If the concentration of the analyte in the sample solution (?) is 0.070 µg/mL, what's the concentration of the pesticide in the sample (ppm)? Show solution/explanation. 0.52 ppm0.70 ppm0.99 ppm2.2 ppm2.9 ppm A 0.2422 g sample of a certain fruit was analyzed for the content of a certain pesticide. After a series ofextraction steps, the final extract obtained was diluted to 10.00 mL, which was then subjected tochromatographic analysis. Four standard solutions of the pesticide were also prepared, processed, andanalyzed in the same manner. The data obtained are shown in the table below.SolutionConcentration (µg/mL)Peak AreaStandard 10.01234.2Standard 20.051101.2Standard 30.102300.1Standard 40.204999.8Sample?1654.3

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If the concentration of the analyte in the sample solution (?) is 0.070 µg/mL, what's the concentration of the pesticide in the sample (ppm)? Show solution/explanation. 0.52 ppm 0.70 ppm 0.99 ppm 2.2 ppm 2.9 ppm

If the concentration of the analyte in the sample solution (?) is 0.070 µg/mL, what's the concentration of the pesticide in the sample (ppm)? Show solution/explanation. 0.52 ppm 0.70 ppm 0.99 ppm 2.2 ppm 2.9 ppm

Appl Of Ms Excel In Analytical Chemistry

2nd Edition

ISBN:9781285686691

Author:Crouch

Publisher:Crouch

Chapter14: Chromatography

Section: Chapter Questions

Problem 9P

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250 mL of water sample was collected from a river and sent to a laboratory to determine the concentration of the total suspended solid. In the laboratory, 5 mL of the water sample was transferred to a 100 mL volumetric flask before diluted with distilled water to 100 mL. After that, 50 mL of the diluted water sample was transferred to a funnel containing a glass fiber filter and attached to a glass flask with the sidearm connected to a operating vacuum pump. Transferred diluted water sample was left drained through the glass filter before glass fiber was transferred to an oven to heat at 105 oC under a constant weight of 0.2654 g was obtained. If the initial weight of the glass fiber filter is 0.2431 g. Calculate the amount of the suspended solid in mg unit contained in the 250 mL collected water sample.
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A 0.2422 g sample of a certain fruit was analyzed for the content of a certain pesticide. After a series of extraction steps, the final extract obtained was diluted to 10.00 mL, which was then subjected to chromatographic analysis. Four standard solutions of the pesticide were also prepared, processed, and analyzed in the same manner. The data obtained are shown in the table below. Solution Concentration (µg/mL) Peak Area Standard 1 0.01 234.2 Standard 2 0.05 1101.2 Standard 3 0.10 2300.1 Standard 4 0.20 4999.8 Sample ? 1654.3
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