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Chapter 28: Q5P (page 790)

Explain 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?

Short Answer

Expert verified

The sample retained by\(170\) mesh sieve would have a size between\(0.090\;{\rm{mm}}\) and\(0.125\;{\rm{mm}}\), it would be called\(120/170\) mesh

Step by step solution

01

Definition of Standard deviation.

  • The standard deviation is a measure of how far something deviates from the mean (for example, spread, dispersion, or spread). A "typical" variation from the mean is represented by the standard deviation.
  • Because it returns to the data set's original units of measurement, it's a common measure of variability.
  • The standard deviation, defined as the square root of the variance, is a statistic that represents the dispersion of a dataset relative to its mean.
02

Determine the particle mesh size be designated.

  • In this task we will explain the preparation of a powder with an average particle diameter\(100\mu {\rm{m}}\)using the sieves from Table\(28 - 2\).

  • First we will convert \(100\mu {\rm{m}}\) to\({\rm{mm}}\)in order to find the mesh:

\(100\mu {\rm{m}}= 0.1\;{\rm{mm}}\)

  • Which would be between\(120 - 140 - 170\) sieve number (considering that values of screen opening from Table\(28 - 2\)are\(0.125 - 0.106 - 0.090\), which are all \(0.1\) mm)
  • From Table \(28 - 2\) we can say that we would use a\(120\)mesh sieve, through which we would pass the powder and then proceed to the\(170\)mesh sieve

Therefore, the sample retained by\(170\)mesh sieve would have a size between\(0.090\;{\rm{mm}}\)and\(0.125\;{\rm{mm}}\), it would be called \(120/170\)mesh

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Most popular questions from this chapter

When you flip a coin, the probability of its landing on each side is \(p = q = \frac{1}{2}\)in Equations 28-2 and 28-3. If you flip it \(n\)times, the expected number of heads equals the expected number of tails \( = np = nq = \frac{1}{2}n.\)The expected standard deviation for \(n\)flips is\({\sigma _n} = \sqrt {npq} \). From Table 4-1, we expect that \(68.3\% \)of the results will lie within \( \pm 1{\sigma _n}\) and \(95.5\% \)of the results will lie within\( \pm 2{\sigma _n}\).

(a) Find the expected standard deviation for the number of heads in \({\bf{1000}}\) coin flips.

(b) By interpolation in Table 4-1, find the value of \(z\)that includes \(90\% \)of the area of the Gaussian curve. We expect that \(90\% \)of the results will lie within this number of standard deviations from the mean.

(c) If you repeat the\({\bf{1000}}\)coin flips many times, what is the expected range for the number of heads that includes\(90\% \) of the results? (For example, your answer might be, "The range \({\bf{490}}\) to \({\bf{510}}\) will be observed \(90\% \)of the time.")

To pre-concentrate cocaine and benzoylecgonine from river water described at the opening of this chapter, solid-phase extraction was carried out at \({\rm{pH}}\,\,2\) using the mixed-mode cation-exchange resin in Figure 28-19. After passing \(500\;{\rm{mL}}\)of river water through \(60{\rm{mg}}\)of resin, the retained analytes were eluted first with \(2\;{\rm{mL}}\)of \({\rm{C}}{{\rm{H}}_3}{\rm{OH}}\)and then with \(2\;\,\,{\rm{mL }}of\,\,\,2\% \) ammonia solution in\({\rm{C}}{{\rm{H}}_3}{\rm{OH}}\). Explain the purpose of using \({\rm{pH}}2\) for retention and dilute ammonia for elution.

Acid-base equilibria of Cr(III) were summarized in Problem 10-36. Cr(VI) in aqueous solution above pH 6 exists as the yellow tetrahedral chromate ion, \({\rm{CrO}}_4^{2 - }.\)Between\({\rm{pH}}2\)and \(6,{\rm{Cr}}\)(VI) exists as an equilibrium mixture of\({\rm{HCrO}}_4^ - \) and orange-red dichromate,\({\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 - }.{\rm{Cr}}({\rm{VI}})\) is a carcinogen, but \({\rm{Cr }}(III)\)is not considered to be as harmful. The following procedure was used to measure\({\rm{Cr }}({\rm{VI}})\) in airborne particulate matter in workplaces.

