Rb'(aq) + ¢ = Rb(s) ~2.98
2.76
Mg (aq) + 2¢° = Ma(s) 20
AI (aq) + 3e = Als)
Ti"(aq) +2€ = Tis)
Ti;O;(s) + 2H” + 2e” y 2TiO(s) + HyO
[Tỉ(@g)tố TS)
Mn””(aq) + 2e" = Mn(s)
VỀ@g+2 GS)
Sn(s) + 4H” + 4eˆ = SnH,(g)
SiO;(s) + 4H” + 4c” = Si(s) + 2H,O j
B(OH);(aq) + 3H” + 3e" w= B(s) + 3H,0 Ề
T¡O””(aq) +2H” + 4e” = Ti(s)+H,O :
2 H;O() + 2e” œ H;(g) + 2 OH (aq)
Zn* (aq) + 2e" = Zn(s) !
|Cr@g)ti Cs) | c0.
t3 <
+|£|*|+ >|a|8|#
:3
Au(CN); (aq) + e = Au(s) +2 CN (aq)
“056
Ga’ (aq) + 3e” = Ga(s)
“050
Fey ere) — | 9m1“050
2C0:G)* 21 Gq) 728 2H00200Nm) | -043 | Cleave acres =e
-0.41
SVTH: Nguyễn Thi Phuong Vi 99
Knead luận tốt nghiệp GVHD: ThS. Ngé Tan Lạc
Cd””(aq) + 2e” = Cd(s)
PbSO,(s) + 2e” et Pb(s) +SO4"(aq)
Ge(s) + 4H” + 4e” + GeH(g) Co ”(aq) + 2e = Co(s)
Ni” (aq) + 2e° = Ni(s)
WO,(s) + 4H + 4e" = W(s)
CO;(g) + 2H” + 2eˆ = HCOOH(aq) ~0.11
010
2H(q)+2 A) | 000_ 003
[H;MoO,(ag)+6H tốc — =Mo(g)+4HO — [ 0.11 | [CG)+4H +4 CH) | 3013, [HCHO&q)+2H +2 — ==CHạOH@g) — | 3013|
SG+2H +2 SS) — — | +04 | Sn@g+2c Snag) — — | +015|
[Cu G@q+c Cag) | +016_
[HSO,(ag)+3H +2 — =§OjAp | 016 - SbO'+2Ht3ec —— œSb@+HO —— | 3020)
[H›AsOy(sa)+3H +ọc — 2Asg)!3HO — | 3024|
GeO(s) + 2H” + 2e” ee Ge(s) + HạO | +0.26 - Bí (aq) +3e- = Bi(s) | 40.32 |
SVTH: Nguyễn Thi Phương Vi 100
Khoá luận tốt nghiệp GVHD: ThS. Ngô Tan Lạc
VO"'(aq) + 2H" +e = V"'(aq)
Cu(aq) + 2e” = Cu(S)
[Fe(CN%]” (aq) + cˆ = [Fe(CN)ô]" (aq)
O2(g) + 2H,0(1) + 4e” = 4OH (aq)
HạMoO¿ + 6H” + 3e” = Mo ”(aq)
CH;OH(aq) + 2H” + 2e” = CH,(g) + HạO
SO;(aq) + 4H” + 4e” = S(s) + 2HyO
Cu (aq) +e = Cu(s)
CO(g) + 2H” + 2e” = C(s) + HạO
la(s) + 2e =>2[ (aq)
I; (aq) +2” = 31 (aq)
[Aul;] (aq) + 3e_ = Au(s) + 4I (aq)
H3AsO,(aq) + 2H” + 2e” = HạAsO;(aq) + H;ạO
[Aul;] (aq) + e” = Au(s) + 2I'(aq)
MnO, (aq) + 2H,O(1) +3e° =MnO;(s) + 4 OH (aq)
~+ 6H" + 4e" = 2S(s) + 3H,0 02(g) + 2H” + 2e7 = H;O;(aq)
TI” (aq) + 3c” =
Fe “(aq) +e = Fe (aq)
Hg**(aq) + 2e” = 2Hg(I)
= Ag(s)
NO; (aq) + 2H (ag) +e = NO;(g) + H;O(I)
[AuBr;] (aq) + 3e” = Au(s) + 4Br (aq)
F = He()
MnO, (aq) + H” +e = HMnO, (aq) 2Hg” (aq) + 2e” = Hg;ˆ (aq)
+0.34
ch | th | + Rl lWlw ce|C|ItL›|C|c|+
+|+|+ +|+ MLO | Ca | Ch | Cha lainCID nA! SiS] rir
œnN©bee
>ga+ ơ + ot
| =—,2
w)ooloo}alalriqiu<=|E|SE|ơ|s|cel|clse
c=|+©|*+e©|*©|*+ec|.‹c|œ mịG@|CŒG:|C2|— |C|ỊtUAa:
.85
