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{"id":25195,"date":"2018-04-04T01:11:05","date_gmt":"2018-04-03T18:11:05","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=25195"},"modified":"2018-04-04T01:11:05","modified_gmt":"2018-04-03T18:11:05","slug":"chuong-5-bai-24-dieu-che-kim-loai","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/chuong-5-bai-24-dieu-che-kim-loai\/","title":{"rendered":"Ch\u01b0\u01a1ng 5 – B\u00e0i 24: \u0110i\u1ec1u ch\u1ebf kim lo\u1ea1i"},"content":{"rendered":"

B\u00e0i 1 (trang 140 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0D\u00e3y c\u00e1c ion kim lo\u1ea1i n\u00e0o sau d\u00e2y b\u1ecb Zn kh\u1eed th\u00e0nh kim lo\u1ea1i?<\/strong><\/p>\n

A. Cu2+<\/sup>, Mg2+<\/sup>, Pb2+<\/sup><\/p>\n

B. Cu2+<\/sup>, Ag+<\/sup>, Na+<\/sup>.<\/p>\n

C. Sr2+<\/sup>, Pb2+<\/sup>, Cu2+<\/sup><\/p>\n

D. Pb, Ag+,Al3+<\/sup>.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

\u0110\u00e1p \u00e1n C<\/p>\n

\n
\n

B\u00e0i 2 (trang 140 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0Ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc n\u00e0o sau \u0111\u00e2y ch\u1ec9 \u0111\u01b0\u1ee3c th\u1ef1c hi\u1ec7n b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p \u0111i\u1ec7n ph\u00e2n?<\/strong><\/p>\n

A. Fe+ CuSO4<\/sub>\u00a0\u2192 FeSO4<\/sub>\u00a0+ Cu.<\/p>\n

B. CuSO4<\/sub>\u00a0+ H2<\/sub>O \u2192 Cu + O2<\/sub>\u00a0+ H2<\/sub>SO4<\/sub><\/p>\n

C. CuSO4<\/sub>\u00a0+ NaOH \u2192 Cu(OH)2<\/sub>\u00a0+ Na2SO4<\/sub>.<\/p>\n

D. Cu + AgN03<\/sub>\u00a0\u2192Ag + Cu(NO3<\/sub>)O2<\/sub>.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

\u0110\u00e1p \u00e1n B<\/p>\n

\n
\n

B\u00e0i 3 (trang 140 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0T\u1eeb m\u1ed7i h\u1ee3p ch\u1ea5t sau: Cu(OH)2<\/sub>, NaCl, Fe2<\/sub>S2<\/sub>. h\u00e3y l\u1ef1a ch\u1ecdn nh\u1eefng ph\u01b0\u01a1ng ph\u00e1p th\u00edch h\u1ee3p \u0111\u1ec3 \u0111i\u1ec1u ch\u1ebf kim lo\u1ea1i t\u01b0\u01a1ng \u1ee9ng. Tr\u00ecnh b\u00e0y c\u00e1c ph\u01b0\u01a1ng ph\u00e1p \u0111\u00f3.<\/strong><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

T\u1eeb Cu(OH)2<\/sub>\u00a0\u2192 Cu: Ph\u01b0\u01a1ng ph\u00e1p nhi\u1ec7t luy\u1ec7n<\/p>\n

Nung Cu(OH)2<\/sub>: Cu(OH)2<\/sub>\u00a0to<\/sup>\u2192 CuO + H2<\/sub>O (1)<\/p>\n

Kh\u1eed CuO b\u1eb1ng H2<\/sub>\u00a0: CuO + H2<\/sub>\u00a0to<\/sup>\u2192 Cu + H2<\/sub>O (2)<\/p>\n

* T\u1eeb NaCl \u2192 Na: Ph\u01b0\u01a1ng ph\u00e1p \u0111i\u1ec7n ph\u00e2n<\/p>\n

2NaCl (\u0111pnc) \u2192 2Na + Cl2<\/sub> (3)<\/p>\n

*T\u1eeb Fe2<\/sub>S2<\/sub>\u00a0\u2192 Fe2<\/sub>: Ph\u01b0\u01a1ng ph\u00e1p nhi\u1ec7t luy\u1ec7n<\/p>\n

