/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q5P Explain how to prepare a powder ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Referring to Table 28-7, explain how an anion-exchange resin can be used for absorption and analysis of \({\bf{S}}{{\bf{O}}_2}\)released by combustion.

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\(11.0 \times {10^2}\;{\rm{g}}\)?

To pre-concentrate cocaine and benzoylecgonine from river water described at the opening of thischapter, solid-phase extraction was carried out at2mL pH2using the mixed-mode cationexchange resin in Figure 28-19. After passing 500mLof river water through  60mgof resin, the retained analytes were eluted first with2mLof localid="1663594337127" CH3OHand then with localid="1663594104084" 2mLof2% ammonia solution inCH3OH. Explain the purpose of using pH2for retention and dilute ammonia for elution.

What mass of sample in Figure 28-3 is expected to give a sampling standard deviation of \( \pm 6\% \)?

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.")

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.