7 NGÀY lấy gốc CHINH PHỤC 7+ hóa

Th y Ph m Minh Thu n https://www.facebook.com/minhthuan.chemistry I / , y = a hay a/ / , b/ ; x= 1; y = SO2 Vd 2: (II) x=2&y=1 2SO4 S d y h t D y online t i Mclass Th y Ph m Minh Thu n https://www.facebook.com/minhthuan.chemistry II S d y h t D y online t i Mclass Th y Ph m Minh Thu n S d y h t https://www.facebook.com/minhthuan.chemistry D y online t i Mclass Th y Ph m Minh Thu n S d y h t https://www.facebook.com/minhthuan.chemistry D y online t i Mclass Th y Ph m Minh Thu n https://www.facebook.com/minhthuan.chemistry : Riêng H2SO4 2(+1) + x + 4(-2) = x = +6 K2Cr2O7 2(+1) + 2x + 7(-2) = x = +6 ion Mg2+ -2) = -1 x = +7 B1 B2 - ne B3 B4 kim phi hidro oxi + 6e 2e Fe2O3 VD: Cu + HNO3 3+ Fe 2H+ + 3H2 q trình oxi hố H2 2Fe + 3H2O Cu(NO3)2 + NO + H2O - S d y h t D y online t i Mclass Th y Ph m Minh Thu n proton 10 11 12 13 14 15 16 17 18 19 20 24 25 26 29 30 35 47 56 80 82 phi kim https://www.facebook.com/minhthuan.chemistry Heli Liti Beri Bo Cacbon Oxi Flo Neon Natri Magie Nhôm Silic Photpho Clo Agon Kali Canxi Crom Mangan Brom Bari Chì H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca 11 12 14 16 19 20 23 24 27 28 31 32 35,5 39,9 39 40 Cr Mn Fe Cu Zn Br Ag Ba Hg Pb 52 55 56 64 65 80 108 137 201 207 I I II III IV, II II I I II III IV III, V II, IV, VI I II II, III I, II II I II I, II II, IV Tên nhóm 3) Sunfat( SO4), Cacbonat( CO3) Photphat( PO4) S d y h t I II III D y online t i Mclass Th y Ph m Minh Thu n https://www.facebook.com/minhthuan.chemistry Câu 1: 2, NaOH, CuSO4 Câu 2: NaOH HCl H2SO4 CuSO4 HCl Ba(OH)2 Câu 3: a b FeCl3 Fe2(SO4)3 CuO Fe(OH)3 Fe2O3 Cu(OH)2 Câu a Magiê oxit (MgO) axit nitric (HNO3) b Nhôm oxit (Al2O3) axit sunfuric (H2SO4) Câu 5: Axit H2SO4 A FeCl3, MgO, Cu, Ca(OH)2 C Mg(OH)2, HgO, K2SO4, NaCl Câu 6: A B C B NaOH, CuO, Ag, Zn D Al, Al2O3, Fe(OH)3, BaCl2 D Câu 3)2, a dd NaOH b dd HCl CuCl2 c dd AgNO3 Câu 1: Câu 2: S d y h t D y online t i Mclass Th y Ph m Minh Thu n CuSO4 HCl Ba(OH)2 https://www.facebook.com/minhthuan.chemistry NaOH x x HCl 0 x H2SO4 0 x Câu 3: Câu Câu 5: H2SO4 kh H2SO4 Câu 6: S d y h t D y online t i Mclass Th y Ph m Minh Thu n https://www.facebook.com/minhthuan.chemistry ( xan ) Câu )2 Mg(OH)2 + CuCl2 S Cu(OH)2 d y h t D y online t i Mclass Th y Ph m Minh Thu n https://www.facebook.com/minhthuan.chemistry I V: t CM ) S d y h t D y online t i Mclass Th y Ph m Minh Thu n https://www.facebook.com/minhthuan.chemistry B BTO ta có: m S ( amol) N2 (b mol) = d y h t D y online t i Mclass Th y Ph m Minh Thu n https://www.facebook.com/minhthuan.chemistry , 0,09g H2O 3 2,7g H2 CTPT : 224 cm3 CO2 0.24 g H2 Tìm CTPT A 2H2 sinh lít khí CO2 A.C2H6 = 1.75 nCO2 A C3H6O2N B C2H4 D C3H8 , H O N2 2 2 B C2H7O2N2 C C2H7O2N xHyNtSr x:y:t:r= C CH4 t nH2O D C3H5ON2 ta có : =0 ) 4N2S)n CH4N2S nC = nCO2 = 0.22/44 = 0.005 mol nH2 = nH2O = 0.09/18*2 = 0.01 mol nAgCl = nCl x : y : v = 0.005 : 0.01 : 0.01 = 1:2:2 n= 2Cl2 2Cl2)n Ta có MA = 5*17 = 85 xHyOz mA = mCO2 + mH2O A + mO = mCO2 + mH2O mO = 2.24/22.4*44 + 2.7 0.15*32 = 2.3 g Ta có mC = 