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MCAT(The Medical College Admission Test)美国医学院入学考试超全模拟试题及全部答案。
MCAT Subject Tests
Dear Future Doctor,
The following Subject Test and explanations contains questions not in test format
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practice in your own http://doc.guandang.net/bbca35c11081d34250955e480.htmlhome as
a courtesy and privilege. Practice today so that you can perform on test day; this
material was designed to give you every advantage on the MCAT and we wish you the
best of luck in your preparation.
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Executive Director, Pre-Health Research and Development Kaplan Test Prep
? 2003 Kaplan, Inc.
All rights reserved. No part of this book may be reproduced in any form, by Photostat,
microfilm, xerography or any other means, or incorporated into any information
retrieval system, electronic or mechanical without the written permission of Kaplan,
Inc. This book may not be duplicated, distributed or resold, pursuant to the terms
of your Kaplan Enrollment Agreement.
_________________________________________________________
GENERAL CHEMISTRY SUBJECT TEST 5
General Chemistry Subject Test 5
1.
What
is
the
ahttp://doc.guandang.net/bbca35c11081d34250955e480.htmlpproximate
F-Xe-F bond angle in the compound XeF4? A. 60° B. 90° C. 109.5° D. 120° E. 160°
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2.
For which one of the following mixtures is the ideal gas law most likely to produce
an accurate prediction of volume at STP?
A. B. C. D. E.
3.
HCl and HBr HCl and NH3 HF and HCl HF and CH4 HCl and CH4
6. 5.
At STP, a 29.0 g sample of an unknown gas occupies a volume of 11.2 liters. This gas
could be which of the following? A. CO B. Ar C. B2H6 D. Kr E. C4H10
A mixture of gases is produced at a total pressure of 234 torr. If the mixture contains
1.35 moles of H2(g) and 2.07 moles of CO(g), what is the partial pressure of CO(g)
in the mixture? A. B. C. D.
1.352.07) 234 torr 2.07)234 torr 1.35
2.07
torr 234
2.071.35 ) torr
∞ 234
The equilibrium constant for Reaction 1 is K. What is the equilibrium constant for
Reaction
2?
A.
K
B.
–K
C.
1/K
1http://doc.guandang.net/bbca35c11081d34250955e480.html/K2
2C
7.
D
A
2B
D.
K2
E.
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2.07
E.
1.35
2.07)234 torr
For a certain chemical reaction, ?H = 128 kJ/mol and ?S = 64 J/mol?K. What is the
minimum temperature at which this reaction will be spontaneous? A. 500 K B. 2000 K
C. 2 K D. 2000°C E. 2°C
8.
Iron reacts with element X to form the compounds FeX and Fe3X2. When copper reacts
with element X, the expected compound(s) should be A. CuX and Cu3X2. B. Cu2X and Cu6X.
C. Cu3X only. D. CuX2 only.
E. Cu3X and Cu3X2.
4.
The value of Kc for the reaction below is 2.0.
2 CO(g)
C(s)
CO2(g
What is the concentration of CO at equilibrium if the CO2 concentration is 0.50 M?
A. 2.0 M B. 1.5 M C. 1.41 M
D. 1.0 M E. 0.50 M
KAPLAN__________________________________________________________________________
______________1
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GENERAL
CHEMISTRY
SUBJECT
TEST
5
_______________http://doc.guandang.net/bbca35c11081d34250955e480.html___________
______________________________________
9.
In which of the following is bromine in a
5 oxidation state? A. NaBr B. BrO4- C.
NaBrO D. HBrO3 E. BrO2-
13. For the reaction below, Keq = 96.2 at 450oC. What is the
equilibrium concentration of PCl3 at 450oC when the equilibrium concentrations of
Cl2 and PCl5 are 3.50 M and 2.50 M, respectively?
PCl5(aq) A. B. C. D. E.
14. Which of the following pairs of elements would form
the most ionic compound?
A. B. C. D. E.
15. Silver hydroxide, AgOH, is a slightly soluble ionic
compound whose Ksp = 1.5 x 10-8 at 20oC. What is the minimum pH at which AgOH will
precipitate from a solution, 0.010 M in Ag ?
