V i3 chat CO so mol bkng nhau nen so nguyen tuf H trong m6i chat la
Bai 4: CJiyO ^X CO 2+ ^H^O •
+ Nhom I : Lam mat mau dung dich brom gom S O 2 va H2S
+ Nhom I I : Khong lam mat mau dung dich brom gom C 0 2 v a N2
* Dung Ca(0H)2 ta nhan biet dirge S O 2 va C O 2 c6 xuat hien ket tua tring Chon dap an C. Chon dap an C.
Bai 58:
* Phirong phap thong thiTdng Dat cong thiJc A la CnH2n+20x Dat cong thiJc A la CnH2n+20x
C„H2„.20. ^ ( 6 n + 2 - 2 x ^ Q ^ _^ ^ ^ ^ ^ ^ (n+l)H20
4
+ Theo de: ( 6" + 2 - 2x ^ = 2,5 -> 6n - 2x = 8 ^ n = 2 va x = 2 4 4
* Phirong phap kinh nghiem Vi ^ = 2,5 -> So C = So O = 2 Vi ^ = 2,5 -> So C = So O = 2
" A
Chon dap an B.
Cau 59:
* PhiTdng phap kinh nghiem
+ Ma„.i„ = 3ig8°_2o = 62,5
36,5
+ Goi X la amin c6 phan tijf khoi nho nhat. Theo gia thiet ta c6 X + 10.(X + 14) + 5.(X + 28) = 62,5.(1 + 10 + 5) -> X = 45 X + 10.(X + 14) + 5.(X + 28) = 62,5.(1 + 10 + 5) -> X = 45 + Suy ra 3 amin tren la: C 2 H 7 N , C 3 H 9 N , C 4 H 1 1 N
+ Ap dung cong thufc tinh so dong phan: T = 2 ^ 2^ + 2^ = 14 Chon dap an B. Chon dap an B.
Bai 60:
m^Oj = 15 - 5,1 = 9,9 gam -> n^ô = 0,225 mol Suy ra m = 0,5.0,225.180.100/90 = 22,5 gam Suy ra m = 0,5.0,225.180.100/90 = 22,5 gam
Chon dap an B.
DAP AN MA D E 115
Cau DA Cau DA Cau DA Cau DA
1 C 16 A 31 A 46 A 2 D 17 A 32 D 47 D 3 B 18 A 33 D 48 D 4 C 19 D 34 C 49 C 5 A 20 A 35 C 50 C 6 C 21 D 36 B 51 D 7 A 22 B 37 A 52 C 8 B 23 A 38 C 53 C 9 A 24 D 39 B 54 D 10 D 25 A 40 C 55 B 11 B 26 B 41 A 56 D 12 B 27 B 42 B 57 A 13 B 28 D 43 B 58 C 14 D 29 B 44 B 59 B 15 A 30 A 45 D 60 A
Hiidng dan gidi
Bai 1:
2CH3CH(NH3Cl)COOH + 2Ba(OH)2 ^ [CH3CH(NH2)COO]2Ba + BaCl2 + 4 H 2 O
0,1 -> 0,1 -> 0,2 Phan ling chi tao thanh H 2 O la chat khi c6 can thi bay hoti, cac chat Phan ling chi tao thanh H 2 O la chat khi c6 can thi bay hoti, cac chat
con lai (san pham va bazcf dif) deu of dang r^n
^ mrin = m^^uiu + m„ , , o H ) , - ^ . . , 0 = 12,55 + 0,15.171 - 0,2.18 = 34,6gam
Chon dap an C.
Bai 4: CJiyO ^ X C O 2 + ^H^O • 2 2
Theo d l : ^ = o = => = i => X chi CO thé la C H 4 O niH^o 9 9y 9 y 4