Answer:
1.76 g/mL
Explanation:
You need to find the volume. You can do this by subtracting the volume of the water and the rock by the volume of the water.
72.7 mL - 50 mL = 22.7 mL
Now that you have volume, divide the mass by the volume to find the density.
39.943 g/22.7 mL = 1.76 g/mL
- The mass percent of
Pentane in solution is 16.49%
- The mass percent of
Hexane in solution is 83.51%
<u>Explanation</u>:
- Take 1 kg basis for the vapor: 35.5 mass% pentane = 355 g pentane with 645 g hexane.
-
Convert these values to mol% using their molecular weights:
Pentane: Mp = 72.15 g/mol -> 355g/72.15 g/mol = 4.92mol
Hexane: Mh = 86.18 g/mol -> 645g/86.18 g/mol = 7.48mol
Pentane mol%: yp = 4.92/(4.92+7.48) = 39.68%
Hexane mol%: yh = 100 - 39.68 = 60.32%
Pp-vap = 425 torr = 0.555atm
Ph-vap = 151 torr = 0.199atm
-
From Raoult's law we know:
Pp = xp
Pp - vap = yp
Pt (1)
Ph = xh
Ph - vap = yh
Pt (2)
-
Since it is a binary mixture we can write xh = (1 - xp) and yh = (1 - yp), therefore (2) becomes:
(1 - xp)
Ph - vap = (1 - yp)
Pt (3)
-
Substituting (1) into (3) we get:
(1-xp)
Ph - vap = (1 - yp)
xp
Pp - vap / yp (4)
xp = Ph - vap / (Pp - vap/yp - Pp - vap + Ph - vap) (5)
-
Subbing in the values we find:
Pentane mol% in solution: xp = 19.08%
Hexane mol% in solution: xh = 80.92%
-
Now for converting these mol% to mass%, take 1 mol basis for the solution and multiplying it by molar mass:
mp = 0.1908 mol
72.15 g/mol
= 13.766 g
mh = 0.8092 mol
86.18 g/mol
= 69.737 g
-
Mass% of Pentane solution = 13.766/(13.766+69.737)
= 16.49%
-
Mass% of Hexane solution = 83.51%
Answer:
you gotta rotate it 90 degrees and get that screen fixed
Explanation:
Low
Water is better at sticking together with other water molecules than it is with the "waxy surfaces".
Answer:
22.67 L of PH₃
Explanation:
The balanced equation is:
![P_4 (s) + 6H_2(g) \to 4PH_3(g)](https://tex.z-dn.net/?f=P_4%20%28s%29%20%2B%206H_2%28g%29%20%5Cto%204PH_3%28g%29)
From the equation:
![34 L \times \dfrac{1 \ mol \ of H_2 }{22.4 \ L \ H_2} \times \dfrac{4 \ mol \ of \ PH_3}{6 \ mol \ H_2} \times \dfrac{22.4 \ L \ PH_3}{1 \ mol \ PH_3}](https://tex.z-dn.net/?f=34%20L%20%5Ctimes%20%5Cdfrac%7B1%20%5C%20mol%20%5C%20of%20H_2%20%7D%7B22.4%20%5C%20L%20%5C%20H_2%7D%20%5Ctimes%20%5Cdfrac%7B4%20%5C%20mol%20%5C%20of%20%5C%20PH_3%7D%7B6%20%5C%20mol%20%5C%20H_2%7D%20%5Ctimes%20%5Cdfrac%7B22.4%20%5C%20L%20%5C%20PH_3%7D%7B1%20%5C%20mol%20%5C%20PH_3%7D)
= 22.67 L of PH₃