Answer:
![K_{goal}=1.793*10^{-33}](https://tex.z-dn.net/?f=K_%7Bgoal%7D%3D1.793%2A10%5E%7B-33%7D)
Explanation:
N2(g)+O2(g)⇌2NO(g), ![K_1 = 4.10*10^{-31}](https://tex.z-dn.net/?f=K_1%20%3D%204.10%2A10%5E%7B-31%7D)
N2(g)+2H2(g)⇌N2H4(g), ![K_2 = 7.40*10^{-26}](https://tex.z-dn.net/?f=K_2%20%3D%207.40%2A10%5E%7B-26%7D)
2H2O(g)⇌2H2(g)+O2(g), ![K_3 = 1.06*10^{-10}](https://tex.z-dn.net/?f=K_3%20%3D%201.06%2A10%5E%7B-10%7D)
If we add above reaction we will get:
2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)
Equilibrium constant for Eq (1) is ![K_1*K_3*K_3](https://tex.z-dn.net/?f=K_1%2AK_3%2AK_3)
Divide Eq (1) by 2, it will become:
N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)
Equilibrium constant for Eq (2) is ![(K_1*K_3*K_3)^{1/2}](https://tex.z-dn.net/?f=%28K_1%2AK_3%2AK_3%29%5E%7B1%2F2%7D)
![Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}](https://tex.z-dn.net/?f=Equilibrium%20constant%20%3DK_%7Bgoal%7D%3D%20%28K_1%2AK_2%2AK_3%29%5E%7B1%2F2%7D%5C%5CK_%7Bgoal%7D%3D%20%284.10%2A10%5E%7B-31%7D%20%2A7.40%2A10%5E%7B-26%7D%2A1.06%2A10%5E%7B-10%7D%29%5E%7B1%2F2%7D%5C%5CK_%7Bgoal%7D%3D1.793%2A10%5E%7B-33%7D)
Answer:
27) Double replacement
28) 2NaOH + H2SO4 --> Na2SO4 + 2H2O
Answer:
21.16 MPa
Explanation:
Partial pressure of oxygen = 5.62 MPa
Total gas pressure = 26.78 MPa
But
Total pressure of the gas= sum of partial pressures of all the constituent gases in the system.
This implies that;
Total pressure of the system = partial pressure of nitrogen + partial pressure of oxygen
Hence partial pressure of nitrogen=
Total pressure of the system - partial pressure of oxygen
Therefore;
Partial pressure of nitrogen= 26.78 - 5.62
Partial pressure of nitrogen = 21.16 MPa
Answer:
volume
v = 4/3π r^3
Explanation:
it isn't specific enough but that is the equation of how to get any volume
volume equals four thirds times pi times radios to the power of three