De Thi Casio Mon Hoa Co Huong Dan Giai

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<p> 1 </p> <p>Mn ho hc</p> <p>I. Ni dung thi</p> <p>- Tt c cc kin thc trong chng trnh trung hc ph thng</p> <p>- Cc php tnh c s dng:1. Php tnh cng, tr, nhn, chia thng thng</p> <p>2. Php tnh hm lng phn trm</p> <p>3. Php tnh cng tr cc phn s</p> <p>4. Php tnh bnh phng, s m, khai cn</p> <p>5. Php tnh logarit (log; ln) v i logarit</p> <p>6. Gii phng trnh bc nht mt n</p> <p>7. Php tnh cc hm s lng gic sin, cos, tg, cotg</p> <p>8. Gii h hai phng trnh bc nht mt n</p> <p>9. Gii h ba phng trnh bc nht mt n</p> <p>10. Gii phng trnh bc hai mt n</p> <p>11. Gii phng trnh bc ba mt n </p> <p>12. Cc php tnh v vi phn, tch phn, o hm</p> <p>II. Cu trc bn thiPhn th nht: HS trnh by li gii v ni dung ha hcPhn th hai: HS trnh by cch s dng my tnh b ti gii phng trnh v tnh ton</p> <p>Phn th ba: HS trnh by kt quIII. Hng dn cch lm bi v tnh im</p> <p> gii mt bi ton Ho hc, th sinh phi ghi tng ng tm tt li gii v ni dung ha hc, cch s dng my tnh b ti gii phng trnh v tnh ton kt qu vo cc phn tng ng c sn trong bn thi.</p> <p>Mi bi ton c chm im theo thang im 5. im ca mt bi ton bng tng im ca 3 phn trn. </p> <p>im ca bi thi l tng im th sinh lm c (khng vi phm qui ch thi) ca 10 bi ton trong bi thi.</p> <p>IV. V d bi ton v cch trnh by bi gii</p> <p>V d 1:</p> <p>Hai nguyn t ha hc X v Y iu kin thng u l cht rn. S mol ca X c trong 8,4 gam X nhiu hn so vi s mol ca Y c trong 6,4 gam Y l 0,15 mol. Bit khi lng mol nguyn t ca X nh hn khi lng mol nguyn t ca Y l 8 gam. Xc nh k hiu ha hc ca X v Y?Phn th nht: HS trnh by li gii v ha hc</p> <p>K hiu khi lng mol nguyn t ca X v Y l x v y </p> <p>So snh s mol:</p> <p>nA = ; nB = ta c phng trnh - = 0,15</p> <p>Theo gi thit: x + 8 = y</p> <p>Ghp hai phng trnh cho: 0,15x2 - 0,8x - 67,2 = 0</p> <p>Phn th hai: HS trnh by cch s dng my tnh b ti gii phng trnhBm MODE hai ln ( mn hnh my tnh hin ln EQN </p> <p> 1</p> <p>Bm nt s 1 ( mn hnh my tnh hin ln Unknowns</p> <p> 2 3</p> <p>Bm nt chuyn sang phi ( mn hnh my tnh hin ln Degree?</p> <p> 2 3</p> <p>Bm 2 ( chn PT bc 2) ( mn hnh my tnh hin a? th bm 0,15</p> <p>Bm = ( mn hnh my tnh hin b ? th bm (-) 0,8</p> <p>Bm = ( mn hnh my tnh hin c ? th bm (-) 67,2</p> <p>Bm = ( mn hnh my tnh hin x1 = 24</p> <p>Bm = ( mn hnh my tnh hin x2 = - 18,6666...