B GIO DC O TO CHNH THC ( thi cú 05 trang) THI TUYN SINH I HC NM 2013 Mụn : HểA HC, Khi A Thi gian lm bi : 90 phỳt, khụng k thi gian phỏt Mó : 374 H v tờn thớ sinh : S bỏo danh : Cho bit nguyờn t ca cỏc nguyờn t : H = 1; C = 12; N = 14; O = 16; Na = 23; Mg = 24; Al = 27; S = 32; Cl = 35,5; P = 31; Ca = 40; Cr = 52, Fe = 56; Cu = 64; Zn = 65; Br = 80; Ag=108; Ba = 137 PHN CHUNG CHO TT C TH SINH (40 cõu, t cõu n cõu 40) Cõu : Hn hp X gm 3,92 gam Fe, 16 gam Fe 2O3 v m gam Al Nung X nhit cao iu kin khụng cú khụng khớ, thu c hn hp cht rn Y Chia Y thnh hai phn bng Phn mt tỏc dng vi dung dch H2SO4 loóng (d), thu c 4a mol khớ H Phn hai phn ng vi dung dch NaOH d, thu c a mol khớ H2 Bit cỏc phn ng u xy hon ton Giỏ tr ca m l A 5,40 B 3,51 C 7,02 D 4,05 S mol electron trao i : m 16 16 3,92 m 16 ( ì 2) ì + ì 2ì + ì = 4ì ( ì 2) ì 27 160 160 56 27 160 m=7,02 Cach khỏc : Al d > khụng cũn Fe2O3 S mol Al mi phn =0,2a:3 s mol Fe mi phn =4aa=(3,92:56+16:160ì2):2 >a=0,045 m=(0,2a:3ì2+16:160ì2)ì27=7,02 Cõu 2: Trong iu kin thớch hp, xy cỏc phn ng sau (a) 2H2SO4 + C 2SO2 + CO2 + 2H2O (b) H2SO4 + Fe(OH)2 FeSO4 + 2H2O (c) 4H2SO4 + 2FeO Fe2(SO4)3 + SO2 + 4H2O (d) 6H2SO4 + 2Fe Fe2(SO4)3 + 3SO2 + 6H2O Trong cỏc phn ng trờn, phn ng xy vi dung dch H2SO4 loóng l A (a) B (c) C (b) D (d) Cõu 3: Trong mt bỡnh kớn cha 0,35 mol C2H2; 0,65 mol H2 v mt ớt bt Ni Nung núng bỡnh mt thi gian, thu c hn hp khớ X cú t so vi H bng Sc X vo lng d dung dch AgNO NH3 n phn ng hon ton, thu c hn hp khớ Y v 24 gam kt ta Hn hp khớ Y phn ng va vi bao nhiờu mol Br2 dung dch? A 0,10 mol B 0,20 mol C 0,25 mol D 0,15 mol Bo ton s mol liờn kt 0,35 ì 26 + 0,65 ì 24 0,35 ì (0,35 + 0,65 ) ì = 0,15 16 240 Cõu 4: iu kin thớch hp xy cỏc phn ng sau: (a) 2C + Ca CaC2 (b) C + 2H2 CH4 (c) C + CO2 2CO (d) 3C + 4Al Al4C3 Trong cỏc phn ng trờn, tớnh kh ca cacbon th hin phn ng A (c) B (b) C (a) D (d) Cõu 5: Lờn men m gam glucoz to thnh ancol etylic (hiu sut phn ng bng 90%) Hp th hon ton lng khớ CO2 sinh vo dung dch Ca(OH)2 d, thu c 15 gam kt ta Giỏ tr ca m l A 15,0 B 18,5 C 45,0 D 7,5 S mol glucoz bng na s mol ancol etylic 0,15 ì 180 : 0,9 = 15 Cõu 6: Hn hp X gm Ba v Al Cho m gam X vo nc d, sau cỏc phn ng xy hon ton, thu c 8,96 lớt khớ H2 (ktc) Mt khỏc, hũa tan hon ton m gam X bng dung dch NaOH, thu c 15,68 lớt khớ H2 (ktc) Giỏ tr ca m l A 29,9 B 24,5 C 19,1 D 16,4 Phng phỏp cng gp : 8,96 137 27 15,68 8,96 ì( + )+ :1,5 ì 27 = 24,5 22,4 22, Cõu 