ChE 190 - LE1 - Answer Key
ChE 190 - LE1 - Answer Key
ChE 190 - LE1 - Answer Key
Instructions: Encircle/Write the letter of your answer for each question. For problem solving items, include
your complete solution. No solution = no point even if you chose the correct answer.
1. Carbon dioxide gas dissociates at high temperature according to the following equation:
2𝐶𝑂2(𝑔) ↔ 2𝐶𝑂(𝑔) + 𝑂2 (𝑔)
At 3000˚Cand 1 atm, CO2 is 40% dissociated. If 25 L of CO2 at 20˚C and 1 atm are heated at 3000˚C
at constant pressure, the volume of the resulting gas is:
A. 335 L B. 285 L C. 205 L D. 312 L
Can also be solved by solving the initial moles using ideal gas equation, calculate moles in the
final condition, then use again the ideal gas equation to solve for final volume
6. An element consists of 60.10% of an isotope with an atomic mass of 68.926 amu and 39.90 % of an
isotope with an atomic mass of 70.925 amu. What is the atomic weight of the element?
A. 34.86 amu B. 13.13 amu C. 41.42 amu D. 69.72 amu
𝑀𝑊𝑎𝑣𝑒 = ∑ 𝑀𝑊𝑖𝑠𝑜𝑡𝑜𝑝𝑒 × %𝑎𝑏𝑢𝑛𝑑𝑎𝑛𝑐𝑒𝑖𝑠𝑜𝑡𝑜𝑝𝑒
7. What is the IUPAC name for the molecule having a methyl group attached to an 8-carbon alkane ring?
A. Cyclooctanol B. Methylcyclooctane C. Cyclooctane D. Boric acid
8. A 0.5 M solution of H2SO4 is the same as _______ H2SO4.
A. 2 N B. 1 N C. 0.5 N D. 0.1 N
𝑁 = 2(0.5) = 1 𝑁
𝐻𝐴𝑐 ⇌ 𝐻 + + 𝐴𝑐 −
𝐻𝐴𝑐 𝐻+ 𝐴𝑐 −
Initial 0.1 0 0
Equil 0.1(1 − 0.0134) = 0.09866 0.1(0.0134) = 0.00134 0.1(0.0134) = 0.00134
[𝐻 + ][𝐴𝑐 − ] 0.00134(0.00134)
𝐾𝑎 = [𝐻𝐴𝑐]
= = 1.82 × 10−5
0.09866
𝐶6 𝐻5 𝐶𝑂𝑂𝐻 ⇌ 𝐻 + + 𝐶6 𝐻5 𝐶𝑂𝑂 −
𝐶6 𝐻5 𝐶𝑂𝑂𝐻 𝐻+ 𝐶6 𝐻5 𝐶𝑂𝑂 −
Initial 0.3 0 0
Equil 0.3 − 𝑥 𝑥 𝑥
[𝐻 + ][𝐶6 𝐻5 𝐶𝑂𝑂− ] 𝑥2
𝐾𝑎 = [𝐶6 𝐻5 𝐶𝑂𝑂𝐻]
6.6 × 10−5 = 𝑥 = 4.4168 × 10−3
0.3−𝑥
11. An aqueous solution of gold (III) nitrate is electrolyzed with a current of 0.555 ampere until 1.32 g of
Au has been deposited on the cathode. If the atomic weight of Au is 197, determine the duration of
the electrolysis.
A. 65.43 min B. 23.67 min C. 58.28 min D. 60 min
3+
𝐴𝑢(𝑎𝑞) + 3𝑒 − → 𝐴𝑢(𝑠)
𝑚𝐴𝑢 𝑚𝑜𝑙 𝑒 − 1.32 3
𝑄 = 𝐼𝑡 = ( )𝐹 0.555 𝑡 = ( ) (96485.3365)
𝑀𝑊𝐴𝑢 𝑚𝑜𝑙 𝐴𝑢 197 1
𝐻2 𝐼2 𝐻𝐼
Initial 5 5 0
Equil 5−𝑥 5−𝑥 2𝑥
2𝑥 2
[𝐻𝐼]2 [10] 5−3.99
𝑘 = [𝐻 62.5 = 5−𝑥 5−𝑥 𝑥 = 3.99 [𝐻2 ] = = 0.10 𝑚𝑜𝑙/𝐿
2 ][𝐼2 ] [ 10 ][ 10 ] 10
15. A C5H8 hydrocarbon is reacted with BH3 in THF, followed by oxidation with alkaline hydrogen
peroxide. Treatment of the resulting product with PCC in CH 2Cl2 produces a chiral ketone, formula
C5H8O. What hydrocarbon best fits these facts?
