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Chemistry 9701/P22

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chiral centerx were 2 am sure,i wrote carbonyl group in last part.iz it ryt?plx tel me i thnk thea waz C=O group
 
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hassam said:
well wat was that empirical formula......18.9/16 : 81.1/56......ne we get the simplest ratio abt 1.22 which is approximately 1...so formula was BaO


Ahhh Hassam Mr of BArium is not 56.. 137.33.. 55.5(56) is for ferrous or what you call as Iron........
hence
81.1/137.33 =0.59
18.9/16 = 1.18

and 1.18(or 1.2)/0.59(or 0.6)
gives us 2:1
hence

BaO2
 
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BILALrox01 said:
Use ur god damned mind.. 0.04 is less then 0.1 How can u use -.6?? :crazy:
If it would have been 0.4 than it would have been bigger than it...if u ming using ur ming u will get to know that 0.1-0.004=0.06... :%) :%) :%) :%) :%)
 
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Yaar if naoh was added after equillibrium then y didnt it hydrolysed ester ?? I dont remember the question.
 
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m2rulz143 said:
Yaar if naoh was added after equillibrium then y didnt it hydrolysed ester ?? I dont remember the question.

Yea...I'm a bit doubtful about that too....

I just wrote that it wasn't ethanolic, So the reaction didn't occur (somethin like that..)
 
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m2rulz143 said:
Yaar if naoh was added after equillibrium then y didnt it hydrolysed ester ?? I dont remember the question.

Because NaOh was not in excess and whatever mols i.e. 0.045 mol and out of them 0.005 reacted with HCl and the other with Carboxyillic acid i.e. there was no excess mol in flask which stayed to hydolise th4e ester formed.
 
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yeah guyzz.. the paper was easy but still it wont have a too high GT.........
and Kc was 2.25 not 0.44.. :)
 
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hey guys, in question 5 there was a part where we had to draw the structural formula for the reaction of glycerol with some carboxylic acid....i drew the structural formula but glycerol was in chained form and i drew the ester in straight chain.... :(.... is this correct?...please do reply!
 
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Serenity said:
hey guys, in question 5 there was a part where we had to draw the structural formula for the reaction of glycerol with some carboxylic acid....i drew the structural formula but glycerol was in chained form and i drew the ester in straight chain.... :(.... is this correct?...please do reply!
yes its correct he asked fr the structural formula
 
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Hey guys, how did you get 485 Kj/mol energy of C----F bond. I know it is right but i cant figure how you guys solved that? :evil:

change in H= h2-h1

H2 was CH3-CH2-F which means 5(C--H) + (C--C) + (C--F)
= 2400 + (C-F)

H1 was CH2--CH2 and HF 4(C--H) + (C==C) + (H--F)
= 2812

so -73= (2400 + (C--F)) - 2812

By solving this, you get (C--F) = 339 Kj/mol
 
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what was the group in the last part of the last question????? And cud we write ammonium sulphate for barium question wala???? :p
 
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armaghanaqib said:
what was the group in the last part of the last question????? And cud we write ammonium sulphate for barium question wala???? :p
.
Group in the last question , I believe, is ester.
I'm not sure about the other one/.
 
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kc was 0.444.................same responses wre outside the centre sum eere saying 2.25 and sum were 0.444...................
it was Bao..............not Ba02 coz 18.1% was given the mass of oxygen hece 32 sud be used..............n btw i never say an oxide of barium with +4 oxidation state hence Bao and it aslo justified the condition of 1:1 mol with h2so4..............
the gas was co2 which will produce baco3 from ba(oh)2...........
btw why ester wiill not be hydrolysed with naoh................i wrote reflux conditions wre not present hence it ill not hydrolyse........... anu one can confirm this?????
and yeas the chiral carbon wree 2 .............
and the last grup was ester............
Hf bond energy = 485...............
 
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