</p> <p>Phn th ba: HS trnh by kt qu</p> <p>Theo iu kin ha hc: x &gt; 0 nn ch chn x = x1 = 24 ( X l Mg</p> <p> y = 24 + 8 = 32 ( Y l S </p> <p>V d 2:</p> <p>Ha tan 15,8 gam hn hp A gm Na2CO3; K2CO3 v Na2O bng dung dch HCl thot ra 1,68 lt CO2 (ktc) v thu c dung dch B. C cn dung dch B c 22,025 gam hn hp cha hai mui khan. Tnh thnh phn % hn hp A. </p> <p>Phn th nht: HS trnh by li gii v ha hcTheo u bi ta c cc phng trnh ha hc:</p> <p>Na2O + 2HCl ( 2NaCl + H2O</p> <p>Na2CO3 + 2HCl ( 2NaCl + CO2( + H2O</p> <p>K2CO3 + 2HCl ( 2KCl + CO2( + H2O</p> <p>Hn hp hai mui khan l NaCl v KCl</p> <p>Kh thot ra l CO2 = = 0,075 (mol)t s mol Na2O; Na2CO3 v K2CO3 ln lt l x, y, z</p> <p>Ta c cc phng trnh:</p> <p>* khi lng A: 62x + 106y + 138z = 15,8</p> <p>* khi lng hai mui khan: 58,5(2x + 2y) + 74,5 x 2z = 22,025</p> <p> hay 117x + 117y + 149z = 22,025</p> <p>* s mol kh CO2: y + z = 0,075</p> <p>Phn th hai: HS trnh by cch s dng my tnh b ti gii phng trnh</p> <p>Bm MODE hai ln ( mn hnh my tnh hin ln EQN </p> <p> 1</p> <p>Bm nt s 1 ( mn hnh my tnh hin ln Unknowns</p> <p> 2 3</p> <p>Bm 3 ( chn h PT 3 n) ( mn hnh my tnh hin a1? th bm 62</p> <p>Bm = ( mn hnh my tnh hin b1 ? th bm 106</p> <p>Bm = ( mn hnh my tnh hin c1 ? th bm 138</p> <p>Bm = ( mn hnh my tnh hin d1 ? th bm 15,8</p> <p>Bm = ( mn hnh my tnh hin a2 ? th bm 117</p> <p>Bm = ( mn hnh my tnh hin b2 ? th bm 117</p> <p>Bm = ( mn hnh my tnh hin c2 ? th bm 149</p> <p>Bm = ( mn hnh my tnh hin d2 ? th bm 22,025</p> <p>Bm = ( mn hnh my tnh hin a3 ? th bm 0</p> <p>Bm = ( mn hnh my tnh hin b3 ? th bm 1</p> <p>Bm = ( mn hnh my tnh hin c3 ? th bm 1</p> <p>Bm = ( mn hnh my tnh hin d3 ? th bm 0,075</p> <p>Bm = ( mn hnh my tnh hin x = 0,1</p> <p>Bm = ( mn hnh my tnh hin y = 0,021</p> <p>Bm = ( mn hnh my tnh hin z = 0,054</p> <p>Bm 62 ( 0,1 : 15,8 SHIFT = (%) ( mn hnh my tnh hin 0,3924</p> <p>Bm 106 ( 0,024 : 15,8 SHIFT = (%) ( mn hnh my tnh hin 0,1360</p> <p>Bm 138 ( 0,054 : 15,8 SHIFT = (%) ( mn hnh my tnh hin 0,4716</p> <p>Phn th ba: HS trnh by kt qu% Khi lng Na2O = 39,24%</p> <p>% Khi lng Na2CO3 = 13,6%</p> <p>% Khi lng K2CO3 = 47,16%</p> <p>V d 3:Al(OH)3 l mt hidroxit lng tnh c th tn ti 2 cn bng sau:</p> <p>Al(OH)3 Al3+ + 3OH- (1) = 10-33</p> <p>Al(OH)3 + OH- AlO2- + 2H2O (2) = 40</p> <p>Vit biu thc biu th tan ton phn ca Al(OH)3 (S) = [Al3+] + [AlO] di dng mt hm ca [H3O+]. pH bng bao nhiu th S cc tiu. Tnh gi tr S cc tiu. </p> <p>Phn th nht: HS trnh by li gii v ha hc( Xt 2 cn bng: </p> <p>Al(OH)3 Al 3+ + 3OH - Tt(1) = [Al3+].