Download Chemistry 2008 Multiple Choice

Multiple Choice
1.
A
Noble gases are the least able to lose electrons
(oxidation) in their period.
2.
D
High atomic mass atoms (# > 82) tend to be
radioactive.
3.
C
Halogens pick up electrons and become anions.
6.
B
8.
D
9.
D
10.
C
11.
A
ln(k1/k2) = (Ea/R)(1/T2 – 1/T1): Ea is determined using the
reaction rates ( k) and T.
12.
C
13.
A
14.
C
15.
B
16.
D
Cations act as Lewis acids when combine with ligands
(Lewis base) to form complex ion.
17.
A
Metal loses electrons to nonmetal (redox) producing a
single compound (synthesis).
18.
C
19.
C
Ions of elements # 33-39 are isoelectric with Kr.
21.
B
1  ½  ¼  1/8  3 t½: 120 s/3 = 40 s
2 H2 + O2  2 H2O
0.30 + 0.15  0.30  0.40 – 0.30 = 0.10 mol H2 (0.2 g)
Ca3(PO4)2 + 4 H3PO4  3 Ca(H2PO4)2
35.
Shift to the right  add reactant (I), increase T (II),
pressure has no effect.
36.
2.8 g C(2 mol C/28 g)(6 x 1023 atom/1 mol) = 1.2 x 1023
37.
A
SiO2 is covalent network.
38.
D
mol dilute = mol concentrate
0.5 L(2 mol/L) = (X L)(10 mol/L)  X = .1 L
39.
From first exp. SnSO4 is soluble  ZnF2 is ppt.
40.
41.
42.
D
Excited state produces gap in electron order.
d = m/V = 5 sig fig/4 sig fig  4 sig fig
34.
C
London forces between non-polar molecules.
S: 30/32  1, F: 70/19  3.5  SF4
33.
A
20.
D
C
D
1 L O2(1 mol O2/22.4 L)(2 mol KClO3/3 mol O2)
m = mol/kg = (¼ mol)/0.1 kg = 2.50 m
32.
A
Ions combine to form solid product.
C2H6 fits CnH2n+2  saturated (all single bonds).
31.
D
Exothermic (H < 0), more gas (S > 0).
Mass: 238 + 1 = 239, Charge: 92 + 0 = 92  D.
30.
D
Break bonds (H > 0), more disorder (S > 0).
Compared to exp. 1, the rate for exp. 2 is ½R because
[½X]1, but 4R because [2Y]2  (4/2)R.
29.
A
Lower energy (H < 0), greater disorder (S > 0).
P = nRT/V = (1.85)(0.0821)(308)/(3.00)
28.
B
Entropy (S) is zero when motion stops (0 K).
p = e = 55; n = 134 – 55 = 79
27.
C
Large molecules (HBr) have slow effusion rates
(escape a small opening).
55Cs:
26.
B
HBr is a strong acid = strong electrolyte.
134
25.
C
NH3 has a trigonal pyramid shape.
Most transition metals form 2+ ions (lose s e-).
24.
D
Buffer is formed with equal concentration of acid and
conjugate, which is along flat section.
7.
A
B
D
Adding more base raises pH closer to pure base.
F2 and CO2 are nonpolar, but only CO2 has polar bonds
(bonds between unlike atoms).
23.
A
Equal moles of acid and base occur at equivalence,
which is midpoint along vertical.
5.
D
C
B
4.
C
22.
All are dispersion forces  largest (Br2) has highest
melting point.
N2 + 3 H2  2 NH3
1 N2 + 3 H2  2 NH3 with 3 – 1 = 2 N2 left over
Ne is 5 x heavier than He  equal masses have 5 He/1
Ne  PHe = (5 mol He/6 mol total)Ptot = 5 atm
43.
C
64.
Boiling occurs when Pvapor = Patmosphere.
44.
B
Cl2 takes electron form
 stronger ox. agent
C
D
67.
t4 is at the beginning of freezing  max. liquid.
B
47.
B
"Do what you aughta, add acid to wata"—mixing is
exothermic, need water to absorb heat.
68.
NH2- + H+  NH3
69.
48.
B
According to the solubility rules, column 1 metals,
ammonia and nitrates are soluble  Phosphate.
70.
10 Br-  5 Br2 + 10 e-
71.
50.
D
51.
C
B
A
49.