xaa ^ề— + Noœ
[AuClJ (aq) + 3e” == Au(s) + 4Cẽ (aq)
[AuBr;] (aq) + e” = Au(s) + 2Br (aq)
Br,(1) + 2e" = 2Br (aq)
IO; (aq) + 5H” + 4e = HIO(aq) + 2HạO
[AuCl;] (aq) + &° = Au(s) + 2Cẽ (aq)
AgaO(s) + 2H” + 2e = 2Ag(s)
210; (aq)+ 12H" +10e =I;(s)+6H;O
ClO, (aq) + 2H” + 2c = CIO; (aq) + HạO
Os(g) + 4H” + 4e" =2H;O
SVTH: Nguyễn Thi Phuong Vi 101
Rea luận tốt nghiệp GVHD: ThS. Ngé Tan Lạc
MnO¿(s) + 4H” + 2e” Mnf (aq) + 2H,0
TI ”(aq) + 2e = TI(s)
Cl(g)+2e = 2CT (aq)
Cr,07 (aq)+14H'+6e œ+2Cr(aq)+ 7H;O
2HIO(aq) + 2H” +2eˆ = I,(s) + 2HạO
BrO; (aq) + SH” + 4e = HBrO(aq) + 2HyO 2BrO;' + 12H” + 10c” = Br,(l) + 6HạO
2CIO;” + 12H” + 10c” œ Clog) + 6H,O
MnO, (aq) + 8H’ + Se = MnF (aq) + 4H;O
HO, +H" +e H;O;(aq)
Au’ (aq) + 3e” = Au(s)
NiO;(s) + 4H +2e = Ni* (aq)
2HCIO(aq) + 2H” + 2c” = Cl;(g) + 2H;O
HCIO;(aq) + 2H” + 2e” = HCIO(aq) + H,0
Pb” (aq) +2c_ = Pb*’(aq)
MnO, (aq) + 4H” + 3e” + MnO;(s) + 2HạO
H;O;(aq) + 2H” + 2e” = 2H,0
AgO(s) + 2 Ä = Ag (aq) + H,0
Au’ (aq) +e = Au(s)
BrO, (aq) + 2H” + 2c” = BrO; (aq) + HạO Co” (aq) + & = Co* (aq)
Ag’ (aq) +e = Ag (aq)
8,0," + 2e€ = 2SO/¿ˆˆ
HMnO, (aq) + 3H” + 2e” = MnO,(s) + 2H,0
F(g) + 2e = 2F(aq)
(Theo Wikipedia )
+ o
SVTH: Nguyễn Thi Phương Vi 102
Khoá luận tốt nghiệp GVHD: ThS. Ngô Tan Lạc
KET QUA PHAN TICH BAI TRAC NGHIEM
# Trac nghiem : KT15'-DAI CUONG VE PU OXI HOA KHU
# Ten nhom : KHOI 10
* So cau TN ô 10
* So bai TN = 102
Thuc hien xu ly luc 13g12ph Ngay 15/ 5/2007
* CAC CHI SO VE TRÔNG BINH va DO KHO tinh tren diem TOAN BAI TRAC NGHIEM
Trung Binh - 7.696 Do lech TC - 1.705 Do Kho bai TEST = 77,0%
Trung binh LT " 6,250 Do Kho Vua Phai = 62.5%
* HE SO TIN CAY cua BAI TEST
(Theo cong thuc Kuder-Richardson co ban}
He so tin cay ô 0.509
* Sai so tieu chuan cua do luong : SEM ô= 1.195
ee i a EAE
* BANG DO KHO VA DO PHAN CACH TONG CAU TRAC NGHIEM
Cau TDcau MEAN(cau) SDicau) | Mp Mg Rpbis
1 60 0.588 0.495 | 8.267 6.881 0.400 **
2 86 0.843 0.365 | 8.081 5.625 0.524 **
3 192 1,000 0,000 | 7.696 0.000 0.000
x 49 0.480 0.502 | 8.510 6.943 0.459-*++
5 84 0.824 0.383 | 8.060 6.000 0.461 **
6 19 0.775 0.420 | 7.975 6.739 0.303 **
? 80 0.184 0,413 | 8.012 6.545 0.354 **
8 88 0.863 0.346 | 8.091 5.214 0.581 **
9 90 0.882 9.324 | 8.033 5.167 0.542 **
10 6? 0.657 0.477 | 8.343 6.457 0.525 **
eae HET ...
