Phân tử - Đơn chất - Hợp chất

Home Lớp 7 Khoa học tự nhiên 7 - Sách kết nối tri thức Bài 5: Phân tử - Đơn chất - Hợp chấtBài tập trung bình

{"common":{"save":0,"post_id":"6045","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"9134","post_id":"6045","mon_id":"1158750","chapter_id":"1158757","question":"Ch\u1ea5t nào d\u01b0\u1edbi \u0111ây không ph\u1ea3i là \u0111\u01a1n ch\u1ea5t?<\/p>","options":["A. <\/strong>Khí oxygen ","B.<\/strong> Kim c\u01b0\u01a1ng","C. <\/strong>Khí carbon dioxide","D.<\/strong> Mu\u1ed1i \u0103n"],"correct":"[\"3\",\"4\"]","level":"2","hint":"\u0110\u01a1n ch\u1ea5t \u0111\u01b0\u1ee3c t\u1ea1o nên t\u1eeb m\u1ed9t <\/strong>nguyên t\u1ed1 hóa h\u1ecdc.<\/p>H\u1ee3p ch\u1ea5t \u0111\u01b0\u1ee3c t\u1ea1o nên t\u1eeb hai nguyên t\u1ed1 hóa h\u1ecdc tr\u1edf lên<\/strong>.<\/p>","answer":"A. <\/strong>Khí oxygen     (\u0111\u01a1n ch\u1ea5t<\/span> vì \u0111\u01b0\u1ee3c t\u1ea1o thành t\u1eeb m\u1ed9t nguyên t\u1ed1 hóa h\u1ecdc là oxygen)<\/p>B.<\/strong> Kim c\u01b0\u01a1ng     (\u0111\u01a1n ch\u1ea5t<\/span> vì \u0111\u01b0\u1ee3c t\u1ea1o thành t\u1eeb m\u1ed9t nguyên t\u1ed1 hóa h\u1ecdc là carbon)<\/p>C. <\/strong>Khí carbon dioxide      (h\u1ee3p ch\u1ea5t <\/span>vì \u0111\u01b0\u1ee3c t\u1ea1o thành t\u1eeb hai nguyên t\u1ed1 hóa h\u1ecdc là carbon và oxygen)<\/p>D.<\/strong> Mu\u1ed1i \u0103n     (h\u1ee3p ch\u1ea5t<\/span> vì \u0111\u01b0\u1ee3c t\u1ea1o thành t\u1eeb hai nguyên t\u1ed1 hóa h\u1ecdc là sodium và chlorine)<\/p>Các \u0111áp án \u0111úng là <\/p>C. <\/strong>Khí carbon dioxide<\/span><\/p>D.<\/strong> Mu\u1ed1i \u0103n<\/span><\/p>","type":"checkbox","extra_type":"classic-checkbox","time":"0","user_id":"131","test":"0","date":"2022-08-12 13:31:55","option_type":"txt","len":2},{"id":"9136","post_id":"6045","mon_id":"1158750","chapter_id":"1158757","question":"Hình d\u01b0\u1edbi mô t\u1ea3 m\u1ed9t phân t\u1eed ch\u1ee9a carbon, hydrogen và oxygen. S\u1ed1 nguyên t\u1eed c\u1ee7a m\u1ed7i nguyên t\u1ed1 carbon, hydrogen và oxygen có trong m\u1ed9t phân t\u1eed ch\u1ea5t này l\u1ea7n l\u01b0\u1ee3t là<\/p><svg height=\"300\" width=\"300\"> <g><title><\/title><rect fill=\"#fff\" height=\"302\" id=\"canvas_background\" width=\"302\" x=\"-1\" y=\"-1\"><\/rect> <g display=\"none\" id=\"canvasGrid\"> <rect fill=\"url(#gridpattern)\" height=\"100%\" id=\"svg_10\" stroke-width=\"0\" width=\"100%\" x=\"0\" y=\"0\"><\/rect> <\/g> <\/g> <g><title><\/title><ellipse cx=\"78.5125\" cy=\"73.45\" fill=\"#005fbf\" id=\"svg_1\" rx=\"26\" ry=\"26\" stroke=\"#000\" stroke-width=\"1.5\"><\/ellipse> <ellipse cx=\"155.5125\" cy=\"95.45\" fill=\"#005fbf\" id=\"svg_2\" rx=\"26\" ry=\"26\" stroke=\"#000\" stroke-width=\"1.5\"><\/ellipse> <ellipse