五、answer: Add HCI into the mixture of Ag",Fe Co2 Cr3,you get AgCl precipitate that is insoluble in HNO3 while soluble in NH3.H2O (Ipoint); Reaction of Fe Co2,Cr with NH:.H2O will produce Cr(OH)a and Fe(OH)asolid, which can be separated from Co(NH3)and CrOions.Acidified Co2solution can be used to indentify Co2*in the presence of acetone and F"using SCNions (blue solution (2 points);Addition of NaOH and H2O2 solution to Cr(OH)and Fe(OH)3 can bring about CrOthat can be separated from Fe(OH)3(s)(1 point);Fe(III)can be identified using SCN',and Cr(Vi)can be identified using H2O2 and ether (I point) 六、answer: 1、 C0+Cl2-→C0C1 (2) 4Gm=4GmC0C)-4Gm(C0F-204.6-(-137.2=-67.4 kJ-mol 4,Gm-RTlnK°，K=6.52×10l (2 4Hm°=4HmC0C)-4HmC0)-218.8-(-110.5)=-108.3 kJ-mol(1.5) 1nK2/K1=-(4,Hm/R)(1/T2-l/T) (1.5) K2=9.9x×107 2 answer: According to pV=nRT,PV=nRT,we get: (3) p(H2)=5.8kPa:p(I2)=145kPa:p(HI)=11.6kPa: [(11.6+x)/P2-[5.8-x)145-x)/P2]=K=54.5 x=5.76kPa (2.5) p(HI,at equilibria)=17.4 kPa nHl)=6×10-3 (2.5) 五、answer: Add HCl into the mixture of Ag+ 、Fe3+、Co2+、Cr3+, you get AgCl precipitate that is insoluble in HNO3 while soluble in NH3.H2O（1point）; Reaction of Fe3+、Co2+、Cr3+ with NH3.H2O will produce Cr(OH)3 and Fe(OH)3solid， which can be separated from Co(NH3)4 2+ and CrO4 2- ions. Acidified Co2+ solution can be used to indentify Co2+ in the presence of acetone and F￾using SCN- ions（blue solution ）(2 points); Addition of NaOH and H2O2 solution to Cr(OH)3 and Fe(OH)3 can bring about CrO4 2- that can be separated from Fe(OH)3(s)（1 point）; Fe(III) can be identified using SCN- , and Cr(Vi) can be identified using H2O2 and ether（1 point） 六、answer: 1、 CO＋Cl2→COCl （2） ⊿rGm o =⊿fGm o (COCl)－⊿fGm o (CO)=-204.6－(-137.2)=-67.4kJ·mol-1 ⊿rGm o =-RTlnKo , Ko =6.52×1011 （2） ⊿rHm o =⊿fHm o (COCl)－⊿fHm o (CO)=-218.8－(-110.5)=-108.3 kJ·mol-1 （1.5） lnKo 2/ Ko 1=-(⊿rHm o /R)·(1/T2－!/T1) （1.5） Ko 2=9.9××107 2 answer: According to pV=nRT, PiV=niRT, we get： （3） p(H2)=5.8kPa；p(I2)=145kPa；p(HI)=11.6kPa； [(11.6＋x)/Po ] 2 ÷[(5.8－x)(145－x)/Po2]=K=54.5 x=5.76kPa （2.5） p (HI, at equilibria)＝17.4 kPa n(HI)=6×10-3 （2.5）