Could someone please help me out with this chemistry question?
The question reads:
Mercury poisoning is a debilitating disease that is often fatal. In the human body, mercury reacts with essential enzymes leading to irreversible inactivity of these enzymes.
If the amount of mercury in a polluted lake is 0.4 ug Hg/ml, what is the total mass in kilograms of mercury in the lake? (The lake has a surface area of 100mi^2 and an average depth of 20ft.)
(NOTE: ug = micrograms)
I thought the problem was a simple conversion, changing micrograms to kilograms, but apparently my book gives me a totally different answer to this question, being: 7 x 10^5 kg.
Obviously I'm doing something wrong, and I have no idea how to incorporate the data given about the lake.
I would highly appreciate it if you help me out here to show the proper work for this question. Thank you in advance.
Answer:
Figure out the volume of the lake.
A mL = cm^3
Convert the volume of the lake to cm^3
Multiply the 0.4 ug/cm^3 x volume of t he length
That gives you ug of Hg, THEN convert to kg of Hg
OK firstly is get all units sorted - who uses miles and feet anymore???
Anyway if I remember back to the dark ages 1 mile = 5280 ft,
1 ft = 12 in, 1 in = 2.54 cm, 1 cu. cm = 1 ml
so mult. out going to have some big numbers!
vol of lake in cubic ft = (10 x 5280) x 10 x 5280 x 20
i.e. 20 x 100 x 5280 x 5280 cu ft
1 cu ft = (12 x 2.54) **3 = 28316.85 cu.cm
So vol lake = 28316.85 x 20x100x(5280)**2 cu.cm
(You do the arith.) my calc. can't handle it.
Mult all this by 0.4
Divide by 10 **9 for kg (1000g in kg & 1000000ug in a g)
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