Oxford_Revise_AQA_GCSE_Chemistry_Higher_Chapter1_Answers-test

01.1

Niels Bohr—electrons orbit the nucleus at
James Chadwick—the nucleus contains neutrons

1

1

AO1

4.1.1.3

01.2

any six from:

• alpha particles/helium nucleus fired at gold foil

• Most passed through the gold foil

• So most of the atom is empty space

• a small number bounced back

• so must have collided with something/mass/nucleus

• a small number passed through but were deflected/changed direction

• positively-charged alpha particles passed near positively charged nucleus and were repelled

6

AO1

4.1.1.3

01.3

19

1

AO2

4.1.1.4

02.1

fractional distillation

do not accept distillation or simple distillation

1

AO2

4.1.1.2

02.2

mixture is heated

both liquids will give off vapours before their boiling point

vapours enter fractionating column (with glass beads)

water will condense on glass beads as it has a higher boiling point (than isopropanol)

isopropanol will continue to rise and pass into condenser

will then condense and can be collected in separate vessel

1

1

1

1

1

1

AO1

4.1.1.2

02.3

boiling point too similar

1

AO2

4.1.1.2

03.1

similarities:

• both suggest that atoms are spherical

• both suggest different elements have different atoms of different masses

differences:

• earlier model states that atoms cannot be divided

• plum pudding suggests that negative electrons are embedded in a ball of positive charge

award 1 mark per point

3

AO1

4.1.1.3

03.2

Level 3: A detailed and coherent explanation is given. All three observations of the experimental evidence are explained.

5–6

AO1

4.1.1.3

Level 2: A coherent explanation is given, but not all observations are linked to aspects of the model.

3–4

Level 1: Some correct points are made. At least tone observation is given and linked to aspects of the model.

1–2

No relevant content

0

Indicative content

• positively-charged alpha particles fired at gold foil

• most alpha particles travelled straight through the foil

• suggesting most of the atom is empty space

• a small number of alpha particles bounced back

• suggesting mass of the atom was concentrated in the centre of the atom

• a small number of alpha particles were deflected

• suggested the central mass had positive charge

• the positively-charged nucleus repelled positive alpha particles, causing the deflection

allow answer in terms of plum pudding model, for example, alpha particles would not have passed through if plum pudding model is correct etc.

04.1

16

1

AO2
4.1.1.4

04.2

17

1

AO2

4.1.1.4

04.3

Y

1

AO3

4.1.1.5

04.4

X and Z

both required for the mark

1

AO3

4.1.1.5

05.1

17

accept same number of electrons

1

AO1
4.1.1.4

05.2

17

1

AO1
4.1.1.4

05.3

chlorine has (two) isotopes with different abundance

relative atomic mass is an average

students do not need to list the isotopes of chlorine to gain mark

accept calculation for 2 marks

1

1

AO1

4.1.1.5

05.4

\(\frac{{{\rm{(62}}{\rm{.9}} \times {\rm{63)}} + {\rm{(30}}{\rm{.8}} \times {\rm{65)}}}}{{{\rm{100}}}}\)

= 63.616

= 63.6

1

1

1

AO2

4.1.1.6

05.5

other isotopes of copper exist

1

AO3

4.1.1.6

06.1

number of protons = number of electrons = 11

mass number = number of protons + number of neutrons
= 11 + 12 = 23

1

1

AO2×1

AO3×1

4.1.1.5

06.2

in both atoms, the electrons are arranged in shells/energy levels

in both atoms, there are three shells/energy levels

in both atoms, the shell/energy level nearest the nucleus has two electrons/is full

in both atoms, the shell/energy level second from the nucleus has eight electrons/is full

in sodium, the outer shell/energy level has one electron only and is not full

in argon, the outer shell/energy level has eight electrons and is full

1

1

1

1

1

1

AO2

4.1.1.7

06.3

\(\frac{{{\rm{71}}}}{{{\rm{1}}{{\rm{0}}_{\rm{ }}}{\rm{000}}}}\)
= 0.0071

7.1×10⁻³ (pm)

1

1

AO2

4.1.1.5

07.1

mass numbers:

L = 14 + 14 = 28 M = 14 + 15 = 29 N = 14 + 16 = 30

percentage abundance of N = 100 – (92.2 + 4.68) = 3.12%

relative atomic mass = (92.20×28)+(4.68×29)+(3.12×30)100$\frac{(92.20\times 28)+(4.68\times 29)+(3.12\times 30)}{100}$
$\frac{{{\rm{(92}}{\rm{.20}} \times {\rm{28)}} + {\rm{(4}}{\rm{.68}} \times {\rm{29)}} + {\rm{(3}}{\rm{.12}} \times {\rm{30)}}}}{{{\rm{100}}}}$

