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Calculati a)raportul atomic de masa al MgCO3, b) masa molara si c)compozitia procentuala

Răspuns :

a)Mg:C:O=24:12:48=2:1:4
b)M=24+12+3*16=84g/mol
c)84g MgCO3..........24gMg.................12gC...............48gO
   100.....................X.........................Y.....................Z
X=24*100/84=28,57%
Y=12*100/84=14,28%
Z=48*100/84=57,14%