| Question 1 |
(0.5 points for each of the following) - N2 carrier gas hit by a ß-particle, causes a cascade of electrons - electrons can be absorbed by an electronegative group on a molecule - takes place in a chamber with a potential across it, loss of absorbed electrons causes a current difference that is measured |
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| Question 2 |
(0.5 points for each of the following) - capillary has a large surface area-to-volume ration, therefore heat is rapidly dissipated - if resolution is directly proportional to voltage, then higher voltage gives better resolution - higher voltage can be applied to a capillary because the temperature buildup that is inevitable with increasing the voltage is dissipated more quickly with a capillary |
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| Question 3 |
NOTE: value of this question should have been 2 points
(not 3 points that was given in the question) |
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| (a) could use a number of types of chromatography,
I gave full value for reverse-phase (non-polar stationary phase coated on
the packing material) and size-exclusion (stationary phase is a gel with
pores in it). Normal phase would not be appropriate, nor would ion
exchange. |
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| (b) a detector that can detect saturated compounds
is necessary. Some examples are mass spectrometry, refractive index,
or IR absorption. UV-vis and fluorescence would not be appropriate. |
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| Question 4 |
NOTE: there is a type here as well. The width
is supposed to be the full width as half height, which I misleadingly labelled
half-width (which is slang for full width at half-height, but many of you
took it literally). It turns out that the question can be done correctly
either way (resolution will be below 1.5). |
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| There are a few ways to do this question, but I did
it as follows: - calculate the number of plates using 5.54(tr/W1/2)2 - do for each peak and average: ave. N = 1193 kA = (3.2-1.1)/1.1 = 2.364 kB = (3.7 - 1.1).1.1 = 1.909 a = kB/kA = 1.238 Rs = (sqrt(N)/4)((a-1)/a)(kB/(1+ kB)) = 1.168 (other methods will get different values but all are around this amount) Then, if Rs is proportional to (N)1/2, then it al also proportional to (L)1/2 Rs(known)/(Lk1/2) = Rs(1.5)/(L1.51/2) or Rs(1.5) requires L = 16.5 cm. |
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Last updated, February 15, 2006
Page maintained by Craig MacKinnon