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Winter 2017 - Enderle
CHEMISTRY 2A
Exam I
Multiple Choice (circle one)
Instructions:
CLOSED BOOK EXAM! No books, notes, or additional scrap paper are
permitted. All information required is contained on the exam. Place all work
in the space provided. If you require additional space, use the back of the exam.
(1) Read each question carefully.
(2) There is no partial credit for the problems in Part I and Part II. You will
lose 10 points if you do not circle your multiple choice answers on the
front page or if you do not write your TA’s name or section in the space
above.
(3) The last page contains a periodic table and some useful information. You
may remove this for easy access.
(4) Graded exams will be returned in the laboratory sections next week.
(5) If you finish early, RECHECK YOUR ANSWERS!
1.
A B C D E
2.
A B C D E
3.
A B C D E
4.
A B C D E
5.
A B C D E
6.
A B C D E
7.
A B C D E
8.
A B C D E
9.
A B C D E
10.
A B C D E
1–15. 45 points
11.
A B C D E
16–17. 08 points
12.
A B C D E
18–20. 24 points
13.
A B C D E
21. 13 points
14.
A B C D E
22–23. 18 points
15.
A B C D E
U.C. Davis is an Honor Institution
Possible Points
Total Score (108)
Points
Exam I
Page 2 of 9
Part I: Multiple Choice
Circle the correct answer here and on the cover – No partial credit (3 points each)
1. You react chemical A with chemical B to make one product. It takes 100 g of A to react
completely with 20 g of B. What is the mass of the product?
A. less than 20 g B. between 20 g and 100 g C. between 100 g and 120 g
D. 120 g
E. more than 120 g
2. What is the approximate percent by mass of fluorine in BF3?
A. 16
B. 28
C. 72
D. 84
E. 100
3. Which of the following can be determined by knowing the number of protons in a neutral
element?
A. The number of neutrons in the neutral element
B. The number of electrons in the neutral element
C. The name of the element
D. Both B and C
E. All of the above
4.
What multipliers are needed for the three respective simultaneous reactions below, in order
to obtain this overall reaction: CO2 (g) → C (s) + O2 (g)?
Find the multipliers (α, β, γ).
[H2O (l) → H2 (g) + ½ O2 (g)] x α
[C2H6(g) → 2 C (s) + 3 H2 (g)] x β
[2 CO2 (g) + 3 H2O (l) → C 2H6 (g) + 7/2 O2 (g)] x γ
A. 2, ½, ½
B. 1.5, 1, 1
C. 1, -½, ½
D. 1.5, ½, ½
E. 1, 1, 1
5. Which of the following is soluble in aqueous solution?
A. calcium acetate
B. ferric hydroxide
C. silver iodide
D. barium phosphate
6. All samples of the same compound have the same composition according to what law?
A. law of conservation of mass
B. law of constant composition
C. law of multiple proportions
D. law of conservation of energy
7. Calculate the answer with the correct number of significant figures.
1.5 x 103 – 98.8 =
3
A. 1.4013 x 10 B. 1.401 x 103 C. 1.40 x 103 D. 1.4 x 103 E. 1 x 103
8. How many carbon atoms are in a 40.1 g sample of C?
A. 2.01 x 1024 B. 2.41 x 1025 C. 2.9 x 1026 D. 1.5 x 1022
9. Which of the following is not a strong base?
A. NaOH
B. KOH C. C2H5OH D. Sr(OH)2
E. LiOH
E. 1.3 x 1021
Exam I
Page 3 of 7
10. 100.0 mL of an aqueous solution of compound A is mixed with an equal volume of an aqueous
solution of compound B. Which of the following is true?
A. The molarity of both compounds will not change after mixing
B. Only the molarity of compound A will change after mixing
C. Only the molarity of compound B will change after mixing
D. The molarity of both compounds (A & B) will decrease by half after mixing
E. The molarity of both compounds (A & B) will double after mixing
11. The average mass of a carbon atom is 12.011. Assuming you were able to pick up only one
carbon atom, what is the chance that you would randomly get one with a mass of 12.011?
A. 0% B. 0.011% C. about 12% D. 12.011% E. greater than 50%
12. All of the following pieces of information, except one, are needed to calculate the molarity of
a salt solution. Which piece of information is not needed?
A. the mass of the salt added
B. the molar mass of the salt
C. the volume of the water added D. the total volume of the solution
13. How many mL of 0.760 M HNO3 are needed to dissolve 7.18 g of CaCO3?
2 HNO3(aq) + CaCO3(s) → Ca(NO3)2(aq) + H2O(l) + CO2(g)
A. 26 mL B. 94 mL C. 143 mL D. 189 mL E. impossible to determine
14. A new element is discovered. It has two isotopes. The relative abundance of the isotopes and
their masses are 18% isotope 1, mass 350.0 u and 82% isotope 2, mass 352.0 u. What is the
atomic mass of the element?
A. 351.6 B. 351.0 C. 350.4
D. 352.0
E. 350.0
15. The total numbers of neutrons, protons, and electrons in 138Ba2+ are ________.
A. 138 neutrons, 56 protons, 54 electrons B. 82 neutrons, 56 protons, 54 electrons
C. 56 neutrons, 82 protons, 80 electrons
D. 82 neutrons, 56 protons, 58 electrons
E. 82 neutrons, 82 protons, 82 electrons
Part II: Short Answer
Fill in the blank. No partial credit for work shown.
16. (4 points) Order the following molecules from lowest to highest oxidation state of the N
atom: HNO3, NH4Cl, N2O, NaNO2.
