Answer:
Following are the solution to these question:
Explanation:
Calculating the mean:


Calculating the standardn:

Please find the correct equation in the attached file.

For point a:

For point b:

For point c:

It is calculated by using the slope value that is
. The slope value
is ambiguous.
Answer:
For large rivers the problem is not simply a matter of deduction of consumptive use from runoff: it is more complex and the complexity is related to the changes in .
Explanation:
<span>NH3 is the answer, if u look up its ph its a weak base</span>
- 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%
<span>0.6 = mass/1.2 </span>
<span>mass = 0.6 x 1.2 </span>
<span>= 0.72 g </span>