Ôn tập chương 5

Home Lớp 10 Hóa học 10 - Sách Kết nối tri thức Bài 18: Ôn tập chương 5Bài tập nâng cao

{"common":{"save":0,"post_id":"8600","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"7894","mon_id":"1158798","chapter_id":"1158879","post_id":"8600","question":"\u0110\u01b0\u1eddng sucrose (C12H22O11) là m\u1ed9t \u0111\u01b0\u1eddng \u0111ôi. Trong môi tr\u01b0\u1eddng acid \u1edf d\u1ea1 dày và nhi\u1ec7t \u0111\u1ed9 c\u01a1 th\u1ec3, sucrose b\u1ecb th\u1ee7y phân thành \u0111\u01b0\u1eddng glucose và fructose, sau \u0111ó b\u1ecb oxi hóa b\u1edfi oxygen t\u1ea1o thành CO2 và H2O. Khi 1 mol \u0111\u01b0\u1eddng sucrose b\u1ecb \u0111\u1ed1t cháy hoàn toàn v\u1edbi m\u1ed9t l\u01b0\u1ee3ng v\u1eeba \u0111\u1ee7 oxygen \u1edf \u0111i\u1ec1u ki\u1ec7n chu\u1ea9n to\u1ea3 ra m\u1ed9t l\u01b0\u1ee3ng nhi\u1ec7t là 5645 kJ. N\u1ebfu 10,00 gam \u0111\u01b0\u1eddng sucrose \u0111\u01b0\u1ee3c \u0111\u1ed1t cháy hoàn toàn \u1edf cùng \u0111i\u1ec1u ki\u1ec7n nh\u01b0 trên thì bi\u1ebfn thiên enthalpy quá trình là:<\/p>","options":["A. + 5645kJ.","B. – 165 kJ.","C. – 5645 kJ.","D. + 165 kJ."],"correct":"2","level":"3","hint":"","answer":"\u0110áp án \u0111úng: B. – 165 kJ.<\/strong><\/p>Ph\u1ea3n \u1ee9ng \u0111\u1ed1t cháy \u0111\u01b0\u1eddng sucrose:C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(l)<\/p>Khi 1 mol \u0111\u01b0\u1eddng sucrose b\u1ecb \u0111\u1ed1t cháy hoàn toàn v\u1edbi m\u1ed9t l\u01b0\u1ee3ng v\u1eeba \u0111\u1ee7 oxygen \u1edf \u0111i\u1ec1u ki\u1ec7n chu\u1ea9n to\u1ea3 ra m\u1ed9t l\u01b0\u1ee3ng nhi\u1ec7t là 5645 kJ.<\/p>→ N\u1ebfu 5,00 gam \u0111\u01b0\u1eddng sucrose \u0111\u01b0\u1ee3c \u0111\u1ed1t cháy hoàn toàn \u1edf cùng \u0111i\u1ec1u ki\u1ec7n nh\u01b0 trên thì bi\u1ebfn thiên enthalpy quá trình b\u1eb1ng: (10\/342).(−5645) = −165 kJ.<\/p>","type":"choose","extra_type":"classic","user_id":"139","test":"0","date":"2025-03-11 10:50:00","option_type":"txt","len":2},{"id":"7895","mon_id":"1158798","chapter_id":"1158879","post_id":"8600","question":"Bi\u1ebfn thiên enthalpy chu\u1ea9n c\u1ee7a quá trình “H2O(s) → H2O(l)” là 6,020 kJ. Bi\u1ebft r\u1eb1ng \u0111\u1ec3 làm cho nhi\u1ec7t \u0111\u1ed9 c\u1ee7a 1 mol n\u01b0\u1edbc l\u1ecfng thay \u0111\u1ed5i 1°C c\u1ea7n m\u1ed9t