Soccerboi's questions thread

[charmanderp]

The number of moles of hydrogen gas is 3.75mol. This is reacted completely with 1.25mol of a fat. A fat (aka triglyceride) is composed of three fatty acids so 1 mole of hydrogen gas reacts with one mole of fatty acids.

So the fatty acids are monounsaturated, as there is only 1 double bond to be broken. Which of those fatty acids is monounsaturated?

[soccerboi]

Thanks guys

Another question: In ethanoic anhydride, is the only fuctional group present and ester? Just wondering if you could call the COC ether or not cause it's also two esters sharing one O. Also is the CH3 considered a methyl functional group in this case(it's at the end of the chain though)?

[charmanderp]

I think that it's only an ester functional group if the COO is linked via oxygen to an alkyl group, but don't quote me on that. Also methyl groups aren't functional groups (to my knowledge), as they don't really have any functions.

Supposedly the functional group in acetic anhydride is called 'anhydride' but I've never seen any evidence of that, it's just what my teacher told me.

[soccerboi]

Trial exam 3 Q1MC of ATAR notes study guide: Why isn't the answer B? Isn't the NH2 an amino group? Why is it option D, hows the N with 2 CH3 a amino group? Someone help please?

[soccerboi]

Was reading about GC on Wikipedia, and it said:"The samples are also required to be salt-free; they should not contain ions." Can someone explain to me why this is the case?

[Mao]

Was reading about GC on Wikipedia, and it said:"The samples are also required to be salt-free; they should not contain ions." Can someone explain to me why this is the case?

The sample needs to be volatile. If it contains salt, the sample won't be completely vapourised, and so the sample injector will eventually clog up with these salts.

[thushan]

Trial exam 3 Q1MC of ATAR notes study guide: Why isn't the answer B? Isn't the NH2 an amino group? Why is it option D, hows the N with 2 CH3 a amino group? Someone help please?

B is an amide (CONH2), and amides are not basic. It's kinda analogous to COOH (which contains an OH group, but doesn't behave like a simple alcohol).

D is a tertiary amine, amines don't need to be NH2 (that'd be a primary amine). Amines can be C-NH-C (secondary) or N bonded to three C's. The N still has a lone pair of e- and can still accept a proton.

[soccerboi]

This is from VCAA 2006 Q4 SA, and only 6% of the state got it right.

I don't understand why it's at 400nm. The typical green dye absobs strongly there, why does the examiners report only consider the absorbance of the bright blue?

[charmanderp]

Because the green dye is made up of 'bright blue' and 'curcumin'. That's why it absorbs where it does. The curcumin which absorbs is part of the green dye, it doesn't absorb light separately.

Think of the line indicating the 'typical green dye mixture' as an addition (or subtraction, Methods makes no sense to me :p) of ordinates of the graph of curcumin and the graph of bright blue.

[soccerboi]

I kinda get it, so the curve for green dye was kinda there to trick students? So the actual light absorbed under analysis was just curcumin and bright blue, and we pick 400nm because less of bright blue is absorbed here?

[charmanderp]

Yup - remember, we're analysing the green dye, not curcumin on its own. But by knowing where curcumin does absorb light, we can calculate the concentration of curcumin the dye by noting the absorbance of the dye at a wavelength where curcumin absorbs very strongly and no other components do.

[soccerboi]

Thanks, i'm stuck on the next part of the question.

In one analysis, a stock solution of curcumin was prepared by dissolving 0.100 g of curcumin (C21H20O6;
molar mass = 368.0 g mol–1) in 250.0 mL of water.

The stock solution was then diluted to make the following three standard solutions.

standard solution curcumin concentration in g L–1
standard 1 1.00 × 10–3
standard 2 5.00 × 10–3
standard 3 1.00 × 10–2

Calculate the concentration of curcumin in the stock solution in mol L–1.
What volume of water must be added to 10.0 mL of the stock solution to make standard 3?

[Lasercookie]

So we have:


250ml --> 1 litre, factor of 4





Since c = n/v and v = 1L:

[soccerboi]

Oh woops, i put the wrong question, sorry.

here's what i'm stuck on: What volume of water must be added to 10.0 mL of the stock solution to make standard 3?

[soccerboi]

What's stronger, ion dipole or dipole dipole?