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9 months ago

Bộ đề thi thử THPT QG 2018 Các môn TOÁN - LÍ - HÓA Các trường THPT Cả nước CÓ HƯỚNG DẪN GIẢI (Lần 8) [DC11042018]

https://app.box.com/s/nhxs7vnw9na94twmj7hc0inbdjzwv940

https://twitter.com/daykemquynhon plus.google.com/+DạyKèmQuyNhơn www.facebook.com/daykem.quynhon http://daykemquynhon.blogspot.com http://daykemquynhon.ucoz.com Nơi bồi dưỡng kiến thức Toán - Lý - Hóa cho học sinh cấp 2+3 / Diễn Đàn Toán - Lý - Hóa 1000B Trần Hưng Đạo Tp.Quy Nhơn Tỉnh Bình Định ♦ Thủy phân X thu được 2 muối nên cấu tạo của nó phải chứa cả 2 gốc oleat và linoleat. dù cái nào thì triglixerit cũng chứa 3 gốc C 17 H ???? COO– ||→ số C của X = 18 × 3 + 3 = 57. thấy ngay đáp án D sai. tạo X do 1 gốc oleat với 2 gốc linoleat hoặc 2 gốc oleat với 1 gốc linoleat ||→ số π C=C là 4 hoặc 5. làm gì có 3 π C=C được → phát biểu B cũng sai luôn. Hiđro hóa hoàn toàn X thì rõ rồi, gốc no C 17 là C 17 H 35 COO là stearat → C sai nốt. Loại trừ đã đủ để chọn đáp án A rồi. Giải thì sao? À, cũng khá đơn giản, như biết trên, X có dạng (C 17 H ??? COO) 3 C 3 H 5 hoặc gọn hơn C 57 H ???? O 6 . n CO2 = 1,71 mol → n X = 0,03 mol; n O trong X = 0,18 mol, biết n O2 cần đốt = 2,385 mol. ||→ bảo toàn O có n H2O = 1,53 mol ||→ m X = m C + m H + m O = 26,46 gam. Rõ hơn nữa, số H của X = 102 cho biết X được tạo từ 2 gốc oleat (C 17 H 33 ) và 1 gốc linoleat (C 17 H 31 ). Tóm lại đáp án cần chọn là A. Câu 38: Đáp án B ► Quy M về C₂H₂NO, CH₂ và H₂O. m bình tăng = mCO₂ + mH₂O = 24,62(g). Bảo toàn khối lượng: mN₂ = 10,74 + 0,495 × 32 - 24,62 = 1,96(g) ⇒ nN₂ = 0,07 mol ⇒ nC₂H₂NO = 0,07 × 2 = 0,14 mol nO₂ = 2,25nC₂H₂NO + 1,5nCH₂ ⇒ nCH₂ = 0,12 mol ⇒ nH₂O = (10,74 - 0,14 × 57 - 0,12 × 14) ÷ 18 = 0,06 mol ► Dù lấy bao nhiêu mol thì %m muối Gly vẫn không đổi ⇒ xét về 10,74(g) M cho tiện :P Muối gồm 0,14 mol C₂H₂NO₂Na và 0,12 mol CH₂ ⇒ m muối = 0,14 × 97 + 0,12 × 14 = 15,26(g) ⇒ nGly-Na = 15,26 × 38,14% ÷ 97 = 0,06 mol ● Đặt nAla = x; nVal = y ⇒ nC₂H₂NO = 0,06 + x + y = 0,14 mol nCH₂ = x + 3y = 0,12 mol ||⇒ giải hệ có: x = 0,06 mol; y = 0,02 mol ⇒ %mVal-Na = 0,02 × 139 ÷ 15,26 × 100% = 18,22% Câu 39: Đáp án B Câu 40: Đáp án A Y gồm 2 muối natri của axit stearic và oleic. Hai phần chia ra bằng nhau nên: n muối axit oleic = n C17H33COONa = n Br2 = 0,12 mol. Lại thêm m C17H33COONa + m C17H35COONa = 54,84 gam DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN ||→ n C17H35COONa = 0,06 mol ||→ n C17H33COONa ÷ n C17H33COONa = 2 ÷ 1. ||→ triglyxerit X được tạo bởi 2 gốc oleic và 1 gốc stearic MỌI YÊU CẦU GỬI VỀ HỘP MAIL DAYKEMQUYNHONBUSINESS@GMAIL.COM HOTLINE : +84905779594 (MOBILE/ZALO) https://daykemquynhonofficial.wordpress.com/blog/ DẠY KÈM QUY NHƠN OFFICIAL ST> : Đ/C 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN Trang 11 Đóng góp PDF bởi GV. Nguyễn Thanh Tú www.facebook.com/daykemquynhonofficial www.facebook.com/boiduonghoahocquynhonofficial

https://twitter.com/daykemquynhon plus.google.com/+DạyKèmQuyNhơn www.facebook.com/daykem.quynhon http://daykemquynhon.blogspot.com ||→ M X = 886. http://daykemquynhon.ucoz.com Nơi bồi dưỡng kiến thức Toán - Lý - Hóa cho học sinh cấp 2+3 / Diễn Đàn Toán - Lý - Hóa 1000B Trần Hưng Đạo Tp.Quy Nhơn Tỉnh Bình Định ----- HẾT ----- DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN MỌI YÊU CẦU GỬI VỀ HỘP MAIL DAYKEMQUYNHONBUSINESS@GMAIL.COM HOTLINE : +84905779594 (MOBILE/ZALO) https://daykemquynhonofficial.wordpress.com/blog/ DẠY KÈM QUY NHƠN OFFICIAL ST> : Đ/C 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN Trang 12 Đóng góp PDF bởi GV. Nguyễn Thanh Tú www.facebook.com/daykemquynhonofficial www.facebook.com/boiduonghoahocquynhonofficial

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CHƯƠNG 8 AXIT CACBOXYLIC VÀ CÁC DẪN XUẤT
Bộ đề luyện thi THPT Quốc Gia Môn Ngữ Văn (2016)
CÁC TRẠNG THÁI VẬT LÝ CỦA POLYMER
ĐỘC HỌC MÔI TRƯỜNG VÀ SỨC KHỎE CỘNG ĐỒNG