Chapter 28: Q4P (page 790)
What mass of sample in Figure 28-3 is expected to give a sampling standard deviation of \( \pm 6\% \)?
Short Answer
The mass of the sample which produces the relative standard deviation of \(1\% \)
Chapter 28: Q4P (page 790)
What mass of sample in Figure 28-3 is expected to give a sampling standard deviation of \( \pm 6\% \)?
The mass of the sample which produces the relative standard deviation of \(1\% \)
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Get started for freeExplain how to prepare a powder with an average particle diameter near \(100\mu m\) by using sieves from Table 28-2. How would such a particle mesh size be designated?
From their standard reduction potentials, which of the following metals would you expect to dissolve in \({\rm{HCl}}\)by the reaction\({\rm{M}} + n{{\rm{H}}^ + } \to {{\rm{M}}^{n + }} + \frac{n}{2}{{\rm{H}}_2}:{\rm{Zn}},{\rm{Fe}},{\rm{Co}},{\rm{Al}},{\rm{Hg}},{\rm{Cu}},{\rm{Pt}}\),\({\bf{Au}}\)?
(When the potential predicts that the element will not dissolve, it probably will not. If it is expected to dissolve, it may dissolve if some other process does not interfere. Predictions based on standard reduction potentials at \({\bf{2}}{{\bf{5}}^{^{\bf{o}}}}C\) are only tentative, because the potentials and activities in hot, concentrated solutions vary widely from those in the table of standard potentials.)
In analyzing a lot with random sample variation, you find a sampling standard deviation of \({\bf{65}}\% .\)Assuming negligible error in the analytical procedure, how many samples must be analyzed to give \(9{\bf{5}}\% \)confidence that the error in the mean is within\(64\% \)of the true value? Answer the same question for a confidence level of \(90\% \).
An example of a mixture of 1-mm-diameter particles of \({\rm{KCl}}\)and \({\rm{KN}}{{\rm{O}}_3}\)in a number ratio \(1:99\)follows Equation 28-4. A sample containing \({10^4}\)particles weighs\(11.0\;{\rm{g}}\). What is the expected number and relative standard deviation of \({\rm{KCl}}\)particles in a sample weighing\(11.0 \times {10^2}\;{\rm{g}}\)?
Why is it advantageous to use large particles \(\left( {{\bf{50}}{\rm{ }}\mu {\bf{m}}} \right)\) for solid phase extraction, but small particles \(\left( {{\bf{5}}{\rm{ }}\mu {\bf{m}}} \right)\) for chromatography?
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