The question is incomplete, the complete question is;
49.9 g per day of a certain industrial waste chemical P arrives at a treatment plant settling pond with a volume of 300 m^3. P is destroyed by sunlight, and once in the pond it has a half-life of 3.4 h. Calculate the equilibrium concentration of P in the pond. Round your answer to 2 significant digits.
Answer:
0.034 g/m³
Explanation:
From the question, mass of industrial waste = 49.9 g
It takes 24 hours to complete a day.
Mass per hour;
Mass in = (49.9 g / day) * (1 day /24 hr )
= 2.079 g/hr
Mass out = 0
Mass leaving due to sunlight = k CA V
Half life of the industrial waste = 3.4 h
This is a first order reaction;
Hence;
k = ln2/t1/2
Where, t1/2 = half life
k = 0.693 / 3.4 = 0.2038 hr⁻¹
Mass leaving due to sunlight = k CA V
CA = Mass per hr / kV
= 2.079 g/hr / ( 0.2038 hr⁻¹ x 300 m³ )
= 0.034 g/m³