7: Khi lng Ag thu c cho 0,1 mol CH 3CHO phn ng hon ton vi lng d dung dch AgNO3 NH3, un núng l A 10,8 gam B 43,2 gam C 16,2 gam D 21,6 gam 0,1ì2ì108=21,6 Cõu 8: Cho bt Fe vo dung dch gm AgNO v Cu(NO3)2 Sau cỏc phn ng xy hon ton, thu c dung dch X gm hai mui v cht rn Y gm hai kim loi Hai mui X v hai kim loi Y ln lt l: A Cu(NO3)2; Fe(NO3)2 v Cu; Fe B Cu(NO3)2; Fe(NO3)2 v Ag; Cu C Fe(NO3)2; Fe(NO3)3 v Cu; Ag D Cu(NO3)2; AgNO3 v Cu; Ag Kim loi cú tớnh kh yu v ion kim loi cú tớnh oxi hoỏ yu u tiờn li! Cõu 9: Cho 100 ml dung dch amino axit X nng 0,4M tỏc dng va vi 80 ml dung dch NaOH 0,5M, thu c dung dch cha gam mui Cụng thc ca X l A H2NC3H6COOH B H2NC3H5(COOH)2 C (H2N)2C4H7COOH D H2NC2H4COOH S nhúm COOH=0,08ì0,5:(0,1ì0,4)=1 M=(5:0,04)-22=103 Cõu 10: Cho 1,37 gam Ba vo lớt dung dch CuSO 0,01 M Sau cỏc phn ng xy hon ton, lng kt ta thu c l A 3,31 gam B 2,33 gam C 1,71 gam D 0,98 gam S mol Ba=0,01; s mol CuSO4=0,01 lng kt ta= 0,01ì98+0,01ì233=3,31 Cõu 11: Khi c chiu sỏng, hirocacbon no sau õy tham gia phn ng th vi clo theo t l mol : 1, thu c ba dn xut monoclo l ng phõn cu to ca nhau? A isopentan B pentan C neopentan D butan CCCCC Cõu 12: Oxi húa hon ton 3,1 gam photpho khớ oxi d Cho ton b sn phm vo 200 ml dung dch NaOH 1M n phn ng xy hon ton, thu c dung dch X Khi lng mui X l A 14,2 gam B 11,1 gam C 16,4 gam D 12,0 gam S mol H3PO4= S mol P=3,1:31=0,1, s mol NaOH=0,2 S mol NaOH:s mol H3PO4=2:1 >Ch to mui Na2HPO4 Khi lng mui X = 0,1ì (46 + 96) = 14,2 Cõu 13: Cho X l hexapeptit, Ala-Gly-Ala-Val-Gly-Val v Y l tetrapeptit Gly-Ala-Gly-Glu Thy phõn hon ton m gam hn hp gm X v Y thu c amino axit, ú cú 30 gam glyxin v 28,48 gam alanin Giỏ tr ca m l A 77,6 B 83,2 C 87,4 D 73,4 Goi x l s mol Ala-Gly-Ala-Val-Gly-Val; y l s mol Gly-Ala-Gly-Glu 2x+2y=30:75=0,4 v 2x+y=28,48:89=0,32 x=0,12 v y=0,08 m=0,12ì(75ì2+89ì2+117ì2-18ì5)+0,08ì(75ì2+89+1473ì18)=83,2 Cõu 14: trng thỏi c bn, cu hỡnh electron ca nguyờn t Na( Z = 11) l A 1s22s22p53s2 B 1s22s22p43s1 C 1s22s22p63s2 D 1s22s22p63s1 Cõu 15: Hn hp X cha ba axit cacboxylic u n chc, mch h, gm mt axit no v hai axit khụng no u cú mt liờn kt ụi (C=C) Cho m gam X tỏc dng va vi 150 ml dung dch NaOH 2M, thu c 25,56 gam hn hp mui t chỏy hon ton m gam X, hp th ton b sn phm chỏy bng dung dch NaOH d, lng dung dch tng thờm 40,08 gam Tng lng ca hai axit cacboxylic khụng no m gam X l A 15,36 gam B 9,96 gam C 18,96 gam D 12,06 gam CnH2nO2 (x mol); CmH2m-2O2 (y