OCR Gateway Chemistry B - C4 C5 C6 - Unofficial Mark Scheme

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some-student
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Starting the mark scheme for C4 C5 C6.

How did you all find it? I found it pretty difficult but not as bad as Biology.
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OCR Gateway Chemistry B
C4 C5 C6 + Section D
Unofficial Mark Scheme
Monday 20 June 2016

Module C4 - 25 marks
1) Table of atoms
a) Complete the table [3]
atomic number: 12 (1)
number of neutrons: 4 (1)
electronic configuration: 2.4 (1)

b) Atoms W and X are both chlorine atoms but different. Name what types of atom they are and explain why [2]
isotopes (1)
same atomic number but different mass number (or equivalent in terms of neutrons/protons) (1)

2. Superconductors
a) Write about two advantages [2]
any two of
• loss free power transmission (1)
• super-fast electronic circuits (1)
• powerful electromagnets (1)

b) Explain one disadvantage [1]
it is very hard / expensive to get them to the low temperatures required (1)

3. Chlorine fluoride (formula: ClF)
a) covalent, dot and cross [2]
shared pair of electrons between Cl and F atoms (1)
Cl and F atoms both having 6 other outer electrons (1)

b) it is a simple molecular structure - predict two properties [2]
any two of
low boiling / melting point (1)
does not conduct electricity (1)
gas or liquid at room temperature (1)

4. Testing for ions [6 QWC]
Marking points
balanced symbol equation: 3AgNO3 + FeCl3 -> 3AgCl + Fe(NO3)3

compound A: copper(II) chloride (CuCl2) - Cu2+ ions in sodium hydroxide solution form a blue precipitate; Cl- ions in silver nitrate solution make a white precipitate

compound B: iron(II) bromide (FeBr2) - Fe2+ ions in sodium hydroxide solution form a green precipitate; Br- ions in silver nitrate solution make a cream precipitate

5. Group 7 elements
a) Complete the table [3]
can't remember this completely
iodine: grey solid (1)
prediction in a certain range (1)
prediction in a certain range (1)

b) Explain the reactivity in group 7 [2]
as you go down the group: increased atomic radius / more nuclear shielding (1)
therefore less attraction from the nucleus to electron to be gained (1)

c) Describe how Mendeleev contributed to the periodic table [2]
any two of
order of atomic mass (1)
left gaps for undiscovered elements (1)
in vertical groups, elements had similar properties (1)

Module C5 - 25 marks
6. Water testing [3]
Sample A: bromide (Br- ions) as cream precipitate formed in lead nitrate (1)
I can't remember if the precipitate was cream (Br- ions) or white (Cl- ions)

Sample B: iodide (I-) ions as yellow precipitate formed in lead nitrate, and sulfate (SO42- ions) as white precipitate formed in barium chloride (1)
Sample C: sulfate (SO42- ions) as white precipitate formed in barium chloride / nitrate (NO3-) ions due to no reaction with lead nitrate (1)

7. Strong and weak acids
a) Explain the difference between strong and weak acids [2]
• acid strength is a measure of thedegree of ionisation of the acid (1)
• acid concentration is a measure of the number ofmoles of acid in one dm3 (1)

b)i) Explain why the reaction is faster for the strong acid [1]
in hydrochloric acid the hydrogen ions have a higher collision frequency with reactant particles as they are in a higher concentration (1)

b)ii) Explain why the same volume of strong and weak acids used in a reaction will still produce the same amount of carbon dioxide gas [1]
the amount of products is determined by the amounts ofreactants present not the acid strength / all the atoms in the gas are from the calcium carbonate (1)

8. Contact process
a) List another condition apart from the pressure or temperature [1]
V2O5 catalyst (or any equivalent name) (1)

b) Explain all the conditions used in the contact process [3]
• increasing the temperature moves the positionof equilibrium to the left and increases rate ofreaction so a compromise temperature is used (1)
• addition of catalyst increases rate but does notchange position of equilibrium (1)
• even at low pressure, the position of equilibriumis already on right so expensive high pressure isnot needed (1)

9. Reaction/calculations
a)i) How much gas is produced? [1]
75 (1)

a)ii) How long did the reaction take before it stopped? [1]
51 - 53 (1)

b) Draw on the graph to show the reaction for 0.005g of magnesium [2]
line half height of original line (1)
correct shape of line (1)

c) Calculate the mass of magnesium sulfate produced with 0.5g magnesium, explaining your answer [2]
calculation (1)
correct answer: 50g (1)

d)i) Calculate the number of moles [1]
0.08 mol (1)

d)ii) Calculate the volume of gas [1]
1920 dm3 (1)

