su dung chuc nang solve giai nhanh trac nghiem

8/7/2019 Su Dung Chuc Nang Solve Giai Nhanh Trac Nghiem

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Dng chc nng Solve ca my tnh Casio fx570ES gii trc nghim ha hc

Nhiu bi tp ha hc dn n thit lp phng trnh i s 1 n. Vi my tnh c chc nng Solve, ta cth dng n tm nghim thay v phi chuyn v, bin i v c th dn n nhm ln.Di y l mt s v d s dng chc nng solve ca my tnh Casio FX570ES gii trc nghim hahc.Phn 1 : mt s thao tc v phm1. S dng phm th hin phn s :

n phm trn mn hnh xut hin mu phn s nhp vo, dng phm dichuyn khi nhp s v n .2.Th hin n s X :

nhn ln lt cc phm3. Th hin du = :

nhn ln lt cc phm4. Thc hin chc nng solve :

nhn ln lt cc phm , sau nhp vo 1 s ban u cho Solve for X (thng l s 0,tuy nhin vi phng trnh bc 2 th nn chn X ph hp nu khng s khng c kt qu nh mun).

Mt s lu :Biu thc khng qu di v 2 l do : th nht l khng ch trn mn hnh hoc tc x l ca mytnh s chm.

Nn u tin n s trn t s hoc chuyn v t s my tnh x l nhanh hn.Th d :

thay v :(X 71) 100

17.15X 2 36.5 100 10 2

+ =

+

c th chuyn thnh :(X+71) 100=17.15 (X+2 36.5 100 102)

i vi trng hp phng trnh c nhiu nghim , cn gn gi tr gn vi X .Th d :

X2 ((0.300 10X) (0.300 10X))=1.873Nu nhn :

0=KQ(X=0.1113943609;LR=0) (1)

Nu nhn :

0.03=KQ(X=0.0173341614;LR=0) (2)Php tnh y xut pht t bi tp v hng s cn bng nn iu kin X 0.300 10=0.03nn khng th chn p n (1), do khi gn gi tr tm X chn 0.03 l hp l (d nhin p n vn ngkhi bn gn gi tr tm X khc min l hp l )


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Phn 2 : Minh hc c th :Bi tp 1 : Cho 9,12 gam hn hp gm FeO, Fe2O3, Fe3O4 tc dng vi dung dch HCl (d). Sau khi cc

phn ng xy ra hon ton, c dung dch Y; c cn Y thu c 7,62 gam FeCl2 v m gam FeCl3. Gitr ca m l : A. 9,75 B. 8,75 C. 7,80 D. 6,50Xem nh hn hp u ch gm FeO v Fe2O3Cch 1 : bo ton khi lng (gi X l khi lng)khi lng hn hp u+khi lng HClkhi lng H2O=khi lng FeCl2+khi lng FeCl3

7.62 X9.12 ( 2 3) (36.5 18 2) 7.62 X

56 71 56 36.5 3+ + = +

+ +

0 =KQ(X=9.75;LR=0)Cch 2 :khi lng FeCl3= (khi lng hn hp ukhi lng FeO) 160 2 (56+35.5 3)=(9.127.62 (56+71) 72) 160 2 (56+35.5 3)=KQ=9.75Bi tp 2 : Ha tan m gam kim loi M ha tr II bng dung dch HCl 10% va thu c dung dchmui c nng l 17,15%. M l :A. Mg B. Ca C. Zn D. Ba

Cch 1 : Lp phng trnh tm nguyn t khi ca M (Gi s ban u c 1 mol M) dng chcnng Solve (gi X l kim loi cn tm(thay cho M) v gi s s mol ban u ca X l 1 mol)

(X 71) 10017.15

X 2 36.5 100 10 2

+ =

+

0 =KQ(X=64.9993965;LR=0)Cch 2 : Dng phng php th s (th) dng chc nng Calc

(X 71) 100

X 2 36.5 100 10 2

+

+

24=KQ=12.6329....

40=KQ=14.453....

65=KQ=17.15006....

