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.
In formation of a Type II Binary Compound, the metal atom present is<span>
NOT</span> found in either Group 1 or Group 2 on the periodic table. For the choices, Ba is under Group 2 on the periodic table, which makes it the atom not involved in formation of type II compounds. The answer is B.
Answer is: cesium sulfide.
Formula for ionic compound is: Cs₂S.
Ionic compound <span>electrically </span>neutral, cesium has charge +1 (2· (+1) = +2) and sulfur has charge -2.
Cesium sulfide is whide powder, soluble in water, pyrophoric in dry air, reductant.
Not sure if this is what you mean but l= A/w
(length equals area divided by width)