Answer:
A. elements
<em>I</em><em> hope</em><em> it's</em><em> helps</em><em> you</em>
Answer:
C) 2 H₂ + O₂ → 2 H₂O
Explanation:
4 atoms of hydrogen on reactant side
2 atoms of oxygen on reactant side
4 atoms of hydrogen on product side
2 atoms of oxygen on product side
Answer:
Equilibrium concentrations of the gases are



Explanation:
We are given that for the equilibrium

Temperature, 
Initial concentration of



We have to find the equilibrium concentration of gases.
After certain time
2x number of moles of reactant reduced and form product
Concentration of



At equilibrium
Equilibrium constant
![K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7Bproduct%7D%7BReactant%7D%3D%5Cfrac%7B%5BH_2%5D%5E2%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
Substitute the values



By solving we get

Now, equilibrium concentration of gases



Static Friction, Sliding Friction, Rolling Friction, and finally Fluid Friction.