1.84 g mixture of Caco, and MgCoz was heated to a constant weight till 0.96 g residue formed. % (by mass) of MgCoz in sample was (1) 45.66% (2) 54.34% (3) 30% (4) 70%

Solution

Share

60

4.0 (1 ratings)

Download App for Answer

Let man of ( a a c_{3}=x g ) Man of ( m g(0,3)=(1.84-x) ) ( x ) gram of ( operatorname{cacos} ) will give ( quad operatorname{cat}=56 x ) 100 ( | log _{2}(1.84-x) ) g will gine ( m g 0=frac{40(1.84-x)}{84} ) Total man ( =0.96 ) ( 0 cdot 56 x+frac{40(1-84-x)}{84}=0.96 ) ( x=1 ) ( % ) of ( operatorname{cac} 0,3=frac{1}{1.84}=54 cdot 35 ) " of ( m g cos =100-54.35=45.65 )

60

4.0 (1 ratings)

Rate Solution

Share