Year 11 Paper 6: Extended Practice
Covers organisation, quantitative chemistry, chemical changes, energy, electricity, particle model and forces.
Year 11 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 11, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.
Questions
Question 1 [4 marks]
Quantitative Chemistry
Ammonium nitrate, NH4NO3, is used as a fertiliser.
Calculate the percentage by mass of nitrogen in ammonium nitrate. Relative atomic masses: H = 1, N = 14, O = 16.
Question 2 [4 marks]
Particle Model of Matter
A metal cube has sides of length 4 cm and a mass of 288 g.
Calculate the density of the metal, in g/cm^3.
Question 3 [5 marks]
Energy
A box is pushed 8.0 m across a floor by a constant horizontal force of 45 N.
Calculate the work done by the force pushing the box. Use work done = force x distance.
Given that 250 J of this work is usefully transferred to the kinetic energy store of the box, calculate the energy wasted to the thermal energy store due to friction.
Question 4 [5 marks]
Chemical Changes
50 cm^3 of copper sulfate solution, concentration 0.20 mol/dm^3, is reacted with excess zinc powder: Zn + CuSO4 -> ZnSO4 + Cu.
Calculate the maximum mass of copper that could be displaced. Relative atomic mass of Cu = 64.
Question 5 [6 marks]
Forces
A skydiver of mass 80 kg is falling and experiences air resistance of 600 N acting upwards.
The gravitational field strength is 9.8 N/kg.
Calculate the resultant force on the skydiver, and calculate their acceleration at this moment. Use weight = mass x gravitational field strength and F = m a.
Question 6 [6 marks]
Energy
A homeowner wants to reduce the rate of unwanted energy transfer by heating from their house in winter.
Explain how cavity wall insulation reduces the rate of energy transfer through the walls of a house, and explain why increasing the thickness of insulation gives a smaller and smaller additional benefit as more is added.
Question 7 [6 marks]
Electricity
In a circuit, a 4 ohm resistor is connected in series with a parallel combination of a 6 ohm resistor and a 3 ohm resistor. The circuit is connected to a 9 V supply.
Calculate the combined resistance of the parallel section, calculate the total resistance of the circuit, and calculate the total current supplied by the battery.
Question 8 [6 marks]
Energy
An electric heater with a power of 1.5 kW is used to heat 2.0 kg of water. The specific heat capacity of water is 4200 J/kg degrees C.
Calculate the energy transferred by the heater in 4 minutes, and calculate the resulting temperature rise of the water, assuming all the energy is usefully transferred to the water. Use energy = power x time and change in thermal energy = m c (change in temperature).
Question 9 [6 marks]
Organisation
The heart has four chambers: the left atrium, right atrium, left ventricle and right ventricle.
Explain why the wall of the left ventricle is thicker and more muscular than the wall of the right ventricle.
Question 10 [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.
Question 11 [6 marks]
Organisation
Cardiac output is the volume of blood pumped by the heart in one minute.
A patient has a cardiac output of 4.8 dm^3 per minute and a heart rate of 80 beats per minute.
Calculate the patient's stroke volume, in cm^3. Use cardiac output = heart rate x stroke volume.
