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Chapter 5: Q4P (page 114)

Distinguish raw data, treated data, and results.

Short Answer

Expert verified

Raw data are the unprocessed measurements collected from the specimens being studied.
Treated data are the concentrations found from the raw data with the use of calibration procedures.
Results are the reported data after the application of various statistical analyses to the treated data.

Step by step solution

01

Raw data, Treated data and results

Raw data:

Raw data are data's which can be measured directly.

Examples such as peak area in a chromatogram or volume from a burette.

Treated data:

When the raw data applied using calibration methods the resultant data obtained is known as treated data.

Results:

Results are ultimate reports that we report after the application of statistical methods (mean, standard deviation) to the Treated data.

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Most popular questions from this chapter

Standard addition. An unknown sample of ${Ni}^{2 +}$ gave a current of $2.36 μA$ in an electrochemical analysis. When $0.500 mL$ of solution containing role="math" localid="1654761474124" $0.0187 {MNi}^{2 +}$ was added to $25.0 mL$ of unknown, the current increased to $3.79 μA$ .

(a) Denoting the initial, unknown concentration as $[{Ni}^{2 +}]$ , write an expression for the final concentration, ${[{Ni}^{2 +}]}_{f}$ , after role="math" $25.0 mL$ of unknown were mixed with $0.500 mL$ of standard. Use the dilution factor for this calculation.

(b) In a similar manner, write the final concentration of added standard ${Ni}^{2 +}$ , designated as ${[S]}_{f}$ .

(c) Find $[{Ni}^{2 +}]$ in the unknown.

Why is it desirable in the method of standard addition to add a small volume of concentrated standard rather than a large volume of dilute standard?

Chloroform is an internal standard in the determination of the pesticide DDT in a polarographic analysis in which each compound is reduced at an electrode surface. A mixture containing $0.500 mM$ chloroform and $0.800 mM$ DDT gave signals of $15.3 μA$ for chloroform and $10.1 μA$ for DDT. An unknown solution $(10.0 mL)$ containing DDT was placed in a $100 - mL$ volumetric flask and $10.2 μL$ of chloroform (FM 119.39, density $= 1.484 g / mL)$ were added. After dilution to the mark with solvent, polarographic signals of and $8.7 μA$ were observed for the chloroform and DDT, respectively. Find the concentration of DDT in the unknown.

How can you validate precision and accuracy?

A solution containing $3.47 mMX$ (analyte) and $1.72 mMS$ (standard) gave peak areas of $3473$ and $10222$ ,respectively, in a chromatographic analysis. Then $1.00 mL$ of $8.47 mMS$ was added to $5.00 mL$ of unknown $X$ ,and the mixture was diluted to $10.00 mL$ . This solution gave peak areas of $5428$ and $4431$ for $X$ and $S$ , respectively.

(a) Calculate the response factor for the analyte.

(b) Find the concentration of $S (mM) in the 10.0 - mL mixture .$

(c) Find the concentration of $X (mM) in the 10.0 - mL mixture .$

(d) Find the concentration of $X in the orignal unknown .$

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