A
B
D
Gases are most ideal at high T and low P.
Volume changes with temperature (thermal expansion)
 only molarity (mol/L) is affected.
53.
A
Smallest molecule has the weakest London forces.
[OH-] = ½(0.002 M) = 0.001 M
pOH = -log(1 x 10-3) = 3  pH = 14 – 3 = 11
At the same temperature both gases have the same
kinetic energy (K = 3/2RT).
Amino acids: NH2–C(R)H–COOH (I hope you
remember your biology).
CO32- + 2 H+  CO2(g) + H2O
Zn + 2 H+  H2(g) + Zn2+
Ba2+ + SO42-  BaSO4(s)
B
O–O=O  O=O–O
There is an additional pair of electrons around the
center oxygen, which makes the molecule bend (120o).
C
Least polar bond would be between atoms that are
closest to each other in the periodic table.
74.
Molarity is mol/L. The number of moles is given, so
volume is the additional information.
55.
D
3(NO2- + H2O  NO3- + 2 H+ + 2 e-)
Cr2O7 + 14 H+ + 6 e-  2 Cr3+ + 7 H2O
8 H+ + 3 NO2- + Cr2O72-  2 Cr3+ + 3 NO3- + 4 H2O
2-
73.
54.
D
Collision theory: reactants collide frequently, with
sufficient force and correct orientation.
72.
52.
C
To equalize the pressure inside the tube to equal
atmospheric pressure, the water levels must be equal.
66.
mol CH3OH = (16 g/32 g) = 0.5
mol H2O = (90 g/18 g) = 5 mol  0.5/5.5  0.1
46.
D
pH > 7 is basic. Bases are anions other than from
strong acids bonded to non-H+ cation.
65.
I-
45.
A
D
D
Vapor pressure only depends on temperature and
type of liquid, which are the same for both containers.
75.
Pvapor = (mol H2O/mol total)Po
Pvapor = (9/10)25.0 = 22.5 mm Hg
B
K = [Z]3/[X]2[Y] = (4)3/(2)2(.5) = 32
1.
1 mol Ni x 2 mol e- x 96,500 C x 1 s = 193,000 s
1 mol Ni2+ 1 mol e- 1 C
56.
E
Free Response
Kp = PCO2/PCO2
57.
A
2  bonds and no unshared electron pairs  sp.
58.
D
mol C9H8O4 = 0.360 g/180 g = 0.002
volume solution = 0.2 L  M = 0.002/0.2 = 0.01
59.
B
K = [N2O4]/[NO2]2 = 3/62 < 1 (but never neg.)
60.
B
NH3 is a weak base, so the initial precipitate is Ni(OH)2,
but with excess NH3, the complex Ni(NH3)42+ is formed.
61.
D
–C–O–C=O– is an ester group.
||
O
62.
B
25 mL pipet has accurate graduation marks and a
capacity closest to the volume to be measured.
63.
B
a.
F- has one more proton compared to O2-, which attracts
the electrons closer to the nucleus.
b.
PV = nRT
(5.00 atm)(2.00 L) = n(0.0821L•atm/mol•K)(1,160 K)
 n = 0.105 mol CO2
c. (1)
PCO = Ptotal – PCO2 = 8.37 atm – 1.63 atm = 6.74 atm
(2)
Kp = PCO2/PCO2 = (6.74 atm)2/1.63 atm = 27.9
d.
Equal to because the catalyst would increase the
forward and reverse reaction rates, which would
decrease the time to reach equilibrium, but the not the
equilibrium position.
e.
Decrease because Qp = PCO2/PCO2 = (2.00 atm)2/2.00
atm = 2.00 < Kp  the reaction will proceed to the right,
which will increase the partial pressure of CO and
decrease the partial pressure of CO2.
2.
a.
No additional mass was lost during the third heating,
indicating that all the water of hydration has been
driven off.
b. (1)
g H2O = 25.825 g – 22.977 g = 1.848 g H2O
1.848 g H2O x 1 mol/18.02 g H2O = 0.1026 mol H2O
(2)
g MgCl2 = 23.977 g – 22.347 g = 1.630 g MgCl2
1.630 g MgCl2 x 1 mol/95.20 g MgCl2 = 0.01712 mol MgCl2
0.1026 mol H2O/0.01712 mol MgCl2 = 6 MgCl2•6 H2O
c.
