资源简介 — -is aS4.zhiWj^ ww2024—2025 T)r mNaOHkw .l O-S ,OlP 4E aiB/Kd=i A. NO2 b. WlB C. MS NaOHNO D. #ZJt^±aft:5.4E^-W mjva. n ho^45 *H HO. CH3 .OH01 B. lf M 8 H,c | H3CMnO2C. M 1-5 mol H2 HO O’_ko ^AA-KX-tztsto M N3. 4f . oT^JOWasO.-f^ioH 1 C 12 N 14 0 16 F 19 Sr 88 Zr 91i&Jj% wD[P- 14 /J^, $/h 3 #, i 42 ^o tH W E'N JJ54’, RW-^ AB Wett:H2O>NH3 M -T SttMgfah,H—0 W&W-fett1. T 0W^W1E$iW h-n mw,H—oA. 9WTMBIWO C ^H,l4:K>NaB. W 1 jW'K W.g A AS WaC. ok W W4>^sffi, D S^:HF>H2O HF AMam h20 fflA.HFD. WSdlW^W^A^ 7.MBPAM Fe- POM ffiO!H iW-55 h2o2 x,gJiffi ^*fflinToH20 CH3CH=CH22. / — Ep. \ HH\ yH oA-o- \ C'-CHjA. V o/ H.C7 >)Q/ 'o 0 O'RoB. CaC2 :Ca2+ [: C : : C : ]2“ Okj±di-R 0J H H.0,H^s^C^Q oX-oC. jAh °'V' Nil, ^N/frH hacvh^ >yHD. CO’ VSEPR o'rV- yv°ixj V~- xFf -~O3. H,0. O'°' x X0n O x T^iJi^iESAKJgn b. ;wpm sij 4’ NaOH a. imwmwc. B. iMasa1 Hltafitt. jf / w wwtafi , owl Ajtsm C. 21 gCH3CH=CH2 nmg m iJx^sk) M2JJ(±h8Jl){#{QQABAQCg5wAQ0gRACZ46A0kUCgmQkJKQLSoOBQAUOA5DgYFIBAA=}#}8. miHIRWWi^JW 84 A. a bCla. :2 OH + C0|- ^2 0" + h2o + co2 T + 4e’ +2H + // \ 0H+2C1"B.^iIfaMnAA'i3E tMW+:Ca2+ +2C10- +2SO2’ =CaSO4 1 +2Cr +S0' Cl OHc. M fi:5CH3CH2OH + 4MnO4 + 12H + 5CH3COOH + C. H‘thAfaA^fiMHAT-AAK4Mnn +11H2O D. 34^9.4 g^ flit,3HTfc±M t K^M*M^'4.4 gd. AfWni2SO4)i§ j!H'HJJ g5aiS i ; ffl&-MftA:C2O;- + 2C10; +4H + 12. At jAiWg .IWiglilk, nTJWit (SrCO3) fll YK-ft I ZrO2) Oil2CO2 f +2C1O2 t +2H2O OiW-W o i)4@m4gteAA0W(ffl2)4Smn9. M Y2X6Z5W2,15TffiT Friedel - Crafts SliJraA XJ.Z.W Jg W fai^.XA.Z x UtTMtfrMa #AA A res2nP2"+1 Y Jg s p 2 fg,Z w JM oY 4 A. >X >W k oa SrB. w YW. 00 eZr o2C. a^:YX4>WZ3 1^1 S2D. JtiS Y^Z 5 W a. co2 JSTMfeSHS;*: * 5^310. SjTF.S&o. i mol L-1 nh4hso3 it-SWOL WJ^^^lEii Bgi4SiJ^g WiW B. 0 2 ,corw °o.3 11 1 11 33)hso; fa 2 mL nh4hso3 4 4 4 ’AKMnO4 D. ^0 2 0B0 d g cm-3 JW 0 X 3/ 227aK -d X 107 nmnh4hso3 fa nh4hso3 X*BBaci2SS I :2N0(g) +2H2(g)—N2(g) +2H2O(g) AT/,ft2 mLO.l mol l< NH4HSO3 ,c R&. n :2N0(g) +H2(g)^N2O(g) +H2O(g) AH2nh4+ in A 2 mLO. 1 mol L-1 NaOH , XW. J figA # 1 mol M) fit 2 mol H2 JffiAflW&'HAAA(* 10 kPa) A±ii g>, W#WfBi- NO a2 a2o SMtt+fitiWBW bb^ Ci.s 6W fa nh4hso3 jOt4> xn(M(J) +n.