— -is aS
4.
zhiWj^ ww
2024—2025 T)
r mNaOH
kw .l O-S ,
OlP 4E aiB/K
d=i A. NO2 b. WlB C. MS NaOH
NO D. #ZJt^±aft:
5.4E^-W mjv
a. n ho^45 *H HO. CH3 .OH
01 B. lf M 8 H,c | H3CMnO2
C. M 1-5 mol H2 HO O’
_ko ^AA-KX-tztsto M N
3. 4f . o
T^JOWasO.-f^ioH 1 C 12 N 14 0 16 F 19 Sr 88 Zr 91
i&Jj% w
D[P
- 14 /J^, $/h 3 #, i 42 ^o tH W E'N JJ54’, RW-^ A
B Wett:H2O>NH3 M -T SttMgfah,H—0 W&W-fett
1. T 0W^W1E$iW h-n mw,H—o
A. 9WTMBIWO C ^H,l4:K>Na
B. W 1 jW'K W.g A AS Wa
C. ok W W4>^sffi, D S^:HF>H2O HF AMam h20 fflA.HF
D. WSdlW^W^A^ 7.MBPAM Fe- POM ffiO!H iW-55 h2o2 x,gJiffi ^*fflinTo
H20 CH3CH=CH2
2. / — Ep. \ H
H\ yH oA-o- \ C'-CHj
A. V o/ H.C7 >)
Q/ 'o 0 O'Ro
B. CaC2 :Ca2+ [: C : : C : ]2“ Ok
j±d
i-R 0
J H H.0,H^s^C^Q oX-o
C. jAh °'V
' Nil, ^N/fr
H hacvh^ >yH
D. CO’ VSEPR o'rV- yv°
ixj V~- xFf -~O
3. H,0. O'°' x X0n O x
T^iJi^iESAKJg
n b. ;wpm sij 4’ NaOH a. imwmw
c. B. iMasa1 Hltafitt
. jf / w wwtafi , owl Ajtsm C. 21 gCH3CH=CH2 n
mg m iJx^sk) M2JJ(±h8Jl)
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8. miHIRWWi^JW 84 A. a b
Cl
a. :2 OH + C0|- ^2 0" + h2o + co2 T + 4e’ +2H + // \ 0H+2C1"
B.^iIfaMnAA'i3E tMW+:Ca2+ +2C10- +2SO2’ =CaSO4 1 +2Cr +S0' Cl OH
c. M fi:5CH3CH2OH + 4MnO4 + 12H + 5CH3COOH + C. H‘thAfaA^fiMHAT-AAK
4Mnn +11H2O D. 34^9.4 g^ flit,3HTfc±M t K^M*M^'4.4 g
d. AfWni2SO4)i§ j!H'HJJ g5aiS i ; ffl&-MftA:C2O;- + 2C10; +4H + 12. At jAiWg .IWiglilk, nTJWit (SrCO3) fll YK-ft I ZrO2) Oil
2CO2 f +2C1O2 t +2H2O OiW-W o i)4@m4gteAA0W(ffl2)4Smn
9. 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 o
Y 4 A. >X >W k o
a Sr
B. w YW. 00 eZr o2
C. a^:YX4>WZ3 1^1 S2
D. JtiS Y^Z 5 W a. co2 JSTMfeSHS;*: * 5^3
10. SjTF.S&o. i mol L-1 nh4hso3 it-SWOL WJ^^^lEii Bgi4SiJ^g W
iW B. 0 2 ,corw °o.
