Unofficial Mark scheme: Advanced Higher Chemistry SQA Paper 2019 Watch

Relentas
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I'll be attaching a power point document of the exam when I can.. Thanks to: @KateM04 and @leavemeblank
Please ignore any working, we don't have any fresh copies.
Marking instructions (from what I think the answer is) will be added here: Please correct any mistakes you see.
Multi choice:
Spoiler:
Show


1)B
2)D
3)C
4)A
5)D
6)D
7)B
8)C
9)A
10)C
11)C
12)A
13)A
14)B
15)C
16)A
17)C
18)C
19)D
20)B
21)C
22)B
23)B
24)A
25)C
26)A
27)D
28)C
29)D
30)A

Paper 2:
Spoiler:
Show

1a) the 5th electron total
b) Sub into G=H-TS = 103.3 (trick = you need to divide the entropy value of S by 1000 to get it to kJ)
c)gives K = 3.2 x 10^-5

2a) Order of a reaction is the power to which a concentration is raised in the rate equation.
bi) I said order 2.
ii) rate = k[h2o2][I-]
c) No clue.

3a) Standard solution rant: to create a 0.01mol (dilution factor of 10) standard solution from 0.1moll-1 thats has a volume of 50cm^3, first pipette out 5ml of your initial solution, add that to the 50cm^3 flask, then rinse your pipette with distilled water, then make up to the mark using distilled water. Invert 3 times and wham.
b) Nitric acid of the required concentration.
bii) The sample solution was diluted in order for an accurate reading to be made, - the absorbance/conc must lie within the boundaries of the pre made standard graph.
ii) I ended up getting ~ 35.5%, someone said there's a dilution factor of 2 somewhere "according to their teacher" but I don't see it.

4ai) A Bronsted-Lowry base is anything that readily accepts protons/Hydrogen ions.
ii) Acid - H2O2, conj base = HO2-
b) I dont know - Somehow talk about bonding pairs and make it make sense.
4c)I think its the more Cl atoms, the stronger it gets.
iiA)= A - 0.08 mol l-1 B - pH = 1.95

5) Possibly excited due to the heat generated in the firework. (MAY also accept the reaction causes the excitation)
bi)HexaaquaZinc(II)
ii) d-orbitals are no longer degenerate and therefore when electrons transition from the lower orbitals to the higher orbitals, the energy absorbed does not correspond to a wavelength in the visible spectrum.
5c)E = 2.04 x 10^-16 J ii) 18.96 eV

6)10.5 mol l-1
b) Acidified dichromate was added in excess to firstly increase K, giving more product, it also ensures all the ethanol reacted.
c) Why was experimentally determined conc of ethanol higher - Possibly due to additives in the vodka (eg flavorings which reacted with the dichromate.) - other things can be added.
d) Test with known conc of ethanol, compare to the standard graph made.

7a) do the filtration under vacuum.
bi) the horizontal overlap of 2 S or and S and a P orbital. - OO or Ooo < meant to be diagrams.
ii) When two p orbitals and an s orbital merge to form 3 hybrid orbitals.
c) Talk about electrons transitioning from HOMO to LUMO, length of the conjugated system determines it too, etc.
di) Methanol has similar polarity.

7diiA) Infrared vibrates the bonds, the bonds absorb some energy, this absorption is dependent on the type of bond, allowing you to deduce what bond is there.
B) circle any OH group I think (if I remember right)
7d)ii)(C) (I) 0.029m (II) 4.065 x 10-3 kJ mol-1

8- Knowledge of chemistry.

9) Chiral centre is below the OH. Has H, Oh, benzene, and R groups.
ii) chiral molecules that are non-superimposable mirror images of eachother
b) Step 2 - substitution.
ii) mass is = 9.5/134 = n, 1:1 mol ratio. m= n*gfm = (9.5/134) *149 *0.718(percentage yield)= 7.58g

10a) ether
bi) A plus sign with an elongated right arm with an O between it. (my best description.)
ii) 2-methoxy-2-methylpropane
c) methanol
ii) curly arrow from C to the C-Cl bond, carbocation intermediate (with pos charge) O- attacks it.
iii) due to the tertiary structure of 2chloromethlypropane - its a tertiary haloalkane. or Provides a more stable carbocation due to the inductive stabilisation of the carbocation. - possibly accept numerous things, 1 marker so as long as theres a brief understanding it's fine.
d) Mirror image wouldn't be optically active. Butan-2-ol however is an isomer that is chiral and hence optically active.
e) A line between 0.9-1.5 that's 3x taller.
Last edited by Relentas; 6 days ago
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Relentas
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Well, I cant seem to post the powerpoint, but here's a link to the advanced chemistry form where the photos are:

https://www.thestudentroom.co.uk/sho...204310&page=11


^ and Page 7 has multi choice.
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Labrador99
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(Original post by Relentas)
I'll be attaching a power point document of the exam when I can.. Thanks to: @KateM04 and @leavemeblank
Please ignore any working, we don't have any fresh copies.
Marking instructions (from what I think the answer is) will be added here: Please correct any mistakes you see.
Multi choice:
Spoiler:
Show



1)B
2)D
3)C
4)A
5)D
6)D
7)B
8)C
9)B
10)C
11)C
12)B
13)A
14)B
15)C
16)A
17)C
18)C
19)D
20)B
21)C
22)B
23)B
24)A
25)D
26)A
27)D
28)C
29)D
30)A


Paper 2:
Spoiler:
Show


1a) the 5th electron total
b) Sub into G=H-TS = 103.3 (trick = you need to divide the entropy value of S by 1000 to get it to kJ)
c)sub into the formula and rearrange.. honestly no point in me doing it. I can't read the equation anyways.

2a) Order of reaction is _
bi) I said order 2.
ii) rate = k[h2o2][I-]
c) No clue.

3a) Standard solution rant: to create a 0.01mol (dilution factor of 10) standard solution from 0.1moll-1 thats has a volume of 50cm^3, first pipette out 5ml of your initial solution, add that to the 50cm^3 flask, then rinse your pipette with distilled water, then make up to the mark using distilled water. Invert 3 times and wham.
b) Nitric acid of the required concentration.
bii) The sample solution was diluted in order for an accurate reading to be made, - the absorbance/conc must lie within the boundaries of the pre made standard graph.
ii) I ended up getting ~ 35.5%, someone said there's a dilution factor of 2 somewhere "according to their teacher" but I don't see it.

4ai) A Bronsted-Lowry base is anything that readily accepts protons/Hydrogen ions.
ii) Acid - H2O2, conj base = what ever the negative molecule was (I cant read it)
b) I dont know - Somehow talk about bonding pairs and make it make sense.
4c)I think its the more Cl atoms, the weaker it gets. (the smaller the Ka, the higher the pKa, hence more alkali/weaker.)
iiA)=(1.89/GFM/0.25)
B) follow through, 1/2Ka - log10c (the c being from the previous question)
- Knowlege of chemistry question

5) Possibly excited due to the heat generated in the firework. (MAY also accept the reaction causes the excitation)
bi)HexaaquaZinc(II)
ii) Zn has a full D shell. this means no transitions can take place. Possibly more needs to be added.
5c)F is in s-1, use e=hf, find E.
ii) divide E by 6.24x10^18 gives eV (1 mark) then sub into EQ giving correct answer (2nd mark)

6)I don't think I got this calculation one about the vodka.
b) Acidified dichromate was added in excess to firstly increase K, giving more product, it also ensures all the ethanol reacted.
c) Why was experimentally determined conc of ethanol higher - Possibly due to additives in the vodka (eg flavorings which reacted with the dichromate.)
d) Test with known conc of ethanol, compare to the standard graph made.

7a) do the filtration under vacuum.
bi) the horizontal overlap of 2 S or and S and a P orbital. - OO or Ooo < meant to be diagrams.
ii) when 2 p and an s orbital overlap. ?
c) Talk about electrons transitioning from HOMO to LUMO, length of the conjugated system determines it too, etc.
di) Methanol has similar polarity.

7diiA) Infrared vibrates the bonds, the bonds absorb some energy, this absorption is dependent on the type of bond, allowing you to deduce what bond is there.
B) circle any OH group I think (if I remember right)
ci) peak is in cm^-1. 1/3395 = wavelength in cm, divide by 100 to make it into meters.
ii) kJmol-1 = E=lhf. using wavelength from previous question.

8- Knowledge of chemistry.

