Higher Tier - Year 10

Year 10 Paper 4: Combined Practice

Covers cell biology, atomic structure, bonding, energy and electricity.

15 questions - 80 marks - calculator allowed

Year 10 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 10, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.

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Questions

Question 1 [4 marks]

Atomic Structure and the Periodic Table

Chlorine has two naturally occurring isotopes: chlorine-35, with an abundance of 75%, and chlorine-37, with an abundance of 25%.

Calculate the relative atomic mass of chlorine. Give your answer to 1 decimal place.

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]

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]

Bonding, Structure and the Properties of Matter

A sample of poly(ethene) contains chains that are, on average, made from 500 repeating -CH2-CH2- monomer units.

The relative formula mass of one -CH2-CH2- unit is 28.

Calculate the average relative molecular mass of a poly(ethene) chain in this sample, and explain why this value is described as an average.

Question 5 [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 6 [5 marks]

Bonding, Structure and the Properties of Matter

Iodine is a simple molecular substance. Silicon dioxide has a giant covalent structure. Both substances contain strong covalent bonds.

Explain why iodine has a much lower melting point than silicon dioxide.

Question 7 [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 8 [6 marks]

Atomic Structure and the Periodic Table

An ion of an element has the electronic structure 2,8 and an overall charge of 2+.

Deduce the atomic number of this element, state its identity, and state the group of the periodic table it belongs to.

Question 9 [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 10 [6 marks]

Energy

A 1.5 kg block of an unknown metal is heated from 20 degrees C to 95 degrees C using 87750 J of energy.

Calculate the specific heat capacity of the metal, in J/kg degrees C. Use change in thermal energy = m c (change in temperature).

Question 11 [6 marks]

Atomic Structure and the Periodic Table

Group 7 of the periodic table contains the halogens, which react by gaining one electron to form a negative ion.

Explain, in terms of electronic structure, why reactivity decreases going down Group 7 of the periodic table.

Question 12 [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 13 [5 marks]

Cell Biology

A student places identical potato chips into salt solutions of different concentrations to investigate osmosis.

One potato chip has a mass of 5.20 g before the experiment. After 24 hours in a 0.4 mol/dm^3 salt solution, the potato chip has a mass of 4.68 g.

Calculate the percentage change in mass of the potato chip, and state whether this is an increase or a decrease. Give your answer to 1 decimal place.

Question 14 [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 15 [6 marks]

Bonding, Structure and the Properties of Matter

Sodium chloride is an ionic compound with a giant ionic lattice structure.

Explain why solid sodium chloride does not conduct electricity, but molten sodium chloride and an aqueous solution of sodium chloride both conduct electricity.