[OH-]3 =10-33 </p> <p>Al(OH)3 + OH - AlO + 2H2O Tt(2) == 40 </p> <p>T Tt(1): [Al3+] = = = 10 9[H3O+]3; </p> <p>v t Tt(2): [AlO] = 40[OH -] = 40</p> <p>Do S = [Al 3+] + [AlO] = 109[H3O+]3 + 40</p> <p>S cc tiu khi o hm = 3.10 9[H3O+]2 - = 0 </p> <p>( [H3O+]4 = ( [H3O+]4= 133,33. 10-24 [H3O+] = ?</p> <p> ( pH = - lg[H3O+] = ?</p> <p> pH = - (- 6) - lg3,4= ?</p> <p>Phn th hai: HS trnh by cch s dng my tnh b ti gii phng trnhBm 4 SHIFT 133,33 = 3,4</p> <p>Bm log 3,4 = 0,53</p> <p>Phn th ba: HS trnh by kt qu[H3O+] = 3,4. 10-6 </p> <p>pH = 5,47</p> <p>Smin = 10 9.(3,4. 10-6) + 40 = 1,5. 10-7 mol/l</p> <p>V d 4:Hy xc nh khong cch gia 2 nguyn t iot trong 2 ng phn hnh hc ca C2H2I2 vi gi thit 2 ng phn ny c cu to phng. (Cho di lin kt C I l 2,10 v C=C l 1,33 ). </p> <p>Phn th nht: HS trnh by li gii v ha hc</p> <p>( ng phn cis- :</p> <p>dcis = d C= C + 2 d C - I ( sin 300.</p> <p>ng phn trans-:</p> <p>d trans = 2( IO</p> <p>IO == </p> <p>Phn th hai: HS trnh by cch s dng my tnh b ti gii phng trnh</p> <p>Bm MODE mn hnh hin COMP SD REG</p> <p> 1 2 3</p> <p>Bm 1 sin 30 = 0,5</p> <p>Bm 2,1 x2 + 0,67 x2 4,2 ( 0,67 cos 120 = 2,5</p> <p>Phn th ba: HS trnh by kt qudcis = dC = C + dC- I = 1,33 + 2,1 = 3,43 dtrans = 2( 2,5 = 5,0 </p> <p>V. Mt s bi tp c hng dn gii</p> <p>Bi 1. Cho nng lng lin kt ca:</p> <p>N - HO = ON ( NH - ON - O</p> <p>kJ/mol389493942460627</p> <p>Phn ng no d xy ra hn trong 2 phn ng sau?</p> <p>2NH3 + 3/2 O2 ( N2 + 3 H2O (1)</p> <p>2NH3 + 5/2 O2 ( 2NO + 3H2O (2)</p> <p>( Hng dn gii:</p> <p>Tnh hiu ng nhit:</p> <p>E1 = (6EN-H +EO=O) - (EN(N + 6EO-H) </p> <p> = 6( 389 + ( 493 - 942 - 6( 460 = - 626,5 kJ</p> <p>E2 = (6EN-H +EO=O)- (2EN-O + 6EO-H) </p> <p> = 6( 389 + ( 493 - 2( 627 - 6( 460 =- 447,5 kJ</p> <p>- Phn ng (1) c (H m hn nn p (1) d xy ra hn.</p> <p>Bi 2. Cho V lt kh CO qua ng s ng 5,8 gam oxit st FexOy nng mt thi gian th thu c hn hp kh A v cht rn B. Cho B tc dng ht vi axit HNO3 long thu c dung dch C v 0,784 lt kh NO. C cn dung dch C th thu c 18,15 gam mt mui st (III) khan. Nu ha tan B bng axit HCl d th thy thot ra 0,672 lt kh. (Cc kh o iu kin tiu chun).a) Xc nh cng thc ca oxt st b) Tnh % theo khi lng cc cht trong B.( Hng dn gii:</p> <p>a) S mol Fe trong FexOy = s mol Fe trong Fe(NO3)3 = 0,075</p> <p>( s mol oxi trong FexOy = = 0,1 ( </p> <p>Vy cng thc ca B l Fe3O4.