Answer: The approximate equilibrium partial pressure of
is 3.92 atm
Explanation:
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.
The given balanced equilibrium reaction is,

![K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
![1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=1.5%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
On reversing the reaction:

initial pressure 4.00atm 2.00 atm 0
eqm (4.00-2x)atm (2.00-x) atm 2x atm
![K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%5BH_2S%5D%5E2%7D%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D)


![0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}](https://tex.z-dn.net/?f=0.67%5Ctimes%2010%5E5%3D%5Cfrac%7B2x%5D%5E2%7D%7B%5B4.00-2x%5D%5E2%5Ctimes%20%5B2.00-x%5D%7D)

![[H_2S]=2x=2\times 1.96=3.92 atm](https://tex.z-dn.net/?f=%5BH_2S%5D%3D2x%3D2%5Ctimes%201.96%3D3.92%20atm)
Thus approximate equilibrium partial pressure of
is 3.92 atm
I think that is called a Metamorphic rock.
Answer:1
Explanation:cause castle told me
Answer:
4.66 x 10^8 yr
Explanation:
The age of the rock can be calculated using the equation:
ln (N/N₀) = - kt where N is the quantiy of radioisotope decayed and N₀ is the initially quantity present of the radioisotope; k is the decay constant, and t is the time.
Now from the data , we have 78 argon-40 atoms for every 22 potassium-40 atoms, we can deduce that originally we had 22 + 78 = 100 atoms of potassium-40 so this is our N₀.
When we look at the equation, we see that k is unknown, but we can calculate it from the half-life which is given by the equation:
k = 0.693/ t half-life = 0.693/ 1.3 x 10⁹ yr = 5.33 x 10⁻¹⁰ yr⁻¹
Now we are in position to answer the question.
ln ( 78/100 ) = - (5.33 x 10⁻¹⁰ yr⁻¹ ) t
- 0.249 = - 5.33 x 10⁻¹⁰ yr⁻¹ t
0.249/ 5.33 x 10⁻¹⁰ yr⁻¹ = t
4.66 x 10^8 yr