\u0110\u1ed1t Fe2<\/sub>S2<\/sub>\u00a0trong oxi: 4Fe2<\/sub>S2<\/sub>\u00a0+11O2<\/sub>\u00a0\u21922Fe2<\/sub>2O3<\/sub>\u00a0+ 8SO2<\/sub>\u00a0(4)<\/p>\n

Kh\u1eed Fe2<\/sub>2O3<\/sub>\u00a0b\u1eb1ng Co Fe2<\/sub>2O3<\/sub>\u00a0+ Co \u2192 2Fe2<\/sub>\u00a0+ CO2<\/sub>\u00a0(5)<\/p>\n

B\u00e0i 4 (trang 140 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0\u0110i\u1ec1u ch\u1ebf Cu b\u1eb1ng c\u00e1ch \u0111i\u1ec7n ph\u00e2n dung d\u1ecbch Cu(NO3<\/sub>)2<\/sub>.<\/strong><\/p>\n

a. Tr\u00ecnh b\u00e0y s\u01a1 \u0111\u1ed3 \u0111i\u1ec7n ph\u00e2n.<\/strong><\/p>\n

b. Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh \u0111i\u1ec7n ph\u00e2n.<\/strong><\/p>\n

c. Cho bi\u1ebft vai tr\u00f2 c\u1ee7a n\u01b0\u1edbc trong qu\u00e1 tr\u00ecnh \u0111i\u1ec7n ph\u00e2n.<\/strong><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a, S\u01a1 \u0111\u1ed3 Catot(-) \u2190 Cu(NO3<\/sub>)2<\/sub>\u00a0dung d\u1ecbch \u2192 Anot(+)<\/p>\n

Cu2+<\/sup>,H2<\/sub>O\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 NO3<\/sub>–<\/sup>\\ H2<\/sub>O<\/p>\n

Cu2+<\/sup>\u00a0+ 2e \u2192 Cu<\/p>\n

2H2<\/sub>O \u2192 O2<\/sub>\u00a0+ 4H+<\/sup>\u00a0+ 4e<\/p>\n

b, Ph\u01b0\u01a1ng tr\u00ecnh \u0111i\u1ec7n ph\u00e2n<\/p>\n

2Cu(NO3<\/sub>)2<\/sub>\u00a0+ 2H2<\/sub>O \u2192 2Cu + 4HNO3<\/sub>\u00a0+ O2<\/sub><\/p>\n

c, H2<\/sub>O l\u00e0 ch\u1ea5t nh\u01b0\u1eddng e => Ch\u1ea5t kh\u1eed.<\/p>\n

d, N\u1ed3ng \u0111\u1ed9 c\u1ee7a Cu2+<\/sup>\u00a0gi\u1ea3m<\/p>\n

N\u1ed3ng \u0111\u1ed9 c\u1ee7a N03+ kh\u00f4ng thay \u0111\u1ed5i N\u1ed3ng \u0111\u1ed9 c\u1ee7a H+<\/sup>\u00a0t\u0103ng.<\/p>\n

B\u00e0i 5 (trang 140 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0C\u00f3 h\u1ed7n h\u1ee3p c\u00e1c b\u1ed9t kim lo\u1ea1i Ag, Cu. B\u1eb1ng nh\u1eefng ph\u01b0\u01a1ng ph\u00e1p h\u00f3a h\u1ecdc n\u00e0o ta c\u00f3 th\u1ec3 thu \u0111\u01b0\u1ee3c Ag t\u1eeb h\u1ed7n h\u1ee3p? Gi\u1ea3i th\u00edch v\u00e0 vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc.<\/strong><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Ng\u00e2m h\u1ed7n h\u1ee3p b\u1ed9t Ag – Cu v\u00e0o dung d\u1ecbch AgNO3<\/sub>\u00a0d\u01b0, l\u1ecdc l\u1ea5y ch\u1ea5t r\u1eafn l\u00e0 Ag.<\/p>\n

