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Paladinen [302]
3 years ago
9

Someone please help me!!!!!!!

Chemistry
1 answer:
Tresset [83]3 years ago
5 0

Answer:

Hydrogen Chloride Gas is heavier than air and sinks to the ground after being released. Firefighters closer to the ground will be subjected to significantly greater concentrations of hydrogen chloride gas than the firefighters walking on top of railcars.

Explanation:

The molar mass of air is 28.97 g/mol. Air is approximately 78% N2, which has a molar mass of 28.014 g/mol and 21% oxygen, which has a molar mass of 32 g/mol; the remaining constituents of air include CO2 and other trace amounts of gases.

The molar mass of HCl is 36.458 g/mol.

Because HCl is heavier than air, it sinks toward the ground, displacing the lighter air.

The height of railway cars is reported to vary between 13.5 ft to 15.5 ft. So firefighters on tank cars are generally above the HCl gas (although localized atmospheric conditions could cause the HCl to move higher up in the air column).

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Mkey [24]
The mass change, or the mass defect, can be calculated by the formula that is very known to be associated with Albert Einstein. 

E = Δmc²
where
E is the energy gained or released during the reaction
c is the speed of light equal to 3×10⁸ m/s
Δm is the mass change

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5 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
3 years ago
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Answer:

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