Principles of Instrumental Analysis
7th Edition
ISBN: 9781305577213
Author: Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher: Cengage Learning
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Chapter 14, Problem 14.9QAP
Interpretation Introduction
(a)
Interpretation:
The concentrations of A and B in solution should be determined that yielded absorbance 0.439 at 475 nm and 1.025 at 700 nm.
Concept introduction:
The Beer-Lambert Law is:
A − absorbance
l − length of the solution light passes through (cm)
c − concentration of solution (mol/L)
Interpretation Introduction
(b)
Interpretation:
The concentrations of A and B in solution should be determined that yielded absorbance 0.662 at 475 nm and 0.815 at 700 nm.
Concept introduction:
The Beer-Lambert Law is:
A − absorbance
l − length of the solution light passes through (cm)
c − concentration of solution (mol/L)
Expert Solution & Answer
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Students have asked these similar questions
When measured in a 1-cm cell, an 8.50 x 10-5 M solution of species A exhibited absorbances of
0.129 and 0.764 at 475 and 700 nm, respectively. A 4.65 x 10-5 M solution of species B gave absorbances of 0.567 and 0.083 under the same circumstances. Calculate the concentrations of
A and B in solutions that yielded the following absorbance data in a 1.25 cm cell : 0.502 at 475
nm and 0.912 at 700 nm.
Pls answer need help.
When measured in a 1.00 cm cuvette, a 4.1x10M solution of species X exhibited absorbances of 0.1025 and 0.533 at 530 nm and 325 nm, respectively. A8.20x10
M solution of species Y gave absorbances of 0.230 and 0.508 at 530 nm and 325 nm, respectively. Both species were dissolved in the same solvent, and the solvent's
absorbance was 0.000 and 0.000 at 530 nm and 325 nm, respectively, in a 1 cm cuvette.
Calculate the concentrations of X and Y in an unknown solution that yielded absorbance data of 0.683 at 530 nm and 1.351 at 325 nm in a 1.0 em cuvete
O A.X-4.20 x10 M
Y=3.20 x104 M
OBX-1.25 X10M
Y420 x10-SM
OCX-3.75x1oM
Y-1 A10M
OD.X-3.80 x 10M
Y=2.10x10M
When measured in a 1.00 cm cuvette, a 4.1x10M solution of species X exhibited absorbances of 0.1025 and 0.533 at 530 nm and 325 nm, respectively. A 8.20x10
M solution of species Y gave absorbances of 0.230 and 0.508 at 530 nm and 325 nm, respectively. Both species were dissolved in the same solvent, and the solvent's
absorbance was 0.000 and 0.000 at 530 nm and 325 nm, respectively, in a 1 cm cuvette.
Calculate the concentrations of X and Y in an unknown solution that yielded absorbance data of 0,683 at 530 nm and 1.351 at 325 nm in a 1.0 cm cuvette
A. X= 4.20 X10M
Y=3.20 x10+M
O B. X1.25 X10M
Y= 4.20 x10-5M
OCX-3.75X10M
Y-1 4x10M
OD.X-3.80 x 10 M
Y 2.10K10M
Chapter 14 Solutions
Principles of Instrumental Analysis
Ch. 14 - Prob. 14.1QAPCh. 14 - A 0.4740-g pesticide sample was decomposed by wet...Ch. 14 - Sketch a photometric titration curve for the...Ch. 14 - Prob. 14.4QAPCh. 14 - Prob. 14.5QAPCh. 14 - The accompanying data (1.00-cm cells) were...Ch. 14 - A 3.03-g petroleum specimen was decomposed by wet...Ch. 14 - Prob. 14.8QAPCh. 14 - Prob. 14.9QAPCh. 14 - The acid-base indicator HIn undergoes the...
Ch. 14 - Prob. 14.11QAPCh. 14 - Prob. 14.12QAPCh. 14 - Copper(II) forms a 1:1 complex with the organic...Ch. 14 - Aluminum forms a 1:1 complex with...Ch. 14 - Prob. 14.15QAPCh. 14 - Prob. 14.16QAPCh. 14 - Prob. 14.17QAPCh. 14 - Prob. 14.18QAPCh. 14 - Prob. 14.19QAPCh. 14 - Given the Information that...Ch. 14 - Prob. 14.21QAPCh. 14 - Mixing the chelating reagent B with Ni(II) forms...Ch. 14 - Prob. 14.23QAP
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