Answer:
D. 0.3 M
Explanation:
NH4SH (s) <--> NH3 (g) + H2S (g)
Initial concentration 0.085mol/0.25L 0 0
Change in concentration -0.2M +0.2 M +0.2M
Equilibrium 0.035mol/0.25 L=0.14M 0.2M 0.2M
concentration
Change in concentration (NH4SH) = (0.085-0.035)mol/0.25L =0.2M
K = [NH3]*[H2S]/[NH4SH] = 0.2M*0.2M/0.14M ≈ 0.29 M ≈ 0.3M
<u>Answer:</u> The solubility of oxygen at 682 torr is 
<u>Explanation:</u>
To calculate the molar solubility, we use the equation given by Henry's law, which is:

Or,

where,
are the initial concentration and partial pressure of oxygen gas
are the final concentration and partial pressure of oxygen gas
We are given:
Conversion factor used: 1 atm = 760 torr

Putting values in above equation, we get:

Hence, the solubility of oxygen gas at 628 torr is 