137=KQ=20.0462....Bi tp 3 : Khi cho 100ml dung dch Ba(OH)2 1M vo 200ml dung dch HCl thu c dung dch c cha20,43 gam cht tan. Nng mol (hoc mol/l) ca HCl trong dung dch dng l (Cho H = 1; O = 16;Cl = 35,5; Ba = 137) A. 1,0M. B. 0,9M. C. 0,5M. D.0,8M0.1 (137+71)=KQ=20.8>20,43 d Ba(OH)2 dng chc nng Solve (gi X l nng mol ca dung dch HCl)0.2 X 2 (137 34) (0.1 0.2 X 2) (137 34) 20.43 + + + =

0 =KQ(X=0.9;LR=0)Bi tp 4 : Khi cho 100ml dung dch Ba(OH)2 1M vo 200ml dung dch HCl thu c dung dch c cha22,99 gam cht tan. Nng mol (hoc mol/l) ca HCl trong dung dch dng l (Cho H = 1; O = 16;


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Cl = 35,5; Ba = 137) A. 1,30M. B. 1,20M. C. 0,95M. D. 0,86M0.1 (137+71)=20.88KQ=20.88


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0=KQ(X=75.571428....;LR=0)Bi tp 10 : Khi oxi ha khng hon ton ancol no, n chc X trong phn t c phn trm khi lngcacbon bng 64,865% thu c xeton Y c phn trm khi lng cacbon bngA. 64,865%. B. 40,00%. C. 67,67%. D. 66,67%.12X (14X+18) 100=64.865

0=KQ(X=4.000034....;LR=0)X 12 100 (14X+16)=KQ=66.66679....Bi tp 11 : m gam phi bo st ngoi khng kh, sau mt thi gian bin thnh hn hp X c khilng 12 gam gm Fe, FeO, Fe3O4, Fe2O3. Cho X tc dng hon ton vi dung dch HNO3 thy gii

phng ra 2,24 lt kh NO duy nht ( ktc). Gi tr ca m l

A. 11,8. B. 10,08. C. 9,8. D. 8,8.

X 56 2 160=(12+2.24 22.4 3 2 16)

0=KQ(X=10.08;LR=0)Bi tp 12 : Ngm mt inh st khi lng 10 gam trong 200 ml dung dch CuSO 4. Sau khi phn ngkt thc ly inh st ra ra sch, lm kh ri em cn thy khi lng inh st bng 10,8 gam. Nng dung dch CuSO4 l A. 0,05M. B. 0,0625M. C. 0,5M. D. 0,625M.

0.2X (6456)=10.810

0=

KQ(X=0.5;LR=0)

Bi tp 13 : em nung hn hp A, gm hai kim loi: x mol Fe v 0,15 mol Cu, trong khng kh mtthi gian, thu c 63,2 gam hn hp B, gm hai kim loi trn v hn hp cc oxit ca chng. em hatan ht lng hn hp B trn bng dung dch H2SO4 m c, d th thu c 0,3 mol SO2. Gi tr ca xl: A. 0,6 mol B. 0,4 mol C. 0,5 mol D. 0,7 mol

80X+0.15 80=63.2+0.3 16

0=

KQ(X=0.7;LR=0)Bi tp 14 : Nung 35,532 gam mui nitrat ca kim loi M n khi lng khng i thu c 15,12gam cht rn v hn hp kh X c t khi hi so vi H2 bng 21,6. Cng thc ca mui nitrat l :

A. Mg(NO3)2 B. Zn(NO3)2 C. Cu(NO3)2 D. AgNO335.532 (X+62 2)=15.12 (X+16)

0=KQ(X=64;LR=0)Bi tp 15 : Cho 0,1 mol este to bi axit 2 ln axit v ancol 1 ln ancol tc dng vi NaOH thu c6,4 gam ancol v 1 lng mui (g) nhiu hn lng este l 13,56% (so vi lng este). Khi lng camui l : A. 15,8 gam B. 25,6 gam C. 21,3 gam D. 13,4 gam

X+6,4=0,1 2 40+X 100 113,56

0=KQ(X=13.3994....;LR=0)


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Bi tp 16 : Ha tan hon ton mt khi lng kim loi R ha tr II vo dung dch HCl 14,6 % va c dung dch mui c nng 18,19 %. R l kim loi no sau y ?A.Zn. B.Fe C.Mg. D.Ca

(X+71) (X+2 36.5 100 14.62)=18.19 100

0=KQ(X=23.94108....;LR=0)Bi tp 17 : Hn hp A gm hai ankan ng ng lin tip. t chy ht m gam A cn dng 9,968 lt