A. 1-methylcyclobutene C. vinylcyclopropane
B. methylenecyclobutane D. cyclopentene
16. Which of the following is the closest to the C-O-C bond angle in CH3 – O – CH3?
A. 180° B. 120° C.109.5° D. 160°
17. A carbonyl compound is
A. one carbon and one oxygen bonded by a double bond
B. one carbon and one oxygen bonded by a single bond
C. one carbon and two oxygen bonded by one single and one double bond
D. one carbon and two oxygen bonded by two double bonds
18. A strip of electrolytically pure copper weighing 3.178 g is strongly heated in a stream of oxygen until
it is all converted to 3.978 g of the black oxide. What is the percent copper of this oxide?
A. 79.9 % B. 20.1% C. 65.2 % D. 34.8 %
Since it is a pure copper and conversion is 100%
3.178
%𝐶𝑢 = (100) = 78.89%
3.978
19. What is the pH of a solution containing 0.01 M acetic acid and 0.01 M sodium acetate?
A. 9.26 B. 4.74 C. 3.25 D. 10.75
[𝑐𝑏]
𝑝𝐻 = − log 𝐾𝑎 + log [𝑎]
𝐾𝑎 𝑜𝑓 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 = 1.8 × 10−5
0.01
𝑝𝐻 = − log(1.8 × 10−5 ) + log = 4.74
0.01
20. Calculate the temperature in ˚C that must be maintained in a gas carrier tank in the form of a
horizontal cylinder with hemispherical heads if it carries 100 g of oxygen gas at 2000 Pa. The total
length of the tank is 10 m and its diameter is 2 m.
A. 1984 B. 1888 C. 1521 D. 1990
2m
10 m
4 4 28
𝑉𝑡𝑎𝑛𝑘 = 𝑉𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 + 𝑉𝑠𝑝ℎ𝑒𝑟𝑒 = 𝜋𝑟 2 𝐿 + 𝜋𝑟 3 = 𝜋(1)2 (10 − 2) + 𝜋(1)3 = 𝜋 𝑚3
3 3 3
2000 28
𝑃𝑉 ( )( 𝜋)(1000)
101325 3
𝑇= = 100 = 2256.93 𝐾 = 1983.78 °𝐶
𝑛𝑅 (0.08206)
32
23. Primary alcohol can be derived from the reaction of Grignard reagent with?
A. methanol B. any ketone C. acetone D. ethylene
24. Which of the following used as a suffix of carbaldehyde in naming?
A. B. C. D. all of these
𝜌𝑅𝑇 1.1705(0.08206)(27+273.15)
𝑀𝑊𝑎𝑣𝑒 = = 750 = 29.2142
𝑃
760
Conjugate base will come from 𝐶6 𝐻5 𝐶𝑂𝑂𝑁𝑎 which is determined by amount of 𝑁𝑎𝑂𝐻
29. A student is titrating 50 ml of 0.2 N HCl solution of 0.2 N KOH. He accidentally adds one ml too much
titrant. What is the pH of the resulting solution?
A. 10.3 B. 11.3 C. 2.7 D. 7.3
30. Carbon-carbon double bonds do not undergo rotation as do carbon-carbon single bonds. The reason
is that
A. the double bond is much stronger and thus more difficult to rotate
B. overlap of the sp2 orbitals of the carbon-carbon sigma bond would be lost
C. overlap of the p orbitals of the carbon-carbon pi bond would be lost
D. The shorter bond length of the double bond makes it more difficult for the attached groups to pass
one another
31. What volume of 0.214 M (NH4)2HPO4 is necessary to precipitate calcium as Ca3(PO4)2 from 838 mg of
a sample that is 9.74% Ca?
