Chú ý: Để đảm bảo quyền lợi và bảo vệ tài khoản của mình
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{"common":{"save":0,"post_id":"7288","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5497","post_id":"7288","mon_id":"1158924","chapter_id":"1159168","question":"<p>\u0110\u1ed9 pH c\u1ee7a m\u1ed9t dung d\u1ecbch \u0111\u01b0\u1ee3c tính theo công th\u1ee9c <span class=\"math-tex\">$pH=-\\log[H^+]$<\/span> trong \u0111ó <span class=\"math-tex\">$[H^+]$<\/span> là n\u1ed3ng \u0111\u1ed9 <span class=\"math-tex\">$H^+$<\/span> c\u1ee7a dung d\u1ecbch \u0111ó tính b\u1eb1ng mol\/L. N\u1ed3ng \u0111\u1ed9 <span class=\"math-tex\">$H^+$<\/span> trong dung d\u1ecbch cho bi\u1ebft \u0111\u1ed9 acid c\u1ee7a dung d\u1ecbch \u0111ó. Dung d\u1ecbch acid A có \u0111\u1ed9 pH b\u1eb1ng 1,9. Bi\u1ebft n\u1ed3ng \u0111\u1ed9 <span class=\"math-tex\">$H^+$<\/span> trong dung d\u1ecbch acid A cao h\u01a1n trong dung d\u1ecbch acid B là <span class=\"math-tex\">$10^{0,6}$<\/span> l\u1ea7n. Tính n\u1ed3ng \u0111\u1ed9 <span class=\"math-tex\">$H^+$<\/span> trong dung d\u1ecbch acid B.<\/p>","options":["<strong>A.<\/strong> <span class=\"math-tex\">$10^{2,5}$<\/span>","<strong>B.<\/strong> <span class=\"math-tex\">$10^{-2,5}$<\/span>","<strong>C.<\/strong> <span class=\"math-tex\">$10^{25}$<\/span>","<strong>D.<\/strong> <span class=\"math-tex\">$10^{-25}$<\/span>"],"correct":"2","level":"3","hint":"","answer":"<p>Dung d\u1ecbch acid A có n\u1ed3ng \u0111\u1ed9 <span class=\"math-tex\">$H^+$<\/span> là <span class=\"math-tex\">$[H^+]_A=10^{-1,9}$<\/span>.<\/p><p>Dung d\u1ecbch acid B có n\u1ed3ng \u0111\u1ed9 <span class=\"math-tex\">$H^+$<\/span> là<\/p><p><span class=\"math-tex\">$[H^+]_A=10^{0,6}.[H^+]_B$<\/span> ⇔ <span class=\"math-tex\">$[H^+]_B=\\dfrac{10^{-1,9}}{10^{0,6}}=10^{-2,5}$<\/span>.<\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>B.<\/strong> <span class=\"math-tex\">$10^{-2,5}$<\/span>.<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-04-07 06:15:53","option_type":"math","len":0},{"id":"5499","post_id":"7288","mon_id":"1158924","chapter_id":"1159168","question":"<p>Gi\u1ea3 s\u1eed quá trình nuôi cáy vi khu\u1ea9n tuân theo quy lu\u1eadt t\u0103ng tr\u01b0\u1edfng t\u1ef1 do. Khi \u0111ó, n\u1ebfu g\u1ecdi <span class=\"math-tex\">$N_o$<\/span> là s\u1ed1 l\u01b0\u1ee3ng vi khu\u1ea9n ban \u0111\u1ea7u và <span class=\"math-tex\">$N(t) $<\/span> là s\u1ed1 l\u01b0\u1ee3ng vi khu\u1ea9n sau <span class=\"math-tex\">$t$<\/span> gi\u1edd thì ta có <span class=\"math-tex\">$N(t)=N_o.e^{rt}$<\/span>, trong \u0111ó <span class=\"math-tex\">$r$<\/span> là t\u1ec9 l\u1ec7 t\u0103ng tr\u01b0\u1edfng vi khu\u1ea9n m\u1ed7i gi\u1edd. Gi\u1ea3 s\u1eed ban \u0111\u1ea7u có 500 con vi khu\u1ea9n và sau m\u1ed9t gi\u1edd t\u0103ng lên 800 con. Tính s\u1ed1 l\u01b0\u1ee3ng vi khu\u1ea9n sau 3 gi\u1edd.<\/p>","options":["<strong>A.<\/strong> 2048 con","<strong>B.<\/strong> 3276 con","<strong>C.<\/strong> 3275 con","<strong>D.<\/strong> 2084 con"],"correct":"1","level":"3","hint":"","answer":"<p>Ban \u0111\u1ea7u có 500 con vi khu\u1ea9n, sau 1 gi\u1edd t\u0103ng lên 800 con nên ta có<\/p><p><span class=\"math-tex\">$500.e^{r}=800$<\/span> ⇔ <span class=\"math-tex\">$e^r=1,6$<\/span> ⇔ <span class=\"math-tex\">$r=\\ln1,6$<\/span>.<\/p><p>S\u1ed1 l\u01b0\u1ee3ng vi khu\u1ea9n t\u0103ng lên sau 3 gi\u1edd là <span class=\"math-tex\">$N(3)=500.e^{3.\\ln1,6}=2048$<\/span> con.<\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>A.<\/strong> 2048 con.<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-04-07 06:22:26","option_type":"txt","len":1},{"id":"5502","post_id":"7288","mon_id":"1158924","chapter_id":"1159168","question":"<p>Cho <span class=\"math-tex\">$\\log_{ab}b=3$<\/span>, a > 0, b > 0, ab <span class=\"math-tex\">$\\ne$<\/span> 1. Tính <span class=\"math-tex\">$\\log_\\sqrt{ab}\\bigg(\\dfrac{a}{b^2}\\bigg)$<\/span>.<\/p>","options":["<strong>A.<\/strong> 5","<strong>B.<\/strong> –4","<strong>C.<\/strong> –10","<strong>D.<\/strong> –16"],"correct":"4","level":"3","hint":"","answer":"<p>Ta có <span class=\"math-tex\">$\\log_{ab}b=3$<\/span> <\/p><p>⇔ <span class=\"math-tex\">$\\log_bab=\\dfrac{1}{3}$<\/span><\/p><p>⇔ <span class=\"math-tex\">$\\log_ba+1=\\dfrac{1}{3}$<\/span><\/p><p>⇔ <span class=\"math-tex\">$\\log_ba=-\\dfrac{2}{3}$<\/span> ⇒ <span class=\"math-tex\">$\\log_ab=-\\dfrac{3}{2}$<\/span>.<\/p><p><span class=\"math-tex\">$\\log_\\sqrt{ab}\\bigg(\\dfrac{a}{b^2}\\bigg)$<\/span> <span class=\"math-tex\">$=2(\\log_{ab}a-\\log_{ab}b^2)$<\/span><span class=\"math-tex\">$=\u200b\u200b\\dfrac{2}{\\log_aab}-4\\log_{ab}b$<\/span><span class=\"math-tex\">$=\\dfrac{2}{1+\\log_ab}-4.3$<\/span><span class=\"math-tex\">$=\\dfrac{2}{1-\\dfrac{3}{2}}-12=-16$<\/span>.<\/p><p>Ch\u1ecdn <span style=\"color:#16a085;\"><strong>D.<\/strong> –16.<\/span><\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-04-07 06:35:18","option_type":"txt","len":1}]}