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. Liquid Nitrogen is much colder than Water.
4. 2Li + 2H2O -> 2LiOH + H2
5. C6H12O6 + 6O2 -> 6CO2 + 6H2O
6. Zn + 2HCl -> ZnCl2 + H2
9. H2SO4 + Pb -> PbSO4 + H2
10. Ca(OH)2 + NH4Cl -> NH4 + CaCl2 + H2O
thats all i know
The new volume when pressure increases to 2,030 kPa is 0.8L
BOYLE'S LAW:
The new volume of a gas can be calculated using Boyle's law equation:
P1V1 = P2V2
Where;
- P1 = initial pressure (kPa)
- P2 = final pressure (kPa)
- V1 = initial volume (L)
- V2 = final volume (L)
According to this question, a 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, the volume is calculated as:
406 × 4 = 2030 × V2
1624 = 2030V2
V2 = 1624 ÷ 2030
V2 = 0.8L
Therefore, the new volume when pressure increases to 2,030 kPa is 0.8L.
Learn more about Boyle's law calculations at: brainly.com/question/1437490?referrer=searchResults