Model solutions
| Question 1[4 marks] | |
|---|---|
| Answer or working | Marks |
| finding the relative formula mass of NH4NO3 = 80 | M1 |
| finding the total mass of nitrogen in the formula, 28 | M1 |
| (28 / 80) x 100 | M1 |
| 35% | A1 |
| Question 2[4 marks] | |
|---|---|
| Answer or working | Marks |
| volume = 4^3 | M1 |
| 64 cm^3 | A1 |
| density = mass / volume, 288 / 64 | M1 |
| 4.5 g/cm^3 | A1 |
| Question 3[5 marks] | |
|---|---|
| Answer or working | Marks |
| using work done = force x distance | M1 |
| substituting 45 x 8 | M1 |
| 360 J | A1 |
| wasted energy = total work done - useful energy, 360 - 250 | M1 |
| 110 J | A1 |
| Final answer: 360 J of work done; 110 J wasted to the thermal energy store due to friction | |
| Question 4[5 marks] | |
|---|---|
| Answer or working | Marks |
| moles of CuSO4 = 0.20 x (50 / 1000) | M1 |
| 0.01 mol | A1 |
| using the 1:1 mole ratio between CuSO4 and Cu from the equation | M1 |
| mass = moles x relative atomic mass, 0.01 x 64 | M1 |
| 0.64 g | A1 |
| Question 5[6 marks] | |
|---|---|
| Answer or working | Marks |
| weight = mass x gravitational field strength, substituting 80 x 9.8 | M1 |
| a weight of 784 N | A1 |
| resultant force = weight - air resistance, substituting 784 - 600 | M1 |
| a resultant force of 184 N, acting downwards | A1 |
| rearranging F = m a to a = F / m, substituting 184 / 80 | M1 |
| an acceleration of 2.3 m/s^2 (2 sf) | A1 |
| Final answer: Resultant force 184 N downwards; acceleration 2.3 m/s^2 | |
| Question 6[6 marks] | |
|---|---|
| Answer or working | Marks |
| cavity wall insulation containing trapped pockets of air (or foam) within the material | B1 |
| trapped air being a poor conductor of heat, reducing energy transfer by conduction through the wall | B1 |
| trapped pockets of air also reducing energy transfer by convection, by preventing convection currents forming within the cavity | B1 |
| stating the rate of energy transfer through a wall depends on the thickness of the insulating layer | B1 |
| stating doubling the thickness of insulation approximately halves the rate of energy transfer (up to a point) | B1 |
| stating each additional layer of insulation therefore saves progressively less energy than the layer before it, giving diminishing returns | B1 |
| Final answer: Cavity wall insulation traps air, which is a poor conductor and prevents convection, reducing heat loss; but each extra layer reduces the rate of loss by a smaller amount than the last, giving diminishing returns | |
| Question 7[6 marks] | |
|---|---|
| Answer or working | Marks |
| using 1 / Rparallel = 1/6 + 1/3 | M1 |
| a parallel resistance of 2 ohms | A1 |
| total resistance = 4 + 2 | M1 |
| 6 ohms | A1 |
| using current = V / R | M1 |
| substituting 9 / 6 to give 1.5 A | A1 |
| Final answer: Parallel resistance 2 ohms; total resistance 6 ohms; current 1.5 A | |
| Question 8[6 marks] | |
|---|---|
| Answer or working | Marks |
| converting 4 minutes to seconds, 4 x 60 | M1 |
| energy = power x time, 1500 x 240 | M1 |
| 360000 J | A1 |
| rearranging E = m c (change in temperature) to change in temperature = E / (m c) | M1 |
| substituting 360000 / (2.0 x 4200) | M1 |
| a temperature rise of 42.9 degrees C (3 sf) | A1 |
| Final answer: 360000 J transferred; a temperature rise of about 42.9 degrees C | |
| Question 9[6 marks] | |
|---|---|
| Answer or working | Marks |
| the left ventricle pumping blood around the whole body (the systemic circulation) | B1 |
| the right ventricle only pumping blood to the lungs (the pulmonary circulation), a much shorter distance | B1 |
| blood needing to be pumped at a higher pressure to reach the whole body than to reach only the lungs | B1 |
| the left ventricle therefore needing to contract with more force than the right ventricle | B1 |
| a thicker, more muscular wall allowing the left ventricle to contract with greater force | B1 |
| this generating the higher pressure needed to pump blood all the way around the body | B1 |
| Final answer: The left ventricle pumps blood around the whole body, much further than the right ventricle pumps to the lungs, so it needs a thicker wall to generate higher pressure | |
| Question 10[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 | |
| Question 11[6 marks] | |
|---|---|
| Answer or working | Marks |
| converting cardiac output to cm^3, 4.8 x 1000 | M1 |
| 4800 cm^3 per minute | A1 |
| rearranging to stroke volume = cardiac output / heart rate | M1 |
| substituting 4800 / 80 | M1 |
| 60 cm^3 | A1 |
| stating the unit as cm^3 (per beat) | B1 |
| Final answer: 60 cm^3 stroke volume | |