Answer:
13.7 moles of O₂ are needed
Explanation:
In order to find the moles of reactants that may react to make the products we need to determine the reaction:
Reactants are hydrogen and oxygen
Product: Water
2 moles of hydrogen can react to 1 mol of oxygen and produce 2 moles of water.
Balanced reaction: 2H₂(g) + O₂(g) → 2H₂O(l)
If 2 moles of hydrogen need 1 mol of oxygen to react
Therefore, 27.4 moles of H₂ must need (27.4 .1) / 2 = 13.7 moles of O₂
1 N₂ + 3 H₂ = 2 NH₃
Mole ratio <span> of hydrogen to ammonia :
3 moles H</span>₂ : 2 moles NH₃ or 3 : 2
hope this helps!
Answer:
The answer is C. The partial pressure of hydrogen will be unchanged.
Explanation:
⇒ 
Argon with electronic configuration
(that is atomic number 18) is an inert gas making it unreactive and it's addition to the reaction has no effect on the partial pressure of either the reactant or production or the state of the system.
The partial pressure of hydrogen will remain unchanged on the addition of Argon.
Apples and bananas guys apples and bananas
The question is incomplete, complete question is :
The first two steps in the industrial synthesis of nitric acid produce nitrogen dioxide from ammonia:
Step 1 : 
Step 2 : 
The net reaction is:

Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants
and
. If you need to include any physical constants, be sure you use their standard symbols
Answer:
Equation that gives the overall equilibrium constant K in terms of the equilibrium constants
:

Explanation:
Step 1 : 
Expression of an equilibrium constant can be written as:
![K_1=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BNO%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E5%7D)
Step 2 : 
Expression of an equilibrium constant can be written as:
![K_2=\frac{[NO_2]^2}{[NO]^2[O_2]}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BNO%5D%5E2%5BO_2%5D%7D)
The net reaction is:

Expression of an equilibrium constant can be written as:
![K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E7%7D)
Multiply and divide
;
![K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO]^4}{[NO]^4}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E7%7D%5Ctimes%20%5Cfrac%7B%5BNO%5D%5E4%7D%7B%5BNO%5D%5E4%7D)
![K=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO_2]^4}{[NO]^4}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E7%7D%5Ctimes%20%5Cfrac%7B%5BNO_2%5D%5E4%7D%7B%5BNO%5D%5E4%7D)
![K=K_1\times \frac{[NO_2]^4}{[O_2]^2[NO]^4}](https://tex.z-dn.net/?f=K%3DK_1%5Ctimes%20%5Cfrac%7B%5BNO_2%5D%5E4%7D%7B%5BO_2%5D%5E2%5BNO%5D%5E4%7D)
![K=K_1\times (\frac{[NO_2]^2}{[O_2]^1[NO]^2})^2](https://tex.z-dn.net/?f=K%3DK_1%5Ctimes%20%28%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BO_2%5D%5E1%5BNO%5D%5E2%7D%29%5E2)

So , the equation that gives the overall equilibrium constant K in terms of the equilibrium constants
:
