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 i
1 answer:
Answer:
The concentration of P in the pond at equilibrium is 0.034 g/m³
Explanation:
Given the total mass = 49.9 g
1 day = 24 hours
mass per hour;
Incoming mass = (49.9 g / day) * (1 day /24 hr )
= 2.079 g/hr
Outgoing mass = 0
Mass lost due to sunlight = kGiven the half-life = 3.4 hours
For a first-order reaction; k, the rate constant = ln2/t, where t is the half-time
ln 2= 0.693, V= volume
k = 0.693 / t_half = 0.693 / 3.4 = 0.2038 hr⁻¹
Substituting all parameters into the equation kMass lost to sunlight = k V
= Incoming mass per hour / kV
= 2.079 g/hr / (0.2038 hr⁻¹ x 300 m³)
0.034 g/m³
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The subscript denotes the atomic number, which is 58. This notation is illustrated in the attached image.