Question

42 g of MgCO3 was heated strongly to give residue
weighing

Answer 1

Answer:

redeuse weight is 44 g

please mark brainlest

Answer 2

According to Question the Equation formed will be

MgCO3 -------> MgO + CO2

Mole of MgCO3 = mole of MgO

Mole of MgCO3 = 42/84 = 0.5

0.5 = mass of MgO / 40

Mass of MgO comes out to be 20 gram.