Koliko grama sumpora ima u 1 kg nikl(II)-sulfat-heptahidrata? Rješenje je kao 144 g, ali meni ne uspijeva da to izračunam. Help
n(NiS4*7H2O)=n(S) M(NiS4*7H2O)=280,77kg/kmol m(S)=32,06kg/kmol m(NiS4*7H2O)/M(NiS4*7H2O)=m(S)/M(S) 1kg/280,77kg/kmol=m(S)/32,06kg/kmol 0,00356kg/kmol=m(S)/32,06kg/kmol m(S)=0,11418kg=114,18g
n(NiSO4*7H2O)=n(S) M(NiSO4*7H2O)=280,77kg/kmol m(S)=32,06kg/kmol m(NiSO4*7H2O)/M(NiSO4*7H2O)=m(S)/M(S) 1kg/280,77kg/kmol=m(S)/32,06kg/kmol 0,00356kg/kmol=m(S)/32,06kg/kmol m(S)=0,11418kg=114,18g
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nik(II)-sulfat-heptahidrat
n(NiS4*7H2O)=n(S)
M(NiS4*7H2O)=280,77kg/kmol m(S)=32,06kg/kmol
m(NiS4*7H2O)/M(NiS4*7H2O)=m(S)/M(S)
1kg/280,77kg/kmol=m(S)/32,06kg/kmol
0,00356kg/kmol=m(S)/32,06kg/kmol
m(S)=0,11418kg=114,18g
nik(II)-sulfat-heptahidrat
n(NiSO4*7H2O)=n(S)
M(NiSO4*7H2O)=280,77kg/kmol m(S)=32,06kg/kmol
m(NiSO4*7H2O)/M(NiSO4*7H2O)=m(S)/M(S)
1kg/280,77kg/kmol=m(S)/32,06kg/kmol
0,00356kg/kmol=m(S)/32,06kg/kmol
m(S)=0,11418kg=114,18g