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Chemistry: Post your doubts here!

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Hey anyone has a chart or table of all the reactions, reagents and conditions for organic chem for A2?
Like I need to know which reagents and conditions to use when and where ... I am really bad at organic!
Thanks in advance! :)
 
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How many H atoms are present in 25.6g of urea, (NH2)2CO? How to solve such questions?
Find the Mr of the compound, (NH2)2CO ==> 2(14+2)+12+16=60
Unitary method:
2g of H in 60g of Urea
Xg of H in 25.6g of Urea
X=(25.6*2)/60=0.853g of H
Moles of H present= (0.853)/1 =0.853 mol
1 mole has 6.02x10^23 atoms
0.853 moles has X atoms
X=0.853*6.03x10^23
X= 5.14x10^23 H atoms in 25.6 g of Urea
 
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Guys, I just took the paper yesterday, what exactly is the contaminant in carbonaceous fuels again?
Also, what reagent to reduce carboxylic acid to primary alcohol? NaBH4 or H2/Ni can? Because I heard it's LiAlH4..
And the reaction between Al2O3 with HCl can form Al2Cl6 or not ya? Some said it was AlCl3 only..
I wonder if acyl chloride is also a possible reagent for making the amide?

Sorry for retarded questions.. First post.. xD
The impurity is sulfur.
Both, NaBH4 and LiAlH4, can be used to reduce carboxylic acid to alcohol.
AlCl3 and Al2Cl6 both can be formed because Al2Cl6 is a dimer of AlCl3.
Acyl chloride could be used, I guess.

And most importantly WELCOME ABOARD. :);)
 
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Naturally occurring copper is composed of 63Cu with an atomic mass of 62.94 amu and 65Cu with an atomic mass of 64.93 amu. Find the abundance of 63Cu isotope.
 
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Naturally occurring copper is composed of 63Cu with an atomic mass of 62.94 amu and 65Cu with an atomic mass of 64.93 amu. Find the abundance of 63Cu isotope.

62.94 ---> X%
64.93 ---> (100-X)%
(62.94*X)+(64.93*(100-X))
---------------------------------- = 63.50
100
62.94X -64.93X + 6493 = 6350
-1.99X = -143
X = 71.8%
 
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w13_qp_43.pdf

I have a few problems with question 6.
First one is 6 ei) How many different dipeptides is it possible to synthesise, each containing two of the three amino acids alanine, serine and lysine?
I first thought 6 (which is correct) but then i looked at the structure of lysine. it has two NH2 groups so if peptide bonds can form at either NH2 wouldn't that make it 12?? Wouldn't joining the NH2 joined to (CH2)4 would form a diff molecule if joined to the NH2 that is from the CH???

Second one is f ii) Which of the structures G, H or J is identical to structure F?
I thought it was G. But its J. I have no idea why. No matter how you flip J it doesn't look like F.

EDIT: i dont get f iii) either, why cant i just switch the positions of NH2 and COOH. that looks different

Please help and thanks in advance
 
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Can somebody draw the enthalpy cycle for part (d) please and explain how to do its calculation.
Answer to b (iii) is -32.6kJ per mole
Answer to c (iii) is +23.2kJ per mole
 

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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w13_qp_43.pdf

I have a few problems with question 6.
First one is 6 ei) How many different dipeptides is it possible to synthesise, each containing two of the three amino acids alanine, serine and lysine?
I first thought 6 (which is correct) but then i looked at the structure of lysine. it has two NH2 groups so if peptide bonds can form at either NH2 wouldn't that make it 12?? Wouldn't joining the NH2 joined to (CH2)4 would form a diff molecule if joined to the NH2 that is from the CH???

Second one is f ii) Which of the structures G, H or J is identical to structure F?
I thought it was G. But its J. I have no idea why. No matter how you flip J it doesn't look like F.

EDIT: i dont get f iii) either, why cant i just switch the positions of NH2 and COOH. that looks different

Please help and thanks in advance

anyone? please?
 
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