A scientist is investigating a cell. The cell has a membrane thickness of 7.×10−8 m and has a surface area of 4.×10−8. The initial concentration inside the cell is n(0)=0.6 M, and the scientist later observes that an(0.2)=0.77 M. Assume that the concentration outside the cell is 1. M and that it will not change appreciably as the compound in the cell undergoes diffusion. What is the value of the parameter k in the equation for Fick's Law

Step-by-step explanation:

we will begin with a step by step process to make this as concise as possible.

From our basic knowledge of Fick's Law of diffusion

we have that;

dm/dt = KA(m₀-m) / x

given that m = concentration at t

                m₀ = outside concentration (1)

                 A = area (4 *10-8)

                  x = thickness (7 *10-8)

therefore we have that from the expression;

dm/dt = (K *4 *10-8 (1 -m)) / 7 *10-8

dm/1-m = 0.57 Kdt  

integrating between limits

∫dm/1-m (0.2 to 0.77) = 0.57 ∫ dt (0 to 0.6)

x is greater than 5

step-by-step explanation:

2x+7 is greater than 17

    -7                           -7

2x is greater than 10

divide by 2 each side

x is greater than 5