物理化学模拟试卷(下册)

1*A(1020)*1.kj/E=BzBFrBcB/EzBcBBkD(A)rBk(B)Ek(C)jk(D)rBEjk2.AgCl(B)(a)0.1mol·dm-3NaNO3(b)0.1mol·dm-3NaCl(c)H2O(d)0.1mol·dm-3Ca(NO3)2(e)0.1mol·dm-3NaBr(A)(a)(b)(c)(d)(e)(B)(b)(c)(a)(d)(e)(C)(c)(a)(b)(e)(d)(D)(c)(b)(a)(e)(d)*3.zBrBcBBB(tB)?D(A)zBtB(B)zBrBtB(C)zBrBcBtB(D)ABC4.298KD(A)La3+(B)Mg2+(C)NH4+(D)H+5.0.001mol·kg-1K3[Fe(CN)6]A2BBB12Izm=∑(A)6.010-31molkg−⋅(B)5.010-31molkg−⋅(C)4.510-31molkg−⋅(D)3.010-31molkg−⋅6.0.31molkg−⋅Na2HPO4A2BBB12Izm=∑24HPO−(A)0.91molkg−⋅(B)1.81molkg−⋅(C)0.31molkg−⋅(D)1.21molkg−⋅7.c1HClc2BaCl2H+(B)(A)λm(H+)/[λm(H+)+λm(Ba2+)+2λm(Cl-)](B)c1λm(H+)/[c1λm(H+)+2c2λm(½Ba2+)+(c1+2c2)λm(Cl-)](C)c1λm(H+)/[c1λm(H+)+c2λm(Ba2+)+λm(Cl-)](D)c1λm(H+)/[c1λm(H+)+2c2λm(Ba2+)+2c2λm(Cl-)]8.10cm31mol·dm-3KOH10cm3B(A)(B)(C)(D)(A)(B)(C)(D)9.0.0011molkg−⋅K2SO40.0031molkg−⋅Na2SO4298KD224K,Na,SO++−2BBB12Izm=∑(A)0.0011molkg−⋅(B)0.0031molkg−⋅(C)0.0021molkg−⋅(D)0.0121molkg−⋅10.()m2HO,291KΛ∞4.8910-2-12molmS⋅⋅(291K)m(H+)=m(OH-)=7.810-8mol·kg-1κBmcκ∞=Λ(A)3.8110-9S·m-1(B)3.8110-6S·m-1(C)7.6310-9S·m-1(D)7.6310-6S·m-1(918)11.NaClCaCl2LaCl3CuSO4γNaCl12.298K0.100mol·dm-3NaCl,U+(Na+)=4.2610-8m2·V-1·s-1,U-(Cl-)=6.8010-8m2·V-1·s-1,1.0710-2Ω-1·m2·mol-113.14.46K[Fe(CN)][]{}[]{}4m46mm6KFe(CN)4(K)Fe(CN)Λλλ−∞∞+∞=+*15.(ti)BaCl2t(Ba2+)t(Cl-)t(Ba2+)t(Cl-)16.298KH2SO40.01mol·kg-10.1mol·kg-1κΛmκΛm()17.()m24YSOΛ∞=2.7210-2-12molmS⋅⋅()m24HSOΛ∞=8.6010-2-12molmS⋅⋅()m4YHSOΛ∞=25.6610−×-12molmS⋅⋅18.298K0.005mol·kg-1KCl0.005mol·kg-1NaAcγ1γ2,1,2γγ±±=19.m1-32-2I1,I2I1/I21.5(540)20.298KAgCl2.6810-4-1mS⋅0.8610-4-1mS⋅298K(-12molmS⋅⋅)4.2110-24.2610-21.3310-2AgCl[]41(AgCl)()()1.8210Smκκκ−=−=×⋅2-12-23mm3mmmolmS101.38)HNO(-HCl)()AgNO(AgCl)(⋅⋅×=+=∞∞∞∞ΛΛΛΛ233-3-3-