Fall 08. Exam3

Fall 08. Exam3 - NAME : NOTE WELL! You Must Show Your Work...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: NAME : NOTE WELL! You Must Show Your Work And A Logical Path To The Solution Of Each Problem To Receive Any Full Or Partial Credit. N0 Credit Will Be Given For Numerical Answers Alone. The relative response it'actor for an individual compound is defined as follows: (signal) X (relative response factor) = (relative amount of component) Where response factor and amount of component have consistent units (as used in lecture). I) 16 points 2) 20 points 3) 20 points 4) 10 points 5) 14 points 6) 20 points (Homework) TOTAL 100 points _ l) 16 points: The weight % nC6H14 in a complex solution of organic compounds is measured using the method of internal standard. Gas chromatography with a thermal conductivity detector is used. nC9H20 is used as the internal standard. Bod: compounds give fully resolved peaks. The relative weight response factor for hexane is 1.25 and the relative weight response factor for nonane is 1.50. 12 grams of nonane are added to 100 grams of the original sample containing hexane, and the chromatogram is run. In this chromatogram, the AREA % hexane is 16,0 and the % nonane is 19.0 Calculate the weight % hexane in BOTH the original solution and in the solution containing the added nonane. 2 2) 20p0in13: ThefipflowhsgisatypicalplotoffivasuspobminedusingtheVan DmEqnafimm==A+Bip+Cuy a) Onthisplotcluniylabelandmarktheporfimofthcmprfinmflycomfled byflm“B”termANDtheportionofflmcurvemhnmflyountmfledbyflw“C” - 14. b) Onfileplatbelow,CLEARLYdrawfl1cnewchmgcdcuweflntwiflbeobsened iffimvaiueoffi”isdonbledmfi“A”and‘C”ruminumflmpd.Shuwcleafiy whexefilewrvachmgesfimnflnmigimlplotmflwhueitdoesm Showtime changes (ifany)inflleminimlm\vahle ofHandlmwihatpointmoveswifllmspeato p. onthe plot. a“. c) Onflleplotbekrw,CLEARLYdrawfllenewdnngedcnrveflmefllbeobsuved ifflnvaimofvisqmltofioffisodgimlvflueand“A”and"B”rwmin mohanged. Showclemiywhflethccurvechmgesfmmiheofig'mlplotandwlmeit doesnot. Showflaemfifanyfinflnmifimmnvameofflmdhowflmmim moveswifluespecttopmflmeplot d) Onflaepkytbeiow,CLEARLYdmwfllcnewdungdmifflaevalueof“A”is doubledand“B”ami“C”remainmdmged. Slmwcleaflywherefiwmedmnges fromflnofigimlplotmdwhaeitdownot Showthemfifmy)h1flw mirfimmnvalw'ofinmflhowfimmhnmwifilmmpmfinplot \/ WM _..—._.._ 3 3) 20 points: Calculate the relative weight response factors for CsFu and C4Hm given the following data: A solution containing 3.15 grams of C5F12 and 1.45 grams of C4Hm was separated by gas chromatography. The AREA % of the CsFu peak was 40.0%. The AREA% of the C4Hm peak was 60.0%. b) Normalize these two response factors so that the response factor for Cd-Im z 1. 4 4) 10 points: The ztbsorbance of a sample containing the absorbing compound ML: at 0.01 molesfliter was measured and found to be 0.655 Absorbance units. The absorbence of the blank (a solution containing everything except MLz) was found to be 0.042 Absorbanm units. A cell with a 5.00 cm path length was used. :1) Calculate the value of the absorptivity coefficient (“a”) for ML; fi'om Beer’s Law. b) Give the units for “a” in this case. S 5) 14 points: Given that the uncertainty in T is +/— 0.005 , Calculate both AA and AMA for the following values of A or T: a) Absorbance = 0.75 6 6) 20 points: Homework Problem 19-13: Finding a 13K,l by spectrophotometry. Consider an indicator, HJn which dissociates according to the equation KHln Him 6—) H’*+1n‘ The molar absorbtivity, s , is 2080 M”l cm" for Hin and 14200 M“1 cm'1 for In', at a wavelength of 440 nm. a) Use Beer’s Law to write an expression for the absorbance at 440 nm of a solution in a 1 cm cuvet containing I-IIn at a concentration [HIn] and In“ at a concentration [10‘]- b) A 1.84 X 10 4M solution ofI-Iin is adjusted to pH 6.23 to give a mixture ome and In' with a total a total concentration of 1.84 X 10". The measured absorbence of the solution at 440 nm is 0.868. Using your expression from (a) and the fact that [Hill] + 111'] = 1.34 X 10“, calculate the pKHm for this indicator. ...
View Full Document

Page1 / 7

Fall 08. Exam3 - NAME : NOTE WELL! You Must Show Your Work...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online