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F334 (19/06/2013 9:00AM) - Salters Chemistry (Chemistry of Materials) Revision Thread

Look for C=O as this is the main functional group that is used to represent perm-dipole-dipoles. You are right in saying that there may be an electronegative atom present, but beware that if it is bonded to hydrogen it hydrogen bonds instead!

thanks alot, I think I feel a abit more confident now :P Yeah I'm okay at recognising H bonds

I've lost my AS book, can anyone summarise acid base titrations?

Reply 181

Tikara

Original post by Dude Ranch

Can someone please explain the whole of question 1h) on the june 2012 paper. Ill thumbs up you!!

do you mean 2h?

Reply 182

stealth_writer

4

Original post by Tikara

whats the question? if it was the end of jan 13 I'm going through the paper now :P

Well since your talking about Jan 13, then how do you do the following:

Q2 b(ii), c(iii) (surely three moles are made?) and the very final question about rate of Iodine disappearing or what not

cheers!

Reply 183

Dude Ranch

2

Original post by Tikara

do you mean 2h?

Yeah sorry 2h

Reply 184

stealth_writer

4

Also guys... How do you dilution type Q's on these papers?

Reply 185

Dualcore

Original post by stealth_writer

Also guys... How do you dilution type Q's on these papers?

Any specific questions?

Reply 186

stealth_writer

4

Help with Q 2 (b) (ii) please guys

Reply 187

stealth_writer

4

Original post by Dualcore

Any specific questions?

Help with Q 2 (b) (ii) Jan 13 please

Reply 188

MTI

Original post by Dude Ranch

Can someone please explain the whole of question 1h) on the june 2012 paper. Ill thumbs up you!!

Original post by Dude Ranch

Yeah sorry 2h

h) i) from fiii we know that the ligand H₂NCH₂CH₂NH₂ is a bidentate ligand because the lone pairs on the nitrogens form 2 dative bonds(if you need more info on this let me know) in the question is gives the formula of the complex as [Cr(H₂NCH₂CH₂NH₂)₃] so there are 3 ligands each forms 2 bonds with the Cr therefore the coordination number is 2x3=6. coordination number is the number of dative bonds to the central ion in the complex

h) ii) and iii) not sure how to explain this but if you look at the mark scheme it might help. basically it just the normal octahedral shape with bond angle of 90 because there are 6 sets of electrons around the Cr however 2 of the bonds come from one ligand molecule/2 bonds are kinda linked. like a loop...

h) iv) sorry i didnt understand this myself :/

Reply 189

Dude Ranch

2

Original post by MTI

h) i) from fiii we know that the ligand H₂NCH₂CH₂NH₂ is a bidentate ligand because the lone pairs on the nitrogens form 2 dative bonds(if you need more info on this let me know) in the question is gives the formula of the complex as [Cr(H₂NCH₂CH₂NH₂)₃] so there are 3 ligands each forms 2 bonds with the Cr therefore the coordination number is 2x3=6. coordination number is the number of dative bonds to the central ion in the complex

h) ii) and iii) not sure how to explain this but if you look at the mark scheme it might help. basically it just the normal octahedral shape with bond angle of 90 because there are 6 sets of electrons around the Cr however 2 of the bonds come from one ligand molecule/2 bonds are kinda linked. like a loop...

h) iv) sorry i didnt understand this myself :/

Ah thanks so much its so simple now!

Reply 190

hannah1994

7

If a polymer is stronger, does it have a higher or lower Tg??
Please can someone explain this too!!

Reply 191

stealth_writer

4

Guys can someone help me out with jan 12 http://www.ocr.org.uk/Images/79818-question-paper-unit-f334-chemistry-of-materials.pdf

Q 1 d (iii) (iv).Im struggling with mathsy titration type Q's and its really frustrating.

Reply 192

Courtz

Original post by stealth_writer

Well since your talking about Jan 13, then how do you do the following:

Q2 b(ii), c(iii) (surely three moles are made?) and the very final question about rate of Iodine disappearing or what not

cheers!

I'm stuck on the rate of disappearing too! I've never seen it come up before that paper...

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Reply 193

super121

5

Original post by hannah1994

If a polymer is stronger, does it have a higher or lower Tg??
Please can someone explain this too!!

Higher, because the chains are closer together so form stronger intermolecular bonds

Reply 194

Tikara

Original post by Dude Ranch

Ah thanks so much its so simple now!

glad someone else answered :P Just finished working through the question

iv) is your stock enantiomer question - if it was monodentate then it wouldn't have this but due to the bidentate nature if you mirror the image it's nonsuperimposable (not the same when put on top of each other)

my problems in that question were that I put diaminoethane rather than 1,2-diaminoethane
and didn't know that electronegativity is only in a bond grrrr

Reply 195

Tikara

Original post by stealth_writer

Help with Q 2 (b) (ii) Jan 13 please

I'll have a look now- but OCR number questions alllways catch me out but I'll have a go haha

Reply 196

Dualcore

Original post by stealth_writer

Help with Q 2 (b) (ii) Jan 13 please

Okay to start, 16cm3 of the NaOH reacted therefore using n=cv (16/1000)*0.25=4x10^-3 mol of NaOH reacted, it is a one to one ratio so 4x10^-3 mol of the acid reacted, this is was in 25cm^3 of the diluted acid solution.

From here you can take 2 approaches
1) you could work out the concentration of the diluted solution and then scale it:
using c=n/v c=4x10^-3/(25/1000)=0.16moldm^-3, you know the ratio between the volumes of the diluted and undiluted solutions 250/14=17.85..., so the original solution is 17.85... time more concentrated so the original concentration is 17.85...x0.16=2.86moldm^-3

2) Or another way of doing it is this, you know that number of moles of the acid in 25cm^3 solution 4x10^-3, so in 250cm^3 diluted solution you got the 25cm^3 from must contain 10 times as much moles of acid 4x10-3x10=4x10^-2, this must contain the same number of moles of acid as the diluted solution because no moles of solution is lost, so the original concentration must be this number of moles over the original volume 4x10^-2/(14/1000)=2.86moldm^-3

I'm assuming you can take it from here, give us a shout if you need further help :biggrin:

Reply 197

Mtheodore

hey can someone upload the Jan 2013 Paper please?

I got the mark scheme if anyone needs it

F334 Jan 2013.pdf97.4KB

Reply 198

MTI

Original post by Dude Ranch

Ah thanks so much its so simple now!

Original post by Tikara

glad someone else answered :P Just finished working through the question

iv) is your stock enantiomer question - if it was monodentate then it wouldn't have this but due to the bidentate nature if you mirror the image it's nonsuperimposable (not the same when put on top of each other)

my problems in that question were that I put diaminoethane rather than 1,2-diaminoethane
and didn't know that electronegativity is only in a bond grrrr

your welcome :smile:

and ah i see iv now!