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Fluoride Ions needed to treat water

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Fluoride ion is poisonous in relatively low amounts: 0.2 g of F - per 70 kg body weight can cause death. Nevertheless, in order to prevent tooth decay, F - ions are added to drinking water at a concentration of 1 mg of F - ion per L of water. How many kilograms of sodium fluoride is needed to treat a 7.50 x 10^2 gallon reservoir?


    2. Relevant equations
    1mg = 1.0 x 10^-6kg
    1 gal = 283,905,884 liters


    3. The attempt at a solution
    I did 7.50 x 10^2 gallons and converted it to liters. Knowing how many liters is in the reservoir and knowing it takes 1.0 x 10^-6 kg to treat 1 liter of water, I got my answer as 284 kg. Anyone know what I did wrong? I'm fairly confident the answer is 284 kg.
     
  2. jcsd
  3. Sep 15, 2009 #2

    Borek

    User Avatar

    Staff: Mentor

    Doesn't look OK.

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  4. Sep 15, 2009 #3
    I figured it out, I forgot a step and yeah I just looked at it again and realized I put the wrong conversion.
     
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