zadanie 1 H2O mH:mO = 2*1u : 16u = 2:16 = 1:8 H2S mH:mS = 2*1u : 32u = 2:32 = 1: 16 W Cl2 nie ma różnych pierwiastków, więc nie można podać stosunku zadanie 2 mH2O = 18u m2H = 2u %H = 2u/18u * 100% = 11% %O = 100%-11% = 89% mH2S = 34u m2H = 2u %H = 2u/34u * 100% = 6% %S = 100%-6% = 94% W Cl2 100% masy stanowi chlor.
1) mHCl=1u+35,5u = 36,5u mH : mCl = 1 : 35,5 mH2O=2u+16u=18u mH : mO = 2 : 16 mH : mO = 1 : 8 mH2S=2u+32u=34u mH : mS = 2 : 32 mH : mS = 1 : 16 2) mHCl=1u+35,5u = 36,5u %H = (1/36,5)*100% = 2,7% %Cl = (35,5/36,5)*100% = 97,3% mH2O=2u+16u=18u %H = (2/18)*100% = 11,1% %O = (16/18)*100% = 88,9% mH2S=2u+32u=34u %H = (2/34)*100% = 5,9% %S = (32/34)*100% = 94,1%
