Help with A2 chem question

OP

Original post by charco

Yes, they almost all (except glycine) have optical isomers

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For part b) why is my method wrong?

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Reply 22

OP

Original post by charco

Yes, they almost all (except glycine) have optical isomers

One more thing that's confusing me...

The standard hydrogen electrode is always placed on the left

But it contradict the fact that when working out emf, Erhs - Elhs (Ered - Eox), therefore, the hydrogen electrode is ALWAYS oxidised????

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Original post by ps1265A

For part b) why is my method wrong?

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if:

1 + x = 0.5

then
x = -0.5 not +0.5

However, I can't comment on your method without seeing the question...

Reply 24

OP

Original post by charco

if:

1 + x = 0.5

then
x = -0.5 not +0.5

However, I can't comment on your method without seeing the question...

ImageUploadedByStudent Room1433695896.773863.jpg

Reply 25

OP

Original post by charco

if:

1 + x = 0.5

then
x = -0.5 not +0.5

However, I can't comment on your method without seeing the question...

ImageUploadedByStudent Room1433695962.140527.jpg

Reply 26

OP

Original post by charco

if:

1 + x = 0.5

then
x = -0.5 not +0.5

However, I can't comment on your method without seeing the question...

One other question; is there a mathematical approach into determining how many isomers a particular compound has rather than just drawing each one? Say for example a compound has 8 isomers, is there anyway I can find out it has 8 without drawing 8?

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Original post by ps1265A

Your method is fine.

P <==> 2Q

initially P = 1 mol
at equilibrium

P = 1-x, Q = 2x

Total moles = 1.5 = 1 + x

hence x = 0.5

Hence mole fraction of Q = 1/1.5 = 0.67

Original post by ps1265A

One other question; is there a mathematical approach into determining how many isomers a particular compound has rather than just drawing each one? Say for example a compound has 8 isomers, is there anyway I can find out it has 8 without drawing 8?

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Depends on the molecule. Maybe you can apply logic ... maybe not!

Reply 29

OP

Oh, that's what they mean by mole fraction!

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Reply 30

OP

Original post by charco

Your method is fine.

P <==> 2Q

initially P = 1 mol
at equilibrium

P = 1-x, Q = 2x

Total moles = 1.5 = 1 + x

hence x = 0.5

Hence mole fraction of Q = 1/1.5 = 0.67

Thanks

The standard hydrogen electrode is always placed on the left

But it contradict the fact that when working out emf, Erhs - Elhs (Ered - Eox), therefore, the hydrogen electrode is ALWAYS oxidised????

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Reply 31

OP

ImageUploadedByStudent Room1433700361.021059.jpg125.0KB

Original post by charco

Depends on the molecule. Maybe you can apply logic ... maybe not!

And...

Why can't the answer be D?

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(edited 8 years ago)

Reply 32

OP

Original post by charco

Depends on the molecule. Maybe you can apply logic ... maybe not!

for....

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Original post by ps1265A

for....

2 makes 3

therefore

1 makes 1.5

Original post by ps1265A

And...

Why can't the answer be D?

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The conditions are elimination and you can only make the double bond either side of the halogen on the chain.

Reply 35

OP

Original post by charco

Your method is fine.

P <==> 2Q

initially P = 1 mol
at equilibrium

P = 1-x, Q = 2x

Total moles = 1.5 = 1 + x

hence x = 0.5

Hence mole fraction of Q = 1/1.5 = 0.67

I've got a question regarding mole ratios

Say if I react sulfuric acid with potassium hydroxide, the ratio in which they react is 1:2

1) But isn't that "actual" ratio in which they react is 1:1, because when sulfuric acid dissociates, there's 2H+ in solution and there will be 2OH- in solution?

2) One other thing, can I not just times the moles of sulfuric acid by 2, to get the moles of KOH?
ImageUploadedByStudent Room1433835683.079970.jpg

ImageUploadedByStudent Room1433835683.079970.jpg

(I understand how to get to the answer, but it's just the theory)

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Reply 36

OP

Original post by charco

The conditions are elimination and you can only make the double bond either side of the halogen on the chain.

Also, do I remove the OH from the acid or the alcohol when forming an ester? Because the repeating units of polyesters would be different if either I removed OH from acid or acid

I've come across both, so I'm not sure

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Original post by ps1265A

Also, do I remove the OH from the acid or the alcohol when forming an ester? Because the repeating units of polyesters would be different if either I removed OH from acid or acid

I've come across both, so I'm not sure

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It doesn't matter, the repeating unit is still the same ...

Reply 38

OP

Original post by charco

It doesn't matter, the repeating unit is still the same ...

Could you help me with the moles questions I just sent before the alcohol one?

Also, why is benzene more stable if it's enthalpy of hydrogenation is less ectothermic than cyclohexatriene? If it's less endothermic, wouldn't that mean less energy is used to break the bonds?

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Reply 39

OP