GCSE Combined Science Higher Paper 2
Covers Cell Biology, Organisation, Infection and Response and 9 more.
Questions
Question 1 [4 marks]
Bonding, Structure and the Properties of Matter
Diamond is a form of carbon with a giant covalent structure.
Explain why diamond has a very high melting point.
Question 2 [4 marks]
Particle Model of Matter
A fixed mass of gas has a volume of 0.80 m^3 at a pressure of 150 kPa.
The gas is compressed at constant temperature until its pressure increases to 400 kPa.
Calculate the new volume of the gas. Use p1 V1 = p2 V2.
Question 3 [4 marks]
Infection and Response
A single bacterium divides by binary fission every 20 minutes on a nutrient agar plate.
Starting with 1 bacterium, calculate the number of bacteria present after 3 hours, assuming unlimited nutrients.
Question 4 [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 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 [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 7 [5 marks]
Atomic Structure and the Periodic Table
Mendeleev produced an early periodic table by arranging the known elements mainly in order of atomic (relative atomic) mass.
Explain how Mendeleev's approach differed from a simple listing of elements by atomic mass, and explain why his periodic table was later accepted as a useful method of classifying elements.
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 [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 10 [6 marks]
Cell Biology
A potato chip is placed in distilled water.
Water moves into the chip by osmosis, increasing its mass by 8% over 24 hours.
The mass of the chip after 24 hours is 5.94 g.
Calculate the original mass of the potato chip before it was placed in the water, and explain why the chip gained mass.
Question 11 [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 12 [6 marks]
Bioenergetics
A grower wants to increase the rate of photosynthesis in a greenhouse of tomato plants.
Explain how each of light intensity, carbon dioxide concentration and temperature can limit the rate of photosynthesis, and explain why increasing all three at once is not usually a cost-effective strategy for a grower.
Model solutions
| Question 1[4 marks] | |
|---|---|
| Answer or working | Marks |
| diamond having a giant covalent (macromolecular) structure | B1 |
| each carbon atom being bonded to four other carbon atoms | B1 |
| there being many strong covalent bonds throughout the structure | B1 |
| a large amount of energy being needed to break these bonds, giving a very high melting point | B1 |
| Final answer: Many strong covalent bonds must be broken, needing a large amount of energy | |
| Question 2[4 marks] | |
|---|---|
| Answer or working | Marks |
| using p1 V1 = p2 V2 | M1 |
| substituting 150 x 0.80 = 400 x V2 | M1 |
| rearranging to V2 = (150 x 0.80) / 400 | M1 |
| 0.30 m^3 | A1 |
| Question 3[4 marks] | |
|---|---|
| Answer or working | Marks |
| finding the number of 20-minute divisions in 3 hours, 180 / 20 = 9 | M1 |
| using 2 raised to the power of the number of divisions | M1 |
| 512 | A1 |
| stating the unit as bacteria (bacterial cells) | B1 |
| Final answer: 512 bacteria | |
| Question 4[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 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[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 7[5 marks] | |
|---|---|
| Answer or working | Marks |
| Mendeleev leaving gaps in his table for elements that had not yet been discovered | B1 |
| Mendeleev sometimes switching the order of two elements from strict atomic mass order, to keep elements with similar properties in the same group | B1 |
| Mendeleev using the gaps to predict the properties of undiscovered elements | B1 |
| stating that elements were later discovered with properties matching these predictions | B1 |
| stating this gave scientists confidence that Mendeleev's table was a valid and useful method of classifying elements | B1 |
| Final answer: Mendeleev left gaps and reordered a few elements to keep similar elements together, then used the gaps to predict undiscovered elements; when these were found with matching properties, his table was accepted | |
| 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[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 10[6 marks] | |
|---|---|
| Answer or working | Marks |
| recognising the final mass is 108% of the original mass | M1 |
| original mass = final mass / 1.08 | M1 |
| substituting 5.94 / 1.08 | M1 |
| 5.5 g | A1 |
| stating distilled water has a higher water potential than the contents of the potato cells | B1 |
| stating water moved into the cells by osmosis, down the water potential gradient, increasing the mass | B1 |
| Final answer: Original mass 5.5 g; water moved in by osmosis because distilled water has a higher water potential than the cell contents | |
| Question 11[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 12[6 marks] | |
|---|---|
| Answer or working | Marks |
| low light intensity limiting the rate at which light energy can be absorbed for photosynthesis | B1 |
| low carbon dioxide concentration limiting the amount of a raw material available for photosynthesis | B1 |
| low temperature reducing the rate of the enzyme-controlled reactions of photosynthesis, and too high a temperature denaturing the enzymes involved | B1 |
| stating whichever factor is in shortest supply relative to the others (the limiting factor) restricts the overall rate | B1 |
| stating increasing a factor that is not currently limiting will not increase the rate of photosynthesis | B1 |
| stating heating, lighting and enriching a greenhouse with carbon dioxide all cost money, so a grower must balance this cost against the increase in yield gained | B1 |
| Final answer: Whichever of light, CO2 or temperature is in shortest supply limits the rate; increasing a factor that is not limiting wastes money without raising the rate | |