Report table 2: Determination of Molar absorptivity of K:CrO4 ppm of K:CrO4 in "DILUTE STOCK (5mL/50mL)"=1.5012 Molarity of K-Croin "DILUTE STOCK(5mL/50mL)" = 7.7319x 10^-3 Peak wavelength Peak Absorbance From spectrum λmax=372 nm A=0.487 8=> 1 2 3 4 A=0.370 A=0.434 A=0.487 A=0.347 From single wavelength values 8= 8= 1.2864×10 6 1.5171×10^6 1.7023×10 6 1.20642×10^6 Average & (for the 4 readings) Standard deviation of 1.428055×10^6 95% Confidence interval of ε Show sample calculation (for solution 1): = A/b.c = Molar Absorption= 0,370/(3.72×10^-5)(7.7319× 10^-3)= 1.2864×10^6 M^-1 cm^-1
Report table 2: Determination of Molar absorptivity of K:CrO4 ppm of K:CrO4 in "DILUTE STOCK (5mL/50mL)"=1.5012 Molarity of K-Croin "DILUTE STOCK(5mL/50mL)" = 7.7319x 10^-3 Peak wavelength Peak Absorbance From spectrum λmax=372 nm A=0.487 8=> 1 2 3 4 A=0.370 A=0.434 A=0.487 A=0.347 From single wavelength values 8= 8= 1.2864×10 6 1.5171×10^6 1.7023×10 6 1.20642×10^6 Average & (for the 4 readings) Standard deviation of 1.428055×10^6 95% Confidence interval of ε Show sample calculation (for solution 1): = A/b.c = Molar Absorption= 0,370/(3.72×10^-5)(7.7319× 10^-3)= 1.2864×10^6 M^-1 cm^-1
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:Report table 2: Determination of Molar absorptivity of K2CrO4
ppm of K2CrO4 in "DILUTE STOCK (5mL/50mL)"=1.5012
Molarity of K:CrO4 in "DILUTE STOCK(5mL/50mL)" = 7.7319× 10^-3
Peak wavelength
λmax=372 nm
From single
wavelength
values
Peak Absorbance
From spectrum
A = 0.487
8=>
1
2
3
4
A= 0.370
A=0.434
A= 0.487
A= 0.347
8=
8=
8=
1.2864×10^6
1.5171×10^6
1.7023×10 6 1.20642×10^6
Average & (for the 4
readings)
Standard deviation of a
95% Confidence interval
of a
= 1.428055×10^6
Show sample calculation (for solution 1): = A/b.c =
Molar Absorption= 0,370/(3.72*10^-5)(7.7319× 10^-3)= 1.2864×10^6 M^-1cm^-1
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