The van Deemter equation decribes how the column, stationary phase, and linear velocity influence plate height. The equation is B Ux Match each symbol with its term in the van Deemter equation. H=A+ 00000 + Cux plate height linear velocity multiple path (eddy diffusion) equilibration time (mass transfer) longitudinal diffusion Answer Bank longitudinal diffusion equilibration time and linear velocity linear velocity and multiple paths linear velocity multiple paths and longitudinal diffusion H B A Ux For an open tubular column (not capillary electrophoresis), which term(s) in the van Deemter equation does(do) not affect the plate height? O multiple paths

The van Deemter equation decribes how the column, stationary phase, and linear velocity influence plate height. The equation is B Ux Match each symbol with its term in the van Deemter equation. H=A+ 00000 + Cux plate height linear velocity multiple path (eddy diffusion) equilibration time (mass transfer) longitudinal diffusion Answer Bank longitudinal diffusion equilibration time and linear velocity linear velocity and multiple paths linear velocity multiple paths and longitudinal diffusion H B A Ux For an open tubular column (not capillary electrophoresis), which term(s) in the van Deemter equation does(do) not affect the plate height? O multiple paths

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

2. Consider the diffusion of oxygenthrough alow density polyethylene (LDPE) sheet 15 mm thick. The pressures of oxygen at the two faces are 2000 kPa and 150 kPa, which are maintained constant. Assuming conditions of steady state, what is the diffusion flux (in cm³ STP/ cm2-s) at 298 K?
Estimate the diffusion coefficient for chloroform in air at 25 °C using Boltzmann's equation. If the reported value (of chloroform diffusivity in air at 25 °C air) is 9.5 X 10-2 cm²/s, what is the % error?
A small crystal of potassium permanganate (orange) and methylene blue (blue crystal) are placed on top of an agar medium in a petri dish. The agar is similar in consistency to gelatin. After one hour, each crystal has diffused away from the center of the dish. The permanganate shows a greater diffusion than methylene blue. The molecular weight of KMnO4 is 158.037 grams per mole, and the methylene blue molecular weight is 319.85 grams per mol. 1. Why did the KMnO4 diffuse faster than the methylene blue crystal? 2. If the plate were cooler, would it change which crystal would diffuse faster? Why
Q(3): Figure below shows a plot of the logarithm (to the base 10) of the diffusion coefficient versus reciprocal of absolute temperature for the diffusion of carbon in iron (a-Fe). Determine values for the activation energy and the preexponential? 10-10 C in a-Fe 10-12 10-14 10-16 10-18 0.5 1.0 1.5 Reciprocal temperature (1000/K) Diffusion coefficient (m2/s)
The sedimentation coefficient for a protein in solution is determined to be 4.3 S, and the diffusion coefficient to be 5.5x10-11 m2s-1 (vb = 0.74 mLg-1; density = 1.0 gmL-1; and viscosity = 1.0x10-3 Pas-1). Use this information to calculate the molar mass (kgmol-1) of the protein. Assume temperature of 298 K.
A sheet of steel 1.8 mm thick has nitrogen atmospheres on both sides at 1200°C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6 x 10-1" m²/s, and the diffusion flux is found to be 1.2 x 107 kg/m²-s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m’. How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m³? Assume a linear concentration profile.
Question 3 Describe the differences between steady-state diffusion and nonsteady-state diffusion
One of the most important reactions in biology is photosynthesis. In this complex chemical process, plants combine molecules of carbon dioxide and water to create glucose and oxygen. The reaction looks something like this: CO2 + H20 – C6H1206 + 02 Of course the number of atoms don't balance on the two sides. You might be able to intuit your way to the correct counting (stoichiometry), but let's do it systematically by setting up equations. (a) Write the reaction as a CO2 + b H20 -→ c CGH1206 + d O2 where a, b, c, and d are meant to be integers adjusted so that the number of each kind of atoms balance on each side. For this reaction, we have three kinds of atoms that must balance: carbon (C), hydrogen (H), and oxygen (0). Write three equations for a, b, c, and d that, if satisfied, will guarantee that the number of atoms of each kind are the same on each side of the reaction. (b) Find the smallest set of integers that will satisfy these equations. First, solve for a. (c) Solve for b.
DAB(PB1-PB2) O1: Is the equation NB dimensionally homogeneous? Na is R+T+ AZ Mole flux in Kmol/m*s. DAB is diffusion coefficient or diffusivity of A in B m2/s AZ distance of transport (m) R is the gas constant in (N*m/mol*K) T temperature in (K) Pa is partial pressure in (Pa).
Nitrogen from a gaseous phase is to be diffused into pure iron at 700°C. If the surface concentration is maintained at 0.13 wt% N, what will be the concentration (in weight percent) 2.5 mm from the surface after 6.4 h? The diffusion coefficient for nitrogen in iron at 700°C is 1.8 × 10-10 m²/s. erf(z) erf(z) erf(z) 0.00 0.0000 0.55 0.5633 1.3 0.9340 0.025 0.0282 0.60 0.6039 1.4 0.9523 0.05 0.0564 0.65 0.6420 1.5 0.9661 0.10 0.1125 0.70 0.6778 1.6 0.9763 0.15 0.1680 0.75 0.7112 1.7 0.9838 0.20 0.2227 0.80 0.7421 1.8 0.9891 0.25 0.2763 0.85 0.7707 1.9 0.9928 0.30 0.3286 0.90 0.7970 2.0 0.9953 0.35 0.3794 0.95 0.8209 2.2 0.9981 0.40 0.4284 1.0 0.8427 2.4 0.9993 0.45 0.4755 1.1 0.8802 2.6 0.9998 0.50 0.5205 1.2 0.9103 2.8 0.9999
Nitrogen from a gaseous phase is to be diffused into pure iron at 700°C. If the surface concentration is maintained at 0.27 wt% N, what will be the concentration (in weight percent) 3.7 mm from the surface after 5.2 h? The diffusion coefficient for nitrogen in iron at 700°C is 4.7 × 10-10 m²/s. erf(z) erf(z) erf(z) 0.00 0.0000 0.55 0.5633 1.3 0.9340 0.025 0.0282 0.60 0.6039 1.4 0.9523 0.05 0.0564 0.65 0.6420 1.5 0.9661 0.10 0.1125 0.70 0.6778 1.6 0.9763 0.15 0.1680 0.75 0.7112 1.7 0.9838 0.20 0.2227 0.80 0.7421 1.8 0.9891 0.25 0.2763 0.85 0.7707 1.9 0.9928 0.30 0.3286 0.90 0.7970 2.0 0.9953 0.35 0.3794 0.95 0.8209 2.2 0.9981 0.40 0.4284 1.0 0.8427 2.4 0.9993 0.45 0.4755 1.1 0.8802 2.6 0.9998 0.50 0.5205 1.2 0.9103 2.8 0.9999 i weight percent
As a particle geometry deviates from that of a sphere (MW unchanged), the result on the sedimentation coefficient measured under otherwise identical conditions would be: Sedimentation coefficient increases Sedimentation coefficient decreases Sedimentation coefficient is independent of geometry Cannot be predicted without conducting the experiment