<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
No doubt it's B. solid, liquid, and gas.
Hope this helps!
You can establish a system of two equation with two variables.
Varibles are:
V1 = volume of the 50% sugar solution
V2 = volumen of the 80% sugar solution
Equations:
Balance of sugar:
Sugar from 50% solution: 0.5*V1
Sugar from 80% solution: 0.8*V2
Sugar in the final solution (mix): 0.6 * 105 = 63
1) 0.5V1 + 0.8V2 = 63
Final volume = volume of 50% solution + volume of 80% solution
2) V1 + V2 = 105
From (2) V1 = 105 - V2
Substitue in (1)
0.5 (105 - V2) + 0.8 V2 = 63
52.5 - 0.5V2 + 0.8V2 = 63
0.3 V2 = 63 - 52.5
0.3 V2 = 10.5
V2 = 10.5/0.3
V2 = 35mL
V1 = 105 - 35 = 70 mL
Answer: 70 mL of the 50% solution and 35 mL of the 80% solution.
No beginning or end I believe
