Chemistry 338 2 Semester Final Exam Review Sheet

Chemistry 338 2 nd Semester Final Exam Review Sheet
Unit 1 - Matter & Measurement – Chapters 1 & 2
1. A student experimentally determines the density of an object to be 4.2 g/mL. If the
accepted value is 5.0 g/mL, calculate the % error.
% error = % error = = -16% error
Unit 2 – Periodic Table – Chapter 5
2.
Element Symbol Atomic
#
Protons Neutrons Electrons Mass # Charge
74 74 110 72 184 +2
Tungsten
ion
3. Using the following atomic radius values, predict the atomic radius of potassium (K):
Li – 145 pm, Na – 180 pm, K - ?, Rb – 235 pm
K – 220 pm
Unit 3 – Reactions/Compounds – Chapters 7 & 8
4. Classify the compound as ionic or covalent. Name the compound
a. Fe 2 O 3 - iron (III) oxide (ionic - roman numerals for transition metals, change ending to -ide)
b. Cl 2 O 8 - dichlorine octaoxide (covalent – use prefixes, change ending to –ide)
5. Classify the compound as ionic or covalent. Write the formula
a. Trisulfur Dinitride – S 3 N 2 (covalent)
b. Zinc Hydroxide – Zn(OH) 2 (ionic – look up ion charges and criss-cross) Zn +2 OH -1
c. Chromium (III) nitrate – Cr(NO 3 ) 3 (ionic – criss-cross) Cr +3 NO 3
-1
6. Write the equation, balance it, and classify the type of reaction:
Iron (II) sulfate reacts with sodium hydroxide to yield iron(II) hydroxide and sodium
sulfate
1 FeSO 4 + 2 NaOH 1 Fe(OH) 2 + 1 Na 2 SO 4
double displacement (two ionic compounds reacting)
7. Balance the equation and classify the type of reaction:
1 Cu + 2 AgNO 3 1 Cu(NO 3 ) 2 + 2 Ag
Single displacement (element and a compound reacting)
Unit 4 – Moles – Chapter 3
8. Calculate the molar mass of Fe 2 (SO 4 ) 3 .
Fe (2)(55.9 g/mol) = 111.8 g/mol
S (3)(32.1 g/mol) = 96.3 g/mol
+ O (12)(16.0 g/mol) = + 192 g/mol
= 400.1 g/mol

9. How many molecules are in 36 g of Li 2 O
x x = 7.27x10 23 molecules Li 2 O
Unit 5 – Stoichiometry – Chapter 9
10. Using the following balanced equation: 4 NH 3 + 5 O 2 4 NO + 6 H 2 O
a. If 10 moles of H 2 O get produced, how many moles of NO will also be produced?
x
b. How many grams of O 2 are needed to react 824 g of NH 3 ?
x x
Unit 6 – Gases – Chapter 11
11. The pressure of a gas is 843 mm Hg at 24 o C. What will the pressure be at 100 o C?
Volume is held constant.
P 1 = 843 mm Hg
T 1 = 297 K (24 o C + 273)
P 2 = ?
T 2 = 373 K (100 o C + 273)
P 2 = 1058.7 mm Hg
12. The volume of a gas is 81.7 mL when the pressure is 119 kPa. What is the volume when
the pressure is changed to 500 kPa?
P 1 = 119 kPa
V 1 = 81.7 mL
P 2 = 500 kPa
V 2 = ?
P 1 V 1 = P 2 V 2 (119 kPa)(81.7 mL) = (500 kPa)(V 2 ) V 2 = 19.4 mL
Unit 7 - Solutions – Chapters 12-13
13. What precipitate forms when you combine CaCl 2 and Na 2 CO 3 ?
CaCl 2 + Na 2 CO 3 CaCO 3 (s) + 2 NaCl (aq)
CaCO 3 (s)
14. A solution is made by dissolving 30 g of NaCl into water creating a solution with a total
volume of 2 L. What is the molarity?
30gNaCl
1moleNaCl
x 0.513molNaC
l
1 58.5gNaCl

