6.488g of lead combine directly with 1.002g of oxygen to form lead peroxide. Lead peroxide is also produce by heating lead nitrate and it was found that percentage of oxygen in lead peroxide is 13.38% use this information to illustrate law of definite composition
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In the first reaction:
Pb + O₂ ⇒ PbO₂
6.488g Pb combines with 1.002g of O₂
So mass of PbO₂ formed is
= (6.488 + 1.002)g = 7.490g
7.490g of lead peroxide combines with 1.002g of oxygen
Therefore, percentage of oxygen =
(1.002 ÷ 7.490) × 100 = 13.38 %
In the second reaction also, the percentage of oxygen present in PbO₂ will be 13.38%, which is already given.
Hence the percentage composition of oxygen in both the reactions is identical. Thus, above data illustrates the law of definite composition.
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Pb + O₂ ⇒ PbO₂
6.488g Pb combines with 1.002g of O₂
So mass of PbO₂ formed is
= (6.488 + 1.002)g = 7.490g
7.490g of lead peroxide combines with 1.002g of oxygen
Therefore, percentage of oxygen =
(1.002 ÷ 7.490) × 100 = 13.38 %
In the second reaction also, the percentage of oxygen present in PbO₂ will be 13.38%, which is already given.
Hence the percentage composition of oxygen in both the reactions is identical. Thus, above data illustrates the law of definite composition.
(Brainliest Answer Please)
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Answer:
In the first reaction,
In the first reaction,Pb + O2 --------> PbO2
In the first reaction,Pb + O2 --------> PbO26.488g Pb combines with 1.002g of O2 so, mass of PbO2 formed=6.488+1.002)g = 7.490g
In the first reaction,Pb + O2 --------> PbO26.488g Pb combines with 1.002g of O2 so, mass of PbO2 formed=6.488+1.002)g = 7.490g7.490 g of lead peroxide combines with 1.002g of oxygen
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