de thi casio hoa 2012

7/23/2019 De Thi Casio Hoa 2012

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UBND TNH BC GIANGS GIO DC V O TO

THI CHN HC SINH GII CP TNHGII TON TRN MY TNH CM TAY CASIO, VINACAL

NM HC 2011 2012MN THI: HA HC LP 12

Thi gian lm bi: 150 pht (khng k thi gian giao )

im ton bi thi Cc gim kho(H tn v ch k)

S phch(Ch tch Hi ng thi ghi)

Bng s Bng ch Gim kho 1:

Gim kho 2:

Lu : - thi ny c 6 trang, 5 im/1 cu.- Th sinh khng cn ghi thao tc bm my, kt qu c lm trn n 5 ch s thp phn.- Cho nguyn t khi ca cc nguyn t: H = 1; C = 12; N = 14; O = 16; Na = 23; Cl = 35,5; Ca = 40; Ba = 137.- Cho s Avogadro NA = 6,022.1023; R = 8,314 J/mol.K; F = 96500

Cu 1. (5 im) Trong phn t hp cht X2Y2 c tng s ht (p, n, e) bng 116, trong s ht mang innhiu hn s ht khng mang in l 36. Tng s ht (p, n, e) trong nguyn t X nhiu hn trong nguyn t

Y l 10. S khi ca X ln hn s khi ca Y l 7. Xc nh tn hp cht X2Y2.Li gii im

Cu 2.(5 im) Mui NaCl kt tinh theo mng lp phng tm din. mng c s c di mi cnh l

5,58 0A . Cho bn knh ca Cl- l 1,81 0A . Tnh bn knh ca ion Na

+ v khi lng ring ca tinh th NaCl

(g/cm3).Li gii im


2/20

Cu 3. (5 im) Nguyn t R l phi kim thuc nhm A trong bng tun hon. T l gia thnh phn phn trm

nguyn t R trong oxit cao nht vi thnh phn phn trm nguyn t R trong hp cht kh vi hiro l73

183

.

a. Xc nh nguyn t R.b. Cho vo bnh kn 0,15 mol n cht ca R v 7 gam bt st. Nung nng phn ng xy ra hon

ton, ngui ri cho nc vo, khuy u cc cht tan c tan hon ton ta thu c 2 lt dung dch X.Tnh nng mol ca cht tan trong dung dch X.

Li gii im


3/20

Cu 4 (5 im) Trn 10 ml dung dch H2SO4 10-3M vi 40 ml dung dch axit tactric H2C4H4O6 2.10-3M thu

c dung dch X, tnh pH ca X? Bit 41,99

a(HSO )K 10 ,

=hng s phn li axit ca H2C4H4O6 l K1 = 10-3,04,

K2 = 10-4,37.Li gii im

Cu 5. (5 im) Cho cn bng sau: 2 4(k) 2(k )N O 2NO , Kp = 4 atm ( 1000C). Tnh thnh phn phn trms mol ca hn hp khi p sut chung ca h ln lt l 4 atm v 25 atm. T hy rt ra kt lun v nhhng ca p sut n s chuyn dch cn bng.

Li gii im


4/20

Cu 6. (5 im) Khi phn tch mt loi khong cht ngi ta thy trong khong cha 0,257 g kim loi206

82Pb v 1g kim loi 23892 U. Thi gian bn r ca

238

92U n 20682 Pb l 4,5.10

9 nm. Gi thit khong cht ban

u khng c ch, trong qu trnh phng x ch xy ra phn r v -.a. Tnh s ln phn r v -.

b. Xc nh tui ca khong vt trn.Li gii im

Cu 7. (5 im)

1. Hy chng minh rng ( 250C) Ag c kh nng y c H2 ra khi dung dch HI 1M. Bit tch stan ca AgI = 8,3.10-17 ;

0

Ag /AgE

+ = 0,8V.


5/20

2. Ngi ta thit lp mt pin gm hai na pin sau : Zn | Zn(NO3)2 0,1M v Cu | CuSO4 0,1M c thchun tng ng l -0,76V v 0,34V. Tnh sut in ng ca pin.