(JN2) +^(1\2O)D iaAnwiw +, w lOO%]B^SW1T(v)W^On0^o BO1 ftT, #0^(N2) =3n(N2O);A1r-/11. W 2,4 - 2MWM M8WBWi5o TW Bl T, I—la hi— ,#Wj±WvSo T iJWiE mg a.9 A. a =80% pNOM _L_N oZt I ' B. I 20% N20OH c.i>^T, WH4 H2IWI-W* 31.25% 6.25%0*2 1 1 -Jr/T-'C10H7 D.^HFE 8.25 kPa' *2{#{QQABAQCg5wAQ0gRACZ46A0kUCgmQkJKQLSoOBQAUOA5DgYFIBAA=}#}14. MT, |nj Cu (N03) 2 #gB HCN w > HNO2 Px9 TaWzK'f’ Cr202- 6T’ NaOH pX[pX = -]g X.Xfta 8 A k 8.07 lq — — 7.02t c(CN-) c(NO2~) 6 Q/(7,5.7) ec -■-cl ;'c(HCN)'c(HN02) 5 4 :=S= - 6.03 N(7,2.2) 5.0tSo B MT,X.(HN02) A > Ka( HCN) ,1g 2 =0. 3 , T <52 XM(2’L2)/ o 4.0I0 + 3.01 2 3 4 5 6 7 8 9 10 11“ pH ■i 2.0A. L2 ft* p c(CN~) * c(HCN) M pH 33 1.00B. MT,X!p[Cu(0H)2] =6.0 xl0~2° 1 2 3 4 5 6 7 8 9 10pHNaN02 HN02 mg-^ T,c(HN02) +c(H;,) > pH <4 04,Na2SO3 NaHSO3 BS^fclCT Cr202‘ (±JUEc(OH’) +c(NO2“)^i )oD. MT.fiiW^O. 1 mol L-‘K HCN § a,HNO2 NaOH iKtt^{4T FeSO4 Cr2o2- oW pH^IW, c(N02-) ,c(CN~) oc(HNO2) (HCN) = 1.0xl06c- 4 /Jv ,it 58 16. (i3^)wim Cl15. (15^)TO^(Bi0.5Na0.5Ti03)>-^mO^M i§ M,*Wtt#6j ■MWT # Ti02 FeOTe2O3,CaO>MgO,BaTiO3 4ta. UK* W4, iWttOHWfflTMXHjSO, MiH,0 NaF 2S + Cl2 =SS2C12,TS2C12 + 33C12 +2SO2 =4SOCI2, glJS^T Cl2 + S02 SO2C12ISM /Kg H ttas i Hf^T aw |—* FeSO4 7H2O (: MttsmazkM## ) =AiSS I iSO 70% Q lx | MillIV da b eA hHE M cBi0JNa05TiO3----- /K&fe - -----TiO -- ---- — TiO(OH)2 am r■f Si:MnO,$ - 'rg"ANaOH . Bi(N03), 5H2O M< k7k ' . WJXWJXB^J: lamWiJlW:Fe2 + ,Fe3+ .TiO2 + ffi Na + W s o MfW:Fe3+ >TiO2+ >H + O(2) ___ (i*W) =(3) _______________________(1) SSKFIWJS oo (4) ^3Witf .ftfl-JHWI Q IWl'C, WTJHX^ N g M(2) “ Kg” ( - o B.(3) at n o(±M4t^S;)o(4 WJ x ,^tOJ Xi±*,jnij Ti02*^T^^Ti3+o g (NH2 OH) i5T iw Ti3 * TiO2 + , nh2oh Jg ti2(so4)3xO XzKCaCljS0C1,(5) Bi(NO3)3 -5H2O 3EWffi|Na2[Sn(OH)4]|*WSSaHtt.lS NaOH ^ft^4’f^Bi(NO3)3 fflOBi, § WfO Na2[Sn(OH)6] , Ao WM aW-Bg;;;(6) glj^tl FeSO4 PrffiT'&SJtzKo Na2SO3 . NallSO, FeSO4 pH (5) W mw^n W BSWT o{#{QQABAQCg5wAQ0gRACZ46A0kUCgmQkJKQLSoOBQAUOA5DgYFIBAA=}#}(6)iOJ Y>o AS(kj mor1 k-1) .wigsoci2o17. (14 (“ b “■aM”‘Sc“c ”)oH,cR’H.CH, C I HjC-o/y=1 0 (3) CH4 - co2 CO w h2 , iSSlSinT:CH3COC1 