2,28836•10²⁵ electroni
Date:
[tex]n_K=2 \: mol[/tex]
[tex]e⁻=?[/tex]
Formulă:
[tex]N=n\cdot N_A[/tex], unde
N=nr. de atomi (atomi)
n=nr. de mol (mol)
[tex]N_A = Numărul \: lui \: Avogadro = 6,022 \cdot {10}^{23} \frac{atomi}{mol} [/tex]
Rezolvare:
[tex]N_K = 2 \: mol \cdot 6,022 \cdot {10}^{23} \frac{atomi}{mol} = 12,044 \cdot {10}^{23} atomi \: K = 1,2044 \cdot {10}^{24} atomi \: K[/tex]
[tex]Z_K=19 \implies e⁻=19 \implies \: 1 \: atom \: de \: K \: \: are \: \: 19 \: e⁻[/tex]
1 at. K……………19 e⁻
1,2044•10²⁴ at. K………………x
x=19•1,2044•10²⁴=22,8836•10²⁴ e⁻=2,28836•10²⁵e⁻
