Download 105Exam1.W06 answers

Chem 105 Exam #1
Chapters 7-8
Winter, 2006
Show all work, including any unit conversions for full credit. Use correct number of significant
figures.
Soluble
K , Li , Na+, NH4+, NO3-, CH3COOHalides (Halogens)
Sulfates
+
Exceptions:
none
Ag, Pb, Hg
Ba, Sr, Pb, Hg
+
Insoluble
Exceptions:
2-
Carbonates
(CO3 )
Group 1A and NH4+
Phosphates
(PO43-)
Group 1A and NH4+
For colligative properties, t = nKM,
for water, Kfreezing = 1.86 C/M, Kboiling = 0.52C/M
Osmotic pressure = nMRT,
R= 0.0821 L atm/(K moles)
Answers and explanations are in red
The # points for each problem are found in blue
Notes where similar problems can be found are in green. The letter in front refers to Homework
(H), Worksheet (W), or Practice Test (P). The number before the decimal is the chapter number,
after the decimal is the problem number. For instance H5.30 is homework problem #30 in
chapter five. This is not a comprehensive list, but indicates where to go for some similar
problems with perhaps lengthier explanations.
Name_______________________
Chem 105 Exam #1 Chapters 7-8
Winter, 2006
1. While working on an experiment, you find that you need some 10.48 M sulfuric acid. After
checking the stock, there is only 18.00 M sulfuric acid. Tell me how you would mix up
500.0 mL of 10.48 M sulfuric acid from the 18.00 M stock solution. (12 pts) H7.26, H7.48,
W7ii.10, W7ii.11, P7.3, P7-10.5
C1V1 = C2V2
V2 = C1V1/C2
Keep in mind that the 500mL goes with the 10.48M soln. It doesn’t matter which you call C1
and which C2 as long as you keep the correct volume with the concentration. In this case, we
will call the 18 M stuff C2 and solve for how much C2 we need (this will be V2). If you reverse
which V goes with which C, you will get a Volume of 18 M stuff greater than the 500 mL of
10.48 M soln. This would be an answer that does not make sense. You’ve made a mistake; go
back and start again.
V2 = C1V1/C2 = 10.48 M x 0.5000 L/ 18.00 M = 0.2911 L
Take 291.1 mL of 18.00 M sulfuric acid and add distilled water until the total volume = 500.0
mL
2. What, if anything, will be the precipitate when the following solutions are mixed together?
For those that produce no reaction, write NR. Chose one that gave a precipitate and write
balanced molecular, total ionic, and net ionic equations for it. Include physical states (s, l, g,
aq) for all species. (16 pts) H5.30, W7.1, W7.2, P7-10.1
a. SrCl2 (aq)
+
MgSO4 (aq)
 SrSO4 (s) + MgCl2 (aq)
SrSO4 (s) is the precipitate
Sr2+ (aq) + 2 Cl- (aq) + Mg2+ (aq) SO42- (aq)  SrSO4 (s) + Mg2+ (aq) + 2 Cl- (aq)
Sr2+ (aq) + SO42- (aq)  SrSO4 (s)
b. AgCH3COO
(aq)
+
NaNO3 (aq)
NR (Neither acetate (CH3COO-) nor nitrate(NO3-) forms a precipitate with any counter ion.
c. mercury (I) nitrate (aq) + potasium carbonate (aq)
potassium has a plus one charge (group I), as does Hg (roman numeral I). Nitrate has a
minus one charge, and carbonate has a minus 2 charge (see potentially useful
information).
2 HgNO3 (aq) + K2CO3 (aq)  Hg2CO3 (s) + 2 KNO3 (aq)
so Hg2CO3 (s) is the precipitate
2 Hg+(aq) + 2 NO3-(aq) + 2 K+ (aq) + CO32- (aq)  Hg2CO3 (s) + 2 K+ (aq) + 2 NO3- (aq)
2 Hg+(aq) + CO32- (aq)  Hg2CO3 (s)
3. What are the boiling and freezing points of the solutions below? (12 pts) H7.66, P7.4
a. 3.00 M NaBr
Calculate T, then add to the boiling pt and subtract from the solvent’s freezing pt (a solute
increases the temperature range on both ends at which the solvent remains a liquid)
Tf = nKfM = 2 x 1.86 C/M x 3.00 M = 11.2 C
Tf = 0.00 - 11.2 C = -11.2 C
Tb = nKbM = 2 x 0.52 C/M x 3.00 M = 3.1 C
Tb = 100.00 + 3.1 C = 103.1 C
b. 1.49 M Fe(NO3)3
Tf = nKfM = 4 x 1.86 C/M x 1.49 M = 11.1 C
Tf = 0.00 - 11.2 C = -11.1 C
Tb = nKbM = 4 x 0.52 C/M x 1.49 M = 3.1 C
Tb = 100.00 + 3.1 C = 103.1 C
c. 0.27 M glucose (C6H12O6)
Tf = nKfM = 1 x 1.86 C/M x 0.27 M = 0.50 C
Tf = 0.00 – 0.50 C = -0.50 C
Tb = nKbM = 1 x 0.52 C/M x 0.27 M = 0.14 C
Tb = 100.00 + 0.14 C = 100.14 C
4. Draw an energy diagram for a very exothermic reaction (large energy of reaction) with a
small activation energy. Next to it, draw one for a slightly endothermic reaction with a large
activation energy. Indicate which is which. On both diagrams, label both axes, reactants,
products, activation energy (Ea), and energy of reaction (Erxn, also called heat of reaction).
