CHEMISTRY 102B Name Hour Exam II March 19, 2015 Signature

CHEMISTRY 102B
Hour Exam II
March 19, 2015
Dr. D. DeCoste
Name ______________________________
Signature ___________________________
T.A. _______________________________
This exam contains 22 questions on 9 numbered pages. Check now to make sure you have a
complete exam. You have one hour and thirty minutes to complete the exam. Determine the
best answer to the first 20 questions and enter these on the special answer sheet. Also, circle
your responses in this exam booklet.
Show all of your work and provide complete answers to questions 21 and 22. The answers
must be kept within the spaces provided.
1-20
(40 pts.)
_________
21
(22 pts.)
_________
22
(18 pts.)
_________
(80 pts)
_________
Total
Useful Information:
Always assume ideal behavior for gases (unless explicitly told otherwise).
PV = nRT {R = 0.08206 Latm/molK}
Root mean square velocity =
K = °C + 273
1 nm = 10-9 m
E = −2.178 x 10-18 J (Z2/n2)
c = 2.998 x 108 m/s
h = 6.62608 x 10-34 Js
E=
hc
λ
3RT
M
CHEMISTRY 102B
Hour Exam II
1.
March 19, 2015
Page No. 1
You have neon gas in the left side, and helium gas in the right side, of a two-bulb container
connected by a valve as shown below. Initially the valve is closed.
Ne
3.00 L
1.00 atm
He
1.00 L
3.00 atm
The left bulb has a volume of 3.00 L and the Ne gas is at a pressure of 1.00 atm. The right
bulb has a volume of 1.00 L and the He gas is at a pressure of 3.00 atm.
After the valve is opened, what is true about the relative partial pressures of helium and
neon? Assume constant temperature.
a)
b)
c)
d)
e)
2.
For which of the following constant volume cases is the number of collisions of gas
particles with the walls of the container the greatest? Volume is the same in all cases.
a)
b)
c)
d)
e)
3.
The partial pressure of neon is 3.00 times as great as the partial pressure of helium.
The partial pressure of neon is 1.33 times as great as the partial pressure of helium.
The partial pressure of helium is 3.00 times as great as the partial pressure of neon.
The partial pressure of helium is 1.33 times as great as the partial pressure of neon.
The partial pressures of helium and neon are equal.
1.0 mol of helium gas at 25 °C.
1.0 mol of helium gas at 125 °C.
1.0 mol of oxygen gas at 25 °C.
1.0 mol of oxygen gas at 125 °C.
Choices a and b have equal collision rates, and these are higher than those in c and d.
For metals that form a 2+ charge in an ionic compound, the general equation for such a
metal reacting with hydrochloric acid is
M(s) + 2HCl(aq) → MCl2(aq) + H2(g)
3.72 g of a metal is reacted completely with an excess of HCl and all of the hydrogen gas
produced is collected in a balloon at 25°C and 1.00 atm. The volume of the balloon is
measured and found to be 3.74 L. Identify the metal that was added to the HCl.
a) Ni
b) Ba
c) Ca
d) Zn
e) Mg
CHEMISTRY 102B
Hour Exam II
4.
What mass of helium gas will exert twice the pressure of 150.0 g of nitrogen gas at the
same conditions of volume and temperature?
a) 21.43 g
5.
b) 42.86 g
c) 75.00 g
d) 85.72 g
e) 300.0 g
For a particular process, q = 31 kJ and w = –25 kJ. How many of the following statements
must be true?
I.
II.
III.
IV.
Heat flows from the surroundings to the system.
The surroundings do work on the system.
∆E = 6 kJ
∆H = 31 kJ
a) 0
6.
March 19, 2015
Page No. 2
b) 1
c) 2
d) 3
e) 4
A 1.00-g sample of potato chips is burned in a bomb calorimeter. The heat capacity of the
bomb calorimeter is 11.3 kJ/°C and the temperature rises by 1.98°C. If a serving of potato
chips is considered to be 28.4 g, determine the caloric content of 1 serving of potato chips.
1 food Calorie = 4184 J
a) 5.35 Cal.
7.
b) 84.2 Cal.
c) 134 Cal.
d) 152 Cal.
e) 268 Cal.
The goal of the Chemistry 103 lab this week was to determine the identity of an unknown
salt by determining the ΔH value when it dissolved in water. For this problem, you are to
determine the final temperature of the solution when you dissolve a known salt in water.
