普通化學(xué)趙士鐸第三版習(xí)題答案

（2）土壤溶液濃度過(guò)大，滲透壓大于植物根細(xì)胞液的滲透壓（3）溶液的凝固點(diǎn)下降?Tb=Kbb=0.52K?kg?mol-1?1.1?T=Kb=1.86K?kg?mol-1?1.17mol?kg-1=2.18KfV誤=(1/4)[-1107kJ?mol-1-(-1150kJ?mol-1)](1/4){-143kJ?mol-1-(-286kJ?2.4?H?=2?H?(CO,g)+3?H?(HO,l)+(-1)??H?(CHOCH,l)+(-3)?H?(O,g)??H?(CHOCH,l)=2?H?(CO,g)+3?H?(HO,l)-?H?=-183kJ?mol-122.6C(s)+HO(g)=CO(g)+H(g)-110.5kJ?mol-1-(-)241.8kJ?mol-1=131.3kJ?mol-1(g)+HO(g)2(g)+HO(g)222rmrmrmrmrmrm?mBrmmmrmmm2m2m2/cΘ}{p(H{KΘ(3)}4{KΘ(3)}4{KΘ(1)}2{KΘ(2)}2{KΘ(1)}2{KΘ(2)}2mmmmV222Θ32mm22-1309.74216.4m-1011115mol52==2m~238t+nx10=rn30=r=rr=2.71tk3k1kkln2kk1ETTETT1mmsm00000n2222245.13He>H,Ba>Sr,Ca>Sc,Cu>Ni,La>Y,Zr>Ti,Zr>Hf232雜化,正三角型,非極性;3433522CHCCH:sp3spsp;36.7HFHClHBrHI;HFHClHBrHI;HIHBrHClHF;HFHIHBrHCl6.8(1)~c;(2)~d;(3)~a2HO+,OH-35H++OH-=HO,1/KHB+OH-=B-+HO,K/K76.4×10-5,1.7×10-13,小于KΘb4bc(H+)/cΘ5KaCHNH+H+=CHNH+c(CHNH+)=0.020mol·L-1K(CHNH+)=K/K(CHNH)=2.2×1+)=6.6×10-4mol/LpH=3.2Θ4KΘb1b24c(CHN)=0.050mol·L-1-1.9×10-4mol·Lc(CHNH+)=c(OH-)=1.9×10-4mol·L-1c(CHNH2+)=cK=1.4×10-11mol·L-12-)=K/K(HS)=1.4b1wa222-)=K/K(HS)=7.7×10-8b2wa12經(jīng)判斷，可忽略第二步離解，依近似式計(jì)算S2-+HO=HS-+OH-2x2KΘCeq/mol·L-10.10-xxx-)=0.094mol·L-1c(S2-)=0.10mol·L-1-0.094mol·L-1=0.01mol·L-1c(HS)=cK=7.7×10-8mol·L-1-202-)=3.7×10-20mol·L-1a1c(HCit-)a1c(HCit-)2c(HCit-)2c(HCit-)2c(HCit)c(HCit)c(HCit-)2a2c(HCit2-)a2c(HCit2-)c(HCit2-)c(HCit2-)c(HCit-)c(HCit-)c(HCit2-)44c(NH+)c(Nc(NH+)c(N4a1c(HA-)pH=pKΘlgca1c(HA-)--a2c(HA-)+a2c(HA-)c(HA-)a22a1a2c(HA-)c(HA-)a22a1a2c(HA-)27.17甘氨酸：不移動(dòng)；谷氨酸：向正極運(yùn)動(dòng)；賴(lài)3-)/c}3KΘ4mol/L42c(OH-c(OH-)=mol/Lc(Mg2+)/cΘ3KΘ39mol/Lmol/L3aF－+H+=HFc(HF)c(F-):c(HF)c(F-)KΘ(CaCO)c(CO2-)/cΘ sp3=3KΘ(CaF){c(F-)/cΘ}24b}Θ{Mg(OH)}2232+)≤10-5mol/L,g故可被沉淀完全的硫化物溶度積的最大32KΘ{Fe(OH)}KΘ{Fe(OH)}-8pH所以無(wú)沉淀生成，不能用硝酸鈣溶液代替氫氧化鈣溶液來(lái)檢驗(yàn)CO。原因是溶液2-4-33對(duì)于兩價(jià)離子M2+的氫氧化物:-3mol/L3++Cu=2Fe2++Cu2+-+10Cl-+16H+=2Mn2++5Cl+8HO(4)HO+2Fe2++2H+=2Fe3++2(5)PbO+2Cl-+4HO=Pb2++Cl(6)HgCl+Sn2+=2Hg-+3Mn2++2HO=5MnO2.3032.303RT{c(CrO2-)/cΘ}{c(H+)/cΘ}142F22F22F{c(Fe2+)/cΘ}F{c(Fe2+)/cΘ}27276F{c(Cr3+)/cΘ}227276F{c(Cr3+)/cΘ}21462+/2+/Sn)<?(Pb2+/Pb)-/NO)=0.957V33+/Fe2+)=0.771V3+/Fe2+)}=-3×96500C/mol×{0.957V-0.771V)}=-5.38×104J/mcΘ}{c(H+)/cΘ}4cΘ}{c(H+)/cΘ}4)43-/NO)-?(Fe3+/Fe2+)}=-3×96500C/mol×{0.721V-0.771V)}=1.45×4J/mol-/NO)<?(Fe3+/Fe2+)312>K323+/Fe2+)=0.771V4J/mol3J/mol3+/Fe2+)23<KPb2++2e-=Pb4Θ(Br/Br-)22+)=0.054mol/L+)=0.187mol/L-14)4)/cΘ}2222Fp(H)/pΘ24mol/L/cΘ=KΘxc(HB)/cΘ}a6aHg++e-=HgFFF222F22F22F計(jì)算結(jié)果與實(shí)驗(yàn)數(shù)據(jù)相符，故可知亞汞離子為Hg2+。2KΘ=KΘ=32:=:2Hg2+=Hg2++Hg22+)=1.455V2?(HO/HO)=1.76V（1）PbO+HO+2H+=Pb2++O+2HO（2）Pb2++HO=PbO+2H+2HO=2HO+O2++CrO2－+HO=2PbCrO+2H+(9)2Cr(OH)－+3HO－=2CrO2－+5HO+OH-(10)2MnO－+3Mn2++2HO=5MnO(14)CrO2－+3HO+8H+=2Cr3++3O+7HO(15)CrO2－+4