{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Kim lo\u1ea1i n\u00e0o sau \u0111\u00e2y kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi oxi d\u00f9 \u1edf nhi\u1ec7t \u0111\u1ed9 cao?","select":["A. Al; Cu","B. Zn; Fe","C. Au; Ag","D. Mg; Pb"],"hint":"","explain":"<span class='basic_left'>Au v\u00e0 Ag kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi oxi<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2131},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" C\u00f3 th\u1ec3 \u0111i\u1ec1u ch\u1ebf nh\u00f4m b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p:","select":["A. \u0110i\u1ec7n ph\u00e2n dung d\u1ecbch mu\u1ed1i nh\u00f4m.","B. \u0110i\u1ec7n ph\u00e2n n\u00f3ng ch\u1ea3y nh\u00f4m oxit c\u00f3 criolit l\u00e0m x\u00fac t\u00e1c.","C. Kh\u1eed nh\u00f4m oxit b\u1eb1ng $CO$ ho\u1eb7c $H_2$ ","D. Kh\u1eed oxit nh\u00f4m b\u1eb1ng cacbon"],"hint":"","explain":"<span class='basic_left'>Al \u0111\u01b0\u1ee3c \u0111i\u1ec1u ch\u1ebf b\u1eb1ng c\u00e1ch \u0111i\u1ec7n ph\u00e2n n\u00f3ng ch\u1ea3y nh\u00f4m oxit:<br\/>$2A{{l}_{2}}{{O}_{3}}\\xrightarrow[criolit]{\\text{\u0111pnc}}4Al+3{{O}_{2}}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2132},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Co\u0301 ca\u0301c kim loa\u0323i sau: Ag, Na, Cu, Al, Fe. Hai kim loa\u0323i d\u00e2\u0303n \u0111i\u00ea\u0323n t\u00f4\u0301t nh\u00e2\u0301t trong s\u00f4\u0301 \u0111o\u0301 l\u00e2\u0300n l\u01b0\u01a1\u0323t la\u0300","select":["A. Ag, Al ","B. Ag, Fe","C. Cu, Na ","D. Ag, Cu"],"hint":"","explain":"<span class='basic_left'>\u0110\u1ed9 d\u1eabn \u0111i\u1ec7n c\u1ee7a c\u00e1c kim lo\u1ea1i \u0111\u01b0\u1ee3c s\u1eafp x\u1ebfp theo th\u1ee9 t\u1ef1:Ag > Cu > Au > Al > Fe<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2133},{"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 c\u1eb7p ch\u1ea5t sau \u0111\u00e2y, c\u1eb7p ch\u1ea5t n\u00e0o x\u1ea3y ra ph\u1ea3n \u1ee9ng ","select":["A. $Cu$ + $ZnSO_4$","B. $Ag$ + $HCl$","C. $Ag$ + $CuSO_4$","D. $Zn$ + $Pb(NO_3)_2$"],"hint":"","explain":"<span class='basic_left'>kim lo\u1ea1i m\u1ea1nh h\u01a1n \u0111\u1ea9y \u0111\u01b0\u1ee3c kim lo\u1ea1i y\u1ebfu h\u01a1n ra kh\u1ecfi mu\u1ed1i<br\/>$Zn+Pb{{(N{{O}_{3}})}_{2}}\\to Zn{{(N{{O}_{3}})}_{2}}+Pb$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2134},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho c\u00e1c kim lo\u1ea1i Fe, Cu, Ag, Al, Mg. Trong c\u00e1c k\u1ebft lu\u1eadn sau \u0111\u00e2y, k\u1ebft lu\u1eadn n\u00e0o sai","select":["A. Kim lo\u1ea1i t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $HCl, H_2SO_4$ lo\u00e3ng: $Cu, Ag$ ","B. Kim lo\u1ea1i t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $NaOH$: $Al$","C. Kim lo\u1ea1i kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi $H_2SO_4$ \u0111\u1eb7c ngu\u1ed9i: $Al, Fe$","D. Kim lo\u1ea1i kh\u00f4ng tan trong n\u01b0\u1edbc \u1edf nhi\u1ec7t \u0111\u1ed9 th\u01b0\u1eddng: T\u1ea5t c\u1ea3 c\u00e1c kim lo\u1ea1i tr\u00ean"],"hint":"","explain":"<span