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Properties of Electric Charges-review prob

  1. Jan 25, 2008 #1
    1. The problem statement, all variables and given/known data

    (a). Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 10g. Silver has 47 electrons per atom, and its molar mass is 107.87 g/mol.
    (b). Imagine adding electrons to the pin until the negative charge has the very large value of 1.00 mC. How many electrons are added for every 10^9 electrons already present?

    2. The attempt at a solution

    This problem seems rather simple, but for some reason part b is not making any sense to me... maybe i messed up part a, even though i've checked it through a few times...

    (a) 10g/107.87(g/mol) = .0927 mol
    .0927 x (6.022 x 10^23)(avgr #) x 47 electrons, which gives me 2.62 x 10^24 electrons in this 10g of silver.

    (b). When I multiplied 2.62 x 10^24 (electrons) by -1.602 x 10^-19, which is the charge for a single electron, the final charge i got was 419814.96, which is obviously much larger than 1 mC.

    Any idea where i went wrong on this old review problem?
     
  2. jcsd
  3. Jan 25, 2008 #2

    Doc Al

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    Staff: Mentor

    Instead of figuring out the total charge of all the electrons in the neutral silver pin (which was not asked), start by figuring out how many electrons must be added to give the pin the extra charge of -1.00 mC.
     
  4. Apr 2, 2008 #3
    hey, im stuck on this problem too.

    i found the number of electron to give that extra charge...but im confused when the questions asks "how many electrons are added for every 10^9 electrons already present"...i just don't understand what that means...are we suppose to divide out answer by 10^9 to get some type of fractional answer?
    thanks for any help! :)
     
  5. Apr 2, 2008 #4

    Doc Al

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    Staff: Mentor

    I agree that it's a bit confusing. Here's an analogous problem: Say you have 5 million dollars in the bank. You add 10 dollars. How many dollars did you add for every million dollars already in the account? Answer $2.

    Do this problem the same way. How many billions (10^9) of electrons are already present?
     
  6. Apr 2, 2008 #5
    wow. thanks so much, that analogy really helped! :biggrin:
     
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