Silver Pin Charged

  • Thread starter jessedevin
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Homework Statement



(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?

The Attempt at a Solution


I got the answers for both (a) and (b) (answers in back of book), but I still don't understand how to get (b). Heres my steps:

For (a), all I did was dimensional analysis to find the amount of electrons in 10 g of Ag, whic is about 2.62 x 10^24 e-. For part (b), I am confused, but what I found out how many electrons in 1 mC, which was 6.25 x 10^24 e-, and divided it by 2.62 x 10^24 e-, and I got 2.38 e- for every 10^9 electrons already present, but still this does not make sense to me. Can someone explain it to me or give me some helpful hints?
 

Answers and Replies

  • #2
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How many elementary electrical charges are there in one milliCoulomb?
 
  • #3
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How many elementary electrical charges are there in one milliCoulomb?
6.25 x 10^24 e-... so what do I do with that?
 
  • #4
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How many groups of 10^9 electrons are there in the silver pin?
 
  • #5
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2.62 x 10^24 e-/10^9 e-=2.62 x 10^15
 
  • #6
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So if I divide the number of electrical charges in one milliCoulomb by the number of groups of 10^9 electrons, won't that supply the answer to the question?
 
  • #8
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It seems to me that's exactly what they are asking for...

How many electrons are added for every 10^9 electrons already present?
 

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