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Answers To Difficult Questions (1 Viewer)

nooj

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the flame does not excite the electrons, if it did u wouldnt need the light, the light it what excites the electrons the flame is to merely vaporise it. i wud agree with c but the book explicitly states that the cathode is chnaged depending on the metal.
 
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jtyler

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Armenikum said:
The way I learnt it was that when electrons go from normal to excited state, they need to ABSORB energy, and when they fall back, they EMIT energy.

Therefore, I don't know why you talk about electrons falling back into normal states. The point of the flame is to excite the electrons. When these elctrons are excited they move up from normal state to excited state, ABSORBING energy, in the form of light. The wavelengths to which these atoms absorb, are characteristic of a particular element, and the more they absorb, the greater the concentration of them.

And maybe this is a stretch, but C is a more OVERALL statement about AAS, whereas D may be true, but is only one part of AAS. C is the general idea of AAS, and if you read the question again, maybe it is asking about AAS as a whole process....if that mae sense:wave:
Not sure if you're talking about flame tests or AAS here. I'll assume you were talking about AAS, in which case when I talked about the return to ground states I was referring to the flame test. The role of the flame in AAS, I was always told, was simply to volatilise the sample, not necesserily to excite the atoms. I was also always told that in a flame test, the energy excited electrons (moved them away from the nucleus), and as they fell back to their ground state, light of characteristic frequencies were emitted (the light you see are the photons emitted).

I do Quanta to Quarks in Physics too, but the fact remains that the qualitative observation of light in a flame test is the result of EMITTED photons, whilst AAS relies upon comparison of an undisturbed beam of light with one that passes through the sample. The spectrometer compares them to determine how many photons were ABSORBED by the sample, in order to determine relative concentrations... That's what I was taught, and that's what I see as the fundamental difference that precluded D (assuming I remember it correctly).

also for question 15 i wouldn't really consider it an equilibrium, whilst the forward and backward reaction are occurring i think the fact there is continuous energy (UV rays) being used to break the bonds kind of means there really isn't a natural equilibrium and its not really a closed system.
Yeah my bad, equilibrium was a bad term to use, what I meant was that the presence of Chlorine radicals causes the total rate of the decomposition of ozone to far exceed that of the formation.

Can someone post the exact question (Q15 with options) please? I might be talking total shit because I can't remember the Q word for word :/
 
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jah_lu

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which term best describes the role of chlorine free radicals in this process?

the equations are one way, so therefore the examiners did not intend for people to view it as a catalyst, in other words, this reaction GOES TO COMPLETION
therefore, you can ignore anything about increased yield etc.

a catalyst decreases the activation energy of a reaction, often by providing an alternate mechanism for the reaction
yes ozone plus oxygen radical reaction does occur, as does ozone decomposing to oxygen radical and oxygen gas, and then oxygen radicals can react with each other to form oxygen gas
in both cases o3+o=2o2

the chlorine radical process does the same thing but provides an alternate mechanism with a lower activation energy

catalysts can increase the yield of reactions, in that say the activation energy of a reaction was 1000kj and you could only supply 800 then the reaction could not occur-ie no yield
however if you add the catalyst, then u decrease the activation energy, allowing it to occur

plus aas is definitely d
when electrons are excited in a flame test, they will emit a few wavelengths of light. this is known as their emission spectrum.

aas uses one of the wavelengths emitted, and analyses HOW MUCH is absorbed

THEREFORE, C IS WRONG FOR 2 REASONS
one, it is not white light, otherwise the same lamp could be used for any cation
two, you don't see which one is absorbed, you choose the wavelength that you think might be absorbed and measure how much is absorbed. answer c implies a qualitative experiment, not a quantitative one which aas is
 

joshuajspence

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jtyler said:
Not sure if you're talking about flame tests or AAS here. I'll assume you were talking about AAS, in which case when I talked about the return to ground states I was referring to the flame test. The role of the flame in AAS, I was always told, was simply to volatilise the sample, not necesserily to excite the atoms. I was also always told that in a flame test, the energy excited electrons (moved them away from the nucleus), and as they fell back to their ground state, light of characteristic frequencies were emitted (the light you see are the photons emitted).

I do Quanta to Quarks in Physics too, but the fact remains that the qualitative observation of light in a flame test is the result of EMITTED photons, whilst AAS relies upon comparison of an undisturbed beam of light with one that passes through the sample. The spectrometer compares them to determine how many photons were ABSORBED by the sample, in order to determine relative concentrations... That's what I was taught, and that's what I see as the fundamental difference that precluded D (assuming I remember it correctly).



Yeah my bad, equilibrium was a bad term to use, what I meant was that the presence of Chlorine radicals causes the total rate of the decomposition of ozone to far exceed that of the formation.

Can someone post the exact question (Q15 with options) please? I might be talking total shit because I can't remember the Q word for word :/
i also do quanta and what you are saying sounds correct. but you are looking too much into the question. for each sample in AAS you use a lamp which emits light of a wavelength known to be absorbed by the element you are testing for..

since you do Q2Q you should know that the wavelength absorbed is the same as the wavelength emitted if the element is excited (for example by a flame as in the flame test).

therefore the answer is D.

As for the chlorine in the decomposition of ozone:

Wikipedia said:
Ozone can be destroyed by a number of free radical catalysts, the most important of which are the hydroxyl radical (OH·), the nitric oxide radical (NO·) and atomic chlorine (Cl·) and bromine (Br·).
Chlorine is a catalyst. It is not used up in the reaction but rather is collected at the end. Also, all the chlorine does is assist in the natural decomposition of ozone (to an oxygen atom and an oxygen molecules).

Therefore the answer is B (catalyst)
 

codyjh

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jah_lu said:
THEREFORE, C IS WRONG FOR 2 REASONS
one, it is not white light, otherwise the same lamp could be used for any cation
two, you don't see which one is absorbed, you choose the wavelength that you think might be absorbed and measure how much is absorbed. answer c implies a qualitative experiment, not a quantitative one which aas is
wat was C and D? ie: anyone have the paper so they can tell me lol
 

jtyler

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since you do Q2Q you should know that the wavelength absorbed is the same as the wavelength emitted if the element is excited (for example by a flame as in the flame test).
I didnt think that this was necesserily true. If it is, I sure as hell wasn't taught that :/
 

Jibby123

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OK lets just get our definitions straight. Anyone who did Bio would know that a catalyst is "A substance which increases the rate of a chemical reaction, but does not itself enter into the reaction and remains unchanged at the end of the reaction" I believe that in the two equations provided to us in the paper the chlorine actually did both the things that, by definition, a catylist should never do. Hence, I answered initiator lol :)
 

Chemfreak

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Jibby123 said:
OK lets just get our definitions straight. Anyone who did Bio would know that a catalyst is "A substance which increases the rate of a chemical reaction, but does not itself enter into the reaction and remains unchanged at the end of the reaction" I believe that in the two equations provided to us in the paper the chlorine actually did both the things that, by definition, a catylist should never do. Hence, I answered initiator lol :)
okok..i know these are really hard questions but can somebody tell me a comparison of thermodynamic and kinetic stability.....i know its kinda easy but i really need it for a project i'm doing!!!!!!!!!
 

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