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
Answer:
Compound.
Explanation:
That would be compound as it consists of a number of elements bonded together. It is inorganic not organic.
Answer:
The volume of NaOH required is - 0.01 L
Explanation:
At equivalence point
,
Moles of 
= Moles of NaOH
Considering
:-
Given that:
So,
<u>The volume of NaOH required is - 0.01 L</u>
Answer:
4Fe + 3O2 + 6H2O → 4Fe(OH)3
Explanation:
The chemical formula for rust is Fe2O3 and is commonly known as ferric oxide or iron oxide. The final product is a series of chemical reactions simplified below as- The rusting of the iron formula is simply 4Fe + 3O2 + 6H2O → 4Fe(OH)3. The rusting process requires both the elements of oxygen and water.
Answer:
I honestly dont know but its cool problably from water fill or from the waves going to much
Explanation:
