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a fost răspuns

nr deelecroni din
a.3g de aur
b.5 moli de CO2

Răspuns :

Kidingcrisbestp6ivat
MAu=AAu=197g/mol
197gAu....6.023×10la23 electroni
3gAu........x electroni=>x=0.09×10la23 electroni
MCO2=AC+2×AO=12+2×16=44g/mol
1 mol de CO2......6.023×10la23 electroni
5moli de CO2......x electroni=>x=30.115×10la23 electroni