Unofficial Mark scheme: Advanced Higher Chemistry SQA Paper 2019
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:
Paper 2:
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
Paper 2:
Spoiler
(edited 4 years ago)
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/showthread.php?t=5600320&p=83204310&page=11
^ and Page 7 has multi choice.
https://www.thestudentroom.co.uk/showthread.php?t=5600320&p=83204310&page=11
^ and Page 7 has multi choice.
Thanks for making this! 
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.
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.
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.
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.
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.
(edited 4 years ago)
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.
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.
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.
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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