*Чтобы посмотреть этот PDF файл с форматированием и разметкой, скачайте его и откройте на своем компьютере.*

dentification of Ions

1-1 For example

1-2 For example

1-3-1 BY

= B

+ 2 Y

Ksp = (S

)(2S

2S

4S

, S

(solubility of BY

) = 2.0

1-3-2 CY

= C

+ 2 Y

Ksp = (S

)(2S

S

2S

4S

, S

(solubility of CY

1-4-1 Plot of Absorbance (A) vs Volume (V

) of L added as follows:

From the volume of L at Break point B (all B

ions form complex with L) in the plot, n can

be calculated:

n/1 = (# of moles of L) / (# of moles of B

= (5.1 mL

) / (2.0 mL

It means that B

forms BL

complex with L.

1-4-2 (1) Calculation of Molar Absorption Coefficient

At Break point, A = 0.66 =

(concentration of BL

And

= 0.66 / (2.0 mL

/ 50 mL) = 2.01

(2) Choose a point in the curve of the plot, for example:

At Point P (2.0 mL of L added): A = 0.26

A = 0.26 =

[BL

[BL

] = 0.26 /

= 0.26 / (2.01

[B

] = (2.0 mL

M) / 50 mL

[B

[ L ] = (2.0 mL

[ L ] = 1.3

[Calculation of formation constant]

So K

= [BL

] / ([B

][ L ]

= 8.74

1-5-1 For CY

: Ksp = [C

-]2 = 2.56

[Y

) / 0.05)

M When CY

begins to form

For BY

: Ksp = [B

-]2 = 3.20

[Y

] = ((3.20

M When BY

CY

forms first

1-5-2 The precipitation of C

ions of Radioisotopes

2-1

2-2-1

A =

ε λ

N, A

ε λ

and N = N

A / A

= N / N

and A = A

2-2-2

N = N

If N = 1/2 N

, t =

then 1/2 N

1/2

= 0.693 / t

For C-14,

Also

A = A

10.2 = 16.5 e

-1.2 x 10

and t = 4008 years

2-3-1

A = Rp Rd = N

) = N

)

= 0.693 / (14.3 x 24 x 60 x 60) = 5.61 x 10

(-3) / 98] x 6 x 10

A = N

) = (6.12 x 10

)(1.00 x 10

)(0.9 x 10

)(1-e

-5.61 x 10

x60x60

and A = 1.11 x 10

cps = 1.11 x 10

/ (3.7 x 10

2-3-2 Total amount of P-32 is constant

after and before the injection,

so, V

= V

(V = volume, A = Activity, x for pool water)

2.0 x 1.0 = V

(12.4 / (3.7 x 10

and V

(pool water) = 5.97 x 10

Answer 3: Ion Exchangers

3-1-1 2 RNa + Ca

or 2 RNa + CaCl

Ca + 2NaCl

3-1-2(a) The tap water contains tr

ace HCl after the

adsorption of Ca

by the ion exchanger R

2 RH + Ca

Ca + 2H

or 2 RH + CaCl

Ca + 2HCl

3-1-2(b) RNa is suitable for drinking purpose

. Because the product of

the adsorption of Ca

by

RNa is Na

or NaCl while the product is H

3-2-1 The removal of H

can be achieved by using

the anionic ion exchanger R

with the

equation:

3-2-2 Firstly, the anionic ion exchanger R

is used to adsorb the SO

ion with the

Secondly, a standard solution of HCl c

an be used for the titration of the OH

in the

solution after adsorption of SO

by the anionic ion exchanger R

O (Acid-Base Titration)

3-3 RH + M

, Kc = [RM][H

] / ([M

][RH]) (3-3-1)

= [RM] / [M

] (3-3-2)

)× 10-3 (3-3-3)

We substitute Equations (3-3-1) and (3-3-2)

into Equation (3-3-3) and obtain:

+] + [RM][H

+]) × 10-3 = (Kd[M+] + Kd[M

+] / Kc[M+]) × 10-3 ð = (Kd[M

] + Kd[H

] / Kc )

S Kc 10

= Kd Kc [M

] + Kd[H

1 / Kd = [M

+] / (S Kc (10

)) (3-3-4)

3-4-1 N

= 16 (10 / 1.0)

= 16 (14 / 1.5)

