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(a). Sketch a graph of the van Deemnter equation (plate height versus linear flow rate).what would the curve look like if the multiple path term were 0? If the longtitundinal diffusinal diffusion term were 0?

(b). Explain why the van Deemter curve for 1.8渭尘particles in figure 25-3is nearly flat at high flow rate.what can you say about each of the terms in the van Deemeter equation for 1.8渭尘particles.

(c). Explain why the 2.7渭尘superficially porous particle enables separations similar to those achieved by 1.8渭尘totally porous particles,but the superficially porous particle requires lower pressure.

Short Answer

Expert verified

a) Sketch a graph of the van Deemnter.

b) Comparing the van Deemter curve of the particles in Figure to the graph

c) Superficially porous particles have a diameter non porous silica core.

Step by step solution

01

Sketch a graph of the van Deemnter 

The van Deemter equation was graphed to help visualise each term in the equation. Figure 24-3 was then examined using the van Deemter equation and its implications.

The figure below shows the graph of the van Deemter equation when the multiple paths term ( A) is , when the longitudinal diffusion term ( B) is , when the equilibration time term (C ) is constant, and when all terms are .

02

Comparing the van Deemter

b) why the van Deemter curve

Comparing the van Deemter curve of the 1.8渭尘particles in Figure 24-3to the graph in (a), we can see that the graph that it most resembles is the graph of c=0. Although not obvious from the graphs, the equation of the curve of 1.8渭尘particles has a small multiple paths term (A ) relative to that of the other larger particles in the graph.

Based on the graph, the curve of the particles plateaus at the smallest plate height ( H) compared to the other graphs. The reduction in plate heights can be related to the decrease in the band spreading due to multiple flow paths.

Given this relationship, the decrease in the H plateau value of the particles means that there is a decrease in A as the particle size decreases.

c) superficially porous particle

Superficially porous particles have a 5渭尘diameter non porous silica core. Compared to other fully porous particles with a larger radii, superficially porous particles can perform at the same level or better.

This is due to the structure of superficially porous particles, having a 0.25 thick porous silica layer on the outside of its 5渭尘diameter non porous silica core

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Most popular questions from this chapter

The antitumor drug gimatecan is available as nearly pure (S)-enantiomer. Neither pure (R)-enantiomer nor a racemic (equal) mixture of the two enantiomers is available. To measure small quantities of (R)-enantiomer in nearly pure (S)-gimatecan, a preparation was subjected to normal-phase chromatography on each of the enantiomers of a commercial, chiral stationary phase designated (S,S)- and (R,R)-DACH-DNB. Chromatography on the (R,R)-stationary phase gave a slightly asymmetric peak at tr 5 6.10 min with retention factor k 5 1.22. Chromatography on the (S,S)- stationary phase gave a slightly asymmetric peak at tr 5 6.96 min with k 5 1.50. With the (S,S) stationary phase, a small peak with 0.03% of the area of the main peak was observed at 6.10 min.

Chromatography of gimatecan on each enantiomer of a chiral stationary phase. Lower traces have enlarged vertical scale. [Data from E. Badaloni, W. Cabri, A. Ciogli, R. Deias, F. Gasparrini, F. Giorgi, A. Vigevani, and C. Villani, 鈥淐ombination of HPLC 鈥業nverted Chirality Columns Approach鈥 and MS/MS Detection for Extreme Enantiomeric Excess Determination Even in Absence of Reference Samples.鈥 Anal. Chem. 2007, 79, 6013.]

(a) Explain the appearance of the upper chromatograms. Dashed lines are position markers, not part of the chromatogram. What Problems 709 would the chromatogram of pure (R)-gimatecan look like on the same two stationary phases?

(b) Explain the appearance of the two lower chromatograms and why it can be concluded that the gimatecan contained 0.03% of the (R)-enantiomer. Why is the (R)-enantiomer not observed with the (R,R)-stationary phase?

(c) Find the relative retention (a) for the two enantiomers on the (S,S)-stationary phase.

(d) The column provides N 5 6 800 plates. What would be the resolution between the two equal peaks in a racemic (equal) mixture of (R)- and (S)-gimatecan? If the peaks were symmetric, does this resolution provide baseline separation in which signal returns to baseline before the next peak begins?

In monolithic columns60 the stationary phase is a single porous piece of silica or polymer filling the entire column and synthesized within the column from liquid precursors. Monolithic columns offer similar plate height to HPLC particles, but with less resistance to flow. Therefore, faster flow or longer columns can be used. The figure shows separation of isotopic molecules on a long monolithic column. Packed columns have too much resistance to flow to be made so long.

Separation of isotopic molecules on a 440-cm-long monolithic C18-silica column eluted withCH3CN/H2O(30: 70 vol/vol) at 308C. [Data from K. Miyamoto, T. Hara, H. Kobayashi, H. Morisaka, D. Tokuda, K. Horie, K. Koduki, S. Makino, O. Nu帽ez, C. Yang, T. Kawabe, T. Ikegami, H. Takubo, Y. Ishihama, and N. Tanaka, 鈥淗igh-Efficiency Liquid Chromatographic Separation Utilizing Long Monolithic Silica Capillary Columns,鈥 Anal. Chem. 2008, 80, 8741.]

(a) Unretained thiourea is eluted in 41.7 min. Find the linear velocity ux (mm/s).

(b) Find the retention factor k forC6D6

(c) Find the plate number N and plate height forC6D6

(d) Assuming that the peak widths forC6H5Dand C6D6are the same as that of C6D6, find the resolution of C6H5Dand C6D6.

(f) If we just increased the column length to increase N, what value of N and what column length would be required for a resolution of 1.000?

(g) Without increasing the length of the column, and without changing the stationary phase, how might you improve the resolution?

(h) When the solvent was changed fromCH3CN/H2O(30:70 vol/vol) toCH3CN/CH3OH/H2O(10:5:85), the relative retention for C6H5D andC6D6increased to 1.0088 and the retention factor for C6H6 changed to 17.0. If the plate number were unchanged, what would be the resolution?

To which kinds of analytes do these liquid chromatography detectors respond?

(a) ultraviolet

(b) refractive index

(c) evaporative light scattering

(d) charged aerosol

(e) electrochemical

(f) fluorescence

(g) nitrogen chemiluminescence

(h) conductivity

(a) List ways in which the resolution between two closely spaced peaks might be changed.

(b) After optimization of an isocratic elution with several solvents, the resolution of two peaks is 1.2How might you improve the resolution without changing solvents or the kind of stationary phase?

(a) You wish to use a wide gradient from 5 vol % to 95 vol %B for the first separation of a mixture of small molecules to decide whether to use gradient or isocratic elution. What should be the gradient time, tG,for a 150.46-cmcolumn containing 3-渭尘particles with a flow of 1.0mL/min

You optimized the gradient separation going from 20vol%to 34 vol% Bin 11.5min at 1.0mLFindk*for this optimized

separation. To scale up to a 151.0cmcolumn, what should be the gradient time and the volume flow rate? If the sample load on

the small column was 1mgwhat sample load can be applied to the large column? Verify that k*is unchanged.

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