{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Trong c\u00e1c h\u1ee3p ch\u1ea5t sau h\u1ee3p ch\u1ea5t c\u00f3 trong t\u1ef1 nhi\u00ean d\u00f9ng l\u00e0m ph\u00e2n b\u00f3n ho\u00e1 h\u1ecdc:","select":["A. $CaCO_3$","B. $Ca_3(PO_4)_2$ ","C. $Ca(OH)_2$ ","D. $CaCl_2$ "],"hint":"","explain":"<span class='basic_left'> $Ca_3(PO_4)_2$ l\u00e0 th\u00e0nh ph\u1ea7n ch\u00ednh c\u1ee7a ph\u00e2n l\u00e2n khi ch\u01b0a qua ch\u1ebf bi\u1ebfn<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1931},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Trong c\u00e1c lo\u1ea1i ph\u00e2n b\u00f3n ho\u00e1 h\u1ecdc sau lo\u1ea1i n\u00e0o l\u00e0 ph\u00e2n \u0111\u1ea1m ","select":["A. $KCl$ ","B. $Ca_3(PO_4)_2$","C. $K_2SO_4$","D. $(NH_2)_2CO$"],"hint":"","explain":"<span class='basic_left'>$(NH_2)_2CO$: ph\u00e2n \u0111\u1ea1m cung c\u1ea5p dinh d\u01b0\u1ee1ng cho c\u00e2y d\u01b0\u1edbi d\u1ea1ng nguy\u00ean t\u1ed1 N<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1932},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Ph\u00e2n \u0111\u1ea1m cung c\u1ea5p N cho c\u00e2y d\u01b0\u1edbi d\u1ea1ng ion: ","select":["A. $NO_3^{-}$ v\u00e0 $NH_4^ +$","B. $NH_4^ +$ v\u00e0 $PO_4^{3-}$","C. $K^+$ v\u00e0 $PO_4^{3-}$ ","D. $K^+$ v\u00e0 $NH_4^ +$"],"hint":"","explain":"<span class='basic_left'>Ph\u00e2n \u0111\u1ea1m cung c\u1ea5p dinh d\u01b0\u1ee1ng cho c\u00e2y d\u01b0\u1edbi d\u1ea1ng nguy\u00ean t\u1ed1 N d\u01b0\u1edbi d\u1ea1ng : $NO_3^ -$ v\u00e0 $NH_4^ +$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1933},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Lo\u1ea1i \u0111\u1ea1m n\u00e0o d\u01b0\u1edbi \u0111\u00e2y \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 \u0111\u1ea1m 2 l\u00e1:","select":["A. $NaNO_3$","B. $NH_4NO_3$","C. $Ca(NO_3)_2$ ","D. $(NH_4)_2CO_3$"],"hint":"","explain":"<span class='basic_left'>\u0110\u1ea1m $NH_4NO_3$ \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 \u0111\u1ea1m 2 l\u00e1<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1934},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Ph\u00e2n \u0111\u1ea1m ur\u00ea c\u00f3 c\u00f4ng th\u1ee9c l\u00e0:","select":["A. $(NH_4)_2CO_3$ ","B. $NH_4NO_3$","C. $(NH_2)_2CO$ ","D. $(NH_4)_2SO_4$"],"hint":"","explain":"<span class='basic_left'>\u0110\u1ea1m ur\u00ea: $(NH_2)_2CO$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1935},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Ph\u00e2n l\u00e2n cung c\u1ea5p dinh d\u01b0\u1ee1ng cho c\u00e2y tr\u1ed3ng d\u01b0\u1edbi d\u1ea1ng nguy\u00ean t\u1ed1:","select":["A. N","B. P","C. K ","D. C"],"hint":"","explain":"<span class='basic_left'>Ph\u00e2n l\u00e2n cung c\u1ea5p dinh d\u01b0\u1ee1ng cho c\u00e2y tr\u1ed3ng d\u01b0\u1edbi d\u1ea1ng