0.513molNaCl
Molarity 0.256M
2LSolution
15. A solution is made by dissolving 2 moles of NaCl into 4 kg of water. What is the
molality?
2molNaCl
molality 0.5m
4kgH 2O
16. A student has 200 mL of a 3 M HCl solution. The student adds water until the total
volume of the solution becomes 600 mL. What is the new molarity of the HCl solution?
M 1 V 1 = M 2 V 2
(3M)(200mL) = (M 2 )(600mL)
M 2 = 1.0 M HCl
17. What is the boiling point elevation of a 2 m NaCl solution? (K b H 2 O = 0.52 o C/m)
ΔT b = K b x m x i
ΔT b = (0.52 C / m) x(2 m) x(2) 2.08 C
Unit 8 - Equilibrium – Chapter 18
18. Write the equilibrium expression for the following equation:
2 A (g) + 3 B (g) ↔ 2 C (g)
K = __[C] 2 __
[A] 2 [B] 3
19. In the equation above, which direction does the equilibrium shift if more A is added?
Shift to the right
20. What happens to the concentration of C when the equilibrium shifts right?
Concentration of C increases
21. The K sp of BaCO 3 is 2.58x10 -9 . What is the concentration of Ba +2 at equilibrium? What
-2
is the concentration of CO 3 at equilibrium?
1BaCO 3 (s) ↔ 1 Ba +2 (aq) + 1 CO -2 3 (aq)
x
x
K sp = [Ba +2 ][ CO 3 -2 ]
2.58x10 -9 = [x][x]
2.58x10 -9 = x 2
X = 5.1x10 -5 M Ba +2 , CO 3
-2

Unit 9 - Acids and Bases _Chapters 14-15 & 18
22. Calculate the pH of the following and classify acid/base/neutral:
a. 1.0x10 -2 M = [H + ] b 3.7 x 10 -5 M = [OH - ]
pH = 2 - Acid
pH = 9.57 – base
23. A 0.11 M weak acid has a [H + ] = 0.0004 M at equilibrium. What is the Ka of the acid?
1 HA (aq) ↔ 1 H + (aq) + 1 A - (aq)
I 0.11 0 0
C -0.0004 +0.0004 +0.0004
E 0.1096 0.0004 0.0004
Ka = [H+][A-] Ka = [0.0004][0.0004] Ka = 1.46x10 -6
[HA] [0.1096]
24. How many mL of a10 M HCl are needed to neutralize 380 mL of a 7 M NaOH?
1 HCl + 1 NaOH 1 NaCl + 1 H 2 O
380 mL _ NaOH 1 L _ NaOH 7 mol _ NaOH 1 mol _ HCl 1 L _ HCl 1000 mL _ HCl
x x x x x
1 1000 mL _ NaOH 1 L _ NaOH 1 mol _ NaOH 10 mol _ HCl 1 L _ HCl
= 266 mL HCl
25. In the following reaction, identify the acid and base and its conjugate pairs (Bronsted-
Lowry): H 2 O + NH 3 → NH 4 + + OH -
A B CA CB
26. Name the following acids:
a) H 2 S – hydrosulfuric acid
b) HNO 2 – nitrous acid
c) HClO 3 – chloric acid
27. What is the salt that forms when Sr(OH) 2 reacts with HNO 3 ?
Sr(OH) 2 + 2 HNO 3 Sr(NO 3 ) 2 + 2 H 2 O
Sr(NO 3 ) 2
Unit 10 - Electrons – Chapter 4, 6
28. Draw the lewis dot structure, state the shape, and hybridization
a. CH 4 - tetrahedral, sp 3