A. B. C. D. E.
6 - log(1.5) 8
log(1.5) 8 - log(1.5) 14 - log(1.5) 14
Cl Pt and Cl La and Br La and Cl
log(1.5) Au and S Au and
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2
.
5096.
2
×
3
.5096.
2
×
2
.
503
.
502
.
50×
3 .http://doc.guandang.net/bbca35c11081d34250955e480.html50
96. 2
96. 2 2 . 50× 3 .50
96. 2 2 . 50× 3 .50
PCl3(aq)
Cl2(g)
M
M
M
M
M
10. The unbalanced redox reaction shown below occurs in a
solution of aqueous base. Which species is the oxidizing agent?
K2Cr2O7(aq)
KClO(aq) ? Cr(OH)3(s)
KClO3(aq) A. Cr in Cr2O72– B. Cl in ClO– C.
Cr in Cr(OH)3 D. Cl in ClO3– E. K
11. If 25 ml of 0.75 M NaOH(aq) is required to completely
neutralize 150 mL of HCl(aq), what is the molarity of the acidic solution?
A. 0.125 M B. 0.15 M C. 0.25 M D. 0.75 M E. 4.5 M
12. The pH of a solution of a weak, monoprotic acid 4.5. The molarity of the acid
solution is:
A. B.
(Ka = 4 x 10-6)
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4 × 10? 6 10(
? 4.
5 2
)
4 . 5 4 × 10) (
? 6 2
C. ? log D. E.
. 5 ? ? 4
?
? 4 × 10? 6 ?
16. What is thttp://doc.guandang.net/bbca35c11081d34250955e480.htmlhe oxidation
number on the arsenic atom in
arsenous acid, H3AsO3?
A. -3 B. -2 C. 0 D.
(
3 E.
5
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5 10? 4.
)
2
4 × 10? 6
? log 4 .52 log 4 × 10? 6
( )
2
________________________________________________________________________________
________KAPLAN
_________________________________________________________
GENERAL CHEMISTRY SUBJECT TEST 5
17. Which of the following pairs of compounds could be
mixed to produce a buffer solution?
A. CH3NH3I and HI
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B. CH3NH2 and CH3NH3I C. CH3NH3I and NH4I D. HI and NH4I
E. CH3NH2 and NH3
22. How much 2 M H2SO4 is required to neutralize 200 mL
of 2 M NaOH ?
A. 75 mL B. 100 mL C. 150 mL D. 200 mL E. 500 mL
23. 20 mL of an ideal gas is placed in a piston with an initial pressure of 700
tohttp://doc.guandang.net/bbca35c11081d34250955e480.htmlrr. If the pressure is
decreased to 630 torr, and the temperature is held constant, what is the final volume
of the gas?
A. 14.4 mL B. 18.8 mL C. 22.2 mL D. 26.4 mL E. 31.8 mL
24. Which of the following could be characteristic of a reaction with a positive ?H
and a positive ?S?
A. B. C. D. E.
25. Why is it more difficult to compress a liquid to a solid
than it is to compress a gas to a liquid?
A. B. C. D.
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van der Waals forces exist within a liquid. Gas particles are close together. Liquid
molecules are close together.
Liquid molecules possess more kinetic energy than gas molecules.
E. Liquid molecules move at a slower speed than gas
particles.
26. Which of the following processes involves an increase
in entropy?
A. Condensation B. Freezing C. Sublimation D. Cooling
E. Solidification
The reaction is shttp://doc.guandang.net/bbca35c11081d34250955e480.htmlpontaneous.
The reaction is nonspontaneous. The reaction is at equilibrium. The reaction is
exothermic. Two of the above
18. A voltaic cell is constructed by connecting a nickel
electrode in a Ni2 (aq) solution to a tin electrode in a Sn2 (aq) solution. Given
that the standard reduction potentials for Ni2
and Sn2
are -0.28V and -0.14V
respectively, what is the cell EMF?
A.
0.84V B.
0.42V
C.
0.14V D. -0.14V E. -0.42V
19. How many seconds will it take to produce 11.2 liters of
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Cl2(g), measured at STP, by electrolysis of molten NaCl with a 12 amp current? (F
= 96500 C/mol)
A. 96500/12 B. 12/96500 C. (12)(96500) D. (2)(12)/(96500) E 96500/(2)(12)
20. 10.0 mL of 0.40 M CuSO4 solution contains what mass
of CuSO4?