nguy\u00ean t\u1ed1: P<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1936},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" T\u1ec9 l\u1ec7 nguy\u00ean t\u1ed1 K trong ph\u00e2n KCl l\u00e0:","select":["A. 45% ","B. 47%","C. 52% ","D. 57%"],"hint":"","explain":"<span class='basic_left'>$\\%K=\\dfrac{39}{74,5}.100\\%\\approx 52\\%$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1937},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Ph\u00e2n l\u00e2n ch\u01b0a qua ch\u1ebf bi\u1ebfn c\u00f3 th\u00e0nh ph\u1ea7n ch\u00ednh l\u00e0:","select":["A. $Ca(H_2PO_4)_2$","B. $NH_4NO_3$","C. $Ca_3(PO_4)_2$ ","D. $KNO_3$"],"hint":"","explain":"<span class='basic_left'>Photphat t\u1ef1 nhi\u00ean l\u00e0 ph\u00e2n l\u00e2n ch\u01b0a qua ch\u1ebf bi\u1ebfn h\u00f3a h\u1ecdc, th\u00e0nh ph\u1ea7n ch\u00ednh l\u00e0: $Ca_3(PO_4)_2$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1938},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Ph\u00e2n b\u00f3n c\u00f3 c\u00f4ng th\u1ee9c l\u00e0 $KNO_3$ c\u00f3 ch\u1ee9a c\u00e1c nguy\u00ean t\u1ed1 dinh d\u01b0\u1ee1ng n\u00e0o:","select":["A. K, N , O ","B. K, O ","C. K, N ","D. N,O"],"hint":"","explain":"<span class='basic_left'>C\u00e1c nguy\u00ean t\u1ed1 dinh d\u01b0\u1ee1ng trong ph\u00e2n b\u00f3n h\u00f3a h\u1ecdc l\u00e0 N, P, K. Trong $KNO_3$ c\u00f3 ch\u01b0a N, K cung c\u1ea5p dinh d\u01b0\u1ee1ng cho c\u00e2y.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1939},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" D\u00e3y ph\u00e2n b\u00f3n ho\u00e1 h\u1ecdc ch\u1ec9 ch\u1ee9a to\u00e0n ph\u00e2n b\u00f3n ho\u00e1 h\u1ecdc \u0111\u01a1n l\u00e0:","select":["A. $KNO_3 , NH_4NO_3 , (NH_2)_2CO$","B. $KCl , NH_4H_2PO_4 , Ca(H_2PO_4)_2$","C. $(NH_4)_2SO_4 , KCl, Ca(H_2PO_4)_2$ ","D. $(NH_4)_2SO_4 ,KNO_3 , NH_4Cl$"],"hint":"","explain":"<span class='basic_left'>Ph\u00e2n b\u00f3n h\u00f3a h\u1ecdc \u0111\u01a1n ch\u1ec9 ch\u1ee9a 1 nguy\u00ean t\u1ed1 dinh d\u01b0\u1ee1ng N ho\u1eb7c K ho\u1eb7c P. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n C th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1940},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" C\u00e1c lo\u1ea1i ph\u00e2n b\u00f3n \u0111\u1ec1u ch\u1ee9a","select":["A. C\u00e1c nguy\u00ean t\u1ed1 dinh d\u01b0\u1ee1ng c\u1ea7n thi\u1ebft cho c\u00e2y","B. Nguy\u00ean t\u1ed1 N v\u00e0 m\u1ed9t s\u1ed1 nguy\u00ean t\u1ed1 kh\u00e1c","C. Nguy\u00ean t\u1ed1 P v\u00e0 m\u1ed9t s\u1ed1 nguy\u00ean t\u1ed1 kh\u00e1c","D. Nguy\u00ean t\u1ed1 K v\u00e0 m\u1ed9t s\u1ed1 nguy\u00ean t\u1ed1 kh\u00e1c"],"hint":"","explain":"<span class='basic_left'>Ph\u00e2n b\u00f3n h\u00f3a h\u1ecdc ch\u1ee9a c\u00e1c nguy\u00ean t\u1ed1 dinh d\u01b0\u1ee1ng N, P, K cung c\u1ea5p dinh d\u01b0\u1ee1ng cho c\u00e2y tr\u1ed3ng.