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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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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