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B GIO DC V O TO CHNH THC ( thi c 02 trang)
THI TUYN SINH I HC, CAO NG NM 2006 Mn: HO HC, khi B Thi gian lm
bi: 180 pht, khng k thi gian pht
PHN CHUNG CHO TT C CC TH SINH Cu I (2 im) 1) Tng s ht mang in
trong ion AB32 bng 82. S ht mang in trong ht nhn ca nguyn t A nhiu
hn s ht mang in trong ht nhn ca nguyn t B l 8. Xc nh s hiu nguyn t
ca hai nguyn t A v B. Vit cu hnh electron ca hai nguyn t A v B. Xc
nh v tr (, chu k, nhm) ca hai nguyn t A v B trong bng tun hon cc
nguyn t ho hc. 2) Ha tan Fe3O4 vo dung dch HCl, c dung dch D. Chia
dung dch D thnh ba phn. Thm dung dch NaOH d vo phn th nht, c kt ta
E. Ly kt ta E ra ngoi khng kh. Cho bt ng kim loi vo phn th hai. Sc
kh clo vo phn th ba. Vit phng trnh ha hc ca cc phn ng xy ra. 3) Cho
dung dch G cha cc ion Mg2+, SO42, NH4+, Cl. Chia dung dch G thnh
hai phn bng nhau. Phn th nht tc dng vi dung dch NaOH d, un nng, c
0,58 gam kt ta v 0,672 lt kh (ktc). Phn th hai tc dng vi dung dch
BaCl2 d, c 4,66 gam kt ta. Vit phng trnh ha hc ca cc phn ng xy ra
(di dng phng trnh ion rt gn). Tnh tng khi lng ca cc cht tan trong
dung dch G. Cu II (2 im) OH 1) Cho hai cht sau: CH2OH , CH3 Vit
phng trnh ha hc ca cc phn ng xy ra (nu c) khi cho tng cht trn tc
dng vi kim loi Na, vi dung dch NaOH v vi axit CH3COOH (ghi iu kin
phn ng, nu c). 2) Hirocacbon X mch h, l cht kh iu kin thng. Khi
hirat ho X trong iu kin thch hp, c mt sn phm duy nht Y (khng cha
lin kt trong phn t). Y phn ng vi Na d, sinh ra hiro c s mol bng mt
na s mol ca Y. a) Xc nh cc cng thc cu to c th c ca X v Y; b) Y1 l
ng phn cng chc ca Y v c quan h vi Y theo s chuyn ha: Y1 X1 Y. Xc nh
cng thc cu to ca X, Y, Y1 v vit phng trnh ho hc ca cc phn ng xy ra
theo s trn. 3) Vit phng trnh ho hc ca cc phn ng xy ra theo s chuyn
ho sau: C2H4 C2H4Br2 C2H6O2 C2H2O2 C2H2O4 C4H6O4 C5H8O4 (Cc cht hu
c vit di dng cng thc cu to thu gn. Ghi iu kin phn ng, nu c). Cu III
(2 im) Cho 5,15 gam hn hp bt A gm Zn v Cu vo 140 ml dung dch AgNO3
1M. Sau khi phn ng xong, c 15,76 gam hn hp kim loi v dung dch B.
Chia dung dch B thnh 2 phn bng nhau. Thm mt lng d dung dch KOH vo
phn th nht, c kt ta. Lc ly kt ta, em nung n khi lng khng i, c m gam
cht rn. 1) Vit phng trnh ha hc ca cc phn ng xy ra v tnh gi tr ca m.
2) Cho bt Zn ti d vo phn th hai ca dung dch B, thu c dung dch D.
