De Thi Hsg Bac Bo Lan Thu Nam

7/30/2019 De Thi Hsg Bac Bo Lan Thu Nam

1/14

HI CC TRNG CHUYNKHU VC DUYN HI BC B

HI THI HC SINH GII DUYN HI BC BLN TH IV

THI CHNH THC

Mn: HA HC 10Ngy thi: 23/4/2011Thi gian lm bi: 180 pht.(khng k thi gian giao )

Ch k gim th 1:

...............................

Ch k gim th 2:

...............................

( thi ny c 3 trang)

Cu 1:(2 im):1. Tnh nng lng ca electron trng thi c bn trong cc nguyn t v ion

sau: H, He+. (Cho ZH = 1; ZHe = 2).2. Tnh nng lng ion ha ca H v nng lng ion ha th 2 ca He.3. Mi phn t XY2 c tng cc ht proton, ntron, electron bng 178; trong

, s ht mang in nhiu hn s ht khng mang in l 54, s ht mang in caX t hn s ht mang in ca Y l 12.

a, Hy xc nh k hiu ho hc ca X,Y v cng thc phn t XY2 .b, Vit cu hnh electron ca nguyn t X,Y v xc nh cc s lng t ca

electron cui cng c in vo.Cu 2:(2 im):

Vit cng thc Lewis, d on dng hnh hc ca cc phn t v ion sau (c

gii thch) v trng thi lai ha ca nguyn t trung tm?SO2; SO3; SO42- ; SF4; SCN-

Cu 3:(2 im):1. Cho gi tr ca bin thin entanpi v bin thin entropi chun 300K v 1200K

ca phn ng:

CH4 (kh) + H2O (kh) CO ( kh) + 3H2 ( kh)Bit:

H0 (KJ/mol) S0 J/K.mol

3000

K - 41,16 - 42,412000K -32,93 -29,6a) Hi phn ng t din bin s theo chiu no 300K v 1200K?

b) Tnh hng s cn bng ca phn ng 300K2. Nng lng mng li ca mt tinh th c th hiu l nng lng cn thit tch nhng ht trong tinh th ra cch xa nhau nhng khong v cc.

Hy thit lp chu trnh tnh nng lng mng li tinh th CaCl2 bit:Sinh nhit ca CaCl2: H1 = -795 kJ/ mol

Nhit nguyn t ho ca Ca: H2 = 192 kJ / mol


2/14

Nng lng ion ho (I1 + I2) ca Ca = 1745 kJ/ molNng lng phn ly lin kt Cl2: H3 = 243 kJ/ moli lc vi electron ca Cl: A = -364 kJ/ mol

Cu 4:(2 im):1.Tnh pH ca dung dch A gm KCN 0,120 M; NH3 0,150 M v KOH 5,00.10-3 M.Cho bit pKa ca HCN l 9,35; ca NH4+ l 9,24.

2. C dung dch A cha hn hp 2 mui MgCl2(10-3

M) v FeCl3(10-3

M)Cho dung dch NaOH vo dung dch A.a) Kt ta no to ra trc, v sao?b) Tm pH thch hp tch mt trong 2 ion Mg2+ hoc Fe3+ ra khi dung dch.

Bit rng nu ion c nng = 106 M th coi nh c tch ht.( Cho tch s tan ca Fe(OH)3 v Mg(OH)2 ln lt l: 10 39 v 10 11 )

Cu 5:(2 im):Mt pin in ha c to bi 2 in cc. in cc th nht l tm ng nhng

vo dung Cu(NO3)2 0,8M. in cc 2 l mt a Pt nhng vo dung dch cha hnhp Fe2+ v Fe3+ (trong [Fe3+] = 4[Fe2+]. Th in cc chun ca Cu2+/ Cu v

Fe3+/Fe2+ ln lt l 0,34V v 0,77V.1. Xc nh in cc dng, in cc m. Tnh sut in ng khi pin bt u lm

vic.

2. Tnh t l][

][2

3

+

+

Fe

Fekhi pin ht in (coi th tch ca dung dch Cu(NO3)2 0,8M l

rt ln).Cu 6:(2 im):

Cho s bin ha:

A FeCl3

X Y Z

T M N

Hon thnh phng trnh ha hc khc nhau trong s bin ha trn. Bit: X lmt n cht, Y, Z, M l cc mui c oxi ca X, T l mui khng cha oxi ca X, Nl axit khng bn ca X.