3 11 1 11 33)
hso; fa 2 mL nh4hso3 4 4 4 ’
A
KMnO4 D. ^0 2 0B0 d g cm-3 JW 0 X 3/ 227aK -d X 107 nmnh4hso3 fa nh4hso3 X*
B
Baci2
SS 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#Wf
Bi- 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 BO
1 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-'
C10H
7 D.^HFE 8.25 kPa
' *2
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14. MT, |nj Cu (N03) 2 #gB HCN w > HNO2 Px
9 TaWzK'f’ Cr202- 6T’ NaOH pX[pX = -]g X.Xfta 8 A k 8.07 lq — — 7.0
2t c(CN-) c(NO2~) 6 Q/(7,5.7) ec -■-
cl ;'c(HCN)'c(HN02) 5 4 :=
S= - 6.0
3 N(7,2.2) 5.0
tSo B MT,X.(HN02) A > Ka( HCN) ,1g 2 =0. 3 , T <52 XM(2’L2)/ o 4.0
I
0 + 3.01 2 3 4 5 6 7 8 9 10 11“ pH ■i 2.0
A. L2 ft* p c(CN~) * c(HCN) M pH 33 1.0
0
B. MT,X!p[Cu(0H)2] =6.0 xl0~2° 1 2 3 4 5 6 7 8 9 10
pH
NaN02 HN02 mg-^ T,c(HN02) +c(H;,) > pH <4 04,Na2SO3 NaHSO3 BS^fclCT Cr202‘ (±JUE
c(OH’) +c(NO2“)
^i )o
D. MT.fiiW^O. 1 mol L-‘K HCN § a,HNO2 NaOH iKtt^{4T FeSO4 Cr2o2- o
W pH^IW, c(N02-) ,c(CN~) oc(HNO2) (HCN) = 1.0xl06c
- 4 /Jv ,it 58 16. (i3^)wim Cl
15. (15^)TO^(Bi0.5Na0.5Ti03)>-^mO^M i§ M,*Wtt#6j ■
MWT # Ti02 FeOTe2O3,CaO>
MgO,BaTiO3 4ta. UK* W4, iWttOHWfflT
MXHjSO, MiH,0 NaF 2S + Cl2 =SS2C12,TS2C12 + 33C12 +2SO2 =4SOCI2, glJS^T Cl2 + S02 SO2C12
ISM /Kg H ttas i Hf^T aw |—* FeSO4 7H2O (: MttsmazkM## ) =
A
iSS I iSO 70% Q
lx | MillIV da b e
A h
HE M c
Bi0JNa05TiO3----- /K&fe - -----TiO -- ---- — TiO(OH)2 am r■
f Si:
MnO,
$ - 'r
g
"A
NaOH . Bi(N03), 5H2O M< k7k ' . WJX
WJX
B^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)3
xO XzKCaClj
S0C1,
(5) Bi(NO3)3 -5H2O 3EWffi|Na2[Sn(OH)4]|*W
SSaHtt.lS NaOH ^ft^4’f^Bi(NO3)3 fflOBi, § WfO Na2[Sn(OH)6] , A
o WM aW-Bg;;;
(6) glj^tl FeSO4 PrffiT'&SJtzKo Na2SO3 . NallSO, FeSO4 pH (5) W mw^n W BSWT o
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(6)iOJ Y>o AS(kj mor1 k-1) .wig
soci2
o
17. (14 (“ b “■aM”‘Sc“c ”)o
H,c
R’H.CH, C I HjC-o/y=1 0 (3) CH4 - co2 CO w h2 , iSSlSinT:CH3COC1 CH3I
A(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 H20
CHO
0 o
u T) Zn-Hg Cl K( ) PPA) J(C15HinO2Cl2) iggEm a 12 kPa ,M-gmffiT.g.isz n wwa kp=mp aws
0- C HC1^^lC l Cl = & x o
B 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 iW
J!1 M M CO-H GDL A
o F
iIj: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)co2
CO (g)m2 (g)> A AB(1) ft \ H Oah - 283. 0 kJ mor\
-285.8 kJ mol’1. -1 118.8 kJ moP’o
MSt 2CO(g) +O2(g) +5H2(g) HOCH2CH2OH (1) + 2H2O (1) AH = Bl
0
(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 Pi
o2 ______ go
a
0
1 1 1 1
r2 ' t3 ta -R-1
S8WI)
{#{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.B
2.A
3.A
4.C
5.B
6.D
7.C8.C
9.B
10.A
11.D
12.D
13.C14.D
15.(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,
0
CI
CH.
催化剂，△
+H0(2分)
OH
OH
CHO
CHs
(6)13(2分)
(2分)
O=C一CH
18.(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