Răspuns:
Explicație:
A.
Mmalachit= 64 + 12 + 3.16 + 64 + 2.16 + 2
= 124 + 98
=222------> 222g/moli
222g malachit-----------124 gCuCO3--------98g Cu(OH)2
100g--------------------------x--------------------------y
x=100 . 124 : 222=55,86%CuCO3
y=100 .98 : 222 =44,14% Cu(OH)2
B.
M calcopirita= 64 + 56 + 2.32
=184-----> 184g/moli
184g-----------------64gCu--------------56gFe-----------64gS
100g----------------------x------------------------y---------------------z
x=100.64 : 184=34,78%Cu
y=100. 56 : 184=30,44%Fe
z= 100.64: 184=34,78%S
C.
M=3.(24 + 16 ) + 4. ( 28 + 32) +18
=120 + 240 +18
=378
378g talc--------120g MgO---------240 g SiO2------------18g H2O
100------------------x----------------------------y---------------------z
x=100.120 : 378=31,75 % MgO
y=100.240 : 378=63,5% SiO2
z=100.18 : 378=4,76% H2O