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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 Câu 27: Đáp án B + Đặt n NH4NO3 = a || n N2 = b ⇒ 10n NH4NO3 + 8n N2 = 3n Al = 1,62 (1) ⇒ 12n NH4NO3 + 10n N2 = n HNO3 = 2 (2) + Giải hệ (1) và (2) ⇒ b = n N2 = 0,05 mol ⇒ V N2 = 1,12 lít Câu 28: Đáp án B X làm dung dịch I 2 hóa xanh tím ⇒ X là hồ tinh bột ⇒ Loại A và C. Z có phản ứng màu biure ⇒ Z có thể là lòng trắng trứng ⇒ Loại D Câu 29: Đáp án C (d) Sai, chẳng hạn HCOOCH=CH 2 + NaOH ° ⎯⎯→ t HCOONa + CH 3 CHO. (e) Sai vì peptit chứa từ 3 mắt xích trở lên mới có phản ứng màu biure. ⇒ Chỉ có (4) và (5) sai Câu 30: Đáp án B Để ý CTPT: andehit formic: HCHO axit axetic, C H O = ( CH O ) , glucozo = ( ) 2 4 2 2 2 Đề yêu cầu tính glixerol nên coi hh trên chỉ có HCHO và C3H8O 3 . Đốt cháy hỗn hợp ta có hệ phương trình về số mol CO 2 và H 2 O như sau: ⎧x + 3y = 0,65 ⎧x = 0,35 mol ⎨ ⇔ ⎨ ⎩x + 4y = 0,75 ⎩y = 0,1mol Vậy thành phần phần trăm khối lượng của glixerol là: 0,1× 92 %Glixerol = ≈ 46,7% 0,1× 92 + 0,35× 30 Câu 31: Đáp án D Ta có ∑n e trao đổi = 0,2 mol ● Ở Catot: 2H 2 O +2e → H 2 ↑ + 2OH – ⇒ n H2↑ = 0,1 mol và n OH – sinh ra = 0,2 mol. ● Ở anot: 2Cl – → Cl 2 ↑ + 2e + Sau khi dừng điện phân có phản ứng: Mg 2+ + 2OH – → Mg(OH) 2 ↓ ⇒ m Dung dịch giảm = m H2↑ + m Cl2↑ + m Mg(OH)2↓ ⇔ m Dung dịch giảm = 0,1×2 + 0,1×71 + 0,1×58 = 13,1 gam Câu 32: Đáp án B Nhận thấy n HCl pứ = n Amin đơn chức . C H O CH O . 6 12 6 2 6 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 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 Ta có H trung bình = 0,8.2 3,2 0,5 = ⇒ Chất có số nguyên tử H < 3,2 chỉ có thể là HCOOH. + Đặt n HCOOH = a và n Amin = b mol. n n H 2O − CO2 1 ⇒ n Amin = = b = ⇒ VHCl = 1 : 0,1 = 2 M 1,5 15 15 3 Ta có n HCOOH = a = 0,5 – 1 = 13 15 30 + Bảo toàn C ta có: Amin có số C = Câu 33: Đáp án D (1) Zn + 2AgNO 3 → Zn(NO 3 ) 2 + 2Ag ⇒ Chọn. (2) Fe + Fe 2 (SO 4 ) 3 → 3FeSO 4 . (3) 2Na + 2H 2 O → 2NaOH + H 2 2NaOH + CuSO 4 → Cu(OH) 2 + Na 2 SO 4 (4) CuO +CO ° ⎯⎯→ t Câu 34: Đáp án D Ta có sơ đồ sau: Cu + CO 2 ⇒ Chọn. ⎧C2H 4NO2K : a Peptit + KOH → ⎨ + H 2 O 0,85 0,15 ⎩C5H 7NO4K 2 : b ( 0,15+ ) 95,15 g 13 0,7 − 30 = 4 ⇒ Amin là C 4 H 11 N 1 15 b mol Bảo toàn nguyên tố K ta có: a + 2b = 0,85 (1) Theo tổng khối lượng 2 muối ta có: 113a + 223b = 95,15 (2). + Giải hệ (1) và (2) ⇒ a = 0,25 và b = 0,3 mol ⇒ n H2O = 0,45 mol. + Áp dụng định luật bảo toàn khối lượng ta có: 0,15 mol peptit có khối lượng = 95,15 + 0,45×18 – 0,85×56 = 55,65 gam. ⇒ 0,1 mol hỗn hợp peptit có khối lượng = 55,65.0,1 = 37,1 gam 0,15 Câu 35: Đáp án A DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN 2Z ⇌ T + H 2 O (t = 140 o C ⇒ Tạo ete). M Ete = M R2O = 23×2 = 46 ⇒ 2R + 16 = 46 ⇒ R = 15 ⇔ Z là CH 3 OH. 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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