Tiêu đề Đáp án đề tham khảo - Môn Hóa học.pdf ✅

s0 cIeo DUC vA DA TRUONG DAI HQC SII oAp aN - DE THI (Ddp dn - Thang ri rru cHgN HSG THpr cAp rntloxc DHSr HA NeI NAu :aec zoz4 - 2o2s ruON: HoA Hec Hgc H Thang iIi6m Cflu Dip 6n :z 0,25 f) a) Khi trQn ur6 v6i vdi: (NH2)2CO + 2H2O-+ (NH4)2CO3 CaO + H2O -r Ca(OH)2 Ca(OH)z + (NH4)2CO: + CaCO3J + 2NH3t+ zHzO Lim mAt di lusng dam (do tho6t ra NHr) vi lim dAt rin lai (do t4o ra CaCO3). 0,25 b) Vi kim lo4i Mg ch6y trong khi CO2: 2Mg + CO2 -5 2MgO + C 2) a) Na2S + CuClz 0,25 -r 2NaCl + CuS J b) SO2 + Br2 + 2H2O 0,25 -+ 2HBr + H2SOa c) NazCO: + H2SOa 0,25 -+ NazSOo + H2O + CO21 d) CO2 + Ca(OH)2 0,25 -r CaCOr, + H2O e 4F +02+ --r 4F 0,25 0,25 u, Mg chiy m4nh do phin ring v6i 02 cta kh6ng khi f) Lirc to 0,25 Khi 02 da Mg chSy im i do phan ring v6i N2 a J +N --) to 0,25 sau phin ring xanh gi6y qu $S:Nz vi MgO, MgrN, phin ring v6i nudc t4o khi NH: lirm y tim dm: N + 6HrO J +ZNH 1 -> 0,25 Ciu I (4,5 d) 2No2 *1o, + 2NaoH---+ 2NaNor + H2o (1) '2 Co thiS c6: 2NO2 + 2NaOH-+ 2NaNO3 +NaNOz + H2O (2) I(hi tho6t ra chi lir N2 g)D hqp khi viro dung dich NaOH du: Vi dung dich Y lirrn m6t mau dung dich KMnO+/HzSo+ nen D phni co N,aNO2 ---) Y NaN NaN NaOH du -) co ra dun Trang ll7 aar * 8H* 2Mg+ Oz+ ?IMoO
0,25 - Dung dich Y + dung dich KMnOa/H2SOa: 2NaOH + H2SOa---+ Na2SOa + 2H2O (3) 5NaNO2 +2KMnO+ + 3H2SO4 ---+ 5 NaNOr+ 2MnSOa +K2SOa +3H2O (4) Mdu dung dich KMnOa mit do (4) - dung dich Z gdm: NaNO3, MnSO+, K2SOa, Na2SOa, HzSO+ du. - Dung dtchZ * Cu * H2SOa: 3Cu + 2NaNO3 + 4 H2SO4 --- 3CuSO+(xauh) f Na2SOa + 2NO + 4H2O 2NO -' Oz - 2NO2 Kh6ng mdu NAu do 0,25 3) a) Nguy6n tri hydro gen (Z: 1) d tr4ng th6i ring v6i n : 5 c6 th6 chuy€n v6 c6c mric ring v6i il : l, 2,3, 4. Bu6c s6ng tuong img v6i cilc chuy6n dich niy li: ^E = + = Es - En = 8,6(i-+)" r, 602x ro-'|e 25n2 " et-;, ^ 6.625x10-rox3xl08 25n2 13, 6 x l, 602 x l0-'' (25 - n') -,_ hc '" 13,611,602 x 10-'e 0,25 ).,,o* ftng voi chuy6n dlch n : 5-) n:4: l,n u* :405414 nm 1,,"i, ring voi chuyiSn dich n:5-+ n:1il,nin:95102 nm 0,25 b) Nguy0n tu hydrogen khi chuy6n tir trpng th6i kich thich ring voi n :5 vO tr4ng th6i img v6i n:2 thl ph6t ra birc xa c6 ndng luo-ng: n : 5 -+ n: 2'.Es - Ez : -2,18x10-rs r(+-j) 0,25 V6i ion He+ (2,11":2): khi chuy6n tu trang th6i kich thich ring voi n" th6i fng yor nts6 ph6t ra 6nh s6ng co n6ng lucrng gi6ng nhu vfy tuc E =-13,6x22 xl,6O2xlO-te "r*-i, : -2,18x 22 xt0-t8 "r* - *r= -2,18x l0-r8 x ( .l l. I .l__1\= I _1 -*t- 2)=E*l?-V)= tt-T He*chuy6n til mri'c n : 10 vd n : 4,hay a : 10, b = 4. vd t 4ng ld: v0v I l. -_-l 52 2" 0,25 Li* + le -) Li Et : - Ir : - 5,390eV c) Do 19 eV Li -+ + le 11 :5,390 eV n6n 2+ L,t L + 2e 8 I eV -) + + I 5 90 7 a -) I e E 8 9 J ,00 Trang2lT tr^:
