Question

a metal forms two oxides one contain 46.67% of the metal and another 63.94% of the metal show that these results are in accordance with the law of multiple proportion​

Answer 1

dear mate find out the quantity of oxygen at fixed rate of metal as follow

Oxide 1 = M:O = 53.33 :46.67 = 1:0.8 = 1:1 

Oxide 2= M:O = 36.06:63.94 = 1: 1.77 = 1:2

so the oxygen is in ratio of 1:2 so this is a example of multiple proportion..

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Answer 2

Answer:

LAW OF MULTIPLE PROPORTIONS: Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number.

In the first oxide:

Mass of metal = 46.67 g/mol

Mass of oxygen = 53.33 g/mol

In the second oxide:

Mass of metal = 63.94 g/mol

Mass of oxygen = 36.06 g/mol

63.94 g of metal combines with = 36.06 g of oxygen

46.67 g of metal combines with =$\frac{36.06}{63.94}\times 46.67=26.32g$ g of oxygen

Mass ratio of metal and oxygen in second oxide- 46.67: 26.32

As, the mass of metal is fixed, so the mass ratio of oxygen in both the compounds must be in whole number ratio.

Mass ratio of oxygen in both the oxides= 53.33 : 26.32 = 2: 1

Hence, these two oxides obeys Law of multiple proportions.