class='basic_left'>$Ag, Cu$ kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi axit $HCl, H_2SO_4$ lo\u00e3ng<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":2135},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" D\u01b0\u0303 ki\u00ea\u0323n na\u0300o d\u01b0\u01a1\u0301i \u0111\u00e2y cho th\u00e2\u0301y nh\u00f4m hoa\u0323t \u0111\u00f4\u0323ng ho\u0301a ho\u0323c ma\u0323nh h\u01a1n s\u0103\u0301t:","select":["A. Al, Fe \u0111\u00ea\u0300u kh\u00f4ng pha\u0309n \u01b0\u0301ng v\u01a1\u0301i $HNO_3$ \u0111\u0103\u0323c ngu\u00f4\u0323i va\u0300 $H_2SO_4$ \u0111\u0103\u0323c ngu\u00f4\u0323i.","B. Al co\u0301 pha\u0309n \u01b0\u0301ng v\u01a1\u0301i dung di\u0323ch ki\u00ea\u0300m","C. Nh\u00f4m \u0111\u00e2\u0309y \u0111\u01b0\u01a1\u0323c s\u0103\u0301t ra kho\u0309i dung di\u0323ch mu\u00f4\u0301i s\u0103\u0301t ","D. Chi\u0309 co\u0301 s\u0103\u0301t bi\u0323 nam ch\u00e2m hu\u0301t"],"hint":"","explain":"<span class='basic_left'>Ph\u1ea3n \u1ee9ng gi\u1eefa kim lo\u1ea1i v\u00e0 mu\u1ed1i s\u1ebd cho bi\u1ebft kim lo\u1ea1i n\u00e0o m\u1ea1nh h\u01a1n kim lo\u1ea1i n\u00e0o. Kim lo\u1ea1i m\u1ea1nh h\u01a1n (tr\u1eeb kim lo\u1ea1i ki\u1ec1m, ki\u1ec1m th\u1ed5) s\u1ebd \u0111\u1ea9y kim lo\u1ea1i y\u1ebfu h\u01a1n ra kh\u1ecfi mu\u1ed1i.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2136},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Kim lo\u1ea1i c\u00f3 nh\u1eefng t\u00ednh ch\u1ea5t v\u1eadt l\u00ed chung n\u00e0o ","select":["A. T\u00ednh d\u1eabn \u0111i\u1ec7n, t\u00ednh d\u1eabn nhi\u1ec7t ","B. T\u00ednh d\u1ebbo, t\u00ednh d\u1eabn nhi\u1ec7t, \u00e1nh kim.","C. T\u00ednh d\u1eabn nhi\u1ec7t, c\u00f3 \u00e1nh kim.","D. T\u00ednh d\u1ebbo, t\u00ednh d\u1eabn \u0111i\u1ec7n, t\u00ednh d\u1eabn nhi\u1ec7t, \u00e1nh kim "],"hint":"","explain":"<span class='basic_left'>H\u1ea7u h\u1ebft kim lo\u1ea1i \u0111\u1ec1u c\u00f3 c\u00e1c t\u00ednh ch\u1ea5t v\u1eadt l\u00ed chung: T\u00ednh d\u1ebbo, t\u00ednh d\u1eabn \u0111i\u1ec7n, t\u00ednh d\u1eabn nhi\u1ec7t, \u00e1nh kim <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":2137},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Tr\u1eadt t\u1ef1 t\u0103ng d\u1ea7n m\u1ee9c \u0111\u1ed9 ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc c\u1ee7a c\u00e1c kim lo\u1ea1i l\u00e0:","select":["A. Al, Mg, Cu ","B. Cu, Fe, Na","C. Zn, Fe, Ag ","D. Pb, K, Cu"],"hint":"","explain":"<span class='basic_left'>D\u00e3y ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc c\u1ee7a kim lo\u1ea1i: (\u0111\u1ed9 ho\u1ea1t \u0111\u1ed9ng h\u00f3a h\u1ecdc gi\u1ea3m d\u1ea7n)K, Na, Mg, Al, Zn, Fe, Pb, H, Cu, Ag, Au<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2138},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cu c\u00f3 th\u1ec3 cho \u0111\u01b0\u1ee3c ph\u1ea3n \u1ee9ng v\u1edbi","select":["A. Dung d\u1ecbch $HCl$ ","B. Dung d\u1ecbch $H_2SO_4$ lo\u00e3ng","C. Dung d\u1ecbch $H_2SO_4$ \u0111\u1eb7c, n\u00f3ng ","D. Dung d\u1ecbch $NaOH$."],"hint":"","explain":"<span