Answer Sheet for Sample Problems for
Chemistry Level 1 Final Exam 2013-2014 Study Guide
Electrons in Atoms – Chapter 5
1. What is the frequency of green light, which has a wavelength of 4.90 x 10-7 m?
 = 4.90 x 10-7 m, c = 3.00 x 108 m/s
c=
 = ? Hz

c


3.00 x108 m / s
4.90 x107 m

3.00 x108 m / s
1.15 x1010 m
6.12 x 1014 Hz
2. An x-ray has a wavelength of 1.15 x 10-10 m. What is its frequency?
 = 1.15 x 10-10 m, c = 3.00 x 108 m/s
c=
 = ? Hz

c

18
2.61 x 10 Hz
3. Which type of electromagnetic radiation has the most energy: Microwaves, X-Rays, or Visible light?
Why?
Ans: X-Rays, b/c they have the shortest wavelength and highest frequency of the types
listed and frequency and energy are directly related.
4. What is the energy of a 9.50 x 1013 Hz radiation?
E=h
 = 9.50 x 1018 Hz, h = 6.626 x 10-34 J.s
=?J
E = h = (6.626 x 10-34 J.s )( 9.50 x 1018 s-1) = 6.29 x 10-15 J
5. The element with the electron configuration of [Kr]5s24d105p1 is indium (In)
6. Barium (Ba)
7. Write out the electron configurations for the following elements: (For the exam make sure you know
how to do both the complete and the noble gas short cut for electron configurations)
a. Tc
Ans: 1s22s22p63s23p64s23d104p65s24d5 or [Kr]5s24d5
b. Aluminum
Ans: 1s22s22p63s23p1 or [Ne] 3s23p1
8. Why are Cr and Cu exceptions to writing the electron configurations of elements?
They are exceptions because instead of following the pattern of filling up the d orbitals Cr
actually has a half filled s and d orbitals and Cu has a half filled s and completely filled d
orbitals. The reason why this occurs is because the electrons in the d-sublevel are more stable
with this electron configuration than the expected configuration.
9. a. Ca
b. As
.
a. 2
Ca.
c. Sn
..
b. 5
.As.
.
.
c. 4
. Sn .
.
Periodic Table and Periodic Trends – Chapter 6
1. Compare & contrast how Mendeleev’s periodic table was organized to how Moseley organized it.
Ans: Mendeleev used atomic mass & properties of elements to order the elements. This
resulted in some elements being out of order and not exactly fitting in with the others in a
group. Moseley used the atomic number of each element as well as the properties of the
elements and this fixed Mendeleev’s table.
2. What is the periodic law? Ans: When elements are arranged by increasing atomic number, there
is a repeating pattern to their chemical and physical properties.
3. Identify each of the following as a metal, nonmetal, or metalloid and as representative elements or
transition elements:
a. Neon
b. Arsenic
c. Zinc
d. Magnesium
Ans: a. Nonmetal,
b. metalloid,
c. metal,
d. metal,
representative
representative
transition
representative
4. Ans:
a.
b.
c.
d.
Alkali metals =
Noble gases =
Alkaline earth metals =
Halogens
=
4.
1.
3.
2.
group 1
group 18
group 2
group 17
Chapters 8 & 9 Chemical Names and Formulas Practice Problems:
1. Write the symbol and name for the ion formed when:
a. A potassium atom loses one electron
K+
Potassium ion
b. An aluminum atom becomes an ion
Al3+
Aluminum ion
c. A fluorine atom becomes an ion
F-
Fluoride ion
d. A nitrogen atom gains three electrons
N3-
Nitride ion
2. Write the formula (including charge) for each polyatomic ion:
a. Hydroxide OH-
b. Nitrate NO3-
c. Sulfate SO42-