Li gii im

Cu 8. (5 im) Dng 16,8 lt khng kh ktc (20% O2 v 80% N2 v th tch) t chy hon ton 3,21gam hn hp A gm hai aminoaxit no, mch h, phn t c mt nhm cacboxyl, mt nhm amino k tipnhau trong dy ng ng. Hn hp thu c sau phn ng em lm kh c hn hp kh B, cho B i quadung dch Ba(OH)2 d thy khi lng dung dch gim 14,535 gam.

a. Tm cng thc cu to v khi lng ca hai aminoaxit.b. Nu cho kh B vo bnh dung tch 16,8 lt, nhit 136,50C th p sut trong bnh l bao nhiu? Bit

rng khi t chy aminoaxit to kh N2.Li gii im

Cu 9. (5 im) Tnh hiu ng nhit ca phn ng sau 250C: CO(NH2)2(r) + H2O(l) CO2(k) +2NH3(k)Bit trong cng iu kin c cc i lng nhit sau y:

CO (k) + H2O (h) CO2 (k) + H2 (k) H1 = -41,13 kJ/molCO (k) + Cl2 (k) COCl2 (k) H2 = -112,5 kJ/molCOCl2 (k) + 2NH3 (k) CO(NH2)2(r) + 2HCl(k) H3 = -201,0 kJ/mol

Nhit to thnh HCl (k) H4 = -92,3 kJ/molNhit ha hi ca H2O(l) H5 = 44,01 kJ/mol

Li gii im


6/20

Cu 10. (5 im) Cho 15,2 gam mt hp cht hu c X cha C, H, O vi 300ml dung dch NaOH 1M (va) thu c dung dch Y, c cn Y c 23,6 gam hn hp cht rn khan Z. t chy hon ton Z ri choton b sn phm kh ln lt qua bnh 1 cha H 2SO4 c, bnh 2 cha dung dch Ca(OH)2 thy bnh 1khi lng tng 6,3 gam, cn bnh 2 thu c 55 gam kt ta. un nng phn dung dch cn li bnh 2li thu c 5 gam kt ta na.

a. Xc nh cng thc phn t ca X, bit cng thc phn t trng vi cng thc n gin nht.b. Xc nh cng thc cu to ca X, bit X khng cha nhm -COOH.

Li gii im

Cn b coi thi khng gii thch g thm

UBND TNH BC GIANGS GIO DC V O TO


7/20


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14/20

Lu : - Th sinh khng cn ghi thao tc n my, kt qu c lm trn n 5 ch s thp phn.

- S Avogadro NA = 6,022.1023

.Cu 1. (5 im) Trong phn t hp cht X2Y2 c tng s ht (p, n, e) bng 116, trong s ht mang innhiu hn s ht khng mang in l 36. Tng s ht (p, n, e) trong nguyn t X nhiu hn Y l 10. S khica X ln hn s khi ca Y l 7. Xc nh tn hp cht X2Y2.

Li gii imT u bi, ta c h

X X Y Y

X Y X Y

X X Y Y

X X Y Y

X

Y

2(2Z N ) 2(2Z N ) 116

2.2Z 2.2Z 2N 2N 36

2Z N 2Z N 10

Z N Z N 7Z 11

Z 8

+ + + = + =

+ =

+ ==

=

2,0

1,0


15/20

Hp cht : Na2O2Tn gi : Natri peoxit.

1,01,0

Cu 2. (5 im) Mui NaCl kt tinh theo mng lp phng tm din. mng c s c di mi cnh l

5,58 0A . Cho bn knh ca Cl- l 1,81 0A . Tnh bn knh ca ion Na

+ v khi lng ring ca NaCl (tinh th).

Li gii imS mng tinh th :

S ion Cl- trong mt c s: 8.1/8 + 6.1/2 = 4S ion Na+ trong mt c s: 12.1/4 + 1.1 = 4S phn t NaCl trong mt c s l 4- Ta c: 2.(rNa+ + rCl-) = a = 5,58.10-8 cm rNa+ = 0,98.10-8 cm.

- Khi lng ring ca NaCl l:3

23 8 3

A 1

n.M 4.(23 35,5)D 2,23652 (g/cm )

N .V 6,022.10 .(5,58.10 )+

= = =

1,0

0,50,51,0

1,0

1,0

Cu 3. (5 im) Nguyn t R l phi kim thuc nhm A trong bng tun hon. T l gia thnh phn phn trm

nguyn t R trong oxit cao nht vi thnh phn phn trm nguyn t R trong hp cht kh vi hiro l73

183.

