UNIT TEST
1 answer:
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Answer:
a) The concentration in ppm (mg/L) is 5.3 downstream the release point.
b) Per day pass 137.6 pounds of pollutant.
Explanation:
The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.
Million Gallons per day
We have one flow of wastewater released into a stream.
First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.
Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.
After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:
Replacing every value in L/d and mg/L
a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.
Finally, we have to calculate the pounds of substance per day (Mp).
We have the total flow F3 = F1 + F2 and the final concentration . It is required to calculate per day, let's take a time of t = 1 day.
After that, mg are converted to pounds.
b) A total of 137.6 pounds pass a given spot downstream per day.
If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .
will all the alcohol evaporate? or none at all?
Answer:
Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.
Explanation:
Given that:
The volume of alcohol which is placed in a small laboratory = 1.0 L
Vapor pressure of ethyl alcohol at 25 ° C = 59 mmHg
Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;
Then, we have:
=
= 0.078 atm
Temperature = 25 ° C
= ( 25 + 273 K)
= 298 K.
Density of the ethanol = 0.785 g/cm³
The volume of laboratory = l × b × h
= 3.0 m × 2.0 m × 2.5 m
= 15 m³
Converting the volume of laboratory to liter;
since 1 m³ = 100 L; Then, we have:
15 × 1000 = 15,000 L
Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:
PV = nRT
Making n the subject of the formula; we have:
n = 47. 88 mol of ethanol
Moles of ethanol in 1.0 L bottle can be calculated as follows:
Since numbers of moles =
and mass = density × vollume
Then; we can say ;
number of moles =
number of moles =
number of moles =
number of moles = 17.039 mol
Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.