# prep problems icho 2005 solutions

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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
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
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
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
= 1.8�10
-11
= [Mn
][CO
[Mn
= 1.0�10
M and [Mn
left
= 1.0 M with E
= 0.0 V (both are Mn)
cell
= 0.0 - (0.0592 / 2) log (1.0�10
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.
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)
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)

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