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Charge on a penny

  1. Sep 6, 2008 #1
    1. The problem statement, all variables and given/known data
    A 3.0-g copper penny has a positive charge of 67 mC.
    What fraction of its electrons has it lost?


    2. Relevant equations
    I don't think im donig this right. i need guidance


    3. The attempt at a solution
    i take 3.0 grams * 63.5amu * 6.022 * 10^23 * 29 (electrons) = 3.3*10 ^ 27

    67 * 10^-6C / 1.6 * 10 ^ -19C = 4.2 * 10^14

    3.3*10 ^ 27/4.2 * 10^14 = WRONG ANSWER. HELP!!!
     
  2. jcsd
  3. Sep 6, 2008 #2

    LowlyPion

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    A mole of copper atoms will weigh 63.54 grams. 3 grams is what portion of the mole? Then you can go wild with Avogadro and all the orbiting electrons.
     
  4. Sep 8, 2008 #3
    ok. would this be correct?

    3.0 grams / 63.5amu * 6.022 * 10^23 * 29 (electrons) = 8.2 *10^23

    67 * 10^-6C / 1.6 * 10 ^ -19C = 4.2 * 10^14

    (4.2 * 10^14)/(8.2 * 10^23) = 5.1*10^-10

    Is this correct?
     
  5. Sep 9, 2008 #4

    LowlyPion

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    That looks more better as far as your method. I didn't check your math.
     
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