a. Xc nh nguyn t R.b. Cho vo bnh kn 0,15 mol n cht ca R v 7 gam bt st. Nung nng phn ng xy ra hon

ton, ngui ri cho nc vo, khuy u cc cht tan c tan hon ton ta thu c 200ml dung dch

X. Tnh nng mol ca cht tan trong dung dch X,Li gii im

a. Oxit: R2Ox2 xR(trong R O )

2R%m

2R 16x =

+

Hp cht kh ca R vi hiro: H8-xR8 xR( trong H R)

R%m

R 8 x =

+ Theo bi ra, ta c:

2R R 73 767x 1464: R

2R 16x R 8 x 183 110

= =

+ + x 4 5 6 7

R 14,58182 21,55454 28,52727 35,50000Kt lun Loi Loi Loi Tha mnNguyn t R l Cl.b. nFe = 0,125 molPhn ng :

ot

2 32Fe + 3Cl 2FeCl (1)

Mol : 0,1 0,15 0,1

Sau (1), nFe d = 0,025 mol.Khi cho nc vo bnh, ta c phn ng

3 2Fe + 2FeCl 3FeCl (2)

Mol: 0,025 0,05 0,075

Dung dch X gm:

0,5

0,5

1, 0

1,0

1,0


16/20

2

3

M(FeCl )2

3M(FeCl )

0,075C 0,0375(M)FeCl : 0,075 (mol) 2

FeCl : 0,05 (mol) 0,05C 0,025M

2

= = = =

1,0

Cu 4 (5 im) Trn 10 ml dung dch H2SO4 10-3M vi 40 ml dung dch axit tactric H2C4H4O6 2.10-3M thu

c dung dch X, tnh pH ca X, bit 41,99

a(HSO )K 10 ,

=hng s phn li axit ca H2C4H4O6 l K1 = 10-3,04, K2

= 10-4,37.Li gii im

Sau khi trn:

2 4

2 4 4 6

34

M(H SO ) 1

33

M(H C H O ) 2

10.10C 2.10 (M) C

10 40

40.2.10C 1,6.10 (M) C

10 40

= = =+

= = =+

Axit tactric c dng H2A.Ta c cc qu trnh sau:

2 4 4

2 1,99

4 4 a

3,04

2 1

2 4,37

2

14

2 w

H SO H HSO

HSO H SO (1) K 10

H A H HA (2) K 10HA H A (3) K 10

H O H OH (4) K 10

+

+

+

+

+

+

+ =

+ =+ =

+ =

V Ka.C1 K1.C2 K2.C2 >> Kw Ta c th b qua cn bng ca H2O.Theo s bo ton in tch

2 2

4 4

a 1 1 21 22

a 1 1 2

h [H ]=[HSO ] 2[SO ]+[HA ]+2[A ]

h 2K K h 2K K h= C C

K h h K h K K

+ = +

+ + +

+ + +Gn K1, K2, Ka, C1, C2 vo my, dng chc nng SOLVE gii vi gi tr h khong 10 -3, kt qu

thu c h = 1,13418.10-3 => pH = 2,94532.

0,5

0,5

2,0

2,0


17/20

Cu 5. (5 im) Cho cn bng sau: 2 4(k) 2(k )N O 2NO Kp = 4atm ( 1000C). Tnh thnh phn phn trm smol ca hn hp khi p sut chung ca h ln lt l 4atm v 25 atm. T hy rt ra kt lun v nhhng ca p sut n s chuyn dch cn bng.

Li gii im- Khi P = 4 atm, ta c :

2

2

2 4

2 4

2 2 4

2

2 4

2

NO

NOp

N ON O

NO N O

NO

N O

PP 2,47214 (atm)K 4

PP 1,52786 (atm)P P 4

2,47214%n .100% 61,80350%

4%n 38,19650%

== =

= + =

= =

=- Khi P = 25 atm

2

2

2 4

2 4

2 2 4

2

2 4

2

NO

NOp

N O

N O

NO N O

NO

N O

PP 8,19804 (atm)K 4

PP 16,85096(atm)

P P 258,19804

%n .100% 32,59616%25

%n 67,40384%

== = =

+ = = =

=

Nhn xt : Khi p sut tng t 4 atm n 25 atm th phn trm s mol NO2 gim t 61,80350%xung cn 32,59616% => Cn bng chuyn dch theo chiu nghch.