(12 pts) H8.26, W8.1, W8.2, P8.2
I have no program that will allow me to draw this. Maybe I can draw by hand and scan this and
add it. I will look into that. If not done by Weds, than I can do this in class. BTW, if, after you
read these answers and explanations, you still have questions, feel free to ask. It is the only
way you will learn.
5. 23.48 mL of a 6.00 M solution of HCl is used to react with 50.00 mL of a solution of sodium
carbonate of unknown concentration. What is the concentration of sodium carbonate
(Na2CO3)? (12 pts)
H7.56, W7ii.15
2 HCl (aq) +
1 Na2CO3 (aq)  1 H2O (l) + 2 NaCl (aq) + 1 CO2 (g)
We have molarity and volume of HCl so we start there, and end up with sodium carbonate,
which is what the question asked for
L HCl 
moles HCl

moles Na2CO3  molarity Na2CO3
0.2348 L HCl x 6.00 moles HCl/L HCl x 1 Na2CO3/ 2 HCl x 1/ 0.05000L = 1.41 M Na2CO3
6. For each of the situations below, the two solutions are initially at the same level on opposite
sides of a U-shaped tube. Then a semipermeable membrane is placed between the solutions.
Which solution will rise? Don’t forget to show your work. (12 pts) W7ii.14
a. 1.6 M glucose (C6H12O6)
and
0.6 M Li2O
Osmotic pressure (o.p.) = nMRT. Since both liquids are in the same container, T is the same.
Since R is a constant, R is the same for both solutions. To see which solution has higher
osmotic pressure, just compare the osmolarity (nM). Water will flow from low concentration
to high concentration (from low o.p. to high o.p.). I am ignoring sig figs in my molarity
calculations because I am not required to report any numbers. No numbers, hence no s.f.’s
Glucose o.p. = nM = 1 x 1.6 M = 1.6 M
Li2O o.p. = nM = 3 x 0.6 = 1.8 M
The lithium oxide side will rise
b. 0.10 M NaCl
and
nM = 2 x 0.1 M = 0.2M
0.19 M NH4Br
nM = 2 x 0.19 M = 0.38 M
the ammonium bromide side will rise
c. 0.26 M FeI3
nM = 4 x 0.26 = 1.04 M
the iron (III) iodide side will rise
and
0.30 M Ca(NO3)2
nM = 3 x 0.30 M = 0.90 M
7. What is the osmotic pressure of 2.00 L of a 4.78 M solution of Fe2(SO4)3 solution at room
temperature (25.0C)? How many moles of the iron (III) sulfate are present in this solution?
(12 pts) H7.70
OP = nMRT = 5 x 4.78 M x 0.0821 L atm/mole K x 298 K = 585 atm
Molarity = moles/L; so moles = Molarity x L = 4.78 moles/L x 2.00 L = 9.56 moles
8. The reaction below is exothermic and has reached equilibrium. What happens when the
following changes occur? (12 pts) H8.52, H8.56, W8.5, P8.3
H2 (g)
+
Br2 (g)
↔
2 HBr (g)
a. the temperature is decreased
Le Chatelier’s principle says that if you add to one side of an equilibrium, you tilt to the other. If
you remove from one side, you draw the equilibrium towards the side that has stuff removed.
Since the rxn is exothermic, heat can be considered a product. As we are removing product, we
force the equilibrium to the right, towards product.
b. some Bromine gas is added
add to reactant side and we force the equilibrium towards the product side (or right)
c. some Hydrogen gas is removed
remove from reactant side and we force the equilibrium towards the reactant side (or left)
d. a catalyst is added
this has no effect on the equilibrium. Yes, the activation energy is lowered, but the activation
energy for BOTH the forward and reverse direction is lowered. This has no net effect on
equilibrium
Extra Credit:
Invariably, the student studies something that is not asked on the exam. Here is a chance
to make good on that. Write a question on some aspect of chemistry covered in chapters 7
and/or 8 but was not on the exam today. Then, solve the problem. Extra credit points will be
based on clarity, complexity and originality of the problem. (up to 10 pts)