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.
Answer:
Because it keeps track of all the elements
The rate of disappearance of chlorine gas : 0.2 mol/dm³
<h3>Further explanation</h3>
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.
For reaction :

The rate reaction :
![\tt -\dfrac{1}{a}\dfrac{d[-A]}{dt}= -\dfrac{1}{b}\dfrac{d[-B]}{dt}=\dfrac{1}{c}\dfrac{d[C]}{dt}=\dfrac{1}{d}\dfrac{d[D]}{dt}](https://tex.z-dn.net/?f=%5Ctt%20-%5Cdfrac%7B1%7D%7Ba%7D%5Cdfrac%7Bd%5B-A%5D%7D%7Bdt%7D%3D%20-%5Cdfrac%7B1%7D%7Bb%7D%5Cdfrac%7Bd%5B-B%5D%7D%7Bdt%7D%3D%5Cdfrac%7B1%7D%7Bc%7D%5Cdfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D%5Cdfrac%7B1%7D%7Bd%7D%5Cdfrac%7Bd%5BD%5D%7D%7Bdt%7D)
Reaction for formation CCl₄ :
<em>CH₄+4Cl₂⇒CCl₄+4HCl</em>
<em />
From equation, rate of reaction = rate of formation CCl₄ = 0.05 mol/dm³
Rate of formation of CCl₄ = reaction rate x coefficient of CCCl₄
0.05 mol/dm³ = reaction rate x 1⇒reaction rate = 0.05 mol/dm³
The rate of disappearance of chlorine gas (Cl₂) :
Rate of disappearance of Cl₂ = reaction rate x coefficient of Cl₂
Rate of disappearance of Cl₂ = 0.05 x 4 = 0.2 mol/dm³