1,0

1,0

1,0

1,0

1,0

Cu 6. (5 im) Khi phn tch mt loi khong cht ngi ta thy trong khong cha 0,257 g 20682 Pb v 1g238

92U. Thi gian bn r ca 23892 U n

206

82Pb l 4,5.109 nm. Gi thit khong cht ban u khng c ch,

trong qu trnh phng x ch xy ra phn r v -.

a. Tnh s ln phn r v -.b. Xc nh tui ca khong vt trn.Li gii im

a. Gi x v y ln lt l s ln phn r v -. Ta c phng trnh sau238 206 4 0

92 82 2 1U Pb x He y e (1)

238 206 4x x 8

92 82 2x y y 6

+ +

= + =

= + = b. Gi mt l khi lng

238

92U cn li, mo l khi lng

238

92U ban u.

Theo (1)

238 20692 82

238 20692 82

20682

23892

U Pb

U Pb

Pb

U

m mn n238 206

mm 238

206

= =

=

phn r

phn r

phn r

20682

238 238 238238 92 92 92U92

Pb

U U U

mm m m 238 m

206 = + = +

ban u phn r cn li cn li

Tui ca khong vt l20682

23892

23892

Pb

Uo 1/2

t U

m238 mm T1 206t ln ln

m ln 2 m

+= =

cn li

cn li

2,0

1,0

1,0

1,0


18/20

99

0,257238 14,5.10 206ln 1687914370 1,68791.10

ln 2 1

+= = n m n m

Cu 7. (5 im)1. 250C, kim loi Ag c kh nng y c H2 ra khi dung dch HI 1M khng ? Ti sao? Bit tch

s tan ca AgI = 8,3.10-17 ;0

Ag /AgE

+ = 0,8V.

2. Ngi ta thit lp mt pin gm hai na pin sau : Zn | Zn(NO3)2 0,1M v Cu | CuSO4 0,1M c thchun tng ng l -0,76V v 0,34V. Tnh sut in ng ca pin.

Li gii im

0 0

AgI/Ag Ag /Ag

17s(AgI)0

Ag /Ag

RT1. E E + ln[Ag ]

F

KRT 8,314.298 8,3.10= E + ln 0,8 ln 0,15066V

F [I ] 96500 1

+

+

+

=

= + =

=> 0AgI/AgE Epin = 2 2Zn /Zn Cu /CuE E+ + = 0,31044 + 0,78956 = 1,10000V

1,0

1,0

1,01,0

1,0

Cu 8. (5 im) Dng 16,8 lt khng kh ktc (20% O2 v 80% N2 v th tch) t chy hon ton 3,21gam hn hp A gm hai aminoaxit no, mch h, phn t c mt nhm cacboxyl, mt nhm amino k tipnhau trong dy ng ng. Hn hp thu c sau phn ng em lm kh c hn hp kh B, cho B i quadung dch Ba(OH)2 d thy khi lng dung dch gim 14,535 gam.

a. Tm cng thc cu to v khi lng ca hai aminoaxit.b. Nu cho kh B vo bnh dung tch 16,8 lt, nhit 136,50C th p sut trong bnh l bao nhiu? Bit

rng khi t chy aminoaxit to kh N2.Li gii im

2 2O N

1 16,8n 0,15(mol) n 4.0,15 0,6(mol)

5 22, 4= = = =

t CTTB ca hai aminoaxit l

2 2n 2n m 2m 1NH C H COOH hay C H NO

+

ot

2 2 2 2 2m 2m 1

6m 3 2m 1 1C H NO O mCO H O N (1)

4 2 2+ +

+ + +

Hn hp kh B gm CO2, N2, O2 khi qua bnh Ba(OH)2 xy ra phn ng :CO2 + Ba(OH)2 BaCO3 + H2O. Gi s mol CO2 l x mol.

3 2BaCO CO

aminoaxit

aminoaxit

m m 14,535 197x 44x 14,535

0,095x 0,095(mol) n (mol)

m3,21.m

M 14m 47 m 2,3750,095

= =

= =

= = + =

=> CTPT ca hai aminoaxit: C2H5NO2 v C3H7NO2.=> CTCT ca hai aminoaxit: NH2 CH2 COOH v NH2 CH2 CH2 COOH

hoc NH2 CH2 COOH v NH2 CH(CH3) COOH

Gi s mol C2H5NO2 : x mol v C3H7NO2 : y mol

0,5

0,5

0,5

0,5

0,5

0,5


19/20

2x 3y 0,095x 0,025

0,095x y 0,04 y 0,015

2,375

+ == + = = =

=>2 5 2

3 7 2

C H NO

C H NO

m 0,025.75 1,875(g)

m 0,015.89 1,335(g)

