Respuesta :
Answer:
The fraction of the copper’s electrons removed is [tex]9.076\times 10^{-13}[/tex].
Explanation:
Mass of copper ball = 50.0 g
Moles of copper = [tex]\frac{50.0 g}{63.5 g/mol}[/tex]
1 mole = [tex]N_A=6.022\times 10^{23}[/tex]
Number of copper atoms =[tex]\frac{50.0 g}{63.5 g/mol}\times 6.022\times 10^{23} [/tex]
1 atom of copper has 29 protons
Total number of protons in 50.0 g of copper =[tex]\frac{50.0 g}{63.5 g/mol}\times 6.022\times 10^{23} \times 29=1.3751\times 10^{25}[/tex]
Since an atom is a neutral specie which means number of protons are equal to number of electrons.
Total number of electrons = [tex]1.3751\times 10^{25}[/tex]....(1)
Net charge on the copper ball = [tex]2.00/mu C=2.00\times 10^{-6} C[/tex]
Q=Ne
Q = Total charge
N = Number of electrons
e = charge on an electron = [tex]1.602\times 10^{-19} C[/tex]
[tex]2.00\times 10^{-6} C=N\times 1.602\times 10^{-19} C[/tex]
[tex]N =1.248\times 10^{13} [/tex]
Total number of electrons removed = N = [tex]1.248\times 10^{13} [/tex]
Fraction of the copper’s electrons has been removed:
[tex]\frac{\text{Number of electrons removed}}{\text{Total electrons}}[/tex]
[tex]\frac{1.248\times 10^{13}}{1.3751\times 10^{25}}=9.076\times 10^{-13}[/tex]
The fraction of the copper’s electrons removed is [tex]9.076\times 10^{-13}[/tex].
