Respuesta :
Answer: a) [tex]2.58\times 10^{27}[/tex] electrons are present is 7.74 L of water.
b)[tex]2.7\times 10^{8} Coulombs[/tex] is the net charge of all these electrons
Explanation:
a) Molar mass of the water = 18 g/mol
Volume of water = 7.74 L
Density of water = 1000 g/L
Mass of water in 7.74 L of water :
[tex]density=1000 g/L=\frac{mass}{volume}=\frac{mass}{7.74 L}[/tex]
Mass of the water = 7740 g
Moles of water =[tex]\frac{\text{mass of water}}{\text{molar mass of water}}=\frac{7740 g}{18 g/mol}=430 moles[/tex]
[tex]1 mol=6.022\times 10^{23} [/tex]molecules
Number of water molecules in 430 moles:
[tex]430\times 6.022\times 10^{23} =2.58\times 10^{26}[/tex]molecules
1 molecule water contains = 10 electrons
Then[tex]2.58\times 10^{26} molecules=2.58\times 10^{26}\times 10=2.58\times 10^{27}[/tex] electrons
[tex]2.58\times 10^{27}[/tex] electrons are present is 7.74 L of water.
b) The net charge of all these [tex]2.58\times 10^{27}[/tex] electrons:
Charge on 1 electron =[tex]-1.602\times 10^{-19} Coulombs[/tex]
The net charge of all these [tex]2.58\times 10^{27}[/tex] electrons:
[tex]2.58\times 10^{27}\times 1.602\times 10^{-19} C=2.7\times 10^{8} Coulombs[/tex]
[tex]2.7\times 10^{8} Coulombs[/tex] is the net charge of all these electrons