</p> <p>b) B c th cha Fe, FeO (a mol) v Fe3O4 d (b mol)</p> <p>3Fe3O4 + 28HNO3 ( 9Fe(NO3)3 + NO + H2O</p> <p>3FeO + 10HNO3 ( 3Fe(NO3)3 + NO + 5H2O</p> <p>Fe + 4 HNO3 ( Fe(NO3)3 + NO + 2H2O</p> <p>Fe + 2HCl ( FeCl2 + H2 , </p> <p>ta c : </p> <p>v </p> <p>Bi 3. Ra c chu k bn hu l 1590 nm. Hy tnh khi lng ca mt mu Ra c cng phng x = 1Curi (1 Ci = 3,7. 1010 Bq)?</p> <p>( Hng dn gii:</p> <p>Theo biu thc v = - = kN = 3,7.1010 Bq </p> <p>(trong N l s nguyn t Ra, cn k = ( N = . T1/2) </p> <p>v T1/2 = 1590.365.24.60.60 = 5,014.1010mRa = = = 1 gam </p> <p>Bi 4. Nung FeS2 trong khng kh, kt thc phn ng thu c mt hn hp kh c thnh phn: 7( SO2; 10( O2; 83( N2 theo s mol. un hn hp kh trong bnh kn (c xc tc) 800K, xy ra phn ng:</p> <p> 2SO2 + O2 2SO3</p> <p>Kp = 1,21.105.</p> <p>a) Tnh chuyn ho (( s mol) SO2 thnh SO3 800K, bit p sut trong bnh lc ny l 1 atm, s mol hn hp kh ban u (khi cha un nng) l 100 mol.</p> <p>b) Nu tng p sut ln 2 ln, tnh chuyn ho SO2 thnh SO3, nhn xt v s chuyn dch cn bng.</p> <p>( Hng dn gii:</p> <p>a) Cn bng: 2SO2 + O2 2SO3Ban u: 7 10 0 (mol)</p> <p>lc cn bng: (7-x) (10 - 0,5x) x (x: s mol SO2 phn ng).</p> <p>Tng s mol cc kh lc cn bng: 100 - 0,5x = n.</p> <p>p sut ring ca cc kh: </p> <p>= (7-x). ; = (10 - 0,5x). ; = x . </p> <p>Kp = = = 1,21. 105 </p> <p> do K &gt;&gt; ( x ( 7 ( Ta c : = 1,21. 105 Gii c x = 6,9225.</p> <p>Vy chuyn ha SO2 ( SO3: = 98,89(.</p> <p>b) Nu p sut tng 2 ln tng t c: 7- x(= = 0,0548 ( x( = 6,9452.</p> <p>( chuyn ho SO2 ( SO3: (6,9452 . 100)/7 = 99,21(Kt qu ph hp nguyn l Lsatlie: tng p sut phn ng chuyn theo chiu v pha c s phn t kh t hn.</p> <p>Bi 5. t chy hon ton 3 gam mt mu than c cha tp cht S. Kh thu c cho hp th hon ton bi 0,5 lt dung dch NaOH 1,5M c dung dch A, cha 2 mui v c xt d. Cho kh Cl2 (d) sc vo dung dch A, sau khi phn ng xong thu c dung dch B, cho dung dch B tc dng vi dung dch BaCl2 d thu c a gam kt ta, nu ho tan lng kt ta ny vo dung dch HCl d cn li 3,495 gam cht rn.</p> <p>a) Tnh ( khi lng C; S trong mu than, tnh a.</p> <p>b) Tnh nng mol/lt cc cht trong dung dch A, th tch kh Cl2 (ktc) tham gia phn ng.</p> <p>( Hng dn gii:</p> <p>a) Phng trnh phn ng: C + O2 ( CO2(1) </p> <p> x x</p> <p> S + O2 ( SO2(2)</p> <p> y y</p> <p>Gi s mol C trong mu than l x, s mol S trong mu than l y </p> <p>( 12x + 32y = 3.