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9966 [12]
3 years ago
12

a metal forms two oxides. The higher oxide contains 80% metal. 0.72g of the lower oxide gave 0.8g of the higher oxide when oxidi

sed.show that the data illustrate the law of multiple proportions.​
Chemistry
2 answers:
zzz [600]3 years ago
6 0

Answer:

0.72 g of the lower oxide gave 0.8 g of higher oxide when oxidised. ... Thus, 90g of lower oxide contains as much metal as 100g of higher oxide, i.e., 80g (given). Hence, 80g of metal combines with 10g of oxygen in the lower oxide and 20g of oxygen in the higher oxide.

Sunny_sXe [5.5K]3 years ago
6 0

Answer:

Solution given:

For higher oxide:

contains 80% metal

weight =80% 0f 0.8g=0.64g.

weight of Oxygen=0.8-0.6=0.16g

<u>For</u><u> </u><u>lower</u><u> </u><u>oxide</u><u>:</u>

the metal's weight is fixed.

weight of oxygen=0.72g-0.64g=0.08g

Now

Ratio of oxygen combined with fixed amount of metal is :

0.16:0.08

=2:1

since it is a simple ratio .

so;

<u>It follows the law of multiple proportions</u>.

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