3. Write the formulas for the following compounds – they are a mixture of ionic and molecular
compounds:
a. Potassium sulfate
K2SO4
d. Magnesium hydroxide Mg(OH)2
b. Carbon tetrabromide CBr4
e. Diphosphorus trioxide P2O3
c. Silver chloride
d. Iron(II) oxide
AgCl
FeO
4. Name the following compounds – they are a mixture of ionic and molecular compounds:
a. CS2
Carbon disulfide
d. Cu2O
Copper(I) oxide
b. NaI
Sodium iodide
e. K2S
Potassium sulfide
f. N2O5
Dinitrogen pentoxide
c. NH4Br Ammonium bromide
Chapter 9 – Covalent Bonding Sample Problems:
1. Draw electron dot structure and state the shape of each of the following:
e. PCl3 trigonal pyramidal
c. BF3 trigonal planar
f. CO2
linear
d. CO32-
trigonal planar but remember it has
resonance
2. State the type of bonds (nonpolar covalent, polar covalent, or ionic) found in the following compounds,
and circle the polar molecules.
a. CO
b. NH3
polar covalent
polar covalent
c. SF6
polar covalent
d. CaO
ionic
Chapter 10 Chemical Reactions Sample problems:
Balance the following & state what type of reaction it is:
1. 2 AgNO3 + _____H2S  _____Ag2S + 2 HNO3
2. 2 HCOOH + _____O2  2 CO2 + 2 H2O
3. ALREADY BALANCED
4. 2 SO2 + ______O2  2 SO3
5. Predict the products in the following chemical reactions; make sure to balance the reactions.
a. _____ Zn (s) + 2 HCl (aq)  H2 (g) + ZnCl2 (aq)
b. _____ H2SO4 (aq) + 2 KOH (aq)  2 H2O (l) + K2SO4 (aq)
6. Write a balanced net ionic equation for the following reaction and state what the spectator ions are:
Pb(NO3)2 (aq) + H2SO4 (aq)  PbSO4 (s) + HNO3 (aq)
Complete ionic equation:
Pb2+ + 2NO3- + 2H+ + SO42-  PbSO4 (s) + 2H+ + 2NO3Net ionic equation:
Pb2+ + SO42-  PbSO4 (s)
Spectator ions: H+ and NO3Identify what is oxidized (O), what is reduced (R), and the oxidation numbers on all the elements.
1.
0
0
+2 -2
2Zn + O2  2ZnO
0
0
+2 -1
2. Mg + I2  MgI2
zinc was oxidized, oxygen was reduced
magnesium was oxidized, iodine was reduced
+1 -2 0
0 +1 -1
3. H2S + Cl2  S + 2HCl
sulfur was oxidized, chlorine was reduced
Chapter 13 – States of Matter Sample Problems:
1. At 25°C, the vapor pressure of compound A is 40 mmHg and the vapor pressure of compound B is 50
mmHg. Which compound will evaporate faster? Which compound has the higher boiling point?
Explain.
Compound B will evaporate faster because there is more vapor above the solution to begin with so
it takes less energy to make the compound turn into a gas.
Compound A has the higher boiling point because it has the lower vapor pressure so, less of the
compound is in the gaseous state so it will take more energy to cause A to boil therefore you need a
higher temperature.
2. Name two changes of state (e.g. freezing, melting, etc.) that require energy.
Melting, boiling (vaporizing), sublimation
3. State the three kinds of intermolecular attractive forces. Rank these forces from weakest to strongest
(for molecules of about the same size).
Dispersion, dipole-dipole, hydrogen bond
4. Which compound would you expect to have the higher boiling point, CH4 or NH3? Explain.
NH3 because these molecules can hydrogen bond so it takes more energy (higher temperature) to
break those IMF and cause ammonia to boil.
Chapter 14 – Gases and Section 13.1 (14.4 covered with Stoichiometry) Sample Problems:
