A metal oxide has the formula `X_(2)O_(3)`. It can be reduced by hydrogen to give free metal and water. 0.159g of metal oxide requires 6 mg of hydrogen for complete reduction. The atomic mass of metal is amu is

A metal oxide has the fomula M_2O_3 . It can be reduced by hydrogen to give free metal and water. 0.16g of the metal requires 6mg of H_2 for complete reduction. The atomic mass of the metal is .n? times the atomic number of calcium. The value of .n. is

0.5g of a metal on oxidation gave 0.79g of its oxide. The equivalent mass of the metal 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

0.635 g of a metal gives on oxidation 0.795g g of its oxide. Calculate the equivalent mass of the metal.

The reduction of 1.49 of a metal oxide required 560 ml of H_(2) at STP. If atomic mass of metal is 40, formula of its chloride will be MCl_(x)x= ______________.

One mole of chlorine combines with certain weight of a metal giving 111 g of its chloride. The same amount of metal can displace 2 g of hydrogen from an acid. The atomic weight of the metal is:

0.548 g of the metal reacts with dilute acid and liberates 0.0198 g of hydrogen at S.T.P. Calculate the equivalent mass of the metal.

10 g of a metal carbonate on heating given 5.6 g of its oxide. The equivalent amont metale 5x. What is x.

1 mole of chlorine combines with a certain mass of a metal giving 111g of its chloride. The atomic mass of the metal (assuming its valency to be 2) is