What is the emprical formula of vanadium oxide , if `2.74g` of the metal oxide contains `1.53g` of metal ?

What is the empirical formula of vanadium oxide, if 2.74 g of the metal oxide contains 1.53 g of metal ? (Atomic mass of vanadium=52 u)

What is the empirical formula of an oxide of vanadium, if 2,74 gm, of the oxide contains 1.53 gm metal (Atomic mass of V = 52 amu)

metal combines with oxygen to form two oxides, having the following composition: (i) 0.398 g of the first metal oxide contains 0.318g of metal. (ii) 0.716 g of the second oxide contains 0.636g of the metal. Show that the above data agrees with the law of multiple proportions.

metal combines with oxygen to form two oxides, having the following composition: (i) 0.398 g of the first metal oxide contains 0.318g of metal. (ii) 0.716 g of the second oxide contains 0.636g of the metal. Show that the above data agrees with the law of multiple proportions.

Metal X forms wo oxides. Formula of the first oxide is XO_(2) . The first oxide contains 50% of oxygen. If the second oxide contains 60% of oxygen, the formula of the second oxide is

On reduction with hydrogen, 3.6 g of an oxide of metal (M) left 3.2 g of the metal. If the atomic weight of the metal is 64. the formula of the oxide is

4 g of a metal oxide contains 1.6 g-oxygen, then equivalent mass of the metal is