NH4Cl (-3) < N2O (+1) < NaNO2 (+3) < HNO3 (+5)
17. (4 points) Balance the following reaction with whole numbers.
____ KMnO4 + ____ KI + ____ H2SO4 → ____ K2SO4 + ____ MnSO4 + ____ I2 + ____ H2O
Answer: 2,10,8,6,2,5,8
Exam I
Page 4 of 7
18. (7 total points) A stock solution of NaOH (aq) has a concentration of 6 M. A 100.0 mL
aliquot of stock solution was diluted to 2.0 L. How much of this diluted solution must be
combined with enough water to make 300.0 mL of 0.1M NaOH (aq) solution?
M1V1 = M2V2, thus (6M)(100.0mL)/(2.0L) = 300 mM
Let M2 = M3 = 300mM
M3V3 = M4V4, thus V3 = (100mM)(300mL)/(300mM) = 100mL
Answer:
19. (7 total points) Write the balanced net ionic chemical equation for the following reactions
(include states for each species). If no reaction takes place, write NR.
A. (3 points) Cesium chloride reacts with potassium nitrate in water.
CsCl(aq) + KNO3(aq) → NR
B. (3 points) Sulfuric acid reacts with barium hydroxide in water.
H2SO4 (aq) + Ba(OH)2 (aq) → BaSO4 (s) + 2 H2O (l)
2H+ (aq) + SO42- (aq) + Ba2+ (aq) + 2OH- (aq) → BaSO4 (s) + 2 H2O (l)
C. (1 point) What type of reaction is in part B of this question? Double Replacement
20. (10 points) Fill in the blanks of the table giving the chemical name or chemical formula.
Chemical Formula
Chemical Name
P4O10
tetraphosphorus decoxide (decaoxide ok!)
SnF4
tin(IV) (or stannic) fluoride
H2SO3
sulfurous acid
Ca(OH)2
calcium hydroxide
decane
Exam I
Page 5 of 7
Part III: Long Answer
Please show all work – Partial credit – Use the correct number of significant figures
21. (13 total points) 100.0 g of N2H4 reacts with 200.0 g of N2O4 according to the following
reaction: 2 N2H4 + N2O4 → 3 N2 + 4 H2O. Remember to include units when writing the answer
in the box. If you cannot solve part A, use 130.0 g as your answer to complete parts B and C.
A. (5 points) What mass of N2 would theoretically be produced?
N2H4: (100 g)(mol/32.05 g)(3 mol/2 mol)(28.02 g/mol) = 131.1 g N2
N2O4: (200 g)(mol/92.01 g)(3 mol/1 mol)(28.02 g/mol) = 182.7 g N2
N2H4 is the limiting reactant, N2O4 is in excess.
Answer:
B. (5 points) What mass of the reactant in excess remains at the end of the reaction if the
reaction goes to completion?
(131.1 g)(mol/28.02 g)(1 mol/3 mol)(92.01 g/mol) = 143.5 g used
200.0 g given – 143.5 g used = 56.5 g remains
{if using 130, then 57.5 g}
Answer:
C. (3 points) If 13.11 g of N2 is actually produced, what is the percent yield?
(13.11 g/131.1 g)(100%) = 10.00%
{if using 130, then 10.08%)
Answer:
Exam I
Page 6 of 7
22. (11 total points) Consider the following chemical equation:
Br2 (l) + Mn2+ (aq) → MnO2 (s) + Br- (aq)
A. (2 points) Determine which element is oxidized and which is reduced.
Element being oxidized: Mn
Element being reduced: Br
B. (3 points) Balance the oxidation half reaction including state symbols (acidic conditions).
Mn2+ (aq) + 2H2O (l) → MnO2 (s) + 4H+ (aq) + 2e-
C. (3 points) Balance the reduction half reaction including state symbols (acidic conditions).
Br2 (l) + 2e- → 2Br-(aq)
D. (3 points) Give the overall balanced redox equation including state symbols (acidic conditions).
Br2 (l) + Mn2+ (aq)+ 2H2O (l) → MnO2 (s) + 2Br- (aq) + 4H+ (aq)
23. (7 total points) A compound has the following composition; 40.9% carbon, 4.58% hydrogen
and 54.5% oxygen. Determine the empirical formula of this compound and put your answer in the
box.
Moles C = 40.9 g/(12.01g/mol C) = 3.41 mol C
Moles H = 4.58 g/(1.0079 g/mol H) = 4.54 mol H
Moles O = 54.5 g/(16.00 g/mol O) = 3.41 mol O
Carbon: 3.41 mol / 3.41 mol = 1.00x3 = 3
Hydrogen: 4.54 mol / 3.41 mol = 1.33x3 = 4
Oxygen: 3.41 mol / 3.41 mol = 1.00x3 = 3
Empirical Fmla: C3H4O3
Empirical
Formula
Exam I
Avogadro’s number = 6.022 x 1023 / mol
K = C + 273.15
M1V1 = M2V2
d = m/V
2.54 cm = 1.00 in (exact)
Page 7 of 7
F = 9/5 C + 32
Solubility Rules:
Compounds that are soluble or mostly soluble
• Group 1, NH4+, chlorates, acetates, nitrates
• Halides (except Pb2+, Ag+, and Hg22+)
• Sulfates (except Ca2+, Sr2+, Ba2+, Pb2+, and Hg22+)
Compounds that are insoluble
• Hydroxides, sulfides (except above rule, and group 2 sulfides)
• Carbonates, phosphates, chromates (except above rules)