Organic: sugar
inorganic: salt
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
B) False- It has seven
A hexagon would have 6.
Answer:
Rate expression has been given below
Explanation:
According to the given equation, 1 molecule of A reacts with 1 molecule of B and produces 2 molecules of B at a time.
So, rate of disappearance of both A and B are one half of rate of appearance of B
Hence rate expression can be represented as:
![Rate=\frac{-\Delta [A]}{\Delta t}=\frac{-\Delta [B]}{\Delta t}=\frac{1}{2}\frac{\Delta [C]}{\Delta t}](https://tex.z-dn.net/?f=Rate%3D%5Cfrac%7B-%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B-%5CDelta%20%5BB%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D)
where
is rate of disappearance of A,
is rate of disappearance of B and
rate of appearance of C
The Answer is D: Reactor products charge huge batteries.