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SSSSS [86.1K]
3 years ago
7

When CL was 2.3×10−6 M and CR was zero, the flux across a microporous membrane was found to be 0.234 pmol cm−2 s−1. The free dif

fusion coefficient of the material being measured was 0.8×10−5 cm2 s−1. A. What is the permeability of the membrane to the material? B. If the thickness of the membrane is 10×10−6 m, what is the equivalent relative area available for diffusion of the material?
Chemistry
1 answer:
aksik [14]3 years ago
3 0

Answer:

A) 8.08*10^-10 cm^-1

B)  1.27 * 10^-5

Explanation:

A) PERMEABILITY OF THE MEMBRANE to the material = K * D/(dx) ---- (1)

K = J / (Cl - Cr )

  = 0.234 * 10^-12 / (2.3*10^-6 M - 0 )

  = 10 *10^-6 M

dx = 10*10^-6 m * 100 cm

    = 10 *10^-4 cm

D = 0.8 * 10^-5 cm^2s^-1

input the calculated values into equation 1

permeability of the membrane = 8.08*10^-10 cm^-1

B) equivalent relative area available for the diffusion of the material

at thickness of membrane = 10*10^-6 m

can be calculated using this relation = \frac{flux * thickness}{diffusion coefficient * ( Cl -Cr)} ---------- (2)

flux  = 0.234 * 10^-12 mol.cm^-2.s^-1

thickness = 10*10^-4 cm

diffusion coefficient = 0.8 * 10^-5 cm^2.s

Cl = 2.3*10^-6 m

input the above values into equation 2

equivalent relative area available for the diffusion of the material

= 1.27 * 10^-5

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Make notes on why did the levels of carbon dioxide in the atmosphere decrease?​
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Read 2 more answers
a substance is analyzed to have a percent composition of 74.186% sodium and 25.814% fluorine. calculate the empirical formula
Novosadov [1.4K]

Answer:

Na₂₆F₁₁

Explanation:

We find the moles of the substance assuming 100 g of the substance is present. Why do we take 100 g? Because then the percent of sodium/fluorine, would be the g of sodium/fluorine respectively:

74.186 g Sodium | 1 mol Sodium/23 g              =>       3.2255 mol Na    

25.814 g Fluorine | 1 mol Fluorine/19 g             =>       1.3586 mol F

Divide each by smallest number of moles:

3.2255/1.3586 = 2.37

1.3586/1.3586 = 1

Multiply by common number to get a smallest whole number:

2.37*11 = 26,

1*11 = 11

The empirical formula is Na₂₆F₁₁

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Explanation:

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