The chemical equation given is:
<span>2x(g) ⇄ y(g)+z(s)</span>
Answer: the higher the amount of x(g) the more the forward reacton will occur and the higher the amounts of products y(g) and z(s) will be obtained at equilibrium.
Justification:
As Le Chatellier's priciple states, any change in a system in equilibrium will be compensated to restablish the equilibrium.
The higher the amount, and so the concentration, of X(g), the more the forward reaction will proceed to deal witht he high concentration of X(g), leading to an increase on the concentration of the products y(g) and z (s).
Tropical or warm air masses form in the tropics and have low air pressure.
polar or cold air masses form north of 50 degrees north latitude and south of 50 degrees south latitude
<u>Answer:</u> The equilibrium concentration of 
is 0.332 M
<u>Explanation:</u>
We are given:
Initial concentration of = 2.00 M
The given chemical equation follows:
<u>Initial:</u> 2.00
<u>At eqllm:</u> 2.00-2x x x
The expression of 
for above equation follows:
![K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BCF_4%5D%7D%7B%5BCOF_2%5D%5E2%7D)
We are given:
Putting values in above expression, we get:
Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible
So, equilibrium concentration of ![COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M](https://tex.z-dn.net/?f=COF_2%3D%282.00-2x%29%3D%5B2.00-%282%5Ctimes%200.834%29%5D%3D0.332M)
Hence, the equilibrium concentration of is 0.332 M
Molar mass H₂SO₄ = 98.079 g/mol
1 mol -------- 98.079 g
? mole ------ 0.0960 g
moles = 0.0960 * 1 / 98.079
= 0.0960 / 98.079
= 9.788 x 10⁻⁴ moles
hope this helps!
The correct option is C. The amount of MgCl2. we know this because <span>no matter how much you increase KOH, if you dont increase Mgcl2, the amount of Mg(OH)2 remains the same. Hope this works for you</span>