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1941},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Trong c\u00e1c lo\u1ea1i ph\u00e2n b\u00f3n sau, ph\u00e2n b\u00f3n ho\u00e1 h\u1ecdc k\u00e9p l\u00e0:","select":["A. $(NH_4)_2SO_4$","B. $Ca(H_2PO_4)_2$ ","C. $KCl$ ","D. $KNO3$"],"hint":"","explain":"<span class='basic_left'>Ph\u00e2n b\u00f3n h\u00f3a h\u1ecdc k\u00e9p l\u00e0 ph\u00e2n b\u00f3n ch\u01b0a 2 ho\u1eb7c 3 nguy\u00ean t\u1ed1 dinh d\u01b0\u1ee1ng. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n D ch\u1ee9a 2 nguy\u00ean t\u1ed1 dinh d\u01b0\u1ee1ng N, K.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1942},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" \u0110\u1ec3 nh\u1eadn bi\u1ebft 2 lo\u1ea1i ph\u00e2n b\u00f3n ho\u00e1 h\u1ecdc l\u00e0: $NH_4NO_3$ v\u00e0 $NH_4Cl$. Ta d\u00f9ng dung d\u1ecbch","select":["A. $NaOH$","B. $Ba(OH)_2$","C. $AgNO_3$","D. $BaCl_2$"],"hint":"","explain":"<span class='basic_left'>D\u00f9ng dung d\u1ecbch $AgNO_3$. Ch\u1ea5t n\u00e0o c\u00f3 k\u1ebft t\u1ee7a tr\u1eafng l\u00e0 $NH_4Cl$, ch\u1ea5t n\u00e0o kh\u00f4ng c\u00f3 hi\u1ec7n t\u01b0\u1ee3ng g\u00ec l\u00e0 $NH_4NO_3$ <br\/>$N{{H}_{4}}Cl+AgN{{O}_{3}}\\to N{{H}_{4}}N{{O}_{3}}+AgCl\\downarrow $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1943},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Ph\u1ea7n tr\u0103m v\u1ec1 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a nguy\u00ean t\u1ed1 N trong (NH2)2CO l\u00e0 ","select":["A. 32,33% ","B. 31,81% ","C. 46,67% ","D. 63,64%"],"hint":"","explain":"<span class='basic_left'>$\\%N=\\dfrac{14.2}{60}.100\\%=46,67\\%$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1944},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Cho 0,1 mol $Ba(OH)_2$ v\u00e0o dung d\u1ecbch $NH_4NO_3$ d\u01b0 th\u00ec th\u1ec3 t\u00edch tho\u00e1t ra \u1edf \u0111ktc l\u00e0","select":["A. 2,24 l\u00edt","B. 4,48 l\u00edt","C. 22,4 l\u00edt","D. 44,8 l\u00edt"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& Ba{{(OH)}_{2}}+2N{{H}_{4}}N{{O}_{3}}\\to Ba{{(N{{O}_{3}})}_{2}}+2N{{H}_{3}}\\uparrow +2{{H}_{2}}O \\\\ & {{n}_{N{H}_{3}}}=2{{n}_{Ba{(OH)_{2}}}}=0,2\\,mol \\\\ & V_{N{H}_{3}}=4,48\\,l \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1945},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Ph\u00e2n \u0111\u1ea1m ure th\u01b0\u1eddng ch\u1ee9a 46% N. Kh\u1ed1i l\u01b0\u1ee3ng (kg) ur\u00ea \u0111\u1ee7 cung c\u1ea5p 70 kg N:","select":["A. 152,2","B. 145,5","C. 160,9 ","D. 200,0"],"hint":"","explain":"<span class='basic_left'>${{m}_{\\text{ur}e}}=\\dfrac{70.100\\%}{46\\%}\\approx 152,2\\,kg$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1946},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a