Some Chem Qs

[prsndwg]

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Can you help me out with these please

Why Hoffman? what kind of reaction is this? ps. E is wrong

I need some explanation on this. I am tried of narrowing down answers 🙄

 

[hausee]

first Q, 3/4 gone and only 1/4 left so went through (1/2)*(1/2) total of two half life. time =24*2=48.

Better go through book first b4 doing Qs. 😎

 

[hausee]

not sure the second one though. might involve double bond on least substituted carbon or some kind of rearrangement. Would Like to know the answer as well.

 

[sfoksn]

Second one, its Hoffman because the attacking base is hella bulky...

For third one, I don't know how exact they want to be, but I think its C.

You have 20ml of 1M NaOH< meaning that you have 0.02 mole of OH. But then after 0.02 mole of OH added, you are still at 2 pH, which means your [H+] = 0.01M

Initial H added by the HCl must be 0.03 mole in order for this to be true... So 30 mL around there.. I think they said 30.4mL b/c they account for the solutions being added.

I am not sure.

 

[hausee]

20 ml 1M HCl is needed to bring pH to 7. So now total volume is about 40 ml. pH=2 means [H+] is 0.01 mole/liter. So extra 1M HCl need to be diluted 100 times in the final solution. 40ml/100 is about 0.4 ml. So you probably need 20.4 ml HCl total. 20.2 ml could be the best answer because we didnt count volume changes.

 

[hausee]

Second one, its Hoffman because the attacking base is hella bulky...

For third one, I don't know how exact they want to be, but I think its C.

You have 20ml of 1M NaOH< meaning that you have 0.02 mole of OH. But then after 0.02 mole of OH added, you are still at 2 pH, which means your [H+] = 0.01M

Initial H added by the HCl must be 0.03 mole in order for this to be true... So 30 mL around there.. I think they said 30.4mL b/c they account for the solutions being added.

I am not sure.

Hi, could you please explain more on the 3rd Q? thanks

 

[sfoksn]

hello hausee,

I think i shoudl change my answer.

umm i didn't really care about the overall solution volume, but just the moles of H+ and OH- that may be reacting.

you have 20ml of 1M NaOH, so you have 0.02 mole of OH-.

at the end of titration, you have 0.01M [H+].

so.. I guess you really do need to take the volume into account.

well, so (x-0.02)/(0.02+x) = 0.01..

So I think you are right with 20mL.