Cho t t V ml dung dch NaOH 2M vo dung dch D, c 2,97 gam kt ta. Tnh
gi tr ca V. Gi thit cc phn ng xy ra hon ton. Cu IV (2 im) Hn hp X
gm axit cacboxylic n chc Y v este n chc Z (phn t cc cht ch cha C,
H, O). un nng m gam hn hp X vi 400 ml dung dch NaOH 1M. C cn dung
dch sau phn ng, thu c p gam mt ru (hay ancol) R v 24,4 gam hn hp rn
khan E gm hai cht c s mol bng nhau. Cho p gam ru R tc dng vi Na d,
thot ra 0,56 lt kh. 1) Xc nh cng thc phn t ca ru R v tnh gi tr ca
p. Bit trong phn t R, phn trm khi lng C v H tng ng bng 52,17% v
13,04%. 2) Xc nh cng thc cu to ca Y, Z. Tnh gi tr ca m. 3) Trn u
24,4 gam hn hp rn khan E vi CaO, sau nung nng hn hp, thu c V lt mt
kh G. Tnh gi tr ca V. Gi thit hiu sut ca cc phn ng l 100%. Th tch
cc kh o iu kin tiu chun.1/2(1) (2) (3) (4) (5) (6)
V. PHN T CHN: Th sinh chn cu V.a hoc cu V.b Cu V.a. Theo chng
trnh THPT khng phn ban (2 im) 1) Dung dch CH3COONa, dung dch
(NH4)2SO4 c pH ln hn, nh hn hay bng 7? Vit phng trnh ho hc ca cc
phn ng gii thch. 2) Vit cng thc phn t ca cc cht ng vi cc k hiu X1,
X2, X3, X4, X5 v hon thnh phng trnh ha hc ca cc phn ng sau: a) X1
b) X2 c) X2 d) X4 + H2O + X4 + X3 + X5pmnx
to
X2 BaCO3 X1 BaSO4
+ + + +
X3 K2CO3 KClO3 CO2
+ H2 + H2O + H2O + H2O
(pmnx: in phn c mng ngn xp).
3) Cc cht hu c n chc Z1, Z2, Z3, Z4 c cng thc phn t tng ng l
CH2O, CH2O2, C2H4O2, C2H6O. Chng thuc cc dy ng ng khc nhau, trong c
hai cht tc dng c vi natri sinh ra kh hiro. a) Vit cng thc cu to v
gi tn Z1, Z2, Z3, Z4. b) T l cht hu c n chc, ng phn ca Z3. Trnh by
phng php ho hc nhn bit cc cht lng Z2, Z3, Z4 v T ng trong cc l ring
bit. Vit phng trnh ho hc ca cc phn ng xy ra minh ha. c) Vit phng
trnh ho hc ca cc phn ng iu ch Z3, Z4 t kh metan v cc cht v c cn
thit. Cu V.b. Theo chng trnh THPT phn ban th im (2 im) 1) Cho s
chuyn ha sau: A1+ dd HNO3 long(1)
A2
to(3)
A4
+ NH3, to(5)o
A1
+ dd HCl + O2(6)
A6
+ dd NaOH(7)
A7
+ dd NH3(8)
A8
(2) + dd H2S
(4) + A1, t
A3
A5
Vit phng trnh ha hc ca cc phn ng xy ra theo s chuyn ha trn. Bit
cc cht t A1 n A8 l ng v cc hp cht ca ng. 2) Vit cng thc cu to v gi
tn cc ipeptit c cng thc phn t C5H10O3N2. 3) Trnh by phng php ha hc
nhn bit 3 dung dch glucoz, fructoz v glixerol ng trong ba l ring
bit. Vit phng trnh ha hc ca cc phn ng xy ra. Cho: H = 1; C = 12; N
= 14; Ag = 108; Ba = 137. O = 16; Na = 23; Mg = 24; S = 32; Cl =
35,5; Cu = 64; Zn = 65;
----------------------------- Ht ----------------------------Cn
b coi thi khng gii thch g thm. H v tn th sinh
................................................................ s
bo
danh...............................................................
