Excess electrons on a lead sphere

AI Thread Summary
The problem involves calculating the number of excess electrons on a lead sphere with a mass of 8.00g and a net charge of -3.20 x 10^-9 C. The solution identifies that there are 2.00 x 10^10 excess electrons. Given that the atomic number of lead is 82 and its atomic mass is 207 g/mol, the number of lead atoms in the sphere is determined to be approximately 2.31 x 10^22 atoms. The calculations lead to the conclusion that the number of excess electrons per lead atom can be derived from the total number of excess electrons and the total number of lead atoms. The discussion effectively combines charge and atomic properties to solve the problem.
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Homework Statement


Excess electrons are placed on lead sphere with mass m=8.00g so that net charge becomes q=-3.20*10^-9 C.

How many excess electrons are there per lead atom?


Homework Equations





The Attempt at a Solution



Have found:

number of excess electrons= 2.00*10^10

atomic number of lead=82
atomic mass of lead= 207g/mol
 
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how many lead atoms are in a lead sphere of 8.00g ?
 
207g=6.623x10^23 atoms... 8g =? atoms. You have the excess electrons, and now the number of atoms of lead in an 8 g sphere...
 

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