Key for Study Guide Exam 2
Spring 2015
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. State the tenets of the Kinetic molecular theory. See notes.
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
6. Draw a picture of three molecules of water
connected by hydrogen bonds.
= 88.08707703 L = 88.1 L
H
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
H
C
H
C
C
O
H
a.
H
C
H
H
C
C
b.
C
H
H
C
C
H
H
H
H
H
H
C
H
O
H
H
c.
H
H
i) a – (propanol) –is polar and can form hydrogen bonds. It also 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
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
H
H
H
electron pair on O.
H
Structural representation of H-bonding:
H
C
C
C
O
H
H
H
H
H
H
H
C
O
C
H
C
H
H
H
Molecule b – has dipole-dipole forces (It does not have a hydrogen bond donor, so no hydrogen bonding.)
The O is more electronegative than the
H
H
H
H
H
H
rest of the molecule.
Structural representation
C
C
C
 H
C
C
C
O   H
O 
of dipole-dipole interations:
H
H
H
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) → NaCl (aq) +
Combustion
e) C5H12 +
Single replacement
f) 2 K + 2 H2O → 2 KOH + H2
3 KCl
+ Fe
8 O2 → 5 CO2 + 6 H2O
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 Cl- (2 Cl- & 2 K+ are spectator ions & cancel)
Net ionic eq:
Mg2+ (aq) + CO32- (aq)  MgCO3 (s)
b) 2Cs(s) + CaBr2(aq)  Ca(s)
Ionic eq: 2Cs + Ca2+ + 2Br1- 
Net ionic eq:
2Cs(s) + Ca2+(aq) 
+ 2CsBr(aq)
Ca + 2Cs+ + 2Br1-
(2 Br- is a spectator ion)
Ca(s) + 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
1 mole
)(
2 anions
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. Given the following reaction between nitrogen and oxygen to form nitric oxide: N 2(g) + O2 (g)  2 NO(g)
How many grams of NO are obtained when 8.6 g of N2 are completely reacted?
Plan: g N2  mol N2  mol NO  g NO
First need to determine molecular weight of N2:
8.6 g x
1 mole
28.0134 g
= 0.3069959 mol N2
2 x 14.0067 = 28.0134 g in 1 mole of N2.
0.3069959 mol N2 x
2 mol NO
1 mol N2
= 0.613399 mol NO
MW of NO 14.0067 + 15.9994 = 30.0061 g
0.613399 mol NO x
30.0061g
1 mol
30.0061 g = 1 mol NO
= 18.4057 g NO  18 g NO
14. 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.
15. Define reduction
The gain of electrons, the gain of H atoms, and/or the loss of O atoms
K2 Cr2 O7
16. 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.
17. A sample of 235.00 g of Mg reacts completely with HCl (aq) to produce an explosive gas.
a) Write a balanced chemical equation for the reaction.
b) How many grams of gaseous product are produced?
a) Mg + 2 HCl → MgCl2 + H2
b) Use Mg’s weight to convert g to moles: (235.00 g Mg)(
(9.6687924 mole Mg) (
(9.6687924 mole H2)(
1 mol Hydrogen gas
1 mole Mg
2.01588 g
1 mole
1 𝑚𝑜𝑙𝑒
24.305 𝑔
) = 9.6687924 mole Mg
) = 9.6687924 mole H2
) = 19.49112528 g H2 or 19.491 g H2
(You could also do these three steps as one continuous problem if you wanted.)
18. 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?
a) Ca → Ca2+ + 2e-
Fe3+ + 3e-  Fe
b) Ca is being oxidized, because it is losing electrons. Fe3+ is being reduced; it gains electrons.