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?
Answers
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.
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
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%
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%
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%
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.