jaką liczbę moli stanowi
a) 1 tona Fe
b)1kg CaCO3
c)23 miligramy CO2
d) 15 kg CaCO3
a)
1t=1000kg=1000000g
M (Fe) = 56 g/mol
n=m/M
n=1000000g/56g/mol≈17857 moli
b)
1kg=1000g
M (CaCO3) = 100g/mol
n=1000g/100g/mol=10moli
c)
23mg=0,023g
M (CO2) = 44 g/mol
n=0,023g/44g/mol≈0,00052 mola
d)
15 kg=15000g
n=15000g/100g/mol=150 moli
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a)
1t=1000kg=1000000g
M (Fe) = 56 g/mol
n=m/M
n=1000000g/56g/mol≈17857 moli
b)
1kg=1000g
M (CaCO3) = 100g/mol
n=1000g/100g/mol=10moli
c)
23mg=0,023g
M (CO2) = 44 g/mol
n=0,023g/44g/mol≈0,00052 mola
d)
15 kg=15000g
M (CaCO3) = 100g/mol
n=15000g/100g/mol=150 moli