2/2
B GIO DC V O TO CHNH THC
P N - THANG IM THI TUYN SINH I HC, CAO NG NM 2006 Mn: HA HC, khi
B (p n - Thang im c 05 trang)
CU NI DUNG I 1 Xc nh s hiu nguyn t ca hai nguyn t A v B. Vit cu
hnh electron ca hai nguyn t A v B. Xc nh v tr ca hai nguyn t A v B
(0,75 im). + Xc nh s hiu nguyn t ca hai nguyn t A v B: Gi s proton,
electron trong hai nguyn t A v B tng ng l PA, EA v PB, EB. Trong
nguyn t: PA = EA, PB = EB. Theo bi, ta c: 2(PA + 3PB) + 2 = 82 (a)
PA PB = 8 (b) Gii h 2 phng trnh (a) v (b), c PA = 16, PB = 8 in tch
ht nhn ca nguyn t A l 16+ v ca B l 8+ S hiu nguyn t ca A l ZA = 16
v ca B l ZB = 8. + Vit cu hnh electron ca hai nguyn t A v B: ZA =
16 cu hnh electron ca A l 1s22s22p63s23p4 ZB = 8 cu hnh electron ca
B l 1s22s22p4 + Xc nh v tr (, chu k, nhm) ca hai nguyn t A v B
trong bng tun hon: Da vo cu hnh electron ca nguyn t A v B, suy ra:
- A th 16, chu k 3, nhm VI; - B th 8, chu k 2, nhm VI. 2 Vit PTHH
cc phn ng (0,50 im). Fe3O4 + 8HCl = FeCl2 + 2FeCl3 + 4H2O - Thm
dung dch NaOH d vo phn th nht: NaOH + HCl = NaCl + H2O 2NaOH +
FeCl2 = Fe(OH)2 + 2NaCl 3NaOH + FeCl3 = Fe(OH)3 + 3NaCl - Ly kt ta
ra ngoi khng kh: 4Fe(OH)2 + O2 + 2H2O = 4Fe(OH)3 - Cho bt ng kim
loi vo phn th hai: Cu + 2FeCl3 = CuCl2 + 2FeCl2 - Sc Cl2 vo phn th
ba: + 2FeCl2 = 2FeCl3 Cl2 3 Vit PTHH cc phn ng di dng ion rt gn v
tnh tng khi lng ca cc cht tan trong dung dch G (0,75 im) + PTHH cc
phn ng di dng phng trnh ion rt gn: to NH4+ + OH = NH3 + H2O (1)
Mg2+ + 2OH = Mg(OH)2 (2) Ba2+ + SO42 = BaSO4 (3) + Tng khi lng ca
cc cht tan trong dung dch G: T (1), (2), (3) suy ra s mol cc ion
trong mt na dung dch G:
IM 2,00
0,25
0,25
0,25
0,25
0,25
0,25
0,672 = 0,03 (mol) 22,4 n Mg2+ = n Mg(OH)2 = 0,58 = 0,01 (mol)
58 4,66 n SO2- = n BaSO = = 0,02 (mol) 4 4 233
n NH+4
=
n NH
3
=
0,25
V dung dch trung ha v in, ta c:
n Cl - = n NH+ + 2 n Mg2+ 2 n SO2- = 0,03 + 2.0,01 2.0,02 = 0,01
(mol). 4 4Tng khi lng cc cht tan trong dung dch G = tng khi lng cc
ion trong dung dch G: 2(0,03 . 18 + 0,01 . 24 + 0,02 . 96 + 0,01 .
35,5) = 6,11 (gam).1/5
0,25
II 1 Vit PTHH cc phn ng (0,50 im). 2C6H5 CH2 OH + 2Na C6H5 CH2
OH + CH3 COOHH2SO4 c, to
2,00
2C6H5 CH2 ONa + H2 CH3 COO CH2 C6H5 + H2O
0,25 0,25
2CH3 C6H4 ONa + H2 2CH3 C6H4 OH + 2Na CH3 C6H4 OH + NaOH CH3
C6H4 ONa + H2O 2 Xc nh cc CTCT c th c ca X v Y. Xc nh CTCT ca X, Y,
Y1 v vit PTHH cc phn ng (0,75 im). a) Xc nh cc CTCT c th c ca X v
Y: Gi CTPT ca X l CxHy (iu kin x 4). Khi hirat ho X, thu c mt sn
phm duy nht Y (khng cha lin kt trong phn t), Y phn ng vi Na d, sinh
ra hiro c s mol bng mt na s mol ca Y Y l ru n chc v X l anken i
xng. Cc CTCT c th c ca X l: CH2 = CH2 hoc CH3 CH = CH CH3; Cc CTCT
c th c ca Y l: CH3 CH2 OH hoc CH3 CH2 CH CH3.OH b) Xc nh CTCT ca X,
Y, Y1 v vit PTHH cc phn ng xy ra: + CTCT ca X, Y, Y1: V Y1 l ng phn
cng chc ca Y v c quan h vi Y theo s chuyn ha: Y1 X1 Y, nn CTCT: -
Ca Y l: CH3 CH2 CH CH3; OH - Ca Y1 l: CH3 CH2 CH2 CH2 OH; - Ca X l:
CH3 CH = CH CH3; + Vit PTHH cc phn ng xy ra theo s chuyn ho:
0,25
0,25
CH3 CH2 CH2 CH2 OH H+ , to CH3 CH2 CH = CH2 + H2O
H2SO4 c, to
CH3 CH2 CH = CH2 + H2O CH3 CH2 CH CH3
0,25
OH3 Xc nh PTHH cc phn ng xy ra theo s chuyn ho (0,75 im) (1) CH2
= CH2 + Br2 Br CH2 CH2 Br (2)t Br CH2 CH2 Br + 2NaOH HO CH2 CH2 OH
o o
+ 2NaBr
0,25
t (3) HO CH2 CH2 OH + 2CuO O = CH CH = O + 2Cu + 2H2O NH3 , to
(4) O = CH CH = O + 2Ag2O HOOC COOH + 4Ag (hoc O = CH CH = O + 2Br2
+ 2H2O HOOC COOH + 4HBr)
0,25
(5) (6) III
HOOC COOH + C2H5 OH HOOC COO C2H5 + CH3 OH
H2SO4 , to H2SO4 , to
HOOC COO C2H5 + H2O CH3 OOC COO C2H5 + H2O .
