Question

calculati numarul de electroni ce pot fi cedati de 5,4 g Al

Answer 1

Al Z=13
Al:1s22s22p63s23p1 - 3e -> 1s22s22p6  Al3+
 n=m/M=5,4/27=0,2 moli Al
n=N/Na=>N=n*Na=0,2*6,022*10 la puterea 23=1,2044*10 la puterea 23
1 atom Al.......cedeaza 3electroni
1,2044*10 la puterea 23.........cedeaza x electroni
x=3*1,2044*10la puterea 23=3,6132*10 la puterea 23 

n-nr moli:m=masa Al:M=masa molara:N-nr atomi:Na-nr lui Avogadro