Consider the dissolution of ammonium nitrate:
NH4NO3(s) → NH4+(aq) + NO3–(aq)
ΔH = 25.7 kJ
A 5.014-g sample of NH4NO3(s) is dissolved in 50.00 mL of water with both substances
initially at room temperature (24.7°C). Calculate the final temperature of the solution.
You may make the following assumptions: no heat loss to the surroundings; the specific
heat capacity of the solution is 4.18 J/g°C; and the density of the solution is 1.00 g/mL.
a) 17.0°C
8.
b) 17.7°C
c) 24.7° °C
d) 31.7°C
e
32.4°C
Use the following enthalpies of combustion to calculate the enthalpy of formation, ΔHof, of
methanol (CH3OH) from its elements.
2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(l)
C(graphite) + O2(g) → CO2(g)
2H2(g) + O2(g) → 2H2O(l)
a)
b)
c)
d)
e)
–238.7 kJ/mol
+243. kJ/mol
–477.4 kJ/mol
–487.7 kJ/mol
+487.7 kJ/mol
ΔH°rxn = –1452.8 kJ
ΔH°rxn = –393.5 kJ
ΔH°rxn = –571.6 kJ
CHEMISTRY 102B
Hour Exam II
9.
March 19, 2015
Page No. 3
What color of light is emitted when an excited electron in the hydrogen atom falls from the
4th energy level to the 2nd energy level?
Visible light spectrum
red (625 nm to 700 nm)
orange (585 nm to 625 nm)
yellow (565 nm to 585 nm)
green (470 nm to 565 nm)
blue (420 nm to 470 nm)
a) Red
10.
e) Blue
b) 2
c) 3
d) 4
e) 5
A ground state electron in the hydrogen atom absorbs enough energy to get to n=2. Which
orbital will the electron occupy?
a)
b)
c)
d)
e)
The 2s orbital.
The 2px orbital.
The 2py orbital.
The 2pz orbital.
Each of the above is equally likely.
How many electrons can be described by the quantum numbers n = 3, l = 1?
a) 0
13.
d) Green
Bohr’s atomic model works for hydrogen and helium, but no other atoms.
An excited atom can return to its ground state by absorbing electromagnetic radiation.
The 1s orbitals for different elements have different radii.
Each energy level has only one type of orbital.
The ground state phosphorus atom has one unpaired electron.
a) 1
12.
c) Yellow
How many of the following statements is/are true?
I)
II)
III)
IV)
V)
11.
b) Orange
b) 2
c) 6
d) 10
e) 18
Which of the following statements (a-c) is false?
a) An electron configuration for an excited state of the carbon atom could be 1s22s 22 p 3.
b) The ground state electron configuration for the most stable ion of sodium in a
compound is 1s22s22p 6.
c) The ground state electron configuration for the valence electrons of the halogens
(Group 7A) is ns2np5.
d) At least two of the above statements (a-c) are false.
e) All of the above statements (a-c) are true.
CHEMISTRY 102B
Hour Exam II
14.
March 19, 2015
Page No. 4
Which of the following concerning second ionization energy values of K and Ca is true?
a) The second ionization energies are equal for K and Ca since the elements are in the
same row of the periodic table.
b) That of Ca is higher than that of K because as you go across a row on the periodic table
ionization energy increases.
c) That of Ca is lower than that of K because Ca wants to lose the second electron, so it is
easier to take the second electron away.
d) That of Ca is higher than that of K because the Ca atom has one more proton than the K
atom and thus has greater attraction for the electrons.
e) That of Ca is lower than that of K because the second electron taken from K is from a
lower energy level.
15.
Which of the following best evaluates the statement “The 1st ionization energy for an
oxygen atom is lower than the 1st ionization energy for a nitrogen atom”?
a) It is consistent with the general trend relating changes in ionization energy across a
period from left to right because it is easier to take an electron from an oxygen atom
than from a nitrogen atom.
b) It is inconsistent with the general trend relating changes in ionization energy across a
period from left to right and due to the fact that oxygen has one doubly occupied 2p
orbital and nitrogen has all unpaired electrons.
c) It is consistent with the general trend relating changes in ionization energy across a
period from left to right because it is harder to take an electron from an oxygen atom
than from a nitrogen atom.
d) It is inconsistent with the general trend relating changes in ionization energy across a
period from left to right and it is due to the fact that the oxygen atom has two doubly
occupied 2p orbitals and nitrogen has only one.
e) The given statement is incorrect.
16.
