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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 Amin no bậc hai (đính với hai gốc ankyl) có tính bazơ mạnh hơn bazơ bậc một: 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 Amin thơm Amin thơm có nguyên tử N ở nhóm amin đính trực tiếp vào vòng benzen. Gốc phenyl có tác dụng làm suy giảm tính bazơ, do vậy amin thơm có lực bazơ rất yếu, yếu hơn amoniac: Câu 13: Đáp án B Ta có dãy điện hóa: Li K Ba Ca Na Mg Al Mn Zn Cr Fe Ni Sn Pb H Cu Fe Hg Li K Ba Ca Na Mg Al Mn Zn C F Ni Sn Pb H Cu F Hg + + 2+ 2+ + + 3+ 2+ 2+ 3+ 2+ 2+ 2+ 2+ + 2+ 3+ 2+ … 2+ r e 2 e 2 3 Ag Pt Au …. Ag Pt Au + + + Từ dãy điện hóa dễ thấy ion Ag + có tính oxi hóa mạnh nhất Câu 14: Đáp án A + Bài học phân loại các hợp chất gluxit: 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 Câu 15: Đáp án D 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 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 Ta có n Fe = 0,12 mol ⇒ ∑n e cho max = 0,36 mol và ∑n e cho min = 0,24 mol. Vì NO là sản phẩm khử duy nhất của N +5 ⇒ n NO max = n HNO3 ÷ 4 = 0,1 mol. ⇒ n e nhận tối đa = 3×n NO = 0,3 mol. Nhận thấy 0,24 < 0,3 < 0,36 ⇒ Fe tan hết tạo 2 muối. ⇒ m Muối = m Fe + m NO3 – = 6,72 + 3n NO ×62 = 25,32 gam. Câu 16: Đáp án B Câu 17: Đáp án C Câu 18: Đáp án B Câu 19: Đáp án C Câu 20: Đáp án D Khối lượng dung dịch sau phản ứng = ∑m Các chất ban đầu – ∑m Kết tủa – ∑m Bay hơi . ⇔ m Dung dịch sau pứ = m Hỗn hợp kim loại + m Dung dịch H2SO4 – m H2 0,1× 98 ⇔ m Dung dịch sau pứ = 3,68 + × 100 – 0,1×2 = 52,48. 20 Câu 21: Đáp án C Nhận thấy các chất trong X đều có số nguyên tử H gấp đôi nguyên tử C. ⇒ Khi đốt cháy luôn luôn cho ta n CO2 = n H2O . Mà ∑n H2O = 0,8 ⇒ ∑n CO2 = 0,8 mol ⇒ m CO2 = 0,8×44 = 35,2 gam Câu 22: Đáp án D Ta có: n Axit glutamic = 0,09 mol, n HCl = 0,2 mol ⇒ ∑n COOH + H + = 0,09×2 + 0,2 = 0,38 mol. + n NaOH = 0,34 mol < ∑n COOH + H + = 0,38 mol ⇒ n H2O tạo thành = 0,38 mol. Bảo toàn khối lượng ta có: m Chất rắn = 13,23 + 0,2×36,5 + 0,4×40 – 0,38×18 = 29,69 gam Câu 23: Đáp án B Câu 24: Đáp án B Câu 25: Đáp án D Chất rắn X có thể có Al, Cu và chắc chắn có Ag. Cho X vào HCl mà tạo khí thì trong X chắc chắn có Al (do Cu và Ag không phản ứng với HCl). DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN Do có Al nên Cu2+ và Ag+ phản ứng hết. Số mol Al ban đầu: 2n 2+ + n + + 2n Cu Ag H n 2 Al = = 0,04 ⇒ m1 = 1,08 3 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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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