Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄
Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.
For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:
Kp = 
where:
P(N₂O₄) and P(NO₂) are the partial pressure of each gas.
Calculating constant:
Kp = 
Kp = 0.0104
After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.
P(N₂O₄) + P(NO₂) = 200
P(N₂O₄) = 200 - P(NO₂)
Kp = 
0.0104 = ![\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B200%20-%20P%28NO_%7B2%7D%29%20%20%7D%7B%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D%7D)
0.0104
+
- 200 = 0
Resolving the second degree equation:
=
= 98.7
Find partial pressure of N₂O₄:
P(N₂O₄) = 200 - P(NO₂)
P(N₂O₄) = 200 - 98.7
P(N₂O₄) = 101.3
The partial pressures are
= 98.7 MPa and P(N₂O₄) = 101.3 MPa
Answer:
The correct answer is from areas of high concentration to low concentration.
Explanation:
A concentration gradient exists for these molecules, so they have the potential to diffuse into (or out of) the cell by moving down it.
Hope this Helps!
Answer:
The amount left after 49.2 years is 3mg.
Explanation:
Given data:
Half life of tritium = 12.3 years
Total mass pf tritium = 48.0 mg
Mass remain after 49.2 years = ?
Solution:
First of all we will calculate the number of half lives.
Number of half lives = T elapsed/ half life
Number of half lives = 49.2 years /12.3 years
Number of half lives = 4
Now we will calculate the amount left after 49.2 years.
At time zero 48.0 mg
At first half life = 48.0mg/2 = 24 mg
At second half life = 24mg/2 = 12 mg
At 3rd half life = 12 mg/2 = 6 mg
At 4th half life = 6mg/2 = 3mg
The amount left after 49.2 years is 3mg.