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A2 OCR Chemistry B F334 - June 2015

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Original post by misteltain
That paper is the reason I am doing a third year lmfao

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My friend was gonna resit as well. It is a pretty nasty exam tbh i especially they threw a lot of people at beginning of the paper.
Original post by chococup123
Can someone explain to me why the formula of the complex ion in june 2014 Q1 (d)(ii) is (Fe(C6H507))-??
don't get the negative 1


I think it's because the charge on the ligand is -3 and the charge on the iron ion is +2, giving the complex an overall charge of -1


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Original post by tinoah
There are no delta positive atoms in line with them to form H bonds.
As a general rule you should just remember that:
Polyester polymer chains= pd-pd
Polyamide polymer chains= H bonds
and then for polymers like poly(ethene) they'll have id-id



'There are no delta positive atoms in line with them to form H bonds.'

I dont understand that.

I get it now that if there are no amide bonds- its a polyester so it will have Pd-Pd and i forgot to check for amide bonds in this question but still about the delta postive in line..

like cant the O at the end on the lhs bond with the delta H of another which is connected to the O of that monomer as the O of the second monomer will make the H delta positive.
Reply 223
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Daisybw
Original post by docdoc02
could i have help with the last part of this q
i thought the rate of formation = rate of disappearance but apparently it is not :/


2 moles of iodide ions are needed to make one mole of iodine, so the rate of disappearance is double the rate of formation
Original post by MathsAddict
My friend was gonna resit as well. It is a pretty nasty exam tbh i especially they threw a lot of people at beginning of the paper.


LOL, same here -just looked around 5mins into the exam to see wtf looks on everyone's face..:eek4:...;P

Anyways, I'm sorry to you & everyone else using my notes, cos there was a minor error about E cells that I made; here's the revised version! (-Couldn't edit original attachment for some reason so deleted that post:P)
Chem A2 -edited.docx145.1KB
Original post by suyoof123
'There are no delta positive atoms in line with them to form H bonds.'

I dont understand that.

I get it now that if there are no amide bonds- its a polyester so it will have Pd-Pd and i forgot to check for amide bonds in this question but still about the delta postive in line..

like cant the O at the end on the lhs bond with the delta H of another which is connected to the O of that monomer as the O of the second monomer will make the H delta positive.


Polyesters have no hydrogen atoms attached to a highly electronegative element (O, N or F), so you won't get hydrogen bonding. No hydrogen atoms in polyesters will have a slightly positive charge in other words, which is needed for hydrogen bonding.
Original post by tealover96
Dear Ocr salters B,
I know I've said that I hate you many,many times over the past 2 years
And that I wish I was doing Ocr A instead
I didn't meant any of that
Please be nice tomorrow (and I mean actually nice, not 'salters nice':wink:.
Thanks

On another note...can someone explain naming amines to me please!
Because sometimes it's 'x'-yl amine, and other times it is -amino-'x'


OCR use the amino-X way of naming for amines, I'd stick to that.
Reply 227
Avatar for tinoah
tinoah
1
Original post by suyoof123
'There are no delta positive atoms in line with them to form H bonds.'

I dont understand that.

I get it now that if there are no amide bonds- its a polyester so it will have Pd-Pd and i forgot to check for amide bonds in this question but still about the delta postive in line..

like cant the O at the end on the lhs bond with the delta H of another which is connected to the O of that monomer as the O of the second monomer will make the H delta positive.


I didn't word that well but what I meant was that when the polymer chains are lined up, the C=O bonds at the end of the chain will be in line of each other meaning there is no H for the lone pair on O to bond with.


polyester (4College).png
Original post by Anonymous696061
LOL, same here -just looked around 5mins into the exam to see wtf looks on everyone's face..:eek4:...;P

Anyways, I'm sorry to you & everyone else using my notes, cos there was a minor error about E cells that I made; here's the revised version! (-Couldn't edit original attachment for some reason so deleted that post:P)


Thanks a lot XD even if they do a bad one tbh there has to be someway to do it its just not as a direct answer as usual.

I am adding thread of life notes on to yours after i finish these last two papers.
Original post by docdoc02
could i have help with the last part of this q
i thought the rate of formation = rate of disappearance but apparently it is not :/


its double the rate of formation because of the molar ratio. 2I- go to 1 mole of iodine
Original post by BrokenS0ulz
Polyesters have no hydrogen atoms attached to a highly electronegative element (O, N or F), so you won't get hydrogen bonding. No hydrogen atoms in polyesters will have a slightly positive charge in other words, which is needed for hydrogen bonding.


I forgot that the O is attached as the center of the ester bond in the formed polymer so it attaches to carbons on each side :smile:

Thank you!!!!!
Original post by tinoah
I didn't word that well but what I meant was that when the polymer chains are lined up, the C=O bonds at the end of the chain will be in line of each other meaning there is no H for the lone pair on O to bond with.


polyester (4College).png

Is it due to that or the fact that there are no delta H+ because none of the H's are bonded to an electronegative element such as N/O to make it delta H+
image.jpg
Just the bond angles, I always forget them somehow
january 2010 quest 5 c ii calculation how are you meant to implement half life in this?
Reply 234
Avatar for tinoah
tinoah
1
Original post by suyoof123
Is it due to that or the fact that there are no delta H+ because none of the H's are bonded to an electronegative element such as N/O to make it delta H+


The second one, my brain is just not functioning properly (which is great the night before an exam)
So I'm not ready.
#gettinganE
who is with me?
June 12 4g- I don't get how you do the first step. The mark scheme says:
If 0.6625g of Na2CO3 is made into a 1dm^3 solution (and Mr is 106) then why is the conc (0.6625/106)moldm^3

Anyone explain/?
Original post by tinoah
The second one, my brain is just not functioning properly (which is great the night before an exam)


^^
Original post by Welbeck
June 12 4g- I don't get how you do the first step. The mark scheme says:
If 0.6625g of Na2CO3 is made into a 1dm^3 solution (and Mr is 106) then why is the conc (0.6625/106)moldm^3

Anyone explain/?


Is there something I'm not seeing? If you know the grams and the Mr and given that its for 1dm^3, then you can work out the conc in moldm^3 by doing mass/mr?

Can someone explain this to me please
Original post by BrokenS0ulz
OCR use the amino-X way of naming for amines, I'd stick to that.


Okay thanks
Also,
Jan 11 3b
C3H7NH2
Answer: propylamine/1-aminopropane do not allow: aminopropane
I'm not sure why aminopropane would be wrong?

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