Answer:
Step-by-step explanation:
Atomic weight of strontium ( Sr ) is 88 .
Atomic weight of Beryllium ( Be) is 9 .
Both are bivalent .
2 x 96500 coulomb is required to decompose 88 g of Sr .
88 g of Sr requires 2 x 96500 coulomb of charge
20 g of Sr will require 2 x 96500 x 20 / 88 coulomb of charge
coulomb of charge required = 43863.63 coulomb .
9 g of Be requires 2 x 96500 coulomb of charge
2 x 96500 coulomb of charge gives 9 g of Be
43863.63 coulomb .of charge gives 9 x 43863.63 / (2 x 96500 ) g of Be
= 2.04 g.