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7 months ago

Bộ đề thi thử THPT QG 2018 Các môn TOÁN - LÍ - HÓA Các trường THPT Cả nước CÓ HƯỚNG DẪN GIẢI (Lần 7) [DC09042018]

https://app.box.com/s/ndw8wijyug5fr425ktyaixesjd9264hw

https://twitter.com/daykemquynhon plus.google.com/+DạyKèmQuyNhơn www.facebook.com/daykem.quynhon https://daykemquynhon.blogspot.com + Bài học phân loại các hợp chất gluxit: 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 P/s: cần chú ý chương trình thi 2017-2018, Mantozơ thuộc phần giảm tải.! Câu 10: Đáp án C Với H 2 SO 4 đặc nóng thì sản phẩm khử có thể là SO 2 , S hoặc H 2 S. HNO 3 có thể tạo ra các sản phẩm khử: NO 2 , NO, N 2 O, N 2 hoặc NH 4 NO 3 . Câu 11: Đáp án C Vì este không tác dụng với kim loại Natri ⇒ Chọn C CH 3 COOCH=CH 2 + H 2 O ⇌ CH 3 COOH + CH 3 CHO CH 3 COOCH=CH 2 + H 2 Ni, t° ⎯⎯⎯→ CH 3 COOCH 2 CH 3 CH 3 COOCH=CH 2 + NaOH → CH 3 COONa + CH 3 CHO. Câu 12: Đáp án D Ta có các phản ứng: (NH 4 ) 2 CO 3 + Ca(OH) 2 → CaCO 3 ↓ + 2NH 3 ↑ + 2H 2 O. (NH 4 ) 2 CO 3 + MgCl 2 → MgCO 3 + 2NH 4 Cl. (NH 4 ) 2 CO 3 + FeSO 4 → FeCO 3 ↓ + (NH 4 ) 2 SO 4 (NH 4 ) 2 CO 3 + 2NaOH → Na 2 CO 3 + 2NH 3 ↑ + 2H 2 O. Câu 13: Đáp án B Một số hiđroxit lưỡng tính phổ biến gặp như Al(OH 3 và Zn(OH) 2 ...⇒ Chọn B DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN Câu 14 Đáp án A Ta có n Este = 0,1 mol. 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 + Phản ứng: C 2 H 5 COOCH 3 + KOH → C 2 H 5 COOK + CH 3 OH. Trang 9 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 https://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 + Ta có: n CH3OH = 0,1 mol. ⇒ Bảo toàn khối lượng ta có: m Chất rắn = 8,8 + 0,12×56 + 0,1×32 = 12,32 gam. Câu 15: Đáp án A Ta có n Mg = n H2 = 0,12 mol. ⇒ m MgO/hỗn hợp = 6,88 – 0,12×24 = 4 gam Câu 16: Đáp án D Vì NH 4 NO 3 ° ⎯⎯→ t Câu 17: Đáp án C N 2 O + 2H 2 O ⇒ D sai Vì C 6 H 11 OH không chứa nhân thơm ⇒ Loại Câu 18: Đáp án B Ta có n Zn = 0,05 ⇒ m Zn = 0,05×65 = 3,25 gam ⇒m Cu = 10 – m Zn = 10 – 3,25 = 6,75 gam ⇒ %m Cu/hh = 67,5% Câu 19: Đáp án A Gọi công thức chung của 2 amin là: RNH 2 ⇒ RNH 2 NH 2 + HCl → RNH 2 NH 3 Cl. + Bảo toàn khối lượng: m HCl = 62,04 – 35,76 = 26,28 gam ⇒ n HCl = 0,72 mol. ⇒ M Amin = 35,76 = 49,66 ⇒ R = 49,66 – 16 = 33,66. 0,72 Vì (–C 2 H 5 ) 29 < 33,66 < 43 (–C 3 H 7 ). ⇒ 2 amin là C 2 H 5 NH 2 và C 3 H 7 NH 2 ⇔ CTPT của 2 amin là C 2 H 7 N và C 3 H 9 N Câu 20: Đáp án D Khi thu khí để ngửa bình chứa ⇒ Khí đó nặng hơn không khí. Mà CO 2 có M = 44 > 29 (Không khí). ⇒ Có thể dùng bộ dụng cụ trên để điều chế khí CO 2 . Câu 21: Đáp án C Ta có phản ứng: N 2 + O 2 3000° ⎯⎯⎯→ C 2NO. Trong đó N có số oxh = 0. Sau phản ứng N tăng lên +2 ⇒ N 2 thể hiện tính khử. DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN C sai Câu 22: Đáp án B 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 10 Đóng góp PDF bởi GV. Nguyễn Thanh Tú www.facebook.com/daykemquynhonofficial www.facebook.com/boiduonghoahocquynhonofficial

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