Alkane - nâng cao

Home Lớp 9 Khoa học tự nhiên 9 - sách kết nối tri thức Bài 23: AlkaneBài tập nâng cao

{"common":{"save":0,"post_id":"7879","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"8144","post_id":"7879","mon_id":"1159533","chapter_id":"1159554","question":"\u0110\u1ed1t cháy hoàn toàn 12,395 lít khí methane (\u0111kc). Th\u1ec3 tích khí carbon dioxide t\u1ea1o thành là:<\/p>","options":["A. <\/strong>1239,5 lít","B.<\/strong> 24,79 lít","C.<\/strong> 1,12 lít","D. <\/strong>12,395 lít"],"correct":"4","level":"3","hint":"- Quá trình \u0111\u1ed1t cháy hoàn toàn methane (CH\u2084) s\u1ebd t\u1ea1o ra carbon dioxide (CO\u2082) và n\u01b0\u1edbc (H\u2082O).<\/p>- Cân b\u1eb1ng ph\u01b0\u01a1ng trình ph\u1ea3n \u1ee9ng.<\/p>- Th\u1ec3 tích khí \u1edf \u0111i\u1ec1u ki\u1ec7n chu\u1ea9n (\u0111kc): 24,79 lít.<\/p>","answer":"\u0110áp án \u0111úng: D. 12,395 lít<\/strong><\/span><\/p>n$_{CH_{4}}$<\/span> = ${12,395 \\over 24,79}$<\/span> = 0,5 mol<\/p>PTHH: CH$_{4}$<\/span> + 2O$_{2}$<\/span> → CO$_{2}$<\/span> + 2H$_{2}$<\/span>O<\/p>n$_{CO_{2}}$<\/span> = n$_{CH_{4}}$<\/span> = 0,5 mol<\/p>V$_{CO_{2}}$<\/span> = 0,5 . 24,79 = 12,395 lít<\/strong><\/p>","type":"choose","extra_type":"shape1","time":"0","user_id":"127","test":"0","date":"2024-10-26 02:58:05","option_type":"txt","len":0},{"id":"8145","post_id":"7879","mon_id":"1159533","chapter_id":"1159554","question":"Phát bi\u1ec3u nào sau \u0111ây là không \u0111úng?<\/p>","options":["A. <\/strong>Trong phân t\u1eed hydrocarbon, s\u1ed1 nguyên t\u1eed hydrogen luôn là s\u1ed1 ch\u1eb5n","B. <\/strong>Trong phân t\u1eed alkene, liên k\u1ebft \u0111ôi g\u1ed3m m\u1ed9t liên k\u1ebft σ và m\u1ed9t liên k\u1ebft π","C. <\/strong>Hydrocarbon no là hydrocarbon mà trong phân t\u1eed ch\u1ec9 có liên k\u1ebft \u0111\u01a1n","D.<\/strong> Công th\u1ee9c chung c\u1ee7a hydrocarbon no, m\u1ea1ch h\u1edf có d\u1ea1ng C$_{n}$<\/span>H$_{2n}$<\/span>"],"correct":"4","level":"3","hint":"- S\u1ed1 nguyên t\u1eed H trong hydrocarbon<\/strong>: Có ph\u1ea3i luôn là s\u1ed1 ch\u1eb5n?<\/p>- Alkene<\/strong>: Liên k\u1ebft \u0111ôi có \u0111\u1eb7c \u0111i\u1ec3m th\u1ebf nào?<\/p>- Hydrocarbon no<\/strong>: Ch\u1ec9 có liên k\u1ebft \u0111\u01a1n?<\/p>- Công th\u1ee9c chung c\u1ee7a hydrocarbon no<\/strong>: có áp d\u1ee5ng cho m\u1ecdi tr\u01b0\u1eddng h\u1ee3p không?<\/p>","answer":"\u0110áp án \u0111úng: D. Công th\u1ee9c chung c\u1ee7a hydrocarbon no, m\u1ea1ch h\u1edf có d\u1ea1ng C$_{n}$<\/span>H$_{2n}$<\/span><\/strong><\/span><\/p>Công th\u1ee9c \u0111úng c\u1ee7a hydrocarbon no (alkane) là CnH\u2082n+2<\/strong>.<\/p>CnH\u2082n<\/strong> là công th\u1ee9c c\u1ee7a alkene<\/strong>, không ph\u1ea3i alkane.<\/p>","type":"choose","extra_type":"cloud","time":"0","user_id":"127","test":"0","date":"2024-10-26 03:02:02","option_type":"math","len":0},{"id":"8146","post_id":"7879","mon_id":"1159533","chapter_id":"1159554","question":"\u0110\u1ed1t cháy hoàn toàn 6,4 gam khí methane, d\u1eabn toàn b\u1ed9 s\u1ea3n ph\u1ea9m qua dung d\u1ecbch n\u01b0\u1edbc vôi trong d\u01b0. Kh\u1ed1i l\u01b0\u1ee3ng k\u1ebft t\u1ee7a thu \u0111\u01b0\u1ee3c là:<\/p>","options":["A. <\/strong>20 gam","B. <\/strong>40 gam","C.<\/strong> 80 gam","D. <\/strong>10 gam"],"correct":"2","level":"3","hint":"- Vi\u1ebft PTHH c\u1ee7a ph\u1ea3n \u1ee9ng<\/p>- Tính n$_{CH_{4}}$<\/span> ⇒ n$_{CO_{2}}$<\/span> ⇒ n$_{CaCO_{3}}$<\/span> <\/p>- Tính kh\u1ed1i l\u01b0\u1ee3ng m$_{CaCO_{3}}$<\/span> <\/p>","answer":"\u0110áp án \u0111úng: B. 40 gam<\/strong><\/span><\/p>n$_{CH_{4}}$<\/span>= ${6,4 \\over 16}$<\/span> = 0,4 (mol)<\/p>PTHH: <\/strong><\/p>CH$_{4}$<\/span> + 2O$_{2}$<\/span> → CO$_{2}$<\/span> + 2H$_{2}$<\/span>O (1)<\/p>CO$_{2}$<\/span> + Ca(OH)$_{2}$<\/span> → CaCO$_{3}$<\/span> + H$_{2}$<\/span>O (2)<\/p>Theo 2 ph\u01b0\u01a1ng trình hoá h\u1ecdc, ta có:<\/p>n$_{CaCO_{3}}$<\/span> = n$_{CO_{2}}$<\/span> = n$_{CH_{4}}$<\/span> = 0,4 (mol)<\/p>⇒ m$_{CaCO_{3}}$<\/span> = 0,4 . 100 = 40 (gam)<\/strong><\/p>","type":"choose","extra_type":"shape2","time":"0","user_id":"127","test":"0","date":"2024-10-26 03:08:25","option_type":"txt","len":0}]}