Question

Two metallic oxide contain 27.6% and 30% o respectively if the formula of the first oxide is x3o4 that of the sec. Will be

Answer 1

Explanation:

$\bf\large\star\underline{solution♀️}$

Let mass of the metal = x

Let mass of the metal = x% of metal in M

Let mass of the metal = x% of metal in M 3

Let mass of the metal = x% of metal in M 3

Let mass of the metal = x% of metal in M 3 O

Let mass of the metal = x% of metal in M 3 O 4

Let mass of the metal = x% of metal in M 3 O 4

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100But as given percentage = (100-27.6) = 72.4 %

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100But as given percentage = (100-27.6) = 72.4 %so, (3x/3x+64)×100=72.4

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100But as given percentage = (100-27.6) = 72.4 %so, (3x/3x+64)×100=72.4or x = 56.

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100But as given percentage = (100-27.6) = 72.4 %so, (3x/3x+64)×100=72.4or x = 56.In 2nd oxide,

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100But as given percentage = (100-27.6) = 72.4 %so, (3x/3x+64)×100=72.4or x = 56.In 2nd oxide,oxygen = 30%....so metal = 70%

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100But as given percentage = (100-27.6) = 72.4 %so, (3x/3x+64)×100=72.4or x = 56.In 2nd oxide,oxygen = 30%....so metal = 70%so, the ratio is

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100But as given percentage = (100-27.6) = 72.4 %so, (3x/3x+64)×100=72.4or x = 56.In 2nd oxide,oxygen = 30%....so metal = 70%so, the ratio is M : O

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100But as given percentage = (100-27.6) = 72.4 %so, (3x/3x+64)×100=72.4or x = 56.In 2nd oxide,oxygen = 30%....so metal = 70%so, the ratio is M : O 70/56 : 30/16

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100But as given percentage = (100-27.6) = 72.4 %so, (3x/3x+64)×100=72.4or x = 56.In 2nd oxide,oxygen = 30%....so metal = 70%so, the ratio is M : O 70/56 : 30/161.25: 1.875

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100But as given percentage = (100-27.6) = 72.4 %so, (3x/3x+64)×100=72.4or x = 56.In 2nd oxide,oxygen = 30%....so metal = 70%so, the ratio is M : O 70/56 : 30/161.25: 1.8752 : 3

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100But as given percentage = (100-27.6) = 72.4 %so, (3x/3x+64)×100=72.4or x = 56.In 2nd oxide,oxygen = 30%....so metal = 70%so, the ratio is M : O 70/56 : 30/161.25: 1.8752 : 3so, 2nd oxide is M

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100But as given percentage = (100-27.6) = 72.4 %so, (3x/3x+64)×100=72.4or x = 56.In 2nd oxide,oxygen = 30%....so metal = 70%so, the ratio is M : O 70/56 : 30/161.25: 1.8752 : 3so, 2nd oxide is M 2

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100But as given percentage = (100-27.6) = 72.4 %so, (3x/3x+64)×100=72.4or x = 56.In 2nd oxide,oxygen = 30%....so metal = 70%so, the ratio is M : O 70/56 : 30/161.25: 1.8752 : 3so, 2nd oxide is M 2

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100But as given percentage = (100-27.6) = 72.4 %so, (3x/3x+64)×100=72.4or x = 56.In 2nd oxide,oxygen = 30%....so metal = 70%so, the ratio is M : O 70/56 : 30/161.25: 1.8752 : 3so, 2nd oxide is M 2 O

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100But as given percentage = (100-27.6) = 72.4 %so, (3x/3x+64)×100=72.4or x = 56.In 2nd oxide,oxygen = 30%....so metal = 70%so, the ratio is M : O 70/56 : 30/161.25: 1.8752 : 3so, 2nd oxide is M 2 O 3

Let mass of the metal = x% of metal in M 3 O 4 =(3x/3x+64)×100But as given percentage = (100-27.6) = 72.4 %so, (3x/3x+64)×100=72.4or x = 56.In 2nd oxide,oxygen = 30%....so metal = 70%so, the ratio is M : O 70/56 : 30/161.25: 1.8752 : 3so, 2nd oxide is M 2 O 3