Question

formula moleculara a substantei cu compozitia procentuala 42,39%Na, 11,32% C si 45,28% O este:
a) NaCO3
b) Na2CO
c) Na2CO3
(ajutati-ma va rooog)

Answer 1

42.36:11.32:45:28=(42.36/23):(11.32/12):(45.28/16) = 1.84:0.94:2.83 = (1.84/0.94):(0.94/0.94):(2.83/0.94) = 2 : 1 : 3 => Na2CO3