O2 (ktc). Cho hp th sn phm chy vo bnh ng nc vi d. Sau th nghim, khi lng bnhnc vi tng thm 18,26 gam. Cng thc hai cht trong hn hp A l:A. C3H8, C4H10 B. C4H10, C5H12 C. C5H12, C6H14 D. C6H14, C7H16(1.5X+0.5) (9.968 22.4)=(62X+18) 18.26

0=KQ(X=5.6;LR=0)Bi tp 18 : Mt loi m cha 40% triolein, 20% tripanmitin v 40% tristearin. X phng ha hon tonm gam m trn thu c 138 gam glixerol. Gi tr ca m l :A. 1,209 kg B. 1,304 kg C. 1,326 kg D. 1,335 kg

Gi X l khi lng (thay cho m)S mol cht bo bng s mol glixerol :0.4X ((17 12+33+44) 3+41)+0.2X ((15 12+31+44) 3+41)+0.4X ((17 12+35+44) 3+41)=138 92

0=KQ(X=1304.273145;LR=0)Bi tp 19 : Cho anken X i qua 1 lng d dung dch KMnO4 thu c kt ta c khi lng bng2,07 ln khi lng X tham gia. Cng thc phn t ca X l :A. C2H4 B. C3H6 C. C4H8 D. C5H10

1

(42X)=2.07

(87

2)

0=KQ(X=2.001380262;LR=0)Bi tp 20 : Trn V1 ml dung dch HCl c pH=1,8 vo V2 ml dung dch HCl c pH=3,6 thu c dungdch c pH=3,0. T l V2:V1 l : A. 18,64 B. 19,83 C. 16,48 D. 15,84(X 103,6+101,8) (X+1)=103,0

0=KQ(X=19.83....;LR=0)Bi tp 21 : Thm m gam CuSO4.5H2O vo 360 gam dung dch CuSO4 10% thu c dung dch cnng 16%. Gi tr ca m l :A. 90 gam B. 45 gam C. 75 gam D.30 gam(X 160 250+360 10 100) (X+360)=16 100

0=KQ(X=45;LR=0)

Bi tp 22 : Cho bit tan ca AgNO3 60oC l 525 gam v 10oCl 170 gam. Khi lm lnh 2500gam dung dch AgNO3 bo ha 60oC xung cn 10oC th khi lng AgNO3 kt tinh l:A. 1420 gam B. 1320 gam C. 1520 gam D. 1220 gam(2500 525 625X) (2500X)=170 270

0=KQ(X=1420;LR=0)Bi tp 23 : Ha tan a gam CuSO4.5H2O vo b gam dung dch CuSO4 8% thu c 560 gam dung dchCuSO4 16%. Gi tr ca a v b l


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A. a = 48 v b = 8 B. a = 480 v b = 80 C. a = 80 v b = 480 D. a = 8 v b = 48(X 160 250+(560X) 8 100)=560 16 100

0=KQ(X=80;LR=0)Bi tp 24 : Clorin l mt loi t si tng hp c to ra do s Clo ha PVC. Mt loi t Clorin chm lng Clo l 63,964% (phn trm khi lng). Bao nhiu n v mt xch PVC phn ng cvi 1 phn t Cl2 to ra loi t ny?A. 1 n v mt xch B. 2 n v mt xch C. 3 n v mt xch D. 4 n v mt xch(X+1) 35.5 100 (62.5X+34.5)=63.964

0=KQ(X=2.999....;LR=0)Bi tp 25 : Trn 100 ml dung dch c pH = 1 gm HCl v HNO3 vi 100 ml dung dch NaOH nng a (mol/l) thu c 200 ml dung dch c pH = 12. Gi tr ca a l (bit trong mi dung dch [H +][OH-]= 10-14)A. 0,15. B. 0,30. C. 0,03. D. 0,12.X 0.10.1 0.1=0.2 102

0=KQ(X=0.12;LR=0)Bi tp 26 : Cn thm bao nhiu gam CuSO4 vo 400 gam dung dch CuSO4 5% thu c dungdch c nng l 7%?A. 9,124 g. B. 8,408 g. C. 12,105 g. D. 8,602 g.(X+400 5 100) 100 (400+X)=7

