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:
By taking away research funds if certain standards ar not met
Explanation:
Answer:
Please find the structure attached as an image
Explanation:
Based on the characteristics ending name (-ene) of the organic compound above, it belongs to the ALKENE GROUP. Alkenes are characterized by the possession of a carbon to carbon double bond (C=C) in their structure.
- But-3-ene tells us that the organic compound has four straight carbon atoms with the C=C (double bond) located on the THIRD carbon depending on if we count from right to left or vice versa.
- 2 methyl indicates that the methyl group (-CH3) is located as an attachment on the second carbon (carbon 2).
N.B: In the structure attached below, the counting is from the left to right (→).
Hello!!! I would like to give you my answer!!!
<span>%error= (-890kJ-0.07kJ)/(-890kJ)x100 = 110.02%
I hope this helps!!! :-D</span>
Answer:
1.8 moles of O₂
Explanation:
The balance chemical equation for said double replacement (photosynthesis) reaction is as follow;
6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂
According to balance chemical equation,
6 moles of O₂ are produced by = 6 moles of CO₂
So,
1.8 moles of O₂ will be produced by = X moles of O₂
Solving for X,
X = 1.8 mol × 6 mol / 6 mol
X = 1.8 moles of O₂
Stoichiometric problems in which moles are given and moles or other reactant or product asked are the simplest problems. One should only write the balanced chemical equation and perform above method to find the required moles.
