Oxford_Revise_AQA_GCSE_Chemistry_Higher_Chapter1_Answers-test
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Here you’ll find all the answers to the activities and exam-style practice questions featured in Oxford Revise AQA A Level Geography.
Chapter 1: [Practice answers]
Question |
Answers |
Extra information |
Mark |
AO / Specification reference |
01.1 |
Niels Bohr—electrons orbit the nucleus at |
|
1 1 |
AO1 4.1.1.3 |
01.2 |
any six from:
|
|
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:
differences:
|
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 |
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No relevant content |
0 |
|||
Indicative content
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 |
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 |
05.2 |
17 |
|
1 |
AO1 |
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 |
|
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 |
\(\frac{{{\rm{71}}}}{{{\rm{1}}{{\rm{0}}_{\rm{ }}}{\rm{000}}}}\) 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 = =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 |
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 |
= 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 |
10.3 |
2.8.8.2 |
accept 2,8,8,2 |
1 |
AO2 |
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
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
|
|
|||
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 |