A. 0.40 mg B. 4.0 mg C. 384 mg D. 400 mg E.
638 mg
21. If 625 ml of 0.75 M NaOH(aq) is diluted to 0.45 M
NaOH(aq), the finahttp://doc.guandang.net/bbca35c11081d34250955e480.htmll volume,
in mL, will be:
A. (625)(0.75/0.45) B. (625)(0.45/0.75) C. (1/625)(0.75/0.45) D. (1/625)(0.45/0.75)
E. (625)(0.75 - 0.45)
KAPLAN__________________________________________________________________________
______________3
GENERAL
CHEMISTRY
SUBJECT
TEST
5
________________________________________________________________
27. 23 g of ethanol (C2H5OH) is dissolved in 250 g of
water. What is the approximate freezing point of this solution?
1.86_C_kg/mol)
(kf of water =
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A. –9_C B. –1.9_C C. –2.7_C D. –3.7_C E. –7.6_C
28. What is the boiling-point elevation when 60 g of
glucose (C6H12O6) is dissolved in 667 g of water?
(kb of water = 0.51_C_kg/mol)
A. B. C.
( 0 . 005 ) ( 1. 8
× 10-5) 0 .01( 180 ) ( 667 )
29. Which of the following solutions will have the greatest
boiling point?
A.
0.5
m
NaCl
B.
1.5
m
C6H12O6
C.
http://doc.guandang.net/bbca35c11081d34250955e480.html1.0 m Na3PO4 D. 1.5 m KCl E.
1.0 m K2CO3
30. The concentration of Pb2
is 1.15 ∞ 10–3 M in a
saturated solution of PbF2. Calculate the Ksp of PbF2.
A. 1.3 ∞ 10–6 B. 5.2 ∞ 10–6 C. 1.5 ∞ 10–9 D. 6.0 ∞ 10–9 E. 1.5 ∞ 10–10
? ( 0 . 51)
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( 667 ) ( 60) ( 1000)
D. 3 - ? ( 0 . 51)
( 667 ) ( 60) ( 1000) E. ? ( 0 .51)
( 180 ) ( 667 )
( 6 ) ( 1000) ( 180 ) ( 1000)
? ( 0 . 51)
STOP!
END OF TEST.
4
________________________________________________________________________________
________KAPLAN
_________________________________________________________
GENERAL CHEMISTRY PRACTICE TEST 1
THE ANSWER KEY AND EXPLANATIONS BEGIN ON THE FOLLOWING PAGE.
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KAPLAN__________________________________________________________________________
_____________5
PREPARhttp://doc.guandang.net/bbca35c11081d34250955e480.htmlATION FOR THE DENTAL
ADMISSION TEST _____________________________________________________
GENERAL CHEMISTRY SUBJECT TEST 5
ANSWER KEY
1. B
2. E
7. B
13. B
3. C
4. D
8. E
5. E
16. D
17. B
18. C
19. A
20. E
21. A
14. E
6. E
22. B
9. D
15. B
23. C
10. A
24. E
11. A
12. D
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25. C
26. C
27. D
28. E
29. C
30. D
6
________________________________________________________________________________
________ KAPLAN
_________________________________________________________
GENERAL CHEMISTRY PRACTICE TEST 1
EXPLANATIONS
1. B
This is an example where an inert gas forms a compound with another element. The Lewis
structure is:
Note that Xe has to expand its octet, which it can do because it is
behttp://doc.guandang.net/bbca35c11081d34250955e480.htmlyond the second period.
According to VSEPR theory, the six pairs of electrons (4 bonding, 2 non-bonding) will
arrange themselves in an octagonal structure. The two lone pairs will occupy opposite
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positions to be away from each other as far apart as possible, leaving the fluorine
atoms in
the corners of a square planar molecular geometry:
The F—Xe—F bond angle, then, would be about 90°.
