GCSE Combined Science Higher Paper 4
Covers Cell Biology, Organisation, Infection and Response and 9 more.
Questions
Question 1 [4 marks]
Chemical Changes
Sulfuric acid reacts with sodium hydroxide according to the equation H2SO4 + 2NaOH -> Na2SO4 + 2H2O.
25.0 cm^3 of sodium hydroxide solution of concentration 0.40 mol/dm^3 is exactly neutralised by 20.0 cm^3 of sulfuric acid.
Calculate the concentration, in mol/dm^3, of the sulfuric acid.
Question 2 [4 marks]
Infection and Response
In a school of 800 students, 6% show symptoms of a viral infection during an outbreak.
Calculate the number of students affected, and state one reason the actual number of infected students could be higher than this figure.
Question 3 [4 marks]
Energy
A crane lifts a load of mass 250 kg through a height of 12 m.
Calculate the gravitational potential energy gained by the load. Use GPE = m g h, with g = 9.8 N/kg.
Question 4 [5 marks]
Electricity
Two resistors, of resistance 15 ohms and 25 ohms, are connected in series with a 20 V supply.
Calculate the current flowing in the circuit, and calculate the potential difference across the 15 ohm resistor.
Question 5 [5 marks]
Organisation
A sample of blood has a total volume of 5.0 cm^3.
The sample contains 2.2 cm^3 of red blood cells, white blood cells and platelets, and the rest of the sample is plasma.
Calculate the percentage of the blood sample's volume that is plasma.
Question 6 [4 marks]
Bonding, Structure and the Properties of Matter
A cube-shaped nanoparticle has sides of length 20 nm.
Calculate the surface area to volume ratio of the nanoparticle.
Question 7 [5 marks]
Particle Model of Matter
Ice at 0 degrees C can be melted by supplying energy to it.
The specific latent heat of fusion of ice is 334000 J/kg.
Calculate the energy needed to melt 0.40 kg of ice at 0 degrees C. Use energy = mass x specific latent heat.
Question 8 [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 9 [6 marks]
Forces
A stationary object of total mass 5.0 kg explodes into two fragments.
One fragment, of mass 2.0 kg, moves off at a velocity of 9.0 m/s.
Calculate the velocity of the second fragment (of mass 3.0 kg) immediately after the explosion. Use conservation of momentum: total momentum before = total momentum after.
Question 10 [6 marks]
Cell Biology
A single cell divides by mitosis, producing two daughter cells at the end of every cell cycle. Each cell cycle takes 18 hours to complete.
Starting with a single cell, calculate the number of cells present after 5 complete cell cycles, and calculate the total time this takes, giving your answer in days and hours.
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]
Atomic Structure and the Periodic Table
The noble gases in Group 0 exist as single, monatomic atoms, and their boiling points increase going down the group.
Explain, in terms of intermolecular forces, why the boiling point of the noble gases increases going down Group 0.
Model solutions
| Question 1[4 marks] | |
|---|---|
| Answer or working | Marks |
| moles of NaOH = 0.40 x (25 / 1000) | M1 |
| using the 2:1 mole ratio between NaOH and H2SO4 from the equation | M1 |
| concentration = moles / volume, substituting 0.005 / (20 / 1000) | M1 |
| 0.25 mol/dm^3 | A1 |
| Question 2[4 marks] | |
|---|---|
| Answer or working | Marks |
| using number affected = percentage x total number of students | M1 |
| substituting 0.06 x 800 | M1 |
| 48 | A1 |
| a valid reason, e.g. some infected students may not yet be showing symptoms (still in the incubation period) or may be asymptomatic | B1 |
| Final answer: 48 students; some infected students may not yet show symptoms (incubation period) or may be asymptomatic | |
| Question 3[4 marks] | |
|---|---|
| Answer or working | Marks |
| using GPE = m g h | M1 |
| substituting 250 x 9.8 x 12 | M1 |
| correct evaluation | M1 |
| 29400 J (29.4 kJ) | A1 |
| Question 4[5 marks] | |
|---|---|
| Answer or working | Marks |
| total resistance = 15 + 25 = 40 ohms | M1 |
| using I = V / R, substituting 20 / 40 | M1 |
| 0.5 A | A1 |
| using V = I R for the 15 ohm resistor, substituting 0.5 x 15 | M1 |
| 7.5 V | A1 |
| Final answer: 0.5 A current; 7.5 V across the 15 ohm resistor | |
| Question 5[5 marks] | |
|---|---|
| Answer or working | Marks |
| plasma volume = 5.0 - 2.2 | M1 |
| 2.8 cm^3 | A1 |
| using percentage = (plasma volume / total volume) | M1 |
| substituting (2.8 / 5.0) x 100 | M1 |
| 56% | A1 |
| Final answer: 56% of the blood sample's volume is plasma | |
| Question 6[4 marks] | |
|---|---|
| Answer or working | Marks |
| surface area = 6 x 20^2 | M1 |
| volume = 20^3 | M1 |
| dividing surface area by volume | M1 |
| a surface area to volume ratio of 0.3 (or 3:10) | A1 |
| Final answer: 0.3 (surface area to volume ratio) | |
| Question 7[5 marks] | |
|---|---|
| Answer or working | Marks |
| using energy = mass x specific latent heat | M1 |
| substituting 0.40 x 334000 | M1 |
| correct evaluation | M1 |
| 133600 J | A1 |
| expressing the answer as 134 kJ (3 sf) | B1 |
| Final answer: 133600 J (134 kJ) | |
| Question 8[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 9[6 marks] | |
|---|---|
| Answer or working | Marks |
| total momentum before the explosion = 0 (the object is stationary) | M1 |
| momentum of the first fragment after the explosion, 2.0 x 9.0 | M1 |
| 18 kg m/s | A1 |
| using conservation of momentum, 0 = 18 + (3.0 x v2) | M1 |
| rearranging to v2 = -18 / 3.0 | M1 |
| -6.0 m/s (i.e. 6.0 m/s in the opposite direction to the first fragment) | A1 |
| Final answer: 6.0 m/s, in the opposite direction to the 2.0 kg fragment | |
| Question 10[6 marks] | |
|---|---|
| Answer or working | Marks |
| using 2 raised to the power of the number of cell cycles | M1 |
| evaluating 2^5 | M1 |
| 32 cells | A1 |
| total time = 5 x 18 | M1 |
| 90 hours | A1 |
| converting 90 hours to 3 days and 18 hours | B1 |
| Final answer: 32 cells; a total time of 90 hours (3 days and 18 hours) | |
| 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 |
| the noble gases existing as simple, monatomic particles held together by weak intermolecular forces | B1 |
| stating these intermolecular forces are the only forces overcome when a noble gas boils, not any covalent bonds within an atom | B1 |
| the number of electrons in an atom increasing going down the group | B1 |
| this increasing the strength of the intermolecular forces between atoms | B1 |
| stating more energy is therefore needed to overcome these stronger forces | B1 |
| stating this means the boiling point increases going down the group | B1 |
| Final answer: Boiling point increases down Group 0 because larger atoms have more electrons, giving stronger intermolecular forces that need more energy to overcome | |