Ôn tập chương 2

Home Lớp 12 Hoá học Lớp 12 - Sách KNTT Bài 7: Ôn tập chương 2: CarbohydrateBài tập nâng cao

{"common":{"save":0,"post_id":"10100","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"9047","post_id":"10100","mon_id":"1160346","chapter_id":"1159849","question":"\u0110em thu\u0309y phân 1 kg khoai ch\u01b0\u0301a 20% tinh bô\u0323t trong môi tr\u01b0\u01a1\u0300ng acid. Nê\u0301u hiê\u0323u suâ\u0301t pha\u0309n \u01b0\u0301ng la\u0300 75% thi\u0300 l\u01b0\u01a1\u0323ng glucose thu \u0111\u01b0\u01a1\u0323c la\u0300:<\/p>","options":["A. 166,67 gam","B. 145,70 gam","C. 210,00 gam","D. 123,45 gam"],"correct":"1","level":"3","hint":"G\u1ee3i ý: Tinh b\u1ed9t th\u1ee7y phân hoàn toàn → glucose v\u1edbi t\u1ec9 l\u1ec7 kh\u1ed1i l\u01b0\u1ee3ng 162 g tinh b\u1ed9t → 180 g glucose.<\/p>","answer":"\u0110áp án \u0111úng: A. 166,67 gam<\/strong><\/span><\/p>Gi\u1ea3i:<\/strong><\/p><ul data-end=\"279\" data-start=\"114\"> <li data-end=\"174\" data-start=\"114\">Kh\u1ed1i l\u01b0\u1ee3ng tinh b\u1ed9t trong 1 kg khoai: 1000 × 20% = 200 g<\/p><\/li> <li data-end=\"225\" data-start=\"175\">Glucose lý thuy\u1ebft: 200 × (180 ÷ 162) = 222,2 g<\/p><\/li> <li data-end=\"279\" data-start=\"226\">Glucose th\u1ef1c t\u1ebf (H = 75%): 222,2 × 0,75 ≈ 166,7 g<\/p><\/li><\/ul>","type":"choose","extra_type":"classic","user_id":"139","test":"0","date":"2025-09-28 15:56:03","option_type":"txt","len":1},{"id":"9048","post_id":"10100","mon_id":"1160346","chapter_id":"1159849","question":"Th\u1ee7y phân m gam tinh b\u1ed9t trong môi tr\u01b0\u1eddng acid (gi\u1ea3 s\u1eed s\u1ef1 th\u1ee7y phân ch\u1ec9 t\u1ea1o glucose). Sau ph\u1ea3n \u1ee9ng, \u0111em trung hòa acid b\u1eb1ng ki\u1ec1m r\u1ed3i th\u1ef1c hi\u1ec7n ph\u1ea3n \u1ee9ng tráng b\u1ea1c thu \u0111\u01b0\u1ee3c m gam Ag. Hi\u1ec7u su\u1ea5t c\u1ee7a ph\u1ea3n \u1ee9ng th\u1ee7y phân tinh b\u1ed9t là:<\/p>","options":["A. 66,67%.","B. 80%.","C. 75%.","D. 50%."],"correct":"3","level":"3","hint":"G\u1ee3i ý: 1 mol m\u1eaft xích tinh b\u1ed9t (C\u2086H\u2081\u2080O\u2085, M = 162) → 1 mol glucose (M = 180). 1 mol glucose kh\u1eed \u0111\u01b0\u1ee3c 2 mol Ag (216 g Ag).<\/p>","answer":"\u0110áp án \u0111úng: C. 75%<\/strong><\/span><\/p>Gi\u1ea3i:<\/strong><\/p><ul data-end=\"365\" data-start=\"136\"> <li data-end=\"199\" data-start=\"136\">m g tinh b\u1ed9t → t\u1ed1i \u0111a m × (180\/162) = m × (10\/9) g glucose.<\/p><\/li> <li data-end=\"261\" data-start=\"200\">S\u1ed1 mol glucose t\u1ed1i \u0111a = [m × (10\/9)] ÷ 180 = m ÷ 162 mol.<\/p><\/li> <li data-end=\"312\" data-start=\"262\">S\u1ed1 mol Ag t\u1ed1i \u0111a = 2 × (m ÷ 162) = m ÷ 81 mol.<\/p><\/li> <li data-end=\"365\" data-start=\"313\">Kh\u1ed1i l\u01b0\u1ee3ng Ag t\u1ed1i \u0111a = (m ÷ 81) × 108 = 4\/3 m g.<\/p><\/li> <li data-end=\"365\" data-start=\"313\">Th\u1ef1c t\u1ebf thu \u0111\u01b0\u1ee3c m g Ag<\/strong>, nên:<\/p><\/li> <li data-end=\"365\" data-start=\"313\">Hi\u1ec7u su\u1ea5t = (m ÷ (4\/3 m)) × 100% = 75%.<\/p><\/li><\/ul>","type":"choose","extra_type":"classic","user_id":"139","test":"0","date":"2025-09-28 15:56:51","option_type":"txt","len":0},{"id":"9049","post_id":"10100","mon_id":"1160346","chapter_id":"1159849","question":"Cellulose trinitrat \u0111\u01b0\u01a1\u0323c \u0111iê\u0300u chê\u0301 t\u01b0\u0300 cellulose va\u0300 nitric acid \u0111\u0103\u0323c co\u0301 xu\u0301c ta\u0301c sulfuric acid \u0111\u0103\u0323c, no\u0301ng. \u0110ê\u0309 co\u0301 29,7 kg cellulose trinitrate, câ\u0300n du\u0300ng m kg nitric acid (hiê\u0323u suâ\u0301t 90%). Gia\u0301 tri\u0323 cu\u0309a m la\u0300:<\/p>","options":["A. 42 kg.","B. 30 kg.","C. 10 kg.","D. 21 kg."],"correct":"4","level":"3","hint":"G\u1ee3i ý: 1 m\u1eaft xích cellulose (M=162) + 3 HNO\u2083 → 1 m\u1eaft xích trinitrat (M=297) + 3 H\u2082O → c\u1ea7n 189 g HNO\u2083 cho 297 g s\u1ea3n ph\u1ea9m. V\u1edbi hi\u1ec7u su\u1ea5t 90% ph\u1ea3i tính s\u1ea3n ph\u1ea9m lý thuy\u1ebft r\u1ed3i quy v\u1ec1 HNO\u2083.<\/p>","answer":"\u0110áp án \u0111úng: D. 21 kg<\/span><\/strong><\/p>Gi\u1ea3i thích ng\u1eafn:<\/p><ul data-end=\"346\" data-is-last-node=\"\" data-is-only-node=\"\" data-start=\"236\"> <li data-end=\"290\" data-start=\"236\">HNO\u2083 c\u1ea7n (lý thuy\u1ebft) = 29,7 × (189\/297) = 18,9 kg.<\/p><\/li> <li data-end=\"346\" data-is-last-node=\"\" data-start=\"291\">Do hi\u1ec7u su\u1ea5t 90%, ph\u1ea3i l\u1ea5y 18,9 ÷ 0,9 = 21,0 kg HNO\u2083.<\/p><\/li><\/ul>","type":"choose","extra_type":"classic","user_id":"139","test":"0","date":"2025-09-28 15:57:34","option_type":"txt","len":0}]}