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Chemistry: Post your doubts here!

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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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Can anyone show me to how to solve this??
Assuming ur doubt to be
bii)
Its just tricky. We have to reverse the process. Its the same way we use to do to find Ar, we have Ar and now we have to find relative isotopic abundance.
Let percentage abundance of 80.92Br be x and of 78.92 be 100-x
Ar = 79.2
79.2 = [(80.92x) + (78.92(100-x)]/100
x = 49 = relative isotopic abundance of 81Br
100-49 = 51 = relative isotopic abundance of 79Br

c)
This is also tricky, we have to reverse the process.
ABr3 has 1:3 ratio
A : 4.31 / Ar = 1
Br : 95.69/79.9 = 3

A : Br
1 : 3
4.31/Ar : 95.69/(79.9*3)
Ar = 4.3/(95.69/239.7) = 10.79 :)
 
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What are observations on all the reactions of period 3 elements with H2O, Cl and any other reagent in the syllabus?
 
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What are observations on all the reactions of period 3 elements with H2O, Cl and any other reagent in the syllabus?
Okay so only 3 elements in P3 react with H2O i.e Na,Mg and Cl. Na would float on H2O,and catch fire. An exotherminc reaction will occur. Na+H20- NaOH+H2 Ph is 14
Mg's reaction will be slow, bubbles of H2 will form after a long time. But it's reaction with steam will be faster. PH is 9
Cl2 will react with H2O to prduce HClO and HCl. Ph is 2-6 . That's just for H2O
 
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Assuming ur doubt to be
bii)
Its just tricky. We have to reverse the process. Its the same way we use to do to find Ar, we have Ar and now we have to find relative isotopic abundance.
Let percentage abundance of 80.92Br be x and of 78.92 be 100-x
Ar = 79.2
79.2 = [(80.92x) + (78.92(100-x)]/100
x = 49 = relative isotopic abundance of 81Br
100-49 = 51 = relative isotopic abundance of 79Br

c)
This is also tricky, we have to reverse the process.
ABr3 has 1:3 ratio
A : 4.31 / Ar = 1
Br : 95.69/79.9 = 3

A : Br
1 : 3
4.31/Ar : 95.69/(79.9*3)
Ar = 4.3/(95.69/239.7) = 10.79 :)

Thanks for replying! :)
 
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I have a question in organic chemistry, what is the role of conct. H2SO4? I know that it forms a double bond and stuff, but what else does it do?
 
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