cx=\"93.0125\" cy=\"13.45\" fill=\"#ffff00\" id=\"svg_3\" rx=\"8\" ry=\"8\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\"><\/ellipse> <ellipse cx=\"15.0125\" cy=\"57.45\" fill=\"#ffff00\" id=\"svg_4\" rx=\"8\" ry=\"8\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\"><\/ellipse> <ellipse cx=\"83.0125\" cy=\"134.45\" fill=\"#ffff00\" id=\"svg_5\" rx=\"8\" ry=\"8\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\"><\/ellipse> <ellipse cx=\"181.0125\" cy=\"39.45\" fill=\"#ffff00\" id=\"svg_6\" rx=\"8\" ry=\"8\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\"><\/ellipse> <ellipse cx=\"175.0125\" cy=\"153.45\" fill=\"#ffff00\" id=\"svg_7\" rx=\"8\" ry=\"8\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\"><\/ellipse> <ellipse cx=\"276.0125\" cy=\"111.45\" fill=\"#ffff00\" id=\"svg_8\" rx=\"8\" ry=\"8\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\"><\/ellipse> <ellipse cx=\"225.5125\" cy=\"100.95\" fill=\"#ff0000\" id=\"svg_9\" rx=\"14.5\" ry=\"14.5\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\"><\/ellipse> <rect fill=\"#ffffff\" height=\"35\" id=\"svg_11\" stroke=\"#000000\" stroke-width=\"1.5\" transform=\"rotate(13.498016357421875 87.51249694824219,37.94999694824218) \" width=\"5\" x=\"85.0125\" y=\"20.45\"><\/rect> <rect fill=\"#ffffff\" height=\"35\" id=\"svg_13\" stroke=\"#000000\" stroke-width=\"1.5\" transform=\"rotate(-77.697265625 38.51250076293943,62.94999694824217) \" width=\"5\" x=\"36.0125\" y=\"45.45\"><\/rect> <rect fill=\"#ffffff\" height=\"35\" id=\"svg_14\" stroke=\"#000000\" stroke-width=\"1.5\" transform=\"rotate(-76.0435562133789 113.51249694824219,84.9500045776367) \" width=\"5\" x=\"111.0125\" y=\"67.45\"><\/rect> <rect fill=\"#ffffff\" height=\"35\" id=\"svg_15\" stroke=\"#000000\" stroke-width=\"1.5\" transform=\"rotate(-9.819310188293457 78.51249694824209,108.95001220703125) \" width=\"5\" x=\"76.0125\" y=\"91.45\"><\/rect> <rect fill=\"#ffffff\" height=\"35\" id=\"svg_16\" stroke=\"#000000\" stroke-width=\"1.5\" transform=\"rotate(-91.87791442871094 197.51249694824222,104.95001220703126) \" width=\"5\" x=\"195.0125\" y=\"87.45\"><\/rect> <rect fill=\"#ffffff\" height=\"35\" id=\"svg_17\" stroke=\"#000000\" stroke-width=\"1.5\" transform=\"rotate(-155.2389373779297 170.51249694824222,62.95000457763672) \" width=\"5\" x=\"168.0125\" y=\"45.45\"><\/rect> <rect fill=\"#ffffff\" height=\"35\" id=\"svg_18\" stroke=\"#000000\" stroke-width=\"1.5\" transform=\"rotate(-76.0435562133789 255.51251220703128,105.95001220703125) \" width=\"5\" x=\"253.0125\" y=\"88.45\"><\/rect> <rect fill=\"#ffffff\" height=\"35\" id=\"svg_19\" stroke=\"#000000\" stroke-width=\"1.5\" transform=\"rotate(-18.091873168945312 167.51249694824213,128.9499969482422) \" width=\"5\" x=\"165.0125\" y=\"111.45\"><\/rect> <ellipse cx=\"35.5125\" cy=\"266.45\" fill=\"#005fbf\" id=\"svg_21\" rx=\"26\" ry=\"26\" stroke=\"#000\" stroke-width=\"1.5\"><\/ellipse> <ellipse cx=\"33.0125\" cy=\"180.45\" fill=\"#ffff00\" id=\"svg_22\" rx=\"8\" ry=\"8\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\"><\/ellipse> <ellipse cx=\"33.5125\" cy=\"214.95\" fill=\"#ff0000\" id=\"svg_23\" rx=\"14.5\" ry=\"14.5\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\"><\/ellipse> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_12\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"70.0125\" xml:space=\"preserve\" y=\"187.45\">: hydrogen<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_20\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"70.0125\" xml:space=\"preserve\" y=\"227.45\">: oxygen<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_24\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"70.0125\" xml:space=\"preserve\" y=\"271.45\">: carbon<\/text> <\/g> <\/svg><\/span><\/p>","options":["A. <\/strong>6, 2, 1","B.<\/strong> 2, 6, 1","C. <\/strong>2, 1, 6","D.<\/strong> 1, 2, 6"],"correct":"2","level":"2","hint":"","answer":"T\u1eeb hình v\u1ebd, ta th\u1ea5y có 2 hình tròn màu xanh, t\u01b0\u01a1ng \u1ee9ng v\u1edbi 2 nguyên t\u1eed carbon;<\/p>6 hình tròn màu vàng, t\u01b0\u01a1ng \u1ee9ng v\u1edbi 6 nguyên t\u1eed hydrogen;<\/p>1 hình tròn màu \u0111\u1ecf, t\u01b0\u01a1ng \u1ee9ng v\u1edbi 1 nguyên t\u1eed oxygen.<\/p>\u0110áp án \u0111úng là B.<\/strong> 2, 6, 1<\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"0","date":"2022-08-12 13:52:42","option_type":"txt","len":0},{"id":"9138","post_id":"6045","mon_id":"1158750","chapter_id":"1158757","question":"Trong các ch\u1ea5t hóa h\u1ecdc: $Na$<\/span>, $N_2$<\/span>, $SO_2$<\/span>, $O_3$<\/span>, $NaCl$<\/span>, $CaCO_3$<\/span>  có các \u0111\u01a1n ch\u1ea5t<\/span> là:<\/p>","options":["A. <\/strong>$Na$<\/span>, $SO_2$<\/span>, $O_3$<\/span>","B.<\/strong> $N_2$<\/span>, $O_3$<\/span>, $NaCl$<\/span>","C. <\/strong>$SO_2$<\/span>, $NaCl$<\/span>, $CaCO_3$<\/span>","D.<\/strong> $Na$<\/span>, $N_2$<\/span>, $O_3$<\/span>"],"correct":"4","level":"2","hint":"\u0110\u01a1n ch\u1ea5t \u0111\u01b0\u1ee3c t\u1ea1o nên t\u1eeb m\u1ed9t <\/strong>nguyên t\u1ed1 hóa h\u1ecdc.<\/p>H\u1ee3p ch\u1ea5t \u0111\u01b0\u1ee3c t\u1ea1o nên t\u1eeb hai nguyên t\u1ed1 hóa h\u1ecdc tr\u1edf lên<\/strong>.<\/p>","answer":"Các \u0111\u01a1n ch\u1ea5t là $Na$<\/span>, $N_2$<\/span>, $O_3$<\/span><\/p>Các h\u1ee3p ch\u1ea5t là  <\/strong>$SO_2$<\/span>, $NaCl$<\/span>, $CaCO_3$<\/span><\/p>\u0110áp án \u0111úng là D.<\/strong> $Na$<\/span>, $N_2$<\/span>, $O_3$<\/span><\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"0","date":"2022-08-12 13:58:12","option_type":"math","len":0}]}