10. Titration [6 QWC]
Marking points
graph: when 20cm3 of acid has been added, neutralisation occurs - this is also the end-point

concentration of sodium hydroxide
calculating number of moles of acid: when the reaction occurred (neutralisation), 20cm3 of acid had been used, which is 0.02dm3 (20 / 1000)

c = n / v
n = c × v = 0.1 × 0.02 = 0.002 mol

as the equation was balanced, there were the same amount of moles of acid as alkali

25cm3 of alkali = 0.025dm3

c = n / v = 0.002 / 0.025 = 0.08 mol/dm3

Module C6 - 25 marks
11. Fats
a) What type of molecule is a fat? [1]
ester (1)

b) Molecular formula of glycerol [1]
C3H8O3(1)

c)i) Explain why the given diagram is unsaturated [1]
carbon to carbon double bond (1)

c)i) Explain how it would be checked if it was unsaturated, and the results expected to be seen [2]
• addition reaction takes place at the carbon-carbon double bond with addition of bromine water
• a colourless dibromo compound is formed (1)

d) Explain how margarine is made [2]
hydrogen gas is bubbled in (using a nickel catalyst at 200°C) (1)
breaking some of the bonds, hardening the oil (1)

12. Displacement reaction [6 QWC]
Marking points
word equation: copper(II) sulfate + iron -> iron(II) sulfate + copper

balanced symbol equation: CuSO4 + Fe -> FeSO4 + Cu

oxidation: Fe - 2e- -> Fe2+ - electrons are lost(iron is the reducing agent as it causes copper's reduction - see below)

reduction: Cu2+ + 2e- -> Cu - electrons are gained (copper is the oxidising agent as it causes iron's oxidation - see above)

13. Ozone layer depletion
a) Pick the reaction to show UV rays breaking down ozone [1]
reaction 4 (1)

b) Explain how a CFC molecule can create a chlorine atom, in terms of bonding and electrons [2]
(UV light) breaks a C-Cl bond (1)
forming a chlorine free radical with an unpaired electron (1)

c) Explain how a single chlorine atom can destroy many ozone molecules, using the reactions given [2]
any two of
reaction 2 produces chlorate radical (1)
which makes a chlorine atom again in reaction 3 (regeneration) (1)
chain reaction in terms of the reactions 2 and 3 (1)

d) Explain why it took a long while for scientists to recommend the ban of CFCs [2]
• initial enthusiasm for the use of CFCs based upon their inertness (1)
• later discovery of ozone depletion and link topresence of CFCs (1)

14. Water hardness [2]
soap affected as lather only 30cm3 with calcium hydrogen carbonate, while for distilled water it is 60cm3 (1)
washing up liquid not affected as lather only 60cm3 with calcium hydrogen carbonate and distilled water (1)

15. Reaction
a) How much magnesium is made after 30 minutes (current: 10.3A)? [1]
7.5g (1)

b) Predict how much magnesium is made - current: 20.6A, time: 120 minutes [2]
7.5 × 2 × 4 (1)
60g (1)

Section D - 10 marks
16. Pollutants
a) Work out, from the graph, the amount of fertilisers used [1]
37 million tonnes (1) I can't remember so this may be wrong

b) Explain why the nitrate restrictions have had an effect from the graph [1]
sudden decrease after 1977 (1)

c)i) Calculate the mass of fertilisers used[1]
1,260,000,000 (1)

c)ii) Explain why a country with a much smaller agricultural area would use more fertilisers[2]
there are likely many answers to this question
more efficient farming (1)
to provide enough food for the population (1)

17. Greenhouse gases
a) Explain which gas is most likely to be produced by the use of fertilisers [2]
nitrous oxide (1)
88% is contributed by farming (in comparison to other gases) (1)

b) Explain why farming has a much higher impact on greenhouse emissions than residential uses [3]
calculation of (average) percentage for farming (1)
calculation of (average) percentage for residential uses (1)
comparison to show farming's greater impact (1)
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GCShElpme
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I've made some predictions for the grade boundaries, if anyone would like them?
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some-student
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(Original post by GCShElpme)
I've made some predictions for the grade boundaries, if anyone would like them?
Yes please
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leopard923
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(Original post by GCShElpme)
I've made some predictions for the grade boundaries, if anyone would like them?
Yes please !
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GCShElpme
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Predictions for grade boundaries.
The last couple of grade boundaries are as follows;
2015- A*: 60, A: 50, B: 40, C: 30
2014- A*: 62, A: 52, B: 41, C: 31
2013- A*: 54, A: 45, B: 36, C: 28
[Links: http://www.ocr.org.uk/i-want-to/convert-raw-marks-to-ums/ ]
Despite my own opinion on the difficulty of the paper, the vast majority thought that this paper was relatively difficult, more so than past papers. On account of the comparative easiness of the first paper, I'd say that these grade boundaries will be lower than 60, though no lower than, say, 55.
Predictions for 2016- A* 56, A: 46, B: 38, C: 30.
Hope this helps. I appreciate feedback! Image
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leopard923
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(Original post by GCShElpme)
Predictions for grade boundaries.
The last couple of grade boundaries are as follows;
2015- A*: 60, A: 50, B: 40, C: 30
2014- A*: 62, A: 52, B: 41, C: 31
2013- A*: 54, A: 45, B: 36, C: 28
[Links: http://www.ocr.org.uk/i-want-to/convert-raw-marks-to-ums/ ]
Despite my own opinion on the difficulty of the paper, the vast majority thought that this paper was relatively difficult, more so than past papers. On account of the comparative easiness of the first paper, I'd say that these grade boundaries will be lower than 60, though no lower than, say, 55.
Predictions for 2016- A* 56, A: 46, B: 38, C: 30.
Hope this helps. I appreciate feedback! Image
I agree with this