9) Chiral centre is below the OH. Has H, Oh, benzene, and R groups.
ii) optical isomers are - (i imagine mirror images)
b) Step 2 - substitution.
ii) mass is = 9.5/134 = n, 1:1 mol ratio. m= n*gfm = (9.5/134) *149 *0.718(percentage yield)= 7.58g

10a) ether
bi) A plus sign with an elongated right arm with an O between it. (my best description.)
ii) god knows.
c) methanol
ii) curly arrow from C to the C-Cl bond, carbocation intermediate (with pos charge) O- attacks it.
iii) due to the tertiary structure of 2chloromethlypropane - its a tertiary haloalkane.
d) Draw it's mirror image.
e) A line between 0.9-1.5 that's 3x taller.


Thanks for making this!
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Relentas
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(Original post by Labrador99)
Thanks for making this!
Yea, I just got a base down of things I think I may know. I didn't check over multi choice and missed a few questions but it should be fine for a start, if anyone wants to correct me they can just reply with the correction
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AHBM
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Hi think you have several mistakes in your multiple choice answers.
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leavemeblank
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Hi, I've just noticed a few different things and got numerical answers to some of the different calculations below, let me know if you disagree.

9 should be A, 10 should be D, 12 should be A, 25 should be C.

1a) sqa probably would accept 5th/7th/9th electron. c) gives K = 3.2 x 10^-5

2a) Order of a reaction is the power to which a concentration is raised in the rate equation.
2c) In Hess' law style from Higher, gives: H2O2 + 2I- + 2H3O+ + I2 + 4H2O

3b)i) Distilled/deionised water is used to set a colorimeter
b)ii) Agreed with you that its 35.5% - think SQA would accept that 35.5% or 71% because the wording lacks a bit of clarity.

4aii) HO2- b) genuinely ahahah
c) ii) A - 0.08 mol l-1 B - pH = 1.95

5bii) This has nothing to do with d shells / special stability as when a transition metal is in a complex, the d-orbitals are no longer degenerate - question does specify that this question is about transitions. (This is according to my teacher) Hence I think they'd be looking for something like stating that d-orbitals are no longer degenerate and therefore when electrons transition from the lower orbitals to the higher orbitals, the energy absorbed does not correspond to a wavelength in the visible spectrum.
c) i) E = 2.04 x 10^-16 J ii) 18.96 eV

6a) I got 10.5 mol l-1. Question is very similar to one from 2016 or 2017 paper about bleach.
6c) could also talk about the indicator being faulty or there being a high uncertainty in the end point.

7b)i) When atomic orbitals overlap sideways. ii) When two p orbitals and an s orbital merge to form 3 hybrid orbitals.
7c) Probably looking for a statement like 'electrons delocalised in a conjugated system'.
7d)ii)(C) (I) 0.029m (II) 4.065 x 10-3 kJ mol-1

9a)ii) chiral molecules that are non-superimposable mirror images of eachother

10bii) 2-methoxy-2-methylpropane
10ciii) Provides a more stable carbocation due to the inductive stabilisation of the carbocation.
10d) Mirror image wouldn't be optically active. Butan-2-ol however is an isomer that is chiral and hence optically active.
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(Original post by AHBM)
Hi think you have several mistakes in your multiple choice answers.
"I didn't check over multi choice" I just copied the person who posted multi choice's answers, I know there are some mistakes but I've not had time to go through it.
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I am going to bolden everything I agree with:
(Original post by leavemeblank)
Hi, I've just noticed a few different things and got numerical answers to some of the different calculations below, let me know if you disagree.

9 should be A, 10 should be D, 12 should be A, 25 should be C.

1a) sqa probably would accept 5th/7th/9th electron. c) gives K = 3.2 x 10^-5

2a) Order of a reaction is the power to which a concentration is raised in the rate equation.
2c) In Hess' law style from Higher, gives: H2O2 + 2I- + 2H3O+ + I2 + 4H2O

3b)i) Distilled/deionised water is used to set a colorimeter
b)ii) Agreed with you that its 35.5% - think SQA would accept that 35.5% or 71% because the wording lacks a bit of clarity.

4aii) HO2- b) genuinely ahahah
c) ii) A - 0.08 mol l-1 B - pH = 1.95

5bii) This has nothing to do with d shells / special stability as when a transition metal is in a complex, the d-orbitals are no longer degenerate - question does specify that this question is about transitions. (This is according to my teacher) Hence I think they'd be looking for something like stating that d-orbitals are no longer degenerate and therefore when electrons transition from the lower orbitals to the higher orbitals, the energy absorbed does not correspond to a wavelength in the visible spectrum.
c) i) E = 2.04 x 10^-16 J ii) 18.96 eV

6a) I got 10.5 mol l-1. Question is very similar to one from 2016 or 2017 paper about bleach.
6c) could also talk about the indicator being faulty or there being a high uncertainty in the end point.

