2022-2023学年高考化学全真模拟预测适应性考试卷(含答案)

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2022-2023

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ᔁᵫIᔁ⌱⚪ᔁ⌱⚪ᡂ஺ὃ
90ᑖᑖ100ᑖ஺ᵨᑮḄ!H-10-16C-12N-14CI-35.5Na-23K-39S-32Iᔁ⌱⚪ᐳ54ᑖ#ஹ⌱⚪
&'⚪3ᑖ&'⚪(ᨵ#*⌱⚗,ᔠ⚪./⌱ஹ┯⌱12ᑖ1ஹ345᛻789Ḅ:;<=஺>ᑡ@ABCḄ7Aஹᜧ!EᵨFGHIᑺKLᩭN⁐PQBஹRSEᵨTUVWḄXḄ7YZRS⏚\Cஹᜧ!]^ᙢ>`abcK`Ḅ◤eDஹ▲ᑴhijk᯼mᨨ7oYjkᓄqjᧇKstuvᓄw2ஹ>ᑡᱥy1z{ᔠᡂᩞᧇḄ7AஹᛛᧇBஹ~NCஹᔠᡂDஹᔠᡂ3ஹ>ᑡᔜᱥy7ᵯḄ7AஹHBrஹHCkBaSO4BNH4CIஹCH3COOH.Na2SsCஹNaOHஹCaOH
2.NH3.H2ODஹHCIOஹNaFஹBaOH
24ஹᨵᨵᐳḄᓄᔠᱥ7AஹNa2OBஹMgCI2CஹNaOHDஹHCI5ஹ>ᑡLḄy┯Ḅ7

1Aஹয++H20;VBஹCH2=CH2HBr2CH2Br-CH2Br;ᡂCஹCH3CH2OH^HCICH3CH2CI+H20;⎽ᓄDஹCH3CH07*H2CH3CH2OH;6ஹᙠy¡¢£¤¥k>ᑡᱥḄ¦§12ᑮ¨ᱥ©ªḄ7Aஹw⏚«BஹUᓄ¬Cஹ⏚«DஹUᓄ«7ஹ(ᵨ#®
ᑺ,°±Na2so4ஹAICI3ஹNH4CIஹMgS04²®¦§³®
ᑺ7AஹHCIBஹBaCI2CஹNaOHDஹAgN038ஹ>ᑡᔜᱥ´stjឋ¶ªḄ7Aஹ¬ᱏ¸vᓄ«¦§Bஹ¹º»⏚Cஹ¼¢Ḅºw⏚Dஹ⏚┛¾⏚9ஹ25࠷,¿¦§y`ᵯstḄC(H+)C(OH-)ḄÀÁYlx10-18,>ᑡ@ABCḄ7Aஹ¨¦§ḄPH77Bஹ¨¦§ḄPH#Â79Cஹ¨¦§ḄPH75Dஹæ§1Äᙠ10.ÆÇÈ>ÉᱥḄ!ºÈḄ>ᑡ¦§yC(NH4+)ᨬ'Ḅ7A.NH4CIB.NaHCO3C.NH4HSO4D.(NH4)2SO411.>ᑡËÌ¢ᓄÍÎÏÐ2ÑḄÒÓBCḄ7A.ÔÕC(q×,s)=C(ØÙqs);AH>0,ᑣØÙqÛq×ÜÂB.ÔÕC(s)+O2(g)=CO2(g);Hl,C(s)+l/2O2(g)=CO(g);