E3(Li3*):- 13,6 "4 =E3 (Li3*):-13,6x I:-rr2,a (eV) Li2* --> Li3*+ le Li -+Li2* + 2e +Li+Li3*+3e Is: - E: :122,4 eY E2 : 81,009 eV E: I: *Ez:203141eY Li2- cfrng ld h0 1e rnQt h4t nhdn: 0,25 T6ng tli6m: 415 1) S : 3,14.(0,2)2 :0,7256 m2 :1256 c-' = Vag: 1256.0,001 : 1,256 cm3 3 mAc : 1,256.10,5 : 13,188 gam = no*: 13,188/108 : 0,1222 mol 0,5 Ta c6: Q=n".F=I.t=, = 0,5 n.jt -0'1222'96500 =5896,5 (s)=1,64(h) I2 0,5 2) a) nH : no : nS : 3',59 ,t'.'!' ]o:1^t - 8:8:l 11632 VQy c6ng thtic ctra A le X(H3O8S)" 0,5 + 57,41o/o + l4,35yo) :24,65oh. V6i n: I =Mx: :Lx24,65 :54,g6g.mol-r 3 X li Manganese (Mn) 14,35 V6i n :2 )Mx : :Lx24,65 : 109,94 g.mol-r = Kh6ng co kim loai ndo 14,35 th6a min. V6i n ) 3 =M" > 164,9 g.mol-r =X thuQc hg lantan ho4c ph6ng xa (Lo4i) Vfly c6ng thfc cua A ld MnHsOsS (trong A chi chira I nguyCn tu kim lopi Mn) 0,5 [ft ma mong tan trong HCl, mir trong A co 1 nguy€n tu S n6n A ld mu6i sulfate MnHsOaSOa. - Vi khi dun n6ng A (chua phdn huy A) thi 32,2gyo khOi luo,ng A m6t di, chinh ld luqng nudc k6t tinh. Ta c6 Ma :222,95 ) 32pgyoMe = 72 g.mol-l tuong ducrng v6i 4 phdn tir nu6c. Vay C6ng thric cua A ld MnSO 4.4HzO - Vi X t5c dung vcri BaCl2 tao 0,25 b) Phucrng trinh phin ring: MnSO+ + BaCl2 + MnCl2 + BaSO+l 2MnSOa+ 5PbO2 + 6HNO3 --+ 2HMnO4 + 3Pb(NO3)2 + 2PbSO4! + 2H2O 0,25 HiQn tuqng quan siit dugc c6 ki5t tua tring *u6t hign. trong thi nghiQm 1: Dung dich chuy0n sang miu tim, T6ng tli6m: 3 Cdu2 (3 d) 1) a) 2FeO (s) 0,25 a----+ 2Fe(s) + Oz (g) (l) Kp, : Po2:4,13.10-2t bar AG:RT 0,25 ln Ke : -8,3 I 4. 1 ooo. ln(4, I 3. 1 o AG >0; phdn irng t.u di6n bi6n theo chi6u nghlch z\ -a9s2r<1 Ciu 3 (4,5 d) Trang3lT
b) Feo(s) + Co(e) r----+ Fe(s) + Co2(g) (2) C6: K ,, - Pco' = x"' Y' =*+ = 0,754 Pco xt-, 0,57 0,25 c) Phuong trinh phAn huy CO2 ld:2COz (g) 0,25 r+ 2CO(g) + Or(g) (3) 0,25 X6t cdn bing: 2COz(g) i-+ 2CO(g) + Oz(g). Gqi o ln dQ phnn huy cua CO2. Ta c6: Kp(r=ff;y o, K. - K p(tt - 4,13.10'2t - 7,26.10-21 bar-l 0 Po.ct Po 'ct 0 P6.0,5u P6.0,5u Po ' Po'o Cdn bdng po.(1- o) Klr, 0,7542 0,25 Vi hing sd phAn huy r6t nh6 n6n coi nhu o kh6ng