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Math Help - Chemistry problems.. dealing somewhat w/ conversions

  1. #1
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    Chemistry problems.. dealing somewhat w/ conversions

    These questions are from my AP chem assignment for this summer. Any help would be much appreciated! Thank you in advance!

    first one..
    "Mercury poisoning is a debilitating diesase 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 micrograms Hg/mL, what is the total mass in kilograms of mercury in the lake? (The lake has a surface area of 100 mi^2 and an average depth of 20 ft.)"

    second one..
    "The contents of one 20. lb bag of topsoil will cover 10. square feet of ground to a depth of 1.0 inch. How many bags are needed to cover a plot which measures 200. by 300. m to a depth of 2.0 cm?"


    I fortunately have the answers to the problems! (They were odd numbered problems, meaning they were in the back of the book, haha.)
    Answer to #1 is 7 x 10^5 kg
    Answer to #2 is 1.0 x 10^5 bags
    Unfortunately, I'm not getting these results when I try the problems. Lol.
    If you could show me the calculations, that'd be great
    Thanks againnnnn!
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  2. #2
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    Quote Originally Posted by chemwhiz View Post
    These questions are from my AP chem assignment for this summer. Any help would be much appreciated! Thank you in advance!

    first one..
    "Mercury poisoning is a debilitating diesase 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 micrograms Hg/mL, what is the total mass in kilograms of mercury in the lake? (The lake has a surface area of 100 mi^2 and an average depth of 20 ft.)"

    second one..
    "The contents of one 20. lb bag of topsoil will cover 10. square feet of ground to a depth of 1.0 inch. How many bags are needed to cover a plot which measures 200. by 300. m to a depth of 2.0 cm?"


    I fortunately have the answers to the problems! (They were odd numbered problems, meaning they were in the back of the book, haha.)
    Answer to #1 is 7 x 10^5 kg
    Answer to #2 is 1.0 x 10^5 bags
    Unfortunately, I'm not getting these results when I try the problems. Lol.
    If you could show me the calculations, that'd be great
    Thanks againnnnn!
    Hmmm for problem #1 I get 6 x 10^5 kg, here are my steps (I only round at the very end)

    V=SA*20\ \text{ft}=(100 mi^2)\left(\frac{5280 ft}{1 mi}\right)^2(20 ft)\left(\frac{12 in}{1 ft}\right)^3 \left(\frac{2.54 cm}{1 in}\right)^3\left(\frac{1 ml}{1 cm^3}\right)=1578856752060825.6 ml

    m=(0.4 \frac{\mu g}{ml})*V=(0.4 \frac{\mu g}{ml})(1578856752060825.6 ml)\left(\frac{1 kg}{10^9 \mu g} \right)= 6\cdot10^5 kg

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  3. #3
    MHF Contributor Unknown008's Avatar
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    1. The first one is simple. Find the volume of the lake. Within 1 mL there are 0.4 micrograms, therefore, within the total volume that you convert into ml, find the mass of mercury present. Just be careful with the units.

    100 miles^2 = 2.59 x 10^10 m^2 (3 sf)
    20 ft = 6.10 m (3 sf)

    Volume of water = 1.58 x 10^11 m^3 ( sf)
    Now, in mL, this gives = 1.58 x 10^17 mL

    Then, the mass of Mercury = 1.58 x 10^17 * 0.4 = 6.3 x 10^ 16 micrograms or 6.3 x 10^7 kg

    I don't see anything I missed there... except if I have taken different conversion factors...

    2. First, find the volume you have in one 20 lb bag.

    10 ft^2 and 1 in gives an volume of (10 * 0.3048^2)(0.0254) = 0.0236 m^3 (3 sf)

    Then, the volume to be covered is (300*200*2) = 120 000 m^3

    The number of bags is given by dividing the total volume by the volume of each bag... 120 000/0.0236 = 5.09 x 10^6 bags, that is 5 x 10^6 bags to 1 sf.

    I'm using:
    1 mile [international] = 1 609.344 metre
    1 foot [international, U.S.] = 0.304 8 metre
    1 inch [international, U.S.] = 0.025 4 metre
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