# Chem alevel help

Guys does anyone understand why in question 8b you have to divide the mol by 2 to work out Y? https://revisionscience.com/sites/revisionscience.com/files/imce/9CH0_02_que_20211013.pdf

working out:
26.10/1000 x 0.320 = 8.352 x 10–3 (mol)
8.352 x 10–3 / 2 = 4.176 x 10–3 (mol) (this one why we dividing?
4.176 x 10–3 x (250/25) = 4.176 x 10–2 (mol)
4.34 / 4.176 x 10–2 = 103.9 / 104 (g mol–1

Scroll to see replies

because Y is a dicarboxylic acid, 2 H+ can dissociate from it to react with the OH from NaOH. 1 mol of Y will therefore react with 2 mol of NaOH in neutralisation titration, so you need to divide the mols of NaOH by 2 to find mols of Y
Original post by jjeeeeeea
because Y is a dicarboxylic acid, 2 H+ can dissociate from it to react with the OH from NaOH. 1 mol of Y will therefore react with 2 mol of NaOH in neutralisation titration, so you need to divide the mols of NaOH by 2 to find mols of Y

Thank you so muchc this made so much sense!!! Can you help me with 8a as well please? I would really appreciate the help
Original post by tasneem.016
Thank you so muchc this made so much sense!!! Can you help me with 8a as well please? I would really appreciate the help

of course, I tried it and did this working, you just need to find the ratio of moles of CO2 to H2O and you can figure out what goes inbetween the 2 COOH groups by findings the ratio of carbon to hydrogen, hope this makes sense lmk if anything is too vague
Original post by jjeeeeeea
of course, I tried it and did this working, you just need to find the ratio of moles of CO2 to H2O and you can figure out what goes inbetween the 2 COOH groups by findings the ratio of carbon to hydrogen, hope this makes sense lmk if anything is too vague

Thanks for answering ur helping me so much! Question: how did you know the empirical formula of oxygen is therefore 4? Btw can i ask you more chem questions i completely understand if you dont want to. Im finding chem so hard rn and just need someones helo
Original post by tasneem.016
Thanks for answering ur helping me so much! Question: how did you know the empirical formula of oxygen is therefore 4? Btw can i ask you more chem questions i completely understand if you dont want to. Im finding chem so hard rn and just need someones helo

it's because you've got 2H2O, so because there are 2 molecules of water and each molecule contains 2 hydrogens you can use 2x2=4 to find that there are 4 hydrogens.
sure you can ask anything I'll try to answer anything I can, I need motivation to revise chemistry anyway lmao so this is great
Original post by jjeeeeeea
it's because you've got 2H2O, so because there are 2 molecules of water and each molecule contains 2 hydrogens you can use 2x2=4 to find that there are 4 hydrogens.
sure you can ask anything I'll try to answer anything I can, I need motivation to revise chemistry anyway lmao so this is great

LOl which year are you in? Im assuming ur in uni?
Original post by tasneem.016
LOl which year are you in? Im assuming ur in uni?

im in y13 but hoping to do chem at uni next year
Original post by jjeeeeeea
im in y13 but hoping to do chem at uni next year

can u explain 6c for me plsss . also im impressed chem at uni is tough but based on how ur answering these qs i know u will do well
Original post by tasneem.016
can u explain 6c for me plsss . also im impressed chem at uni is tough but based on how ur answering these qs i know u will do well

and i mean 6cii
Original post by tasneem.016
can u explain 6c for me plsss . also im impressed chem at uni is tough but based on how ur answering these qs i know u will do well

aw thanks thats really nice of you i hope so!! here's how I worked that one out
Original post by jjeeeeeea
aw thanks thats really nice of you i hope so!! here's how I worked that one out

btw for devise reaction pathways questions how would u answer them? Those questions have always been a struggle
Original post by tasneem.016
btw for devise reaction pathways questions how would u answer them? Those questions have always been a struggle

I made flashcards for all the reaction mechanisms with conditions/reagents you need and just look at it sometimes.
Not sure how much mechanism detail you need for Edexcel bc I do OCR but I looked up the mechanisms for even the reactions that you don't need to know them for, just to at least get a vague understanding of why the reaction works and that way it's not solely memory, makes it SO much easier eventually. Flashcards are the way to go though 100% even if you don't bother with the unneeded mechanisms
Once you get those solid in your mind you just have to try your best to apply them to whatever you're given
Original post by jjeeeeeea
I made flashcards for all the reaction mechanisms with conditions/reagents you need and just look at it sometimes.
Not sure how much mechanism detail you need for Edexcel bc I do OCR but I looked up the mechanisms for even the reactions that you don't need to know them for, just to at least get a vague understanding of why the reaction works and that way it's not solely memory, makes it SO much easier eventually. Flashcards are the way to go though 100% even if you don't bother with the unneeded mechanisms
Once you get those solid in your mind you just have to try your best to apply them to whatever you're given

Thanks for the tip! cramming those in tonight!
Original post by jjeeeeeea
aw thanks thats really nice of you i hope so!! here's how I worked that one out

also can u plsss explain q 9a how do u know peaks in benzene rings?
Original post by tasneem.016
also can u plsss explain q 9a how do u know peaks in benzene rings?

sorry only just saw your msg, but you're looking for the number of carbon environments in each molecule. The second one has 6 carbon environments because there is no symmetry in the molecule, no carbon in there is connected to exactly the same things as another carbon in there.

For the first one, there is a line of symmetry down the middle, meaning there are 4 carbon environments. One directly bonded to NO2 and one directly to OH, the two I coloured yellow and red here are the same carbon environment because of symmetry
Original post by jjeeeeeea
sorry only just saw your msg, but you're looking for the number of carbon environments in each molecule. The second one has 6 carbon environments because there is no symmetry in the molecule, no carbon in there is connected to exactly the same things as another carbon in there.

For the first one, there is a line of symmetry down the middle, meaning there are 4 carbon environments. One directly bonded to NO2 and one directly to OH, the two I coloured yellow and red here are the same carbon environment because of symmetry

ohhh so with carbon environments u need to see the symmetry in order to understand how many there are, that makes senseee. can u explain 9b and 1a pls and i wont ask anymore qs related to this paper. Thank u for ur helppp
Original post by tasneem.016
ohhh so with carbon environments u need to see the symmetry in order to understand how many there are, that makes senseee. can u explain 9b and 1a pls and i wont ask anymore qs related to this paper. Thank u for ur helppp

So for the first question you're given mass and Mr, so you can use Mr = m/n to find the number of moles (26.4/132 = 0.2 moles), avogadro's constant tells you the number of particles in one mole of anything so use 0.2 x 6.0x10^23 to find the number of ammonium sulfate particles, 1.2x10^23.
In one molecule ammonium sulfate there are two NH4+ and one SO42-, so 3 ions. For 1.2x10^23 molecules we therefore have (1.2x10^23 multiplied by 3) 3.6x10^23 ions so the answer is D

For 9b I wrote it out here, again you need to use Mr = m/n and find mole values, you know that the moles of phenol is the same as the moles of paracetamol because in the equation one molecule of phenol can eventually form one mole of paracetamol. Hope this makes sense? Lmk if not
oops sorry!! I just realised question 9 tells you the yields in the question, I did it wrong, will correct it in a second
Original post by jjeeeeeea
oops sorry!! I just realised question 9 tells you the yields in the question, I did it wrong, will correct it in a second

no worries ur honestly amazing for helping me out x