2022年夏季山东省普通高中学业水平合格考试化学仿真模拟试卷03（解析版）

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719•ூ᪆௃⌱B஺$qᓄª«,Éᑖ$/jÊᨵᐗÌḄᓄᔠÍ2ᓄᦑA┯@C1ᐗÌḄᓄᔠÍÏÐᙠ$NaCIChÒᑮᵯª«Fª«ᦑB¸@$SᐗÌḄᓄᔠÍᓣÕᑣNa2s03ª«ᑺᦑC┯@NaCICh$ÖᐗÌḄᓄᔠÍᵫ+5ÍÏÐ=+4ÍᡠÙFᡂImolClO?ÞßImolᵯ6.75gQO2Ḅᱥ-Ḅ.="äC1Oå=6,75g=0.1mol,ᡠÞßO.lmolᵯᦑD┯.267.5g/mol20.ூ᪆௃⌱D஺ZnᜫᵯFqᓄ,<ê᩽ᦑA¸d᩽´AgzOÒᵯFª«ᦑB¸@«ᵯì$ᵯíÂ᩽îïðíᔣê᩽«ᵯì$Znê᩽ஹAg2O᩽ᡠóᵯ5ᵯíÂAgQôïðíᔣZn,ᦑC¸@õឋᩩ÷uøùFᡂZnä0Hå2,ᡠê᩽ú=Zn+2OH--2e-=ZnäOHå2ᦑD┯஺21.ä10ᑖåூýᫀ௃ä1åÿᕜWA2ᑖ2N2ᑖ3S2->O2->Na+2ᑖ4AlOH+3H+=Al3++3HO2ᑖ32ூ᪆௃ᕜAஹFḄᓄᔠ"#$+4%&4,ᡠ)*+VIAᐗ/01ᔠ23456A*Cᐗ/ஹF*Siᐗ/•:ᵫBḄ23%ᓄᔠ"456B*Nᐗ/:Dᓫ?ᑴᡂḄBCD0EFḄGᐝIJKLஹMNO0ᡠ)D*Naᐗ/6MḄQᓄᱥSTUᕈW᝱0MḄYZ[▢][ᨵ18aᵯ[0ᡠ)M*Oᐗ/:GZ[᪶᜜MfgLfi2aᵯ[0ᡠ)GZ[᪶ᑁᨵ16a?[0*Sᐗ/:EḄ+3"□][Ḅ᪶᜜ᵯ[᣸opLZ[qr0ᡠ)E*A1ᐗ/஺1A*Cᐗ/0ᙠᐗ/ᕜ⊤vḄwx*6+ᕜNA:2B*Nᐗ/0BḄyz᪶/v[ᦪg?[ᦪ|1,ᑣ⊤~᪶/ḄZ[*6N஺3M?ஹD+ஹG2][ᑖ*6ஹNa\S2,Sz᪶᜜ᵯ[᣸oᨵ3f023ᨬᜧ002%Na+᪶᜜ᵯ[᣸oᨵ2f0Z[ᦪ6Na>O,236O?>,Na+0ஹNa+ஹS6-23ᜧ$6S2->O2->Na+oG%EḄᨬB"ᓄᱥᓄᱥᑖ*H2s஺4%AlOH3,ὅḄᓄN*2AlOH3+3H2so4=AbS03+6H2O,][N*AlOH4+3H+=Al3++3HO..3222.8ᑖூᫀ௃12ᑖ✊¡I22cH2=CH-CH0+Ch>2cH2=CH-C00H2ᑖ¤oCH-O—C—CH=CHCH2-0H22CH-OHCH-OOC—CH^CH23CH2-OHᡈCH2-OH2ᑖ433.62ᑖூ᪆௃1ᵫCH2=CH—CHOfC3HCh0|©1aZ[0ᑣC3HKḄ1᪀«*CH2=CH-COOH,tᓽCvḄ®¯°*஺2BfCḄᓄN*2cH2=CH-CHO+O2ᓄᑺ,2CH=CH-COOH„2¤OHOHOH3²ᾦ´ᡂBµᾦ¶⏚%ᵞ²0¹ID*ᵞ²0ᐸ1᪀«*»2—¼&%2,¾p1ᑖ[CH2=CH-COOHE´¿ᓄÀ04¯ῃ“CH2OH”ÅḄÆZ[0Ç4¯ῃ“CHOH"ÅḄQZ[0CH-O—C—CH=CHÉ-OH22Oi-OHCH-OOC—CH=CH2ᡠ)5ᑮḄ¿EḄ1᪀«*8H2-OH%CH2-OH஺ভImolᵞ²vᨵ3moiÅ0

