Sagorevanjem 3,27 g Zn oslobodi se 17,4 KJ toplote,izracunaj standardnu entalpiju
2Zn(s)+O2(g)-->2ZnO(s) nadjes broj molova Zn....0,05mol 0,05mol........17,4kJ 2mol..............X X=696kJ/mol delta rH=delta fH*2 -696kJ/mol/2=delta fH dobijes -349kJ/mol
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2Zn(s)+O2(g)-->2ZnO(s)
2Zn(s)+O2(g)-->2ZnO(s) nadjes broj molova Zn....0,05mol 0,05mol........17,4kJ 2mol..............X X=696kJ/mol delta rH=delta fH*2 -696kJ/mol/2=delta fH dobijes -349kJ/mol