Cu + 2AgNO3<\/sub>\u00a0\u2192 Cu(NO3<\/sub>)2<\/sub>\u00a0+ 2Ag<\/p>\n

B\u00e0i 6 (trang 140 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0\u0110i\u1ec7n ph\u00e2n 200ml m\u1ed9t dung d\u1ecbch c\u00f3 ch\u1ee9a hai mu\u1ed1i l\u00e0 Cu(NO3<\/sub>)2<\/sub>\u00a0v\u00e0 AgNO3<\/sub>\u00a0v\u1edbi c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n l\u00e0 0,804 A, \u0111\u1ebfn khi b\u1ecdt kh\u00ed b\u1eaft \u0111\u1ea7u tho\u00e1t ra \u1edf c\u1ef1c \u00e2m th\u00ec th\u1eddi gian \u0111i\u1ec7n ph\u00e2n l\u00e0 2 gi\u1edd, ng\u01b0\u1eddi ta nh\u1eadn th\u1ea5y kh\u1ed1i l\u01b0\u1ee3ng c\u1ef1c \u00e2m t\u0103ng th\u00eam 3,44 g. H\u00e3y x\u00e1c \u0111\u1ecbnh n\u1ed3ng \u0111\u1ed9 mol c\u1ee7a m\u1ed7i mu\u1ed1i trong dung d\u1ecbch ban \u0111\u1ea7u?<\/strong><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

2 gi\u1edd = 7200 s<\/p>\n

G\u1ecdi th\u1eddi gian \u0111i\u1ec7n ph\u00e2n mu\u1ed1i b\u1ea1c l\u00e0 t1<\/p>\n

G\u1ecdi th\u1eddi gian \u0111i\u1ec7n ph\u00e2n mu\u1ed1i \u0111\u1ed3ng l\u00e0 t2<\/p>\n

=> t1 + t2 = 7200 (1)<\/p>\n

Theo \u0111inh lu\u00e2t Faraday: mAg<\/sub>\u00a0= (108.0,804. t1) : (96500) = 9.10-4<\/sup>. t1<\/p>\n

mCu<\/sub>\u00a0= (64.0,804. t2) : (2.96500) = 2,666.10-4<\/sup>. t2<\/p>\n

m\u00e0 mAg<\/sub>\u00a0+ mCu<\/sub>\u00a0= 3,44 (g) => (9 t1 + 2,666 t2 ). 10-4<\/sup>\u00a0= 3,44 (2)<\/p>\n

(1),(2) => t1 = 2400 (s) => mAg<\/sub>\u00a0= 2,16 gam => nAg<\/sub>\u00a0= 0,02<\/p>\n

t2 = 4800 (s) => mCu<\/sub>\u00a0= 1,28 gam => nCu<\/sub>\u00a0= 0,02<\/p>\n

CMCu(NO3)2<\/sub>\u00a0= 0,02 : 0,2 = 0,1M ; CMAgNO3<\/sub>\u00a0= 0,02 : 0,2 = 0,1 M<\/p>\n

B\u00e0i 7 (trang 140 sgk H\u00f3a 12 n\u00e2ng cao):\u00a07. \u0110i\u1ec7n ph\u00e2n ho\u00e0n to\u00e0n 3,33 gam mu\u1ed1i clorua c\u1ee7a m\u1ed9t kim lo\u1ea1i nh\u00f3m IIA, ng\u01b0\u1eddi ta thu \u0111\u01b0\u1ee3c 6,72 l\u00edt kh\u00ed clo (\u0111ktc). H\u00e3y x\u00e1c \u0111\u1ecbnh t\u00ean c\u1ee7a mu\u1ed1i clorua kim lo\u1ea1i.<\/strong><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc c\u1ee7a ph\u1ea3n \u1ee9ng<\/p>\n

MCl2<\/sub>\u00a0\u0111pnc<\/sup>\u00a0\u2192 M + Cl2<\/sub><\/p>\n

S\u1ed1 mol nMCl2<\/sub>\u00a0= nCl2<\/sub>\u00a0= 6,72 : 22,4 = 0,3 (mol)<\/p>\n

Kh\u1ed1i l\u01b0\u1ee3ng mol MCl2<\/sub>\u00a0: MMCl2<\/sub>\u00a0= 33,3 : 0,3 = 111 => M + 71 = 111 => M = 40 (Ca)<\/p>\n

