Answer:
The equilibrium concentration of hydrogen gas is 0.0010 M.
Explanation:
The equilibrium constant of the reaction = 
}
Moles of hydrogen sulfide = 0.31 mol
Volume of the container = 4.1 L
![[concentration]=\frac{moles}{volume (L)}](https://tex.z-dn.net/?f=%5Bconcentration%5D%3D%5Cfrac%7Bmoles%7D%7Bvolume%20%28L%29%7D)
![[H_2S]=\frac{0.31 mol}{4.1 L}=0.076 M](https://tex.z-dn.net/?f=%5BH_2S%5D%3D%5Cfrac%7B0.31%20mol%7D%7B4.1%20L%7D%3D0.076%20M)
Initially
0.076 M
At equilibrium
(0.076-2x) 2x x
The expression of an equilibrium constant :
![K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5D%5E2%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
Solving for x:
x = 0.00051
The equilibrium concentration of hydrogen gas:
![[H_2]=2x=2\times 0.00051 M=0.0010 M](https://tex.z-dn.net/?f=%5BH_2%5D%3D2x%3D2%5Ctimes%200.00051%20M%3D0.0010%20M)
<u>Answer:</u>
<em>The returns of the company as it is given is already large that is 91% and to get an outcome even better is what it is getting now. There should be certain improvements that needs to be applied with the </em><em>help of stoichiometric data gathered.
</em>
<u>Explanation:</u>
Since the cost of materials cannot be affected and have to be bought under similar rate of cost of working or process can be decreased. By using machines calculation and mechanical devices this come would be accurate and the laws would be minimised. Hence the companies outcome would increase.
Hrxn = Q reaction / mol of reaction
mol of reaction = M * V = 10 * 1 = 10 mmol = 0.01 mol
Q water = m * C * (Tf - Ti)
= (10 + 10) (4.184) (26-20) = 502.08 J
Q reaction = - Q water = -502.08 J
Hrxn = -502.08 / (0.01) = - 50208 J = - 50.21 kJ/mol
