Key for Study Guide Exam 2
Spring 2016
Show logic and calculations for all problems. Remember to include units and be careful with sig. fig.
Remember you will need to show your work for full credit. On the real exam always work the problems you know best
first. If you get hung up on a problem, you should move on and come back to it at the end. If you have time, check
over your work. To test your speed, work this study guide as if it was the exam.
Remember, this study guide is not representative of all possible questions!
1. What would be the physical state of a compound at room temp that boiled at 55.9 oC and melted at 10.8 oC.?
Room temperature is roughly 22 oC. This temperature is above the melting point and below the boiling point of the
compound, so the compound would be a liquid.
2. Regarding the Boltzmann molecular simulator program used in lecture, would increasing the number of red spheres in the box
increase or decrease the pressure? Explain your answer using the Kinetic Molecular Theory.
Increasing the number of red spheres in the box will increase the pressure.
Pressure is proportional to the number of collisions that the gas phase atoms or molecules make per unit time
with the walls of the container. Collisions with greater force will also increase pressure.
Increasing the number of atoms in a container will increase the number of collisions per unit time. This will
result in higher pressure. (Note: the collisions will not have greater force.)
3. What pressure is exerted by 0.533 moles of Ar(g) at 24.7 C in a 2.500 L container?
nRT
o
PV = nRT
P=
L-atm
mol-K
(0.533 moles)(0.082058
P =
24.7 C + 273.15 = 297.85 K
V
)(297.85 K)
= 5.2108 atm = 5.21 atm
2.500 L
4. A syringe if filled with 48.0 mL of air at a pressure of 748 mm Hg. What would the pressure in the syringe be when the volume is
decreased to 9.2 mL?
P1V1 = P2V2
P2 =
P1 V1
V2
V1 = 48.0 mL
=
P1 = 748 mm Hg
(748 mm Hg )( 48.0 mL)
9.2 mL
V2 = 9.2 mL
= 3902.6 mm Hg = 3900 mm Hg or 3.9 x 103 mm Hg
5. A weather balloon contains 8.50 moles of gas at 35.0 oC and has a volume of 2.44 L. What is the pressure of the gas in the
balloon?
nRT
PV = nRT
P=
(8.50 moles)(0.082058
P =
Convert temperature into Kelvin. T1 = 35.0 + 273.15 = 308.15
V
L-atm
mol-K
)(308.15 K)
2.44 L
= 88.08707703 L = 88.1 L
H
6. Draw a picture of three molecules of water
connected by hydrogen bonds.
H
O
H
I have drawn five molecules.
O
H
O
H
H
H
O
H
O
H
H
7.(i) For each of the following compounds state all the IMF that would exist between two molecules of that compound. (ii) Which
would have the overall strongest, intermediate, and weakest intermolecular forces (IMF). Explain your answer by drawing such
interactions between those molecules where appropriate.
H
H
H
H
H
C
C
H
C
H
O
H
C
H
C
C
H
C
H
C
C
H
H
H
O
b.
a.
i)
H
H
H
H
H
H
C
H
H
H
H
c.
a – (propanol) –is polar and can form hydrogen bonds and has dipole-dipole forces and London forces
b – (propanal) – is polar and has dipole-dipole forces and London Forces
c – (butane)– is non-polar and thus has only London dispersion forces,
(ii) All three molecules have similar mass and thus similar London forces. London Forces is the only IMF for
molecule c. Molecule a also has dipole-dipole forces and can hydrogen bond. Molecule b also has dipole-dipole
forces. So Molecule a would have the strongest IMF, b is intermediate and molecule c has the weakest IMF. Because
stronger IMF mean a higher boiling point, a would have the highest boiling pt, followed by b and then c
H
H
Molecule a –– has hydrogen bonds. It has
both a hydrogen bond donor (the hydrogen
attached to O) and a
hydrogen bond
acceptor (the non-bonding
electron pair on O.
H
H
H
H
C
C
C
O
H
H
H
H
H
H
H
C
O
C
H
C
H
H
Structural representation of H-bonding:
Molecule b – has dipole-dipole forces (It does not have a
no hydrogen bonding.) The O is more electronegative
the rest of the molecule.

H
H
H
C
H
Structural representation of dipole-dipole interations:
hydrogen bond donor, so
H
H
H than
H
C
C
O
  H
H
C
H
C
C
O

H
8. Determine the type of reaction (like single replacement), predict the products of the following reactions & balance the equations.