2. E
The ideal gas law assumes (among other things) that there are no intermolecular forces
among the gas particles, which hold the molecules tighter together, leading to a
smaller volume than that predicted by the ideal gas law. Intermolecular forces among
neutral particles are due to hydrogen bonding, dipole-dipole interactions and
dispersion forces (in the order of decreasing strength). For the ideal gas law to
give an accurate prediction of the volume, then, we are looking for gases that do
not have ahttp://doc.guandang.net/bbca35c11081d34250955e480.html strong dipole
moment. Methane, CH4, does not have a dipole moment: What little dipole moments exist
from the relatively nonpolar C–H bonds cancel one another because of the tetrahedral
arrangement. Between choices D and E, HF can participate in hydrogen bonding while
HCl cannot. (Hydrogen bonding occurs when a hydrogen atom is bonded to F, O, or N.)
The intermolecular forces will be weaker overall for choice E than for choice D.
3. C
Note that reaction 2 is the reverse reaction of reaction 1. If a reaction has an
equilibrium constant of K, then the reverse reaction must have an equilibrium constant
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of 1/K.
4. D
The expression for the equilibrium constant for the given reaction is:
[CO]2
Kc = [CO]2
KAPLAN__________________________________________________________________________
_____________7
PREPARATION
FOR
THE
DENTAL
ADMISSION
TEST
_______http://doc.guandang.net/bbca35c11081d34250955e480.html___________________
___________________________
The concentration of CO is squared because of the stoichiometric coefficient of 2
in the balanced equation. Carbon does not appear in the expression because it is a
pure solid. We are told that the value of Kc is 2.0, which enables us to solve for
the equilibrium concentration of CO:
[CO] = Kc[CO2] = 2.0 ∞1.0 = 1.0 M
5. E
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One mole of any gas that behaves ideally will occupy a volume of 22.4 liters at STP.
A volume of 11.2 liters thus implies that there is 1/2 mole of gas. The sample has
a weight of 29.0 g, which means that the gas has a molecular weight of 29.0/(1/2)
= 58.0 g/mol. Choice E, C4H10, has a molecular weight of 4 ∞ 12
1 ∞ 10 = 58 which
is what we are looking for.
Choice A, CO, has a molecular weight of 28 g/mol. Choice B, Ar, has a molecular weight
of
40
g/mol.
Choice
C,
B2H6,
has
a
molecular
weight
of
about
28
g/mol.http://doc.guandang.net/bbca35c11081d34250955e480.html Choice D, Kr, has a
molecular weight of about 84 g/mol.
6. E
Dalton’s law of partial pressures states that the partial pressure of a gas in a
mixture is equal to the total pressure of the mixture times the mole fraction. The
mole fraction of gas A is the number of moles of A divided by the total number of
moles of gases present (including A itself). If a mixture of 1.35 moles of hydrogen
and 2.07 moles of CO exert a total pressure of 234 torr, then the partial pressure
of CO is given by:
2.07
PCO= XCOPtot1.35
2.07
7. B
Since _H and _S are both positive, the reaction will be spontaneous only at high
temperatures. The minimum temperature at which this reaction becomes spontaneous
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occurs when _G, the free energy change, is zero. We therefore need to solve for T
in the equation:
_G = _H – T_S = 0 128000 – 64T = 0 T = 128000/64 = 2000 K
8. E://doc.guandang.net/bbca35c11081d34250955e480.htmlpar
Iron forms the
3 or the
2 ion: Fe3
(ferric) and Fe2
(ferrous). The fact that it
reacts with X to form FeX and Fe3X2 suggests that X forms an anion with a –3 charge.
X3– reacts with the ferric ion to form FeX, since both have a –3 charge, but reacts
with the ferrous ion to form Fe3X2 to balance the charge. Copper usually forms either
the
1 or the
2 ion. When the former reacts with X, then, there must be three Cu
ions to balance the charge, resulting in Cu3X. When the latter reacts with X, it does
so in the same way that the ferrous ion does, forming Cu3X2.
8
________________________________________________________________________________
________ KAPLAN
_________________________________________________________
GENERAL CHEMISTRY PRACTICE TEST 1
9. D
[charge][value], e.g.