3-4-2 H = L / N = 30 / 1497 = 0.021 cm

3-4-3 R = 2 (t

) / (

) = 2 (14 10) / (1.0 +1.5) = 3.2

3-4-4

) / (t

) = (14 1) / (10 - 1) = 1.44

3-5-1 Z-Na

3-5-2 Z-Na

10 CO

+ 8 H

[Ca

] = 2.5 x 10

M x 0.02741 L x

/ 0.025 L

= 6.85 x 10

4-3 Mass-balance: [Ca

] = [C

] + [HC

= [C

] (1 + [H

] / K

+ [H

[C

] = [Ca

+] / K

+ [H

/ K

Substituting (1) into [Ca

2O42-] = K

[Ca

4-4 C

2+] + [CaC

(2O4)22-] = K

sp

1242−OC + K

+ K

f1

242−OCddCCa = 0 = - K

1−OC + Ksp

[C

[Ca

2O42-] = 1.3 x 10-6 M 4-5 Charge balance: 2[Ca

] + [H

2O42-] + [HC

-] (1) Mass balance: [Ca

2O42-] + [HC

2C2O4] (2) Because K

is too small, [H

] can be neglected.

Comparing (1), (2), [HC

+] [H+] (3) [C

+]2 K

(4)

[Ca

2O42-] = K

+]2 / (K

+]2) (5) Substituting (3), (4), (5) into (2)

+]5 + (K22 - Ksp) [H+]4 2 K

+]3 2 K22 Kw [H+]2 + K2 Kw2 [H+] + K22 Kw2 = 0 Solving [H

+] = 5.5x10-8 M (or pH = 7.26) Substituting [H

] into (5), [Ca

in Wastewater

5-1-1 NH

5-1-2 NH

5-1-3 H

5-1-4 H

Answer 7: Atomic Orbitals

7-1

1s: 0, 2s: 1 and 3s: 2.

7-2

, 3p

There is one angular node for 2p

; one angular node and one spherical node for 3p

7-3 (0, 2, 4, 1, 3)

Answer 8: Intermolecular Forces

8-1-1

8-1-2

8-2-1

8-2-2

CCH

8-3

Thymine

Adenine

Cytosine

Guanine

Answer 9: Crystal Packing

9-1 Simple cubic: 6, body-centered cubic: 8 and face-centered cubic: 12

9-2

For simple cubic, a = 2r,

For body-centered cubic,

a

4

3

For face-centered cubic,

a

4

2

9-3

a

4

2

r

a

407

2

2

9-4

+2 H

10-1-2 CrO

: + 6, Cr

: +6.

10-1-3 This is not a redox reaction because the oxidation state in each metal center does not

change.

10-1-4 Hydrogen ion concentration is the main

factor to control the equilibrium position.

10-1-5

10-2-1

+12e

+2 Cr (

Cathode

Anode

Overall

10-2-2 1.5 moles of oxygen gas will evolve.

+12e

+2 Cr (

Cathode

Overall

1 mol Cr

3 mol O2

2 mol Cr

1.5 mol O

14 H

10-2-3 16 h

52 g Cr

ol C

mol Cr

1 sec

1 min

60 sec

60 min

10-2-4 Chromium readily forms a thin, adherent, transparent coating of Cr

in air, making the

metal extremely useful as an attractive protective coating on easily corroded metals.

Answer 11: Electrochemistry of Inorganic Compounds

11-1 For the concentration cell: Mn

(s)

| Mn

(1M) || Mn

/ MnCO

| Mn

(s)

cell

= E

(0.0592 / 2) log ([Mn

right

/ [Mn

left

= 1.810

-11

= [Mn

][CO

[Mn

= 1.010

M and [Mn

left

= 1.0 M with E

= 0.0 V (both are Mn)

cell

= 0.0 - (0.0592 / 2) log (1.010

M / 1.0 M) = 0.237 V

11-2

Reduction of O

to H

O is obtained as (0.70V+1.76V) / 2 = 1.23 V,

for O

O E

= 1.23V

The E

value could be obtained directly from the

diagram by dividing the differences (2.46)

of O

and H

O by the differences of the oxidation number (2).

For H

O E

06

� 0.0

The disproportionation reaction is spontaneous.

11-3 The number of electron pair should be 5 (tri

gonal bipyramidal) with three electron pairs in

the equatorial plane, thus the molecular geometry of XeF

is linear.

+

E

= -1.23V

+ 2H

Xe

+ 2HF

E

= 2.32V

2 XeF

+ 2H

2 Xe

+ O

+ 4HF

(aq)

E

= 1.09 V

The decomposition of XeF

in aqueous solution is favored in acidic solution.

2H2

00

frequency because of stronger back donation of the anionic charge to the anti bonding

orbital of CO thus weakening the CO bond. For the neutral species

, absorption bands

appear at the higher frequency.

12-2

W(CO)

NaC

HCCCH

FeSO

Na/Hg

metal

migration

12-3

metal migration

95

Answer 13: Carbocation and Aromaticity

13-1 (CH

13-2 Spectrum I: (CH

13-3 6

13-4 yes

13-7

Answer 14: Photochemical Ring Closure and Opening

14-1 (2

)-octatriene

14-2

14-3

96

14-4

14-5 No.

