OCR A Chemistry F322 Chains, Energy and Resources Thu 19 Jan 2012

Yeah your most probably right about the alternatives. yup i got Butanoic acid!

for the question on the equlibrium for the catalyst i said stays same and is on right hand side?

and for the question on cfc is insoluble in water acceptable ?

Reply 581

faz_341

Original post by dee11

WOO! i got all of those

did u get -134 for the MxCXT question and 94% for percentage yeild?

Reply 582

dee11

Original post by sam06

If it was a ketone wouldn't it be pentan-3-one 'cause a ketone is always in the middle of a chain?

a ketone is one thats attatched to 2 alkyle groups and a haydrogen atom on the carcon with the C=O , so pentan-2-one should work too right?

Reply 583

dee11

Original post by faz_341

did u get -134 for the MxCXT question and 94% for percentage yeild?

yeah i did get -134, although i got 93%, howd' you get 94% did you round it up or something?

Reply 584

Eternal Rest

Wait, how come everyone's saying the mass spec compound was from a secondary alcohol?
I don't remember it saying anything about what type of alcohol it was, just that products such as carboxylic acids, ketones and aldehydes were formed. :s-smilie:

Reply 585

sam06

Original post by Eternal Rest

Wait, how come everyone's saying the mass spec compound was from a secondary alcohol?
I don't remember it saying anything about what type of alcohol it was, just that products such as carboxylic acids, ketones and aldehydes were formed. :s-smilie:

I don't remember it saying anything about the type of alcohol.. I think it tells you in part b not part a

Reply 586

faz_341

Original post by dee11

yeah i did get -134, although i got 93%, howd' you get 94% did you round it up or something?

as my calculator tells me it was 147000000/157082243 x 100 = 93.58 so i rounded up as it wanted it in 2sigfig

Reply 587

dee11

Original post by otrivine

for the question on the equlibrium for the catalyst i said stays same and is on right hand side?

and for the question on cfc is insoluble in water acceptable ?

yeah catalyst had no effect on the equilibrum position

i said CFC'S dont contain O-H groups so dont form hydrogen bonds with the water molecules and also because i dont think they're polar

Reply 588

otrivine

14

Original post by dee11

yeah catalyst had no effect on the equilibrum position

i said CFC'S dont contain O-H groups so dont form hydrogen bonds with the water molecules and also because i dont think they're polar

true

is insoluble in water still acceptable u think or not? shoot !!

Also, on that question about equation i wrote down three equations for the depletion one! how it happens how it works and final product

Reply 589

dee11

Original post by faz_341

as my calculator tells me it was 147000000/157082243 x 100 = 93.58 so i rounded up as it wanted it in 2sigfig

oh right, damn :s-smilie:

i didnt round up for some reason.
but i did right down the calculator display then wrote 93%

Reply 590

dee11

Original post by otrivine

true

is insoluble in water still acceptable u think or not? shoot !!

Also, on that question about equation i wrote down three equations for the depletion one! how it happens how it works and final product

yeah i think you'd get the mark!

ozone depletion you mean?

cl+o3 --> clo + 02
clo + o --> cl + o2
_______________
o3 + o --> 2o2

thats what you did right?

Reply 591

superdentist

Original post by -Simon-

I agree with this person.

P.s. I wrote 8 for the structural isomers, however, I am inclined to believe this is wrong.

Probably something like 12, 16 or 20. :smile:

I WROTE 8 too XD

Reply 592

superdentist

Original post by dee11

yeah i think you'd get the mark!

ozone depletion you mean?

cl+o3 --> clo + 02
clo + o --> cl + o2
_______________
o3 + o --> 2o2

thats what you did right?

do you lose marks fro including a termination step?

Reply 593

otrivine

14

Original post by dee11

yeah i think you'd get the mark!

ozone depletion you mean?

cl+o3 --> clo + 02
clo + o --> cl + o2
_______________
o3 + o --> 2o2

thats what you did right?

yh i got that for the final product and i wrote o2+o to give o3
o3 reversible o2+o
and then i got the same for final product ?
is that acceptable

Reply 594

superdentist

Original post by MEPLUS-->YOU

what did you put?

i put 8

Reply 595

sam06

Original post by dee11

a ketone is one thats attatched to 2 alkyle groups and a haydrogen atom on the carcon with the C=O , so pentan-2-one should work too right?

Yeah I guess it won't make a difference, 'cause it said suggest a possible structure for J so it could be either one I hope

Reply 596

dee11

Original post by superdentist

do you lose marks fro including a termination step?

think they'd ignore it as long as you wrote the first two steps
but there was a question asking about the termination steps want there, after this question

Reply 597

MEPLUS-->YOU

1

Did anyone get CH3OH+?
As peak?

Reply 598

faz_341

Original post by dee11

oh right, damn :s-smilie:

i didnt round up for some reason.
but i did right down the calculator display then wrote 93%

no worries wwhat about the radicals q was it
initiation:
Cl2---> 2Cl* (does UV need to be above the arrow for this????)
propagation:
Cl* +h2 --> HCl +H*
H* +Cl2 --> HCL + CL*
Termination
CL* +H* --> HCl (major)
Cl + Cl --> Cl2 (minor)

Reply 599

superdentist

Original post by dee11

think they'd ignore it as long as you wrote the first two steps
but there was a question asking about the termination steps want there, after this question

eurrh..dont remember , no i dont think so tbh

OCR A Chemistry F322 Chains, Energy and Resources Thu 19 Jan 2012

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