By
sticking
3, and ionic charges are written as [value][charge], e.g. 2–.
to
a
system
yhttp://doc.guandang.net/bbca35c11081d34250955e480.htmlour
you'll
chances
decrease
of
careless errors during redox reactions & oxidation state calculations].
causing
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The sum of the oxidation numbers of the atoms in a molecule or ion have to equal the
charge of the molecule or ion.
charge of the polyatomic ion.
on the atom.
The sum is either zero for a neutral molecule or the
For single atoms, the oxidation state equals the charge
Oxygen always gets a –2 oxidation number, unless it is in the form
of a peroxide (e.g. H2O2), in which case it is –1.
Hydrogen always gets a
it is in the form of a hydride (e.g. LiAlH4), in which case it is
1.
1, unless
The oxidation
states of other atoms has to be deduced using the above rules, with a little help
from the periodic table (knowing how many electrons an atom wants to gain or lose
based on its electron configuration helps determine possible oxidation states).
choice A, Na is
1, so Br must be –1.
In choice B, each O is –2, so four give a
totahttp://doc.guandang.net/bbca35c11081d34250955e480.htmll of –8.
the polyatomic ion is 1–, so Br must be 7 (x
C, there is a Na is at
is a H is at
(–8) = –1; x =
1, an O at –2, and so Br must be
1.
The charge of
7).
In choice
In choice D, there
1, three O's (–2 each) giving a total of –6, and so Br must be
the correct answer.
In
5,
In choice E we have two O's giving a total of –4, the charge
of the ion is 1–, so Br must be
3.
10. A
A reducing agent is the species that gets oxidized in a redox reaction, while an
oxidizing agent is one that gets reduced. Using the mnemonic OIL RIG, we remember
that reduction is gain of electrons, and a decrease in oxidation number. In choice
A, the chromomium goes from
agent (it gets reduced).
6 in Cr2O72-
to
3 in Cr(OH)3, so it is an oxidizing
In choice B, the chlorine goes from
ClO3–, so is a reducing agent (it gets oxidized).
a
d
subshell,
allowing
it
1 in ClO– to
[Cl is in period 3, so it has
to
have
oxidhttp://doc.guandang.net/bbca35c11081d34250955e480.htmlation
multiple
states.]
choice C, Cr(OH)3 is a product of the reaction, so cannot be the answer.
E, K
5 in
In
In choice
does not change during the reaction, as it is just a spectator ion.
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11. A
To solve neutralization questions, we use the formula NAVA = NBVB;
eq
NALxA, then
N = M _ x .
For example, a 3 M H3PO4 is a 9 N H3PO4.
VA is the volume of acid used;
NB is the normality of the base.
then
y.
Bases that bind H
For any given base with the general formula B(OH)y,
are treated similarly, e.g. 2 M Na2CO3 is 4 N.
VB is the volume of base used.
Essentially, the formula states that the molar amount of acid equals the molar amount
of base, which it must if the solution has been neutralized.
Here, both the base and the acid are monoequivalent, so their normalities equal their
normalities.
Solving for the
NV(.75)(25)751
normality
ofhttp://doc.guandang.net/bbca35c11081d34250955e480.html
NAV1506008A
equation, no conversion to liters is necessary.
acid,
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12. D
The chemical equation for the dissociation of a monoprotic acid is HA
The stoichiometry tells us that the concentration of H
of A–.
By solving for the concentration of H
concentrations of both.
be
?
H
A–.
must equal the concentration
using the pH, we will have found the
Since pH = –log [H ], then [H ] = 10-pH (since 10a = b can
expressed
as
log
b=
KAPLAN__________________________________________________________________________
_____________9
N = M _
PREPARATION
FOR
THE
DENTAL
ADMISSION
TEST
_____________________________________________________
[H ][A–]
a[HA]a).
Here the [A–] = [H ] = 10–4.5.
The acid dissociation constant for the
reaction is K[H ][A–](10–4.5)2
concentration of acid, [HA] =
[K]4 x 10a
13. B
This
is
a
straightforward
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queshttp://doc.guandang.net/bbca35c11081d34250955e480.htmltion that asks us to
manipulate the formula for Keq and solve for an unknown
[PCl][Cl]K[Cl]
concentration.