mol) x+y=0,3 14(nx+my)+32x+30y=25,560,3ì22 62(nx+my)18y=40,08 x=0,15; y=0,15; nx+my=0,69 0,15n+0,15m=0,69 S C trung bỡnh =0,69:0,3=2,3 >Axit no n chc mch h : HCOOH hoc CH3COOH HCOOH >m=3,6 CH3COOH >m=2,6 (loi) Khi lng axit khụng no =0,15ì(14ì3,6+30)=12,06 Cõu 16: Dung dch axit axetic phn ng c vi tt c cỏc cht dóy no sau õy? A Na, NaCl, CuO B Na, CuO, HCl C NaOH, Na, CaCO3 D NaOH, Cu, NaCl Cõu 17: Tờn thay th (theo IUPAC) ca (CH3)3C-CH2-CH(CH3)2 l A 2,2,4-trimetylpentan B 2,2,4,4-tetrametylbutan C 2,4,4,4-tetrametylbutan D 2,4,4-trimetylpentan Cõu 18: T nilon-6,6 l sn phm trựng ngng ca A etylen glicol v hexametyleniamin B axit aipic v glixerol C axit aipic v etylen glicol D axit aipic v hexametyleniamin Cõu 19 : Hn hp X gm Na, Ba, Na2O v BaO Hũa tan hon ton 21,9 gam X vo nc, thu c 1,12 lớt khớ H2 (ktc) v dung dch Y, ú cú 20,52 gam Ba(OH) Hp th hon ton 6,72 lớt khớ CO2 (ktc) vo Y, thu c m gam kt ta Giỏ tr ca m l A 23,64 B 15,76 C 21,92 D 39,40 Gi x l s mol NaOH dung dch Y 1,12 20,52 x ì 16 = ì 153 + : ì 62 Qui i v Na2O v BaO : 29,12 + 22,4 171 40 x=0,14; s mol Ba(OH)2=20,52:171=0,12; s mol CO2=0,3 S mol OH=0,38 >S mol CO32=0,08m=0,08ì197=15,76 Cỏch khỏc : Qui i hn hp X v nguyờn t : Na(x mol), Ba(y mol); O(z mol) 23x+137y+16z=21,9 x+2y-2z=1,12:22,4*2=0,1 y=20,52:171=0,12 >x=0,14;z=0,12 S mol OH-=0,12*2+0,14=0,38 1nY ->2,5 >2k2,5 >k< 2,5 S mol CO2>s mol H2O >k=2 mY=(26,88:22,4ì44+19,830,24:22,4ì32) 2,5:(2k)ì0,4ì72=11,4 Cõu 22: Tin hnh cỏc thớ nghim sau (a) Sc khớ etilen vo dung dch KMnO4 loóng (b) Cho hi ancol etylic i qua bt CuO nung núng (c) Sc khớ etilen vo dung dch Br2 CCl4 (d) Cho dung dch glucoz vo dung dch AgNO3, NH3 d, un núng (e) Cho Fe2O3 vo dung dch H2SO4 c, núng Trong cỏc thớ nghim trờn, s thớ nghim cú xy phn ng oxi húa - kh l A B C D Cõu 23: Cho s cỏc phn ng: t t , CaO X +1500 NaOH (dung dch) Y + Z; Y + NaOH (rn) T + P; C t , xt T Q + H2; Q + H2O Z Trong s trờn, X v Z ln lt l A HCOOCH=CH2 v HCHO B CH3COOC2H5 v CH3CHO C CH3COOCH=CH2 v CH3CHO D CH3COOCH=CH2 v HCHO Cõu 24: ng vi cụng thc phõn t C4H10O cú bao nhiờu ancol l ng phõn cu to ca nhau? A B C D Cõu 25: Cho m gam Fe vo bỡnh cha dung dch gm H 2SO4 v HNO3, thu c dung dch X v 1,12 lớt khớ NO Thờm tip dung dch H 2SO4 d vo bỡnh thu c 0,448 lớt khớ NO v dung dch Y Bit c hai trng hp NO l sn phn kh nht, o iu kin tiờu chun Dung dch Y hũa tan va ht 2,08 gam Cu (khụng to thnh sn phm kh ca N +5) Bit cỏc phn ng u xy hon ton