Question

calculari compozitia procentuala hidroxidului de magneziu..pls va roggg​

Answer 1

Comp. proc . a  Mg(OH ) 2

M=24 + 2.16+ 2=58----------> 58g/moli

58 g Mg(OH)2---------24 g Mg--------32 g O---------2g H

100 g--------------------------x----------------y--------------------z

x=100. 24 : 58=41,37 %Mg

y=100 . 32 : 58=55,17 % O

z=100. 2 : 58=3,45 % H