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Chemistry 9701/P22

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Hey, did you people gave 21, 22 or 23. Which paper had this K.C question (0.44 answr) and the rest.
 
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hey the last question of 22 which functional grp was present?
i wrote ETHER because the C--O--C functional grp is ether was this right.....................
i think the inorganic part was tough!!!
 
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yeah it was ester and there were 2 centres
i got these two parts wrong thank God they were for a mark each... not more
 
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i say k minimum 35/40 in p1 ho.
and about ppr 34 bacho ko kafi difficulties hoti hai to mera khayal se above 30 ho to inshallah A a jaye ga
 
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Guys the answer to kc was 0.429 or 0.43 coz after rxn with naoh 0.06 moles of acid will be left and then at the equillibrium half of it will react so 0.03 moles will be there at equillibrium and the table will be something like this
Acid: 0.03 alcohol: 0.07(coz 0.03 reacted to form ester and water) ester 0.03 ( 1:1 ratio) and water 0.03. For those of you sayin 0.06 mol acid at equillibrium this is not possible that ester is being formed without acid being used up. I hope im pretty clear on this..:)
 
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What is the Bond energy for C-F? and the alcohol, is it all 3 -OH react wit the acid?
 
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bod energy was 4 hundred something and yeah all of them did because the question said completely esterified
 
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inshAllah the lower the better
are you done with practical?
would someone tell me a way to get over with p34 with marks 30 or above?
 
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m2rulz143 said:
Guys the answer to kc was 0.429 or 0.43 coz after rxn with naoh 0.06 moles of acid will be left and then at the equillibrium half of it will react so 0.03 moles will be there at equillibrium and the table will be something like this
Acid: 0.03 alcohol: 0.07(coz 0.03 reacted to form ester and water) ester 0.03 ( 1:1 ratio) and water 0.03. For those of you sayin 0.06 mol acid at equillibrium this is not possible that ester is being formed without acid being used up. I hope im pretty clear on this..:)


Sorry dude. I already explained the reason.The value of Kc was 2.25 or smthng not 0.44....as NaOH was 0.045 moles and 0.005 reacted with HCl. 0.04 reacted with Ethanoic Acid, meaning that 0.04 moles of Ethanoic Acid were there at equilibrium. So Both the reactants were of 0.04 and products at 0.06 concentration. 8) ..it just didn't click at that time. I wrote the opposite and got 0.44
 
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