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FinnZ [79.3K]
3 years ago
14

Excess electrons are placed on a small lead sphere with a mass of 7.70 g so that its net charge is −3.35 × 10^−9 C.A) Find the n

umber of excess electrons on the sphere.
B) How many excess electrons are there per lead atom? The atomic number of lead is 82, and its atomic mass is 207 g/mol?
Physics
1 answer:
natka813 [3]3 years ago
5 0

Answer:

a

N  =  2.094*10^{10} \  electrons

b

O  =  9.33*10^{-13} \  electrons

Explanation:

From the question we are told  that

   The  mass of the lead sphere is  m  =  7.70g =  0.0077 \  kg

   The net charge is Q_{net} =  -3.35*10^{-9} \ C

   The  atomic number is  u  =  82

     The  molar mass is  M  =  207 \  g/mol

Generally the excess number of  electron on the sphere is mathematically represented as

     N  =  \frac{Q_{net}}{ e }

Here e is the charge on the electron is e  =  -1.60 *10^{-19} \  C  

     So

    N  =  \frac{-3.35 *10^{-19}}{ -1.60*10^{-19}}

     N  =  2.094*10^{10} \  electrons

Generally the number of atom present is mathematically represented as  

     n  =  N_a  *  \frac{m}{ M}

Here N_a is the Avogadro's number with value  N_a  =  6.0*10^{23} \  atoms

       n  =  6.03 *10^{23} *  \frac{7.70}{ 207}

      n  =  2.24 *10^{22} \  atoms

Generally the electrons are there per lead atom is mathematically represented as

       O =  \frac{N}{n}

=>   O =  \frac{2.24*10^{22}}{2.094*10^{10}}

=>    O  =  9.33*10^{-13} \  electrons

       

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