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Sati [7]
3 years ago
5

Marie and James are bubbling dry pure nitrogen (N2) through a tank of liquid water (H2O) containing ethane (C2H6). The vapor str

eam enters at 25oC, and 14000 mmHg. The tank operates at steady state, and the liquid flow rate enters at 15000 mol/min with a mole fraction of water of 0.999995. The separation of ethane is isothermal and isobaric, and 91% of the ethane is separated into the exiting vapor. Henry's law applies for ethane entering in the liquid and exiting in the vapor. Determine the following three unknowns.
1. Mole fraction of exiting ethane in the vapor phase = Ex: 0.021
2. Flow rate of ethane exiting in the liquid phase = Ex: 15 mol/day
3. Flow rate of entering nitrogen = Ex: 4.3 mol/min
Engineering
1 answer:
tekilochka [14]3 years ago
3 0

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5 0
3 years ago
In the lab, a container of saturated soil had a mass of 113.27 g before it was placed in the oven and100.06 g after the soil had
kipiarov [429]

Answer:

  • Moisture/ water content w = 26%
  • Void ratio , e =  0.73

Explanation:

  • Initial mass of saturated soil w1 = mass of soil - weight of container

                                                 = 113.27 g - 49.31 g = 63.96 g

  • Final mass of soil after oven w2 = mass of soil - weight of container

                                                  = 100.06 g - 49.31 g = 50.75

Moisture /water content, w =   \frac{w1-w2}{w2} = \frac{63.96-50.75}{50.75} = 0.26 = 26%

Void ratio =  water content X specific gravity of solid

                  = 0.26 X 2.80 =0.728

5 0
3 years ago
What does the supply chain management process involve
VMariaS [17]

Answer:

It involves the active streamlining of a business's supply-side activities to maximize customer value and gain a competitive advantage in the marketplace

Explanation:

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5 0
3 years ago
A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The
AleksAgata [21]

Answer:

18 kJ

Explanation:

Given:

Initial volume of air = 0.05 m³

Initial pressure = 60 kPa

Final volume = 0.2 m³

Final pressure = 180 kPa

Now,

the Work done by air will be calculated as:

Work Done = Average pressure × Change in volume

thus,

Average pressure = \frac{60+180}{2}  = 120 kPa

and,

Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³

Therefore,

the work done = 120 × 0.15 = 18 kJ

4 0
3 years ago
If a car sits out in the sun every day for a long time can light from the sun damage the car paint
Reika [66]

Answer:

i think yes it could make the color go lighter

Explanation:

6 0
3 years ago
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