre more frequently | B1 |
| the total force exerted on the inside surface of the tyre increasing as a result | B1 |
| pressure = force / area, so a greater force on the same area gives a greater pressure | B1 |
| Final answer: More particles in the same volume collide with the walls more often, increasing the force and therefore the pressure | |
| Question 5[5 marks] | |
|---|---|
| Answer or working | Marks |
| momentum before the collision, 0.40 x 8.0 | M1 |
| 3.2 kg m/s | A1 |
| momentum after the collision, in the opposite direction, 0.40 x 6.0 = 2.4 kg m/s | M1 |
| change in momentum = 3.2 + 2.4 | M1 |
| a force of 5.6 / 0.02 = 280 N | A1 |
| Final answer: 280 N | |
| Question 6[5 marks] | |
|---|---|
| Answer or working | Marks |
| increasing temperature increasing the average kinetic energy of the gas particles | B1 |
| particles moving faster on average | B1 |
| particles colliding with the walls of the container more frequently | B1 |
| particles colliding with the walls with greater force | B1 |
| this increasing the pressure exerted on the walls, as the volume is fixed | B1 |
| Final answer: Faster particles collide with the walls more often and with greater force, increasing pressure | |
| Question 7[6 marks] | |
|---|---|
| Answer or working | Marks |
| the solution containing Na+, H+ (from water), Cl- and OH- (from water) ions | B1 |
| stating positive ions move to the cathode and gain electrons (reduction) | B1 |
| sodium being more reactive than hydrogen, so H+ ions are discharged in preference to Na+ ions, producing hydrogen gas | B1 |
| stating negative ions move to the anode and lose electrons (oxidation) | B1 |
| stating that in a concentrated solution, chloride ions are discharged in preference to hydroxide ions, producing chlorine gas | B1 |
| stating sodium ions and hydroxide ions remain in solution, forming sodium hydroxide solution | B1 |
| Final answer: Hydrogen forms at the cathode because sodium is more reactive than hydrogen; chlorine forms at the anode because the solution is concentrated in chloride ions; sodium and hydroxide ions remain, forming sodium hydroxide solution | |
| Question 8[6 marks] | |
|---|---|
| Answer or working | Marks |
| using GPE = m g h | M1 |
| substituting 40 x 9.8 x 6.0 | M1 |
| 2352 J | A1 |
| rearranging efficiency = useful output energy / total input energy to input energy = useful output energy / efficiency | M1 |
| substituting 2352 / 0.60 | M1 |
| 3920 J | A1 |
| Final answer: 2352 J of useful GPE gained; 3920 J of total input energy needed | |