GCSE Combined Science Higher Paper 1
Covers Cell Biology, Organisation, Infection and Response and 9 more.
Questions
Question 1 [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 2 [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 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 [5 marks]
Cell Biology
A student investigates the effect of hydrogen peroxide concentration on the rate of the reaction catalysed by catalase in celery extract, measuring the volume of oxygen gas produced in 60 seconds.
Using 1% hydrogen peroxide, 18 cm^3 of oxygen gas is collected in 60 seconds. Using 3% hydrogen peroxide, 54 cm^3 of oxygen gas is collected in 60 seconds.
Calculate the percentage increase in the rate of reaction caused by using the higher concentration of hydrogen peroxide.
Question 5 [5 marks]
Bioenergetics
A culture of yeast produces 720 cm^3 of carbon dioxide gas during 2 hours of anaerobic respiration.
Calculate the mean rate of carbon dioxide production, in dm^3 per hour.
Question 6 [5 marks]
Organisation
A person's heart rate is 60 beats per minute, so each cardiac cycle lasts 1 second.
During each cardiac cycle, the ventricles contract (systole) for 0.3 seconds, and are relaxed (diastole) for the rest of the cycle.
Calculate the time spent in diastole during each cardiac cycle, and calculate this as a percentage of the total cardiac cycle.
Question 7 [5 marks]
Quantitative Chemistry
In an experiment, the theoretical maximum mass of ammonia, NH3, that could be produced from a reaction of nitrogen and hydrogen is 3.4 g.
The actual mass of ammonia produced is 2.55 g.
Calculate the percentage yield of ammonia, and suggest one reason why the actual yield is less than the theoretical yield.
Question 8 [5 marks]
Forces
A car of mass 900 kg travelling at 20 m/s brakes and comes to a complete stop over a distance of 45 m.
Calculate the average braking force needed to stop the car. Use work done by the brakes = kinetic energy lost, and work done = force x distance.
Question 9 [5 marks]
Atomic Structure and the Periodic Table
Group 1 of the periodic table contains the alkali metals.
Explain, in terms of electronic structure, why reactivity increases going down Group 1 of the periodic table.
Question 10 [6 marks]
Bonding, Structure and the Properties of Matter
Graphite is a form of carbon with a giant covalent structure, different from diamond.
Explain, in terms of structure and bonding, why graphite can conduct electricity but diamond cannot, and why graphite is soft and slippery.
Question 11 [6 marks]
Chemical Changes
Copper sulfate solution is electrolysed using inert (graphite) electrodes.
Explain, in terms of the ions present and their discharge at each electrode, the products formed at the cathode and the anode.
Question 12 [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.
Model solutions
| Question 1[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 2[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 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[5 marks] | |
|---|---|
| Answer or working | Marks |
| finding the rate with 1% hydrogen peroxide, 18 / 60 = 0.3 cm^3 per second | M1 |
| finding the rate with 3% hydrogen peroxide, 54 / 60 = 0.9 cm^3 per second | M1 |
| finding the increase in rate, 0.9 - 0.3 = 0.6 cm^3 per second | M1 |
| (0.6 / 0.3) x 100 | M1 |
| 200% | A1 |
| Final answer: 200% increase in the rate of reaction | |
| Question 5[5 marks] | |
|---|---|
| Answer or working | Marks |
| converting 720 cm^3 to dm^3 by dividing by 1000 | M1 |
| 0.72 dm^3 | A1 |
| using rate = volume / time | M1 |
| substituting 0.72 / 2 | M1 |
| 0.36 dm^3 per hour | A1 |
| Question 6[5 marks] | |
|---|---|
| Answer or working | Marks |
| finding the length of the cardiac cycle, 60 / 60 | M1 |
| 1 second | A1 |
| diastole time = 1 - 0.3 | M1 |
| (0.7 / 1) x 100 | M1 |
| 70% | A1 |
| Final answer: 0.7 seconds in diastole, which is 70% of the cardiac cycle | |
| Question 7[5 marks] | |
|---|---|
| Answer or working | Marks |
| using percentage yield = (actual yield / theoretical yield) x 100 | M1 |
| substituting 2.55 / 3.4 | M1 |
| correct evaluation of 2.55 / 3.4 | M1 |
| 75% | A1 |
| a valid reason the actual yield is less than the theoretical yield, e.g. the reaction did not go to completion or product was lost during the experiment | B1 |
| Final answer: 75%; e.g. the reaction did not go to completion or product was lost during the experiment | |
| Question 8[5 marks] | |
|---|---|
| Answer or working | Marks |
| using KE = 1/2 m v^2 | M1 |
| substituting 0.5 x 900 x 20^2 | M1 |
| 180000 J | A1 |
| force = work done / distance, 180000 / 45 | M1 |
| 4000 N | A1 |
| Question 9[5 marks] | |
|---|---|
| Answer or working | Marks |
| atoms having more electron shells further down the group | B1 |
| the outer electron being further from the nucleus | B1 |
| increased shielding by inner electron shells | B1 |
| a weaker attraction between the nucleus and the outer electron | B1 |
| the outer electron being lost more easily, making the atom more reactive | B1 |
| Final answer: Larger atoms with more shielding hold the outer electron less strongly, so it is lost more easily | |
| Question 10[6 marks] | |
|---|---|
| Answer or working | Marks |
| each carbon atom in graphite forming three covalent bonds, leaving one delocalised electron per atom | B1 |
| these delocalised electrons being free to move and carry charge, so graphite conducts electricity | B1 |
| each carbon atom in diamond forming four covalent bonds, so there are no delocalised electrons | B1 |
| diamond therefore not being able to conduct electricity | B1 |
| graphite consisting of layers of carbon atoms with only weak forces between the layers | B1 |
| these weak forces allowing the layers to slide over each other, making graphite soft and slippery | B1 |
| Final answer: Graphite conducts because of delocalised electrons (none in diamond); its layers slide due to weak forces between them, making it soft | |
| Question 11[6 marks] | |
|---|---|
| Answer or working | Marks |
| the solution containing Cu2+, H+ (from water), SO4 2- and OH- (from water) ions | B1 |
| stating positive ions are attracted to the cathode and gain electrons (reduction) | B1 |
| copper being less reactive than hydrogen, so Cu2+ ions are discharged in preference to H+ ions, producing copper metal | B1 |
| stating negative ions are attracted to the anode and lose electrons (oxidation) | B1 |
| OH- ions being discharged in preference to SO4 2- ions (sulfate is not discharged) | B1 |
| this producing oxygen gas at the anode | B1 |
| Final answer: Copper is deposited at the cathode; oxygen gas is produced at the anode | |
| Question 12[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 | |