</p> <p>Khi cho CO2; SO2 vo dung dch NaOH d:</p> <p>CO2 + 2NaOH = Na2CO3 + H2O(3)</p> <p>SO2 + 2NaOH = Na2SO3 + H2O(4)</p> <p>Cho kh Cl2 vo dung dch A (Na2CO3; Na2SO3; NaOH d)</p> <p>Cl2 + 2NaOH = NaClO + NaCl + H2O(5)</p> <p> (d)</p> <p>2NaOH + Cl2 + Na2SO3 = Na2SO4 + 2NaCl + H2O(6)</p> <p>Trong dung dch B c: Na2CO3; Na2SO4; NaCl; NaClO. Khi cho BaCl2 vo ta c:</p> <p>BaCl2 + Na2CO3 = BaCO3( + 2NaCl</p> <p>(7)</p> <p> x x</p> <p>BaCl2 + Na2SO4 = BaSO4( + 2NaCl</p> <p>(8)</p> <p> y y</p> <p>Ho tan kt ta vo dung dch HCl c phn ng, BaCO3 tan.</p> <p>Na2CO3 + 2HCl = 2NaCl + CO2( + H2O</p> <p>Vy : BaSO4 = 3,495 g = 0,015mol</p> <p>Vy y = 0,015 mol ( mS = 0,48 g(S = 16(</p> <p> mC = 2,52 g(C = 84(a gam kt ta = 3,495 + (137 + 60) = 41,37 g</p> <p>b) Dung dch A gm: Na2CO3; Na2SO3; NaOH(d)</p> <p>( Na2CO3 ( = 0,21: 0,5 = 0,12M</p> <p>( Na2SO3 ( = 0,015: 0,5 = 0,03M</p> <p>( NaOH ( = = 0,6M</p> <p> Th tch Cl2 (ktc) tham gia phn ng:</p> <p>MCl2 = 1 . 0,3/2 ( VCl2 = 0,3 . 22,4/2 = 3,36 lt</p> <p>Bi 6. Cho 23,52g hn hp 3 kim loi Mg, Fe, Cu vo 200ml dung dch HNO3 3,4M khuy u thy thot ra mt kh duy nht hi nng hn khng kh, trong dung dch cn d mt kim loi cha tan ht, tip t t dung dch H2SO4 5M vo, cht kh trn li thot ra cho n khi kim loi va tan ht th mt ng 44ml, thu c dd A. Ly 1/2 dd A, cho dd NaOH cho n d vo, lc kt ta, ra ri nung ngoi khng kh n khi lng khng i thu c cht rn B nng 15,6g.</p> <p> a) Tnh % s mol mi kim loi trong hn hp.</p> <p> b) Tnh nng cc ion (tr ion H+, OH-) trong dung dch A.</p> <p>( Hng dn gii:</p> <p>a) Gi x, y, z l s mol Mg, Fe, Cu trong hn hp, ta c : </p> <p>24x + 56y + 64z = 23,52 ( 3x + 7y + 8z = 2,94 (a) </p> <p>ng cn d c cc phn ng:</p> <p>Cho e:</p> <p>Nhn e:</p> <p>Mg - 2e = Mg2+(1)NO3- + 3e + 4H+ = NO + 2H2O (4)</p> <p> Fe - 3e = Fe3+(2) Cu + Fe3+ = Cu2+ + Fe2+ (5)</p> <p>Cu - 2e = Cu2+(3)</p> <p>Phng trnh phn ng ho tan Cu d:</p> <p>3Cu + 4H2SO4 + 2NO3- = 3CuSO4 + SO42- + 2NO + H2O (6)</p> <p>T Pt (6) tnh c s mol Cu d: = = 0,165 mol</p> <p>Theo cc phng trnh (1), (2), (3), (4), (5): s mol cho bng s mol e nhn:</p> <p> 2(x + y + z - 0,165) = (3,4.0,2 - 2(x + y + z - 0,165)(.3</p> <p> ( x + y + z = 0,255 + 0,165 = 0,42 (b)</p> <p>T khi lng cc oxit MgO; Fe2O3; CuO, c phng trnh:</p> <p> .40 + .160 + . 80 = 15,6 (c)</p> <p>H phng trnh rt ra t (a), (b), (c): 3x + 7y + 8z = 2,94</p> <p> x + y + z = 0,42</p> <p> x + 2y + 2z = 0,78</p> <p>Gii c: x = 0,06; y = 0,12; z = 0,24.