Giỏ tr ca m l A 2,40 B 4,20 C 4,06 D 3,92 Bo ton electron : m 2,08 1,12 + 0,448 ì2 + ì2 = ì3 56 64 22,4 m=4,06 Cõu 26: Liờn kt húa hc gia cỏc nguyờn t phõn t HCl thuc loi liờn kt A cng húa tr khụng cc B ion C cng húa tr cú cc D hiro Cõu 27: Thc hin cỏc thớ nghim sau (a) Cho dung dch HCl vo dung dch Fe(NO3)2 (b) Cho FeS vo dung dch HCl (c) Cho Si vo dung dch NaOH c (d) Cho dung dch AgNO3 vo dung dch NaF (e) Cho Si vo bỡnh cha khớ F2 Sc khớ SO2vo dung dch H2S (f) Trong cỏc thớ nghim trờn, s thớ nghim xy phn ng l A B C D Cõu 28: Hũa tan hon ton m gam Al bng dung dch HNO loóng, thu c 5,376 lớt (ktc) hn hp khớ X gm N2, N2O v dung dch cha 8m gam mui T ca X so vi H bng 18 Giỏ tr ca m l A 17,28 B 19,44 C 18,90 D 21,60 Khi lng Al(NO3)3+khi lng NH4NO3=khi lng mui thu c m m 5,376 44 36 36 28 ì 213 + ( ì ì( ì 10 + ì 8)) : ì 80 = 8m 27 27 22,4 16 16 m=21,6 0 0 Cõu 29: Cho hn hp X gm 0,01 mol Al v a mol Fe vo dung dch AgNO n phn ng hon ton, thu c m gam cht rn Y v dung dch Z cha cation kim loi Cho Z phn ng vi dung dch NaOH d iu kin khụng cú khụng khớ, thu c 1,97 gam kt ta T Nung T khụng khớ n lng khụng i, thu c 1,6 gam cht rn ch cha mt cht nht Giỏ tr ca m l A 8,64 B 3,24 C 6,48 D 9,72 3+ 3+ 2+ Dung dch Z cha Al , Fe , Fe Cht rn Y ch cú Ag Gi x l s mol Fe2+ v y l s mol Fe3+ : 90x+107y=1,97 v x+y=1,6:160ì2 >x=0,01; y=0,01 m=(0,01ì3+0,01ì2+0,01ì3) ì108=8,64 Cõu 30: Cht no sau õy khụng to kt ta cho vo dung dch AgNO3? A HCl B K3PO4 C KBr D HNO3 Cõu 31: Phenol phn ng c vi dung dch no sau õy? A NaCl B KOH C NaHCO3 D HCl Cõu 32: Cho cỏc cõn bng húa hc sau: 2HI (k) N O (k) (a) H2 (k) + I2 (k) (b) 2NO2 (k) 2SO (k) 2NH (k) (c) 3H (k) + N (k) (d) 2SO (k) + O (k) 2 2 nhit khụng i, thay i ỏp sut chung ca mi h cõn bng, cõn bng húa hc no trờn khụng b chuyn dch? A (a) B (c) C (b) D (d) (a) S phõn t khớ v bng Cõu 33: Kim loi st tỏc dng vi dung dch no sau õy to mui st(II)? A CuSO4 B HNO3 c, núng, d C MgSO4 D H2SO4 c, núng, d Cõu 34: Hũa tan hon ton 1,805 gam hn hp gm Fe v kim loi X vo bng dung dch HCL, thu c 1,064 lớt khớ H2 Mt khỏc, hũa tan hon ton 1,805 gam hn hp trờn bng dung dch HNO loóng (d), thu c 0,896 lớt khớ NO (sn phm kh nht) Bit cỏc th tớch khớ u o iu kin tiờu chun Kim loi X l A Al B.Cr C Mg D Zn Nu X l kim loi hoỏ tr khụng i v hoỏ tr II : 0,896 1,064 1,064 0,896 1,064 ( ì3 ì 2) ì 56 + ( ( ì3 ì 2)) ì X = 1,805 22,4 22,4 22,4 22,4 22,4 X=18 >X l Al (i hoỏ tr : 18ì3:2=27) Cõu 35: Cỏc cht dóy