bai tap hoa lop 11 chuong1

... học lớp 11 theo chương trình SGK lớp 11 cơ bản. - Hệ thống các bài tập SGK lớp 11, các bài tập ở sách tham khảo và các câu hỏi trong đề thi đại học từ 2007 - 2 011. IV. GIẢ THUYẾT KHOA HỌC ... VÀ BÀI TẬP HÓA HỌC 11 CƠ BẢN VÀ NÂNG BAO, áp dụng cho chương trình hóa học lớp 11 chương trình SGK 11 cơ bản. II. MỤC ĐÍCH NGHIÊN CỨU Xây dựng hệ thống bài tập hóa học 11 từ dể đến khó phù ... Trường THPT Nguyễn Chí Thanh GV: Nguyễn Phú Hoạt Page 1 LÝ THUYẾT VÀ BÀI TẬP HÓA HỌC 11 CƠ BẢN VÀ NÂNG CAO I. LÝ DO CHỌN ĐỀ TÀI Mục tiêu xuyên suốt của người giáo viên trong giảng...

... dd X. cho 20 ml ddXphản ứng với 30 ml dd KOH 0,1M thu đợc ddY. pH của ddY là:A. 1,7 B. 7,0 C .11, 7 D.12,3Câu 52: Trộn 200ml dd gồm HCl 0,1M và H2SO4 0,05M với 300ml dd Ba(OH)2 có nồng ... 160 ml dd chứa đồng thời Ba(OH)2 0,08M và KOH 0,04M. giátrị pH của dd thu đợc là :A. 10 B. 11 C.12 D. 13Câu 54: (đề thi khối B -2008) Câu 55: (đề thi khối B -2008)Câu 56: (đề thi khối ... (tím), Fe2(SO4)3 (xanh) D. K2CO3 (đỏ) , Fe2(SO4)3 (tím), Ba(NO3)2 (xanh)Câu 11: Theo phơng trình ion thu gon thì ion CO32- không thể phản ứng với các ion:A. NH4+, Na+,...

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... không đổi thu được a gam chấtrắn. Giá trị của m và a là:A. 111 , 84g và 157,44g B. 111 , 84g và 167,44g C. 112 , 84g và 157,44g D .112 , 84g và 167,44g Hướng dẫn giảiTa có bán phản ứng:CuFeS2 ... BaSO4 0,48 0,48⇒ m = 0,48 × 233 = 111 ,84 gam.nCu = 0,33 mol; nFe = 0,24 mol.Cu → CuO 2Fe → Fe2O3 0,33 0,33 0,24 0,12⇒ a = 0,33 × 80 + 0,12 ×160 + 111 ,84 = 157,44 gam. (Đáp án A).Ví ... tạo thành m (g) kết tủa. Giá trị của V và m lần lượt làA. 11, 2 lít và 78,8 gam B. 20,16 lít và 78,8 gamC. 20,16 lít và 148,7 gam D. 11, 2 lít và 148,7 gamPhương pháp giải nhanh:Pt ion rút gọn:...

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... 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...

... sung phần nào kiến thức khiếm khuyết để bắt kịp với sự phát triểncủa khoa học kỹ thuật và đáp ứng đợc yêu cầu mới của khoa học giáo dục hiệnđại.II- mục đích và nhiệm vụ của đề tài1- Mục đích:- ... thông tin trong quá trình dạy học, đúc rút kinh nghiệm của bản thân- Nghiên cứu sách giáo khoa lớp 9,10 ,11 và các sách nâng cao về phơng phápgiải bài tập- Trực tiếp áp dụng đề tài đối với học ... dỡng nâng cao kiến thức để làm chủkiến thức tự tin trớc bài giảng và học sinh .- Hoá học là môn khoa học thực nghiệm vì vậy để khắc sâu kiến thứccho học sinh, giáo viên thờng xuyên làm các thí...

... CHƯƠNG II : DÒNG ĐIỆN KHÔNG ĐỔI Lớp 11 . Ban cơ bản Năm học :2010 -2 011 GV:NGUYỄN TIẾN THÀNH.THPT TRẦN PHÚ .KRÔNG NÔ. ĐĂK NÔNG. § 7. Dòng điện ... nguồn điện là acquy. D. Trong mạch điện kín thắp sáng đèn với nguồn điện là pin mặt trời. Câu 11: Hiệu điện thế 1V được đặt vào điện trở 10 trong khoảng thời gian là 20s. Lượng điện tích ... giờ là: A. 2500J B. 2,5 kWh C. 500J D. đáp án khác. Câu 37: Một bếp điện có công suất định mức 110 0W và hiệu điện thế định mức 220V. Điện trở của bếp bằng: A. 0,2  B. 20 C. 44  D. 440...

