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Firlakuza [10]
2 years ago
13

8. Zeolite is used to remove moisture from methane. A vertical column is filled with 1000.0 kg of dry zeolite. The zeolite has t

he capacity to hold 0.100 kg water/kg dry zeolite. Once the zeolite becomes saturated with moisture, it must be regenerated by heating. The inlet moisture content of the methane is 7.00% (by mass) and the outlet moisture content is 0.05% (by mass). How much methane (kg) will be produced before the zeolite must be regenerated
Chemistry
1 answer:
Natasha_Volkova [10]2 years ago
3 0

Answer:

Follows are the solution to this question:

Explanation:

Methane inlet humidity content of =7.00\%

Methane moisture outlet content =0.05\%

Zeolite absorption humidity = 6.95\%

Dry zeolite 1 kg will accommodate water= 0.1000 \ kg

One kilogram of Dry Zeolite will carry water from =0.1000\ kg

The water can contain 1000 kg of zeolite = 100 \ kg

Methane which would be made =\frac{100}{6.95\%}= 1,439 \ kg

That's why it will be producing 1439 kg of methane.

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Air in a closed cylinder is heated from 25 degree to 36 degree. If the initial pressure is 3.80 atm what is the final pressure ?
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Aww man, I love these types of problems!!!
So, it's a simple formula:
P1*V1/T1=P2*V2/T2

In this case, we remove V(volume) as there is none listed, leaving this:
P1/T1=P2=T2

Think you got it from here? All you gotta do is plug in the numbers and solve
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3 years ago
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A cylinder with a moving piston expands from an initial volume of 0.350 L against an external pressure of 1.90 atm. The expansio
myrzilka [38]

Answer:

1.84 L

Explanation:

Using the equation for reversible work:

W = -P*(V_{2} - V_{1})

Where:

W is the work done (J) = -287 J.

Since the gas did work, therefore W is negative.

P is the pressure in atm = 1.90 atm.

However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K);  R = 0.0821 (L*atm)/(mol*K)

Therefore, the conversion unit is 0.0821/8.314 = 0.00987 (L*atm)/J

V_{1} is the initial volume = 0.350 L

V_{2} is the final volume = ?

Thus:

(-287 J)*0.00987 (L*atm)/J = -1.9 atm*(V_{2} - 0.350) L

V_{2} = [(287*0.00987)+(1.9*0.350)]/1.9 = (2.833+0.665)/1.9 =1.84 L

7 0
3 years ago
4. An organism that creates its own food is called
STatiana [176]

Answer:

producer

Explanation:

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A 2.1−mL volume of seawater contains about 4.0 × 10−10 g of gold. The total volume of ocean water is about 1.5 × 1021 L. Calcula
Fiesta28 [93]

Answer:

Total worth of gold in the ocean = $5,840,000,000,000,000

Explanation:

As stated in the question above, 4.0 x 10^-10 g of gold was present in 2.1mL of ocean water.

Therefore, In 1 L of ocean water there will be,

(4.0 x 10^-10)/0.0021

= 1.9045 x 10^-7 g of gold per Liter of ocean water.

So in 1.5 x 10^-21 L of ocean water, there will be

(1.9045 x 10^-7) * (1.5 x 10^-21)

= 2.857 x 10^14 g of gold in the ocean.

1 gram of gold costs $20.44, that is 20.44 dollars/gram. The total cost of the gold present in the ocean is

20.44 * (2.857 x 10^14)

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Solutions of potassium iodide and lead(II)nitrate are mixed. The resulting lead
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Answer is D
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