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some-student
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(Original post by GCShElpme)
Predictions for grade boundaries.
The last couple of grade boundaries are as follows;
2015- A*: 60, A: 50, B: 40, C: 30
2014- A*: 62, A: 52, B: 41, C: 31
2013- A*: 54, A: 45, B: 36, C: 28
[Links: "]http://www.ocr.org.uk/i-want-to/convert-raw-marks-to-ums/ ]
Despite my own opinion on the difficulty of the paper, the vast majority thought that this paper was relatively difficult, more so than past papers. On account of the comparative easiness of the first paper, I'd say that these grade boundaries will be lower than 60, though no lower than, say, 55.
Predictions for 2016- A* 56, A: 46, B: 38, C: 30.
Hope this helps. I appreciate feedback! Image
Looks great - thanks
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GCShElpme
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(Original post by leopard923)
I agree with this


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I'm glad. How do you think you did?
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GCShElpme
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(Original post by some-student)
Looks great - thanks
It's not a problem.
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jessdstromberg
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In general i didn't think the paper was bad - the only question that stumped me was the c5 6 marker, can someone please tell me the concentration they got for this?
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subhan23
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Tbh that was quite of a easy blessed 6 marker. If u looked through the revision guide in c5 it talks about questions like that. So to find the concentration of sodium hydroxide u find the number of moles of hydrocloroc acid . U find that from the graph the verticle line. Which lies on 20 cm cubed. So from there find the moles of hydrochrloic acid. Then use the nalanced symbol equation tl find the moles in the soidum hydroxide and hse the formula of number of moles divided by the volume which was 0.025 dm cubed. Orignally given 25 cm cubed. And The ANSWER IS 0.08 moles per dm cubed
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BenSidebottom
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Have you got the rest of the mark scheme?
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jessdstromberg
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(Original post by subhan23)
Tbh that was quite of a easy blessed 6 marker. If u looked through the revision guide in c5 it talks about questions like that. So to find the concentration of sodium hydroxide u find the number of moles of hydrocloroc acid . U find that from the graph the verticle line. Which lies on 20 cm cubed. So from there find the moles of hydrochrloic acid. Then use the nalanced symbol equation tl find the moles in the soidum hydroxide and hse the formula of number of moles divided by the volume which was 0.025 dm cubed. Orignally given 25 cm cubed. And The ANSWER IS 0.08 moles per dm cubed
oh no i forgot to convert cm to dm !! i ended up getting 0.16.. oh well, hopefully i did ok on the rest of the paper!
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GCShElpme
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(Original post by BenSidebottom)
Have you got the rest of the mark scheme?
I have a more complete version, if you'd like the link? This mark scheme is great, though!
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BenSidebottom
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I prefer this one as it is more detailed, just need to see the rest
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subhan23
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Which one do u need ben?
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leopard923
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(Original post by GCShElpme)
I'm glad. How do you think you did?
I think it went quite well hopefully I got an a* wbu?

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leopard923
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(Original post by subhan23)
Tbh that was quite of a easy blessed 6 marker. If u looked through the revision guide in c5 it talks about questions like that. So to find the concentration of sodium hydroxide u find the number of moles of hydrocloroc acid . U find that from the graph the verticle line. Which lies on 20 cm cubed. So from there find the moles of hydrochrloic acid. Then use the nalanced symbol equation tl find the moles in the soidum hydroxide and hse the formula of number of moles divided by the volume which was 0.025 dm cubed. Orignally given 25 cm cubed. And The ANSWER IS 0.08 moles per dm cubed
Yayyyy I got this

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some-student
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(Original post by BenSidebottom)
I prefer this one as it is more detailed, just need to see the rest
Just done C5
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