Dekoyl's Questions

[dekoyl]

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[jackinthepatch]

1. Yeah it would be worth mentioning it, because that's not necessarily wrong.

2. I think so...lol.

3. Yep

4. Um...I'll get back to you haha.

5. It brings the absorbance reading on the spectrophotometer back down to zero, so that the only reading you get is the absorbance for the sample (no absorbance or scattering is counted from the solvent and cell, because this absorbance has been accounted for in the calibration process)

6. Um yeah dunno lol sorry.

7. Not sure again sorry

8. I use pressure but I don't think it matters at all.

Sorry about the blanks lol

[TrueTears]

6. We did the prac today at school and yeah the AAS machine measured transmittance, it is just an upside down graph, and absorbance can be easily measured by subtracting the amount of light passed through (transmittance) from 100%. Ie, if you have 40 % transmittance, that means 60% absorbance.

8. High pressure and high performance are just different names for the same spectroscopy technique, ie HPLC

[Mao]

Delivered as requested:

1. Yes, you might as well (in VCE, showing off knowledge is not necessarily a bad thing)

2. Correct

3. Yes, also that the mobile phase doesn't interact much with the stationary phase much at all.

4. Not sure, and it's not a very good analysis

5. Basically, it sets a '0' so the machine knows what value to spit out (on the calibration curve, the line has to pass through the origin, hence it is important that the absorbance are relative to this point)

6. for the purpose of VCE, what your book says is correct. However, some machines may be manufactured differently, the readings are equally valid.

7. no idea, never heard of it, and most definitely not in the course

8. same thing

[nerd]

7. I think since organic substances may be colourless, by looking at them with UV light, they are revealed.

[dekoyl]

Just these weird ones. The answers to this MS section is quite weird. Directly from the answers:



What's the difference in the diagrams of d and e?
And I think it's meant to be and for the last 2 entries in the table.

Thanks

[TrueTears]

Lol I remember doing this question and confirming it with my teacher, you picked up another spastic wrong answer one dekoyl. Answers are wrong, you are right.

[dekoyl]

Damn the spastic wrong answers Thanks TT

[Dark Horse]

Exposing to Uv light, spraying with ninhydrin cause the spots to be visible because it reacts with the component (i think). I know that potassium Permanganate (KMnO4) is used because it reacts with functional groups, causing the molecule to become visible against the paper. 

[dekoyl]

^Thanks Dark Horse.

Another one =\   :

When we're asked to write an equation illustrating the formation of a fragmentation (mass spec), should it be


What I'm asking is, there should be a positively charged part and an uncharged part on the right side of the equation right?

Thanks

[Mao]

The process happen in two steps:

first, the parent molecular ion:



Next, fragmentation occurs



The dot indicates a free radical, not to be confused with a negatively charged ion. A free radical is an atom with unpaired electrons (all neutral group 1, 13, 14, 15, 16 and 17 atoms). These atoms have no charge, but in their electron orbitals the electrons are unpaired in their valence shell. In this case, the H has a single unpaired electron in its valence shell.

[apologies for previous mistake]

[TrueTears]

Just wondering Mao,

When you have , shouldn't that get bombarded with electrons first ie instead of to balance out the charges?

And is it compulsory to write the radical on the non positive fragment? Would marks be deducted if you don't?

[Mao]

Apologies to people who've read the my post earlier, it was a massive massive mistake. Now corrected, credits goes to TT.

And is it compulsory to write the radical on the non positive fragment? Would marks be deducted if you don't?

No it's not, so long as you specify it is a single atom (I did that last year in the exam and got the mark for it).
But, it's easier to just say it's a radical.

[dekoyl]

^Thanks again everyone

Hmm are the solutions correct? Or maybe my understanding is very distorted. I had stuff like 1-chloro-3-methylbutane.

[ed_saifa]

they did 2 isomers twice.