There is process called unit factor method or factor label method for unit conversions that is taught in basically every chemistry program and every engineering program today. I highly recommend you learn this method.
I'll solve your problem first, then provide information about how to handle other unit conversion problems at the end....
here we go...
***** volume of lake... ****
volume = cross sectional area x depth = 100 mi² x 20 ft x (5280 ft / mi)² = 5.58 x 10^10 ft³
**** conversion from ft³ water to kg of Hg ****
(5.58 x 10^10 ft³) x (12 in / ft)³ x (2.54 cm / in)³ x (1 ml / 1 cm³) x (0.40 µg Hg / ml) x (1 g Hg / 10^6 µg Hg) x (1 kg / 10³ g) = 6.32 x 10^5 kg
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"unit factor method"
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rule 1) units on top and bottom of a fraction cancel.
examples:
in / in = 1
ft / ft = 1
cm / cm cancels.
sec / sec cancel
sec² / sec² cancel
m³ / m³ cancels
got that down?
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rule 2) any number x 1 = that number
examples:
4 x 1 = 4
4 in x 1 = 4 in
5280 ft x 1 = 5280 ft
9.8 m / s² x 1 = 9.8 m / s²
easy right?
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rule 3) any equality can be rearranged to = 1
examples:
2.54 cm = 1 in
if we divide both sides by 2.54 we get
2.54 cm / 2.54 cm = 1 in / 2.54 cm
1 = 1 in / 2.54 cm
12 in = 1 ft
12 in / 12 in = 1 ft / 12 in
1 = 1 ft / 12 in
also... since 1/1 = 1....
1 / 1 = 1 / (1 ft / 12 in)
1 = 12 in / 1 ft
meaning for any equality of the form a = b, .....a / b = b / a = 1
a/b and b/a are called unit factors because they = unity and we will be using them in rules 4 and 5 to factor (or more precisely "change" units)
ok? this may be a bit more complicated than the first two rules but if you play around with it for a while I'm sure you will get it.
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rule 4) units can be changed by multiplying by an appropriate "unit factor".
this essentially combines rules 1, 2, and 3
example....
10 in = ? cm
10 in x 1 = 10 in right?
and 2.54 cm / 1 in = 1 right?
substituting
10 in x (2.54 cm / 1 in ) = 10 x 2.54 x in x cm / in = 25.4 cm
because inches cancel....
3 ft = ? in
3 ft x 1 = 3 ft x (12 in / ft) = 36 in
here is an important trick... if the units you want to cancel are on the top, put the matching units for the "unit factor" on the bottom.. and vice versa. if on the bottom, then put the matching ones one the top....
this will definitely take practice...
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rule 5) exponents....
since 1 to any power = 1... any "unit factor" raised to any power = 1
examples:
1² = 1
(2.54 cm / in) = 1
(2.54 cm / in)² = 1² = 1
similiarly
(2.54 cm / in)³ = 1³ = 1
how you use this is like this...
10 in³ = ? cm³
10 in³ x 1 = 10 in³ x (2.54 cm / in)³
= 10 in³ x (2.54 cm)³ / (1 in)³
= 10 in³ x 16.4 cm³ / 1 in³
= 164 cm³
and that's all there is to it...
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if you've made it this far..... here's your problem explained
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volume = cross sectional area x depth = 100 mi² x 20 ft x (5280 ft / mi)² = 5.58 x 10^10 ft³
since 5280 ft = 1 mi
5280 ft / 1 mi = 1
(5280 ft / 1 mi) ² = 1
by multiplying 100 mi² x 20 ft x (5280 ft / 1 mi)² , I changed the units from mi² x ft to ft² x ft = ft³... do you see that?
next....
(5.58 x 10^10 ft³) x (12 in / ft)³ x (2.54 cm / in)³ x (1 ml / 1 cm³) x (0.40 µg Hg / ml) x (1 g Hg / 10^6 µg Hg) x (1 kg / 10³ g) = 6.32 x 10^5 kg
since
12 in = 1 ft → (12in / 1 ft)³ = 1
2.54 cm = 1 in → (2.54 cm / 1 in)³ = 1
1 ml = 1 cm³ → (1 ml / cm³) = 1
(0.40 µg Hg / ml) = 1
1 g = 10^6 µg → (1 g Hg / 10^6 µg Hg) = 1
1 kg = 10³ g → (1 kg / 10³ g) = 1
so in the above conversion, I keep multiplying by different forms of "1" and progressively changing the units...
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final word
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although my answer may look complicated at first... notice the simplicity compared to the other answers. and also notice the difference in results....
volume of lake.....
100 mi² x 20 ft x (5280 ft / mi)² = 5.58 x 10^10 ft³
mass in kg....
(5.58 x 10^10 ft³) x (12 in / ft)³ x (2.54 cm / in)³ x (1 ml / 1 cm³) x (0.40 µg Hg / ml) x (1 g Hg / 10^6 µg Hg) x (1 kg / 10³ g) = 6.32 x 10^5 kg
relatively easy to follow right? each () = a conversion factor..... each conversion factor = 1 right ?
now let's compare to the other answers...
**** chem man ****
this was actually quite well done except that he misread the problem... V = A x depth and area was given to be 100 mi² chem man took A to be 100 mi x 100 mi...
if instead, he did this....
