Interpretation:
An internal standard and its uses needs to be explained.
Concept introduction:
Use of an internal standard before actual sample analysis is an established method in analytical tests and is one important step towards eliminating/reducing the sources of errors and ensuring the accuracy and precision of the results.
Answer to Problem 10.1QAP
An internal standard is a chemical substance, added to the sample, blank or calibration standards to identify and account for uncontrollable variables. The uncontrollable variables are expected to affect the internal standard and the analyte to a similar extent.
Explanation of Solution
Before actual analytical analysis of a substance, a blank and/or a calibration standard is run. This consists of a chemical substance of known composition and purity. It is added in a known quantity. This responds to the analytical and measurement processes as will the actual substance sample.
This is used to draw a calibration curve with the signals of internal standard on Y axis.
The accuracy of the results is increased as the instrument noise gets eliminated. This is because the signal to noise ratio is the same for the internal standard as for the substance under analysis.
Running the internal standards prior to running the analyte, improves the accuracy of the results. The internal standard, however, is to be properly selected.
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Chapter 10 Solutions
Principles of Instrumental Analysis
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- Assignment 4.docx - Word Lorelei C Home Insert Design Layout References Mailings Review View Help O Tell me what you want to do Cut Calibri (Body) |12 -AA Aa, .三,, 划T AaBbCcDc AaBbCcDc AaBbC AaBbCcC AaB AABBCCD AaBbCcDt AaBbCcD AaBbCcD AaE aCopy B I U - abc X, x A、少,A,三=== 加, -. 1 Normal INo Spac. Heading 1 Heading 2 Subtle Em. Emphasis Intense E.. Title Subtitle Str Format Painter ipboard Font Paragraph Styles 8.46. A scuba diver 60 ft below the ocean surface inhales 50.0 mL of compressed air from a scuba tank at a pressure of 3.00 atm and a temperature of 8 C. What is the final pressure of air, in atmospheres, in the lungs when the gas expands to 150.0 mL at a body temperature of 37 C, if the amount of gas does not change? Page 3 of 3 1032 words o de e W Type here to search DELLarrow_forwardAutoSave Off Document1 - Word O Search retroget File Home Insert Draw Design Layout References Mailings Review View Help A Share P Comments O Find - E Replace Calibri (Body) - 16 - A A Aa- A -- v i- v a- v AaBbCcDc AaBbCcDc AaBbC AaBbCcC AaB Paste I U v ab x, x A - ev Av 1 Normal I No Spac. Heading 1 Heading 2 Dictate Editor Reuse В Title A Select Files Clipboard Font Paragraph Styles Editing Voice Editor |Reuse Files Calculate the standard reaction entropy of H2(g)+1202(g) → H2O (1) at 25°C. Given that Smø (H2O,/) = 69.9 J· K-1•mol-1; Smø (H2,g) = 130.7 J·K-1-mol-1; Smø (02,g) = 205.0 J-K-1•mol-1. Page 1 of 1 English (Malaysia) O Focus 41 words 100% 8:59 AM 26°C Mostly sunny A 4)) G 29/06/2021arrow_forwardAc₂0/NaOAc PhCO,H HNO,/H₂SO4 Nowarrow_forward
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- Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning