Hóa trị và công thức hóa trị

Home Lớp 7 Khoa học tự nhiên 7 - Sách kết nối tri thức Bài 7: Hóa trị và công thức hóa trịBài tập nâng cao

{"common":{"save":0,"post_id":"6040","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"9171","post_id":"6040","mon_id":"1158750","chapter_id":"1158757","question":"Công th\u1ee9c hóa h\u1ecdc c\u1ee7a h\u1ee3p ch\u1ea5t t\u1ea1o b\u1edfi l\u01b0u hu\u1ef3nh có hóa tr\u1ecb VI và oxygen là:<\/p>","options":["A. <\/strong>$SO_2$<\/span>","B.<\/strong> $SO$<\/span>","C. <\/strong>$S_2O_6$<\/span>","D.<\/strong> $SO_3$<\/span>"],"correct":"4","level":"3","hint":"","answer":"Vi\u1ebft công th\u1ee9c hóa h\u1ecdc d\u1ea1ng chung là <svg height=\"60\" width=\"60\"> <g> <title><\/title> <rect fill=\"#fff\" height=\"62\" id=\"canvas_background\" width=\"62\" x=\"-1\" y=\"-1\"><\/rect> <g display=\"none\" height=\"100%\" id=\"canvasGrid\" overflow=\"visible\" width=\"100%\" x=\"0\" y=\"0\"> <rect fill=\"url(#gridpattern)\" height=\"100%\" stroke-width=\"0\" width=\"100%\" x=\"0\" y=\"0\"><\/rect> <\/g> <\/g> <g> <title><\/title> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_1\" stroke=\"#000\" stroke-width=\"0\" text-anchor=\"start\" x=\"2.0125\" xml:space=\"preserve\" y=\"37.8375\">S O<\/text> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"17\" id=\"svg_2\" stroke=\"#000\" stroke-width=\"0\" text-anchor=\"start\" x=\"15.0125\" xml:space=\"preserve\" y=\"49.8375\">x<\/text> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"17\" id=\"svg_3\" stroke=\"#000\" stroke-width=\"0\" text-anchor=\"start\" x=\"40.0125\" xml:space=\"preserve\" y=\"49.8375\">y<\/text> <text fill=\"#ff0000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"17\" id=\"svg_4\" stroke=\"#000\" stroke-width=\"0\" text-anchor=\"start\" x=\"4.0125\" xml:space=\"preserve\" y=\"16.8375\">VI<\/text> <text fill=\"#ff0000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"17\" id=\"svg_5\" stroke=\"#000\" stroke-width=\"0\" text-anchor=\"start\" x=\"30.0125\" xml:space=\"preserve\" y=\"16.8375\">II<\/text> <\/g> <\/svg><\/span><\/p>Theo quy t\u1eafc hóa tr\u1ecb, ta có: x . VI = y . II<\/p>Chuy\u1ec3n thành t\u1ec9 l\u1ec7: $\\dfrac{x}{y}=\\dfrac{II}{VI}=\\dfrac{2}{6}=\\dfrac{1}{3}$<\/span><\/p>T\u1ec9 l\u1ec7 s\u1ed1 nguyên t\u1eed trong phân t\u1eed th\u01b0\u1eddng là nh\u1eefng s\u1ed1 nguyên \u0111\u01a1n gi\u1ea3n nh\u1ea5t, vì v\u1eady l\u1ea5y x = 1, y = 3<\/p>V\u1eady công th\u1ee9c hóa h\u1ecdc c\u1ee7a h\u1ee3p ch\u1ea5t là  D.<\/strong> $SO_3$<\/span><\/span><\/p>","type":"choose","extra_type":"shape1","time":"0","user_id":"131","test":"0","date":"2022-08-15 13:28:47","option_type":"math","len":0},{"id":"9172","post_id":"6040","mon_id":"1158750","chapter_id":"1158757","question":"Bi\u1ebft công th\u1ee9c hóa h\u1ecdc c\u1ee7a \u0111\u01b0\u1eddng \u0103n là  $C_{12}H_{22}O_{11}$<\/span>, hãy tính ph\u1ea7n tr\u0103m v\u1ec1 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a carbon trong \u0111\u01b0\u1eddng \u0103n.<\/p>(kh\u1ed1i l\u01b0\u1ee3ng nguyên t\u1eed c\u1ee7a C là 12, H là 1 và O là 16)<\/p>","options":["A. <\/strong>42,11%","B.<\/strong> 60,11%","C. <\/strong>51,46%","D.<\/strong> 6,43%"],"correct":"1","level":"3","hint":"B\u01b0\u1edbc 1: Tính kh\u1ed1i l\u01b0\u1ee3ng phân t\u1eed c\u1ee7a h\u1ee3p ch\u1ea5t.<\/p>B\u01b0\u1edbc 2: % kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a nguyên t\u1ed1 = kh\u1ed1i l\u01b0\u1ee3ng nguyên t\u1eed .<\/strong> s\u1ed1 nguyên t\u1eed . <\/strong>100%<\/span>kh\u1ed1i l\u01b0\u1ee3ng phân t\u1eed<\/span><\/span><\/p>","answer":"Kh\u1ed1i l\u01b0\u1ee3ng phân t\u1eed c\u1ee7a \u0111\u01b0\u1eddng \u0103n là:<\/p>12 . 12 + 22 . 1 + 11 . 16 = 342 (amu)<\/p>Ph\u1ea7n tr\u0103m kh\u1ed1i l\u01b0\u1ee3ng carbon trong \u0111\u01b0\u1eddng \u0103n là:<\/p>%C = $\\dfrac{12 . 12 . 100\\%}{342}$<\/span> ≈ 42,11%<\/p>\u0110áp án \u0111úng là A. <\/strong>42,11%<\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"0","date":"2022-08-15 13:40:53","option_type":"txt","len":0},{"id":"9264","post_id":"6040","mon_id":"1158750","chapter_id":"1158757","question":"Xác \u0111\u1ecbnh công th\u1ee9c hóa h\u1ecdc c\u1ee7a h\u1ee3p ch\u1ea5t potassium phosphate có c\u1ea5u t\u1ea1o t\u1eeb K và nhóm $PO_4$<\/span>. <\/p>","options":["A. <\/strong>$KPO_4$<\/span>","B.<\/strong> $K_2PO_4$<\/span>","C. <\/strong>$K_3PO_4$<\/span>","D.<\/strong> $K(PO_4)_3$<\/span>"],"correct":"3","level":"3","hint":"K có hóa tr\u1ecb là I, nhóm $PO_4$<\/span> có hóa tr\u1ecb là III.<\/p>","answer":"Vi\u1ebft công th\u1ee9c hóa h\u1ecdc d\u1ea1ng chung là <svg height=\"70\" width=\"100\"> <g> <title><\/title> <rect fill=\"#fff\" height=\"72\" id=\"canvas_background\" width=\"102\" x=\"-1\" y=\"-1\"><\/rect> <g display=\"none\" height=\"100%\" id=\"canvasGrid\" overflow=\"visible\" width=\"100%\" x=\"0\" y=\"0\"> <rect fill=\"url(#gridpattern)\" height=\"100%\" stroke-width=\"0\" width=\"100%\" x=\"0\" y=\"0\"><\/rect> <\/g> <\/g> <g> <title><\/title> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"24\" id=\"svg_1\" stroke=\"#000\" stroke-width=\"0\" style=\"cursor: move;\" text-anchor=\"start\" x=\"1.0125\" xml:space=\"preserve\" y=\"45.8375\">K (PO )<\/text> <text fill=\"#bf0000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"17\" id=\"svg_2\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"17.0125\" xml:space=\"preserve\" y=\"56.8375\">x<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"17\" id=\"svg_3\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"62.0125\" xml:space=\"preserve\" y=\"56.8375\">4<\/text> <text fill=\"#ff0000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"17\" id=\"svg_4\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"80.0125\" xml:space=\"preserve\" y=\"57.8375\">y<\/text> <text fill=\"#ff0000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"17\" id=\"svg_5\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"15.0125\" xml:space=\"preserve\" y=\"20.8375\">I<\/text> <text fill=\"#ff0000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"17\" id=\"svg_6\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"49.0125\" xml:space=\"preserve\" y=\"20.8375\">III<\/text> <\/g> <\/svg><\/span><\/p>Theo quy t\u1eafc hóa tr\u1ecb, ta có: x . I = y . III<\/p>Chuy\u1ec3n thành t\u1ec9 l\u1ec7: $\\dfrac{x}{y}=\\dfrac{III}{I}=\\dfrac{3}{1}$<\/span><\/p>T\u1ec9 l\u1ec7 s\u1ed1 nguyên t\u1eed trong phân t\u1eed th\u01b0\u1eddng là nh\u1eefng s\u1ed1 nguyên \u0111\u01a1n gi\u1ea3n nh\u1ea5t, vì v\u1eady l\u1ea5y x = 3, y = 1<\/span><\/p>V\u1eady công th\u1ee9c hóa h\u1ecdc c\u1ee7a h\u1ee3p ch\u1ea5t là  C. <\/strong>$K_3PO_4$<\/span><\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"0","date":"2022-08-19 13:02:32","option_type":"math","len":0}]}