For this reaction the Keq[PCl and solving we get [PCl33[PCl5]5]96.2
x 2.5
14. E
The two elements that would form the most ionic compound will have the greatest
difference in electronegativities.
Electronegativity increases from left to right
and from bottom to top on the periodic table of the elements.
The elements with the
greatest difference in electronegativities will be the furthest apart from one
another on the periodic table.
Lanthanum and chlorine are the furthest apart on the
periodic table.
15. B
To answer this question, we must first solve for the concentration of OH–, then use
that to figure out the pH.
= [Ag ][OH–].
The solubility product constant for this reaction is Ksp
Solving for the concentration of OH– we get [OH–] = K1.5 x 10–8
–6http://doc.guandang.net/bbca35c11081d34250955e480.html–6
[Ag ]1 x 10–2 1.5 x 10. To calculate the pOH we use the formula pOH = –log[OH] =
–log(1.5 x 10) = –(log1.5
log10–6) = –(log1.5 –6) = 6 – log1.5.
use the pOH so determine the pH.
Since pH
Now we can
pOH = 14, then pH = 14 – pOH = 14 –
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(6 – log 1.5) = 8
log 1.5.
It is essential to understand the mathematical rules
involving logarithms, as it goes hand in hand with acid/base chemistry.
16. D
Using the oxidation rules outlined in question 9, we have to determine the oxidation
state of arsenic in the
polyatomic ion AsO33–.
The oxygens give a total oxidation state of –6 (3 x (–2)),
and the oxidation number of arsenic plus that of the oxygens have to sum to the charge
of the ion.
Therefore arsenic must be
3 (x
(–6) = 3–; x = –3).
Alternatively, you could solve for the entire molecule, H3AsO3, with the hydrogens
giving
total
oxidation
state
of
3
(3
x
(
1)),
the
oxygenhttp://doc.guandang.net/bbca35c11081d34250955e480.htmls –6, and arsenic
(x
3
(–6) = 0; x =
3
3).
17. B
Buffers are solutions that resist change in pH.
acid and its conjugate base.
A buffer is a solution of a weak
Weak acids and bases must be used, since a strong acid
or base would completely dissociate and not have any buffering
capacity.
If the acid and base in the buffer were not conjugates, they would react
with one another, producing salt and water with no buffering capacity.
a weak acid and a strong acid, so cannot make a buffer.
Choice A is
Choice B is the correct answer,
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the buffer consists of the weak base methyl amine and the conjugate base
methylammoniumiodide.
Choices C, D and E are not conjugates.
18. C
A voltaic or galvanic cell is a spontaneous cell that has a positive EMF (electromotive
force) and a
negative _G.
When given half reactions you have to determine which
species will be oxidized ahttp://doc.guandang.net/bbca35c11081d34250955e480.htmlnd
which we be reduced such that a positive electrical cell potential is obtained.
The
half reactions with the corresponding reduction potentials are as follows:
10
________________________________________________________________________________
_______ KAPLAN
_________________________________________________________
GENERAL CHEMISTRY PRACTICE TEST 1
Ni2 (aq)
2e–
?
Ni(s)
?
Sn(s) E_ = –0.14 V
E_ = –0.28 V
Sn2 (aq)
2e–
Since the nickel has the more negative reduction potential, it will get oxidized.
The reaction for nickel above will therefore proceed to the left.
the cell is:
The reaction for
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Sn2 (aq)
Ni(s)
?
Sn(s)
Ni2 (aq)
The EMF of the cell is the reduction potential of the species reduced minus the
reduction potential of the species oxidized, as expressed in the formula EMF =
Ecathode
–
Eanode.
The
purpose
of
the
subtraction
is
to
rhttp://doc.guandang.net/bbca35c11081d34250955e480.htmleverse the sign of the
reduction potential since that species is getting oxidized.
Be sure not to make the
mistake of reversing the sign of reaction occurring at the anode, and then insert
it into the preceding equation--you will have reversed the sign twice and gotten the
answer wrong!
For this reaction, EMF = (–0.14) – (–0.28) =
0.14 V.
19. A
This question requires dimensional analysis to solve, as it would take too many
equations and way too much time so solve directly.
In dimensional analysis, a series
of fractions are multiplied together, canceling the units you don't need and
eventually giving the units you are looking for.