0,25 2,00
1 Vit PTHH cc phn ng v tnh gi tr ca m (1,25 im) Khi cho hn hp A
gm Zn v Cu vo dung dch AgNO3, xy ra phn ng: Zn + 2AgNO3 = Zn(NO3)2
+ 2Ag (1) Khi Zn phn ng ht, xy ra tip phn ng: Cu + 2AgNO3 =
Cu(NO3)2 + 2Ag (2) Theo bi, sau phn ng thu c hn hp kim loi, hn hp
ch c th l: Zn d, Ag, Cu (trng hp 1) hoc Ag, Cu d (trng hp 2). Xt
trng hp 1: Khi ch c (1) xy ra v AgNO3 phn ng ht. Gi s mol Zn, Cu
trong 5,15 gam hn hp A tng ng l x v y; s mol Zn phn ng l a, ta c:
Khi lng hn hp A 65x + 64y = 5,15 (a) Khi lng hn hp kim loi: 65(x -
a) + 64y + 108.2a = 15,76 (b) S mol AgNO3: 2a = 0,14.1 = 0,14 hay a
= 0,07 (c) T (b) v (c), suy ra 65x + 64y = 5,19 Mu thun vi (a) Loi
trng hp 1.2/5
0,25
0,25
Xt trng hp 2: Khi c (1), (2) xy ra v AgNO3 phn ng ht. Gi s mol
Cu phn ng l b, ta c: Khi lng hn hp kim loi: 64(y - b) + 108.2(x +
b) = 15,76 (b') S mol AgNO3: 2(x + b) = 0,14 hay (x + b) = 0,07
(c') Gii h 3 phng trnh (a), (b'), (c'), c: x = 0,03 (mol); y = 0,05
(mol); b = 0,04 (mol). Mi phn ca dung dch B c 0,015 mol Zn(NO3)2 v
0,02 mol Cu(NO3)2. Phn ng phn th nht: Cu(NO3)2 + 2KOH = 2KNO3 +
Cu(OH)2 Zn(NO3)2 + 2KOH = 2KNO3 + Zn(OH)2 Zn(OH)2 + 2KOH = K2ZnO2 +
2H2O Khi nung kt ta: Cu(OH)2 = CuO + H2O S mol CuO = s mol Cu(NO3)2
= 0,02 mol m = 0,02.80 = 1,6 (gam). 2 Tnh gi tr ca V (0,75 im) Khi
cho Zn vo phn th hai ca dung dch B: Zn + Cu(NO3)2 = Cu + Zn(NO3)2 S
mol Zn(NO3)2 = s mol Cu(NO3)2 = 0,02 mol Tng s mol Zn(NO3)2 trong
dung dch D = 0,015 + 0,02 = 0,035 (mol). Khi cho dung dch NaOH vo
dung dch D, xy ra phn ng: 2NaOH + Zn(NO3)2 = Zn(OH)2 + 2NaNO3 Nu
NaOH d: Zn(OH)2 + 2NaOH = Na2ZnO2 + 2H2O + Trng hp ch xy ra phn ng
(8): 2,97 S mol Zn(OH)2 = = 0,03 (mol) S mol NaOH = 2.0,03 = 0,06
(mol). 99 0,06.1000 = 30 (ml). Th tch dung dch NaOH: V = 2 + Trng