7b)i) When atomic orbitals overlap sideways. ii) When two p orbitals and an s orbital merge to form 3 hybrid orbitals.
7c) Probably looking for a statement like 'electrons delocalised in a conjugated system'.
7d)ii)(C) (I) 0.029m (II) 4.065 x 10-3 kJ mol-1

9a)ii) chiral molecules that are non-superimposable mirror images of eachother

10bii) 2-methoxy-2-methylpropane
10ciii) Provides a more stable carbocation due to the inductive stabilisation of the carbocation.
10d) Mirror image wouldn't be optically active. Butan-2-ol however is an isomer that is chiral and hence optically active.
Reasoning: (I may be wrong)
Why I disagree with MC:10; I didnt fully know mid exam but my logic was the temperature is measured in kelvin, the higher the temperature, the more likely things are to react. If the reaction is to be feasible below 300k then it surely must also be feasible when its 0K (absolute 0) where theres no movement of the reactants.
MC12; I agree as I remember my teacher saying it wasnt B. :P
MC25; I agree as my teacher said it was C.

Paper 2:
I disagree with 1a, it's possible they will but I assumed the it was only the 5th electron as I know in a p shell the quantum number m can equal -1,0,1 I assigned each box one of these numbers each in order. so the one that would match -1 is the 3rd box overall. I may be wrong but that was my logic. - I'll add a warning in the mark scheme.

3bi) I disagree, you need a small concentration of nitric acid. (Since you use nitric acid to dissolve the screw.) your solution used to zero should be the exact same as your test solution but without the ions. I may be wrong, but I learnt while doing my biology project that the solution used should be the same without the thing you're testing. "Reagent blank is used in order to cancel out or zero the absorbance of all the other components in the sample except the component whose absorbance is to be measured."

bii) I agree, poorly worded/asked questions throughout the paper, I'm sure the answer is 35.5 and they wont accept anything else.

4aii) thanks I cant read that :P
b)dont imagine too many got this mark...

6a) I got no clue tbh. I'm usually quite good at these calculation questions but this one was just a bit mean.
6c) they will probably accept a multitude of answers for this, some they may forget to put down and you'll lose a mark for for no reason, but hey.

7bi) horizontally, sideways, same thing. If you've got a diagram you'll get the mark for sure.
c) they usually look for the rant about how length of conjugated system effects the colour released, transition from HOMO to LUMO etc. There are some questions like this one in past papers and the mark scheme usually looked for this.

10bii) no clue. I had no idea how to name this.
10cii) theres a few things you can say, according to my teacher since its just a 1 marker, stating its a tertiary haloalkane may be good enough.
10d) I had no clue for this one. I didn't even realise there was no chiral center. xD

I'll update the mark scheme.
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(Original post by Relentas)
I am going to bolden everything I agree with:

Reasoning: (I may be wrong)
Why I disagree with MC:10; I didnt fully know mid exam but my logic was the temperature is measured in kelvin, the higher the temperature, the more likely things are to react. If the reaction is to be feasible below 300k then it surely must also be feasible when its 0K (absolute 0) where theres no movement of the reactants.
MC12; I agree as I remember my teacher saying it wasnt B. :P
MC25; I agree as my teacher said it was C.

Paper 2:
I disagree with 1a, it's possible they will but I assumed the it was only the 5th electron as I know in a p shell the quantum number m can equal -1,0,1 I assigned each box one of these numbers each in order. so the one that would match -1 is the 3rd box overall. I may be wrong but that was my logic. - I'll add a warning in the mark scheme.

3bi) I disagree, you need a small concentration of nitric acid. (Since you use nitric acid to dissolve the screw.) your solution used to zero should be the exact same as your test solution but without the ions. I may be wrong, but I learnt while doing my biology project that the solution used should be the same without the thing you're testing. "Reagent blank is used in order to cancel out or zero the absorbance of all the other components in the sample except the component whose absorbance is to be measured."

bii) I agree, poorly worded/asked questions throughout the paper, I'm sure the answer is 35.5 and they wont accept anything else.

4aii) thanks I cant read that :P
b)dont imagine too many got this mark...

6a) I got no clue tbh. I'm usually quite good at these calculation questions but this one was just a bit mean.
6c) they will probably accept a multitude of answers for this, some they may forget to put down and you'll lose a mark for for no reason, but hey.

