Two metallic oxides contain 27.6% and 30% oxygen respectively.If the formula of the first oxide is X3O4,that of the second will be:
1)XO 2)XO2
3)X2O5 4)X2O3
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Answered by
1718
Given that, formula of first oxide = M3O4
Let mass of the metal = x
% of metal in M3O4 = (3x/ 3x+64) *100
but as given % age = (100-27.6) = 72.4 %
so, (3x/ 3x+64)*100 = 72.4
or x = 56.
in 2nd oxide,
oxygen = 30%....so metal = 70%
so, the ratio is
M : O
70/56 : 30/16
1.25 : 1.875
2 : 3
so, 2nd oxide is M2O3
B-) Hope this helps you
Just mark as brilliant!!
Let mass of the metal = x
% of metal in M3O4 = (3x/ 3x+64) *100
but as given % age = (100-27.6) = 72.4 %
so, (3x/ 3x+64)*100 = 72.4
or x = 56.
in 2nd oxide,
oxygen = 30%....so metal = 70%
so, the ratio is
M : O
70/56 : 30/16
1.25 : 1.875
2 : 3
so, 2nd oxide is M2O3
B-) Hope this helps you
Just mark as brilliant!!
Answered by
192
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70/56 : 30/16
1.25 : 1.875