A. 6.36 ml B. 4.24 ml C. 3.18 ml D. 2.12 ml
𝑚𝑜𝑙
𝑉= 𝑜𝑓 (𝑁𝐻4 )2 𝐻𝑃𝑂4 *check for common ions to determine to correct stochiometric ratio
𝑀
1 1 𝑚𝑜𝑙 𝐶𝑎3 (𝑃𝑂4)2 2 𝑚𝑜𝑙 (𝑁𝐻4 )2 𝐻𝑃𝑂4
0.838(0.0974)(40)( )( 1 𝑚𝑜𝑙 𝐶𝑎 (𝑃𝑂 ) )
3 𝑚𝑜𝑙 𝐶𝑎
𝑉= 3 4 2
= 6.36 × 10−3 𝐿 𝑜𝑟 6.36 𝑚𝐿
0.214
A. B. C. D.
34. The percentage of MnO2 in a 0.50-gram sample, which after addition of 50 mL of 0.10 N ferrous
sulfate solution required 16 mL of 0.80 N potassium dichromate for back titration is
A. 16.7 % B. 32.34 % C. 46.51 % D. 64.68 %
Back titration with 𝐾2 𝐶𝑟2 𝑂7 : 𝐶𝑟2 𝑂72− + 6𝐹𝑒 2+ + 14𝐻 + → 2𝐶𝑟 3+ + 6𝐹𝑒 3+ + 7 𝐻2 𝑂
Observe that 16 mL of 0.80 N is too high of a volume and will result to a negative value, so adjust it to
1.6 mL *These errors in decimal places and/or signs are common for board exam problems
0.1 0.8 6 𝑚𝑜𝑙 𝐹𝑒𝑆𝑂4 1 𝑚𝑜𝑙 𝑀𝑛𝑂2
[50( )−1.6( )( )]( )(86.94)
1 6 1 𝑚𝑜𝑙 𝐾2𝐶𝑟2𝑂7 2 𝑚𝑜𝑙 𝐹𝑒𝑆𝑂4
%𝑀𝑛𝑂2 = (100) = 32.34%
0.5(1000)
35. A sample of pyrite, FeS2, contains only inert impurities and weighs 0.508 grams. After the sample has
been decomposed and dissolved, a precipitate of 1.561-gram BaSO4 is obtained. If the calculated
percentage of S in the sample is 42.21 %, what weight of ignited precipitate would have been
obtained if the Fe in the solution had been precipitated as Fe (OH)3 and ignited as Fe2O3?
A. 0.267 g B. 0.217 g C. 0.2985 g D. 0.3025 g
38. A solution of HCl has a specific gravity of 1.12 and contains 23.81% HCl by weight. How many grams
of HCl are present in each milliliter of the solution?
A. 1.2 B. 12.6 C. 13.2 D. 0.27
Δ𝐺° −4.027(1000)
[𝑃𝐶2 𝐻5𝑂𝐻 ] −( 8.314(650) )
𝐾𝑃 = Δ𝐺° = −𝑅𝑇 ln 𝐾𝑃 𝐾𝑃 = 𝑒 − 𝑅𝑇 = 𝑒 = 2.1068
[𝑃𝐶𝐻3𝐶𝐻𝑂 ][𝑃𝐻2 ]
𝐶𝐻3 𝐶𝐻𝑂 𝐻2 𝐶2 𝐻5 𝑂𝐻
Initial 30 40 0
Equil 30 − 𝑥 40 − 𝑥 𝑥
𝑛 𝑇 𝑎𝑡 𝑒𝑞𝑢𝑖𝑙 = 30 − 𝑥 + 40 − 𝑥 + 𝑥 + 20 + 10 = 100 − 𝑥
𝑛𝐶𝐻3 𝐶𝐻𝑂 30−𝑥 40−𝑥 𝑥
𝑃𝐶𝐻3 𝐶𝐻𝑂 = (𝑃𝑇 ) = (2) 𝑃𝐻2 = (2) 𝑃𝐶2 𝐻5𝑂𝐻 = (2)
𝑛𝑇 100−𝑥 100−𝑥 100−𝑥
30−16.3169
𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝐻3 𝐶𝐻𝑂 𝑖𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 = = 0.1635
100−16.3169
45. Which of the following best explains the relative stabilities of the eclipsed and staggered forms of
ethane? The ____ form has the most _____ strain.