Mu\u1ed1i \u0111\u00e3 d\u00f9ng l\u00e0 CaCl2<\/sub>\u00a0: canxi clorua<\/p>\n

B\u00e0i 8 (trang 140 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0\u0110i\u1ec7n ph\u00e2n m\u1ed9t dung d\u1ecbch AgNO3<\/sub>\u00a0trong th\u1eddi gian 15 ph\u00fat v\u1edbi c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n l\u00e0 5 ampe. \u0110\u1ec3 l\u00e0m k\u1ebft t\u1ee7a h\u1ebft ion Ag+ c\u00f2n l\u1ea1i trong dung d\u1ecbch sau \u0111i\u1ec7n ph\u00e2n, c\u1ea7n d\u00f9ng 25 ml dung d\u1ecbch NaCl 0,4M.<\/strong><\/p>\n

a. Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh \u0111i\u1ec7n ph\u00e2n v\u00e0 ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc c\u1ee7a ph\u1ea3n \u1ee9ng x\u1ea3y ra.<\/p>\n

b. T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng Ag thu \u0111\u01b0\u1ee3c \u1edf catot.<\/p>\n

c. T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng AgNO3<\/sub>\u00a0c\u00f3 trong dung d\u1ecbch ban \u0111\u1ea7u.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a. S\u01a1 \u0111\u1ed3 \u0111i\u1ec7n ph\u00e2n v\u00e0 c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh h\u00f3a h\u1ecdc \u0111\u00e3 x\u1ea3y ra :<\/p>\n

S\u01a1 \u0111\u1ed3 :<\/p>\n

Catot(-)\u2190 Cu(NO3<\/sub>)2<\/sub>\u00a0dung d\u1ecbch \u2192 Anot(+)<\/p>\n

Ag+<\/sup>, H2<\/sub>O \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 NO3<\/sub>–<\/sup>, H2<\/sub>O<\/p>\n

Ag+<\/sup>\u00a0+ e \u2192 Ag<\/p>\n

2H2<\/sub>O \u2192 O2<\/sub>\u00a0+ 4H+<\/sup>\u00a0+ 4e<\/p>\n

4AgNO3<\/sub>\u00a0+ 2H2<\/sub>O \u2192 4Ag + 4HNO3<\/sub>\u00a0+ O2<\/sub>\u00a0(\u0111pdd)<\/p>\n

AgNO3<\/sub>\u00a0+ NaCl \u2192 AgCl \u2193 + NaNO3<\/sub><\/p>\n

b. mAg<\/sub>\u00a0= (108 x 5 x 15 x 60) : 96500 = 5,04 gam<\/p>\n

c. nNaCl<\/sub>\u00a0= 0,025 x 0,4 = 0,01 mol<\/p>\n

nAg<\/sub>\u00a0= 5,04 : 108 mol<\/p>\n

Theo (1) nAgNO3<\/sub>\u00a0= nAg<\/sub>\u00a0\u2248 0,0466<\/p>\n

Theo (2) nAgNO3<\/sub>\u00a0= nNaCl<\/sub>\u00a0= 0,01<\/p>\n

=> nAgNO3<\/sub>\u00a0ban \u0111\u1ea7u \u2248 0,0566 => Kh\u1ed1i l\u01b0\u1ee3ng AgNO3<\/sub>\u00a0ban \u0111\u1ea7u : 0,0566.170 \u2248 9,62 gam.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