7bi) horizontally, sideways, same thing. If you've got a diagram you'll get the mark for sure.
c) they usually look for the rant about how length of conjugated system effects the colour released, transition from HOMO to LUMO etc. There are some questions like this one in past papers and the mark scheme usually looked for this.

10bii) no clue. I had no idea how to name this.
10cii) theres a few things you can say, according to my teacher since its just a 1 marker, stating its a tertiary haloalkane may be good enough.
10d) I had no clue for this one. I didn't even realise there was no chiral center. xD

I'll update the mark scheme.
Feel like a tool now, Q10 should be C (got the knowledge just cant read lmao). Course spec states that "Under non-standard conditions any reaction is feasible if ΔG is negative" - so hence the reaction is feasible at 300K and above. I have no clue with Q8 whether its C or D - any reasoning would be appreciated, similarly with Q26 although I put B in the exam, quite a few people on the thread think it could be A?

For the very first question, SQA have tended to not be very specific and accept a wide range of answers in the quantum numbers questions - hence I think they wouldn't mind as long as a 2p electron with a positive spin was circled.
3bi - have to disagree. Distilled water always has an absorbance of 0, did this in my project and my teacher agrees too.
7c - Forgot to finish off! Meant to say because electrons are delocalised they can absorb energy and be promoted from HOMO to LUMO, etc and light from the visible spectrum being absorbed means complementary colour is transmitted.
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Relentas
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(Original post by leavemeblank)
Feel like a tool now, Q10 should be C (got the knowledge just cant read lmao). Course spec states that "Under non-standard conditions any reaction is feasible if ΔG is negative" - so hence the reaction is feasible at 300K and above. I have no clue with Q8 whether its C or D - any reasoning would be appreciated, similarly with Q26 although I put B in the exam, quite a few people on the thread think it could be A?

For the very first question, SQA have tended to not be very specific and accept a wide range of answers in the quantum numbers questions - hence I think they wouldn't mind as long as a 2p electron with a positive spin was circled.
3bi - have to disagree. Distilled water always has an absorbance of 0, did this in my project and my teacher agrees too.
7c - Forgot to finish off! Meant to say because electrons are delocalised they can absorb energy and be promoted from HOMO to LUMO, etc and light from the visible spectrum being absorbed means complementary colour is transmitted.
Question 8 is D as there are 4 bonds on the copper ion, that means it has a coordination number of 4, - C or D, there is only 1 ligand surrounding the copper, this has 4 bonds coming from it meaning its tetradentate (tetra being 4 and dentate being bonds)
Its a thing you should just have been taught when learning about ligands.

26: this question was a trick question.
Definition of an agonist: enhances the body's normal response
Definition of antagonist: blocks the body's normal response.
Pramipexole acts like a natural compound in the body (dopamine), it stimulates nerve cells. - this hence is an agonist as it's stimulating the body and enhancing it's normal response.
Buprenophine is a trick question here, it says: it's a drug used to treat heroin addiction, it stimulates the receptors BUT it produces less of a response COMPARED to heroin. This drug is also an agonist as it's stimulating the body and enhancing it's natural response, yes, compared to heroin it's not as good as stimulating the body but it's still stimulating it and hence it has to be an agonist.
Both agonist = A.

3bi) Im still sticking with it, I'll ask my teacher when I go in for psychology...
I did a biology project where I tested how bleach effects growth of bacteria. The more bacteria grew, the higher the absorbance (in theory.) I had solutions of broth, bleach and bacteria. I was only interested in measuring the absorbance of the bacteria however so everything else had to be the same. (this broth, bleach and bacteria solution is comparable to water, nitric acid and copper).
First time round I tried to zero the colorimeter with water, this gave me values that werent even able to be read on the colorimeter as the broth solution was contributing to the absorption, the technician then called me an idiot and told me I needed to use broth solutions when measuring the colorimeter value as it had to be the same.
I then tested it again with broth, found that the more bleach added, the higher the absorbance (which would indicate the more bleach I added, the more bacteria grew which is just illogical.) I then asked again and they called me an idiot as the solution used to zero had to be the exact same as the solution being tested, otherwise you're just testing the absorbance of the bleach.
They add nitric acid to the flask along with distilled water, if nitric acid has an absorbance (which I'm unsure if it does) then it'd be measuring the absorbance of the nitric acid and the copper ions. This would effect the standard graph, giving you different values then what they should be.

7) I'll update.
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