Type
Single replacement
a) FeCl3 + 3 K →
Double replacement
b) 2 K3PO4(aq) + 3CoCl2(aq) → 6 KCl (aq) + Co3(PO4)2 (s)
Combination or synthesis
c) 2 Ca + O2 → 2 CaO
Double replacement
d) HCl(aq) + NaOH(s)
Combustion
e) C5H12 + 8 O2 → 5 CO2 + 6 H2O
Single replacement
f) 2 K + 2 H2O → 2 KOH + H2
3 KCl
→
+ Fe
NaCl (aq) + H2O (l)
9. Write a net-ionic equation for the following full balanced equations:
a) MgCl2(aq) + K2CO3(aq) → MgCO3(s) + 2KCl(aq)
Ionic eq: Mg2+ + 2 Cl- + 2 K+ + CO32-  MgCO3 + 2 K+ + 2 ClNet ionic eq:
Mg2+ (aq)
+ CO32- (aq)  MgCO3 (s)
b) 2Cs(s) + CaBr2(aq)  Ca(s)
Ionic eq: 2Cs + Ca2+ + 2Br1Net ionic eq:
(2 Cl- & 2 K+ are spectator ions & cancel)
 Ca
+ 2CsBr(aq)
+ 2Cs+ + 2Br1-
2Cs(s) + Ca2+(aq)  Ca(s)
(2 Br- is a spectator ion)
+ 2Cs+ (aq)
10. a) How many moles of propane (C3H8) are present in 500.8 g propane? b) How many propane molecules are
present in this 500.8 g? c) How many C atoms are present in this sample?
a) First find the molecular weight of C3H8.
500.8 g x
1 𝑚𝑜𝑙𝑒
44.09652 𝑔
3 C 3 x 12.011 = 36.033
8 H 8 x 1.00794 = 8.06352
44.09652 g/mole
= 11.3569 mol C3H8. or 11.36 mol C3H8.
b) (11.3569 mol C3H8 )(6.022 x 1023) = 6.839125 x 1024 or 6.839 x 1024 molecules C3H8
c) (6.839125 x 1024 molecules C3H8 )(
3 C atoms
1 molecule
) = 2.051737 x 1025 or 2.052 x 1025 C atoms
11. How many anions are in 0.0135 moles of MgBr2?
First convert moles to number of formula units (what we use for an ionic compound instead of molecule) of
MgBr2 using Avogadro’s number. Then convert to number of anions.
(0.0135 moles MgBr2)(
6.022 x 1023 formula units
2 anions
1 mole
1 formula unit
)(
) = 1.62594 x 1022 anions  1.63 x 1022 anions.
12. Calculate the formula weight (molecular weight) of Ca(NO3)2.
1 Ca
2N
6O
40.07838 g
2 x 14.0067 = 28.0134
6 x 15.9994 = 95.9964
164.08818 g = 1 mole  164.0882 g/mol
13 Magnesium metal reacts with iron (II) chloride according to the equation
Mg(s) + FeCl2(aq)  MgCl2(aq) + Fe(s).
Is the magnesium oxidized or reduced?
Is the iron (II) ion oxidized or reduced?
There are several ways one could proceed with this question. One way is to write the net ionic equation so as to see
charges on the ions.
I.E.
Mg(s) + Fe2+(aq) + 2 Cl-(aq)  Mg2+(aq) + 2 Cl-(aq) + Fe(s)
(The 2 Cl- cancel.)
N.I.E.
Mg(s) + Fe2+(aq)  Mg2+(aq) + Fe(s)
Now you can see that Mg is going from Mg0 to Mg2+. This means Mg is losing electrons and is oxidized.
And you can see that Fe is going from Fe2+ to Fe0. This means Fe is gaining electrons and is reduced.
14. Define reduction
The gain of electrons, the gain of H atoms, and/or the loss of O atoms
K2 Cr2O7
15. In the following reaction CH2CH2OH + O2 →
CH2CH2O, is CH2CH2OH oxidized or reduced? How do you know?
CH2CH2OH is being oxidized. You can tell because it is losing H atoms.
16. For the following oxidation-reduction reaction: Ca + Fe3+  Ca2+ + Fe
a) Write the half-reactions for this reaction.
b) Which reactant is being oxidized (How do you know?) and which is being reduced?
Fe3+ + 3e- 
a) Ca → Ca2+ + 2eFe
b) Ca is being oxidized, because it is losing electrons. Fe 3+ is being reduced; it gains electrons.
17. Briefly describe a human genetic disease based on oxidation-reduction chemistry of hemoglobin.
A full credit answer should include one or more equations for chemical reactions.
There are inherited blood disorders, methemoglobinemia, that exist because a patient has problems with
maintaining the required charge on the Fe ion in hemoglobin. One type occurs due to a mutation in the beta
chain of Hb that makes the iron heme more easily oxidized: Hb(Fe 2+)  Hb(Fe3+) + e-. Hb(Fe3+) doesn’t
bind O2, so the hemoglobin can’t carry oxygen through the body.
Everyone’s Hb(Fe) is occasionally oxidized to the Fe 3+ when it releases oxygen. So another type of
methemoglobinemia results from a lack of an enzyme (cytochrome b5 reductase). This enzyme is necessary
to
convert Fe3+ in methemoglobin back to Fe 2+, so that it can bind oxygen.
18. What is the concentration in molarity units of a solution that has 7.00 g of NaCl dissolved in water to give 520.6 mL of
solution?