5-3-2-mmdmg101.89dmmol101.32mmol101.32)AgCl()AgCl()AgCl()AgCl(⋅×=⋅×=⋅×=≈=∞ΛκΛκc621.AgNO35.43210-5kgAg2.76710-2kgAgNO32.32610-4kg1.94810-2kgAgNO31.32610-4kg,,t(Ag+)t(NO3-)[]n()=5.03410-4mol1n()=1.36910-3mol1n()=1.10710-3mol3n()=2.41410-4mol2t(Ag+)=n()/n()=0.47952t(NO3-)=0.5205122.25AgCl1.2710-5mol·kg-1AgCl=Ag+(aq)+Cl-(aq)mG∆\AgClKNO3I=0.010mol·kg-1:A=0.509(1molkg−⋅)-1/2[]lgγ±(AgCl)=−Az+z-I1/2=−0.0018139γ±=0.996(2)∆Gm\=−RTlnKsp=−RTln(γ±m/m)2=55.8kJ·mol-1(3)KNO3lgγ±'(AgCl)=-Az+z-I1/2=−0.0509γ±'=0.8894(2)Ksp=(γ±m1/m)2=(γ±'m2/m)2m2=1.4210-5mol·kg-1(3)23.298KNaCl0.100mol·dm-3Na+Cl-U(Na+)=42.610-9m2·V-1·s-1U(Cl-)=68.010-9m2·V-1·s-1[]Λm(NaCl)=λm(Na+)+λm(Cl-)=[U(Na+)+U(Cl-)]F=106.73S·m2·mol-1(3)k(NaCl)=Λm(NaCl)c(NaCl)=1.0710-4S·m-1(2)24.291K,CaF2k=38.610-4S·m-1k=1.510-4S·m-1291KΛm∞(NaCl)Λm∞(NaF)Λm∞(12CaCl2)108.910-490.210-4116.710-4S·m2·mol-1210-4mol·dm-3NaFCaCl2[]Λm(CaF2)=Λm∞(½CaCl2)+Λm∞(NaF)-Λm∞(NaCl)=98.010-4S·m2·mol-1k(CaF2)=kk3.7110-3S·m-1c(CaF2)=k/[2Λm∞(½CaF2)]=1.8910-4mol·dm-3Ksp=[Ca2+]/c([F-]/c)2=2.710-10[Ca2+]/c([F-]/c)2=110-12Ksp(2)4(320)25.HClc1KClc2mλ∞(H+)mλ∞(K+)mλ∞(Cl-)t(H+)c1/c2[],H1HHH1212,H,K,Cl,H,K,ClHKCl[][()]mmmmmmmmcctcccccccλλλλλλλλ++++++−++−++−∞∞∞∞∞∞∞∞==+++++5c2,K,Cl12,H,ClH[]1{[()1]}mmmmcctλλλλ+−+−+∞∞∞∞+=−−526.PtCuSO4,Cu[](1)Pt(3)(2)Cu(2)27.(1)NaOHC6H5OH(2)NaOHHCl(3)AgNO3K2CrO4(4)BaCl2Tl2SO4[]5B(1020)1.NaCl10A5minAFaraday(A)0.715g(B)2.545g(C)23g(D)2.08g2.-B(A)(B)(C)(D)3.A(A)(B)(C)(D)4.-B(A)(B)(C)(D)5.10.0A1.5hPbSO4BFaraday(A)(A)84.8g(B)169.6g(C)339.2g(D)(Mr(PbSO4)=303)6.HittorffAgAgNO3AgNO3xmolAgymolAg,Ag+(D)(A)x/y(B)y/x(C)(x-y)/x(D)(y-x)/y7.U∞+2m,+(M)λ∞+FD(A)z+U∞+/m,+λ∞=F(B)z+U∞+m,+λ∞=F(C)z+m,+λ∞U∞+F=1(D)m,+λ∞/z+U∞+=F8.CaCl2(C)(A)Λ∞(CaCl2)=λm(Ca2+)+λm(Cl-)(B)Λ∞(CaCl2)=½λm(Ca2+)+λm(Cl-)(C)Λ∞(CaCl2)=λm(Ca2+)+2λm(Cl-)(D)Λ∞(CaCl2)=2[λm(Ca2+)+λm(Cl-)]9.