0=KQ(X=8.602...;LR=0)Bi tp 27 : Cho hn hp X gm Fe, FeO, Fe3O4 c khi lng 4,04 gam phn ng vi dung dch HNO3d thu c 336 ml kh NO(ktc, sn phm kh duy nht). S mol HNO3 tham gia phn ng l:A. 0,06 (mol). B. 0,036 (mol). C. 0,125(mol). D. 0,18(mol).4.04+X 63=(X336 22400) 3 (56+62 3)336 22400 30X 0.5 18

0=KQ(X=0.18;LR=0)Hoc :(4.04+336 22400 3 2 16)+160 6+336 22400=KQ=0.18Bi tp 28 : Hirocacbon X c cng thc n gin l CH. t chy hon ton 0,01 mol X ri dn sn

phm chy vo bnh ng dung dch NaOH d, thy khi lng bnh NaOH tng 4,24 gam. X c cngthc phn t l A. C2H2. B. C4H4. C. C6H6. D.C8H8.0.01 X (44+0.5 18)=4.24

0=KQ(X=8;LR=0)Bi tp 29 : Cho 11,1 gam axit cacboxylic no, n chc X tc dng hon ton vi 500 ml dung dchgm KOH 0,08M v NaOH 0,1M. C cn dung dch thu c 12,64 gam hn hp cht rn khan. Cngthc phn t ca X l A. C2H5COOH. B. CH3COOH. C. HCOOH. D C3H7COOH

11.1+0.5

(0.08

56+0.1

40)=12.64+11.1

(14X+32)

18

0=KQ(X=3;LR=0)


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Bi tp 30 : Dung dch natri hirocacbonat khi un si to nn dung dch natri cacbonat. C% caNaHCO3 trong dung dch ban u l: (bit sau khi un si c dung dch Na2CO3 5,83%, b qua lngnc mt khi un) A. 4,92%. B. 6,84%. C. 9,02%. D. 10,50%.X 84 0.5 106 100 (1000.5 X 84 44)=5.83

0=KQ(X=9.021...;LR=0)Bi tp 31 : Cho axit oxalic HOOCCOOH tc dng vi hn hp 2 ancol no, n chc, ng nglin tip, thu c 5,28g hn hp 3 este trung tnh. Thy phn lng este bng dung dch NaOH, thuc 5,53 g mui. Hai ancol c cng thc:A. CH3OH v C2H5OH B. C2H5OH v C3H7OHC. C3H7OH v C4H9OH D. C4H9OH v C5H11OH5.28 (90+28X)=5.53 134

0=KQ(X=1.355...;LR=0)Bi tp 32: A l este caglixerol vi axit cacboxylic no n chc mch h . un nng 2,18 gam A vidung dch NaOH cho ti khi phn ng xy ra hon ton thu c 2,46 gam mui. S mol ca A l :A. 0,015 B. 0,02 C. 0,01 D. 0,03

2.18+3X 40=2.46+X 92

0=KQ(X=0.01;LR=0)Bi tp 33: Hn hp A gm 2 ancol. un nng m gam hn hp A vi H2SO4 m c, thu c 3,584lt hn hp 2 olefin k tip nhau trong dy ng ng (ktc). Nu em t chy ht lng olefin ny, richo hp th sn phm chy trong bnh ng dung dch NaOH d, th khi lng bnh tng 24,18g. Cc

phn ng xy ra hon ton. Gi tr ca m l: A. 6,1g B. 8,34g C. 10,58g D. 12,74gX3.584 22.4 18=24.18 (44+18) (12+2)

0=KQ(X=8.34;LR=0)Bi tp 34: Cho 35,12 gam hn hp CH2(COOH)2, HCOOH, CH3COOH v CH2=CHCOOH tc dngvi K d thu c cht rn A v 6,832 lt H2 (ktc). Khi lng cht rn A l 64,15 gam. Thm nc dvo cht rn A thu c V lt H2 (ktc). V c gi tr l :A. 1,12 lt B. 1,68 lt C. 3,36 lt D. 2,24 ltX 22.4 2+6.832 22.4 2=(35.12+6.832 22.4 235.12) 39

0=KQ(X=1.68;LR=0)