BANG PHAN TICH CAC TAN SO LUA CHON TUNG CAU
{item Analysis Results for Observed Responses)
SK EE EK KE s9 0= 46030 460390 68B 36
Trac nghiem : KT15'=-DAI CUONG VE PU OXI HOA KHU
* Ten nhom lam TN : KHOI 10 `
* So cau 2 10
* So nguoi g 102
* Xu ly luc 13g1l4ph * Ngay 15/ 5/2007
*** Cau ứ@ : 1
Lua chon A 8 € D* Missing
Tan so 3 4 7 31 60 0 Tí le $% $ 3.9 6.9 30.4 58.8
Pt-biserial : -0.14 -0.04 -0.34 0.40 Muc xacsuat : NS NS <.01 <.01
*** Cau so : 2
Lua chon A B cf D Missing Tan so 3 2 10 86 4 0
SVTH: Nguyẫn Thi Phuong Vi 103
rv\oq an tot nghiệp GVHD; ThS. Ngé Cấn Loc
Tí le $ : 2.0 9.8 84.3 3.9 Pt-biserial : ~0,22 -0.39 0.52 -0.23 Muc xacsuat : <.05 <.01 <.01 <.05
e** Cau so : 3
Lua chon A B c* D Missing Tan so 5 0 0 102 0 0
TÌ le $ : 0.0 0.0 100.0 0.0 Pt-biserial : NA NA NA NA
Muc xacsuat : NA NA NA NA
*** Cau so : 4
Lua chon At B c D Missing Tan so È 49 9 40 4 0
Ti le % ‡ 48,0 8,8 39.2 3.9 Pt-biserlaì : 0,46 -0.29 ~0.16 -0.35 Muc xacsuat : <.01 <.01 NS <.01
*** Cau so 5
tua chon A B € Be Missing Tan so 8 1 3 4 84 0
Ti le $ 8 10.8 2.9 3.9 82.4
Pt-biserial : -0.40 ~0.17 ~0.11 0.46
Muc xacsuat : <.01 NS NS <.01
*** Cau so : 6
Lua chon A B Cc o* Missing
Tan so D 11 11 1 19 0 Ti le $ 7 10.8 10.8 1.0 77.5
Pt-biserial : -0,22 -0.12 -0.22 0.30 Muc xacsuat : <,05 NS <.05 <.01
ee Cau so : 7
tua chon A Be € D Missing Tan so : 13 80 3 6 0
Tí le % Ế 12.7 78.4 2.9 5.9
Pt-biserial : -0.19 0.35 -0.28 -0.15
Muc xacsuat : ns <.01 <.01 NS
*** Cau so 8
Lua chon A Bt Cc D Missing
Tan so 3 8 98 2 4 0 Ti le $ : 7,8 86.3 2.0 3.9
Pt-biserial : -0.35 0.58 -0.18 -0.41
Mục xacsuat : <.01 <.01 NS <01 `
“**> Cau so : -9
Lua chon at B c D Missing Tan so 3 90 3 $ 4 9
Ti le $ ‡ 88.2 2.9 4.9 3.9 Pt-biserial : 0.54 -0.41 -0.23 -0.29 Muc xacsuat ; <.01 <.01 <,05 <.01
*** Cau sơ : 10
Lua chon A B € or Missing
Tan so ệ 10 9 16 67 0 Ti le % ‡ 9.8 8.8 15.7 65.7
Pt-biserial ; -0.3 -0.25 -0.22 0.53 Muc xacsuat : <.01 <.05 <.05 <.01
SVTH: Nguyễn Thi Phuong Vi 104
Ano luận tốt nghiệp GVHD; ThS. Ngô Cấn Lạc
... AAAÁ...Á... . AC... . Ree
eee HET ....
KET QUA PHAN TICH BAI TRAC NGHIEM
# Trac nghiem : KTITIET-DAI CUONG VE PU OXI HOA KHU
# Ten nhom : KHOI 10
* So cau TN ô= 20
* So bai TN’ = 43
Thuc hien xu ly luc i3g3iph Ngay 15/ 5/2007
=“==——====
* CAC CHI SO VE TRUNG BINH va DO KHO tinh tren diem TOAN BAI TRAC NGHIEM
Trung Binh 14.140 Do lech TC = 2.242 Do Kho bai TEST = 70.7%
Trung binh LT 12.500 Do Kho Vua Phai 62.5%
* HE SO TIN CAY cua BAI TEST
(Theo cong thuc Kuder-Richardson co ban}
He go tin cay = 0.434
* Sai so tieu chuan cua do luong : SEM “= 1.687