=28.1092

= 28.1

1

1

1

1

1

AO1×2

AO2×3

4.1.1.6

07.2

three shells

two electrons in first shell, eight electrons in second shell, and four electrons in third shell

accept one shell shown with four electrons for 1 mark

1

1

AO1

4.1.1.7

07.3

both isotopes will have the same chemical properties

as they have the same number of (outer) electrons

1

1

AO1

4.1.1.5

08.1

−1

any units given negate mark

1

AO1

4.1.1.4

08.2

two crosses in shell nearest centre
eight crosses in next shell
four crosses in outer shell

accept dots for electrons

all electrons must be drawn with the same shape

1

AO2

4.1.1.7

08.3

(Niels) Bohr

1

AO1

4.1.1.3

09.1

atoms of the same element/with the same atomic number/same number of protons but a different number of neutrons

1

AO1

4.1.1.5

09.2

The three isotopes have the same atomic number.

1

AO1

4.1.1.5

09.3

(79.0×24)+(10.0×25)+(11.0×26)100$\frac{(79.0\times 24)+(10.0\times 25)+(11.0\times 26)}{100}$
$\frac{{{\rm{(79}}{\rm{.0}} \times {\rm{24)}} + {\rm{(10}}{\rm{.0}} \times {\rm{25)}} + {\rm{(11}}{\rm{.0}} \times {\rm{26)}}}}{{{\rm{100}}}}$

= 24.32

= 24.3

1

1

1

AO2

4.1.1.6

10.1

zirconium

1

AO2

4.1.1.4

10.2

calcium

1

AO2
4.1.1.5

10.3

2.8.8.2

accept 2,8,8,2

1

AO2
4.1.1.7

10.4

10.5

award 1 mark for every two points plotted correctly

award 1 mark for correct line of best fit

3

1

AO2×3

AO3×1

4.1.1.5

10.6

as the number of protons increases, so does the number of neutrons

not directly proportional (as the number of protons increases, the number of neutrons increases more quickly)

accept not linear/ non-linear

1

1

AO3

4.1.1.5

11.1

triangle

accept drawn symbol

do not accept water

1

AO2

4.1.1.1

11.2

A

1

AO2

4.1.1.1

11.3

C

1

AO2

4.1.1.1

4.1.1.2

11.4

D

1

AO2

4.1.1.1

4.1.1.2

11.5

NaCl

must have capitalisation shown

accept ClNa

1

AO2

4.1.1.1

12.1

silicon atoms have more electron shells

1

AO3

4.1.1.7

12.2

1.22×10⁻¹⁰ (m)

1

AO2

4.1.1.5

12.3

Si : C 0.111 : 0.077

0.1110.077:0.0770.077$\frac{0.111}{0.077}:\frac{0.077}{0.077}$
$\frac{{{\rm{0}}{\rm{.111}}}}{{{\rm{0}}{\rm{.077}}}}{\rm{:}}\frac{{{\rm{0}}{\rm{.077}}}}{{{\rm{0}}{\rm{.077}}}}$

1.5 : 1

3 : 2

1

1

1

AO2

4.1.1.5

12.4

0.077 nm = 7.7×10⁻¹¹ m

\(\frac{{{\rm{7}}{\rm{.7}} \times {\rm{1}}{{\rm{0}}^{ – {\rm{11}}}}}}{{{\rm{2}}{\rm{.7}} \times {\rm{1}}{{\rm{0}}^{ – {\rm{15}}}}}}\)= 28 519

accept answer gained from converting metres to nm

1

1

AO1×1

AO2×1

4.1.1.5

13.1

filtration

1

AO1

4.1.1.2

13.2

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

5–6

AO1

4.1.1.2

Level 2: Basic method provided, identifying that the water needs to evaporate (either by heating or by being left). At least one piece of equipment identified.

3–4

Level 1: Method identifies idea that water needs to evaporate/be heated. No equipment named.

1–2

No relevant content

0

Indicative content

• mixture placed in evaporating dish

• evaporating dish placed on beaker half full of water

• place beaker/evaporating dish on tripod and gauze

• heat the mixture/water

• using Bunsen burner

• until crystals start to form

• remove mixture from the heat

• leave for the rest of the water to evaporate

13.3

chromatography

1

AO1

4.1.1.2

14.1

A

1

AO2

4.1.1.2

14.2

dyes B and C produced spots that overlap with dye A and each other

therefore cannot distinguish whether dye B or C produces the top spot

1

1

AO3

4.1.1.2

14.3

rerun experiment with difference mobile phase/solvent

1

AO3

4.1.1.2