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![[P(NO_{2} )]^{2}](https://tex.z-dn.net/?f=%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D)
+ 
- 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
<h2>ANSWER OF EACH PART ARE GIVEN BELOW</h2>
Explanation:
A)
We know, each mole contains 
atoms.
It is given that mass of one oxygen atom is m= 
.
Therefore, mass of one mole of oxygen, 
.
Putting value of n and 
,
B)
Given,
Mass of water in glass=0.050 kg = 50 gm.
From above part mass of one mole of oxygen atoms = 16.0 gm.
Therefore, number of mole of oxygen equivalent to 50 gm oxygen
LEARN MORE :
Avogadro's number
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Answer chocies .......................
1. Berkelium(Berkeley, CA) 2. Dubnium(Dubna, Russia) 3. Darmstaditum (Darmstadt, Germany) 4. Erbium(Ytterby, Sweden) 5. Strontium(Strontian, Scotland) 6. Terbium(Ytterby, Sweden) 7. Yttebium(Ytterby, Sweden) 8. Yttrium(Ytterby, Sweden)