2H2,ᑣÝ2Þ1C.ÔÕCH4(g)+2O2(g)=CO2(g)+2H2O(g);AH=-802.33kJ/mol,ᑣCH4Ḅjk¢Y802.33kJ/molD.ÔÕNaOH(aq)+HCI(aq)=NaCI(aq)+H2O(I)H=-57.3kJ/mol,ᑣF20gNaOHḄߦ§àß¾⏚áᐰyy¢Y28.65kJ/mol12.>ᑡ@ABCḄ7A.ᡠᨵä¢ᙳ7æç7èéᜧḄB.ᔣAgCI©§ëᔠᱥyìᐭKI⛳¦§ᨵAglïðtᡂ@ñAgCIḄ¦È'{ᨵAglḄ¦ÈC.òàßw⏚ᑴ¸¶¡ᐭS!Ḅw⏚¹óôKEst¸¶ḄõÈR᠒D.ᵨᵯA÷øù¹ù¹úû□᩽þ¹úû▢᩽13.ᑡḄ┯ḄA.ᑴ⁐ᐜ⌨⁐B.ᓄ"#$%ᐭᑮ()*+,(⌨⁐,-.⁐Ḅ)*/0)*1C.ᔣFeCI2)*+3/ᑴ,43/KSCN)*,ᕈ⊈⁐D.ᔣ⛳89⏚;)*+%ᐭCO2#$,)*+<=>14.?@ᑮA⚜CDḄḄEᑁGDḄH

3AᔣⅅJ)*+/ᐭ43K"⏚,LMNOⅅJᔲQRᑖT.Uᓾ,4/ᐭWX)*,Y/0-CI2%ᐭNaBr)*+Z[#\]Ḅᓄឋ_`C-a\bc⏚defᡂḄ#$ᦈi,Ẇklmnᓄop⊝ᵨstᔠᱥUᓾwlḄxyDᑖzᔣ2{|+/ᐭ}~$.}~bẆkᓄᑺnH2O2mḄH2O2)*,4ᔣᐸ+1{/ᐭᑖR᳛ḄxyMnO215.ᨵজNa2c03)*ঝCH3coONa)*ঞNaOH)*25ml,ᑡḄA.ᱥḄbmᙳ0.1mol/L,3)*PHḄᜧঞ,ঝுজBᱥḄbmᙳO.lmol/L,-3)*K}~ᦪ,PHᓄᨬᜧḄঝCᱥḄbmᙳO.lmol/L,ᑖz/ᐭ25ml0.1mol/L⏚PHᨬᜧḄজD.PHᙳ9,3)*ḄᱥḄbmḄᜧঞ>জ>ঝ16.¡¢ᑴ£¤⏚¤¥Ḅ¦§¡஺ᑡᐵª«Ḅ¬?ḄA.ᔣa|+ᐜ/ᐭb"⏚ᯠ®᤯°|®᠒᠒JZ-τ/ᐭ¤⏣,JiJ4/ᐭs³⏚

4B.|b+´#|µ|¶?@·ᐭ*☢Ḅ¹º■¼½¾+¿f᎔ᔾC./0|aḄDḄÂÃÄ-¤⏚¤¥ÅÆ,p⊝ᔣfᡂ¤⏚¤¥ḄÇᔣÈ°D.|b+⛳8Na2CO3)*ḄÉᵨÂÃᔾᦈʤ⏚¤¥ÅÆḄ¤⏚8¤⏣17.ᑡËÌǾÍÎᑏḄA.ÐÑÒfᵯᓄῖ⇨ḄÖ᩽de×Fe-3e-=Fe3+B.ᵯR"⏚a)*Ḅ▢᩽de40H--4e-=2H2O+02C.9⏚Ù;)ª)*ᕈÚឋHC03-+H20=H30++CO32-D.9⏚Ù;⛳8)*\"⏚Û)*tᔠ×AI3++3HCO3-=AI(OH)3+3CO218.2.0molPCI381.0molCI2ᐙᐭ$?ḄÝÞGᘤ+,ᙠÃáᩩãÒfdePCI3(g)+CI2(g)=PCI5(g)Ap⊝,PCI50.4mol᝞åæÃçl.OmolPCI380.50molCI2,ᙠ}~lmo4Aᑮp⊝PCI5ḄᱥḄA.0.40molB.0.20molC.ᜧª0.20mol,ª0.40molD.ª0.20molé¶ᔁ(ì⌱î⚪ᐳ46ᑖ)ஹóô⚪(õ⚪ᒹ÷4⚪,ᐳ24ᑖ)19ஹ(10ᑖ)ᐗùᕜC⊤\ᐗùᕜCüᙠoýஹẆk8f¿þ+ᨵÿ⌕Ḅᵨ஺⊤ᐗᕜ⊤Ḅᑖ⊤⚪஺