Bi tp 35: Kh hon ton m gam hn hp X gm Fe2O3 v CuO hoc m gam Fe3O4 u cn 1 lngkh CO (t0) nh nhau. Phn trm khi lng Fe2O3 trong X l :A. 68.14% B. 75,86% C. 84,16% D. 48,19%X 160 3+(100X) 80=100 232 4

0=KQ(X=75.862...;LR=0)Bi tp 36: Thm m gam bt Fe vo 400 gam dung dch FeCl3 20% thu c dung dch c nng %ca FeCl2 v FeCl3 bng nhau. Gi tr ca m l :A. 5,51g B. 6,35g C. 6,64g D. 4,85g3X 56 (56+35.5 2)=(400 20 100X 56 2 (56+35.5 3))

0=KQ(X=5.5138....;LR=0)


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Bi tp 37: Ha tan 56,8 gam hn hp MgCO3, MgSO3, MgO, CaCO3, CaSO3 bng dung dch HCl16% va thu c dung dch X trong khi lng ca MgCl2 to thnh l 28,5 gam v 11,2 lt hnhp kh (ktc) c t khi so vi hiro l 28. Khi lng dung dch HCl dng l :A. 228,125 gam B. 273,750 gam C. 410,625 gam D. 365,000 gam(56.811.2 22.4 28 228.5 (24+35.5 2) 40) 56 2+28.5 (24+35.5 2) 2=0.16X 36.5

0=KQ(X=273.75;LR=0)Bi tp 38: Ha tan km vo dung dch axit X 20% va thu c dung dch mui c nng l25,77%.Axit X l : A. HCl B. H2SO4 C. HBr D. HI(65+2X) 100 (X+2 (X+1) 100 202)=25,77

0=KQ(X=80.0483...;LR=0)Bi tp 39 : Hn hp X gm 2 anken l ng ng lin tip. Hiro ha hon ton X thu c hn hpkh Y. T khi hi ca Y so vi X bng 37/35. Xc nh cng thc phn t ca 2 anken.A. C2H4 v C3H6 B. C3H6 v C4H8 C. C4H8 v C5H10 D. khng c gi tr xc nh.(14X+2) (14X)=37 35

0=KQ(X=2.5;LR=0)Bi tp 40 : Cho m gam hn hp FeO v Cu2O c t l v s mol l 2: 2 :1 FeO Cu On n = tc dng vidung dch HNO3 long d thu c dung dch X v kh NO(sn phm kh duy nht). C cn dung dchX thu c m+37,2 gam mui khan. Gi tr ca m l :A. 17,28 gam B. 32,.68 gam C. 16,15 gam D. 14,86 gamX (72 2+144) (2 3+2 2) 62=37.2

0=KQ(X=17.28;LR=0)Bi tp 41 : Cho thanh kim loi M ha tr 2 vo dung dch cha hn hp gm 0,02 mol AgNO3 v 0,03mol Cu(NO3)2. Sau khi cc mui tham gia ht ly thanh M ra thy khi lng tng 1,48 gam. Vy M l

A. Fe B. Mg C. Zn D. Be0.01 (216X)+0.03 (64X)=1.48

0=KQ(X=65;LR=0)Bi tp 42: Tin trnh phn ng thun nghch trong bnh khi dung dch 1 lt

CO (k) + Cl2 (k) COCl2 (k)

t

0

C khng i, nng cn bng ca cc cht l [CO] = 0,02M; [Cl2] = 0,01M; [COCl2] = 0,02M.Bm thm vo bnh 0,03 mol Cl2.Tnh nng cc cht trng thi cn bng mi.A. [CO] = 0,01M; [Cl2] = 0,02M; [COCl2] = 0,03M.B. [CO] = 0,03M; [Cl2] = 0,01M; [COCl2] = 0,02M.C. [CO] = 0,01M; [Cl2] = 0,03M; [COCl2] = 0,03M.D. [CO] = 0,02M; [Cl2] = 0,02M; [COCl2] = 0,01M(0.02+X) ((0.02X) (0.01+0.03X))=0.02 (0.01 0.02)

0.02=KQ(X=0.01;LR=0)

Bi tp 43: Xt phn ng sau: H2O (k) + CO (k) H2 (k) + CO2 (k)

7000C phn ng ny c hng s cn bng K = 1,873. Tnh nng H2O trng thi cn bng, bitrng hn hp ban u gm: 0,300 mol H2O v 0,300 mol CO trong bnh 10 lt 7000C.