= =

= =

b. 2 2 2 2B CO N (kk) N (1) O (d )n n n n n= + + + - = 0,095 + 0,6 + 0,04.0,5 + 0,15 -(6.2,375 3)

0,044

= 0,7525 (mol)22,4

0,7525 (273 136,5)nRT 273P 1,505(atm)V 16,8

+= = =

0,5

0,5

0,5

0,5

Cu 9. (5 im) Tnh hiu ng nhit phn ng 250C ca phn ng sau:CO(NH2)2(r) + H2O(l) CO2(k) + 2NH3(k)

Bit trong cng iu kin c cc i lng nhit sau y:CO (k) + H2O (h) CO2 (k) + H2 (k) H1 = -41,13 kJ/mol

CO (k) + Cl2 (k) COCl2 (k) H2 = -112,5 kJ/molCOCl2 (k) + 2NH3 (k) CO(NH2)2(r) + 2HCl(k) H3 = -201,0 kJ/molNhit to thnh HCl (k) H4 = -92,3 kJ/molNhit ha hi ca H2O(l) H5 = 44,01 kJ/mol

Li gii im

2 2 5

2 2 2 1

2 2 2

H O(l) H O(h) H 44,1kJ

CO (k) H O(h) CO (k) H (k) H 41,13kJ

COCl (k) CO (k) + Cl (k) - H

=

+ + =

=

2 2 4

2 2 2 3 3

2 2 2 2 3 5 1 2 4 3

112,5kJ

H (k) Cl (k) 2HCl(k) 2 H 2.92,3kJ

2HCl(k) CO(NH ) (r) COCl (k) 2NH (k) - H 201,0kJCO(NH ) (r) H O(l) CO (k) 2NH (k) H H H H 2 H H =131,

+

+ =

+ + =+ + = + + 87kJ 5,0

Cu 10. (5 im) Cho 15,2 gam mt hp cht hu c X cha C, H, O vi 300ml dung dch NaOH 1M (va) thu c dung dch Y, c cn Y c 23,6 gam hn hp cht rn khan Z. t chy hon ton Z ri choton b sn phm kh ln lt qua bnh 1 cha H 2SO4 c, bnh 2 cha dung dch Ca(OH)2 thy bnh 1khi lng tng 6,3 gam, cn bnh 2 thu c 55 gam kt ta. un nng phn dung dch cn li bnh 2li thu c 5 gam kt ta na.

a. Xc nh cng thc phn t ca X, bit cng thc phn t trng vi cng thc n gin nht.b. Xc nh cng thc cu to ca X, bit X khng cha nhm COOH, khng c kh nng tham gia

phn ng trng bc.

Li gii im- S mol H2O sinh ra khi cho X tc dng vi NaOH l

2H O

15, 2 0, 3.40 23, 6n 0,2(mol)

18

+ = =

- Khi lng bnh 1 tng l khi lng ca H2O sinh ra do phn ng chy

=> nH (trong X) = 2.0,2 + 26,3

18 - 0,3 = 0,8 (mol)

- Khi t chy Z, sn phm rn l Na2CO3

2 3Na CO NaOH

1 0,3n n 0,15(mol)

2 2= = =

- Phn dung dch bnh 2 un nng sinh ra kt ta => sn phm c Ca(HCO3)2

0,25

0,25

0,25


20/20

o

2 3 3

2 2 3 2

2 2 3 2

t

3 2 3 2 2

CO CaCO (1) CaCO (3)

CO Ca(OH) CaCO H O (1)

2CO Ca(OH) Ca(HCO ) (2)

Ca(HCO ) CaCO CO H O (3)

55 5n n 2n 2 0,65(mol)

100 100

+ +

+

+ +

= + = + =

=> nC (trong X) = 2 3 2Na CO COn n+ = 0,15 + 0,65 = 0,8 (mol)

=> nO (trong X) =15, 2 0,8.12 0,8 0,3(mol)

16 =

=> nC : nH : nO = 0,8 : 0,8 : 0,3 = 8 : 8 : 3 => CTPT l C8H8O3 (V CTPT trng vi CTGN).

b. X X NaOH15,2

n 0,1(mol) n : n 1: 3152

= = = , X khng cha nhm COOH, khng c kh

nng tham gia phn ng trng bc => CTCT ca X c th l mt trong s cc cht sau:

0,25

0,25

0,251,0

1,0

1,5

de thi casio hoa 2012

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