M = mol/liter, First convert 7.00 g of NaCl into moles of NaCl using the molecular weight of NaCl.
Na 22.98977 g
Cl 35.4527 g
58.44247 g in 1 mol of NaCl
520.6 mL x
0.1198 mol
0.5206 L
1L
1000 mL
7.00 g x (
= 0.5206 L
= 0.2301 mol/L
or 0.230 M
1 mol NaCl
58.44247 g
) = 0.1197759 mol
19. If 83.00 mL of a 3.075 M solution of NaOH is diluted to a final volume of 12.88 L, what will its concentration be?
M1V1 = M2V2 .
Rearrange to get
M1V1
V2
Convert 83.00 mL to liters. 83.00 mL x
= M2. The volume amounts must be expressed in the same units.
1L
1000 mL
= 0.08300 L; M2 =
3.075 𝑀 𝑥 0.08300 𝐿
12.88 𝐿
= 0.01982 M
20. What volume of a 15% (w/v) solution of LiOH would be produced if you used 8750 g of LiOH?
% (w/v) =
x=
8750
15
g solute
x 100
mL solution
Rearrange to get mL of solution by itself. (mL solution) =
g solute
% (w/v)
x 100
x 100 = 58333 mL or 58,000 mL
21. Based on the following reaction, HCl + NaOH  H2O + NaCl,
determine the concentration of hydrochloric acid, HCl, if 10.00 mL of acid required 15.51 mL of 0.2500 M NaOH to reach the
endpoint of the titration.
Follow the four steps for titration. Step 1. The equation is already balanced.
Step 2. Find the moles of NaOH: (15.51 mL)(
1L
1000 mL
0.2500 mol NaOH
)(
Step 3. Find the moles of HCl: (0.0038775 mol NaOH)(
Step 4. Find the concentration (M) of HCl: (10.00 mL)(
0.0038775 mol HCl
0.01000 L
L
1 mol HCl
1 mol NaOH
1L
1000 mL
) = 0.0038775 mol NaOH
) = 0.0038775 mol HCl
) = 0.01000 L
= 0.38775 mol/L  0.3878 M HCl
21 (the second). For salt X, determine how much will remain undissolved if 500 g is mixed into a liter of pure water at 65 °C.
From the table, it is apparent that at 65 oC, 120 grams of salt will dissolve in 1 liter of water. If 500 g of this
salt was added to a liter than 500-120 or 380 g would remain undissolved
Grams of salt Dissolved per Liter of H2O
Solubility of Salt X in Water at Different Temp.
160
140
120
100
80
60
40
20
0
15
25
35
45
Temp oC
55
65
75
22. a) Determine whether the following would be more soluble in water or CCl4.
First note that water is polar and CCl4 is non-polar. (You can determine this from the Lewis Dot structures.)
Then remember that like dissolves like. So, look at the structure of each solute to determine if it is polar.
Ionic compounds form ions in water, and polar molecules like water form attractive forces with charged
particles.
i) CH3CH3 and iv) I2 are non-polar, so CCl4
ii) HCl, iii) NaCl and v) Na3PO4 form ions in water
polar, so H2O
and are
b) Which of the following are strong electrolytes, weak electrolytes or non-electrolytes?
Remember, strong acids, strong bases and ionic compounds are strong electrolytes.
Strong electrolytes: ii) HCl and iv) NaCl
Weak electrolytes:
i) CH3COOH (acetic acid)
Non-electrolytes :
iii) CH3CH2OH (ethanol)
23. Which of the following compounds would you predict to have the highest, intermediate, and lowest solubility in water? Explain
H
your answer, using structures where appropriate.
H
H
Solubility: a > b> c
H
a.
Molecule c. is a non-polar. It would not be
soluble in polar water.
C
C
H
C
C
O
b.
H
H
H
H
H
H
H
H
C
H
O
H
c.
C
H
C
H
H
C
H
H
C
H
H
Both a. and b. have a non-polar region and a polar region. The non-polar regions have similar surface areas for both
molecules. Next, compare the polar regions.
a. can form hydrogen bonds with water. Each of its propanol molecules has two H-bonding acceptor sites and 1
donor
site.
b. can also form hydrogen bonds w/ water. It has two acceptor sites and no donor sites. (Water would be
supplying all the donors in the hydrogen-bonding.) This has fewer hydrogen bonding sites than molecule a.
So: a. is most soluble. b. has intermediate solubility. c. is least soluble.
Drawings.
Indicates hydrophobic surface area
H
H
H
H
H
C
C
C
C
H
H
H
H
H
O
Indicates hydrophobic surface
H
H
H
C
C
H
H
a. Highest solubility. It has about
the same size non-polar surface as
c, but it has more sites for
hydrogen-bonding with water.
H
H
O
H
O
H
H
O
H
H
c. Lowest solubility. All of
it's surface is non-polar.
O
H
H
H
H
H
C
H
b. Intermediate solubility. It has fewer
hydrogen bonding sites than a.
C
H
O
C
H
H
H
O