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
D. precision. At first glance you can mark out A and B because the answers does not relate to the question. If question had said "what do you call it when the measurement is close to the actual answer" then you would have picked C. So that leaves you D. precision.
Answer:
Percentage yield = 0.49 × 10² %
Explanation:
Given data:
Actual yield of SO₂ = 4.309 ×10² g
Theoretical yield of SO₂ = 8.78 ×10² g
Percentage yield = ?
Solution:
Chemical equation:
2NiS₂ + 5O₂ → 2NiO + 4SO₂
Percentage yield:
Percentage yield = actual yield / theoretical yield × 100
Percentage yield = 4.309 ×10² g / 8.78 ×10² g × 100
Percentage yield = 0.491 × 100
Percentage yield = 49.1%
In scientific notation:
Percentage yield = 0.49 × 10² %
Answer:
the gravity
Explanation:
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