Asked by Kien

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?
Answered by bobpursley
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.
Answered by Sangita
Solve kr ke photo nhi dal skte?
Answered by Bhavya
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
Answered by Anonymous
15.44 '/,
Answered by @nkit kumar
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%
Answered by @nkit kumar
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%
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