A. 0,0173. B. 0,0086. C. 0,0127. D. 0,0136.

X2 ((0.300 10X) (0.300 10X))=1.873


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0.03=KQ(X=0.01733...;LR=0)Bi tp 44 : t chy ht 0,5 mol hn hp X gm CH3CHO, C6H5OH, C2H5OH thu c 61,6 gam CO2v 23,4 gam H2O. Th tch O2 (ktc) cn t chy 21,92 gam hn hp X l :A. 32,256 lt B. 32,704 lt C. 40,320 lt D. 33,152 lt(61.6 44 2 23.4 18 0.5) 2 22.4 X

61.6 44 12 23.4 18 2 0.5 16 21.92

+ =

+ +

0=KQ(X=32.256;LR=0)Bi tp 45 : Hn hp R gmX,Y,Z l 3 cht hu c u cu to t C,H,O v c 2 nguyn t O trong 1

phn t. t chy a mol hn hp R cn b mol O2 thu c 22,4 lt CO2 (ktc)v 22,5 gam H2O. Mtkhc t chy 15,25 gam hn hp R cn 12,6 lt O2 (ktc). a c gi tr l :A. 0,4 mol B. 0,5 mol C. 0,6 mol D. 0,7 mol(1 2 22.5 18 2X) 2 22.4 12.6

12 22.5 18 2 2X 16 15.25

+ =

+ +

0=KQ(X=0.5;LR=0)Bi tp 46: t chy ht hn hp gm 0,4 mol glyxin v 0,2 mol cht X (ng ng vi glyxin) thcn dng 252 lt khng kh ktc. Khng kh cha 20% th tch l oxi. Sn phm chy gm CO 2, H2Ov nit n cht. X l: A. Alanin B. Valin C. Axit glutamic D. C4H9NO2

0.6 (1.5X-3 4)=252 22.4 5 0= CnH2n+1NO2+(1.5n3:4)O2

nCO2+(n+1/2)O2+(1/2)N2KQ(X=3;LR=0)

(2 2+X) 0.3=3 0=KQ(X=5;LR=0)Bi tp 47 : Cho m gam bt Fe vo 800 ml dung dch hn hp gm Cu(NO3)2 0,2M v H2SO4 0,25M.Sau khi cc phn ng xy ra hon ton, thu c 0,6m gam hn hp bt kim loi v V lt kh NO (sn

phm kh duy nht, ktc). Gi tr ca m v V ln lt lA. 17,8 v 4,48. B. 17,8 v 2,24. C. 10,8 v 4,48. D. 10,8 v 2,24.0.8 0.25 2 4 22.4=KQ=2.240.6X=X0.8 0.25 2 3 8 56+0.8 0.2 (6456)

0=KQ(X=17.8;LR=0)Bi tp 48 : Cho 0,1 mol mi axit H3PO2 v H3PO3 tc dng vi dung dch KOH d thu c hai muic khi lng ln lt l 10,408 gam v 15,816 gam. S chc axit ca hai axit trn ln lt l :A. 1 v 3 B. 2 v 3 C. 1 v 2 D. 2 v 210.4080.1 (3+31+16 2)=0.1X 38

0=KQ(X=1.002105...;LR=0)

15.8160.1 (3+31+16 3)=0.1X 38

0=KQ(X=2.0042105...;LR=0)Bi tp 49 : Ha tan 8,45 gam oleum X, dung dch thu c trung ha bi 200 ml dung dch NaOH1M. Cng thc ca X l A. H2SO4.SO3. B. H2SO4.2SO3. C. H2SO4.3SO3. D. H2SO4.4SO3


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8.45(2X 2) 0.2

98 80X + =

+

0=KQ(X=3;LR=0)Bi tp 50 : Khi thy phn 1 protit X thu c hn hp gm 2 amino axit no k tip nhau trong dyng ng. Bit mi cht u cha 1 nhm NH2 v 1 nhm COOH. t chy hon ton 0,2 mol hnhp 2 aminoaxit ri cho sn phm chy qua bnh ng dung dch NaOH d , thy khi lng bnh tng

