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Tamiku [17]
3 years ago
5

Change of a gas to a liquid

Chemistry
2 answers:
erastovalidia [21]3 years ago
7 0

Answer:

vaporization or condensation

Explanation:

google

OLEGan [10]3 years ago
4 0
Equally the same paragraph
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Convert 30°C into kelvin
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Answer:303.15

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How can we make use of acids or bases to Prevent the growth of microorganisms in swimming pools?
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By putting smarticle particles 
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Heart, 5 stars, and Brainiest to first right answer!
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How does the total volume of gas particles compare to the volume of the space between gas particles?
saw5 [17]
Answer:
             There is more space between gas particles than the size of the particles.

Explanation:
                   This scenario can be understand by taking a very simple example. As we know that 1 mole of any gas at standard temperature and pressure occupy 22.4 liters of volume. Lets take Hydrogen gas and Oxygen gas, 1 mole of each gas will occupy same volume. Why it is so? Why same volume although Oxygen is 16 times more heavier? This is because the space between gas molecules is very large. Approximately the distance between gas molecules is 300 times greater than their own diameter from its neighbor molecules. 
7 0
3 years ago
A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
2 years ago
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