5JA0ᕜᓩ1IIAIIIIVVAVIWIAAAA23পᵨᩞᧇḄᐗ"#ᐗ$%&,(ᙠᐗᕜ⊤*Ḅ+,-.ᕜ"/&஺ফ12.3ᕜIA/ᐗḄ456᪀89:ᐗ;<ᕜVIDA/ᐗ=ᡂḄᓄᔠᱥḄᵯ5C஺Dᓄᔠᱥ-E⌕4ᧇᙠFG*HIJK⌕ḄᓄLGMN1ᑡ"ᵨᓄPQRCᩭTU&஺বᙠ.WᕜᐗḄᨬYZ[ᓄᱥ\]Ḅ^ᓄᱥ*⏚ឋᨬaḄ"#ᓄPC<&bᐹᨵeឋḄᐸ;g[ᓄᱥhij]ḄᓄPQRC-஺20ஹ
4ᑖ&ᵫgmn[mj]Fᡂlmol^ompq241.8kJ,j]ḄqᓄPQRC-ros1g^omtᓄᡂi᝱^pq2.444kJ,ᑣgmḄwxq-KJ/mol21ஹ
4ᑖ&ᙠyzḄ{|}ᘤ*᝞ᓄPj]rC02g
+H2g
=COg
+H20g
,ᐸᓄP⊝ᦪKntḄᐵ᝞⊤ᡠ8rt/࠷70080083010001200

6K0.60.91.01.72.6ᑡ⚪rপj]ḄᓄP⊝ᦪ⊤C-K=ফj]-j](#"ᔾq"ᡈ"pq")বᑨj]ᑮᓄP⊝᝱Ḅ()Aஹ}ᘤ*aBஹᔠm*C(CO)CஹV(H2)F=V(H2O)⌮DஹC(CO2)=C(CO)ভ⊝$ᔠCrC(CO2).C(H2)=C(CO).C(H2O),Ḅ-࠷22ஹ(6ᑖ)¡ᨵ25࠷0.1mol/LḄ¢^஺D⚪rপsᔣ¢^*¤ᐭ¦§¨⏚©ªhi*C(OH)/C(NH3.H2O)(#"«ᜧ"ஹ"®"ᡈ"")ফsᔣ¢^*¤ᐭ¯¨⏚°ᐸា²*nᡠ³hiḄPH7(#">"ᡈ"=")ᵨ¶5QRC⊤8ᐸ4·஺বsᔣ¢^*¤ᐭPH=1Ḅ¨⏚¸¢^;¨⏚Ḅz¹-1:1,ᑣᡠ³hi*ᔜ¶5ᱥ»Ḅ§ᵫᜧᑮ®Ḅᐵ஺3ஹ¼½⚪(¾⚪ᒹÀ2®⚪ᐳ14ᑖ)23ஹ(4ᑖ)¼½ÂÃÄᑡÅᑺÇIᑡÈÉÊËḄAஹFeS04hi*pᐭÌÍ┑BஹFeCI3hi*¤ᐭ¦§Ï⏚