class='basic_left'>$Cu+2{{H}_{2}}S{{O}_{4\\text{\u0111}}}\\xrightarrow{{{t}^{0}}}CuS{{O}_{4}}+S{{O}_{2}}+2{{H}_{2}}O$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2139},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" \u0110i\u1ec1u n\u00e0o sai khi n\u00f3i v\u1ec1 kim lo\u1ea1i","select":["A. \u0110\u1ec1u \u1edf th\u1ec3 r\u1eafn ","B. C\u00f3 \u00e1nh kim","C. C\u00f3 t\u00ednh d\u1ebbo ","D. D\u1eabn \u0111i\u1ec7n t\u1ed1t"],"hint":"","explain":"<span class='basic_left'>Hg \u1edf th\u1ec3 l\u1ecfng<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2140},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Nh\u00f4m c\u00f3 th\u1ec3 ph\u1ea3n \u1ee9ng v\u1edbi d\u00e3y ch\u1ea5t n\u00e0o sau \u0111\u00e2y:","select":["A. $H_2O, CuSO_4, H_2SO_4$(\u0111\u1eb7c,ngu\u1ed9i) ","B. $CuO, Ba(OH)_2, AgN0_3$","C. $H_2SO_4$ (\u0111\u1eb7c,ngu\u1ed9i), $CuO, HCl$","D. $O_2 ,MgCl_2,CuSO_4$"],"hint":"","explain":"<span class='basic_left'>Nh\u00f4m kh\u00f4ng t\u00e1c d\u1ee5ng v\u1edbi $H_2SO_4$ (\u0111\u1eb7c,ngu\u1ed9i), $MgCl_2$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2141},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho 26 gam k\u1ebdm ph\u1ea3n \u1ee9ng v\u1eeba \u0111\u1ee7 v\u1edbi 200 gam dung d\u1ecbch $H_2SO_4$. N\u1ed3ng \u0111\u1ed9 ph\u1ea7m tr\u0103m c\u1ee7a $H_2SO_4$ \u0111em d\u00f9ng l\u00e0","select":["A. 19,6%","B. 15%","C. 20% ","D. 25,6%"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& Zn+{{H}_{2}}S{{O}_{4}}\\to ZnS{{O}_{4}}+{{H}_{2}}\\uparrow \\\\ & {{n}_{{{H}_{2}}S{{O}_{4}}}}={{n}_{Zn}}=0,4\\,mol \\\\ & {{m}_{{{H}_{2}}S{{O}_{4}}}}=39,2\\,gam \\\\ & C{{\\%}_{\\text{dd}}}=\\dfrac{39,2}{200}.100\\%=19,6\\% \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2142},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Trong h\u1ee3p ch\u1ea5t oxit c\u1ee7a kim lo\u1ea1i A th\u00ec oxi chi\u1ebfm 17,02% theo kh\u1ed1i l\u01b0\u1ee3ng. Kim lo\u1ea1i A l\u00e0","select":["A. Cu ","B. Zn","C. K ","D. Na"],"hint":"","explain":"<span class='basic_left'>G\u1ecdi c\u00f4ng th\u1ee9c oxit l\u00e0 $A_2O_x$ <br\/>$\\begin{aligned}& \\%{{m}_{O}}=\\dfrac{16x}{2A+16x}.100\\%=17,02\\% \\\\ & =>A=39x \\\\ \\end{aligned}$ <br\/>Th\u1eed x l\u1ea7n l\u01b0\u1ee3t b\u1eb1ng 1, 2, 3 => A l\u00e0 K<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2143},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho 5,6 gam Fe t\u00e1c d\u1ee5ng 100 ml dung d\u1ecbch HCl 1M. Th\u1ec3 t\u00edch kh\u00ed $H_2$ thu \u0111\u01b0\u1ee3c (\u0111ktc) l\u00e0","select":["A. 1,12 l\u00edt","B. 2,24 l\u00edt","C. 22,4 l\u00edt","D. 1 l\u00edt"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{Fe}}=0,1\\,mol \\\\ & {{n}_{HCl}}=0,1\\,mol \\\\ & Fe+2HCl\\to FeC{{l}_{2}}+{{H}_{2}}\\uparrow \\\\ & \\dfrac{{{n}_{Fe}}}{1}>\\dfrac{{{n}_{HCl}}}{2} \\\\ \\end{aligned}$ <br\/>=> Fe d\u01b0, HCl h\u1ebft. T\u00ednh theo HCl <br\/>$\\begin{aligned}& {{n}_{{{H}_{2}}}}=\\dfrac{1}{2}{{n}_{HCl}}=0,05\\,mol \\\\ & {{V}_{{{H}_{2}}}}=1,12\\,l \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2144},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" H\u00f2a tan ho\u00e0n to\u00e0n 6,4 gam \u0111\u1ed3ng v\u00e0o dung d\u1ecbch axit sunfuric d\u01b0 thu \u0111\u01b0\u1ee3c V l\u00edt kh\u00ed sunfur\u01a1 (s\u1ea3n ph\u1ea9m kh\u1eed duy nh\u1ea5t). Gi\u00e1 tr\u1ecb c\u1ee7a V l\u00e0","select":["A. 1,12 l\u00edt","B. 2,24 l\u00edt","C. 4,48 l\u00edt","D. 5,6 l\u00edt"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{Cu}}=0,1\\,mol \\\\ & Cu+2{{H}_{2}}S{{O}_{4\\text{\u0111}}}\\xrightarrow{{{t}^{0}}}CuS{{O}_{4}}+S{{O}_{2}}+2{{H}_{2}}O \\\\ & {{n}_{S{{O}_{2}}}}={{n}_{Cu}}=0,1\\,mol \\\\ & {{V}_{S{{O}_{2}}}}=2,24\\,l \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2145},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Nh\u00fang m\u1ed9t l\u00e1 nh\u00f4m v\u00e0o dung d\u1ecbch $CuSO_4$. Sau m\u1ed9t th\u1eddi gian, l\u1ea5y l\u00e1 nh\u00f4m ra kh\u1ecfi dung d\u1ecbch th\u00ec th\u1ea5y kh\u1ed1i l\u01b0\u1ee3ng dung d\u1ecbch gi\u1ea3m 1,38g. Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a Al \u0111\u00e3 tham gia ph\u1ea3n \u1ee9ng l\u00e0:","select":["A. 0,27g","B. 0,54g","C. 0,81g","D. 1,08g"],"hint":"","explain":"<span class='basic_left'>G\u1ecdi x l\u00e0 s\u1ed1 mol nh\u00f4m \u0111\u00e3 tham gia ph\u1ea3n \u1ee9ng. <br\/>$\\begin{aligned}& 2Al+3CuS{{O}_{4}}\\to A{{l}_{2}}{{(S{{O}_{4}})}_{3}}+3Cu \\\\ & x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,1,5x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0,5x \\\\ \\end{aligned}$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng dung d\u1ecbch gi\u1ea3m b\u1eb1ng kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a $CuSO_4$ tr\u1eeb kh\u1ed1i l\u01b0\u1ee3ng $Al_2SO_4$ <br\/>$\\begin{aligned}& =>1,38=160.1,5x-342.0,5x \\\\ & =>1,38=69x \\\\ & =>x=0,02\\,mol \\\\ & {{m}_{Al}}=0,54\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2146},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho 1,5 gam h\u1ed7n h\u1ee3p (X) g\u1ed3m Mg v\u00e0 MgO t\u00e1c d\u1ee5ng v\u1edbi axit HCl d\u01b0, thu \u0111\u01b0\u1ee3c 336 $cm^3$ kh\u00ed $H_2$ (\u0111ktc). Th\u00e0nh ph\u1ea7n ph\u1ea7m tr\u0103m c\u1ee7a m\u1ed7i ch\u1ea5t trong (X) l\u00e0:","select":["A. 50% Mg v\u00e0 50% MgO","B. 25% Mg v\u00e0 75% MgO","C. 24% Mg v\u00e0 76% MgO","D. 30% Mg v\u00e0 70% MgO"],"hint":"","explain":"<span class='basic_left'>\u0110\u1ed5i 336 $cm^3$ = 0,336 l\u00edt<br\/>$\\begin{aligned}& {{n}_{{{H}_{2}}}}=0,015\\,mol \\\\ & Mg+2HCl\\to MgC{{l}_{2}}+{{H}_{2}} \\\\ & MgO+2HCl\\to MgC{{l}_{2}}+{{H}_{2}}O \\\\ & {{n}_{Mg}}={{n}_{{{H}_{2}}}}=0,015\\,mol \\\\ & {{m}_{Mg}}=0,36\\,gam \\\\ & \\%{{m}_{Mg}}=\\dfrac{0,36}{1,5}.100\\%=24\\% \\\\ & \\%{{m}_{MgO}}=76\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2147},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" M\u1ed9t lo\u1ea1i qu\u1eb7ng Boxit c\u00f3 h\u00e0m l\u01b0\u1ee3ng Al2O3 \u0111\u1ea1t 40%. T\u1eeb 10 t\u1ea5n qu\u1eb7ng tr\u00ean c\u00f3 th\u1ec3 s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c