A standard solution was prepared containing 10.0 ppm of an analyte and 15.0 ppm of an internal standard. Analysis of the sample gave signals for the analyte and internal standard of 0.155 and 0.233 (arbitrary units), respectively. Sufficient internal standard was added to an unknown sample to make it 15.0 ppm in the internal standard. Analysis of the unknown sample yielded signals for the analyte and internal standard of 0.274 and 0.198, respectively. Calculate the concentration of analyte in the unknown sample.
A standard solution was prepared containing 10.0 ppm of an analyte and 15.0 ppm of an internal standard. Analysis of the sample gave signals for the analyte and internal standard of 0.155 and 0.233 (arbitrary units), respectively. Sufficient internal standard was added to an unknown sample to make it 15.0 ppm in the internal standard. Analysis of the unknown sample yielded signals for the analyte and internal standard of 0.274 and 0.198, respectively. Calculate the concentration of analyte in the unknown sample.
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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A standard solution was prepared containing 10.0 ppm of an analyte and 15.0 ppm of an internal standard. Analysis of the sample gave signals for the analyte and internal standard of 0.155 and 0.233 (arbitrary units), respectively. Sufficient internal standard was added to an unknown sample to make it 15.0 ppm in the internal standard. Analysis of the unknown sample yielded signals for the analyte and internal standard of 0.274 and 0.198, respectively. Calculate the concentration of analyte in the unknown sample.
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