no sau õy u to kt ta cho tỏc dng vi dung dch AgNO NH3 d, un núng? A vinylaxetilen, glucoz, anehit axetic B glucoz, imetylaxetilen, anehit axetic C vinylaxetilen, glucoz, imetylaxetilen D vinylaxetilen, glucoz, axit propionic Cõu 36: Tin hnh in phõn dung dch cha m gam hn hp CuSO v NaCl (hiu sut 100%, in cc tr, mng ngn xp), n nc bt u b in phõn c hai in cc thỡ ngng in phõn, thu c dung dch X v 6,72 lớt khớ (ktc) anot Dung dch X hũa tan ti a 20,4 gam Al 2O3 Giỏ tr ca m l A 25,6 B 23,5 C 51,1 D 50,4 CuSO4+2NaCl >Cu+Cl2+Na2SO4 (1) x 2x x *Nu CuSO4 d sau (1) CuSO4+H2O >Cu+(1/2)O2+H2SO4 y 0,5 y y Al2O3+3H2SO4 >Al2(SO4)3+3H2O 0,2 >0,6 y=0,6 >0,5y=0,3(loi) *Nu NaCl d sau (1) 2NaCl+2H2O >2NaOH+H2+Cl2 y y 0,5y 0,5y Al2O3+2NaOH >2NaAlO2+H2O 0,2 0,4 y=0,4 >x=0,1 m=160x+(2x+y)ì58,5= 0,1ì160+(0,1ì 2+0,4)ì58,5=51,1 Cõu 37: Cht no sau õy un núng vi dung dch NaOH thu c sn phm cú anehit? A CH3-COO-C(CH3)=CH2 B CH3-COO-CH=CH-CH3 C CH2=CH-COO-CH2-CH3 D CH3-COO-CH2-CH=CH2 CH3COOCH=CH2+NaOH >CH3COONa+CH3CHO Cõu 38: Dung dch no sau õy lm phenolphtalein i mu? A glyxin B metylamin C axit axetic D alanin Cõu 39: Cho 0,1 mol tristearin ((C17H35COO)3C3H5) tỏc dng hon ton vi dung dch NaOH d, un núng, thu c m gam glixerol Giỏ tr ca m l A 27,6 B 4,6 C 14,4 D 9,2 m=0,1ì92=9,2 Cõu 40: Dóy cỏc cht u tỏc dng c vi dung dch Ba(HCO3)2 l: A HNO3, Ca(OH)2 v Na2SO4 B HNO3, Ca(OH)2 v KNO3 C HNO3, NaCl v Na2SO4 D NaCl, Na2SO4 v Ca(OH)2 II PHN RIấNG (10 cõu) Thớ sinh ch c lm mt hai phn (Phn A hoc Phn B) A Theo chng trỡnh Chun (10 cõu, t cõu 41 n cõu 50) Cõu 41: Cho X v Y l hai axit cacboxylic mch h, cú cựng s nguyờn t cacbon, ú X n chc, Y hai chc Chia hn hp X v Y thnh hai phn bng Phn mt tỏc dng ht vi Na, thu c 4,48 lớt khớ H2 (ktc) t chỏy hon ton phn hai, thu c 13,44 lớt khớ CO (ktc) Phn trm lng ca Y hn hp l A 28,57% B 57,14% C 85,71% D 42,86% Gi x l s mol X, y l s mol Y 0,5x+y=0,2 0,6 0,6 0,6 1,5 = < Soỏ C = < = X : CH3COOH vaứ Y : HOOCCOOH x + 2y x + y 0,5x + y 2x+2y=0,6 x=0,2 v y=0,1 %Y=0,1ì90ì100:( 0,1ì90+0,2ì60)=42,85714286 Cõu 42: t chỏy hon ton hn hp X gm 0,07 mol mt ancol a chc v 0,03 mol mt ancol khụng no, cú mt liờn kt ụi, mch h, thu c 0,23 mol khớ CO2 v m gam H2O Giỏ tr ca m l A 5,40 B 2,34 C 8,40 D 2,70 S mol H2Os mol CO2=s mol ancol no mch h m 0,23 = 0, 07 18 m=5,4 Cõu 43: Dóy cỏc cht u cú kh nng tham gia phn ng thy phõn dung dch H 2SO4 un núng l: A fructoz, saccaroz v tinh bt B saccaroz, tinh bt v