Răspuns:
comp. proc .prin masa molrculara
Explicație:
K2CO3
M= 2.39 +12 + 3.16=138 ----->138 g/moli
138g K2CO3--------78g K-------12 g C -------48 g O
100 g---------------------x-------------y---------------z
x= 100.78 : 138=56,52% K
y=100.12 : 138= 8, 7 %C
z=100.48 : 138=34,78 %O
verifica calculele !!!!!!
