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3600_Ch24_probs - 24-1{a Lew hailing eulutee are separated...

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Unformatted text preview: 24-1. {a} Lew hailing eulutee are separated well at lee-I.r temperature, and the retentien et‘ high huiling selutee ir. redueed to u reuemehle time at high temperamre. {1:} Higher pressure gives higher flew rate. If preeeure is increased during a angel—Minn, retentien til-nee flrlHLE-Elufiflg peeks are refitted. The effect is the same as increasing temperature. hut high temperature-e are net required. Pressure pmgrrmmdng reduce; the likeliheud ef deeemfmeiug merit-telly Hmili‘nle eempeuntls. 24-5. (a) All analytes (b) Carbon atoms bearing hydrogen atoms (0) Molecules with halogens, conjugated C=O, CN, N02 ((1) P and S and other elements selected by wavelength (e) P and N (and also hydrocarbons) (f) S (g) Most elements (selected individually by wavelength) (h) All analytes 24-6. The thermal conductivity detector measures changes in the thermal conductivity of the gas stream exiting the column. Any substance other than the carrier gas will change the conductivity of the He carrier gas. Therefore, the detector responds to all analytes. The flame ionization detector burns eluate in a H2/02 flame to create CH radicals from carbon atoms (except carbonyl and carboxyl carbons), which then go on to be ionized to a small extent in the flame: CH + O ——> CHO+ + e’. Most other kinds of molecules do not create ions in the flame and are not detected. . = _ Ml: Neill-.111 _ 24- 1 8 . Use a longer column (doubling the length increases resolution by «15.). Alternatively, use a narrower column with a thinner stationary phase. If necessary, try a different stationary phase. 234 mg / 88.15 g/mol = 0.2655 M 24-19. (a) S = [pentanol] = 10 0 mL X = [2,3—dimethyl—2—butanol] = 237 “‘16 (1)0315 ““01 = 0.2320 M A; _ 4; 1.00 _ 0.913 _ [X] ‘ F([S]) => [0.2320 M] ‘ F([O.2655 M]) = F — 1253 (b) I estimate the areas by measuring the height and Wm in millimeters. Your answer will be different from mine if the figure size in your book is different from that in my manuscript. However, relative peak areas should be the same. pentanol: height: 40.1 mm; w1/2 = 3.7 mm; area = 1.064 x peak height x w1/2 = 153 mm2 2,3-d1methyl—2-butanol: height = 77.0 mm; w1/2 = 2.0 mm; area = 164 mm2 164 (C) ___._____ = ._1_58_. [2,3-d1methyl-2—butanol] 1.253 ([93-7 mM]) => [2,3—dimethyl-2-butanol] = 77-6 mM ...
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