Cu 7:(2 im):Cho 6,00 gam mu cht cha Fe3O4, Fe2O3 v cc tp cht tr. Ha tan mu

vo lng d dung dch KI trong mi trng axit (kh tt c Fe 3+ thnh Fe2+) to

ra dung dch A. Pha long dung dch A n th tch 50ml. Lng I2 c trong 10ml

dung dch A phn ng va vi 5,50 ml dung dch Na 2S2O3 1,00M (sinh ra

2

4 6S O ). Ly 25 ml mu dung dch A khc, chit tch I2, lng Fe2+ trong dung

(5)

(6)

(1)(3)

(7)

(4)

(2)

(8) (10)

(11) (12)

(9)


3/14

dch cn li phn ng va vi 3,20 ml dung dch KMnO 4 1,00M trong dung

dch H2SO4.

1. Vit phng trnh ha hc ca cc phn ng xy ra (dng phng trnh ion thu

gn).

2. Tnh phn trm khi lng Fe3O4 v Fe2O3 trong mu ban u?

Cu 8:(2 im):

Nguyn t ca mt nguyn t X trong electron cui cng c 4 s lng t

n = 3, l= 1, m = 0, s = -

1) Xc nh tn nguyn t X.

2) Ha tan 5,91 hn hp NaX v KBr vo 100ml dung dch hn hp Cu(NO3)20,1M v AgNO3 cha bit nng , thu c kt ta A v dung dch B.

Trong dung dch B, nng % ca NaNO3 v KNO3 tng ng theo t l 3,4 :

3,03. Cho ming km vo dung dch B, sau khi phn ng xong ly ming km rakhi dung dch, thy khi lng tng 1,1225g.

a) Tnh lng kt ta ca A?

b) Tnh CM ca AgNO3 trong dung dch hn hp.Cu 9:(2 im):1. Mt cht thi phng x c chu k bn hy l 200 nm c cha trong thng knv chn di t. Phi trong thi gian l bao nhiu tc phn r gim t 6,5.1012

nguyn t/pht xung cn 3.10-3 nguyn t/pht.2. Hon thnh cc phn ng ht nhn sau:

a) 12Mg

26

+ ...?

10Ne

23

+ 2He

4

b) 9F19 + 1H1 ...? + 2He4

c) 92U235 + 0n1 3(0n1) +...? + 57La146

d) 1H2 + ...? 2 2He4 + 0n1

Cu 10:(2 im): 270C, 1atm N2O4 phn hu theo phn ng : N2O4 (kh) 2NO2(kh)

vi phn hu l 20%

1. Tnh hng s cn bng Kp.

2. Tnh phn hu mt mu N2O4 (kh) c khi lng 69 gam, cha trong mtbnh c th tch 20 (lt) 270C

------------------------- Ht ---------------------------(Th sinh c s dng bng HTTH-Cn b coi thi khng gii thch g thm)

H v tn th sinh: S bo danh:


4/14

HI CC TRNG CHUYNKHU VC DUYN HI BC B

HI THI HC SINH GII DUYN HI BC BLN TH IV

THI CHNH THC

Mn: HA HC 10Ngy thi: 23/4/2011Thi gian lm bi: 180 pht.(khng k thi gian giao )

CU V P N I1 1. Tnh nng lng ca electron trng thi c bn trong cc

nguyn t v ion sau: H, He+. (Cho ZH = 1; ZHe = 2).2. Tnh nng lng ion ha ca H v nng lng ion ha th 2

ca He.3. Mi phn t XY2 c tng cc ht proton, ntron, electron

bng 178; trong , s ht mang in nhiu hn s ht khng mangin l 54, s ht mang in ca X t hn s ht mang in ca Y l12.

a , Hy xc nh k hiu ho hc ca X,Y v XY2 .b , Vit cu hnh electron ca nguyn t X,Y v xc nh cc

s lng t ca electron cui cng c in vo.

