# Question c7f67

Mar 12, 2015

You have approximately $6 \cdot {10}^{5} \text{kg}$ of mercury in the lake.

Now, before getting started, notice that you have a wide array of units that you must work with, so I suggest deciding what units would be most useful to convert to.

Since density was given to you in micrograms per mL, I'll calculate the volume of the lake in mL by going from cubic miles to cubic meters, and finally to mL.

The volume of the lake can be caulculated by multiplying the surface with the average depth. Go from feet to miles first

$\text{20 ft" * "0.0001893939 miles"/"1 ft" = "0.0037878 miles}$

The volume in cubic miles will be

$V = \text{area" * "average depth" = "100 mi"^(2) * "0.0037878 mi}$

$V = {\text{0.37878 mi}}^{3}$

Now use two conversion factors to get to mL

$\text{0.37878 mi"^(3) * (4.16818183 * 10^(9)"m"^(3))/"1 mi"^(3) * (10^(6)"mL")/("1 m"^(3)) = 1.5788 * 10^(15)"mL}$

The mass of mercury in micrograms will be

rho = m/V => m = rho * V = (0.4mu"g")/"mL" * 1.5788 * 10^(15)"mL"

$m = 6.3152 \cdot {10}^{14} \mu \text{g}$

Now covert to kilograms to get the final answer

$6.3152 \cdot {10}^{14} \mu \text{g" * "1 kg"/(10^(9)mu"g") = 6.3152 * 10^(5)"kg}$

Rounded to one sig fig, the number of sig figs in 100 square miles and in 20 ft, the answer is

m_("mercury") = 6 * 10^(5) "kg"#