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Charge on a gold bar

  1. Jan 16, 2005 #1
    I'm sure im missing something simple here but...

    Suppose a 1.72 g nugget of pure gold has zero net charge. What would be its net charge after it has 1% of its electrons removed?

    So i know that an electron is 9.109 x 10^-31 kg, and has a charge of -1.602 x 10^-19C. However, how do i know the split of electrons and protons in the gold bar?

    I'd assume im trying to get to mass of electrons / mass of electron, that times the charge of an electron divided by 100???
     
  2. jcsd
  3. Jan 16, 2005 #2

    Gokul43201

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    How many electrons in a Gold atom ? How many atoms in 1.72 g ?
     
  4. Jan 16, 2005 #3
    79 electrons in a gold atom

    1.72/196.96654 = .008732 atoms

    so .008732 * .79 * 1.602 x 10 ^-19 ?
     
  5. Jan 16, 2005 #4

    dextercioby

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    Okay,you're on the wrong track...

    197g gold----------------------------->6.023 10^{23} atoms

    1.72g gold----------------------------> x atoms

    1 atom gold---------------------------> 79 electrons
    x atoms gold---------------------------> y electrons


    What do you get for "y"??

    Daniel.
     
  6. Jan 16, 2005 #5
    (1.71 / 197) * 6.023 X 10 ^23 * 79 = 4.154 x 10 ^ 23

    AH!

    and that divided by 100 x the charge is the answer

    thank you

    Ryan
     
    Last edited: Jan 16, 2005
  7. Jan 16, 2005 #6

    Gokul43201

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    Not so fast !

    You've only found the number of electrons removed. Multiply this number by the charge on an electron to get the net positive charge.
     
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