Hng dn1. Nng lng ca electron trong h mt ht nhn v mt electron:

En = (eV)

trng thi c bn: n = 1.* Vi H: E1(H) = -13,6eV;* Vi He+: E1(He+ ) = - 54,4 eV;2. Nng lng ion ha ca hidro l nng lng ti thiu bt e ra

khi nguyn t hoc ion, tc l a e t trng thi c bn ra xa v cng(khng truyn thm ng nng cho e). D thy: I1(H) =13,6eV; I2(He) = 54,4eV.

a , K hiu s n v in tch ht nhn ca X l Zx , Y l Zy ; s ntron(ht khng mang in) ca X l Nx , Y l Ny . Vi XY2 , ta c cc

phng trnh:2 Zx + 4 Zy + Nx + 2 Ny = 178 (1)2 Zx + 4 Zy Nx 2 Ny = 54 (2)

0

0,

0,


5/14

4 Zy 2 Zx = 12 (3)Zy = 16 ; Zx = 26

Vy X l st, Y l lu hunh. XY2 l FeS2 .

b, Cu hnh electron: Fe : 1s2 2s2 2p6 3s2 3p63d64s2 ;S : 1s2 2s2 2p6 3s2 3p4

B 4 s lng t cui ca X: n = 3; l = 2; ml =-2; ms= -1/2.B 4 s lng t cui ca X: n = 3; l = 1; ml =-1; ms= -1/2.

0,

0,

2 Vit cng thc Lewis, d on dng hnh hc ca cc phn t v ionsau (c gii thch) v trng thi lai ha ca nguyn t trung tm?

SO2; SO3; SO42- ; SF4; SCN-

Hng dn

Phn t Cng thc Lewis Cng thccu trc

Dng lai haca NTTT

Dng hnh hc ca phn t

O

S

O

AX2E sp2 Gp khc

OS

O

O AX3 sp2 Tam gic u

O

S

O

O

O

2- AX4 sp3 T din

F

S

F

F F AX4E sp3d Ci bp bnh

S C NAX2 Sp ng thng

Mn0,1im

3 1.Cho gi tr ca bin thin entanpi v bin thin entropi chun

3000K v 12000K ca phn ng:

CH4 (kh) + H2O (kh) CO ( kh) + 3H2 ( kh)

Bit l H0 (KJ/mol) S0 J/K.mol3000K - 41,16 - 42,412000K -32,93 -29,6

a) Hi phn ng t din bin s theo chiu no 3000K v12000K?

b) Tnh hng s cn bng ca phn ng 3000K2. Nng lng mng li ca mt tinh th c th hiu l nng lngcn thit tch nhng ht trong tinh th ra cch xa nhau


6/14

nhng khong v cc.Hy thit lp chu trnh tnh nng lng mng li tinh th

CaCl2 bit:Sinh nhit ca CaCl2: H1 = -795 kJ/ molNhit nguyn t ho ca Ca: H2 = 192 kJ / molNng lng ion ho (I1 + I2) ca Ca = 1745 kJ/ mol

Nng lng phn ly lin kt Cl2:

H3 = 243 kJ/ moli lc vi electron ca Cl: A = -364 kJ/ molHng dn

1.a) Da vo biu thc: G0 = H0 - TS0

3000K ; G0300 = (- 41160) - [ 300.(- 42,4)] = -28440J = -28,44kJ

12000K ; G01200 = (- 32930) - [ 1200.(- 29,6)] = 2590 = 2,59 kJG0300< 0, phn ng cho t xy ra 3000K theo chiu t tri

sang phi.G01200 > 0, phn ng t din bin theo chiu ngc li 12000K

b) + Tnh hng s cn bng ca phn ng 3000KG0 = -2,303RT lgK(-28440) = (-2,303).8,314. 300.lgK

lgK = 28440/ 2,303.8,314.300 = 4,95 K = 10 4,95

2. Thit lp chu trnhChu trnh Born - Haber

Ca(tt) + Cl2 (k) CaCl2(tt)

Ca (k) 2Cl (k)

Ca2+ (k) + 2Cl- (k)

Ta c:

Uml = H2 + I1 + I2 + H3 + 2A - H1

Uml = 192 + 1745 + 243 (2 x 364) - (-795)

Uml = 2247 (kJ/.mol)

0

0

0

H1

H2

H3

I1+I

22A

-Uml


7/14

4 1.Tnh pH ca dung dch A gm KCN 0,120 M; NH3 0,150 M vKOH 5,00.10-3 M.Cho bit pKa ca HCN l 9,35; ca NH4+ l9,24

2.C dung dch A cha hn hp 2 mui MgCl2(10-3M) v FeCl3(10-3M)