7CஹᵨÐÑÒÓᴻÕḄÅᑺÖ×NaSiO3hiDஹNaBrhi*¤ᐭ¦§Ø^24ஹ"10ᑖ&xÙ᪵MÛᨵ¦§;⏚ᵨḄᩖ»,-ÝÞyᐸßDÞyàrAஹáᑴ250mlxÙhi;BஹᵨÙCÞyäI25.00mlxÙhiå┵çÖ*èÞ¤éÞê⏍ìᢣ8ᑺbCஹᙠᜩïðËñIxÙ᪵Mwg,ᙠxò*ᵨoó^hôbDஹõᱥ»Ḅ§-cḄ᪗ð¨⏚hi÷ᐭ⏚CÞyäø᦮i☢ûüýþÿVlml;Eஹᙠ┵⁐ᜫᑖᑁᓄᦪV2ml஺#$%&'ᡂᑡ*+,প./Ḅ12345,6ᵨ89:;ᑏ=AD-“஺-AফBCDEFG☢ḄIJKLMN,oবD12OG☢LP⁚ᑮSᘂUᑖL஺ভWXY%&KZ┵ᵨ[\᪵^_`abDḄcJdIe5dfghi5মk[\᪵^lJḄmno஺pஹmn⚪68ᑖ=25ஹrstᵯ᩽ᵯw400mlWxyz{Ḅ|⏚~GG

8OḄz{ᐰUL▢᩽1.28g□᩽Y224ml6ᢚnᡂ᪗=kz{Ḅ᥅ᵯwGḄPH6ᵯwGᓄᶍm=஺ὃᫀஹ⌱¢⚪,6£¤⚪3ᑖm54ᑖ£¤⚪ᨵ¦⌱⚗¨ᔠ⚪N,«⌱ஹ┯⌱ᑖ=⚪9123456789ᫀDBACCBCAC⚪9101112131415161718᪍1-1¯BBDDACADDxஹ*+⚪,6°⚪ᐳ4¤⚪ᔠm24ᑖ=19ஹ
10ᑖ=পSi1ᑖ=3ᕜ³IVA6´=6ᔜ0.5ᑖ=\\,,ফ
1ᑖ=ᑖ=2NaCI+2H2Oᵯw2NaOH+H2T+CI2T2ᑖ=ᐸ·ᔠᳮᫀᙳºᑖ=(3)HCIO4;(1ᑖ)AI(OH)3(1ᑖ)AI(OH)3+NaOH=NaAIO2+2H2O(2ᑖ)20ஹ(4ᑖ)H2(g)+l/2O2(g)=H2O(g);AH=-241.8kJ-mol-l

92H2g
+O2g
=2H2Og
;AH=-483.6kJ-mol-l2ᑖ=285.792KJ/mol2ᑖ=21ஹ
4ᑖ=i
K={CCO
.CH2O
}/{CCO2
.CH2
}ফᔾ½
3BC
4
830ᔜ1ᑖ=22ஹ6ᑖ=প¤61ᑖ=
2PH71ᑖ=hiNH4++H2O=NH3.H20+H+2ᑖ=
3CNH4+
CSO42-
CH+
COH-
62ᑖ=¾ஹ%&⚪,6°⚪ᐳ2¤⚪ᔠm14ᑖ=23ஹ
4ᑖ=CD6ᔜ2ᑖᨵ┯ºᑖ=24ஹ
10ᑖ=পC-K-B-DE2ᑖ=ফ¿ÀஹÁÀÂÃG☢ᨬfᜐÆÇÈᑗ1ᑖ=ব0ÁJᡈ0rWÁJᐙÌGÍÎ6ᔜ1ᑖ,m2ᑖ=ভdIhi5ÏbᱥḄ«6ᔜ1ᑖm2ᑖ=
5O.8CV2-V1
/W3ᑖ=Ñஹmn⚪,
8ᑖ=25ஹw,Òk|⏚~MSO4,

10□᩽YÓnO2=0.224L/22.4L.mol-lQᑖ=᪷Õ:2MSO4+2H20==2M+02+2H2SO42ᑖ=ᨵ,ᵯw▢᩽᪆Øz{ᱥḄ0.02mol,Yᡂ|⏚0.02mol2ᑖ=kz{Ḅ᥅=1.28g/0.02mol=64g/molQᑖ=LGOÙᱥḄCH+
=0.02x2mol/04L=0.1mol/L1ᑖ=PH=11ᑖ=6ᐸ·ᔠᳮᫀÚÛºᑖ=