Question

Calcule o numero de mols presentes em 578 .5g de K2Cr2O7

Answer 1

K=39
Cr=52
O=16

K₂Cr₂O₇
2*39+2*52+7*16
78+104+112
294g/mol

n=m/M
n=578,5/294
n≈2mols

[CO(NH₂)₆]Br₂
12+16+6*14+12*1+2*80
12+16+84+12+160
284g/mol