Please show all work Y Blackboard @ Texas Tecx Course Homepege place X/. Texas T

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Y Blackboard @ Texas Tecx Course Homepege place X/. Texas Te h University-C × d www.saplinglearning.com/ibiscms/mod/ibis/view.php?id=39743.5 Sapling Learning macmilan learning Jump to... Rohit Paradkar Question 1 of 15 Available From: Due Date: Points Possible: Grade Category: 11/1/2017 06:00 PM 11/13/2017 11:59 PMM 100 Default Mapt Sapling Learning An unknown amount of a compound with a molecular mass of 272.71 volumetric flask. A 1.00-mL aliquot of this solution is transferred to a water is added to dilute to the mark, The absorbance of this diluted solution at 339 nm is 0.490 in a 1.000- cm cuvette. The molar absorptivity for this oompound at 339 nm is a”-6533 M-1 cm-1 l is dissolved in a 10-mL L volumetric flask and enough F14 (a) What is the concentration of the compound in the cuvette? You can check your answers. You can view solutions when you complete or give up on any question You have three attempts per question. You lose 20% of th Number (b) What is the concentration of the compound in the 10-mL lask? e points available to each r in your question for each incorrect Number attempt at that answer eTextbook 10 1 (c) How many milligrams of compound were used to make the 10-mL solution? Help With This Topic Number 12 1 13 1 14 15 1 ng OWeb Help& videos O Technical Support and Bug Reports @ Previous Give Up & View Solution Check Answer Next Exit Hint 2011-2017 Sapling Learning, Inc about uscrsprivacy policy erms of usecotus help 8 50 PM O Type here to search 11/13/2017

Explanation / Answer

MM = 272.71 g/mol

V = 10 mL;

then V = 1 mL is used to a V = 25 mL

A = 0.49

l = 1 cm

e = 6533

A = e*l*C

A = absorbance of sample

e is the molar absorptivity , the typical units are 1/M-cm

l size of cuvette, typically reported in cm

c is the molar concentration, in mol per ltier or M

then

0.49 = 6533*1*C

C = 0.49 /6533

C = 0.00007500 M (in cuvette, this is a)

b)

in the v10 mL

M1*V1 = M2*V2

M1*1 = 0.00007500*25

M1 = 0.00007500*25 = 0.001875 M

this V = 1 mL, is the same concentration as in 10 mL

b) --> 0.001875 M

c)

mg in solution

V = 10 mL

M = 0.001875

mmol = MV = 10*0.001875 = 0.01875

mass = mmol*MW = 0.01875*272.71 = 5.1133 mg

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