Department of Chemistry, Geosciences, and Environmental Science
Chemistry 1054-010 College Chemistry Fall 2006
Exam 2-A-Key
Name:_______________________________________________________
(Please write your name legibly)
Read all directions and questions carefully!! This exam consists of two parts. The first part consists of 10 multiple choice questions worth four points each for a total of 40 points. The second part consists of five numerical problems worth either 10 or 15 points per question for a total of 60 points. Show all your work necessary for the numerical problems as partial credit will be given for those problems.
Possibly Useful Constants
Avogadro’s Number: NA = 6.022 × 1023 particles/mol
q = (heat capacity)×∆T
q = (mass in grams)×(specific heat)×∆T
q = (number of moles)×(molar heat capacity)×∆T
Calorimetry: qrxn = –Ccalorimeter×∆T or
qrxn = –(mass of solution)×(specific heat of solution)×∆T
Score
Part 1 (40 points):_____________________
Part 2 (60 points):_____________________
Total (100 points):_____________________
Don’t forget to put your name on this test!
Good Luck!!
Part 1
Multiple Choice
Please indicate the answer to each question by putting your choice in the space provided. There is only one correct answer for each question. There will be 10 multiple choice questions worth 4 points each.
1. Which of the following 4 compounds are strong electrolytes?
HCl HC2H3O2
NH3 KCl
(a) All of them. (c) HCl, NH3, and KCl
(b) HCl and KCl (d) HCl and HC2H3O2
(e) KCl
Answer: B
HCl is a strong acid and KCl is an ionic compound that dissolves in water. Both are strong electrolytes. HC2H3O2 is a weak acid and NH3 is a weak base which classify them as weak electrolytes.
BaI2(aq) + Na2SO4(aq) ¾¾® BaSO4(s) + 2 NaI(aq)
2. The spectator ions in the molecular equation shown above are:
(a) Ba2+ only (c) Ba2+ and SO42-
(b) Na+ only (d) Na+ and I¯
(e) SO42- and I¯
Answer: D
The total ionic equation is:
Ba2+(aq) + 2 I¯(aq) + 2 Na+(aq) + SO42-(aq) ¾¾® BaSO4(s) + 2 Na+(aq) + 2 I¯(aq)
The species that remain unchanged on both sides of this reaction are Na+ and I¯.
3. Which is of the following salts is not soluble in water according to the solubility rules?
(a) Na2CO3 (c) Fe(NO3)3
(b) K2SO4 (d) ZnS
(e) NH4Cl
Answer: D
Choices (a) and (b) contain group 1A metal cations. They are soluble. Choice (c) contains the nitrate ion which is soluble. Choice (e) contains the ammonium ion which makes it soluble. Choice (d) is a sulfide which tends to be insoluble except for group 1A and ammonium salts. This is zinc sulfide so this is insoluble.
4. Oxidation cannot occur without ____________________.
(a) acid (c) water
(b) oxygen (d) air
(e) reduction
Answer: E
5. The concentration of a solution prepared by diluting 0.200 L of a 2.00 M solution to 0.800 L is:
(a) 0.800 M (c) 0.500 M
(b) 0.200 M (d) 8.00 M
(e) 0.400 M
Answer: C
Using the dilution formula: 
6. The value of DE for a system that performs 213 J of work on the surroundings and loses 79 J of heat is equal to:
(a) 292 J (c) 134 J
(b) –134 J (d) –292 J
(e) 0 J
Answer: D
From the problem: q = –79 J and w = –213 J. DE = q + w = –79 J + – 213 J = –292 J
7. What is the value of DH for the reaction:
N2(g) + O2(g) ¾¾® 2 NO(g)
given the two thermochemical equations below:
N2(g) + 2 O2(g) ¾¾® 2 NO2(g) DH = 66.4 kJ
2 NO(g) + O2(g) ¾¾® 2 NO2(g) DH = –114.2 kJ
(a) 180.6 kJ (c) 47.8 kJ
(b) –47.8 kJ (d) 90.3 kJ
(e) –180.6 kJ
Answer: A
The reaction in the problem is equal to the 1st thermochemical equation plus the reverse of the second thermochemical equation. Using Hess’s Law: DH = 66.4 kJ + 114.2 kJ = 180.6 kJ
8. A _____________ value of DH corresponds to an _______________ process.
(a) negative, endothermic (c) positive, exothermic
(b) zero, exothermic (d) negative, exothermic
(e) zero, endothermic
Answer: D
9. The substance with the highest fuel value is:
(a) coal (c) gasoline
(b) wood (d) hydrogen
(e) natural gas
Answer: D
|
Substance |
DHf° (kJ mol-1) |
|
|
SO2(g) |
–297 |
|
|
SO3(g) |
–396 |
|
10. Given the data in the table above, the DH° for the reaction:
2 SO2(g) + O2(g) ¾¾® 2 SO3(g)
is equal to:
(a) –99 kJ (c) –198 kJ
(b) 198 kJ (d) 99 kJ
(e) The DHf° for O2 needs to be given before DH° can be calculated.