... 2:1. Tính thể tích của hỗn hợp khí X (đktc). Câu 5. Hòa tan hoàn toàn 11 gam hh gồm Fe và Al trong dd HNO3 dư thu được 11, 2 lít hh khí X (đktc) gồm NO và NO2 có khối lượng 19,8 gam. Biết ... dung dịch HNO3 đặc, dư đun nóng sinh ra 11. 2 lít khí NO2 (đktc) là sản phẩm khử duy nhất. Tính % khối lượng mỗi kim loại trong hỗn hợp ban đầu. Câu 11. Cho 19,5 gam một kim loại M hóa trị ... hỗn hợp khí B (đktc) gồm NO2 và NO có tỉ khối so với H2 là 19. Xác định giá trị của V. Câu 11. Thuỷ phân hoàn toàn 8,25 gam một photpho trihalogenua thu được dung dịch X. Để trung hoà X...

... 0.05 (mol) 2oxit Kl CO COm = m + m - m = 2.32 + 44*0.05 - 28*0.05 = 3.12 (gam) Ví dụ 3 (A-2 011) : Hấp thụ hoàn toàn 0,672 lít khí CO2 (đktc) vào 1 lít dung dịch gồm NaOH 0,025M và Ca(OH)2 ... bằng khí CO dư. Khí thu được sục vào dung dịch Ca(OH)2 dư thu được m gam kết tủa. Tính m. Câu 11. Khử hoàn toàn m gam Fe2O3 bằng khí CO dư. Khí thu được sục vào dung dịch Ca(OH)2 dư thu ... 0,04M và Ca(OH)2 0,02M thu được m gam kết tủa. Xác định giá trị của m. Câu 6. Hoà tan hoàn toàn 11, 2 gam CaO vào H2O thu được dung dịch A. Sục V lít khí CO2 (đktc) vào dung dịch A thu được...

... CnH2n (n≥3). - Thí dụ: (xiclopropan) (xiclobutan) b. Danh pháp: Số chỉ vị trí nhánh + tên nhánh + xicol + tên mạch C chính (vòng) + an - Thí dụ: (metylxiclopropan). 2. Tính chất hóa ... đẳng kế tiếp nhau ta thu được 11, 7g H2O và 17,6g CO2. Xác định CTPT của hai hidrocacbon trên. Giải Đặt CTPT của 2 ankan là n 2n 2CH. 22CO H O17.6 11. 7n = = 0.4 (mol); n = 0.65 ... (A-2 011) . Hấp thụ hoàn toàn 0,672 lít khí CO2 (đktc) vào 1 lít dung dịch gồm NaOH 0,025M và Ca(OH)2 0,0125M, thu được x gam kết tủa. Tính giá trị của x. Đáp án: x = 1,25 gam. Câu 15 (B-2 011) ....

... (đktc) thu được 11, 2 lít khí CO2 (đktc). a. Xác định % theo thể tích mỗi khí trong hỗn hợp đầu. b. Tính khối lượng nước sinh ra. Câu 36. Một hỗn hợp gồm hai anken có thể tích 11, 2 lít (đktc) ... hỗn hợp khí Y có tỉ khối so với H2 là 18. Biết M phản ứng hết. Xác định CTPT của M. Câu 11 (A-2 011) . Hỗn hợp X gồm C2H2 và H2 có cùng số mol. Lấy một lượng hỗn hợp X cho qua chất xúc ... (A-2 011) . Cho buta-1,3- đien phản ứng cộng với Br2 theo tỉ lệ mol 1:1. Viết CTCT số dẫn xuất đibrom (đồng phân cấu tạo và đồng phân hình học) thu được trong phản ứng trên. Câu 14 (B-2 011) ....

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... 2 and type 3: no tense changes“I would travel to Ha Long Bay if I had enough money”, said Hoai. Hoai said (that) she would travel Ha Long bay if she had enough money.“If Japan hadn’t attacked ... said my mother. (REMIND) 2. “It’s not a good idea to indulge the children so much”, said Mr. Hoang.(OBJECT) 3. Trong said to me, “I’m very grateful to you for your enthusiasm.” (THANK) ... feelings.(+)Start schoolMoved to a new houseWent to Da Lat with familySaid good bye to a friendAge 6 1 311 15(-)43§9. Infinitive and GerundRevision InfinitiveThere are 2 kinds of infinitive: infinitive...