KY ̀ THI OLYMPIC TRUYÊ ̀ N THÔ ́ NG 30/4 L ̀ N THƯ ́ XIII TA ̣ I THA ̀ NH PHÔ ́ HUÊ ́ ĐÊ ̀ THI MÔN HÓA 10 Chu ́ y ́ : ! " # $ %& # '%&& ( Câu I : I.1)*+,-./*+,0123 45'&6789: 4;<= 5>1*'&?9@A BC$6: +D E: + 1)F4/<$G k k I I + 0 I I 0 H I I H - I I - I I I I J I I X 1,94 4,31 1,31 1,26 1,30 Y 2,17 1,96 1,35 6,08 1,25 KLLMNO5MP)F4/9 I.2Q?+BR<SRTL<U9 2.1 V'NRWT$X1N4'9 2.2Y$6 D F4 Z %[$'%$6?>% T$X9 2.3)5MP5+Y 1 D 9 R \-.HJE H ]% Z \.-^ ]\JH.I]\HI.I9_B\J.0H9 0H 9 I.3`%%V7OT4%Ra b$G U 0H c0 → α → − β Q → − β ` → α → α d eBM3'M152>1b%&9 Câu II: II.1 % f+RYI H >g%h97fMBI O2'% 2>G 0gi%j0i+jDg 0 i+j k$l+m4- 1.1Yn Q 12> 1.2 Y+6<=ol1p O3l'MNB4YA4'9 \09 II.2 65'ai 0 J j $2q4+Yg 0 F4 0 giojX0Ir9 2.1 eB<%fA 5812>O2'%9V'O5MPAf4a F4;<=&+B\g9_B+M65' a 2%<=A4IJ.Ins9e4G t ins9 Z j K&+B ;<=&+B ins9 Z j g 0 ZHcH.I u H-v 0 gij Z0I. u -H g 0 ug -J- g\g -cI 2.2Q2>7twr\Z-Jv.Iins9 Z j9V'YM*B&a% 12>MV a MFPs9 Z 9n Z 9 Câu III: III.1&RRP 4 NH SCN .F4 RRP 3 Fe + .F4 F − 974 Mo1> 2+ FeSCN '+x_B%m4COlA+ 2+ 6 FeSCN C 7.10 M − > F4R RP M<= O 7 M1 MN $@ T > R% O 1 ya i:::j O2' % + M5+N9 H. H ay H F β − − = ] 0 H.H az F β + = i β 4m$6{j9 III.25543m% b=[ Ag + .9 ZH ] 3 NH .F4*9 3 2 7,24 2 Ag( NH ) 10 β + = ] 3 2 12,03 4Cu( NH )4 10 β + = ] 2 0 0 Ag / Ag Cu / Cu E 0,799V;E 0,337V + + = = iX0I j Câu IV: IV.1_BB O7+?&qG | 0D E D \D.Je | ya HD Eya 0D \D.vve | ^ D E^\D.e | D E\D.I0e | ya 0D Eya\Z.--e | : 0 E0: Z \D.I-e V' BA<=fO2'%% 5%<}=$G 1.1 *$hF4 RRP$hi:::j$~ 1.2 *M[F4 RRPM[i::j$~ 1.3 RRPT%F4 RRP$hi::j% 1.4 RRP$hi:::j%F4 RRP+ IV.2 4v.0)g H F4 <SM<=RRP^9ARRP^FSMA @% ZB}MA4'fM<=+ TT F4.vc0Y+YiM+jT ZB}MA40'fM<=.IJY+YiM+j )5MP)F4Y}B:\.cH^9 Câu V: V.1 65' 4 4-.-$~1+ Ti>zj% OR<9l%h$ 2>Ma 4% <=F•M1RRPg H Hv.€l'[M*3%;1 6% RRPM<=4-.v0€9n4TRRP4'f 5%.6%h9 K856%hl'[M*3%;16% RRP4H-.v€9)5MP> 6%h9 V.2 eB5<%f2>O2'%G 2.1 : : Z % n:P O 54: 0 Xya H .g H ]•: 0 O 5M<= 0 z 0 g H 9 2.2: _% Z P O 5X 0 zg -MU ._%g H Z i%<}Oj]•_% 0 T O 5M<=Q4O <>9 2.3 0 g 0 P+?%g 0 i% %<}‚jF4P O 5% RRPng - i% %<}Oj9 Học sinh không được sử dụng bảng tuần hoàn. ---------- Hết ---------- ĐÁP ÁN Đáp án câu 1: 6FS).•: 0 &: H ;M**.FL' ) 0D 7lf1*+YBR M7G )4ƒ^%„-$ 0 iOj (0,5 đ) 6FS/.•: - &: I ;M**.FL' / -D 7lf1*+YBR M7G /4ƒa„0$ 0 0 0 i j (0,5 đ) I.2Q?+BR<SRTL<U9 2.1 V'NRWT$X1N4'9 2.2Y$6 D F4 Z %[$'%$6?>% T$X9 2.3)5MP5+Y 1 D 9 R \-.HJE H ]% Z \.