κΛm1/(molkg)m−⋅κmΛmΛ∞1κ1Λm,1∞m,1Λ0.5κ2Λm,2∞m,2Λ0.1κ3Λm,3∞m,3Λ0.05κ4Λm,4∞m,4ΛA(A)m,1m,2m,3m,4ΛΛΛΛ(B)m,1m,2m,3m,4ΛΛΛΛ(C)1234κκκκ===(D)m,1m,2m,3m,4ΛΛΛΛ∞∞∞∞10.?C(A)(B)(C)(D)6(918)11.10cm31mol·dm-3KOH10cm3()12.0.01mol·kg-1KClCaCl2Na2SO4AlCl3γKCl13.LiCl115.0310-4-12molmS⋅⋅298KLiClLi+0.3364Cl-λm(Cl-)76.3310-4-12molmS⋅⋅()()mmm(LiCl)LiClλλ∞∞+∞−Λ=+()()()mLimmLiLiCltλλλ+∞+∞+∞−=+14.0.001mol·kg-1KCl0.001mol·kg-1K4Fe(CN)6I=0.011mol·kg-125KClγ±=0.88A=0.509(mol·kg-1)-1/215.250.1mol·dm-3KOHK+t10.1mol·dm-3KClK+t212ttOH−Cl−OHCltt−−1tt+−+=16.mNa3PO4γa(Na3PO4)()()4434NaPO27mamγ±⎛⎞=⎜⎟⎝⎠\17.Li+Na+K+Rb+…_18.KClKOHHClHCl19.(A)0.011molkg−⋅KCl(B)0.011molkg−⋅CaCl2(C)0.011molkg−⋅LaCl3(D)0.0011molkg−⋅KClDC(430)20.298K0.1mol·dm-3KClk=0.14114S·m-1525Ω0.10mol·dm-3NH3·H2O2030ΩNH3·H2O210-4S·m-1λm∞(OH-)=1.9810-2S·m2·mol-1λm∞(NH4+)=73.410-4S·m2·mol-1[](1)k(NH3·H2O)=k(KCl)R(KCl)/R(NH3·H2O)=3.6510-2S·m-1Λm(NH3·H2O)=k(NH3·H2O)/c=3.6510-4S·m2·mol-1Λm∞(NH3·H2O)=λm∞(NH4+)+λm∞(OH-)=2.71410-2S·m2·mol-1α=Λm/Λm∞=0.01345(2)(2)R(H2O)=R(KCl)k(KCl)/k(H2O)=3.705105Ω(2)21.298K0.01mol·kg-1NaCl0.003mol·kg-1Na2SO40.007mol·kg-1MgCl2:A=0.509(mol·kg-1)-1/2[]7I=0.041molkg−⋅(2)lgγNa+=(-Azi2I1/2)/(1+I1/2)=-0.0845γNa+=γCl−=0.823lgγ(Mg2+)=(-Az12I1/2)/(1+I1/2)=-0.3380γ(Mg2+)=γ(SO42−)=0.459lgγ±(NaCl)=(-Az+z-I1/2)/(1+I1/2)=-0.0845γ±(NaCl)=0.823lgγ±(MgCl2)=(-Az+z-I1/2)/(1+I1/2)=-0.1691γ±(MgCl2)=γ±(Na2SO4)=0.678(2)22.0.100mol·dm-3KCl0.200mol·dm-3RbClK+Rb+Cl-7410-44.010-47610-4S·m2·mol-1[]Λm(KCl)=Λm(K+)+Λm(Cl-)=1.5010-2S·m2·mol-11Λm(RbCl)=Λm(Rb+)+Λm(Cl-)=1.1610-2S·m2·mo