nhi\u1ec7t l\u01b0\u1ee3ng là 75,4 J. Gi\u1ea3 s\u1eed m\u1ed7i viên n\u01b0\u1edbc \u0111á t\u01b0\u01a1ng \u1ee9ng v\u1edbi 1 mol n\u01b0\u1edbc, s\u1ed1 viên n\u01b0\u1edbc \u0111á t\u1ed1i thi\u1ec3u c\u1ea7n tan ch\u1ea3y \u0111\u1ec3 có th\u1ec3 làm l\u1ea1nh 500 gam n\u01b0\u1edbc l\u1ecfng \u1edf 20°C xu\u1ed1ng 0°C là:<\/p>","options":["A. 1.","B. 7.","C. 14.","D. 15."],"correct":"2","level":"3","hint":"","answer":"\u0110áp án \u0111úng: B. 7.<\/strong><\/p>Giá tr\u1ecb nhi\u1ec7t l\u01b0\u1ee3ng mà 500 gam n\u01b0\u1edbc l\u1ecfng t\u1eeb 20oC gi\u1ea3m xu\u1ed1ng 0oC thu vào là: (500\/18).75,4.(20−0) = 41,8889(kJ).<\/p>Ph\u1ea7n nhi\u1ec7t l\u01b0\u1ee3ng t\u1ecfa ra này \u0111\u01b0\u1ee3c viên n\u01b0\u1edbc \u0111á h\u1ea5p th\u1ee5 \u0111\u1ec3 tan ch\u1ea3y. S\u1ed1 viên n\u01b0\u1edbc \u0111á t\u1ed1i thi\u1ec3u c\u1ea7n là: 41,8889\/6,02 ≈ 7viên.<\/p>","type":"choose","extra_type":"classic","user_id":"139","test":"0","date":"2025-03-12 08:47:17","option_type":"txt","len":0},{"id":"7896","mon_id":"1158798","chapter_id":"1158879","post_id":"8600","question":"Bi\u1ebfn thiên enthalpy chu\u1ea9n c\u1ee7a quá trình “H2O(s) → H2O(l)” là 6,020 kJ. Bi\u1ebft r\u1eb1ng \u0111\u1ec3 làm cho nhi\u1ec7t \u0111\u1ed9 c\u1ee7a 1 mol n\u01b0\u1edbc l\u1ecfng thay \u0111\u1ed5i 1°C c\u1ea7n m\u1ed9t nhi\u1ec7t l\u01b0\u1ee3ng là 75,4 J. Gi\u1ea3 s\u1eed m\u1ed7i viên n\u01b0\u1edbc \u0111á t\u01b0\u01a1ng \u1ee9ng v\u1edbi 1 mol n\u01b0\u1edbc, \u0111\u1ec3 làm l\u1ea1nh 120 gam n\u01b0\u1edbc l\u1ecfng \u1edf 45°C xu\u1ed1ng 0°C c\u1ea7n dùng:<\/p>","options":["A. 57,63 gam n\u01b0\u1edbc \u0111á.","B. 67,63 gam n\u01b0\u1edbc \u0111á.","C. 47,63 gam n\u01b0\u1edbc \u0111á.","D. 77,63 gam n\u01b0\u1edbc \u0111á."],"correct":"2","level":"3","hint":"","answer":"\u0110áp án \u0111úng: B. 67,63 gam n\u01b0\u1edbc \u0111á.<\/strong><\/p>120 gam n\u01b0\u1edbc l\u1ecfng t\u1eeb 45oC gi\u1ea3m xu\u1ed1ng 0oC t\u1ecfa ra nhi\u1ec7t l\u01b0\u1ee3ng có giá tr\u1ecb là: 120\/18.75,4.(45-0) = 22620 J = 22,62 kJ.<\/p>L\u01b0\u1ee3ng n\u01b0\u1edbc \u0111á c\u1ea7n dùng là: (22,62\/6,02).18 = 67,63 (g).<\/p>","type":"choose","extra_type":"classic","user_id":"139","test":"0","date":"2025-03-12 08:48:39","option_type":"txt","len":2}]}