Answer: 5. Positive
6. Negative
9. Oxygen carbon hydrogen nitrogen
Explanation:
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:
The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5.
Explanation:
The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5 because of the diagram.
Explanation:
The O atom is sp3 in a water molecule, with two sigma bonds and two lone pairs of electrons like that in water. The steric integer is thus 4, and its structure is tetrahedral.
The C atom is sp hybridised into two identical bonds and two identical bonds in acetylene.
The steric integer is therefore 2 because only sigma bonds are engaged in deciding hybridization, and its structure is linear.
The C atom is sp2 hybridised in ethene with single pi bond and three sigma identical bonds.
Thus the steric integer is 3, and its structure is planar trigonal.
The C atom is sp2 hybridized in ethene, with one pi bond and three sigma identical bonds.
The steric integer would therefore be 3 and its structure is planar trigonal.
The O atom is sp3 in a water molecule with two bond pairs and two lone pairs of electrons like that. The steric integer is thus 4, and its structure is tetrahedral.
The C atom is sp3 in a methane ring, with 4 bond pairs and no solitary pairs of electrons like that. The steric integer is thus 4, and its structure is tetrahedral.
