GCSE Combined Science Higher Paper 3
Covers Cell Biology, Organisation, Infection and Response and 9 more.
Questions
Question 1 [4 marks]
Atomic Structure and the Periodic Table
An ion of oxygen, O2-, has a mass number of 16 and an atomic number of 8.
State the number of protons, neutrons and electrons in this ion of oxygen.
Question 2 [4 marks]
Bonding, Structure and the Properties of Matter
A polymer chain is made from 250 repeating ethene monomer units.
Each monomer unit has a length of 0.25 nm when incorporated into the polymer chain.
Calculate the total length of the polymer chain, in nm.
Question 3 [4 marks]
Cell Biology
A student views a cheek cell using a light microscope. The actual width of the cell is 0.05 mm. The width of the cell in the micrograph image is 20 mm.
Calculate the magnification of the image. Use magnification = image size / actual size.
Question 4 [5 marks]
Particle Model of Matter
A rectangular block of wood measures 20 cm by 10 cm by 5 cm and has a mass of 750 g.
Calculate the density of the wood, in kg/m^3.
Question 5 [5 marks]
Organisation
A potometer is used to measure the rate of water uptake by a leafy shoot.
In still air, the air bubble in the capillary tube moves 6.0 cm in 5 minutes.
In front of a fan, the air bubble moves 15.0 cm in 5 minutes.
Calculate the percentage increase in the rate of water uptake caused by the fan.
Question 6 [5 marks]
Bioenergetics
A plant respires continuously, releasing 15 cm^3 of carbon dioxide per hour in the dark.
In the light, the plant's leaves take up a net 40 cm^3 of carbon dioxide per hour from the surrounding air, because photosynthesis and respiration are both taking place at the same time.
Calculate the actual (gross) rate of photosynthesis, in cm^3 of carbon dioxide used per hour, assuming respiration continues at the same rate in the light.
Question 7 [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 8 [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 9 [5 marks]
Infection and Response
Vaccination can protect an individual against future infection by a specific pathogen.
Explain how vaccination can protect an individual against future infection by a specific pathogen.
Question 10 [6 marks]
Energy
A washing machine motor is supplied with 1500 J of energy each cycle.
Of this, 1050 J is usefully transferred to kinetic energy of the drum, and the rest is wasted as heat and sound.
Calculate the efficiency of the motor as a percentage, and calculate the energy wasted as heat and sound each cycle.
Suggest one way the amount of wasted energy could be reduced.
Question 11 [6 marks]
Quantitative Chemistry
A sample of an oxide of iron contains 4.48 g of iron and 1.92 g of oxygen only.
Calculate the empirical formula of this compound. Relative atomic masses: Fe = 56, O = 16.
Question 12 [6 marks]
Electricity
A 3.0 kW immersion heater is used for 45 minutes.
A 100 W lamp is left on for 8 hours on the same day.
Electricity costs 30p per kWh.
Calculate the total cost of using both appliances for these times on this day.
Model solutions
| Question 1[4 marks] | |
|---|---|
| Answer or working | Marks |
| protons = atomic number = 8 | M1 |
| neutrons = mass number - atomic number, 16 - 8 | M1 |
| 8 neutrons | A1 |
| electrons = 10, since a 2- ion has 2 more electrons than protons | B1 |
| Final answer: 8 protons, 8 neutrons, 10 electrons | |
| Question 2[4 marks] | |
|---|---|
| Answer or working | Marks |
| using total length = number of units x length per unit | M1 |
| substituting 250 x 0.25 | M1 |
| 62.5 | A1 |
| the unit nm | B1 |
| Final answer: 62.5 nm | |
| Question 3[4 marks] | |
|---|---|
| Answer or working | Marks |
| using magnification = image size / actual size | M1 |
| substituting 20 / 0.05 | M1 |
| magnification = 400 | A1 |
| stating magnification has no unit, written as x400 | B1 |
| Final answer: x400 magnification | |
| Question 4[5 marks] | |
|---|---|
| Answer or working | Marks |
| volume = 20 x 10 x 5 | M1 |
| 1000 cm^3 | A1 |
| converting mass to kg (0.75 kg) and volume to m^3 (0.001 m^3) | M1 |
| density = mass / volume, 0.75 / 0.001 | M1 |
| 750 kg/m^3 | A1 |
| Question 5[5 marks] | |
|---|---|
| Answer or working | Marks |
| finding the rate in still air, 6.0 / 5 = 1.2 cm per minute | M1 |
| finding the rate in front of the fan, 15.0 / 5 = 3.0 cm per minute | M1 |
| finding the increase in rate, 3.0 - 1.2 = 1.8 cm per minute | M1 |
| (1.8 / 1.2) x 100 | M1 |
| 150% | A1 |
| Final answer: 150% increase in the rate of water uptake | |
| Question 6[5 marks] | |
|---|---|
| Answer or working | Marks |
| recognising the measured net rate = gross rate of photosynthesis - rate of respiration | M1 |
| rearranging to gross rate of photosynthesis = net rate + rate of respiration | M1 |
| substituting 40 + 15 | M1 |
| 55 cm^3 of carbon dioxide per hour | A1 |
| explaining that some of the carbon dioxide produced by respiration is used directly in photosynthesis, so it is not detected in the net exchange with the air | B1 |
| Final answer: 55 cm^3 of carbon dioxide per hour; some CO2 from respiration is used directly in photosynthesis, so it does not appear in the net exchange with the air | |
| Question 7[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 8[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 9[5 marks] | |
|---|---|
| Answer or working | Marks |
| the vaccine containing a dead or inactive/weakened form of the pathogen (or its antigens) | B1 |
| this stimulating the immune system to produce antibodies specific to the antigen | B1 |
| the person not suffering symptoms of the disease | B1 |
| memory (lymphocyte) cells remaining in the blood | B1 |
| a faster and greater secondary immune response if infected again, destroying the pathogen before symptoms occur | B1 |
| Final answer: Vaccination stimulates antibody and memory cell production for a faster secondary response | |
| Question 10[6 marks] | |
|---|---|
| Answer or working | Marks |
| using efficiency = useful output energy / total input energy | M1 |
| substituting 1050 / 1500 | M1 |
| 0.7 (70%) | A1 |
| wasted energy = total input - useful output | M1 |
| 450 J | A1 |
| a valid way to reduce wasted energy, e.g. lubricating moving parts to reduce friction | B1 |
| Final answer: 70% efficient; 450 J wasted; e.g. lubrication reduces friction and wasted energy | |
| Question 11[6 marks] | |
|---|---|
| Answer or working | Marks |
| moles of iron = 4.48 / 56 | M1 |
| 0.08 mol | A1 |
| moles of oxygen = 1.92 / 16 | M1 |
| 0.12 mol | A1 |
| simplifying the mole ratio 0.08 : 0.12 to the whole-number ratio 2 : 3 | M1 |
| the empirical formula Fe2O3 | A1 |
| Final answer: Fe2O3 | |
| Question 12[6 marks] | |
|---|---|
| Answer or working | Marks |
| converting 45 minutes to hours, 45 / 60 = 0.75 hours | M1 |
| energy used by the heater, 3.0 x 0.75 = 2.25 kWh | M1 |
| energy used by the lamp, 0.100 x 8 = 0.8 kWh | M1 |
| total energy, 2.25 + 0.8 = 3.05 kWh | M1 |
| total cost = 3.05 x 30 | M1 |
| 91.5p (0.915 pounds) | A1 |
| Final answer: 91.5p (0.915 pounds) total | |