北京化工大学《无机化学》(双语)期末考试模拟试卷-A-答案

北京化工大学ModelofFinalExaminationof《InorganicChemistry》(bi-lingualcourse)Coursecode课程代码CHM2170TClassNo.:NameandID:Items(题号)一二三四五六Totalscore(总分)Score(得分)一、是非题：（判断下列叙述是否正确，正确的在括号中画√，错误的画×。不必写在答题纸上。）（本大题共10小题，每题1分，共10分）1x;2x;3x;4√;5√;6√;7x;8x;9x;10x.二、选择题：（在下列各题中，选择出符合题意的答案，将其代号填入括号内。）（本大题共20题，每题1.5分，共30分）1B;2D;3B;4A;5C;6B;7D,8A;9D;10B;11B;12B;13B;14C;15A;16C;17B;18B;19B;20D.三、Pleasefillintheblankwithcorrectwords,symbolordataaccordingtotherequirements(10questions,2score/question,20scoreintotal)1．六氟合硅（IV）酸四氨合镍(II)；[Co(H2O)2(NH3)4]Cl3.2．(a);(b);3．12.39;0.011.4．HNO3;He2+;H2+,B4H10;H3PO4.5．Formationofnewmatter;breakandformofnewchemicalbonds;changeofenergy.6．Same;miniumenergyprinciple;Paulingincompatibilityprinciple;andHundrule.7．σ1s2σ1s*2σ2s2σ2s*2π2p3；1.5；(a);paramagnetic.8．4.3x10-6;8.63.9．Na2CO3CaCO3MgCO3CuCO3；第1页Al2O3AlF3AlCl3.10．HgO+2e+H2O→Hg+2OH-；Zn-2e+2OH-→Zn(OH)2；NaO;(-)Zn(s)|Zn(OH)2(s)|OH-(aq)|HgO(s)|Hg(s)(+).四、Pleasebalancethefollowingchemicalequations(5equations,2pointsforeachone,10pointsintotal)1．PbCl2＋H2OPb(OH)Cl+HCl⎯→⎯2．B2H6(g)＋6H2O(l)2H⎯→⎯3BO3+6H23．2Ni(OH)2＋ClO-2NiO(OH)+Cl⎯→⎯-+H2O4．2KMnO4＋5H2O2＋3H2SO42MnSO⎯→⎯4+5O2+K2SO4+8H2O5．Hg22+＋4I-（excess）⎯→⎯Hg+HgI42-五、Simplequestions(twoquestions,5pointsforeachone,10pointsintotal)1．Anwser:O,asitsvalenceelectronshellis2s22p4.Whenonee-isremoved,moreenergyisrequiredtoremovetheseconde-whentheporbitalishalf-filledwhichisstableinviewofenergy.2.Answer:AdditionofexcessNaOHtothesolutionofZn2+，Co2+andAl3+resultsinprecipitateofCo(OH)2andformationofsolubleAl(OH)4-andZn(OH)4-complexes.SeparatetheprecipitateCo(OH)2bycentrifugationfromthesolution,thenaddHCl,NH3.H2O,acetone,KSCNsubsequentlytoCo(OH)2,bluecolorwillbeobserved,indicatingthepresenceofCo2+.TothesolutionofAl(OH)4-andZn(OH)4-complexesisaddedHClandNH3.H2O,thenAl(OH)3precipitatewillform,whichisseparatedandidentifiedbyformingredsolidafterreactionwithHAcandAlregeant;theZn2+isidentifiedbyformationofwhiteZnSafterreactionofZn2+withNa2S.六、Answerthefollowingquestionsaccordingtorequests(4questions,20pointsintotal)1．(5points)(1)Anwser:(1)AgClAg++Cl-Atequilibriumx0.05第2页0.05x=Ksp(AgCl)=1.77x10-10,x=3.54x10-9molL-1Ag++2NH3Ag(NH3)2Initial0.1y0Atequilibria3.54x10-9y–2(0.1–3.54x10-9)0.1–3.54x10-9(0.1–3.54x10-9)/3.54x10-9[y-2(0.1–3.54x10-9)]2=Kf(Ag(NH3)2)=1.12x107y=1.79molL-1thetotalconcentrationofNH3is1.79molL-1.(2)AgIAg++I-3.54x10-90.0013.54x10-9x0.001=3.54x10-12Ksp(AgI)=8.52x10-17,SoAgIprecipitates.2．(5points)Anwser:p(SbCl5)initial=nRT/V=0.0125molx8.314J⋅K-1x521K/3.50L=15.47KPaSbCl5(g)SbCl3(g)+Cl2(g)Initial/KPa15.4700Atequilibria15.47-xxx(X/100)2/[(15.47-x)/100]=KPΘ=1.07,x=P(SbCl3)=P(Cl2)=13.71Kpa,P(SbCl5)=15.47-13.71=1.76kpaPtotal=1.76+13.71x2=29.17kpa3.(5points)Anwser:Ag2SO4(s)2Ag++SO42-0.01x0.012x=Ksp(Ag2SO4)=1.4x10-5X=0.14CaSO4(s)Ca2++SO42-0.01y0.01y=Ksp(CaSO4)=9.1x10-6y=9.1x10-4SoCa2+precipitatesfirst.第3页WhenAg+beginstoprecipitate,CaSO4(s)Ca2++SO42-z0.14z2x0.14=Ksp(CaSO4)=9.1x10-6z=6.5x10-5M4.(5points)Anwser:(1)Cu2+(aq)＋2e－===Cu(s)1)ΔrGmΘ1=-zFE1=-2Fx0.3394V;Cu+(aq)＋e－===Cu(s)2)ΔrGmΘ1=-zFE2=-1Fx0.5180V;____________________________________1)-2)x2=3):Cu2+(aq)＋Cu(s)===2Cu+(aq)3)ΔrGmΘ3=-RTlnKΘΔrGmΘ1-2ΔrGmΘ2=ΔrGmΘ3,-2Fx0.3394+2x1Fx0.5180=-RTlnKΘ,2x96485x0.3394+2x96485x0.5180=-8.314x298lnKΘ,KΘ=9.1x10-7(2)Cu2+(aq)＋2e－===Cu(s)1)ΔrGmΘ1=-zFE1=-2Fx0.3394V;Cu+(aq)＋e－===Cu(s)2)ΔrGmΘ1=-zFE2=-1Fx0.5180V;Cu++2Cl-=CuCl2-3)ΔrGmΘ3=-RTlnKfΘ1）-2x2)+2x3)=4):Cu2+(aq)＋Cu(s)＋4Cl-(aq)===2CuCl2-(aq)4)ΔrGmΘ4=-RTlnKΘΔrGmΘ1-2ΔrGmΘ2+2ΔrGmΘ3=ΔrGmΘ4,-2Fx0.3394+2x1Fx0.5180–2x8.314x298ln6.91x104=-RTlnKΘ,KΘ=4.34x103第4页