Oxford Revise AQA GCSE Combined Science Foundation | Chapter C2 answers

Question

Answers

Extra information

Mark

AO / Specification reference

01.1

nitrogen

1

AO1

5.1.1.1

01.2

covalent

1

AO1

5.2.1.1

01.3

one dot and one cross in each of the overlaps between H and N

1

AO2

5.2.1.4

01.4

low melting point

1

AO1

5.2.2.4

02

Level 3: A detailed and coherent comparison is given, demonstrating a sound knowledge of the differences in properties and the reasons for them.

5-6

AO1

5.2.3.1

5.2.3.2

Level 2: A correct description is given of the properties of each allotrope. Some reasons are given, but are not clearly articulated / not clearly linked to the property.

3-4

Level 1: Some correct points are made about each structure. Comparisons and reasons are not included.

1-2

No relevant content.

0

Indicative content:

• graphite conducts electricity but diamond does not
• because graphite includes delocalised electrons but diamond does not
• graphite is soft/slippery but diamond is hard
• because the layers in the structure of graphite can slide over each other, but there are no such layers in diamond/diamond has lots of strong bonds
• both have high melting and boiling points
• because both include strong covalent bonds between their atoms.

03.1

long

lots of

strong

1

1

1

AO1

5.2.2.5

03.2

propene

1

AO2

5.2.2.5

03.3

diagram should match polyethene as shown in Figure 1, but replace one H with Cl

carbon-carbon bond must be single, not double

2

AO3

5.2.2.5

03.3

(C₂H₃Cl)_n

1

AO2

5.2.2.5

04.1

electrons are shared

1

AO1

5.2.1.4

04.2

carbon dioxide – simple molecule

silicon dioxide – giant

polythene – polymer

2

AO2

5.2.1.4

04.3

lower

intermolecular forces

1

AO1

5.2.2.4

05.1

no charged atoms/ions/particles

to carry charge

one mark for each correct bullet point up to a maximum of two marks

1

1

AO1

5.2.2.4

05.2

graphite/graphene

delocalised electrons that can move

1

1

AO1

5.2.2.6

05.3

Level 3: A full description of the method provided, with at least two pieces of equipment named.

5-6

AO1

AO2

5.2.2.1

5.2.2.4

5.2.2.6

Level 2: Basic method provided, identifying that a higher boiling point is needed to break the covalent bonds than the intermolecular bonds. At least one piece of equipment identified.

3-4

Level 1: Method identifies that a high boiling point is needed to break the covalent bonds. No equipment named.

1-2

No relevant content.

0

Indicative content:

• to boil silicon dioxide must break strong covalent bonds
• requires lots of energy, therefore very high boiling point
• to boil octane and hydrogen fluoride, need to break weaker intermolecular forces
• less energy needed to break intermolecular forces than covalent bonds, therefore, much lower boiling points than silicon dioxide
• octane is a larger molecule than hydrogen fluoride so has stronger intermolecular forces
• therefore, more energy needed to break the intermolecular forces in nonane, therefore higher boiling point than hydrogen flouride

05.4

accept any answer between 125.6 and 2230

intermolecular forces need to be broken, not covalent bonds

but much higher boiling point than octane as much larger molecule

1

1

1

AO3

5.2.2.1

5.2.3.3

06.1

one from:

• only shows a small number of the atom in silicon dioxide
• atoms are not spheres/bonds are not sticks
• does not show the electrons being shared

accept any other sensible answer

1

AO3

5.2.1.4

06.1

four

1

AO2

06.2

carbon + oxygen → carbon dioxide

AO1

06.3

simple molecule

1

AO1

06.4

C

1

AO1

06.5

one from:

• diamond is a solid
• diamond is hard
• diamond has a high melting point/boiling point

accept reverse for carbon dioxide

1

AO1

06.6

CH₄

1

AO2

07.1

four

high

hard/strong

1

1

1

AO1

5.2.3.1

07.2

delocalised electron

means it can conduct electricity

1

1

AO1

5.2.3.2

08.1

black

white

covalent

1

1

1

AO1

5.2.1.4

08.2

liquid

1

AO2

5.2.2.1

08.3

B

1

AO1

5.2.2.1

08.4

one cross and one dot should be placed within the overlap between each O and H molecule

1

AO2

5.2.1.4

09.1

carbon

1

AO1

5.2.3.3

09.2

top – buckminster fullerene

middle – carbon nanotube

bottom – graphene

1

1

1

AO1

5.2.3.3

09.3

conductor

electronics/wire

1

1

AO1

5.2.3.3

09.4

lubricants/catalysts/drug delivery/nanotechnology

1

AO1

5.2.3.3

10.1

Z

1

AO3

5.2.3.3

10.2

high (tensile) strength

1

AO1

5.2.2.3

10.3

electronics – because of electricity conduction

reinforcing composite materials – because of high tensile strength

both the use and reason required for each mark

1

1

AO1

5.2.2.3