12.976 g of lead combines with 2.004 g of oxygen to form PbO2 . PbO2 can also be produced by heating lead nitrate and it was found that the percentage of oxygen present in PbO2 is 13.38%. With the help of given information,illustrate the law of constant composition?

Answer 1

find the percent of Oxygen in Lead(IV) oxide.
percent: 2.004/(12.976+2.004)= 0.133778371 or 13.38Percent.

Of course, your mentioning of lead nitrate is silly.
The decomposition of PbO2 and the combination of Pb and O2 reflect the same percentage.

Answer 2

Solve kr ke photo nhi dal skte?

Answer 3

In first experiment
Total mass of compound =12.976+2.004=14.98g
%oygen =2.004/12.976*100
=13.38%
In 2nd experiment %oygen is given that is 13.38%
So it illustrates law of definite proportions

Answer 4

15.44 '/,

Answer 5

In first experiment
Total mass of compound =12.976+2.004=14.98g
%oygen =2.004/12.976*100
=13.38%
In 2nd experiment %oygen is given that is 13.38%

Answer 6

In first experiment
Total mass of compound =12.976+2.004=14.98g
%oxygen =2.004/12.976*100
=13.38%
In 2nd experiment %oygen is given that is 13.38%