Răspuns:
Explicație:
MCu2O=2. 64 + 16=144--------> 144g/moli
MCa(OH)2= 40 + 2.16 +2=74-------> 74g/moli
MH2SO3= 2+32 +3.16=82-------->82g/moli
MCaCO3= 40 + 12 + 3.16=100------> 100g/moli
oxid de cupru I = Cu2O
M.Cu2O = 2x64+16 = 144 g/mol
hidroxid de calciu = Ca(OH)2
M.Ca(OH)2 = 40+2(16+1) = 74 g/mol
acid sulfuros = H2SO3
M.H2SO3 = 2x1+32+3x16 = 82 g/mol
carbonat de calciu = CaCO3
M.CaCO3 = 40+12+3x16 = 100 g/mol