A = 100 mi² x (1609 m / 1 mi)² x (100 cm / m)² = 2.59 x 10^12 cm²
V = V = A x d = (2.59 x 10^12 cm²) x 610. cm = 1.58 x 10^15 cm^3
etc....
then the answer would have been mass Hg = 6.32 x 10^5 kg
***** norrie *****
= 1,579,776,000,000,000mL is correct....
0.4μg/ml / 1,000 = 0.0004μg/L is correct but why did you do this?
1,579,776,000,000,000 mL x 0.0004μg/L does NOT =
= 631,910,400,000μg Hg....
it does = 631,910,400,000μg Hg x ml / L.... the ml / L does NOT cancel...
if you did this instead....
1,579,776,000,000,000 mL x 0.4μg/ml = 6.32 x 10^14 μg = 6.32 x 10^8 g = 6.32 x 10^5 kg
then your answer would agree with the rest of ours...
***** for Steve Y and rmjrenne *******
Steve has the basic idea in that (5280ft x 5280 ft / 1 mile x 1 mile) is the conversion factor.. however (5280 ft / mi)² is generally how this is written for simplicity
rmjrenee.... you converted from 100 mi² → 10 mi x 10mi →
16 km x 16 km → 256 km² → 16 x 10^5 cm x 16 x 10^5 cm →2.59 x 10^12 cm².... that worked out ok but...
100 mi² x (1.6 km / mi)² x (10^6 cm / km)² = 2.59 x 10^12 cm²
is easier to follow isn't it?
******* an update for Norrie *******
Thank you for critiquing my answer. This method, that I keep trying to "teach" in my answers to "how do you do this conversion" questions is in fact the method taught for unit conversions in Universities for at least the past 30 years. I know this because I have studied this... have taught it.... and use it all the time. And yes, it always is difficult at first as you suggest, but then it becomes easier as you use it...
So why is this method taught? here are some examples.... let's start by comparing your method to mine...looking at just the first part...
yours...
100mi² = 10mi x 10mi.
1mi = 5,280ft/mi.
10mi = 52,800ft
52,800ft x 52,800ft x 20ft = 55,756,800,000ft³
mine
100 mi² x 20 ft x (5280 ft / mi)² = 5.58 x 10^10 ft³
ok? notice the (5280 ft / mi)² converts the mi² to ft² ? mi² is on the top and bottom of the fraction and therefore cancels... leaving units of ft² x ft. you may not see the advantage yet... so let's continue....
Now what would your answer look like if your units were cubic?
100mi³ = ³√10 mi x ³√10 mi x ³√10 mi
or perhaps if you think about it for awhile....
100mi³ = 10mi x 10mi x 1 mi
1mi = 5,280ft/mi.
10mi = 52,800ft
52,800ft x 52,800ft x 5,280ft ft = 1.47 x 10^13 ft³
versus
100 mi³ x (5280 ft / mi)³ = 1.47 x 10^13 ft³
which is simplier?
or what if the cube roots were more difficult.. let's say convert 157 mi³ to ft³
your way...
157 mi³ = ³√157mi x ³√157mi x ³√157mi
1mi = 5,280ft/mi.
³√157mi = 5.3947 x 5,280ft/mi = 28484 ft
28484 ft x 28484 ft x 28484 ft= 2.31 x 10^13 ft³
my way
157 mi³ x (5280 ft / mi)³ = 2.31 x 10^13 ft³
or how about this...
157 ft³ /s to m³ / hr
your method
157 ft³ = ³√157ft x ³√157ft x ³√157ft
3.2808 ft = 1 m.
³√157ft / 3.2808ft / 1 m = 1.644 m
1.644 m x 1.644 m x 1.644 m= 4.446 m³
1 hr = 3600 sec
4.446 m³ /s = 4.446 m³/ s/ 1 hr /3600 sec = 1.60 x 10^4 m³/hr
my method...
157 ft³ /s x (1 m / 3.2808 ft)³ x (3600 s/hr) = 1.60 x 10^4 m³/hr
you may also notice that the
"³√157ft / 3.2808ft / 1 m" and the "4.446 m³/ s/ 1 hr /3600 sec" steps are very complicated... do you multiply or divide? you don't have that problem with my method. right?
So what happens if the units are even more complicated? g cm²/s² /ft² to lbsf / in² for example? or to lbsm / in² and so on....
All that said..your original answer was incorrect. The reason you made a mistake was because your method does not clearly carry units through and you cannot watch them cancel. Which is the whole point of the unit factor method. I gave you a thumbs down because your answer was incorrect and I gave you an explanation as to where you went wrong. I could certainly give you a thumbs down without the explanation. Would that be better?
One last thing. This is yahoo answers. People ask questions. People answer questions. There is a rating system for the answers. Thumbs up if you like the others answers, thumbs down if you don't. You gave an incorrect answer and you have really have no reason to take offense for a thumbs down now do you. So instead of getting upset why don't you focus a bit of that energy on trying to understand or learn what the points others are making? I certainly understand that you have spent many years in industry and have much valuable knowledge. But so have I. Take my advice and read this answer again and try to pick up this method.