32,8 gam. Cng thc cu to ca 2 aminoaxit l :A.H2NCH(CH3)COOH v C2H5CH(NH2)COOH B. H2NCH2COOH v H2NCH(CH3)COOHC. H2NCH(CH3)COOH v N2N[CH2]3COOH D. H2NCH2COOH v v H2NCH2CH2COOH0.2 (44X+(X+1.5) 18)=32.8

0=KQ(X=2.209677....;LR=0)Bi tp 51 : Hiro ho hon ton m gam hn hp X gm hai anehit no, n chc, mch h, k tipnhau trong dy ng ng thu c (m + 1) gam hn hp hai ancol. Mt khc, khi t chy hon toncng m gam X th cn va 17,92 lt kh O2 ( ktc). Gi tr ca m lA. 10,5 B. 17,8 C. 8,8 D. 24,8

3X 1 17.920.52 22.4

=

0=KQ(X=1.4;LR=0)0.5 (14X+16)=KQ=17.8Bi tp 52 : Oxi ha hon ton hn hp X gm HCHO v CH3-CHO bng O2 (xt) thu c hn hpaxit tng ng Y. T khi (hi) ca Y so vi X bng 145/97. Tnh % s mol ca mi cht trong X.A. 22,7% HCHO v 77,3% CH3-CHO B. 83,3% HCHO v 16,7% CH3-CHO

C. 50,2% HCHO v 49,8% CH3-CHO D. 80% HCHO v 20% CH3-CHO14X 32 145

14X 16 97

+=

+

0=KQ(X=1.1666.....;LR=0)(2X) 100=KQ=83.333...Bi tp 53 : Hn hp X gm 3 ancol no n chc mch h c t khi hi so vi H2 l 24,4. t chy3,66 gam X thu c bao nhiu gam H2O? A. 4,32 gam B. 4,50 gam C. 8,28 gam D. 3,96 gam

14X+18=24.4 2

0=KQ(X=2.2;LR=0)

3.66(X 1) 18

14X 18 + =

+

KQ=4.32Bi tp 54: Hn hp A gm hai olefin k tip nhau trong dy ng ng. Th nghim cho thy 18,06gam hn hp A lm mt mu va 250 mL dung dch KMnO4 0,64M, ng thi thy c to ra chtkhng tan c mu nu en. Phn trm s mol mi cht trong hn hp A l:

A. 37,5%; 62,5% B. 43,8%; 56,2% C. 33,33%; 66,67% D. 25%; 75%18.061.5 0.25 0.64

14X=

0=


11/13

KQ(X=5.375;LR=0)(6X) 100=KQ=62.5Bi tp 55: Khi ha tan mt oxit kim loi ha tr II bng 1 lng va dung dch H2SO4 9,8% th thuc dung dch mui c nng 14,8 %. Cng thc phn t ca oxit kim loi l :A. CaO B. CuO C. MgO D. BaO

(X 96) 10014.8

X 16 98 100 9.8

+ =

+ +

0=KQ(X=63.81220657;LR=0)Bi tp 56: Hn hp gm hirocacbon X v oxi c t l s mol tng ng l 1:10. t chy honton hn hp trn thu c hn hp kh Y. Cho Y qua dung dch H2SO4 c, thu c hn hp kh Zc t khi i vi hiro bng 19. Cng thc phn t ca X l (cho H = 1, C = 12, O = 16)A. C3H8. B. C3H6. C. C4H8. D. C3H4.

X(10 A ) 32 44A

4 19 2X

104

+

=

3=0=

KQ(X=16;LR=0) Loi

4=0=KQ(X=8;LR=0)Bi tp 57 : Mt hn hp X gm 1 ancol no a chc mch h A v 1 ankanol B c t l s molnA:nB=1:2. t chy 18,72 gam X thu c 26,4 gam CO2 v 17.28 gam H2O. S nhm chc ca A l :A. 2 B. 3 C. 4 D. 5

26.4 17.28 17.28 26.4 16X 2 1612 2 ( ) 18.7244 18 18 44 3

+ + + =

0=

KQ(X=3;LR=0)Bi tp 58 : Cu c 2 ng v l 63Cu v 65Cu. Khi lng nguyn t trung bnh ca Cu l 63,54u.Phn trm khi lng 63Cu trong CuX2 (trong X l 1 loi nguyn t) l 34,439%. S khi ca X l:A. 35. B. 37. C. 80. D. 19.63.54 100 (65 63.54) 63