hp xy ra phn ng (8), (9): S mol NaOH (8) = 2 s mol Zn(NO3)2 =
2.0,035 = 0,07 (mol) S mol NaOH (9) = 2 s mol Zn(OH)2 b tan =
2(0,035 - 0,03) = 0,01 (mol). Tng s mol NaOH cn dng = 0,07 + 0,01 =
0,08 (mol) 0,08.1000 = 40 (ml). Th tch dung dch NaOH: V = 2 IV 1 Xc
nh CTPT ca ru R v tnh gi tr ca p (0,50 im) V este Z n chc nn ru n
chc. t CTPT ca ru l CxHyO.to
0,25
(3) (4) (5) (6)
0,25 0,25
(7)
0,25
(8) (9) 0,25
0,25
2,00
52,17 13,04 34,79 : : = 2 : 6 : 1 x = 2 v y = 6. 12 1 16 CTPT ca
R l C2H6O. Ru R l C2H5OH.Ta c t l: x : y : 1 =
0,25
2C2H5OH + 2Na 2C2H5ONa + H2 S mol C2H5OH = 2 s mol H2 = 20,56 =
0,05 (mol) p = 46.0,05 = 2,3 (gam) 22,40,25
2 Xc nh CTCT ca Y, Z. Tnh gi tr ca m (1,25 im) + Xc nh CTCT ca
Y, Z: Gi CTPT ca axit Y l R1COOH, ca este Z l R2COOC2H5 (R1, R2 l
cc gc hirocacbon); s mol ca Y v Z trong m gam hn hp X l a v b, ta
c: R1COOH + NaOH R1COONa + H2O t R2COOC2H5 + NaOH R2COONa +
C2H5OHo
0,25
Nu NaOH khng d th hn hp rn khan E gm hai mui R1COONa v R2COONa S
mol hai mui = s mol NaOH = 0,4.1 = 0,4 (mol). S mol R2COONa = s mol
C2H5OH = 0,05 (mol)
0,25
S mol R1COONa = 0,35 (mol) 0,05 (mol) Mu thun vi bi. Vy NaOH phi
d.
3/5
Hn hp rn khan E gm NaOH d v R1COONa (R1 = R2) c s mol bng nhau.
Ta c phngtrnh: S mol C2H5OH = b = 0,05 (mol) S mol NaOH d = 0,4 - a
- b = a + b a + b = 0,2 (mol) a = 0,15 (mol). Khi lng hn hp rn khan
E = (R1 + 67).0,2 + 40.0,2 = 24,4 R1 = 15 hay R1 l CH3
0,25
CTCT ca axit Y l CH3COOH v ca este Z l CH3COOC2H5.+ Tnh gi tr ca
m: m = 60.0,15 + 88.0,05 = 13,4 (gam) 3 Tnh gi tr ca V (0,25 im)
PTHH ca phn ng xy ra: CaO, to CH3COONa + NaOH CH4 + Na2CO3 Th tch
kh CH4: V = 22,4. 0,2 = 4,48 (lt). V.a 1 Xc nh pH ca dung dch. Vit
PTHH cc phn ng gii thch (0,50 im) + Dung dch CH3COONa c pH > 7.