xenluloz C glucoz, saccaroz v fructoz D glucoz, tinh bt v xenluloz Cõu 44: Cho cỏc cp oxi húa kh c sp xp theo th t tng dn tớnh oxi húa ca cỏc ion kim loi: Al3+/Al; Fe2+/Fe, Sn2+/Sn; Cu2+/Cu Tin hnh cỏc thớ nghim sau: (a) Cho st vo dung dch ng(II) sunfat (b) Cho ng vo dung dch nhụm sunfat (c) Cho thic vo dung dch ng(II) sunfat (d) Cho thic vo dung dch st(II) sunfat Trong cỏc thớ nghim trờn, nhng thớ nghim cú xy phn ng l: A (b) v (c) B (a) v (c) C (a) v (b) D (b) v (d) Cõu 45: Cho cỏc phỏt biu sau: (a) Trong bng tun hon cỏc nguyờn t húa hc, crom thuc chu kỡ 4, nhúm VIB (b) Cỏc oxit ca crom u l oxit baz (c) Trong cỏc hp cht, s oxi húa cao nht ca crom l +6 (d) Trong cỏc phn ng húa hc, hp cht crom(III) ch úng vai trũ cht oxi húa (e) Khi phn ng vi khớ Cl2 d, crom to hp cht crom(III) Trong cỏc phỏt biu trờn, nhng phỏt biu ỳng l: A (a), (b) v (e) B (a), (c) v (e) C (b), (d) v (e) D (b), (c) v (e) Cõu 46: Thớ nghim vi dung dch HNO3 thng sinh khớ c NO2 hn ch khớ NO2 thoỏt t ng nghim, ngi ta nỳt ng nghim bng: (a) bụng khụ (b) bụng cú tm nc (c) bụng cú tm nc vụi (d) bụng cú tm gim n Trong bin phỏp trờn, bin phỏp cú hiu qu nht l A (d) B (c) C (a) D (b) Cõu 47: Hn hp X gm H2, C2H4 v C3H6 cú t so vi H l 9,25 Cho 22,4 lớt X (ktc) vo bỡnh kớn cú sn mt ớt bt Ni un núng bỡnh mt thi gian, thu c hn hp khớ Y cú t so vi H bng 10 Tng s mol H2 ó phn ng l A 0,070 mol B 0,015 mol C 0,075 mol D 0,050 mol S mol gim bng s mol H2 phn ng 18,5 = 0,075 20 Cõu 48: Trong cỏc dung dch CH3-CH2-NH2, H2N-CH2-COOH, H2N-CH2-CH(NH2)-COOH, HOOC-CH2-CH2-CH(NH2)-COOH, s dung dch lm xanh qu tớm l A.4 B.1 C D.3 cAl(NO3)3 + dNO + eH2O Cõu 49: Cho phng trỡnh phn ng aAl +bHNO3 T l a : b l A : B : C : D : Al(NO3)3 + NO + 2H2O Al +4HNO3 Cõu 50: Cho 25,5 gam hn hp X gm CuO v Al 2O3 tan hon ton dung dch H2SO4 loóng, thu c dung dch cha 57,9 gam mui Phn trm lng ca Al2O3 X l A 40% B 60% C 20% D 80% Gi x l s mol CuO v y l s mol Al2O3 80x+102y=25,5 160x+342y=57,9 x=0,255; y=0,05 %Al2O3=0,05ì102ì100:25,5=20 B Theo chng trỡnh Nõng cao (10 cõu, t cõu 51 n cõu 60) Cõu 51: Cho phng trỡnh phn ng: aFeSO + bK 2Cr2O7 + cH SO dFe (SO )3 + eK 2SO + fCr2 (SO )3 + gH 2O T l a:b l A.3:2 B 2:3 C 1:6 D 6:1 6FeSO + K 2Cr2O7 + 7H SO 3Fe (SO )3 + K 2SO + Cr2 (SO )3 + 7H 2O Cõu 52: Cho cỏc phỏt biu sau: (a) x lý thy ngõn ri vói, ngi ta cú th dựng bt lu hunh (b) Khi thoỏt vo khớ quyn , freon phỏ hy tn ozon (c)Trong khớ quyn, nng CO2 vt quỏ tiờu chun