The calculated mass of water lost by the hydrate will be
too large because the mass remaining in the container
will be less the mass of the spattered solid.
d.
Add excess AgNO3. Mass the filter paper. Filter the
mixture. Wash and dry the AgCl(s). Mass of AgCl +
filter paper. Subtract mass from mass of filter paper.
e. (1)
5.48 g AgCl x 1 mol AgCl x 1 mol MgCl2 = 0.0191 mol MgCl2
143.32 g AgCl 2 mol AgCl
(2)
5.
F(g)  F+(g) + 1 eb.
In both cases the electron removed is from the same
energy level (2p), but fluorine has a greater effective
nuclear charge due to one more proton in its nucleus
(the electron is held more tightly and thus takes more
energy to remove).
c.
Less because the electron removed from fluorine is
from a 2p orbital, whereas the electron removed from
xenon is from a 5p orbital, which is at a higher energy
level and farther from the nucleus than 2p, hence it
takes less energy to remove the electron from xenon.
d.
F
\ ..
O–Xe–O
F–Xe–F
|
.. \
O
F
e. (1)
0.0191 g MgCl2/2.94 g mixture x 100 = 61.9% MgCl2
3.
a.
Ereaction = ECu2+ + ENO3-  ENO3- = Ereaction – ECu2+
ENO3- = 0.62 V – (-0.34 V) = 0.96 V
b.
Go = -nFEo = (6)(96,600 C•mol-1)(0.96 J/C)
Go = -360,000 J•mol-1 = -360 kJ•mol-1
c.
So > 0 because of the formation of two moles of
NO(g) from non-gaseous reactants.
d. (1)
Second order because the reaction rate for exp. 3 is 9
times the reaction rate for exp. 1 (0.767/0.0852) when
[NO]o is tripled and [O2]o remains constant (9 = 32).
(2)
First order because the reaction rate for exp. 2 is 3
times the reaction rate for exp. 1 (0.256/0.0852) when
[O2]o is tripled and [NO]o remains constant (3 = 31).
e.
rate = k[NO]2[O2]
4.
f.
k = rate/[NO]2[O2]
k = 0.0852 mol L-1 s-1/(0.0200 mol L-1)2(0.0300 mol L-1)
k = 7.10 x 103 mol-2 L2 s-1
a. (1)
OH- + Al(OH)3  Al(OH)4(2)
Decrease because the H+ would react with the OH-,
reducing the [OH-] and shifting the equilibrium to the
left, thus reducing the concentration of Al(OH)4-.
b. (1)
HCl + O2  HClO2
(2)
Neither reactant would be present because they react
in a 1 to 1 ratio, thus three moles of HCl would require
3 moles of O2, which is what is available.
c. (1)
K2O + H2O  2 K+ + 2OH-
(2)
The phenolphthalein would turn pink because of the
presence of OH- in solution making it basic.
a.
reduction
potential
Eo for the reduction of NO3- in acidic solution.
Trigonal
pyramid
(2)
sp3d2
6.
f.
XeO3 is polar because the pair on non-bonding
electrons push the bonding oxygens toward each
other resulting in an asymmetrical structure, where the
bond dipoles do not cancel each other out.
a.
Pyridine can form H-bonds with water around the :N,
but benzene can not because it has no polar region.
As a result, pyridine will dissolve in water whereas
benzene will not.
b.
Liquid ethanol and dimethyl ether form London
dispersion and dipole forces, but liquid ethanol will
also form H-bonds with other ethanol molecules
because of the presence of the OH-. Since H-bonds
are stronger than London dispersion and dipole
forces, the energy (temperature) needed to break the
bond is higher for ethanol than it is for dimethyl ether.
c.
In the solid state SO2 forms London dispersion and
dipole forces between distinct molecules whereas,
SiO2, a covalent network solid, forms covalent bonds
throughout. The much stronger covalent bonds,
which are broken during melting of SiO2, require much
more energy (higher temperature) to break compared
to the relatively weak London dispersion and dipole
forces, which are broken during the melting of SO2.
d.
Liquid Cl2 is held together by London dispersion
forces, which although weak increase in strength as
the number of electrons increases. Liquid HCl is held
together by dipole forces in addition to London
dispersion forces, but the addition of dipole forces
between HCl molecules must not make up for the
fewer electrons around the HCl molecule compared to
Cl2.