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If a certain oxide of nitrogen weighing 0.11g gives 56 ml of nitrogen and another oxide of nitrogen weighing 0.15 g gives the same volume of nitrogen (both at STP), show that these results support the law of multiple proportions.

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4 answers
  1. 22400 ml of N2 = 28 gm so,

    56 ml of N2 = 28 x 56/ 22400 = 0.07 gm

    Here weight of oxide (A) =0.11 gm

    After reduction the weight = 0.07gm

    Loss of weight = 0.11-0.07 = 0.04gm (It is the amount of oxygen)--1.

    Similarly weight of oxide(B) =0.15 gm

    Weight after reduction = 0.07 gm

    Hence, in case (A)

    0.07 gm of nitrogen combines with 0.04 gm of oxygen

    so, weight of nitrogen which would combine with 0.08 gm of oxygen should be equal to

    = 0.07 x 0.08/ 0.04 = 0.14 gm--.---2

    Here,in both the cases 1 and 2 the weight of nitrogen which combines with fixed amount of oxygen ( 0.04 gm) are 0.07 gm and 0.14 gm which are in the simple ratio of 2:1

    Hence the oxides follow the law of multiple proportions.

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  2. Hlo bro you are best people in world because you can help other people

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  3. Really liked the answer and I hope it will help other like me

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  4. Sir maine nitrogen ka mass fixed kr diya aur oxygen ka ratio nikala 1:2 aa rha hai kya mera answer shi hai kya please reply

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