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Chemistry: Post your doubts here!

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http://maxpapers.com/syllabus-mater...hment/9701_may-june-2011-all-question-papers/

Variant 22 - Question 1 - c - i
In part b iv
The moles of NaOH that reacted with acid are found to be 0.04 moles
So in part c i
Shouldn't the moles at equilibrium be 0.06 moles for the acid and ROH and 0.04 mol for ester and H2O (Since 0.04 reacts and originally 0.1 is present so left over = 0.1-0.04 = 0.06)
The mark scheme says other way round... how?
 
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For the substitution of alchohols to form halogenoalkanes, Sulphur dichloride oxide (SOCl2) can be used as a reagent. But the problem is what are the conditions for the reaction?? I've tried searching the internet but i got nothing. Can someone please help.
 
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n
For the substitution of alchohols to form halogenoalkanes, Sulphur dichloride oxide (SOCl2) can be used as a reagent. But the problem is what are the conditions for the reaction?? I've tried searching the internet but i got nothing. Can someone please help.
no conditions..
 
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why is the observation sol turns yellow and not that yellow green gas ie chlorine disappears?
 

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why is the observation sol turns yellow and not that yellow green gas ie chlorine disappears?
What year is this?Shouldnt it be that the iodine is displaced from by Cl2,and that would give a brown color to the solution?Maybe if all the I is displaced that would lead to Cl2 going in excess and causing solution to turn yellow maybe?Still what year is this from?
 

NIM

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Need help !!
9701/43/O/N/13
Q#5b
 

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http://maxpapers.com/syllabus-mater...hment/9701_may-june-2011-all-question-papers/

Variant 22 - Question 1 - c - i
In part b iv
The moles of NaOH that reacted with acid are found to be 0.04 moles
So in part c i
Shouldn't the moles at equilibrium be 0.06 moles for the acid and ROH and 0.04 mol for ester and H2O (Since 0.04 reacts and originally 0.1 is present so left over = 0.1-0.04 = 0.06)
The mark scheme says other way round... how?
Moles of NaOH used is 0.045
Moles of NaOH that reacted with HCl is 0.005
Remaining NaOH that will react wid acid = 0.045-0.005 = 0.04
Hence 0.04 0.04 0.06 0.06 :)
 
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Moles of NaOH used is 0.045
Moles of NaOH that reacted with HCl is 0.005
Remaining NaOH that will react wid acid = 0.045-0.005 = 0.04
Hence 0.04 0.04 0.06 0.06 :)
But then 0.04 is the amount of acid with which the NaOH reacted
So would the equilibrium remaining mixture be 0.1 - 0.04 = 0.06 ?
Like I know it's not but please could you explain why :p
 
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But then 0.04 is the amount of acid with which the NaOH reacted
So would the equilibrium remaining mixture be 0.1 - 0.04 = 0.06 ?
Like I know it's not but please could you explain why :p
Moles of NaOH used is 0.045
Moles of NaOH that reacted with HCl is 0.005
Remaining NaOH that will react wid acid = 0.045-0.005 = 0.04
Hence 0.04 0.04 0.06 0.06 :)
Oh wait nevermind! i just got it after reading the question! Thanks! :)
 
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Can anyone show me to how to solve this??
c part?
http://studyguide.pk/Past Papers/CIE/International A And AS Level/9701 - Chemistry/9701_w05_qp_2.pdf
In question 5 part f, shouldn't the aldehyde change into OO-C-R as the reaction of Tollen's reagent woth aldehyde yeilds:
CH3CHO+(Ag(NH3)2)+----> CH3COO- ?? Anyone?
Either is correct in recent paper as i saw in a question both were accepted RCOOH and RCOO- as the product of reaction with tollens.This is an old one so dont worry.
 
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Anyone have any last minute notes, tips and predictions for questions/topics that may come up tomorrow?
 
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