You start with a fraction that has
the units you are solving for in the numerator, and proceed to cancel out the
denominator and subsequent units until you are left with the answer.
This question
is asking for seconds, so we must start with a fraction with seconds in the
nhttp://doc.guandang.net/bbca35c11081d34250955e480.htmlumerator.
proceeds as follows:
(1 sec12 C)
current
The analysis
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x
96500 Cmol efaraday's constant
x 1 mol Cl2
2 mol e 2Cl- ? 2 Cl2
2e-
x
(1 mol Cl22.4 Lx (11.2 L) = 1296500
given volume
gas law
The trick is to recognize that amps are C/sec, and that coulombs can be canceled using
Faraday's constant.
20. E
This is another question that can be solved using dimensional analysis: 160 g
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(1 mol CuSO)
4
x 1 L
0.40 mol
x (.01L) = 0.64 g = 640 mg
21. A
This is an example of a dilution question, where the volume has been changed but not
the molar amount of the substance.
in neutralization questions:
Mi
is
The formula used is very similar to that used
MiVi = MfVf:
the
initial
molarity
of
solutionhttp://doc.guandang.net/bbca35c11081d34250955e480.html.
initial volume of the solution.
Vi
the
is
Mf is the final molarity of the solution.
the
Vf is
the volume volume of the solution.
Essentially, since M x V = moles,
the formula states ni = nf.
This formula can only
be used when the molar amount
MVhas not changed.
This question is looking for the final volume, so solving and
plugging in we get VfM2
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(0.75)(625)
0.45
KAPLAN__________________________________________________________________________
____________11
PREPARATION
FOR
THE
DENTAL
ADMISSION
TEST
_____________________________________________________
22. B
This is a neutralization question that is looking for the volume of acid.
Using the
techniques outlined in question 11,
NV(2)(200)
we rearrange the equation NAVA = NBVB for acid volume and solve;
VAA
23. C
To
solve
this
question
we
have
to
use
Boyle's
http://doc.guandang.net/bbca35c11081d34250955e480.htmlP1V1 = P2V2.
the equation and solving for the final
PV(700)(20)1400
Law,
Rearranging
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volume we get V2 = P630? 602
24. E
The change in enthalpy (?H) and change in entropy (?S) of a reaction are related to
the change in free energy (?G) of a reaction by the formula ?G = ?H – T?S.
Since ?H
is positive and ?S is positive, the reaction will be spontaneous when temperatures
are very high, and will be nonspontaneous when temperatures are very low.
of this is ice melting:
An example
?H is positive because breaking bonds requires energy, and
disorder is increasing (solid _ liquid), so ice will only melt when T > 0_C.
choice C is incorrect because when a reaction is at equilibrium, ?S = 0.
D is incorrect because a
Answer
Answer choice
?H means heat is being absorbed, and is defined as
endothermic.
25. C
As
the
phases
of
matter
change
from
gas
_
liquid
mhttp://doc.guandang.net/bbca35c11081d34250955e480.htmlolecules
closer together.
_
solid,
are
the
becoming
Since
molecules in the liquid phase are already close together, it is more difficult to
compress them further into the solid phase than it is to compress gas phase molecules
(which are far apart). Answer choice A is incorrect because van der Waals forces do
not effect how easily a substance can be compressed.
Choice B is incorrect because
gas molecules are far apart from one another, and are the least dense of all the phases
of matter.
Liquid molecules possess less kinetic energy than gas molecules, since
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liquids exist at lower temperatures than gases, making choice D wrong, and although
choice E is a correct statement, it does not answer the question, as it does not address
a correlaton between solids and liquids.
26. C
An increase in entropy means an increase in disorder.
become
gases,
with
no
liquid
Sublimation is when solids
phase
in
between.
Sincehttp://doc.guandang.net/bbca35c11081d34250955e480.html gas molecules are more
disorder than solids, entropy increases. Condensation is when gases become liquids,
a decrease in entropy.
Freezing and solidifaction are both processes by which
liquids become solids, also a decrease in entropy.
Cooling is also a decrease in
entropy, as the molecules are moving slower and less randomly.