A. Eclipsed; steric C. Eclipsed, torsional
B. Staggered; steric D. Staggered; torsional
46. How many chiral centers are there in 3-bromo-2-pentanol?
A. 2 B. 3 C. 4 D. 1
47. The coolant in a car mobile is 50 % ethylene glycol, C2H6O2 by mass. It has a density of 1.06 g/ml.
What is the molarity of ethylene glycol solution?
A. 4.22 M B. 5.44 M C. 7.88 M D. 8.55 M
100(0.5)
62
Using a basis of 100 g coolant: 𝑀= 100 1 = 8.55
( )
1.06 1000
48. A spoon, with a surface of 45 cm2, is suspended in a cell-filled with a 0.10 M solution of gold(III)
chloride AuCl3. A current of 0.52 A has been passed through the cell, until a coating of gold 0.10 mm
thick, has plated on the spoon. How long did the current run?
A. 3.6 hr B. 6.9 hr C. 11 hr D. 2.3 hr
𝑚𝐴𝑢 𝑚𝑜𝑙 𝑒 −
𝑄 = 𝐼𝑡 = ( )𝐹 𝜌𝐴𝑢 = 19.3 𝑔/𝑐𝑚3 𝑀𝑊𝐴𝑢 = 197
𝑀𝑊𝐴𝑢 𝑚𝑜𝑙 𝐴𝑢
0.1
19.3(45)( 10 ) 3
0.55 𝑡 = ( ) (96485.3365) 𝑡 = 24540.47 𝑠 𝑜𝑟 6.82 ℎ𝑟
197 1
49. A 25 mL sample of 0.025 M HBr is mixed with 25.0 mL of 0.023 M KOH. What is the pH of the resulting
mixture?
A. 1.00 B. 3.00 C. 7.00 D. 11.00
A. 1-ethyl-2-butene C. 3-methyl-2-pentene
B. 3-methyl-3-pentene D. 3-ethyl-2-butene
56. When a sample of impure potassium chloride (0.4500 g) was dissolved in water and treated with an
excess of silver nitrate, 0.8402 g of silver chloride was precipitated. Calculate the percentage KCl in
the original sample.
A. 96.27% B. 92.16% C. 96.12% D. 97.12%
0.8402 1 𝑚𝑜𝑙 𝐾𝐶𝑙
( )(74.45)
143.32 1 𝑚𝑜𝑙 𝐴𝑔𝐶𝑙
% 𝐾𝐶𝑙 = (100) = 96.99%
0.45
57. How many secondary carbon atoms does methyl cyclopropane have?
A. none B. one C. two D. three
58. Which of the following statement is false?
A. In considering chemical equilibrium, the relative stabilities of the products and reactants are
important
B. In considering chemical equilibrium, the pathway from the initial state to the final state is
important
C. In treating reaction rates, the rate at which reactants are converted to products is important
D. In treating reaction rates, the sequence of physical processes by which reactants are converted to
products is important
59. German silver is an alloy of
A. copper, nickel, and zinc C. copper, aluminum, and silver
B. silver, zinc, and aluminum D. silver, nickel, and zinc
60. Calculate the number of gallons of sulfuric acid solution (sp gr. 1.83 and 93 percent by weight H2SO4)
necessary to react completely with 100 lb of borax (Na2B4O7·10H2O) to form boric acid. The
unbalanced equation for this reaction is
𝑁𝑎2 𝐵4 𝑂7 + 𝐻2 𝑆𝑂4 + 𝐻2 𝑂 → 𝐻3 𝐵𝑂3 + 𝑁𝑎𝑆𝑂4
A. 1.2 gal B. 1.8 gal C. 2.4 gal D. 3.1 gal
61. The empirical formula of a commercial ion-exchange resin is C8H7SO3Na. The resin can be used to
soften water according to the reaction.