Many decades ago, a chemist at UIUC reported the discovery of a new element and named it
Illinium. Unfortunately, he could not substantiate its existence and many years later another
chemist claimed it and it is now named Promethium, Pm. What is the expected ground state
electron configuration for the element formerly known as Illinium?
a)
b)
c)
d)
e)
[Xe] 6s25d104f 4
[Xe] 6s25d12f 4
[Xe] 6s25d14f 4
[Xe] 6s26d15f 4
[Xe] 6s26d16f 4
17. Which of the following correctly ranks the atoms from smallest ionization energy to greatest
ionization energy?
a)
b)
c)
d)
e)
Li, B, C, N, Na
Fr, N, P, O, F
K, Na, S, Cl, F
Be, Mg, Ca, Sr, Cs
At least two of the above (a-d) correctly ranks the atoms from smallest ionization energy to
greatest ionization energy.
CHEMISTRY 102B
Hour Exam II
18.
March 19, 2015
Page No. 5
The following graph plots the first, second, and third ionization energies for Mg, Al, and Si.
A
B
C
Which of the following corrects matches the plot to the element?
a)
b)
c)
d)
e)
19.
B = Al
B = Al
B = Mg
B = Si
B = Si
C = Si
C = Mg
C = Al
C = Al
C = Mg
In order to remove an electron from the potassium atom, energy is _______, and in order to
remove an electron from a chlorine atom, energy is _______.
a)
b)
c)
d)
20.
A = Mg
A = Si
A = Si
A = Mg
A = Al
required, required
required, released
released, required
released, released
Which of the following is true about the general trend of atomic size across a row on the
periodic table?
a)
b)
c)
d)
e)
Atomic size increases to the right because with a greater number of electrons, there is
more electron repulsion.
Atomic size increases because with a greater number of protons, the nucleus is bigger
thus the atom is bigger.
Atomic size decreases to the right because with a greater number of protons, the
electrons are more attracted to the nucleus.
Atomic size decreases to the right because with a greater number of electrons, there is
more attraction to the nucleus.
Atomic size is constant since all atoms in a row have the same valence energy level.
CHEMISTRY 102B
Hour Exam II
21.
March 19, 2015
Page No. 6
Consider a steel, rigid tank at 1.00 atm and 25°C that contains equal masses of hydrogen
gas and nitrogen gas.
a.
Circle one of the following choices: The partial pressure of hydrogen gas is
greater than
less than
equal to
the partial pressure of nitrogen gas in the steel tank. [1 point]
b.
Explain your answer to part a above using the ideal gas law. [3 points]
c.
Explain your answer to part a above using the kinetic molecular theory. [4 points]
d.
Determine the ratio of
Partial pressure of hydrogen gas
. Show all work. [3 points]
Partial pressure of nitrogen gas
CHEMISTRY 102B
Hour Exam II
21.
March 19, 2015
Page No. 7
(con’t)
e.
Determine the density of the mixture of nitrogen and hydrogen gases in g/L. Show all
work. [4 points]
f.
You initiate a reaction between the nitrogen and hydrogen gases to form ammonia
(NH3) gas and let the reaction run to completion at constant temperature.
i.
Calculate the new pressure in the tank after the reaction is complete. Show all work.
[4 points]
ii.
How does density compare to that you found in part e? Determine the ratio of
density of gases after the reaction
. Support your answer/show all work.
density of gases before the reaction
[3 points]
CHEMISTRY 102B
Hour Exam II
22.
March 19, 2015
Page No. 8
Consider the combustion of ethanol as represented by the following balanced equation:
C2H5OH(l) + 3O2(g) → 3H2O(l) + 2CO2(g)
a.
Using the enthalpies of formation given below, determine ∆H° for the combustion of ethanol
in units of kJ. Show all work. [3 points]
∆Hf° [C2H5OH(l)] = –278 kJ/mol
∆Hf° [H2O(l)] = –286 kJ/mol
∆Hf° [CO2(g)] = –394 kJ/mol
b.
Explain your work above. Do not just state what you did, but explain the conceptual reasons
behind why what you did works. In your explanation, explain what ∆Hf° values represent
and why you were not given the ∆Hf° value for O2(g). [7 points]
CHEMISTRY 102B
Hour Exam II
22.
March 19, 2015
Page No. 9
(con’t)
c.
You combust 1.50 g of ethanol in an excess of oxygen and all of the energy produced as heat
is transferred to 100.0 g of ice [H2O(s)] at –10°C. Determine the final state (solid/liquid
mixture, solid, or liquid) and temperature of the H2O. Use the information below and show
all work. [8 points]
∆Hfusion (water) = 6.02 kJ/mol
Specific heat capacity of ice = 2.03 J/g°C
Specific heat capacity of water = 4.18 J/g°C
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