chuyen de bai tap toan lop 4

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... RRIR R R R= =+ ++=+ + +(3)Từ (2) rút ra 2 4 12R R= − và thay giá trị này của R2 vào (3):( )( ) ( ) 4 2 4 2 4 4 4 4 4 42 4 4 4 30 1 21 3 330 1 21 12 3 3 123 8 3 03R R R R RR ... +120,61I AI A⇒ ==Ta có:3 4 30 ,4 83 1,80 ,4 86,2 0 ,4. 1,88ACBI I IUx xx xxx x= ++ =−⇔ + =−−⇔ =−2220 ,4 6,2 14, 4 1,8.0 ,4 4 ,4 14, 4 020. 36 0x x xx xx x⇔ − + ... +22 14, 4. 4, 8.5. 6 5. 6CBUx xIR x x= = =+ +Ta có: 1 20 ,4 3,2.(3 ) 4, 8.0 ,4 5. 6 5. 6I I Ix xx x= − =+⇔ − =+ +3,2.(3 ) 4, 8. 0 ,4. (5. 6)x x x⇔ + − = +9,6 3,2. 4, 8. 2. 2 ,4 3,6....

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... p-CH3C6H 4 C2H5 (2); C6H5C2H3 (3); o-CH3C6H 4 CH3 (4) Dãy gồm các chất là đồng đẳng của benzen là: A. (1); (2) và (3). B. (2); (3) và (4) . C. (1); (3) và (4) . D. (1); (2) và (4) . ... CO2 (đktc). Giá trị của m và số tổng số mol của A, B là: A. 4, 59 và 0, 04. B. 9,18 và 0,08. C. 4, 59 và 0,08. D. 9, 14 và 0, 04. Câu 76: Đốt cháy hết 9,18 gam 2 đồng đẳng của benzen A, B thu ... D. đimetyl axetilen. Câu 47 : Stiren không phản ứng được với những chất nào sau đây ? A. dd Br2. B. không khí H2 ,Ni,to. C. dd KMnO 4 . D. dd NaOH. Câu 48 : A + 4H2  etyl xiclohexan....

... 0, 34 mol = 0,17.2 = 0, 34 mol n hhA = 9 ,40 8 : 22 ,4 = 0 ,42 mol n hhA = 9 ,40 8 : 22 ,4 = 0 ,42 mol Goïi x , y , z là số mol CGọi x , y , z laø soá mol C22HH22 , C , C22HH 4 4 ... CC22HH22 + 2Br + 2Br22  C C22HH22BrBr 4 4 CC22HH 4 4 + Br + Br22  C C22HH 4 4BrBr22nCnC66HH66 = 1,56 : 78 = 0,02 mol => ... = 0,5 mol⇒Goïi a mol CGoïi a mol C22HH 4 4 đã phản ứng đã phản ứng Theo PT : n CTheo PT : n C22HH66 = n C = n C22HH 4 4 = a mol = a mol t0Ni Bài giảiBài...

... thu được sắt và khí CO2 .-Nếu cho lượng sắt ở trên vào dd HNO3 đặc , nóng , dư thu được 13 ,44 lít khí NO2 ở đktc và dd chứa Fe(NO3)3 .-Nếu cho khí CO2 hấp thụ hết vào dd Ca(OH)2 ... = a + (5a/68)32 – (6a/68)18 = 120a /68 (g)⇒nNO = mNO : 30 = 4a/68 (mol) Ta coù : mO = mO(NO) + mO(H2O) = (4a/68).16 + (6a/68).16 = 160a/68Maø : mO(O2) = 5a/68 .32 = 160a/68 ... A là hợp chất của N và H . Gọi công thức hóa học của A là NxHy .⇒Ta có tỉ lệ : x : y = (4a/68): (12a/68) x : y = 1 : 3 ⇒ A laø (NH3)n . Maø A chứa 1 nguyên tử N⇒ A là NH3 ...

... khi gọi x, y là số mol của Mg và Zn đã phản ứng . 2,02 – ( 24x + 65y ) + 85x + 136y = 4, 86 => x + y = (4, 86 – 2.02) : 71 = 0, 04 Bài giảiVậy chất rắn không tan ở TN1 và TN2 chỉ gồm ... và Zn vào cốc (1)đựng 200ml dd HCl .Sau phản ứng cô cạn dd được 4, 86g chất rắn . Cho 2,02g hỗn hợp Mg và Zn vào cốc (2)đựng 40 0ml dd HCl như trên , sau phản ứng cô cạn dd được 5,57g chất ... ta coù :  soá mol K2O = ẵ soỏ mol KOH tan = 18:102 (mol)Khoỏi lửụùng K2O = (18:102 ). 94 = 16,59g . Bài tóan về lượng chất dưCho hh A gồm 3 oxit : Al2O3 ; K2O ; CuO Tieán...

... ddKMnO 4 )SO2Mất màu nâuđỏ (hay màu tím)SO2 + Br2 + H2O  2HBr + H2SO 4 SO2 + KMnO 4 + 2H2O 2H2SO 4 +2MnSO 4 +K2SO 4 Dd AgNO3HCl Kết tủa trắng AgNO3 + HCl  AgCl + HNO3 Dd Pb(NO3)2H2S ... hãy nhận biết 4 dung dịch đựng trong các lọ mất nhãnsau : NaOH , Ba(OH)2 , KCl và K2SO 4 (lớp 9) . Ví dụ 4 : Chỉ dùng thêm một chất thử duy nhất (tự chọn) hãy nhận biết 4 dung dịch đựngtrong ... tự lập được sơ đồnhận biết các chất Ví dụ : Nhận biết 4 lọ mất nhãn , mỗi lọ đựng một trong 4 dung dịch sau NaOH ,Na2SO 4 , H2SO 4 loóng v HCl .ã Giỏo viờn cú thể đặt một số câu hỏi...

... chiều cao SO = h và góc SAB = α (α> 45 0). Tính diện tích xung quanh của hình nón đỉnh S và có đường tròn đáy ngoại tiếp hình vuông ABCD.Bài 14: Cho một đường tròn nằm trên mặt phẳng ... một điểm A với OA = 2r. qua A kẻ một tiếp tuyến với mặt cầu tại B. khi đó độ dài đoạn AB bằng: 4 Trường THPT Lê Lợi-Phan Thiết GV : Trần Phú HiếuBài 17: Một khối trụ có bán kính r = 5cm, khoảng ... h, A và B là hai điểm thay đổi trên hai đường tròn đáy sao cho độ dài AB = a không đổi ()224Rhah+<>.a) Chứng minh góc giữa hai đường thẳng AB và OO’ không đổi.b) Chứng minh khoảng...