0.900 y = 0.1215x - 0.0023 R2 = 0.9982 0.800 0.700 0.600 0.500 0.400

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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In a Beer's Law study, a Beer's Law Graph for FeSCN2+ was done at 485 nm and is shown below.
Beer's Law Plot of FeSCN2+
0.900
y = 0.1215x - 0.0023
R2 = 0.9982
0.800
0.700
0.600
0.500
0.400
0.300
0.200
0.100
0.000
0.000
1.000
2.000
3.000
4.000
5.000
6.000
7.000
8.000
-0.100
Concentration (ppm)
If a unknown sample containing FeSCN2+ has an absorbance of 0.542, what is the concentration of
FESCN2+ in the unknown sample? (Note: ppm = parts per million)
The concentration of the unknown sample is
in units of
If the path length of light is 13.5 mm (millimeters), the absorptivity coefficient, ɛ, for FESCN2* is
in units of
(Do not use superscripts, subscripts, or carets. Write 1.2 x 103 as 1.2x10-3 or 1.2e-3.)
Absorbance at 485 nm
Transcribed Image Text:In a Beer's Law study, a Beer's Law Graph for FeSCN2+ was done at 485 nm and is shown below. Beer's Law Plot of FeSCN2+ 0.900 y = 0.1215x - 0.0023 R2 = 0.9982 0.800 0.700 0.600 0.500 0.400 0.300 0.200 0.100 0.000 0.000 1.000 2.000 3.000 4.000 5.000 6.000 7.000 8.000 -0.100 Concentration (ppm) If a unknown sample containing FeSCN2+ has an absorbance of 0.542, what is the concentration of FESCN2+ in the unknown sample? (Note: ppm = parts per million) The concentration of the unknown sample is in units of If the path length of light is 13.5 mm (millimeters), the absorptivity coefficient, ɛ, for FESCN2* is in units of (Do not use superscripts, subscripts, or carets. Write 1.2 x 103 as 1.2x10-3 or 1.2e-3.) Absorbance at 485 nm
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