Answer 15: Stereochemistry

15-1 (

15-2

= PPh

15-3 36%

β

HO

HOH

HOH

15-7 none

97

15-8 99:1

15-9 0

Answer 16: Organic Synthesis

F

C

98

Answer 17: Spectroscopy and Polymer Chemistry

17-1 C

17-2 C=O group

17-3

OCH

17-4

In: initiato

17-5 Organic reactions that could tr

ansform acetate to alcohol such

as acid or base hydrolysis,

alcoholysis, or LiAlH

reduction.

17-6 There are 100 units/molecule. However,

the last one does not c

ontain chiral center,

therefore, there are 99 chiral

centers and each of which would have

or

configuration.

Totally there will be 2

mers and diastereomers.

Therefore, the number of pairs of enantiomers is 2

17-7

17-8

C=CH

17-9

99

O

O

O

O

O

O

E F

O

NH

18-4 (b) A high dilution condition is employed

in order to inhibit polymer formation.

18-5 Curve

; Curve

; Curve

101

Answer 19: Works in Thermodynamics

19-1 Isothermal reversible expansion

We have 100/22.41=4.461 moles, and the final volume is

100

1

10

10

V

1

1

2

(1)

The work done by gas is

(2)

19-2 Adiabatic reversible expansion

(3)

Thus

(4)

and the final temperature is obtained from

(5)

For adiabatic processes,

and

(6)

19-3 Irreversible adiabatic expansion

Since q=0, we have

(7)

102

(8)

10

2

.

273

1

(

)

2

.

273

(

2

3

2

(9)

It follows that

(10)

(11)

where

devotes the initial pressure of NO

20-2 At

2

1

NO

P

P

k

2

/

1

0

l / atm

min

103

.. (1)

.. (2)

(1) divided by (2) gives

,

where

is molar enthalpy of water and

is volume change. The phase diagram

shows that the slope of dP/dT for the liquid-so

lid coexistence region

is negative, indicating

the volume expands when water freezes.

22-4 As pressure is lowered, liquid phase tr

ansforms directly to gas phase at the same

temperature. Thus water may vaporize.

At the same time, the process of water

evaporation is endothermic as to make the

surrounding cooled. The left water becomes

frozen. The solid state will sublime unt

il none is left, if the pump is left on.

22-5 The ice surface, exerted by a pressure mo

re than one atm, turns to liquid state at 0

C.

[

]

[

k

dt

B

d

[

]

[

k

dt

C

d

1

[

]

[

k

k

C

d

B

d

1

.

0

1

1

k

k

C

B

1

[

]

[

k

k

A

B

2

[

]

[

k

k

A

C

1

200

100

0005

.

0

1

.

0

01

.

0

1

]

[

]

[

2

1

1

k

k

k

C

B

104

Answer 23: Standard Deviation in

23-1 average spe&#xv000;ed v:

RT

= 4.45 x 10

standard deviation

=

=2.33 x 10

2

v

v

23-2 average po&#xx000;sition x:

=

= 0

standard deviation

= 1

&#xp000;23-3 p =

dx

x

h

i

2

(

ihxe

4

/

dx

4

(

2

2

2

*

2

23-4

h

105

Answer 24: A Particle in 2-D Box Quantum Mechanics

24-1

= 2

=

2,1

= 8

=

3,1

= 10

=

3,2

= 13

=

4,1

= 17

= 18

=

4,2

= 20

=

4,3

= 25

=

5,1

= 26

where

24-2 The total number of electrons in

the highest occupied energy level is 4.

106

Answer 27: Enzyme Catalysis

27-1 A =

bC;

b;

C x V (volume)

(()3)] x 5 x 10-3 = 2.7 x 10-8 mol/sec 27-2 Four electrons are needed to reduce one mo

lecule of oxygen, therefore, the oxygen

consuming rate is 2.7 x 10

/ 4 = 6.75 x 10

27-3 By definition, the tur

nover number equals 6.75 x 10

(mol/sec) / (2.7 x 10

M x 5 x 10

Therefore, oxidase has a

turnover number of 500.

Answer 30: Identification of Unknown Solutions

€II 1. Use the indicator to find out NaOH, HCl, and H

(confirmed by Pb

S by the odor, and use it to find Cd

(by precipitation. and color).

3. By electrolysis of the four solutions remained

, KI solution can be found by the trace of yellowish

brown (I

) formed in the anode.

4. The color of I

will be disappeared by Na

solution.

5. The concentration of unknown solution is about 0.5 M (mol/L)