Answer:
it answer D because waste is flowing thru the air
Answer:
9.55 grams of SiO2
Explanation:
If the mass you mean by grams:
0.159 mole x 60.08 g (Periodic table by adding both elements)
Cancel moles with moles (Original moles with the 1 mol at the bottom of the grams) and gives you:
9.55 grams of SiO2
Answer : The correct expression for equilibrium constant will be:
![K_c=\frac{[C]^8}{[A]^4[B]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5E8%7D%7B%5BA%5D%5E4%5BB%5D%5E2%7D)
Explanation :
Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.
The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.
As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.
The given equilibrium reaction is,

The expression of
will be,
![K_c=\frac{[C]^8}{[A]^4[B]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5E8%7D%7B%5BA%5D%5E4%5BB%5D%5E2%7D)
Therefore, the correct expression for equilibrium constant will be, ![K_c=\frac{[C]^8}{[A]^4[B]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5E8%7D%7B%5BA%5D%5E4%5BB%5D%5E2%7D)
The equilibrium constant, Kc=0.026
<h3>Further explanation</h3>
Given
1.72 moles of NOCI
1.16 moles of NOCI remained
2.50 L reaction chamber
Reaction
2NOCI(g) = 2NO(g) + Cl2(g).
Required
the equilibrium constant, Kc
Solution
ICE method
2NOCI(g) = 2NO(g) + Cl2(g).
I 1.72
C 0.56 0.56 0.28
E 1.16 0.56 0.28
Molarity at equilibrium :
NOCl :

NO :

Cl2 :

![\tt Kc=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\\\\Kc=\dfrac{0.224^2\times 0.112}{0.464^2}=0.026](https://tex.z-dn.net/?f=%5Ctt%20Kc%3D%5Cdfrac%7B%5BNO%5D%5E2%5BCl_2%5D%7D%7B%5BNOCl%5D%5E2%7D%5C%5C%5C%5CKc%3D%5Cdfrac%7B0.224%5E2%5Ctimes%200.112%7D%7B0.464%5E2%7D%3D0.026)