Explanation:
The given data is as follows.
Volume of lake = =
Concentration of lake = 5.6 mg/l
Total amount of pollutant present in lake =
= mg
= kg
Flow rate of river is 50
Volume of water in 1 day =
= liter
Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are or
Flow rate of sewage =
Volume of sewage water in 1 day = liter
Concentration of sewage = 300 mg/L
Total amount of pollutants = or
Therefore, total concentration of lake after 1 day =
= 6.8078 mg/l
= 0.2 per day
= 6.8078
Hence, =
=
= 1.234 mg/l
Hence, the remaining concentration = (6.8078 - 1.234) mg/l
= 5.6 mg/l
Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.
Nuclear energy comes from splitting of Barium atom to form Krypton atom
Answer:
15.35 g of (NH₄)₃PO₄
Explanation:
First we need to look at the chemical reaction:
3 NH₃ + H₃PO₄ → (NH₄)₃PO₄
Now we calculate the number of moles of ammonia (NH₃):
number of moles = mass / molecular wight
number of moles = 5.24 / 17 = 0.308 moles of NH₃
Now from the chemical reaction we devise the following reasoning:
if 3 moles of NH₃ are produce 1 mole of (NH₄)₃PO₄
then 0.308 moles of NH₃ are produce X moles of (NH₄)₃PO₄
X = (0.308 × 1) / 3 = 0.103 moles of (NH₄)₃PO₄
mass = number of moles × molecular wight
mass = 0.103 × 149 = 15.35 g of (NH₄)₃PO₄
Answer:
63.9 grams. Yes, the Nacl was converted. Maximum possible ppm is 540ppm.
Explanation:
I this is college level chemistry not regular high school chem.