100 kg of simazine is added to a box canyon to control purple loosestrife via cr

Question

100 kg of simazine is added to a box canyon to control purple loosestrife via crop duster. Assume air in the canyon is closed through a temperature inversion long enough for the system to reach equilibrium. The canyon has a total area of 0.5 Times 0.5 km2. There is a pond that is 60 m Times 60m in surface area, averaging 1.5 m deep. It has 40 total kg of fish and a particulate loading of 10 mg/L. The accessible soil layer is 10 cm thick as is the sediment. The air volume trapped in the inversion is 1 km3. Given all that: Calculate z for all compartments Solve for the percentage simazine that ends up in the water. Solve for the percentage in the air. Somebody salts a nearby icy road and the runoff raises the salinity in the pond to 20 ppt. What percentage simazine are in the sediment before and after the salting event? Assume the fish are all dead but their bodies can still take up the chemical. Assume the safe limit of simazine in fish for raptors is 10 mg/kg body weight. If the valley loses the temperature inversion and air begins to exchange at a rate of 100 exchanges/day, and assuming instantaneous equilibrium, how long is it before eagles can safely eat the fish? Based on your calculations, and referring to the constants but without doing any more calculations, estimate which of these chemicals will have the lowest body burden in fish. Explain why ( with numbers, citations and full sentences).

Explanation / Answer

2.

a. You need to mention what do you mean by 'z'; if 'z'is the percent distribution

b.total volume of water in the pond = 60x60x1.5 m3 = 5400 m3 = 5.4 x106 litre (1 m3 = 103 litre)

The pond has a particulate loading of 10 mg/litre; so for 5.4x106 litre, total particulate loading = 5.4x107 mg = 54 kg

So the percentage of Simazine that ends up in water is 54/100 x 100% = 54%

c. Total volume of accessible soil layer in the canyon = 500x500x0.1 m3 =2.5x 104 m3

The air volume trapped in the inversion = 1 km3 = 109 m3

So the ratio soil volume/ air volume = 2.5/105= 0.000025; in terms of percentage, its 0.0025%

So form the remaining (100-54)= 46 kg of Simazine

46x0.0025% = 0.00115 kg will go into the sediment, rest, i.e., 45.99885 kg wll go into the air

So the percentage of Simazine that ends up in air is 45.99885/100 x 100% = 45.99885%

3.

1 ppt =1000 mg/l

the pond earlier had a particulate loading of 10 mg/L

So after the salting, it won't be able to consume any Simazine

So, now, the amount of Simazine in the sediment will be, 100 x 0.0025% = 0.0025 kg

From 2(b), we get that percentage of total Simazine in the sediment before the salting event was= 0.00115/ 100 x 100% = 0.00115%

and, after the salting event, now it'll be 0.0025/100 x 100% = 0.0025%

4.

Since the safe limit of Simazine in fish for raptors is 10 mg/kg of body weight, and total amount of fish in the pond is 40 kg, so eagles can safely eat the fish only after the total Simazine in the pond reduces to at least 10 x40 mg = 400 mg =4 x 10-3 kg, which is 4 x10-3/54 x 100% = 0.0074% of the present value

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