/*! 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} Q28P Which of the following columns w... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Which of the following columns will provide:

(a) highest number of plates?

(b) greatest retention?

(c) highest relative retention?

(d) best separation?

Column1:N=1000;k2=3.9;α=1.16;resolution=0.6Column2:N=5000;k2=3.9;α=1.06;resolution=0.8Column3:N=500;k2=4.7;α=1.31;resolution=1.1Column4:N=2000;k2=2.4;α=1.24;resolution=1.5

Short Answer

Expert verified

The solution is

- a) Column 2 because it has the biggest N value (N = 5000)

- b) Column 3 because it has the largest k2value k2=4.7

- c) Column 3 because it has the largest role="math" localid="1654934451145" αvalue α=1.31

- d) Column 4 because it is the only one with baseline resolution (resolution = 1.5)

Step by step solution

01

of 3

In this task we will determine which of the columns will provide:

a) highest number of plates

b) greatest retention

c) highest relative retention

d) best separation

02

of 3

Column1:N=1000;k2=1.2;α=1.16;resolution=0.6Column2:N=5000;k2=3.9;α=1.06;resolution=0.8Column3:N=500;k2=4.7;α=1.31;resolution=1.1Column4:N=2000;k2=2.4;α=1.24;resolution=1.5

03

of 3

The solution is

- a) Column 2 because it has the biggest N value (N = 5000)

- b) Column 3 because it has the largest k2value k2=4.7

- c) Column 3 because it has the largest αvalue α=1.31

- d) Column 4 because it is the only one with baseline resolution (resolution = 1.5)

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

Butanoic acid has a partition coefficient of 3.0 (favouring benzene) when distributed between water and benzene. Find the formal concentration of butanoic acid in each phase when 100mL of 0.10M aqueous butanoic acid is extracted with 25mL of benzene (a) at pH 4.00 and (b) at pH 10.00.

Which column is more efficient: plate height = 0.1 mm or plate height = 1 mm?

Consider a chromatography experiment in which two components with retention factors k1=4.00and k2=5.00are injected into column with N=1.00×103theoretical plates. The retention time for the less-retained component is tr1=10.0min.

(a)Calculate tmandt12.Find w12(width at half height) and w (at the base) for each peak.

(b)Using graph paper , sketch the chromatogram analogous to figure 23-7,supposing that two peaks have same amplitude(height). Draw the half widths accurately.

(c)Calculate the resolution of the two peaks and compare this value with those drawn in Figure 23-10.

The partition coefficient for a solute in chromatography is K=cs/cm, where csis the concentration in the stationary phase and cmis the concentration in the mobile phase. The larger the partition coefficient, the longer it takes a solute to be eluted. Explain why.

Consider the extraction of Mntfrom aqueous solution into organic solution by reaction with protonated ligand HL : D=Mn+(aq)+nHL(org)â–¡MLn(org)+nH+(aq)Kextraction=[MLn]org[H+]aqn[Mn+]aq[HL]orgn.Rewrite Equation 23 - 13 in terms of role="math" localid="1654863844402" Kextractionand express Kextractionin terms of the constants in Equation 23 - 13 . Give a physical reason why each constant increases or decreases Kextraction

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.