nguy\u00ean t\u1ed1 N c\u00f3 trong 200 g $(NH_4)_2SO_4$ l\u00e0 ","select":["A. 42,42 g ","B. 21,21 g ","C. 24,56 g ","D. 24,56 g "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& \\%{{m}_{N}}=\\dfrac{14.2}{132}.100\\%\\approx 21,2\\% \\\\ & {{m}_{N}}=\\dfrac{200.21,2\\%}{100\\%}=42,42\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1947},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" \u0110\u1ec3 kh\u1eed \u0111\u1ed9 chua c\u1ee7a \u0111\u1ea5t ng\u01b0\u1eddi ta d\u00f9ng ch\u1ea5t n\u00e0o sau \u0111\u00e2y","select":["A. Mu\u1ed1i \u0103n ","B. Th\u1ea1ch cao","C. V\u00f4i s\u1ed1ng ","D. Ph\u00e8n chua"],"hint":"","explain":"<span class='basic_left'>\u0110\u1ec3 kh\u1eed \u0111\u1ed9 chua (\u0111\u1ed9 axit) c\u1ee7a \u0111\u00e1t ng\u01b0\u1eddi ta s\u1ebd s\u1eed d\u1ee5ng ch\u1ea5t c\u00f3 m\u00f4i tr\u01b0\u1eddng baz\u01a1 \u0111\u1ec3 trung h\u00f2a b\u1edbt \u0111\u1ed9 chua c\u1ee7a \u0111\u1ea5t. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n C th\u1ecfa m\u00e3n v\u00ec v\u00f4i s\u1ed1ng $(CaO)$ h\u00f2a tan v\u00e0o n\u01b0\u1edbc t\u1ea1o th\u00e0nh dung d\u1ecbch baz\u01a1 $Ca(OH)_2$.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1948},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Trong c\u00e1c lo\u1ea1i ph\u00e2n b\u00f3n sau, lo\u1ea1i ph\u00e2n b\u00f3n n\u00e0o c\u00f3 l\u01b0\u1ee3ng \u0111\u1ea1m cao nh\u1ea5t ","select":["A. $NH_4NO_3$","B. $NH_4Cl$","C. $(NH_4)_2SO_4$","D. $(NH_2)_2CO$"],"hint":"","explain":"<span class='basic_left'>H\u00e0m l\u01b0\u1ee3ng \u0111\u1ea1m ch\u00ednh l\u00e0 % kh\u1ed1i l\u01b0\u1ee3ng nguy\u00ean t\u1ed1 N <br\/> $\\begin{aligned}& A.\\,\\%N=\\dfrac{28}{80}.100\\%=35\\% \\\\ & B.\\%N=\\dfrac{14}{53,5}.100\\%=26\\% \\\\ & C.\\%N=\\dfrac{28}{132}.100\\%=21\\% \\\\ & D.\\%N=\\dfrac{28}{60}.100\\%=47\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1949},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Ph\u00e2n kali $KCl$ s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c t\u1eeb qu\u1eb7ng sinvinit th\u01b0\u1eddng ch\u1ec9 ch\u1ee9a 50% $K_2O$. H\u00e0m l\u01b0\u1ee3ng % $KCl$ trong ph\u00e2n b\u00f3n \u0111\u00f3:","select":["A. 72,9% ","B. 76.0%","C. 79,2% ","D. 75,5%"],"hint":"","explain":"<span class='basic_left'>Gi\u1ea3 s\u1eed c\u00f3 100 gam ph\u00e2n b\u00f3n <br\/>$\\begin{aligned}& =>{{m}_{{{K}_{2}}O}}=\\dfrac{50\\%.100}{100\\%}=50\\,gam \\\\ & {{n}_{{{K}_{2}}O}}=\\dfrac{25}{47}\\,mol \\\\ & {{n}_{K}}=2{{n}_{{{K}_{2}}O}}=\\dfrac{50}{47}\\,mol \\\\ & {{m}_{KCl}}=\\dfrac{50}{47}.74,5\\approx 79,2\\,gam \\\\ & \\%{{m}_{KCl}}=\\dfrac{79,2}{100}.100\\%=79,2\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1950}],"lesson":{"save":0,"level":2}}