m\u1ed9t l\u01b0\u1ee3ng nh\u00f4m t\u1ed1i \u0111a l\u00e0","select":["A. 2,12 t\u1ea5n","B. 1,055 t\u1ea5n","C. 0,502 t\u1ea5n ","D. 0,25 t\u1ea5n"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{m}_{A{{l}_{2}}{{O}_{3}}}}=\\dfrac{10.40\\%}{100\\%}=4\\,\\text{t\u1ea5n} \\\\ & 2A{{l}_{2}}{{O}_{3}}\\xrightarrow{\\text{\u0111}pnc}4Al+3{{O}_{2}} \\\\ & 2.102\\,\\text{t\u1ea5n}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4.27\\,\\text{t\u1ea5n} \\\\ & 4\\,\\text{t\u1ea5n}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\text{t\u1ea5n} \\\\ & \\text{x=}\\dfrac{4.4.27}{2.102}\\approx 2,12\\,\\text{t\u1ea5n} \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2148},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" \u0110\u1ed1t 1,62 gam kim lo\u1ea1i M c\u00f3 h\u00f3a tr\u1ecb III. L\u1ea5y to\u00e0n b\u1ed9 s\u1ea3n ph\u1ea9m \u0111em t\u00e1c d\u1ee5ng v\u1eeba \u0111\u1ee7 v\u1edbi 180ml dung d\u1ecbch HCl 1M. Kim lo\u1ea1i M l\u00e0: ","select":["A. Fe","B. Al","C. Cr","D. Mn"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& 4M+3{{O}_{2}}\\xrightarrow{{{t}^{o}}}2{{M}_{2}}{{O}_{3}} \\\\ & {{M}_{2}}{{O}_{3}}+6HCl\\to 2MC{{l}_{3}}+3{{H}_{2}}O \\\\ & {{n}_{M}}=2{{n}_{{{M}_{2}}{{O}_{3}}}}=2.\\dfrac{1}{6}{{n}_{HCl}}=\\dfrac{1}{3}.0,18\\,mol=0,06\\,mol \\\\ & {{M}_{M}}=\\dfrac{1,62}{0,18}=27\\,(Al) \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2149},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Ho\u00e0 tan 4,54 gam h\u1ed7n h\u1ee3p Al, Fe, Cu trong dung d\u1ecbch HCl d\u01b0, thu \u0111\u01b0\u1ee3c 1,792 l\u00edt $H_2$ (\u0111ktc) v\u00e0 1,2 gam kim lo\u1ea1i. Ph\u1ea7n tr\u0103m kh\u1ed1i l\u01b0\u1ee3ng m\u1ed7i kim lo\u1ea1i trong h\u1ed7n h\u1ee3p ban \u0111\u1ea7u l\u1ea7n l\u01b0\u1ee3t l\u00e0","select":["A. 11,9%; 61,7% v\u00e0 26,4% ","B. 51,8%; 12,8% v\u00e0 32,4% ","C. 61,8%; 11,8% v\u00e0 26,4% ","D. 32,4%; 28,4% v\u00e0 39,2%"],"hint":"","explain":"<span class='basic_left'>Kh\u1ed1i l\u01b0\u1ee3ng kim lo\u1ea1i kh\u00f4ng tan (1,2 gam) ch\u00ednh l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a \u0111\u1ed3ng. <br\/>$=>{{m}_{Al+Fe}}=4,54-1.2=3,34\\,gam$ <br\/>G\u1ecdi x, y l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 mol c\u1ee7a Al v\u00e0 Fe trong h\u1ed7n h\u1ee3p.<br\/>=> 27x + 56 y = 3,34. (1) <br\/>$\\begin{aligned}& 2Al+6HCl\\to 2AlC{{l}_{3}}+3{{H}_{2}} \\\\ & x\\,mol\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,1,5\\,x\\,mol \\\\ & Fe+2HCl\\to FeC{{l}_{2}}+{{H}_{2}} \\\\ & y\\,mol\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y\\,mol \\\\ & {{n}_{{{H}_{2}}}}=0,08\\,mol \\\\ & =>1,5x+y=0,08\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(2) \\\\ \\end{aligned}$ <br\/>Gi\u1ea3i (1) v\u00e0 (2) ta c\u00f3 x = 0,02; y = 0,05<br\/>$\\begin{aligned}& \\%{{m}_{Cu}}=\\dfrac{1,2}{4,54}.100\\%\\approx 26,4\\% \\\\ & \\%{{m}_{Al}}=\\dfrac{0,02.27}{4,54}.100\\%\\approx 11,9\\% \\\\ & \\%{{m}_{Fe}}=61,7\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2150}],"lesson":{"save":0,"level":2}}