Cho dung dch NaOH vo dung dch A.a) Kt ta no to ra trc, v sao?b) Tm pH thch hp tch mt trong 2 ion Mg2+ hoc Fe3+ ra

khi dung dch.Bit rng nu ion c nng = 106 M th coi nh c tch

ht.( Cho tch s tan ca Fe(OH)3 v Mg(OH)2 ln lt l: 10 39 v 10 11 )

Hng dn

1) Tnh pH ca dung dch:

CN-

+ H2O HCN + OH-

Kb1 = 10- 4,65

NH3 + H2O NH4+ + OH- Kb2 = 10- 4,76

KOH -> K+ + OH-

H2O H+ + OH-

[OH-] = CKOH + [HCN] + [NH4+] + [H+]

t [OH-] = x

x = 5.10-3 + Kb1[CN]/x + Kb2[NH3]/x + KH2O/x

x2 - 5.10-3x - (Kb1[CN-] + Kb2[NH3] + KH2O) = 0

Tnh gn ng coi [CN-] bng CCN- = 0,12M ; [NH3] = CNH3 = 0,15 M .

Ta c: x2 - 5.10-3 . x - 5,29 . 10-6 = 0 -> x = [OH-] = 5,9.10-3M.

Kim li [HCN] / [CN-] = 10-4,65/ 5,9.10-3 = 3,8.10-3 -> [HCN] [NH4+] pH = 11,77.2) MgCl2 Mg2+ + 2Cl v Mg2+ + 2OH Mg(OH)2 (1)

FeCl3 Fe3+ + 3Cl v Fe3+ + 3OH Fe(OH)3 (2)

a) to Fe(OH)3 th [OH ] 3 339

1010

= 10-12 M (I)

to Mg(OH)2[OH ] 311

10

10

= 10-4 M (II)

So snh (I) < (II) thy Fe(OH)3 to ra trc.b) to Mg(OH)2: [OH ] = 10-4[H+] = 10-10 pH = 10 (nu

pH < 10 th khng ) hon ton Fe(OH)3: [Fe3+] 10-6M [OH ]3 > 10-33 [H+]

3

2

0

0

0,5

0,5


8/14

Vy tch Fe3+ ra khi dd th: 3 < pH < 10

5 Mt pin in ha c to bi 2 in cc. in cc th nht l

tm ng nhng vo dung Cu(NO3)2 0,8M. in cc 2 l mt a Pt

nhng vo dung dch cha hn hp Fe2+ v Fe3+ (trong [Fe3+] =

4[Fe2+]. Th in cc chun ca Cu2+/ Cu v Fe3+/Fe2+ ln lt l

0,34V v 0,77V.

1. Xc nh in cc dng, in cc m. Tnh sut in ng khipin bt u lm vic.

2. Tnh t l][

][2

3

+

+

Fe

Fekhi pin ht in (coi th tch ca dung dch

Cu(NO3)2 0,8M l rt ln).Hng dn

1.E(Fe3+/Fe2+) = 0,77 + 0,059/1 . lg4 = 0,8055 VE(Cu2+/Cu) = 0,34 + 0,059/2 . lg0,8 = 0,3371 VVy in cc dng l in cc Pt; in cc m l in cc Cu

Epin = 0,8055 - 0,3371 = 0,4684 V

2. Pin ht in tc l Epin = 0. Khi E (Cu2+/Cu) = E (Fe3+/Fe2+)

V th tch dung dch Cu(NO3)2 rt ln => nng Cu2+ thay i khng

ng k=> E (Cu2+/Cu)=0,3371 V

E (Fe3+/Fe2+) = 0,77 + 0,059/1 . lg ([Fe3+]/[Fe2+]) = 0,3371

=> [Fe3+]/[Fe2+] = 4,5995.10-8.

2

1

1

6Cho s bin ha :

A FeCl3

X Y Z

T M N

Hon thnh phng trnh ha hc khc nhau trong s bin hatrn. Bit: X l mt n cht, Y, Z, M l cc mui c oxi ca X, T lmui khng cha oxi ca X, N l axit khng bn ca X.