CH3IA(C6H6O) aicra B( OR 3) ^gircC( R ) —D(C9H9O2C1) && I :CH4(g) +CO2(g)^2CO(g) +2H2(g)0 ft I :H2(g) +CO2(g)—CO(g) +H2O(g)E( Clx^L y S^HI:CH4(g) +H2O(g)—CO(g) +3H2(g)Cl ■'OH rnAMISTmilnll'm 1 rnol CH4 1.4 mol CO2 3EAM1SWIWW ,I( ) TWHrtm n(CO) =1.8 mol,CH4 H20CHO0 ou T) Zn-Hg Cl K( ) PPA) J(C15HinO2Cl2) iggEm a 12 kPa ,M-gmffiT.g.isz n wwa kp=mp aws0- C HC1^^lC l Cl = & x oB n:CH3CHO + CH3CHO >f-mA >CII,CH —CHCHO + H2O (4) /■ WT3BIW: co2(co)8(1) iwmo o ill ;o(2) B o (5) WEA.SiO:ftiTAfflf W co2 ,iWaa^WfLffiOiJfeW(3) A- II.K^I. [WAA’OJA 0 co2 co mmtiimfWSM i co2 >(4) d _________ o Si >OJJgiWiiU;*KEl3 C02(^).h2o( #)(5) E jg I J O JnS 2 UffzRo(6) Mf C KilW#tW,M Feci3 iWJ!1 M M CO-H GDL Ao FiIj:I:I 1 is. (16 co Ifco2 § |C (1 )C0 jg H2,O2 Hlm^-lSZ.-B ,RliL^J 2C0 + O2 +5H2 =HOCH2CH2OH +2H20o GDL' -M !fifi ,fl ( H co2 (%fa)co2CO (g)m2 (g)> A AB(1) ft \ H Oah - 283. 0 kJ mor\-285.8 kJ mol’1. -1 118.8 kJ moP’oMSt 2CO(g) +O2(g) +5H2(g) HOCH2CH2OH (1) + 2H2O (1) AH = Bl0(2)Xik^3^fiW CO(g) 4-Cl2(g)—COCl2(g) AH0O mW-WM#fcW^77(CO) =80% [7 (CO) = COMMW) 1 mol CO jg 1 mol Cl2 COC12, WjEtf CO M 1SI (y) , X 100% ] ft (lljiEtSteE&EF)OlWsilW/T: CO MMC02)S(CO) =90%[S(CO) = X 100% ]"(iBiSte C02)100 Pio2 ______ goa01 1 1 1r2 ' t3 ta -R-1S8WI){#{QQABAQCg5wAQ0gRACZ46A0kUCgmQkJKQLSoOBQAUOA5DgYFIBAA=}#}天一小高考2024一2025学年(下)高三第三次考试化学·答案1-14小题,每小题3分,共42分。1.B2.A3.A4.C5.B6.D7.C8.C9.B10.A11.D12.D13.C14.D15.(1)3d4s2(2分)(2)粉碎酸钛渣、搅拌、适当升高温度等(答一条,合理即可,2分)(3)MgF2、CaF:(2分)(4)铁或铁粉(2分)NH20H+2Ti3+H,0=NH4+2Ti02+H*(2分)(5)3Na2[Sn(OH)]+2Bi(NO,)+6NaOH =2Bi+3Na2 Sn(OH)]+6NaNO3 (2)(6)①NaHSO3(1分)②Cr,0号”+6Fe2·+14H*=2Cr3·+6Fe5++7H20(2分)16.(1)球形冷凝管(1分)(2)ebahicdfg(2分)(3)吸收S02和Cl2,同时防止空气中的水蒸气进入三颈烧瓶中(合理即可,2分)(4)先由Cl2与S反应生成S2Cl2,减少S02Cl2产生,提高S0Cl2的产率(合理即可,2分)(5)重结晶(2分)】(6)NaOH溶液(合理即可,2分)(7)-0oH+s0C,一◇codl+s0,↑+HC↑(2分)17.(1)4-氯苯甲醛或对氯苯甲醛(2分)(2)羟基、酮羰基(2分)(3)取代反应(1分)还原反应(1分)CI(4)CH;(2分)OCH,0CICH.催化剂,△+H0(2分)OHOHCHOCHs(6)13(2分)(2分)O=C一CH18.(1)-876.2kJ·mol-(2分)(2)①小于(1分)②a点(2分》(3)①80%(2分)②0.643(2分)(4)碳元素的电负性比氧元素小,容易给出孤电子对(合理即可,2分)(5)①C02+2e+H,0C0+20H(2分)②360(3分)一2 展开更多...... 收起↑ 资源列表 化学高三天一小高考(三)简易答案.pdf 河南省鹤壁市天一小高考2025届高三下学期第三次考试化学试卷.pdf