. NH 3 - pyramidal, sp 3
29. Using electronegativity values, determine if the compound has nonpolar, polar, or ionic
bonds:
a. KF
Electronegativity F – 4.0
Electronegativity K – 0.8
4.0 – 0.8 = 3.2 electronegativity difference ionic bond
b. SO 2
Electronegativity S – 2.5
Electronegativity O – 3.5
3.5 – 2.5 = 1.0 electronegativity difference polar covalent bond
c. Cl 2
Electronegativity Cl – 3.0
Electronegativity Cl – 3.0
3.0– 3.0 = 0 electronegativity difference nonpolar covalent bond
30. Rank the following compounds from lowest to highest melting point:
KF, CH 4 , C 3 H 8 , H 2 O
KF – ionic bonds, ionic forces
CH 4 – nonpolar bonds, London forces
C 3 H 8 – nonpolar bonds, London forces (stronger London forces than CH 4 due to bigger
mass)
H 2 O – polar bonds, polar molecule (no symmetry due to bent shape), hydrogen bonds
CH 4 < C 3 H 8 < H 2 O < KF
31. Write the electron configuration (the long and short way) for the following elements and
state the number of unpaired electrons:
a. Silicon (Si)
1s 2 2s 2 2p 6 3s 2 3p 2
[Ne] 3s 2 3p 2
2unpaired e -

. Palladium (Pd)
1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 8
[Kr] 5s 2 4d 8
2unpaired e -
32. Calculate the wavelength (λ) of light which gives off 2.87x10 -19 J of energy
E = hν 2.87x10 -19 J = (6.6262x10 -34 Js)( ν ) ν = 4.33x10 14 s -1
c = λν 3x10 8 m/s = ( λ ) (4.33x10 14 s -1 ) λ = 6.93x10 -7 m
Unit 11 - Redox – Chapters 19-20
33. Balance the following equations using redox and identify the element being oxidized,
reduced, the oxidizing agent, and the reducing agent.
0 +1+6-2 +4-2 +4-2 +1 -2
a. C + H 2 SO 4 CO 2 + SO 2 + H 2 O
0 +4
(C → CO 2 + 4e - ) carbon is oxidized, carbon is the reducing agent
+6 +4
(2e - + H 2 SO 4 → SO 2 )2 sulfur is reduced, sulfur is the oxidizing agent
C → CO 2 + 4e -
4e - + 2H 2 SO 4 → 2SO 2
C + 2 H 2 SO 4 CO 2 + 2 SO 2 + 2 H 2 O
+1 +7 -2 +1 -1 +2 -1 0 +1 -2 +1 -1
b. KMnO 4 + HCl MnCl 2 + Cl 2 + H 2 O + KCl
-1 0
(2 HCl → Cl 2 + 2e - ) 5 chlorine is oxidized, chlorine is the reducing agent
+7 +2
(5e - + KMnO 4 → MnCl 2 )2 manganese is reduced, manganese is the oxidizing agent
10 HCl → 5 Cl 2 + 10 e -
10 e - + 2 KMnO 4 → 2 MnCl 2
2 KMnO 4 + 10 HCl 2 MnCl 2 + 5 Cl 2 + H 2 O + KCl
34. Calculate the voltage of the following voltaic cell, label the anode, cathode, and the
direction of the electron flow for the following voltaic cell:
Cu/Cu +2 and Al/Al +3
Oxidation – Anode Al Al +3 + 3e - 1.66 V
Reduction – Cathode Cu +2 + 2e - Cu
0.34 V
2 Al + 3 Cu +2 2 Al +3 + 3 Cu 2.00 V produced
Electron flow from the Al to Cu (anode to cathode)

Unit 12 - Thermochem –Chapter 16
Specific Heats: Cu ( C = 0.381 J/g o C), Al ( C = 0.907 J/g o C), water (C = 4.18 J/g o C)
Substance Heat of Fusion in cal/g Heat of Vaporization in cal/g
H 2 O 80 540
Cu 49 1150
35. How much heat is needed to raise 100 g of Cu from 10 o C to 80 o C?
q = mcT
q = (100 g)( 0.381 J/g o C)(70 o C)
q = 2667 J
36. How much heat is absorbed when 15g water boils?
Liquid gas Vaporization
q = mass x heat of vaporization
q = (15 g) x (540 cal/g)
q = 8100 cal
37. How much heat is released when 8g of water at 99 o C is cooled to 70 o C?
q = mcT
q = (8g)( 4.18 J/g o C)(-29 o C)
q = -970 J or -232 cal