Two components in an HPLC separation have retention times that differ by 15 s. The first peak elutes in 9.0 min, and the peak widths are approximately equal. The dead time to was 65 s. Use a spreadsheet to find the minimum number of theoretical plates needed to achieve the following resolution. R. values: 0.50 0.75, 0.90, 1.0, 1.10. 1.25, 1.50. L75. 2.0 and 2.5. How would the results change if peak 2 were twice as broad as peak 1
Two components in an HPLC separation have retention times that differ by 15 s. The first peak elutes in 9.0 min, and the peak widths are approximately equal. The dead time to was 65 s. Use a spreadsheet to find the minimum number of theoretical plates needed to achieve the following resolution. R. values: 0.50 0.75, 0.90, 1.0, 1.10. 1.25, 1.50. L75. 2.0 and 2.5. How would the results change if peak 2 were twice as broad as peak 1
Principles of Instrumental Analysis
7th Edition
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Chapter27: Gas Chromatography
Section: Chapter Questions
Problem 27.23QAP
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Two components in an HPLC separation have retention times that differ by 15 s. The first peak elutes in 9.0 min, and the peak widths are approximately equal. The dead time to was 65 s. Use a spreadsheet to find the minimum number of theoretical plates needed to achieve the following resolution. R. values: 0.50 0.75, 0.90, 1.0, 1.10. 1.25, 1.50. L75. 2.0 and 2.5. How would the results change if peak 2 were twice as broad as peak 1?
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