Answer:
0.204.
Explanation:
- For the following reaction:
<em>CO(g) + Cl₂(g) ⇌ COCl₂(g), </em>
Kp = (P of COCl₂)/(P of CO)(P of Cl₂)
P of COCl₂ = 0.22 atm, P of CO = 0.83 atm, P of Cl₂ = 1.3 atm.
∴ Kp = (P of COCl₂)/(P of CO)(P of Cl₂) = (0.22 atm)/(0.83 atm)(1.3 atm) = 0.204.
1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)
=> p1 = 2 p2
Which is easy to demonstrate using ideal gas equation:
p1 = nRT/V = 2.0 mol * RT / 1 liter
p2 = nRT/V = 1.0 mol * RT / 1 liter
=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2
2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.
So, the pressure in both chambers (which form one same vessel) is:
p = nRT/V = 3.0 mol * RT / 2liter
which compared to the initial pressure in chamber 1, p1, is:
p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1
So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.
You can also see how the pressure in chamber 2 changes:
p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
Answer: D. the hardening of liquid magma that leads to igneous rock
Explanation:
<span>These elements are red. They are typically gaseous at normal temperatures: some examples can include oxygen, hydrogen, and the noble gases (neon, argon, krypton). These elements all use the concept of STP (standard temperature and pressure) as their baseline: 1 atmosphere of pressure and 0 degrees Celsius.</span>