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nina_rox

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Re: Homework questions thread
« Reply #675 on: June 09, 2012, 09:27:35 pm »
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Few other general questions
3. How do you find defection (mass spectrum)?
4. How do you distinguish layers of a biofuel?
5. What is the test for amide bonds?
6. What do we need to know about mono/di/tri protic acids?
7 Co-polymers?

Thank you very much!

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Re: Homework questions thread
« Reply #676 on: June 10, 2012, 12:20:42 am »
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its either 3 or 1 im not sure but i recall reading somewhere when something is in symmetry the peak produced = 1 ? let me ask someone lol

The peak for the protons on the 3rd carbon in pentane will be split into a pentet (5), because the two neighbouring carbons are in the same environment as each other but different to carbon 3, and between them they have 4 protons. So by n+1, a pentet will be formed.

The TOTAL number of peaks in pentane will be 3 - a pentet, a triplet of a quartet (dw about why that is, its to do with multiplicity), and a triplet.

What do you mean different to carbon 3???

And what does symmetry do to splitting? So say for butane? What splitting would CH3CH2CH2CH3 have with Carbon 2? Also if the CH3 and CH2 are different environments shouldn't we use (n+1)x(n+1) [I am so confused].

As you can see from my nonsensical ramblings, any help would be greatly appreciated! Thanks!

Comatose

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Re: Homework questions thread
« Reply #677 on: June 10, 2012, 08:58:33 am »
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its either 3 or 1 im not sure but i recall reading somewhere when something is in symmetry the peak produced = 1 ? let me ask someone lol

The peak for the protons on the 3rd carbon in pentane will be split into a pentet (5), because the two neighbouring carbons are in the same environment as each other but different to carbon 3, and between them they have 4 protons. So by n+1, a pentet will be formed.

The TOTAL number of peaks in pentane will be 3 - a pentet, a triplet of a quartet (dw about why that is, its to do with multiplicity), and a triplet.

 i agree with 3 peaks in pentane but isnt it a quintet, sextet and quartet? so confused haha
« Last Edit: June 10, 2012, 09:58:58 am by Comatose »
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Aurelian

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Re: Homework questions thread
« Reply #678 on: June 10, 2012, 09:51:39 am »
+1
is it a pentet or a quintet?
The names following it would be sextet, septet, octet? (Just wanted to clarify :P)

Also, note that there will be three PEAKS, but in the hydrogen atom, it will have five SPLITS.

pentet, sextet, septet and octet

Quintet, not pentet =)
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sahil26

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Re: Homework questions thread
« Reply #679 on: June 10, 2012, 11:53:51 am »
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Hi guys,

Why is sodium salt of aspirin more water-soluble than aspirin?
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destain

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Re: Homework questions thread
« Reply #680 on: June 10, 2012, 12:06:16 pm »
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Hi guys,

Why is sodium salt of aspirin more water-soluble than aspirin?

sodium salt makes ion dipole bonds to water which is a stronger form of bonding.

soccerboi

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Re: Homework questions thread
« Reply #681 on: June 10, 2012, 12:07:49 pm »
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Pure aspirin is not very soluble in water as it is a weak acid with very little dissociation in water. By converting the carboxyl group to sodium salt, we can change the aspirin molecule into an ion, which now makes it much more soluble in water as it can form ion dipole interactions with water molecules.

Edit: He beat me to it  :P
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Aurelian

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Re: Homework questions thread
« Reply #682 on: June 10, 2012, 12:08:06 pm »
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Hi guys,

Why is sodium salt of aspirin more water-soluble than aspirin?

The sodium cations and the "aspirate" anions can form ion-dipole bonds with water, which are stronger than the dipole-dipole/hydrogen bonds which would have otherwise been formed for normal aspirin. Thus, you can view it as water being able to "pull" the aspirin apart more strongly when in its sodium-salt form compared to normal aspirin.


(NB: I just made the word 'aspirate' up then... I do not believe this is the technical term for referring to the anion bit of the sodium salt of aspirin).
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AllAboutTheLGs

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Re: Homework questions thread
« Reply #683 on: June 10, 2012, 12:10:59 pm »
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Because of the charge COO-Na+, it is able to form ion-dipole bonds with the water molecules. Ion dipole bonding is greater than hydrogen bonding, hence it is more soluble in water.
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ICECOLD

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Re: Homework questions thread
« Reply #684 on: June 10, 2012, 12:14:43 pm »
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When working throughout a q which requires 3 sig fig at the end does each step have to be to 3 sig fig?

Aurelian

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Re: Homework questions thread
« Reply #685 on: June 10, 2012, 12:22:17 pm »
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When working throughout a q which requires 3 sig fig at the end does each step have to be to 3 sig fig?

VCAA convention stipulates that you should keep your calculator display running and round at the end.

When actually writing down answers to each step of the working, have three of four sig figs (it doesn't matter too much; just don't excessively round) and then round to three at the end (but be keeping the full decimals running in your calculator throughout the question). Thus each step does not have to be to 3 sig figs.
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Re: Homework questions thread
« Reply #686 on: June 10, 2012, 01:30:09 pm »
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And what does symmetry do to splitting? So say for butane? What splitting would CH3CH2CH2CH3 have with Carbon 2? Also if the CH3 and CH2 are different environments shouldn't we use (n+1)x(n+1) [I am so confused].

Firstly, CH3CH2CH2CH3 will be split into a quartet. A signal can only by split by neighbouring protons in different environments. In this case, the other CH2 group is in the same environment as the bolded group, and thus will not split the bolded group. You will only have splitting from the terminal CH3.

Secondly, (n+1)(n+1) almost never works in simple hydrocarbons, because instruments are not sensitive enough resolve all the fine peaks. Just stick with the simple n+1 rule for VCE.
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Comatose

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Re: Homework questions thread
« Reply #687 on: June 10, 2012, 01:41:54 pm »
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And what does symmetry do to splitting? So say for butane? What splitting would CH3CH2CH2CH3 have with Carbon 2? Also if the CH3 and CH2 are different environments shouldn't we use (n+1)x(n+1) [I am so confused].

Firstly, CH3CH2CH2CH3 will be split into a quartet. A signal can only by split by neighbouring protons in different environments. In this case, the other CH2 group is in the same environment as the bolded group, and thus will not split the bolded group. You will only have splitting from the terminal CH3.

Secondly, (n+1)(n+1) almost never works in simple hydrocarbons, because instruments are not sensitive enough resolve all the fine peaks. Just stick with the simple n+1 rule for VCE.

CH3CH2CH2CH3 ehh isnt there 5 neighbouring hydrogens for that bolded CH2? so is isnt split into 5+1=6 peaks? gahhh
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Re: Homework questions thread
« Reply #688 on: June 10, 2012, 01:46:16 pm »
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^apparently if there's symmetry you only count one side or something

Comatose

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Re: Homework questions thread
« Reply #689 on: June 10, 2012, 01:49:46 pm »
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ohhh i see what you mean now, sorry my mistake! :)
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