B\u00e0i 1 (trang 140 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0D\u00e3y c\u00e1c ion kim lo\u1ea1i n\u00e0o sau d\u00e2y b\u1ecb Zn kh\u1eed th\u00e0nh kim lo\u1ea1i? A. Cu2+, Mg2+, Pb2+ B. Cu2+, Ag+, Na+. C. Sr2+, Pb2+, Cu2+ D. Pb, Ag+,Al3+. L\u1eddi gi\u1ea3i: \u0110\u00e1p \u00e1n C B\u00e0i 2 (trang 140 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0Ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc n\u00e0o […]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1306],"tags":[1436,1435],"yoast_head":"\nCh\u01b0\u01a1ng 5 - B\u00e0i 24: \u0110i\u1ec1u ch\u1ebf kim lo\u1ea1i<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/chuong-5-bai-24-dieu-che-kim-loai\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Ch\u01b0\u01a1ng 5 - B\u00e0i 24: \u0110i\u1ec1u ch\u1ebf kim lo\u1ea1i\" \/>\n<meta property=\"og:description\" content=\"B\u00e0i 1 (trang 140 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0D\u00e3y c\u00e1c ion kim lo\u1ea1i n\u00e0o sau d\u00e2y b\u1ecb Zn kh\u1eed th\u00e0nh kim lo\u1ea1i? A. Cu2+, Mg2+, Pb2+ B. Cu2+, Ag+, Na+. C. Sr2+, Pb2+, Cu2+ D. Pb, Ag+,Al3+. L\u1eddi gi\u1ea3i: \u0110\u00e1p \u00e1n C B\u00e0i 2 (trang 140 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0Ph\u1ea3n \u1ee9ng h\u00f3a h\u1ecdc n\u00e0o […]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/lop12.edu.vn\/chuong-5-bai-24-dieu-che-kim-loai\/\" \/>\n<meta property=\"og:site_name\" content=\"Lop12.edu.vn - C\u1ed9ng \u0111\u1ed3ng h\u1ecdc sinh l\u1edbp 12 l\u1edbn nh\u1ea5t t\u1ea1i Vi\u1ec7t Nam\" \/>\n<meta property=\"article:published_time\" content=\"2018-04-03T18:11:05+00:00\" \/>\n<meta name=\"author\" content=\"H\u00e0 Trang\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"H\u00e0 Trang\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"5 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"WebPage\",\"@id\":\"https:\/\/lop12.edu.vn\/chuong-5-bai-24-dieu-che-kim-loai\/\",\"url\":\"https:\/\/lop12.edu.vn\/chuong-5-bai-24-dieu-che-kim-loai\/\",\"name\":\"Ch\u01b0\u01a1ng 5 - B\u00e0i 24: \u0110i\u1ec1u ch\u1ebf kim lo\u1ea1i\",\"isPartOf\":{\"@id\":\"https:\/\/lop12.edu.vn\/#website\"},\"datePublished\":\"2018-04-03T18:11:05+00:00\",\"dateModified\":\"2018-04-03T18:11:05+00:00\",\"author\":{\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/2da855e5d06b553ac0065157cc261a45\"},\"breadcrumb\":{\"@id\":\"https:\/\/lop12.edu.vn\/chuong-5-bai-24-dieu-che-kim-loai\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/lop12.edu.vn\/chuong-5-bai-24-dieu-che-kim-loai\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/lop12.edu.vn\/chuong-5-bai-24-dieu-che-kim-loai\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/lop12.edu.vn\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Ch\u01b0\u01a1ng 5 – B\u00e0i 24: \u0110i\u1ec1u ch\u1ebf kim lo\u1ea1i\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/lop12.edu.vn\/#website\",\"url\":\"https:\/\/lop12.edu.vn\/\",\"name\":\"Lop12.edu.vn - C\u1ed9ng \u0111\u1ed3ng h\u1ecdc sinh l\u1edbp 12\",\"description\":\"\",\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/lop12.edu.vn\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"Person\",\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/2da855e5d06b553ac0065157cc261a45\",\"name\":\"H\u00e0 Trang\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/545a8100db1147f314111301d261dc33?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/545a8100db1147f314111301d261dc33?s=96&d=mm&r=g\",\"caption\":\"H\u00e0 Trang\"},\"url\":\"https:\/\/lop12.edu.vn\/author\/trangvth2\/\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"Ch\u01b0\u01a1ng 5 - B\u00e0i 24: \u0110i\u1ec1u ch\u1ebf kim lo\u1ea1i","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/lop12.edu.vn\/chuong-5-bai-24-dieu-che-kim-loai\/","og_locale":"en_US","og_type":"article","og_title":"Ch\u01b0\u01a1ng 5 - B\u00e0i 24: \u0110i\u1ec1u ch\u1ebf kim lo\u1ea1i","og_description":"B\u00e0i 1 (trang 140 sgk H\u00f3a 12 n\u00e2ng cao):\u00a0D\u00e3y c\u00e1c ion kim lo\u1ea1i n\u00e0o sau d\u00e2y b\u1ecb Zn kh\u1eed th\u00e0nh kim lo\u1ea1i? 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