Răspuns:
Explicație:
MCa I2= 40 + 2.127=294-------> 294g/moli
II
M Fe3 (PO4)2= 3 .56 + 2. 31 + 8.16=358-------> 358g/moli
III
MFePO 4 = 56 +31 + 4.16=151 ------> 151g/moli [ 1mol= 151g]
MFe(OH)2= 56 + 2.16 +2=90 --------> 90g/moli [ 1mol= 90g]
MCu2O= 2.64 +16= 144------> 144g/moli
[1mol Cu2O are 144 g Cu2O ]