Answer:
DH° =( 2 mol)(DHf°(SO3(g)) – (2 mol)(DHf°(SO2(g)) – (1 mol)(DHf°(O2(g))
=(2 mol)(–396 kJ/mol) – (2 mol)(–297 kJ/mol)
= –792 kJ + 594 kJ = –198 kJ
Since O2(g) is an element in its most stable state, its DHf° is equal to exactly zero by definition.
Part 2
Numerical Problems
Solve the following problems, keeping track of significant figures where applicable. Please show all the work necessary to obtain your answer in order to receive partial credit for possibly wrong answers. Generally, full credit will not be given for the correct answer without any of the work performed to obtain the answer being shown on the paper. Each question is worth either 10 or 15 points.
11. (10 points) How many milliliters of 0.240 M CuSO4 solution contains 2.25 g of the solute?
Convert 2.25 g CuSO4 into moles of CuSO4:
Calculate the volume of the solution required in liters:
Convert to mL’s:
Volume in mL = (0.0587 L)(1000 mL/L) = 58.7 mL (3 sig figs)
12. (15 points) A 1.500 g sample of hexane, C6H14(l), was combusted in a bomb calorimeter with a heat capacity of 34.75 kJ/°C. The temperature of the calorimeter increased from 25.000°C to 27.085°C during the combustion. What is the heat of combustion for 1.500 g of hexane, the heat of combustion per gram of hexane, and the heat of combustion per mole of hexane.
Calculate the temperature change: DT = 27.085°C — 25.000°C = 2.085°C
Calculate the heat of combustion for 1.500 g of hexane using the equation qrxn = –Ccalorimeter×DT:
qrxn = –(34.75 kJ/°C)(2.085°C) = –72.45 kJ (4 sig figs)
To calculate the heat of combustion per gram of hexane, divide the heat of reaction by the number of grams of hexane:
To calculate the heat of combustion per mole of hexane, first calculate the number of moles of hexane and divide the heat of reaction by the number of moles of hexane:
13. (10 points) Calculate the final temperature when 3125 J of heat are added to 22.5 g of water at 20.5°C. The specific heat of water is equal to 4.18 J/(g·°C).
The relationship between heat and specific heat is: q = (mass in grams)×(specific heat)×DT
The following data is given in the problem: q = 3125 J, mass in grams = 22.5 g, specific heat = 4.18 J/(g·°C). Solving for DT:
Calculate the final temperature: DT = Tfinal – Tinitial
33.2°C = Tfinal – 20.5°C
Tfinal = 53.7°C
14. (10 points) Write the net ionic equations for the following reactions:
(a) MgSO4(aq) + Na2CO3(aq) ¾¾® MgCO3(s) + Na2SO4(aq)
The total ionic equation is:
Mg2+(aq) + SO42-(aq) + 2 Na+(aq) + CO32-(aq) ¾¾® MgCO3(s) + 2 Na+(aq) + SO42-(aq)
The spectator ions in this equation are Na+ and SO42-. Removing them yields the net ionic equation:
Mg2+(aq) + CO32-(aq) ¾¾® MgCO3(s)
(b) Mg(s) + SnCl2(aq) ¾¾® MgCl2(aq) + Sn(s)
The total ionic equation is:
Mg(s) + Sn2+(aq) + 2 Cl¯(aq) ¾¾® Mg2+(aq) + 2 Cl¯(aq) + Sn(s)
The spectator ion is Cl¯. Remove it to get the net ionic equation:
Mg(s) + Sn2+(aq) ¾¾® Mg2+(aq) + Sn(s)
(c) Cu(OH)2(s) + 2 HCl(aq) ¾¾® CuCl2(aq) + 2 H2O(l)
The total ionic equation is:
Cu(OH)2(s) + 2 H+(aq) + 2 Cl¯(aq) ¾¾® Cu2+(aq) + 2 Cl¯(aq) + 2 H2O(l)
The spectator ion is Cl¯. Remove it to get the net ionic equation:
Cu(OH)2(s) + 2 H+(aq) ¾¾® Cu2+(aq) + 2 H2O(l)
15. (15 points) How many milliliters of a 0.1158 M NaOH solution are required to completely neutralize 25.00 mL of a 0.05135 M H3PO4 solution? The neutralization reaction is:
H3PO4(aq) + 3 NaOH(aq) ¾¾® Na3PO4(aq) + 3 H2O(l)
Solution stoichiometry problem!!
Known: 25.00 mL of 0.05135 M H3PO4
Unknown: ?? mL of 0.1158 M NaOH
Calculate moles of H3PO4 contained in 25.00 mL of 0.05135 M H3PO4:
Convert moles of H3PO4 into moles of NaOH using the balanced chemical equation:
Calculate how many mL’s of 0.1158 M NaOH solution contain 0.003851 mol of NaOH:
(10 points extra credit) Give the complete definition for 2 out of the 3 terms below. 5 points will be given for each correct complete definition. Clearly note which 2 definitions you want graded, if you write three definitions without a clear notation, only the first two will be graded.
Oxidizing agent: a substance that causes another substance to get oxidized.
Endothermic process: a process that absorbs heat.
Endpoint of a titration: the point in a titration in which an indicator changes color.