Answer:
B. 111 J
Explanation:
The change in internal energy is the sum of the heat absorbed and the work done on the system:
ΔU = Q + W
At constant pressure, work is:
W = P ΔV
Given:
P = 0.5 atm = 50662.5 Pa
ΔV = 4 L − 2L = 2 L = 0.002 m³
Plugging in:
W = (50662.5 Pa) (0.002 m³)
W = 101.325 J
Therefore:
ΔU = 10 J + 101.325 J
ΔU = 111.325 J
Rounded to three significant figures, the change in internal energy is 111 J.
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:
Water-gas A mixture of carbon monoxide (CO) and hydrogen (H2) produced by passing steam over red-hot coke using the endothermic reaction C + H2O → CO + H2.
Answer:
1-C
2-D
3-B
4-A
Explanation:
I think it is this if it is not , sorry!
Ribosomes.ribosomes are small structures where proteins are made.Although they are not enclosed within a membrane,they are frequently considered organelles. Hope it helps :)