M(Ca(C17H35COO)2) = 40 + (12*17 + 1*35+12+16+16)*2) = 606 g/mol
M(Fe2O3) = 56*2 + 16*3 = 160 g/mol
Са40+(С12*17+Н35+С12+О16+О16)*2 = 606
Fe2 56*2+O 16*3= 160
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Answers & Comments
M(Ca(C17H35COO)2) = 40 + (12*17 + 1*35+12+16+16)*2) = 606 g/mol
M(Fe2O3) = 56*2 + 16*3 = 160 g/mol
Verified answer
Са40+(С12*17+Н35+С12+О16+О16)*2 = 606
Fe2 56*2+O 16*3= 160