2.24/22.4*12 = 1.2 g ; mH = 2.7/18*2 = 0.3 g mO = 2.3 - 1.2 0.3 = 0.8 g x : y : z = 1.2/12 : 0.3/1 : 0.8/16 = 2:6:1 2H6O S d y h t D y online t i Mclass Th y Ph m Minh Thu n https://www.facebook.com/minhthuan.chemistry xHy xHy 2 & H2O } + H 2O } nC = 0.09 mol mH2O = 4.86 3.96 = 0.9 nH2 = d 3) : nCO2 = nCaCO3 = 0.09 mol CO2 = 0.09*44 = 3.96 g 0.9/18*2 = 0.1 mol x : y = 0.09 : 0.1 = 9:10 1.08+0.1)/0.01 = 118 9H10 9H10 Ngồi ta có M = m/n = ( , p) n VC = lit; V H = V O = VO2 xHy ta có x:y = 3:8 O2 n=3 C +VH C3H8 X : CnH2n+2 ta có pt sau : C3H8 CxHy có M = 74-16 = 58 Ta có : 12x + y = 58 x =4 y = 58 12x CTPT C4H10O 4H10O ; C3H6O2 ; C2H2O3 xHyOz chia cho 2/75) C ta có mC = 0.12 g ; mH = 2/75 3H8)nOz 44n < 76 n < 1.7 CTPT C3H8O2 n n 2;n 2n C : H : 1: S d y h t x:y = (0.12/12) : ( 2/75 ) = 3/8 ( Vì mO (12*3+8)n + 16z = 76 44n + 16z = 76 16z = 76 44*1 z = CH D y online t i Mclass Th y Ph m Minh Thu n S d y h t https://www.facebook.com/minhthuan.chemistry D y online t i Mclass Th y Ph m Minh Thu n https://www.facebook.com/minhthuan.chemistry theo danh pháp IUPAC VD: C2H5 Cl: Etyl clorua; C2H5 O CH3: Etyl metyl ete - tert- VD: H3C CH3 - CH3: et+an (etan); C2H5 CH=CH Cl: clo+et+an (cloetan); CH3: but-2-en; CH3 CH(OH) CH = CH2: but-3-en-2-ol Chú ý: -COOH>-CHO>-OH>-NH2>-C=C>VD: OHCOHC- -CH2-CH2-C(CH=CH2)=CH-CHO: 3-vinylhept-2-en-6-inal -CH2-CH2-C(CH=CH2)=CH-CHO: 3-vinyloct-2-en-6- Tên Mono Met Et Tri Prop Tetra But Penta Pent Hexa Hex Hepta Hept Octa Oct Nona Non 10 S d y h t D y online t i Mclass Th y Ph m Minh Thu n CH3-: metyl; https://www.facebook.com/minhthuan.chemistry CH3-CH2-: etyl; CH3-CH2-CH2-: propyl; CH3-CH(CH3)-: isopropyl; CH3[CH2]2CH2-: butyl; CH3-CH(CH3)-CH2-: isobutyl; CH3-CH2-CH(CH3)-: sec-butyl (CH3)3C-: tert-butyl; CH3-CH(CH3)-CH2-CH2-: isoamyl -: vinyl; CH2=CH-CH2-: anlyl -: phenyl; C6H5-CH2-: benzyl -xeton: -CHO: fomyl; -CH2-CHO: fomyl metyl; CH3-CO-: axetyl; C6H5CO-: benzoyl ANKAN: CnH2n+2 a) Ankan không phân nhánh ANKAN: CnH2n+2 -CnH2n+1 Tên (Theo IUPAC) Tên CH4 Metan CH3- Metyl CH3CH3 Etan CH3CH2- Etyl CH3CH2CH3 Propan CH3CH2CH2- Propyl CH3[CH2]2CH3 Butan CH3[CH2]2CH2- Butyl CH3[CH2]3CH3 Pentan CH3[CH2]3CH2- Pentyl CH3[CH2]4CH3 Hexan CH3[CH2]4CH2- Hexyl CH3[CH2]5CH3 Heptan CH3[CH2]5CH2- Heptyl CH3[CH2]6CH3 Octan CH3[CH2]6CH2- Octyl CH3[CH2]7CH3 Nonan CH3[CH2]7CH2- Nonyl CH3[CH2]8CH3 CH3[CH2]8CH2- a: b: CH3 CH2 CH | CH CH3 | CH3 | S d y h t D y online t i Mclass Th y Ph m Minh Thu n https://www.facebook.com/minhthuan.chemistry 3-etyl-2-metylpentan M XICLOANKAN: CnH2n (n>=3) - VD: Xiclo+hex+an Metyl+xiclo+pent+an 1,2- +an 1,1,2- trimetyl+xiclo+prop+an (Xiclohexan) (Metylxiclopentan) (1,2- (1,1,2- trimetylxiclopropan) ANKEN: CnH2n (n>=2) CH2=CH2: etilen; CH2=CH-CH3: propilen; CH2=CH-CH2CH3-CH=CH- -butilen; -butilen; CH2=C(CH3)-CH3: isobutilen - - -en - CH2=CHCH2CH2CH3: pent-1-en; CH3CH=CHCH2CH3: pent-2-en; CH2=C(CH3)-CH2CH3: 2-metylbut-1-en; CH3C(CH3)=CHCH3: 2-metylbut-2-en S d y h t D y online t i Mclass Th y Ph m Minh Thu n https://www.facebook.com/minhthuan.chemistry àng cao - H