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puteri [66]
4 years ago
12

Diffusion of Ammonia in an Aqueous Solution Ammonia (A)-water (B) solution ta 278 K and 4 mm thick is in contact with an organic

liquid at this interface. The concentration of ammonia in the organic phase is held constant and is such that the equilibrium concentration of ammonia at the interface is 2.0 wt% ammonia (density of aqueous solution is 991.7 Kg/m) The concentration of ammonia in water 4 mm away is 10 wt% (density of solution is 961.7 Kg/m). Water and the organic phase are insoluble in each other. The diffusion coefficient of NH3 in water is 1.24 × 10-9 m2/s.
(a) At steady state, calculate Na in Kmol/m-s

(b) Calculate NB. Explain.

Engineering
1 answer:
Tom [10]4 years ago
4 0

Answer:

Explanation:

The pictures below shows the whole explanation for the problem

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THE MASS FOR OBJECT 1 is 107.01 grams what is the objects force in Newton’s answer please
Aleksandr [31]

Given that,

Mass of the object 1, m = 107.01 grams

To find,

Force on the object.

Solution,

The force acting on the object is gravitational force. The force is given by the formula as follow :

F = mg

g is acceleration due to gravity

F = 0.10701 kg × 9.8 m/s²

F = 1.048 N

So, the force acting on object 1 is 1.048 N.

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3 years ago
EJERCICIO 6
Ket [755]

Answer:

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Explanation:

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8 0
3 years ago
Choose the person who best completes each statement.
victus00 [196]

The exercise is about finding the missing medical terms from the list below.. See the sentences below for the missing words.

<h3>What are the completed sentences?</h3>

  • The outbreak and spread of epidemic diseases, such as swine flu, are tracked by a <u>Public Health Surveillance Team.</u>
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7 0
2 years ago
Read 2 more answers
While walking across campus one windy day, an engineering student speculates about using an umbrella as a "sail" to propel a bic
makvit [3.9K]

Answer:

Given data:\\While walking across campus one windy day\\Frontal area, \(A=0.3 m ^{2}\)\\Wind speed \(V=24 Km / hr\)\\The drag coefficient \(C_{D, b}=1.2\)\\The combined mass \(m=75 kg\)\\Umbrella diameter, \(D=1.22 m\)\\Velocity of wind \(V=24 \frac{ km }{ hr }\)\\The rolling resistance \(C_{R}=0.75 \%\)

Solution:

Note: Refer the diagram

Basic equation:\\'s law of motion: \(\sum F_{x}=m a_{x}\)\\Lift coefficient, \(C_{L}=\frac{F_{L}}{\frac{1}{2} \rho V^{2} A_{p}}\)\\Drag coefficient, \(C_{D}=\frac{F_{D}}{\frac{1}{2} \rho V^{2} A_{p}}\)

From force balance equation:\\\(\sum F_{x}=F_{D}-F_{R}=0\)\\But \(F_{D}=\left(C_{D, \alpha} A_{u}+C_{D, B} A_{b}\right) \frac{1}{2} \rho\left(V_{\nu}-V_{b}\right)^{2}\)\(F_{R}=C_{R} m g\)\\Area of the Umbrella \(A_{u}=\frac{\pi D_{u}^{2}}{4}\)\(A_{x}=\frac{\pi \times 1.22^{2}}{4} m ^{2}\)\(A_{v}=1.17 m ^{2}\)

Drag coefficient data for selected objects table at

Hemisphere (open end facing flow), C_{D, x}=1.42

Substituting all parameters,

\begin{aligned}&F_{R}=0.0075 \times 75 \times 9.81\\&F_{R}=5.52 N\end{aligned}

Then,

\begin{aligned}&V_{b}=V_{w}-\left[\frac{2 F_{R}}{\rho\left(C_{D, w} A_{w}+C_{D, B} A_{b}\right)}\right]^{\frac{1}{2}} \dots\\&V_{w}=24 \times 1000 \times \frac{1}{3600}\\&V_{w}=6.67 \frac{ m }{ s }\end{aligned}

And the equation becomes,

\begin{aligned}&V_{b}=6.67-\left[\frac{2 \times 5.52}{1.23(1.42 \times 1.17+1.2 \times 0.3)}\right]^{\frac{1}{2}}\\&V_{b}=6.67-2.11\\&V_{b}=4.56 \frac{ m }{ s }\end{aligned}

Thus the floyds travels at 68.3^{\circ}wind speed.

7 0
4 years ago
In the ______ phase of the organizational life cycle, the organization is usually very small and agile, focusing on new products
Vlad [161]

Answer:

Entrepreneurship

Explanation:

Different phases of  organizational life cycles can be observed depending on the stage of the company. These cycles are,

  • Entrepreneurship
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  • Sustained Success
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In the initial stage which is  Entrepreneurship Phase,  all of the founders take part in various activities, no formality is formed between founders and employees. Generally lots of ideas are present and company is actively searching to define correct market and products to focus on.

5 0
3 years ago
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