(5)

(6)

(1)(3)

(7)

(4)

(2)

(8) (10)

(11) (12)

(9)


9/14

Hng dn: S bin ha tha mn l:

HCl FeCl3

X KClO3 KClO4

KCl KClO HClO

C cc phng trnh phn ng:H2 + Cl2 2HCl (1)

(X) (A)6HCl + Fe2O3 2FeCl3 + 3H2O (2)

(A) (Fe3O4,)2Fe + 3Cl2 2FeCl3 (3)

3Cl2 + 6KOH 5KCl + KClO3 + 3H2O (4)

(Y)6HCl + KClO3 3Cl2 + KCl + 3H2O (5)Cl2 + 2KOH KCl + KClO + H2O (6)

(T)2KClO3 2KCl + 3O2 (7)

KCl + 3H2O KClO3 + 3H2 (8)

4KClO3 3KClO4 + KCl (9)

KClO4 KCl + 2O2 (10)

KCl + H2O KClO + H2 (11)(M)

KClO + CO2 + H2O HClO + NaHCO3 (12)(N)

7 Cho 6,00 gam mu cht cha Fe3O4, Fe2O3 v cc tp cht tr.

Ha tan mu vo lng d dung dch KI trong mi trng axit

(kh tt c Fe3+ thnh Fe2+) to ra dung dch A. Pha long dung

dch A n th tch 50ml. Lng I2 c trong 10ml dung dch A

phn ng va vi 5,50 ml dung dch Na 2S2O3 1,00M (sinh ra

2

4 6S O ). Ly 25 ml mu dung dch A khc, chit tch I 2, lng Fe2+

trong dung dch cn li phn ng va vi 3,20 ml dung dch

KMnO4 1,00M trong dung dch H2SO4.

2

to

(5)

(6)

(1)(3)

(7)

(4)

(2)

(8) (10)

(11) (12)

to

p dung dch(80oC)Khng c mnx

300o

p dung dchKhng c mnx

tocao

(9)


10/14

1. Vit phng trnh ha hc ca cc phn ng xy ra (dng phng

trnh ion thu gn).

2. Tnh phn trm khi lng Fe3O4 v Fe2O3 trong mu ban u?

Hng dn

1.

3 2Fe O 8H 2Fe Fe 4H O3 4 2+ + ++ + + (1)

3Fe O 6H 2Fe 3H O2 3 2+ ++ + (2)

3 22Fe 3I 2Fe I3+ + + + (3)

2 22S O I S O 3I2 3 3 4 6 + + (4)

2 3 25Fe MnO 8H 5Fe Mn 4H O4 2+ + + ++ + + + (5)

2.Trong 25 ml: 2

4

3

Fe MnOn 5n 5x3,2x1x10+

= = =0,016 (mol)

trong 10ml 2Fen + = 6,4x10-3(mol)

T (3) v (4): 2Fen + = 22 3S On = 5,5x1x10-3 = 5,5x10-3(mol)

T (3): 3Fen + = 2Fen + =5,5x10-3(mol) =2( 3 4Fe On + 2 3Fe On )

C th xem Fe3O4 nh hn hp Fe2O3.FeO

FeOn =

3 4Fe On = 6,4x10-3 5,5x10-3 = 9x10-4(mol)

2 3Fe On = 3Fe

1n

2+ 3 4Fe On =1,85x10

-3(mol).

Trong 50 ml :3 4Fe O

n =4,5x10-3(mol) 3 4Fe Om =1,044 gam

% khi lng Fe3O4 = 1,044/6 x 100% = 17,4%

2 3Fe On = 9,25x10-3(mol) 2 3Fe Om =1,48 gam

% khi lng Fe2O3 = 1,48/6 x 100% = 24,67%

1

0,

0,

08

Cu 8: Bi tp tng hp(2 )

Nguyn t ca mt nguyn t X trong electron cui cng c 4 slng t n = 3, l= 1, m = 0, s = -

2


11/14

1. Xc nh tn nguyn t X.

2. Ha tan 5,91 hn hp NaX v KBr vo 100ml dung dch hn hpCu(NO3)2 0,1M v AgNO3 cha bit nng , thu c kt ta A vdung dch B.

Trong dung dch B, nng % ca NaNO3 v KNO3 tng ng theot l 3,4 : 3,03. Cho ming km vo dung dch B, sau khi phn ng xong

ly ming km ra khi dung dch, thy khi lng tng 1,1225g.a, Tnh lng kt ta ca A?

B,Tnh CM ca AgNO3 trong dung dch hn hp.

