The concentration of nitrite (NO₂) in the unknown sample was determined by a direct (ordinary) calibration plot as shown below: 0.300 Absorbance at 520nm 0.250 0.200 0.150 0.100 0.050 0 0.4 0.5 0.6 0.7 0.8 0.9 Nirite (µg/ml) If the absorbance of the unknown sample was measured 0.238 at 520 nm, what would be the concentration of nitrite? y = 0.1793x + 0.0763 R² = 0.9969 0.1 0.2 0.564 μg/ml 0.200 μg/ml 0.754 μg/ml 0.902 µg/ml 0.801 μg/ml 0.3 1
The concentration of nitrite (NO₂) in the unknown sample was determined by a direct (ordinary) calibration plot as shown below: 0.300 Absorbance at 520nm 0.250 0.200 0.150 0.100 0.050 0 0.4 0.5 0.6 0.7 0.8 0.9 Nirite (µg/ml) If the absorbance of the unknown sample was measured 0.238 at 520 nm, what would be the concentration of nitrite? y = 0.1793x + 0.0763 R² = 0.9969 0.1 0.2 0.564 μg/ml 0.200 μg/ml 0.754 μg/ml 0.902 µg/ml 0.801 μg/ml 0.3 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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![The concentration of nitrite (NO2) in the unknown sample was determined by a direct (ordinary)
calibration plot as shown below:
0.300
Absorbance at 520nm
0.250
0.200
0.150
0.100
0.050
0
0.564 µg/ml
0.200 μg/ml
0.754 µg/ml
0.902 µg/ml
y = 0.1793x + 0.0763
R² = 0.9969
0.1 0.2
0.801 μg/ml
If the absorbance of the unknown sample was measured 0.238 at 520 nm, what would be the
concentration of nitrite?
0.3
0.4 0.5 0.6 0.7 0.8 0.9 1
Nirite (µg/ml)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F93d9f4fa-0622-4417-979d-12a19703a696%2F7ac55020-0eeb-4388-8bc9-2d82d8eb655d%2Fc2e8uo_processed.png&w=3840&q=75)
Transcribed Image Text:The concentration of nitrite (NO2) in the unknown sample was determined by a direct (ordinary)
calibration plot as shown below:
0.300
Absorbance at 520nm
0.250
0.200
0.150
0.100
0.050
0
0.564 µg/ml
0.200 μg/ml
0.754 µg/ml
0.902 µg/ml
y = 0.1793x + 0.0763
R² = 0.9969
0.1 0.2
0.801 μg/ml
If the absorbance of the unknown sample was measured 0.238 at 520 nm, what would be the
concentration of nitrite?
0.3
0.4 0.5 0.6 0.7 0.8 0.9 1
Nirite (µg/ml)
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