Chemistry: Post your doubts here!

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bumble26

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TheMan123 said:
Well the molecule is a cis-molecule so its either A or C
it has a formula of C20H28O, which is also CnH2n-12O. It has some double bonds, an aldehyde group and a cyclohexene ring
Lets start of with an alkane and change it into the molecule in question
An alkane has a formula of CnH2n+2
If you could change it into an aldehyde, you lose 2hydrogens and gain one Oxygen
so now its CnH2nO
If you could make part of it into a ring, adding a cyclohexane ring, you reduce number of hydrogens by 2. Noticed I added cyclohexane isntead of cyclohexene. We'll take care of double bonds later
now its CnH2n-2O
So now the only group we haven't considered is alkene. Each alkene reduces the H number by 2, so if there are 5 alkenes,
the formula would be CnH2n-12O, which is our molecule

So answer is A

EDIT: Fixed up some mistakes, you should reread it if you read it already
Click to expand...

Thanks! You saved my grade! :)
 
angelgirl:)

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bumble26 said:
please help me with Q20 with explanation.
Click to expand...
it says that there is an aldehyde group and cyclo hexene so C6H10 + CHO = C7H11O
The real moleculer formula is C20H28O...C20 -C 7= C13....
Now,there iz a long aliphatic chain...general formula is CnH2n+2...so C13H28...
for each double bond formation a hydrogen molecule is removed...so if there iz 5 double bond means ...5*2=10 H is removed...
SO OUT OF 28 H in C13H28...28-10=18 is remained...but 1 more H is removed due to attachment...
 
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bumble26

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angelgirl:) said:
View attachment 8865
it says that there is an aldehyde group and cyclo hexene so C6H10 + CHO = C7H11O
The real moleculer formula is C20H28O...C20 -C 7= C13....
Now,there iz a long aliphatic chain...general formula is CnH2n+2...so C13H28...
for each double bond formation a hydrogen molecule is removed...so if there iz 5 double bond means ...5*2=10 H is removed...
SO OUT OF 28 H in C13H28...28-10=18 is remained...but 1 more H is removed due to attachment...
Click to expand...


Thanks so much! :D
 
étudiante

étudiante

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Mustehssun Iqbal said:
Apologies. Q.6...
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no problem ;)

calculate the oxidation nos of Cr in all the equations
A) +6 ----> +3
B) +6 ---->+6
C) +6 ----> +6
D) +6 ----> +3

Answer is D because the no of O (per Cr atom) is also reducing along with the ON.. this was the explanation i was given by someone... I'd like it if u guys could confirm please
 
Student12

Student12

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Use of the Data Booklet is relevant to this question.
Lead(IV) chloride will oxidise bromide ions to bromine. The Pb4+ ions are reduced to Pb2+ ions in
this reaction.
If 6.980 g of lead(IV) chloride is added to an excess of sodium bromide solution, what mass of
bromine would be produced?
A 0.799 g B 1.598 g C 3.196 g D 6.392 g

The answer is C ?? explain someone!
 
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