Odpowiedź :
Odpowiedź:
zadanie 84
FeO Fe= x O(-ll)
x+ (-ll)=0
x= ll => Fe(ll)
Fe2O3 Fe=x O(-ll)
2x + 3×(-ll)=0
2x+ (-6)=0
2x= 6/ : 2
x= 3 => Fe(lll)
PbO Pb=x O(-ll)
x+ (-ll)=0
x= ll => Pb(ll)
PbO2 Pb=x O(-ll)
x+ 2×(-2)=0
x+(-4)=0
x=4=> Pb(IV)
Al2O3 Al=x O(-ll)
2x + 3×(-ll)=0
2x+ (-6)=0
2x= 6/ : 2
x= 3 => Al(lll)
SiO2 Si=x O(-ll)
x+ 2×(-2)=0
x+(-4)=0
x=4=> Si(IV)
H2O H=x O(-ll)
2x+(-2)=0
2x= 2/:2
x= 1=> H(l)
zadanie 85
Cu(+) => (l)
Zn(2+) => Zn(ll)
HSO4(-) H(l) S=x O(-ll)
1+ x+ 4×(-2)= -1
1+ x +(-8)= -1
x+ (-7)= -1
x= 7 +(-1)
x= 6=> S(Vl)
Wyjaśnienie: