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Chemistry: Post your doubts here!

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Ans for question 22 is c.
 

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October/November 2011 (Variant 3), Question 33.
How do I do this question?
 

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U s
October/November 2011 (Variant 3), Question 33.
How do I do this question?[/quote
Boss,this is like a fact H2so4 act as a reducing agent to both br2 and i2 but not cl2.I think that is what they want.Answer should be option 1 and 2.
 
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Ans for question 22 is c.
for q22
a free radical ? well it would be one less hydrogen atom right ?
rule 1 take hydrogen from primary carbon atoms only as the secondary ones are stable very much stable than primary ones
u can take an hydrogen from ethyl group
from 3 methyl groups
so it would be 4 but the answer says 3?
well if u take hydrogen from the methyl at top or bottom it would still be same thing
so there are 3 different forms
 
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for q22
a free radical ? well it would be one less hydrogen atom right ?
rule 1 take hydrogen from primary carbon atoms only as the secondary ones are stable very much stable than primary ones
u can take an hydrogen from ethyl group
from 3 methyl groups
so it would be 4 but the answer says 3?
well if u take hydrogen from the methyl at top or bottom it would still be same thing
so there are 3 different forms
Thank u.:sneaky:
 
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we dont know the reactions of D will be same as that of H(hydrogen ) as the number of electrons and protons is still the same
so lets consider NH4cl+Cao u do know that it will form
CaO(s) + 2 NH4Cl(s) 2 NH3(g) + H2O(g) + CaCl2(s)
D is no exception here so 1 is correct
for option 2
we know that hydrolysis of cyanide gives an carboxyllic acid and ammonia
here is a base hydrolysis we have same number of D atoms as we would have expected for H atoms

for question 3

NH4+ yields NH3
but we dont have 4D atoms here we have only 1D so it is not possible
since 1 and 2 are correct option B seems the right option
i
 
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Guys,I dont really get this question.anyone have a clue?
 

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we dont know the reactions of D will be same as that of H(hydrogen ) as the number of electrons and protons is still the same
so lets consider NH4cl+Cao u do know that it will form
CaO(s) + 2 NH4Cl(s) 2 NH3(g) + H2O(g) + CaCl2(s)
D is no exception here so 1 is correct
for option 2
we know that hydrolysis of cyanide gives an carboxyllic acid and ammonia
here is a base hydrolysis we have same number of D atoms as we would have expected for H atoms

for question 3

NH4+ yields NH3
but we dont have 4D atoms here we have only 1D so it is not possible
since 1 and 2 are correct option B seems the right option
i
Thanks man.
 
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we dont know the reactions of D will be same as that of H(hydrogen ) as the number of electrons and protons is still the same
so lets consider NH4cl+Cao u do know that it will form
CaO(s) + 2 NH4Cl(s) 2 NH3(g) + H2O(g) + CaCl2(s)
D is no exception here so 1 is correct
for option 2
we know that hydrolysis of cyanide gives an carboxyllic acid and ammonia
here is a base hydrolysis we have same number of D atoms as we would have expected for H atoms
for question 3

NH4+ yields NH3
but we dont have 4D atoms here we have only 1D so it is not possible
since 1 and 2 are correct option B seems the right option
i
Boss,sorry but what do u mean by that here is a base hydrolysis we have same number of D atoms as we would have expected for H atoms.
 
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I got to go out know.But generally what u need to do is this.First form a half equation for for pb ion n bromide ion.U will then get the ratio.So then u find the mole of pb ion using the mass given.Finally with the ration value,u can get the value of bromine.
u got it?
 
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Hai guys.Anyone have idea about this question.Just want to ask why option 3 is rejected .Thanks guys.
 

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