A standard sample contains 15.0 mM of analyte and 25.0 mM of internal standard. The analysis of the standard sample gives peak areas for the analyte and internal standard of 1.782 and 3.539, respectively, in a chromatography analysis. Then sufficient internal standard is added to a sample to make its concentration 10.0 mM. The analysis of this new sample yields peak areas for the analyte and internal standard of 0.894 and 2.981, respectively. Name the calibration method and report the analyte's concentration in the new sample.

A standard sample contains 15.0 mM of analyte and 25.0 mM of internal standard. The analysis of the standard sample gives peak areas for the analyte and internal standard of 1.782 and 3.539, respectively, in a chromatography analysis. Then sufficient internal standard is added to a sample to make its concentration 10.0 mM. The analysis of this new sample yields peak areas for the analyte and internal standard of 0.894 and 2.981, respectively. Name the calibration method and report the analyte's concentration in the new 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 sample contains 15.0 mM of analyte and 25.0 mM of internal
standard. The analysis of the standard sample gives peak areas for the analyte
and internal standard of 1.782 and 3.539, respectively, in a chromatography
analysis. Then sufficient internal standard is added to a sample to make its
concentration 10.0 mM. The analysis of this new sample yields peak areas for
the analyte and internal standard of 0.894 and 2.981, respectively. Name the
calibration method and report the analyte's concentration in the new sample.
(e)

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