1. Particles were collected by drawing a known volume of air through a polyvinyl chloride filter with \(5 - \mu {\rm{M}}\)pore size.

2. The filter was placed in a centrifuge tube and \(10\;{\rm{mL}}\)of \(0.05{\rm{M}}{\left( {{\rm{N}}{{\rm{H}}_4}} \right)_2}{\rm{S}}{{\rm{O}}_4}/0.05{\rm{MN}}{{\rm{H}}_3}buffer,{\rm{pH}}8,\) were added. The immersed filter was agitated by ultrasonic vibration for\(30\;{\rm{min}}\)at \({35^\circ }{\rm{C}}\)to extract all \({\rm{Cr }}(III)and{\rm{Cr}}\)(VI) into solution.

3. A measured volume of extract was passed through a "strongly basic" anion exchanger (Table 26-1) in the \({\rm{C}}{{\rm{l}}^ - }\)form. Then the resin was washed with distilled water. Liquid containing \({\rm{Cr}}\)(III) from the extract and the wash was discarded.

4. Cr(VI) was then eluted from the column with\(0.5{\rm{M}}{\left( {{\rm{N}}{{\rm{H}}_4}} \right)_2}{\rm{S}}{{\rm{O}}_4}/0.05{\rm{MN}}{{\rm{H}}_3}\) buffer, \({\rm{pH}}8,\)and collected in a vial.

5. The eluted \({\rm{Cr}}\)(VI) solution was acidified with \({\rm{HCl}}\)and treated with a solution of 1,5 -diphenylcarbazide, a reagent that forms a colored complex with Cr(VI). The concentration of the complex was measured by its visible absorbance.

(a) What are the dominant species of \({\rm{Cr}}\)(VI) and \({\rm{Cr}}\)(III) at\({\rm{pH}}8\)?

(b) What is the purpose of the anion exchanger in step 3 ?

(c) Why is a "strongly basic" anion exchanger used instead of a "weakly basic" exchanger?

(d) Why is Cr(VI) eluted in step 4 but not step 3 ?

Barium titanate, a ceramic used in electronics, was analyzed by the following procedure: Into a Pt crucible was placed \(1.2\;{\rm{g}}\)of \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\) and \(0.8\;{\rm{g}}\)of \({\rm{N}}{{\rm{a}}_2}\;{{\rm{B}}_4}{{\rm{O}}_7}\)plus \(0.3146\;{\rm{g}}\)of unknown. After fusion at \({1000^\circ }{\rm{C}}\)in a furnace for\(30\;{\rm{min}}\), the cooled solid was extracted with \(50\;{\rm{mL}}\)of\(6{\rm{MHCl}}\), transferred to a \(100 - {\rm{mL}}\) volumetric flask, and diluted to the mark. A \(25.00 - {\rm{mL}}\)aliquot was treated with \(5\;{\rm{mL}}\)of \(15\% \)tartaric acid (which complexes \({\rm{T}}{{\rm{i}}^{4 + }}\)and keeps it in aqueous solution) and \(25\;{\rm{mL}}\)of ammonia buffer,\({\rm{pH}}9.5\). The solution was treated with organic reagents that complex\({\rm{B}}{{\rm{a}}^{2 + }}\), and the \({\rm{Ba}}\)complex was extracted into \({\rm{CC}}{{\rm{l}}_4}.\)After acidification (to release the \({\rm{B}}{{\rm{a}}^{2 + }}\) from its organic complex), the \({\rm{B}}{{\rm{a}}^{2 + }}\)was backextracted into\(0.1{\rm{MHCl}}\). The final aqueous sample was treated with ammonia buffer and methylthymol blue (a metal ion indicator) and titrated with \(32.49\;{\rm{mL}}\) of \(0.01144{\rm{M}}\)EDTA. Find the weight per cent of Ba in the ceramic.

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.)
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