-^ ]\JH.I]\HI.I9_B\J.0H9 0H 9 w2G I.2.19(0,5 đ) 2.2(0,75 đ)efL<U& Z XMCG =× Z JUG H 0 J =× Z D X…0TG H - 0 =× D XGO\ D eL'$6?% T$X4- D D- Z \ - 2.3 (0,50 đ) VN MN d A CuCl 9 9 = FSe\ H iG$6?.4TfL<j , ., .,., ),,(, . . H0- 0H ^ H ^-vI cJII 0HJHJ- IHIIJH- R =⇒ = + ==⇒ − i0,25 đ) U+5a fF†7\0% D D0% Z o A ra r JII. 0 -.90-v.I 0 0 = − = − =⇒ − + (0,25 đ) I.3. Z D ⇒ - Z ⇒ - D U 0H c0 → Th 0H- c D He - 0 0,25 Th 0H- c → Pa 0H- c D e − 0,25 Pa 0H- c → U 0H- c0 D e − 0,25 U 0H- c0 → Th 0H c D He - 0 0,25 Th 0H c → Ra 00J D He - 0 0,25 Đáp án câu 2: 1.1(1 đ)0gi%j0i+jDg 0 i+j ƒ„ ƒ„ u0O 0O O 0 0 H 0 H 0 - -9- 9 c.- H H 0v 0v p Hg O K P P P P P = = = = = ÷ 192(1 đ)9z6 ol+\0O9•> g -9.I H .I .09vvH e'\.0 .090J -.IH PV n x x RT g = = = → = = = II.2. 65'ai 0 J j $2q4+Yg 0 F4 0 giojX0Ir9 2.1 eB<%fA 5812>O2'%9V'O5MPA f4aF4;<=&+B\g9_B+M65' a 2% <=A4IJ.Ins9e4G t ins9 Z j K&+B ;<=& +Bins9 Z j g 0 ZHcH.I u H-v 0 g Z0I. u -H g 0 ug -J- g\g -cI 2.2Q2>7twr\Z-Jv.Iins9 Z j9V'YM*B&a% 1 2>MV a MFPs9 Z 9n Z 9 Giải: 2.1. 0 J D 0 v g 0 → 0g 0 D H 0 gt\Z IJ.Ins i0 0 J D vg 0 → -g 0 D J 0 gt\ZH0 nsj 0,5 t < \-t g 0 D Jt 0 gZvt g 0 Z0t 0 J t 0 J \ ( ) ( ) ( ) [ ] 0 H0.0IJI.HcH- −−−+− \ZH.cins9 Z j 0,5 t < \0| u D0| u Dv| g\g Z| \g Z0| ug | \ g \ ( ) [ ] H0-J-O0-cIOv-HO0H-vO0 −−−++ \ HHins9 Z j 0,5 2.2 ∆wr\∆rZ∆zr ∆zr\ ( ) [ ] ( ) 0vH0I I.-JvI.IJ + −−− \Z.H0i+s9 Z n Z j\ZH0 s9 Z .n Z 0,5 Đáp án câu 3: III.1. 7G 3 Fe C + ‡‡ F C ( 1) − = H yay β %lS9 efFL'% RRP.ya HD 5RˆBFSy Z T %> 3 FeF 9 3+ 3 Fe 3F FeF − + → _M3 . z2>‰‰.cv. 0.5đ z+%*FS 4 NH SCN G 3 FeF C \I9 ZH ] F C − \.-I] 2 SCN C 5.10 M − − = yay H ya HD DHy Z ZH. ya HD Dz Z yaz 0D DH.H yay H Dz Z yaz 0D DHy Z n\ Z.v 0,5 đ I9 ZH I9 Z0 .-I ƒ„iI9 ZH ZOj iI9 Z0 ZOjO.-IDHO 3 10,07 3 2 x(0,485 3x) 10 (5.10 x)(5.10 x) − − − + ⇒ = − − 0.5đ eSO‡‡I9 ZH M<=G ( ) JH H vI OvOJ -I OO0I O −− −− <== , , , 0,5 đeL'4Mo1> 2+ FeSCN +OlA.Š4y Z MVa 4 4 ya HD III.2. 5‹5%fO2'%G ZT > 2+ 3 Ag(NH ) i + > ^ H j ^ D D0 H ^i H j 0 D 7,24 2 10 β = .9 ZH . ‰‰.Z0.9 ZH .9 ZH 0.5đ Zn? 2+ 3 Ag(NH ) XG 0O^i H j 0 D ^ D D0 H v.0- 0 β − − = ij 0^ D D0^D 0D I.J K = i0j ZT >1 0D FS 3 NH i + > 0 H j 0D D- H i H j - 0D 0.H - β = iHj Œ=iji0jF4iHjG 0^i H j 0 D D0^Di H j - 0D ] - 0 0 nn ββ= − \ H.J 0,5 đ .9 ZH ZZZZZI.9 Z- QwG 2 3 4 Cu(NH ) + GI.9 Z- ] 3 3 NH :1,0 2.10 1, 0M − − ≈ m i H j - 0D D0^0^i H j 0 D D ZH.J I.9 Z- ƒ„I.9 Z- ZO0O 2 13,16 4 (2x) 10 (5, 0.10 x) − − ⇒ = − O\I. Z- 0O\ -0HJH- OIOI −−−− <= ,, . 