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AQA A level Chemistry Paper 1 5th June 2018 Unofficial Markscheme

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Original post by Stevennick123
what was the one about the s block element with the highest first ionisation energy ?
and wasnt the isomerism EZ beacuse they were unidentate ligands not bidentate


- I put lithium. Since it's in group 1, the second IE removes a 1s electron which is very strongly attracted to the nucleus.

EDIT: Lithium is wrong. The correct answer is beryllium. I misread the question; it was about the 1st IE not the 2nd IE.

- Correct about the isomerism.
(edited 5 years ago)
Original post by uhasan298
Who got berillium


Looking at graphs online berrilyum is right
Original post by 2Anon1234
The partial pressures question was 50kpa 75kpa and 25kpa. As 0.8 moles was the 2NH3, so 3H2 WAS 1.2 and NH2 was 0.4 Then do each one divided by total moles (2.4) and times by 150kpa??


0.8 moles was the mole fraction not moles of NH3, misread the question.
How many marks was it worth? 3 for the question and 2 for calculating Kp?
Original post by brendonurine
It was pretty maths


I just the correct formula but wrote down 87.7 as 87.3 in the beginning, do you think I’d get any marks? My answer was correct for 87.3, so my working was definitely correct


no idea tbh.
Original post by Ali Bomaye!
I got X=5


same but most people got 1
Original post by 2Anon1234
The partial pressures question was 50kpa 75kpa and 25kpa. As 0.8 moles was the 2NH3, so 3H2 WAS 1.2 and NH2 was 0.4 Then do each one divided by total moles (2.4) and times by 150kpa??

No p.p of product= 120
Now 150(total ) - 120=30
You have 1:3 so for 1st reactant its 30/4
And for other one its 30-( 30/4 )
for partial pressure, why isnt is 60, 120? etc
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Original post by ROadmanKys
i got 15% for x so the 88 isotope would get 70%. how did u get 80%? someone tell me, the ram was 87.7 and isotope 86,87 and 88.


I checked the answer of 80 at the end and it was right. You had to go backwards but instead multiply the 86 and 87 each by x, as it’s a 1:1 ratio so we know they are by the same percentage, and then the 88 by y. Because it’s a percentage we then know that y is 100-2x so you substitute that in and work out x which gives you 10, so you can work out the remaining percentage which is 80, if that makes sense!
Original post by Bulletzone
This is my first 10 page thread :biggrin:

:whoo:


10 pages and it doesn't serve its purpose, you're 'mark scheme' is all over the place and absolute nonesense. Other users created proper mark scheme threads but thanks to you, quality stuff was deleted for this bs.
did anyone get 88.9 % abudance
Original post by 2Anon1234
The partial pressures question was 50kpa 75kpa and 25kpa. As 0.8 moles was the 2NH3, so 3H2 WAS 1.2 and NH2 was 0.4 Then do each one divided by total moles (2.4) and times by 150kpa??

so you think a mole fraction can be more than 100% ?
it didnt give u moles it gave you the fractions.
Would I lose 6 marks for saying atomic radius instead of ionic radius in the bonding 6 marker?
Original post by Tommy59375
I put lithium. Since it's in group 1, the second IE removes a 1s electron which is very strongly attracted to the nucleus.


Original post by brendonurine
Looking at graphs online berrilyum is right


http://wwwchem.uwimona.edu.jm/courses/CHEM1902/IonizEnergy.jpg Notice the orange line (2nd IE) peaks at lithium. The question wanted an S block element, not necessarily a group 2 element.

EDIT: Apparently the question was asking for the 1st IE. d'oh!!
(edited 5 years ago)
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Original post by LunaCat
Yh i put that


I was unsure about this one, but I swapped the stat symbols round for the left side
Original post by Anonymouspsych
You were literally so close to the answer then. After you work out moles of Na2Co3.xH20 you can work out Mr of it because you know the mass.
Then subtract the Mr of Na2CO3 from that Mr you just calculated and you'll get 18. You know Mr of water is 18 so x must be 1


How many marks would i get outta 7
Original post by Aqsa11
Guys did anyone find it super hard or super easy? Btw for the partial pressures what did everyone get?


7.5 and 22.5
Original post by 2Anon1234
The partial pressures question was 50kpa 75kpa and 25kpa. As 0.8 moles was the 2NH3, so 3H2 WAS 1.2 and NH2 was 0.4 Then do each one divided by total moles (2.4) and times by 150kpa??


NH3 was 120KPa (0.8*150KPa) so the rest of the gases must've occupied the other 0.2 mol fraction.

it was a 1:3 ratio so the 30KPa 'remaining' would've been 22.5KPa and 7.5KPa (forg
(edited 5 years ago)
Original post by 2Anon1234
The partial pressures question was 50kpa 75kpa and 25kpa. As 0.8 moles was the 2NH3, so 3H2 WAS 1.2 and NH2 was 0.4 Then do each one divided by total moles (2.4) and times by 150kpa??


Their moles changed by 1.2 and 0.4, that wasn’t their eqm moles. They started off in a ratio of 1:3, and continued to be in that ratio, so partial pressures 150-120=30, 30/4=7.5, 7.5x3=22.5
Original post by Benistonw11
How did you get the bond angle as 104.5Wasn’t it tetrahedral repelled into pyramidal by 1 lone pairI managed 107.5 if anyone else did

there were 2 lone pairs because you got 5 electrons from the N + 2 from 2H+ and 1 from the negative charge
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Original post by tora123
so you think a mole fraction can be more than 100% ?
it didnt give u moles it gave you the fractions.


Yeah it gave you the mole fraction, which always adds up to 1 I think!! And then the partial pressures should all add up to the total pressure to make sure

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