</p> <p>( lng Mg = 6,12(; ( lng Fe = 28,57(; ( lng Cu = 65,31(b) Tnh nng cc ion trong dd A (tr H+, OH-) </p> <p>(Mg2+( = = 0,246 M; (Cu2+( = 0,984 M; </p> <p>(Fe2+( = 0,492 M; (SO42-( = 0,9 M; (NO3-( = 1,64 M</p> <p>Bi 7. a) Tnh pH ca dung dch HCl nng 0,5.10-7 mol/lt.</p> <p>b) Tnh pH ca dung dch X c to thnh khi trn 200ml dung dch HA 0,1M (Ka = 10-3.75) vi 200ml dung dch KOH 0.05M; pH ca dung dch X thay i nh th no khi thm 10-3 mol HCl vo dung dch X.</p> <p>( Hng dn gii:</p> <p>a) ( H+(.0,5.10-7 do nng nh ( phi tnh n cn bng ca H2O </p> <p> H2O H+ + OH - HCl ( H+ + Cl -Theo nh lut bo ton in tch: </p> <p>( H+( = ( Cl-( + (OH-( ( ( H+( = 0,5.10-7 + </p> <p>( ( H+( 2 - 0,5.10 - 7( H+( - 10 -14 = 0. </p> <p>Gii c: ( H+( = 1,28.10-7 ( pH ( 6,9</p> <p>b) nHA = 0,1.0,2 = 0,02 mol; nKOH = 0,05.0,2 = 0,01 mol</p> <p> KOH + HA ( KA + H2O</p> <p> 0,01 ( 0,01( 0,01 Theo phng trnh HA cn d = 0,01 mol</p> <p>Trong d2 X: CHA = CKA = = 0,025M. Xt cc cn bng sau: </p> <p>H2O H+ + OH- KW = 10-14 (1) </p> <p>HA H+ + A- KHA = 10-375 (2)</p> <p>A- + H2O HA + OH- KB = KHA-1. KW = 10-10,25 (3) </p> <p>So snh (1) vi (2) ( KHA &gt;&gt; KW ( b qua (1) </p> <p>So snh (2) vi (3) ( KHA &gt;&gt; KB ( b qua (3) ( Dung dch X l dung dch m axit</p> <p>c pH = pKa + lg = 3,75 + lg = 3,75</p> <p>( Khi thm 10-3 mol HCl</p> <p> KA + HCl ( KCl + HA</p> <p> 0,001 ( 0,001 ( 0,001 (mol)</p> <p>(HA( = = 0,0275 M v (KA( = = 0,0225M.</p> <p>Dung dch thu c vn l dung dch m axit.</p> <p>Tng t, pH = 3,75 + lg = 3,66</p> <p>Bi 8. Ho tan 8,862 gam hn hp: Al, Mg trong 500ml dd HNO3 long thu c dd A v 3,316 lt (ktc) hn hp 2 kh khng mu c khi lng 5,18g trong c 1 kh b ho nu trong khng kh.</p> <p>a) Tnh thnh phn ( theo khi lng mi kim loi trong hn hp.</p> <p>b) C cn dd A c bao nhiu gam mui khan.</p> <p>c) Tnh nng mol/lt ca dd HNO3 tham gia phn ng.</p> <p>d) Ho tan dd A vo dd NaOH d tnh khi lng kt ta to thnh.</p> <p>( Hng dn gii:</p> <p>a) = = 37. Do MNO = 30 nn kh th 2 l N2O (c M = 44)</p> <p>Tm c nNO = nN2O = 0,07 mol Theo nh lut bo ton: Al - 3e = Al3+ Mg - 2e = Mg2+t s mol Al = x; Mg = y th tng s mol e nhng = 3x + 2y</p> <p>N+5 + 3e = N+2 2N+5 + 8e = 2N+9 tng s mol e thu = 3.0,07 + 8.0,07 = 0,77</p> <p>Ta c h phng trnh: 3x + 2y = 0,77 </p> <p> 27x + 24y = 8,862</p> <p>Suy ra: x = 0,042 ; y = 0,322</p> <p>(mAl = . 100% = 12,8% v %mMg = 87,2%...</p>