Answer:
8.354 nanometers
Explanation:
To treat a diffusive process in function of time and distance we need to solve 2nd Ficks Law. This a partial differential equation, with certain condition the solution looks like this:
![\frac{C_{s}-C{x}}{C_{s}-C_{o}}=erf(x/2\sqrt{D*t})](https://tex.z-dn.net/?f=%5Cfrac%7BC_%7Bs%7D-C%7Bx%7D%7D%7BC_%7Bs%7D-C_%7Bo%7D%7D%3Derf%28x%2F2%5Csqrt%7BD%2At%7D%29)
Where Cs is the concentration in the surface of the solid
Cx is the concentration at certain deep X
Co is the initial concentration of solute in the solid
and erf is the error function
Then we solve right side,
![\frac{C_{s}-C{x}}{C_{s}-C_{o}}=\frac{1018atoms/cm3-1016atoms/cm3}{1018atoms/cm3}=0.001964](https://tex.z-dn.net/?f=%5Cfrac%7BC_%7Bs%7D-C%7Bx%7D%7D%7BC_%7Bs%7D-C_%7Bo%7D%7D%3D%5Cfrac%7B1018atoms%2Fcm3-1016atoms%2Fcm3%7D%7B1018atoms%2Fcm3%7D%3D0.001964)
And we need to look up the inverse error function of 0.001964 resulting in: 0.00174055
Then we solve for x:
![x=0.00174055*2*\sqrt{D*t} =0.00174055*2*\sqrt{2*10^{-12}cm^{2}/s*8h*3600s/h}=8.35464*10^{-7}cm](https://tex.z-dn.net/?f=x%3D0.00174055%2A2%2A%5Csqrt%7BD%2At%7D%20%3D0.00174055%2A2%2A%5Csqrt%7B2%2A10%5E%7B-12%7Dcm%5E%7B2%7D%2Fs%2A8h%2A3600s%2Fh%7D%3D8.35464%2A10%5E%7B-7%7Dcm)
Answer:
b I my current pls let me known
The volume of a sample of ammonia gas : 5.152 L
<h3>Further explanation</h3>
Given
0.23 moles of ammonia
Required
The volume of a sample
Solution
Assumed on STP
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
So for 0.23 moles :
= 0.23 x 22.4 L
= 5.152 L
Solubility is the maximum amount of a substance that will dissolve in a given amount of solvent at a specific temperature. There are two direct factors that affect solubility: temperature and pressure. Temperature affects the solubility of both solids and gases, but pressure only affects the solubility of gases.