# Question #fb7f1

##### 1 Answer

#### Explanation:

The idea here is that you can treat the lake as a **cylinder** and use the given *diameter* and *depth* to find its approximate **volume**.

As you can see, the diameter is given in *miles* and the average depth in *meters*, so the first thing to do here is to convert the diameter to meters.

To do that, go from miles to *kilometers* first, then from kilometers to meters

#10 color(red)(cancel(color(black)("mi"))) * (1.61color(red)(cancel(color(black)("km"))))/(1color(red)(cancel(color(black)("mi")))) * (10^3"m")/(1color(red)(cancel(color(black)("km")))) = "16100 m"#

Now, the volume of a cylinder can be calculated using the equation

Here the radius of the lake is equal to **half** of the diameter, so

#r = "16100 m"/2 = "8050 m"#

The height of the cylinder is actually the *depth* of the lake, so

#V = pi * ("8050 m")^2 * "50 m"#

#V ~~ 1.02 * 10^10"m"^3#

Notice that the concentration of mercury is given in *micrograms per liter*,

*cubic meters*to

*liters*

#1.02 * 10^(10) color(red)(cancel(color(black)("m"^3))) * (10^3"L")/(1color(red)(cancel(color(black)("m"^3)))) = 1.02 * 10^(13)"L"#

This means that the concentration of mercury in *micrograms* will be

#1.02 * 10^(13)color(red)(cancel(color(black)("L"))) * (0.0125color(white)(a)mu"L")/(1color(red)(cancel(color(black)("L")))) = 1.275 * 10^(11)mu"g"#

To convert this to *kilograms*, go to grams first

#1.275 * 10^(11)color(red)(cancel(color(black)(mu"g"))) * (1color(red)(cancel(color(black)("g"))))/(10^6color(red)(cancel(color(black)(mu"g")))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = color(green)(bar(ul(|color(white)(a/a)color(black)("130 kg")color(white)(a/a)|)))#

I'll leave the answer rounded to two **sig figs**.