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Fraction of Electrons Removed

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data
    A 50.0 g ball of copper has a net charge of 2.00µC. What fraction of the copper’s electrons has been removed? (Each copper atom has 29 protons, and copper has an atomic mass of 63.5.)


    2. Relevant equations
    1.00 C× (1proton)/(1.60×10^−19C)=6.25×10^18 protons
    C = coloumb, the SI unit of charge


    3. The attempt at a solution

    I don't know where to begin. I haven't taken Chemistry in a long time so I'm pretty stumped.
     
  2. jcsd
  3. Jan 27, 2013 #2

    SammyS

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    How many copper atoms are there in 63.5 grams of copper?
     
  4. Jan 27, 2013 #3

    gneill

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    The atomic mass tells you the number of grams making up a mole of a substance. A mole is a specific number of particles (look up Avogadro's Number). So start by determining how many atoms comprise the 50.0 g ball of copper.


    EDIT: Beaten to the punch by SammyS!
     
  5. Jan 27, 2013 #4
    Got it thanks.
     
    Last edited: Jan 27, 2013
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