27. D
The freezing point of a liquid decreases when a substance is dissolved in it according
to the formula ?Tf = kfm ; where kf is the freezing point depression constant (a
constant specific for a given solvent), and m is the molality (mol solute/kg solvent)
of the solution.
Remember that the formula reports the change–if a question asks
for the final
temperature, you must subtact ?T from the initial freezing point of the liquid (Tf
= Ti – ?Tf).
To
We are given 23g ethanol, 250 g H2O (= 1/4 kg), and a kf of 1.86_C_kg/mol.
calculate
the
molality
we
have
to
first
dethttp://doc.guandang.net/bbca35c11081d34250955e480.htmlermine the molar amount
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m23 gmol1/1/2 mol.
Therefore m?Tf = kfm; = (1.86)(2) ?
4
(2)(2) = 4_C.
The freezing point of water is 0_C, so the new freezing point is –4_C.
[Notice: calculations are usually made easier when numbers are used in fractional
format instead of decimals, and values are rounded to the nearest whole number when
possible.]
12
________________________________________________________________________________
_______ KAPLAN
_________________________________________________________
GENERAL CHEMISTRY PRACTICE TEST 1
28. E
Similar to freezing point depression is boiling point elevation.
The boiling point
of a solution increases according to the formula ?Tb= kbm ; where kb is the boiling
point elevation constant.
asks
for
the
final
Remember, the formula reports the change, so if a question
boiling
point,
the
?Tb
has
to
be
added
to
the
ihttp://doc.guandang.net/bbca35c11081d34250955e480.htmlnitial boiling point (Tf =
Ti
?Tb).
Before doing any calculation, you should always look at the answer
choices to help determine how the question is to be solved.
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mol
Here, we are not required to do any math, we need only set up the formula:
kbmkgm60 gMW180 /(0.51_C_kg/mol)kg = (0.51)1kg.
?Tb=
After dropping the units and
rearranging the complex fraction we get:
1000g
(60)(1000)(0.51)_(667)(180)
29. C
Boiling point elevation and freezing point depression are colligative properties.
Colligative properties depend on the amount of the substance, and does not depend
on the properties of the substance itself.
If a molecule dissociates when it
dissolves, the molality has to be multiplied by the number of particles formed upon
dissociation, that way the total number of dissolved particles are taken into
consideration when calculating the new freezing point or boiling point.
We
ahttp://doc.guandang.net/bbca35c11081d34250955e480.htmlre
of
given
molalities and asked which will have the greatest boiling point.
a
list
The solution with
the greatest boiling point will have the greatest number of dissolved particles in
solution.
Let's go through the individual answer choices and determine the
equivalent molality:
A. B. C. D. E.
.5 mol NaCl2 mol ions
_ mol NaClm ions
[1 mol Na
and 1 mol Cl– are formed] kg1.5 mol CHO1 mol
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_
molm
[glucose does not dissociate] kg
1.0 mol NaPO4 mol ions
_m ions [3 mol Na
_ mol KClm ions
and 1 mol PO43- are formed] 341.5 mol KCl2 mol ion
[1 mol K
and 1 mol Cl- are formed] kg
1.0 mol KCO_ (3 ions,mol K2CO3) = 3.0 m ions
[2 mol K
and 1 mol CO32- are formed]
kg
Since Na3PO4 has the greatest amount of particles in solution, it will increase the
boiling point the most.
30. D
The formula for the dissolution of PbF2 is:
PbF
Thttp://doc.guandang.net/bbca35c11081d34250955e480.htmlhe stoichiometry tells us
that for every mole of PbF2 that dissolves, 1 mol of Pb2
is formed and 2 moles of
F- are
formed.
If we let x = [Pb2 ], then the [F-] = 2x.
Plugging this into the formula
we get Ksp = [Pb2 ][F-]2 = (x)(2x)2 = (x)(4x2) = 4x3.
1 x 10–3
D.
Pb2
2F-
The [Pb2 ] = 1.15 x 10–3 ?
and so Ksp = 4x3 = 4(1 x 10–3)3 = 4 x 10–9, which is close to choice
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Ksp = [Pb2 ][F-]2
KAPLAN__________________________________________________________________________
____________13
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