34.43963.54 2X (65 63.54) 63 (63.54 63) 65

=

+ +

0=KQ(X=35.000231....;LR=0)

Bi tp 59 : Hn hp A gm hai ankan ng ng lin tip. t chy ht m gam A cn dng9,968 lt O2 (ktc). Cho hp th sn phm chy vo bnh ng nc vi d. Sau th nghim, khilng bnh nc vi tng thm 18,26 gam. Cng thc hai cht trong hn hp A l:

A. C3H8, C4H10 B. C4H10, C5H12 C. C5H12, C6H14 D. C6H14, C7H161.5X 0.5 62X 18

9.968 22.4 18.26

+ +=

0=

KQ(X=5.6;LR=0)


12/13

Bi tp 60 : t chy hon ton m gam hn hp 2 ancol X v Y thuc dy ng ng ca ancol metylicth thu c 79,2 gam CO2 v 43,2 gam H2O. Gi tr m l:A. 36 gam B. 28 gam C. 20 gam D. 12 gam

43.2 79.2 43.2 79.2X ( 2 ( )) 2 32 43.2 79.2

18 44 18 44+ + = +

0=KQ(X=36;LR=0)

Bi tp 61 : Nu tha nhn rng nguyn t Ca c dng hnh cu sp xp c kht nhau, th th tchchim bi cc nguyn t kim loi ch bng 74 % so vi ton khi tinh th. Hy tnh bn knh nguyn tCa theo n v A0, bit khi lng ring ktc ca Ca th rn l 1,55 g/cm3. Cho nguyn t khi caCa l 40,08. A. 1,28A0 B. 1,96A0 C. 1,43A0 D. 1,5A0

3

23

4 X 40.08 10.74

3 1.55 6.023 10

=

0=

KQ(X=1.9647... 10-8;LR=0)Bi tp 62 : Trung ho 8,2 gam hn hp gm axit fomic v mt axit n chc X cn 100 ml dung dch

NaOH 1,5M. Nu cho 8,2 gam hn hp trn tc dng vi mt lng d dung dch AgNO 3 trong NH3un nng th thu c 21,6 gam Ag. Tn gi ca X lA. axit acrylic B. axit propanoic C. axit etanoic D. axit metacrylic21.6 108 2 46 (0.1 1.5 21.6 108 2)X 8.2 + =

0=KQ(X=72;LR=0)

Bi tp 63 : Khi lng nguyn t ca Clo l 35,5. Clo c 2 ng v l 35Cl v 37Cl. Phn trm khi

lng ca 35Cl c trong HClOn l 31,065%(hiro l 11H v oxi l 168O). Gi tr ca n l :A. 1 B. 2 C. 3 D.4

35.5 100 (37 35.5) 3531.065

1 35.5 16X (37 35.5) 35 (35.5 35) 37

=

+ + +

0=KQ(X=3.000015089;LR=0)Bi tp 64 : un m gam etyl axetat vi 200 ml dung dch hn hp KOH 1M v NaOH 1 M (d) n phnng hon ton thu c dung dch A. C cn dung dch A thu c m+6,32 gam cht rn khan. Gi tr ca ml : A. 26,40 gam B. 24,64 gam C. 28,16 gam D. 22,88 gam

XX 0.2 (56 40) X 6.32 46

88+ + = + +

0=KQ(X=24.64;LR=0)Bi tp 65 : X l mt axit ankenoic, t chy hon ton 1,72 gam X phi dng va ht 2,016 lt O 2(ktc) . Xc nh cng thc phn t ca X ?A. C3H4O2 B. C4H8O2 C. C3H6O2 D. C4H6O2

14X 30 1.5X 1.51.72 2.016 22.4

+ =

0=

KQ(X=4;LR=0)


13/13

Ngi gi :HONG VN CHUNG

Gio vin trng TRUNG HC PH THNG CHUYN BN TRE

S 21, L Qu n phng 2 Th x Bn Tre (thnh ph Bn Tre), tnh Bn Tre

su dung chuc nang solve giai nhanh trac nghiem

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