I. (20%) A lake had a concentration of mercury (Hg) in the water of 2.5 ng/L. Th

Question

I. (20%) A lake had a concentration of mercury (Hg) in the water of 2.5 ng/L. The concentration of Hg in the lake's algae was 375 ug/kg and in fish was 2.8 ppm (mg/kg). What was the bioconcentration factor for algae and fish? 2. (20%) The measured log octanol-water partition coefficient (Kow) for a pesticide was 5.5. The concentration in the water in a lake was 18 ng/L Calculate the concentration of the pesticide in fish that have a fat content of 6%. 3. (20%) An unknown gas has a Henry's Law constant of 9.71 x 10-4 mol/L-atm. Its concentration in the atmosphere is 0.43 ppm. What is its concentration in lake water in equilibrium with the atmosphere? What assumptions did you have to make?

Explanation / Answer

Biocentration factor=concentration biota/conc. Water

For algae=> bioconc.factor=conc. Algae/conc. Water

=(375x10^-6 g/kg)/2.5x10^-9 g/L

= 1.50 x10^5L/Kg

2): for fish. Conc.Hg=2.8ppm= 2.8x10^-6mg/kg

Biocentration factor=(2.8x10^-6)/2.5x10^-9g/L

=1.12x10^3 L/Kg

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