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Chemistry: Post your doubts here!

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Can anyone tell me the products of oxidation of But-1-ene by Hot, Conc, Potassium Manganate(VII)?
6iqeF5d.jpg


Will it be Carboxylic Acid + CO2 + H2O?
Yes
CH3CH2COOH + CO2 + H2O
 
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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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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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