A metal M forms a metal Carbonate M2Co3 of the carbonate contains 48%oxygen by mass then determine the atomic weight of metal
Answers
Answer:
M is 20
Explanation:
Mass % of O=48%
Mass % of O=48%Let the atomic mass of M be x
Total mass = 2y+12+48
Total mass = 2y+12+48=(2y+60)g/mol
Total mass = 2y+12+48=(2y+60)g/molO is 48% of the total mass.
Total mass = 2y+12+48=(2y+60)g/molO is 48% of the total mass.∴48% of (2y+60)
Total mass = 2y+12+48=(2y+60)g/molO is 48% of the total mass.∴48% of (2y+60)100
Total mass = 2y+12+48=(2y+60)g/molO is 48% of the total mass.∴48% of (2y+60)10048
Total mass = 2y+12+48=(2y+60)g/molO is 48% of the total mass.∴48% of (2y+60)10048
Total mass = 2y+12+48=(2y+60)g/molO is 48% of the total mass.∴48% of (2y+60)10048 ×(2y+60)=48
Total mass = 2y+12+48=(2y+60)g/molO is 48% of the total mass.∴48% of (2y+60)10048 ×(2y+60)=4896y+2880=4800
Total mass = 2y+12+48=(2y+60)g/molO is 48% of the total mass.∴48% of (2y+60)10048 ×(2y+60)=4896y+2880=480096y=1920
Total mass = 2y+12+48=(2y+60)g/molO is 48% of the total mass.∴48% of (2y+60)10048 ×(2y+60)=4896y+2880=480096y=1920y=20