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6 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 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 Số chất có thể tác dụng với dung dịch KMnO 4 gồm: axetilen và stiren ⇒ Chọn B Câu 23: Đáp án A Có n X = n HCl = 0,1 mol → X chứa 1 nhóm NH 2 Bảo toàn khối lượng → m HCl = 45,18 – 32,04 = 13,14 gam. → n HCl = n X = 0,36 mol → M X = 32,04 0,36 = 89 ⇔ X là alanin Câu 24: Đáp án A Ta có các phản ứng: Al(NO 3 ) 3 + 3NaOH → Al(OH) 3 + NaNO 3 . Sau đó: Al(OH) 3 + NaOH → Na[Al(OH) 4 )] Na 2 CO 3 + 2HCl → 2NaCl + CO 2 ↑ + H 2 O. NH 4 Cl + NaOH → NaCl + NH 3 ↑ + H 2 O. NaHCO 3 + HCl → NaCl + CO 2 ↑ + H 2 O. Câu 25: Đáp án C Vì số mol kết tủa cực đại = 0,1 mol ⇒ n BaCO3 = 0,1 mol. + Tại thời điểm n CO2 = 0,35 mol ⇒ n BaCO3 = 0,05 mol. + Bảo toàn nguyên tố Ba ⇒ n Ba(HCO3 ) 2 = 0,1 – 0,05 = 0,05 mol. + Bảo toàn nguyên tố cacbon ta có: ∑n CO2 = n BaCO3 + 2n Ba(HCO3)2 + n KHCO3 . ⇒ n KHCO3 = 0,35 – 0,05×1 + 0,05×2 = 0,2 mol. ⇒ n K = 0,2 mol và n Ba = 0,1 mol ⇒ m = 21,5 gam Câu 26: Đáp án C Câu 27: Đáp án D Câu 28: Đáp án A Câu 29: Đáp án B Câu 30: Đáp án B Câu 31: Đáp án A Ta có n CO2 = n H2O = 0,26 mol ⇒ X và Y đều no đơn chức mạch hở. Đặt: n X = a và n Y = b ⇒ a + b = 0,1 (1) DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN + Bảo toàn oxi ta có: a + 2b + 0,31×2 = 0,26×2 + 0,26 ⇔ a + 2b = 0,16 (2) Giải hệ (1) và (2) ⇒ a = 0,04 và b = 0,06. Đặt X là C n H 2n O 2 và Y là C m H 2m O 2 . 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 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 ⇒ n CO2 = 0,04n + 0,06m ⇒ 2n + 3m = 13. Giải PT nghiệm nguyên ⇒ n = 2 và n = 3 ⇒ X là CH 3 CHO. Vì n Ag = 0,1 = 2n X ⇒ Y là HCOOC 2 H 5 . Câu 32: Đáp án C Ta có: C 2 H 5 ⇌ C 2 H 4 + H 2 O. n C2H5OH = n C2H4 = 0,2 mol. + Phản ứng ete hóa: 2C 2 H 5 OH ⇌ C 2 H 5 –O–C 2 H 5 + H 2 O. 0, 2× 74 ⇒ m Ete = = 7,4 gam 2 Câu 33: Đáp án D Vì số nguyên tử oxi lớn hơn nguyên tử cacbon rất nhiều. ⇒ Phân tử phải có CO 3 , NO 3 hoặc HCO 3 . ⇒ CTCT thu gọn ứng với CTPT C 2 H 9 O 6 N 3 là: NO 3 NH 3 –CH 2 –NH 3 HCO 3 NO 3 NH 3 –CH 2 –NH 3 HCO 3 + 3KOH → KNO 3 + CH 2 (NH 2 ) 2 ↑ + K 2 CO 3 + 3H 2 O. ⇒ n KOH dư = 0,1 mol ⇒ m Chất rắn = 0,1×56 + 0,1×101 + 0,1×138 = 29,5 gam Câu 34: Đáp án D Nhỏ từ từ axit vào dung dịch X ⇒ dung dịch Y. ⇒ Y chứa KCl và KHCO 3 . Cho Y + dung dịch Ba(OH) 2 dư ⇒ có phản ứng. KHCO 3 + Ba(OH) 2 → BaCO 3 + KOH + H 2 O. + Mà nKHCO 3 = n BaCO3 = 0,2 mol. + Bảo toàn Cl ⇒ n KCl = n HCl = 0,5 mol. ⇒ Bảo toàn K ⇒ n KOH = n KCl + n KHCO3 = 0,7 mol. ⇒ C M KOH = 0,7 0,4 = 1,75M Câu 35: Đáp án B Vì H 2 SO 4 dư ⇒ Chất rắn Z đó là Cu. ⇒ Trong dung dịch chỉ chứa muối Fe 2+ và Cu 2+ DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN Câu 36: Đáp án C bên anot đầu tiên ra khí Cl 2 , hết sẽ ra O 2 (do H 2 O điện phân). 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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