Answer:
The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)
That is the rate of diffusion of ammonia through the layer is
5.153×10⁻⁴(kmol)/(m²·s)
Explanation:
The diffusion through a stagnant layer is given by

Where
= Diffusion coefficient or diffusivity
z = Thickness in layer of transfer
R = universal gas constant
= Pressure at first boundary
= Pressure at the destination boundary
T = System temperature
= System pressure
Where
= 101.3 kPa
,
,
0.5×101.3 = 50.65 kPa
Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m
R =
T = 298 K and
= 1.18
= 1.8×10⁻⁵
= 5.153×10⁻⁴
Hence the rate of diffusion of ammonia through the layer is
5.153×10⁻⁴(kmol)/(m²·s)
Answer:
The total hole mobility is 41.67 cm²/V s
Explanation:
Data given by the exercise:
hole mobility due to lattice scattering = ul = 50 cm²/V s
hole mobility due to ionized impurity = ui = 250 cm²/V s
The total mobility is equal:

Answer:
As many variables as we can coherently communicate in 2 dimensions
Explanation:
Visualization is a descriptive analytical technique that enables people to see trends and dependencies of data with the aid of graphical information tools. Some of the examples of visualization techniques are pie charts, graphs, bar charts, maps, scatter plots, correlation matrices etc.
When we utilize a visualization on paper/screen, that visualization is limited to exploring as many variables as we can coherently communicate in 2-dimensions (2D).
Solution :
Given :
k = 0.5 per day


Volume, V 
Now, input rate = output rate + KCV ------------- (1)
Input rate 


The output rate 
= ( 40 + 0.5 ) x C x 1000

Decay rate = KCV
∴
= 1.16 C mg/s
Substituting all values in (1)

C = 4.93 mg/L