Gii thch: CH3COONa = CH3COO + Na+ CH3COO + H2O CH3COOH + OH Trong
dung dch c d ion OH, do vy dung dch c pH > 7. + Dung dch
(NH4)2SO4 c pH < 7. Gii thch: (NH4)2SO4 = 2NH4+ + SO42 NH4+ +
H2O NH3 + H3O+ + + Trong dung dch c d ion H3O (hoc H ), do vy dung
dch c pH < 7. 2 Vit CTPT cc cht X1, X2, X3, X4, X5 v hon thnh
PTHH cc phn ng (0,50 im). + CTPT: X1 l KCl, X2 l KOH, X3 l Cl2, X4
l Ba(HCO3)2, X5 l H2SO4. + PTHH cc phn ng: pmnx 2KOH + Cl2 + H2
(pmnx: in phn c mng ngn xp) a) 2KCl + 2H2O b) 2KOH + Ba(HCO3)2
BaCO3 + K2CO3 + 2H2O to c) 6KOH + 3Cl2 5KCl + KClO3 + 3H2O d)
Ba(HCO3)2 + H2SO4 BaSO4 + 2CO2 + 2H2O 3 Vit CTCT v gi tn Z1, Z2,
Z3, Z4. Nhn bit cc cht lng Z2, Z3, Z4 v T. Vit PTHH cc phn ng iu ch
Z3, Z4 (1,00 im). a) Vit CTCT v gi tn Z1, Z2, Z3, Z4 Anehit fomic
Z1: HCHO Z3: HCOOCH3 Metyl fomiat Z2 : HCOOH Z4: CH3CH2OH Axit
fomic Ru etylic
0,25 0,25
0,25 2,00
0,25
0,25
0,25
0,25
0,25
b) Nhn bit cc cht lng Z2, Z3, Z4 v T T l cht hu c n chc, ng phn
ca Z3 T l CH3COOH Nhn bit: HCOOH; HCOOCH3; C2H5OH v CH3COOH. - Dng
qu tm nhn bit c cc axit HCOOH; CH3COOH (lm qu tm). Cn li HCOOCH3 v
C2H5OH khng lm qu tm. - Dng Ag2O trong dung dch NH3 nhn bit c HCOOH
(to kt ta Ag): HCOOH + Ag2ONH3, to
0,25
CO2 + H2O + 2Ag
Cn li l CH3COOH. - Dng Na nhn bit C2H5OH (c kh thot ra): 2C2H5OH
+ 2Na 2C2H5ONa + H2 Cn li l HCOOCH3.
4/5
c) Vit PTHH cc phn ng iu ch HCOOCH3, C2H5OH Cl2 CH4 + CH3Cl +
NaOH CH3OH + CuO HCHO + Ag2O HCOOH + CH3OH 2CH4 C2H2 + H2O CH3CHO +
V.b 1 Vit PTHH cc phn ng xy ra theo s chuyn ha (1,00 im). (1) 3Cu +
8HNO3 3Cu(NO3)2 + 2NO + 4H2O (A1) (A2) (2) Cu(NO3)2 + H2S CuS +
2HNO3 (A2) (A3) (3) (4) 2Cu(NO3)2 (A2) CuO + Cu (A4) 3CuO + 2NH3
(A4) 2Cu + 4HCl + O2 (A1) CuCl2 + 2NaOH (A6) Cu(OH)2 + 4NH3 (A7) to
to
to
askt
CH3Cl CH3OH HCHO HCOOH
+ HCl + NaCl + Cu + H2O + 2Ag 0,25
H2SO4 , to 1500OC
to
NH3, to
HCOOCH3 + H2O 0,25 2,00
HgSO4, 80oC Ni, to
C2H2 + 3H2 CH3CHO C2H5OH
H2
0,25
to
2CuO (A4) Cu2O (A5)
+ 4NO2 + O2
0,25
(5) (6) (7) (8)
3Cu + N2 + 3H2O (A1) 2CuCl2 + 2H2O (A6) Cu(OH)2 + 2NaCl (A7)
[Cu(NH3)4](OH)2 (A8)
0,25
0,25
2 Vit cc CTCT v gi tn cc ipeptit ng vi cng thc phn t C5H10O3N2
(0,50 im).
H2N CH2 C NH OH2N CH CH3
CH
C
OH
Glyxylalanin 0,25
CH3 OC OH OAlanylglyxin
C NH CH2 O
0,25
3 Trnh by phng php nhn bit 3 dung dch glucoz, fructoz v
glixerol. Vit PTHH cc phn ng (0,50 im). * Nhn bit glucoz bng nc
brom qua du hiu nc brom b mt mu: CH2OH[CHOH]4CHO + Br2 +H2O
CH2OH[CHOH]4COOH + 2HBr * Nhn bit c fructoz bng phn ng trng bc, do
trong mi trng kim fructoz chuyn ho thnh glucoz qua cn bng sau: OH
Glucoz Fructoz CH2OH[CHOH]4CHO + 2[Ag(NH3)2]OH CH2OH[CHOH]4COONH4 +
2Ag + 3NH3 + H2O * Dung dch cn li l glixerol: CH2OHCHOHCH2OH.to
0,25
0,25
Nu th sinh lm bi khng theo cch nu trong p n m vn ng th c im tng
phn nh p n quy nh. ---------------- Ht ----------------
5/5