0,5 đ eL'G + -8,23 9 3 2 [Ag(NH ) ]=2x=10 5, 9.10 M − = 2+ -4 3 4 [Cu(NH ) ]=5,0.10 M 0.5đ UR•^ D [TR<SRT> 3 2 Ag(NH ) + <FŽP+? 4 49 Đáp án câu 4: IV.1. 99ef| ya HD Eya 0D \D.vve•| ya 0D Eya\Z.--e Y O7Gya HD Tya 0D Y+?GyaTya 0D Q2>O2'% H 0 0 ya ya H ya + + + → •RP4F4'N$ˆT (0.5đ) 909ef| D E\D.I0e •| 0D E D \D.Je Y O7G D T 0D Y+?G D T Q2>O2'% 0 + + + + → + • M72>P+O2'%Š4 *M[F4 RRP zg - +7A<=f (0.5đ) 9H9ef| ^ D E^\D.e•| ya HD Eya 0D \D.vv Y O7G^ D Tya HD Y+?Gya 0D T^ Q2>O2'% 0 H ya ^ ya ^ + + + + → + •RP4ˆT'N$4F4 (0.5đ) 9-9ef| ya HD Eya 0D \D.vve•| : 0 E0: Z \D.I-e Y O7Gya HD T: 0 Y+?G: Z Tya 0D Q2>O2'% H 0 0 0: 0ya : 0ya − + + + → + •RP+4'N$4 (0.5đ) IV.2 ARRP^G(2đ) H H )g ) g + − → + ¬ ‘ G 0 gu0a →0 D D’g 0 ‘ G) D Da →) FS0'.$6 g 0 \0O.vc0E00.-\.90‡.IJE00.-\.0I 0,5 đ eL'X 7+Y 0 5%G.0IZ.J\.c >o) D MVP+?B ‘ G) D Da →) 0 0 gD0a →0g Z D 0 ‘ G 0 gu0a → 0 D D’g 0 0,5 đ a '&hmaa% LX0MA@G D.c90\.909- iFS4$6 1)g H j ⇒ \.-J '\.-JM<=)\i^j 0,5 đ FS}$'%$6 aa% % MŒG H0 0 J- cJI : , , == cJI9.H0 J ' .cH = = 0,5đ Đáp án câu 5: efg 0 R<&7 5%P l% O 0zDi0D:0jg 0 0 g D0zg 0 (0,25 đ) .I 0 g D0g H 0ig H j D 0 g (0,25 đ) .I n6<=RRPg H \×JH×: Hv.\I: Hij n6<=RRP$2> \DDI: Hij 7iDJ0j: iDI0-: Hj\.-v0 &\.JI (0,50 đ) 8\Hz'%\IJiyaj 7GiDH0j\-.-z'%\.I +6<=yaig H j H 4 \.I×0-0\0.ij n6<=RRP$+6+BG RR \DI0-: Hu.\0.c0ij n6<=yaig H j H •T% RRP4G \0.c0×H-.v: \v.0Ic0-ij n6<=yaig H j H +B \0.Zv.0Ic0-\-.-ij (0,50 đ) U>yaig H j H 9 0 g z'%-.-:0-0×i0-0Dj\.z'%\c yaig H j H 9c 0 g (0,50 đ) V.2. eB5<%f2>O2'%G 2.19: : Z % n:P O 54: 0 Xya H .g H ]•: 0 O 5M<= 0 z 0 g H 9 2.29: _% Z P O 5X 0 zg -MU ._%g H Z i%<}Oj]•_% 0 T O 5 M<=Q4O<>9 2.39 0 g 0 P+?%g 0 i% %<}‚jF4P O 5% RRP ng - i% %<}Oj Giải:V. 2 (Mỗi phương trình 0,25 đ) 2.10n:D0ya H 0ya 0 D0nD: 0 0n:Dg H D 0 g0ngDg 0 D: 0 : 0 D0 0 z 0 g H 0:D 0 z - g J 2.20_% Z D- D Dzg - 0Z iMUj_% 0 Dzg 0 D0 0 g I_% Z D_%g H Z DJ D H_% 0 DH 0 g I_% 0 D0QD 0 g_%D0 H Qg - 2.3 H 0 g 0 D0%g 0 D0g0 0 %g - D- 0 g I 0 g 0 D0ng - DH 0 zg - 0zg - Dn 0 zg - DIg 0 D 0 g . KY ̀ THI OLYMPIC TRUYÊ ̀ N THÔ ́ NG 30/4 L ̀ N THƯ ́ XIII TA ̣ I THA ̀ NH PHÔ ́ HUÊ ́ ĐÊ ̀ THI MÔN HÓA 10 . Jt 0 gZvt g 0 Z0t 0 J t 0 J ( ) ( ) ( ) [ ] 0 H0.0IJI.HcH- −−−+− ZH.cins9 Z j 0,5 t <