(cho Na = 23, N = 14, K = 39, Ag = 108, Br = 80, Zn = 65, Cu = 64)Hng dn

1(0,75) Nguyn t ca nguyn t X c:

n = 3

l= 1

m = 0

s = -

Cu trc hnh e ca X : 1s2 2s2 2p6 3s2 3p5

-> Zx = 17 X l clo

2(1,25).

a/ NaCl + AgNO3 = AgCl + NaNO3

KBr + AgNO3 = AgBr + KNO3Khi cho Zn vo dd B, khi lng ming Zn tng, chng t AgNO 3

d.

Zn + 2AgNO3 = Zn(NO3)2 + 2Ag

Zn + Cu(NO3)2 = Zn(NO3)2 + Cu

NaCl : x mol

KBr : y mol

01,0000.1

1,0.1002)

==3Cu(NO

n

mol

03,3

4,3=

3

3

C%KNO

C%NaNO

-> 03,34,3

=3

3

KNO

NaNO

m

m

xy 75,003,3

4,3=>=

101y

85x(1)

58,5x + 119y = 5,91 (2)

0,

0,25

electron cui cng phn lp 3p

electron ny l e th 5 ca phn lp 3p


12/14

Gii h pt (1), (2)

=

=

03,0

04,0

y

x

mA = 0,04 . 143,5 + 0,03 . 188 = 11,38g

b/ 1 mol Zn -> 2 mol Ag khi lng tng 151g

a mol Zn -> 151a

1 mol Zn -> 1 mol Cu khi lng gim 1g0,01 mol -> 0,01g

151a 0,01 = 1,1225

a = 0,0075=

AgNO3n

b 0,04 + 0,03 + 0,015 = 0,085 mol

M85,0100

1000.085,0) ==3M(AgNOC

0

0

91. Mt cht thi phng x c chu k bn hy l 200 nm c chatrong thng kn v chn di t. phi trong thi gian l bao nhiu tc phn r gim t 6,5.1012 nguyn t/pht xung cn 3.10-3

nguyn t/pht.2. Hon thnh cc P ht nhn sau:

a) 12Mg26 + ...? 10Ne23 + 2He4

b) 9F19 + 1H1 ...? + 2He4

c) 92U235 + 0n1 3(0n1) +...? + 57La146

d) 1H2 + ...? 2 2He4 + 0n1Hng dn

1.

/00347,0200

693,0693,0

2/1

===

tk nm

p dng cng thc: ln 0N

ktN

=

ln12

36,5.10 0,003473.10

t =

t = 1,0176.104 nm hay 10.176 nm

2. T nh lut bo ton in tch v s khi cc ht cn thiu:

a. 0n1 b. 8O16 c. 35Br87 d. 3Li7

2

0,

0,2

0

04


13/14

10 270C, 1atm N2O4 phn hu theo phn ng :

N2O4 (kh) 2NO2(kh)

vi phn hu l 20%

1. Tnh hng s cn bng Kp.2. Tnh phn hu mt mu N2O4 (kh) c khi lng 69 gam, cha

trong mt bnh c th tch 20 (lt) 270C

Hng dn

1.Gi phn hu ca N2O4 270C, 1 atm l , s mol ca N2O4 ban

u l n

Phn ng: N2O4 (k) 2NO2(k)

Ban u: n 0

Phn ly: n 2n

Cn bng n(1- ) 2n

Tng s mol hn hp lc cn bng: n = n(1+ )

Nn p sut ring phn ca cc kh trong hn hp lc cn bng:

PpON

+

=

1

142

; PPNO

+

=

1

22

42

2

2

ON

NO

P

P

PK = =

+

+

P

P

1

1

1

22

= P22

1

4

vi P = 1atm, = 20% hay = 0,2 KP = 1/6 atm2.

42ON

n = 69/92 = 0,75mol

Gi phn hu ca N2O4 trong iu kin mi l

Phn ng: N2O4 (k 2NO2(k)

Ban u: 0,75 0

Phn ly: 0,75 1,5

Cn bng 0,75(1- ) 1,5

Tng s mol hn hp lc cn bng: n = 0,75(1+ )

2 i

1,0

1,0


14/14

p sut hn hp kh lc cn bng:

V

RTnP

'''= =

20

300.082,0).1(75,0'

+

= 0,9225(1+)

KP = P2'2'

1

4

= 1/6

V KP = const nn: 6/1)1(9225,0.1

4 '2'

2'=+

0,19

De Thi Hsg Bac Bo Lan Thu Nam

Documents

Transcript of De Thi Hsg Bac Bo Lan Thu Nam