1. Which gas diffuses faster, nitrogen or carbon monoxide? Why?
Neither because they have the same molar mass.
2. A gas at 155 kPa and 25oC occupies a container with an initial volume of 1.00 L. By changing the
volume, the pressure of the gas increases to 605 kPa as the temperature is raised to 125oC. What is the
new volume?
𝑷𝟏𝑽𝟏
𝑻𝟏
=
𝑷𝟐𝑽𝟐
𝑻𝟐
𝑽𝟐 =
(𝟏𝟓𝟓𝒌𝑷𝒂)(𝟏𝑳)(𝟑𝟗𝟖𝑲)
(𝟐𝟗𝟖𝑲)(𝟔𝟎𝟓𝒌𝑷𝒂)
= 𝟎. 𝟑𝟒𝟐𝑳
3. A sample of air has a volume of 6.00 L at 100 kPa. What volume will it occupy at 25.0 kPa if the
temperature remains constant at 30oC?
𝑷𝟏𝑽𝟏
𝑻𝟏
=
𝑷𝟐𝑽𝟐
𝑻𝟐
𝑽𝟐 =
(𝟏𝟎𝟎𝒌𝑷𝒂)(𝟔𝑳)(𝟑𝟎𝟑𝑲)
(𝟑𝟎𝟑𝑲)(𝟐𝟓.𝟎𝒌𝑷𝒂)
= 𝟐𝟒 𝑳
4. A steel cylinder that has a volume of 20.0 L is filled with nitrogen gas to a final pressure of 25.0 atm at
28oC. How many moles of nitrogen gas does the cylinder contain?
𝑷𝑽 = 𝒏𝑹𝑻 𝒏 =
(𝟐𝟓 𝒂𝒕𝒎)(𝟐𝟎.𝟎𝑳)
(𝟎.𝟎𝟖𝟐𝟏
𝑳𝒂𝒕𝒎
)(𝟑𝟎𝟏𝑲)
𝒎𝒐𝒍𝑲
= 𝟐𝟎. 𝟐 𝒎𝒐𝒍
5. A gas mixture containing oxygen, nitrogen, and carbon dioxide has a total pressure of 32.9 kPa. If the
pressure of oxygen is 6.6 kPa and the pressure of nitrogen is 23.0 kPa, what is the pressure of CO2 in the
mixture?
PT = PO2 + PN2 + PCO2
P = 32.9 kPa – (6.6 kPa + 23.0 kPa) = 3.3 kPa
Chapter 15 – Solutions Sample Problems:
1. 36.0 g of glucose are dissolved to make 2.00 liters of solution. What is the molarity of the solution?
(The molar mass of glucose is 180 g/mol).
𝑴=
𝒎𝒐𝒍
𝑳
𝟏 𝒎𝒐𝒍
36.0g𝟏𝟖𝟎 𝒈 = 𝟎. 𝟐𝟎 𝒎𝒐𝒍
𝑴=
𝟎.𝟐𝟎 𝒎𝒐𝒍
𝟐.𝟎𝟎 𝑳
= 𝟎. 𝟏𝑴
2. How many grams of solute are in 335 mL of 0.425 M KNO3?
𝑴=
𝒎𝒐𝒍
𝑳
→ 𝒎𝒐𝒍 = 𝑴 ∗ 𝑳 = (𝟎. 𝟒𝟐𝟓
𝒎𝒐𝒍
𝑳
𝟏𝟎𝟏.𝟏𝒈
) (𝟎. 𝟑𝟑𝟓 𝑳) = 𝟎. 𝟏𝟒𝟐𝟒 𝒎𝒐𝒍 ( 𝟏 𝒎𝒐𝒍 ) = 𝟎. 𝟎𝟎𝟏𝟒𝟏 𝒈 𝑲𝑵𝑶𝟑
3. If 250 mL of 2.0 M CaCl2 is diluted to a final volume of 700 mL, what is the concentration of the diluted
solution?
𝑴𝟏𝑽𝟏 = 𝑴𝟐𝑽𝟐
𝑴𝟐 =
𝑴𝟏𝑽𝟏
(𝟐. 𝟎𝑴)(𝟐𝟓𝟎 𝒎𝑳)
=
= 𝟎. 𝟕𝟏𝟒 𝑴
𝑽𝟐
𝟕𝟎𝟎𝒎𝑳
4. Calculate the percent by mass of 3.55 g of NaCl dissolved in 88 g water.
𝟑. 𝟓𝟓𝒈
𝒙𝟏𝟎𝟎 = 𝟑. 𝟖𝟖%
(𝟑. 𝟓𝟓𝒈 + 𝟖𝟖𝒈)
5. What mass of water must be added to 255.0g KCl to make a 15.00 percent by mass aqueous solution?
255.0𝑔
𝑥100 = 15% 255𝑔 = 38.25𝑔 + 0.15𝑥
(255.0𝑔 + 𝑥)
216.75𝑔
0.15𝑥
=
𝑥 = 𝟏𝟒𝟒𝟓𝒈
0.15
0.15
6. What is the mole fraction of 3.76 g NaOH in 850 g H2O?
𝑋=
1𝑚𝑜𝑙
𝑚𝑜𝑙𝑁𝑎𝑂𝐻 0.094 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
=
= 𝟎. 𝟎𝟎𝟏𝟗𝟖𝟕
𝑚𝑜𝑙𝑡𝑜𝑡𝑎𝑙
47.316 𝑚𝑜𝑙
3.76g NaOH( 40𝑔 ) = 0.094 𝑚𝑜𝑙
1 𝑚𝑜𝑙
850gH2O( 18.0𝑔 ) = 47.2 𝑚𝑜𝑙
Chapter 19 – Acids and Bases Sample Problems:
1. An aqueous solution tastes bitter and turns litmus blue. Is the solution acidic or basic? BASIC
2. Classify the following as an Arrhenius acid or an Arrhenius base:
a. H2S ACID
c. Mg(OH)2 BASIC
b. RbOH BASIC
d. H3PO4
ACID
3. Given the concentration of either hydrogen ion or hydroxide ion, calculate the pH, pOH and
concentration of the other ion at 298 K:
a. [H+] = 1.0 x 10-4 M
b. [OH-] = 1.3 x 10-2 M
pH = -log(1x10-4) = 4
pOH = -log (1.3x10-2) = 1.89
pOH = 14 – 4 = 10
pH = 14-1.89 = 12.11
[OH-] = 1 x 10-10 M
[H+] = 10-12.11 = 7.76x10-13 M
4. Calculate [H+] and [OH-] in each of the following solutions at 298 K.
a. pH = 3.00
b. pH = 5.24
[H+] = 1 x 10-3 M
[H+] = 10-5.24 = 5.75 x 10-6 M
[OH-] = 1 x 10-11 M
[OH-] = 𝟓.𝟓𝟕 𝒙 𝟏𝟎−𝟔 = 𝟏. 𝟕𝟒 𝒙 𝟏𝟎−𝟗 𝑴
𝟏𝒙𝟏𝟎−𝟏𝟒
5. How many milliliters of 0.225 M HCl would be required to titrate 6.00 g KOH?
HCl + KOH  H2O + KCl
1 𝑚𝑜𝑙 𝐾𝑂𝐻
6.00𝑔 𝐾𝑂𝐻 (
56,1 𝑔
1 𝑚𝑜𝑙 𝐻𝐶𝑙
1𝐿
) (1 𝑚𝑜𝑙 𝐾𝑂𝐻) = 0.107𝑚𝑜𝑙 𝐻𝐶𝑙 (0.225 𝑚𝑜𝑙 𝐻𝐶𝑙) = 0.475 𝐿 = 𝟒𝟕𝟓 𝒎𝑳