Model solutions

Mark scheme for Question 1 [4 marks]
Question 1[4 marks]
Answer or workingMarks
multiplying each isotope's mass by its percentage abundanceM1
(35 x 75) + (37 x 25)M1
dividing the total by 100M1
35.5A1
Mark scheme for Question 2 [4 marks]
Question 2[4 marks]
Answer or workingMarks
using GPE = m g hM1
substituting 250 x 9.8 x 12M1
correct evaluationM1
29400 J (29.4 kJ)A1
Mark scheme for Question 3 [4 marks]
Question 3[4 marks]
Answer or workingMarks
using magnification = image size / actual sizeM1
substituting 20 / 0.05M1
magnification = 400A1
stating magnification has no unit, written as x400B1
Final answer: x400 magnification
Mark scheme for Question 4 [5 marks]
Question 4[5 marks]
Answer or workingMarks
using total mass = number of units x mass per unitM1
substituting 500 x 28M1
14000A1
explaining that different polymer chains in the same sample have different lengths (numbers of monomer units)B1
stating this means the sample contains molecules of different relative molecular mass, so an average value is used to describe the whole sampleB1
Final answer: 14000; different chains in a polymer sample have different lengths, so an average relative molecular mass is used to describe the whole sample
Mark scheme for Question 5 [5 marks]
Question 5[5 marks]
Answer or workingMarks
total resistance = 15 + 25 = 40 ohmsM1
using I = V / R, substituting 20 / 40M1
0.5 AA1
using V = I R for the 15 ohm resistor, substituting 0.5 x 15M1
7.5 VA1
Final answer: 0.5 A current; 7.5 V across the 15 ohm resistor
Mark scheme for Question 6 [5 marks]
Question 6[5 marks]
Answer or workingMarks
iodine consisting of simple molecules held together by weak intermolecular forcesB1
only weak intermolecular forces needing to be overcome to melt iodine, needing little energyB1
the covalent bonds within each iodine molecule remaining intact on meltingB1
silicon dioxide having a giant covalent structure with strong covalent bonds throughoutB1
melting silicon dioxide requiring many strong covalent bonds to break, needing much more energyB1
Final answer: Iodine only needs weak intermolecular forces broken; silicon dioxide needs strong covalent bonds broken
Mark scheme for Question 7 [6 marks]
Question 7[6 marks]
Answer or workingMarks
converting 4 minutes to seconds, 4 x 60M1
energy = power x time, 1500 x 240M1
360000 JA1
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
Mark scheme for Question 8 [6 marks]
Question 8[6 marks]
Answer or workingMarks
finding the number of electrons in the ion, 2 + 8 = 10M1
recognising a 2+ ion has 2 fewer electrons than protonsM1
atomic number = 10 + 2M1
atomic number 12A1
identifying the element as magnesiumB1
stating the element is in Group 2 (it loses its 2 outer electrons to form a stable ion)B1
Final answer: Atomic number 12; magnesium; Group 2
Mark scheme for Question 9 [6 marks]
Question 9[6 marks]
Answer or workingMarks
recognising the final mass is 108% of the original massM1
original mass = final mass / 1.08M1
substituting 5.94 / 1.08M1
5.5 gA1
stating distilled water has a higher water potential than the contents of the potato cellsB1
stating water moved into the cells by osmosis, down the water potential gradient, increasing the massB1
Final answer: Original mass 5.5 g; water moved in by osmosis because distilled water has a higher water potential than the cell contents
Mark scheme for Question 10 [6 marks]
Question 10[6 marks]
Answer or workingMarks
finding the temperature change, 95 - 20 = 75 degrees CM1
rearranging to c = E / (m x change in temperature)M1
substituting 87750 / (1.5 x 75)M1
evaluating 1.5 x 75 = 112.5M1
780 J/kg degrees CA1
stating this value could be used to identify the metal, by comparing it to known specific heat capacitiesB1
Mark scheme for Question 11 [6 marks]
Question 11[6 marks]
Answer or workingMarks
halogen atoms having more electron shells further down the groupB1
the outer shell being further from the nucleus (a larger atomic radius)B1
there being increased shielding by inner electron shellsB1
there being a weaker force of attraction between the nucleus and an incoming electronB1
it being harder for the atom to attract (gain) an extra electron into its outer shellB1
stating the atom is less reactive further down the groupB1
Final answer: Larger atoms with more shielding attract an extra electron less strongly, so reactivity decreases down Group 7
Mark scheme for Question 12 [6 marks]
Question 12[6 marks]
Answer or workingMarks
using efficiency = useful output energy / total input energyM1
substituting 1050 / 1500M1
0.7 (70%)A1
wasted energy = total input - useful outputM1
450 JA1
a valid way to reduce wasted energy, e.g. lubricating moving parts to reduce frictionB1
Final answer: 70% efficient; 450 J wasted; e.g. lubrication reduces friction and wasted energy
Mark scheme for Question 13 [5 marks]
Question 13[5 marks]
Answer or workingMarks
finding the change in mass, 4.68 - 5.20 = -0.52 gM1
dividing the change in mass by the original massM1
multiplying by 100M1
-10.0%A1
stating this is a decrease (loss of mass)B1
Final answer: -10.0% (a decrease in mass)
Mark scheme for Question 14 [6 marks]
Question 14[6 marks]
Answer or workingMarks
cavity wall insulation containing trapped pockets of air (or foam) within the materialB1
trapped air being a poor conductor of heat, reducing energy transfer by conduction through the wallB1
trapped pockets of air also reducing energy transfer by convection, by preventing convection currents forming within the cavityB1
stating the rate of energy transfer through a wall depends on the thickness of the insulating layerB1
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 returnsB1
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
Mark scheme for Question 15 [6 marks]
Question 15[6 marks]
Answer or workingMarks
solid sodium chloride having ions held in fixed positions in a rigid latticeB1
stating the ions cannot move to carry an electric charge in the solidB1
stating solid sodium chloride therefore does not conduct electricityB1
the rigid lattice breaking down when the compound is molten, so the ions are free to moveB1
the ions separating and also being free to move when the compound is dissolved in waterB1
stating these free-moving, charged ions can carry an electric current, so molten or dissolved sodium chloride conducts electricityB1
Final answer: Ions are fixed in place in solid NaCl so cannot carry charge; when molten or dissolved the ions are free to move and can carry an electric current