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Ok, so I was going off on tangents :D It makes sense. So the main thing you are saying is that different atoms in a metal sample randomly absorb different energies :confused: I'm fairly new to this so bear with me... a photon=specific wavelength of light ? :wave:

That's alright. I read that you get a printout after doing AAS, which shows you the results. Maybe the diagram/curve lets you work out the concentration of the metal sample you are analysing. And apparently there are little shadows, like on tv screens, on the curve, if you happen to have errors...

jyu, I'm not sure I follow what you're saying. Can you explain it clearer, eg. examples might be a good idea, or explain it differently? =S hmm... what makes a(- x)^(+/-n) = a(x)^(+/-n) ?? what does this =/= mean?

I desperately need help with this question, so I hope someone cna help me with it. I was given a linear graph, which was f'(x) (the derivative function), so I found its equation. Knowing the f'(x) equation, I found the original equation, f(x)= bla bla + C. I was given one piece of information...

Ah, so each element is VERY specific for the wavelengths it absorbs. K, so back to the other one btw, do you know how a calibration curve corrects for the error that small amounts of the atoms are excited by the flame (rather than the light source)? Something to do with little shadows...

I see... that makes sense, but there lies just a small catch. First, if the atoms excited gave off the same colour, wouldnt that imply that with each atom, the sum of all the photons of light the electrons gave off when going back to the ground state are the same? If the photons are the same...

Yeah, I meant the mobile phase :) ----------- I might have done some of this in the past, but I don't remember (in desperate need of revision) but what makes alcohols and amines polar?

Uh, k, so what I said before was more correct... 'stoichiometrically equivalent'= when you have x moles of a reactant to produce x moles of a product....? Rather than the phrase = amount of products/reactants you have according the mole ratio... ('cause a bit confusing as to why you editted out...

Ah, ok, so I see you've figured it out :) So which equations like these, its a matter of knowing the properties of species. Um, do you know whether all peroxides behave similiarly/ are strong oxidants? I thought that oxidants and reductants act as oxidants and reductants as part of a redox...

Hmmm, I thought that most chemical reactions don't proceed to completion?

I wasnt talking about odd and even functions, but thanks for the explanations all the same ^_^ You know how if you replace x with -x you get a reflection in the y-axis. I'm trying to work out whether trickier graphs are reflected in y, because it seems, just because there is -x doesn't MEAN it's...

No two atoms as in from two different elements or the same...? What'd you mean by 'theoretically'? >.<

So, since flames are not reliable in exciting atoms, do the flame colours that appear mean that only some of the atoms in a sample are excited?

yay ^^ Hang on, just looked back to the question. polar compounds eg. alcohols and amines, are absorbed more strongly and eluted less readily than the less polar compounds such as halogen compounds, aldehydes, ketones, ethers and hydrocarbons. That means less polar compounds are supposed to...

kk :D How do you tell whether things undergo further equilibrium, though...

Hm, maybe...

Thanx guys. ---So, with things that are slightly soluble, it doesn't matter whether you put (s) or (aq) since some of the substances will be (s) and some will be (aq)...? ---So... stoichiometrically equivalent basically means you have to have reactant/ product ammounts in the mole ratio, so...

Ok, this completely puzzled me. For any function with form something like this: y=a(x-b)^n+c ... basically, hyperbolas, truncus, linear, quadratics, cubics, quartics (and higher powers) all apply. So, for any function with this form, apparently if there is a negative in front of the x, and you...

Well, he's smart, I can tell ya that ;) -- Hmm, that makes sense. So each metal not only absorbs light at particular wavelengths, they also emit at those same wavelengths. Question. Can two metals ever absorb/emit the same wavelength of light?

Things that made me scratch my head... -- When you are told that compounds ie. in a solubility table are 'slightly soluble', how are you supposed to predict their states in a chemical equation?---In titrations, we know the 'known volume' goes in the conical flask, but is the exact volume...