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

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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

s 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.

Chemistry

1 answer:

Alekssandra [3K] · Answer 1 · 2026-04-08T06:20:57+03:00

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 = k $C_{A}$ V

Given 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 k $C_{A}$ V;

Mass lost to sunlight = k $C_{A}$ V

$C_{A}$ = Incoming mass per hour / kV

= 2.079 g/hr / (0.2038 hr⁻¹ x 300 m³) $C_{A}$

=

0.034 g/m³ $C_{A}$