𝐶𝑎2+ + 2𝐶8 𝐻7 𝑆𝑂3 𝑁𝑎 → (𝐶8 𝐻7 𝑆𝑂3 )2 𝑁𝑎 + 2𝑁𝑎+
What would be the maximum uptake of Ca2+ by the resin expressed in moles per gram of resin?
A. 0.00225 B. 0.0225 C. 0.225 D. 2.25
62. What is the name given for a species that contains a positively charged carbon atom?
A. carbanion B. carbocation C. methyl radical D. free radical
63. The evidence that the solute does not freeze with the solvent is that.
A. the first crystals formed are precipitates of the solute
B. the freezing point of the solution that remains a liquid is getting lower and lower as freezing
proceeds
C. the crystals formed could clearly be seen as that of the solution
D. the freezing point is a constant
64. Saponification is a reaction in which tricylglycerol reacts with a strong base to form
A. glycerol and three soap molecules C. glycerol and three water molecules
B. three fatty acid molecules and water D. oleic acid and water
65. From the thermal decomposition of a pure solid, we obtained a solid and a gas, each of which is a
pure substance. From this information, we can conclude with certainty that
A. the original solid is not an element
B. the solid is a compound and the gas is an element
C. at least one of the products is an element
D. Both products are elements
66. What volume of 0.214 M (NH4)2HPO4 is necessary to precipitate calcium as Ca3(PO4)2 from 838 mg of
a sample that is 9.74% Ca?
A. 6.36 mL B. 4.24 mL C. 3.18 mL D. 2.12 mL
𝑚𝑜𝑙
𝑉= 𝑜𝑓 (𝑁𝐻4 )2 𝐻𝑃𝑂4 *check for common ions to determine to correct stochiometric ratio
𝑀
1 1 𝑚𝑜𝑙 𝐶𝑎3 (𝑃𝑂4)2 2 𝑚𝑜𝑙 (𝑁𝐻4 )2 𝐻𝑃𝑂4
0.838(0.0974)(40)( )( 1 𝑚𝑜𝑙 𝐶𝑎 (𝑃𝑂 ) )
3 𝑚𝑜𝑙 𝐶𝑎
𝑉= 3 4 2
= 6.36 × 10−3 𝐿 𝑜𝑟 6.36 𝑚𝐿
0.214
67. Calculate the pH of a solution prepared by mixing 10 mL of 0.1 N NaOH and 25 mL of 0.1 N HAc.
A. 2.46 B. 12.96 C. 4.57 D. 8.72
𝑐𝑏
𝑝𝐻 = 𝑝𝐾𝑎 + log 𝐾𝑎 𝑜𝑓 𝐻𝐴𝑐 = 1.8 × 10−5
𝑎
Conjugate base will come from 𝑁𝑎𝐴𝑐 which is determined by amount of 𝑁𝑎𝑂𝐻
68. When U-235 is bombarded with one neutron, fission occurs and the products are three neutrons, Kr-
94 and
A. Ba-139 B. Ba-141 C. Ce-139 D. I-142
69. In the compound KMnO4, what is the oxidation number of Mn?
A. +6 B. +7 C. +4 D. +3
70. Which of the following are soluble? I)BaSO4 II) AgBr III) Sr3(NO3)2 IV) PbS V) Na2CO3
A. all are soluble B. III only C. II and V D. I, II, and IV only
71. If an atom has a diameter of 428 pm, what is the volume of 100 atoms?
A. 4.11 x 10-29 m3 C. 4.11 x 10-27 m3
7 3
B. 3.28 x 10 m D. 4.11 x 108 m3
72. A gaseous compound is composed of 85.7% C and 14.3% H. Its density is 2.28 g/L at 300K and 1.00
atm pressure. Determine the molecular formula of the compound. C=12.01, H=1.01
A. C2H4 B. C3H6 C. C4H8 D. C5H10
𝜌𝑅𝑇 2.28(0.08206)(300)
𝑀𝑊𝑎𝑣𝑒 = = = 56.129 𝑔/𝑚𝑜𝑙
𝑃 1
7.1357 14.1584
𝑛𝑜. 𝑜𝑓 𝐶 = ≈4 𝑛𝑜, 𝑜𝑓 𝐻 = ≈8
1.7816 1.7816