84pp3moiÍÎÏE´xᣚ´ᡂ1.5mol,ᐸᙠ᪗ÓÔÕUUḄÖ×*VH=1.5molx22.4mol/L=233.6L023.12ᑖூᫀ௃1ᑖWÚᧅ2ᑖ:Na2sO3+H2so470%=Na2sO4+SO2t+H2O2ᑖ2BvFÝÞG⁐àá2ᑖ:âãäWå⁐2ᑖ3Ba2++SO42-=BaSOI2ᑖ:æçxᑁḄᓄèᐰêëᐭbvᔾᦈ0■ðñòó᛻2ᑖ4ூ᪆௃*■ðᑴõḄᓄèöÆᓄ0÷øùúûᐜý&þÀÿḄ
᣸Ḅ⏚ᐭ⏚ᡂᓄᓄᐭᓄᡂSᓫ#$ᓄḄᓄឋᐭ&'⏚()*&═⏚(ᓄ$ᓄḄ,-ឋᐭ./)0./1⁐$ᓄḄ34ឋᨬ6ᐭ7ᓄ8)*ᓄᡂ⏚᪷:;<Ḅ=:;>ᔠᡂ⏚=@ABCḄᓄᵨEᓄᔾᦈ஺I1KᵫMᘤaḄ᪀⌼RSaTᑖVᧅA⏚<⏚ᡂᓄYZ[T\Na2sCh+H2so4I70%K=Na2so4+SO2t+m0\I2Kᓄᐭᓄ):;ᓄᡂᓫ#n$op⁐Ḅqr$stᓄ6ᐭ./)0./1⁐$ᐸ34ឋtI3Kᓄ6ᐭ7ᓄ8)*ᓄᡂ⏚᪷:;<Ḅw:;>ᔠᡂ⏚=@AḄ:;YZ[T\2Fe3++SO2+2H2O=2Fe2++4H++SO42-ஹBa2++SO2-=BaSOItBCḄᓄᵨEᓄᔾᦈ~bTEᓄt>446ᐭN2,ᑁḄᓄᐰᐭbᔾᦈ■᛻஺24.I10ᑖKூᫀ௃I1KCI2ᑖKI2K᳛I2ᑖKI3KᡈI2ᑖKI4KHO+2Fe2++2H+=2Fe3++2HOI2ᑖKI5K12I2ᑖK222ூ᪆௃8<⏚ᡂ⏚8E⏚8<ᓄᡂ⏚8ᔣᐭ⏚╆R¢⏚8£¤¥஺I1K¦Ḅ§⌕ᡂᑖ©&ªᾦ¬⏚Ḅᵞ¦®¯ᙠ±ឋᩩ³´)µ¶ᐰᡂRឋḄ&ªᾦ¬⏚ᵞ¦tI2Kᓣ&¸¹0᳛ḄºḄ©;I3K¼½ঝ¸¹¿ᢝ80~95°C,ᡠÆÇᵨᔠ〉ḄY[©ᡈtI4KH2O2Fe2+ᐰᓄTFe3+,8ᐗÊḄᓄᔠËᓣ&ᡠÆᐗÊḄᓄᔠËÌÍ,-Îᱥ©ᦑḄ:;YZ[©\HO+2Fe2++2H+=2Fe3++2HO»I5KnIBaSOK=஻ISO2"K=46-6g=0.2mol,᪷¤SO42ØÙឤR¢22244233g/mol᝞´ᐵÞ[:NHFeISOK2•XHO-2SO2~2BaSO,ᑣ”[NHFeISOK2•xHO]=-nIBaSOK=44244442240.1mol,MLNHFeISOK2•xHO]=48-2g=482g/mol,ᓽ18+56+2x96+18x=482,¢x=12஺4420.Imol