cho phộp gõy hiu ng nh kớnh (d) Trong khớ quyn , nng NO2 v SO2 vt quỏ tiờu chun cho phộp gõy hin tng ma axit Trong cỏc phỏt biu trờn , s phỏt biu ỳng l: A.2 B C D Cõu 53: Cho cỏc phỏt biu sau: (a) Glucoz cú kh nng tham gia phn ng bc (b) S chuyn húa tinh bt c th ngi cú sinh mantoz (c) Mantoraz cú kh nng tham gia phn ng bc (d) Saccaroz c cu to t hai gc -glucoz v -fructoz Trong cỏc phỏt biu trờn , s phỏt biu ỳng l: A.3 B C D Cõu 54: Cho 13,6 gam mt cht hu c X (cú thnh phn nguyờn t C, H, O) tỏc dng va vi dung dch cha 0,6 mol AgNO3 NH3, un núng , thu c 43,2 gam Ag Cụng thc cu to ca X l : A CH C C CHO B CH = C = CH CHO C CH C CH CHO D CH C [ CH ] CHO S mol AgNO3=0,6>s mol Ag=0,4 >Cú ni ba u mch 13,6:(0,4:2)=68 >C3H3CHO Cõu 55: Peptit X b thy phõn theo phng trỡnh phn ng X + 2H 2O 2Y + Z (trong ú Y v Z l cỏc amino axit) Thy phõn hon ton 4,06 gam X thu c m gam Z t chỏy hon ton m gam Z cn va 1,68 lớt khớ O2 (ktc), thu c 2,64 gam CO2; 1,26 gam H2O v 224 ml khớ N2 (ktc) Bit Z cú cụng thc phõn t trựng vi cụng thc n gin nht Tờn gi ca Y l A glyxin B lysin C axit glutamic D alanin Z : S mol C= 0,06; S mol H= 0,14; s mol N= 0,02; S mol O=0,06ì2+0,07-0,15=0,04 Z : C3H7NO2 (0,02 mol) X + 2H2O 2Y + Z 4,06g 0,04 mol 0,04 mol 0,02 mol 4,06+0,04ì18=0,04MY+0,02ì89 >MY=75 Cõu 56: Trng hp no sau õy khụng xy phn ng? t0 (a) CH = CH CH Cl + H O (b) CH3 CH CH Cl + H O t cao,p cao (c) C6 H5 Cl + NaOH ( ủaởc ) ; vi (C6H5- l gc phenyl) t (d) C2 H Cl + NaOH A (a) B (c) C (d) D (b) Cõu 57: Trng hp no sau õy, kim loi b n mũn in húa hc? A t dõy st khớ oxi khụ B Thộp cacbon khụng khớ m C Kim loi km dung dch HCl D Kim loi st dung dch HNO3 loóng Cõu 58: Cho 12 gam hp kim ca bc vo dung dch HNO loóng (d), un núng n phn ng hon ton, thu c dung dch cú 8,5 gam AgNO3 Phn trm lng ca bc mu hp kim l A 65% B 30% C 55% D 45% Phn trm lng ca bc mu hp kim= 8,5:170ì108ì100:12=45 + Cl,dử + dungdũch NaOH,dử X Y Cõu 59: Cho s phn ng Cr t0 Cht Y s trờn l A Na2Cr2O7 B Cr(OH)2 C Cr(OH)3 D Na[Cr(OH)4] Cõu 60: Hn hp X gm ancol metylic, ancol etylic v glixerol t chỏy hon ton m gam X, thu c 15,68 lớt khớ CO (ktc) v 18 gam H 2O Mt khỏc, 80 gam X hũa tan c ti a 29,4 gam Cu(OH)2 Phn trm lng ca ancol etylic X l A 46% B 16% C 23% D 8% Gi x,y,z ln lt l s mol ancol metylic, ancol etylic , glixerol x+2y+3z=15,68:22,4=0,7 2x+3y+4z=18:18=1 32x + 46y + 92z 80 = z 29,4 : 98 ì x=0,05; y=0,1; z=0,15 %C2H5OH=0,1ì46ì100:(0,05ì32+0,1ì46+0,15ì92)=23