Chemistry 1A, Spring 2009 V.2 Final Exam Practice Set and Solutions
Final Note: First off, I’d like to apologize for the longer delay. Word was being very stubborn about not saving my work properly. In the limited amount of time I had, I put together the computationally heavy problems and solutions in this last quarter of the class. There is no review from the previous sections from the course mainly because the resources for those sections are still available. That’s pretty much all I have to say about this practice set. I really enjoyed meeting many of you during this semester and hope to see you guys again, on Sproul or wherever. This will probably also be the last practice set and student review sessions that I’ll hold for the rest of my college career since my time in the upcoming years will be devoted towards club officer duties in NSU. With that, I wish all of you the best of luck for all your finals and hope you guys also have a great summer vacation given that you’re not attending summer sessions like I am ^_^ . ~Stephen Kok
Student Review Session s The review session has already past, but I will be available through the bSpace chatroom and email.
©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
1 Complete the dissociation reaction of the strong acid, HNO3 Nitric Acid . Identify the acid‐base and conjugate acid‐base pairs. __________ ___ __________ ___ ___________ ___ 2 Explain why the usage of the single reaction arrow instead of the equilibrium arrows is properly justified. 3 Consider a 25mL sample of 0.052M HNO3 at 25°C, calculate the pOH of the solution. Calculate the pH of the solution after diluting it with 50mL of pure water H2O at 25°C. Before: After: Circle the most appropriate choice 4 The Kb of HNO3 is: 5 The pKa of HNO3 is: 6 Explain your responses:
©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
7 Complete the dissociation reaction of weak base, NH3 ammonia , in water. Identify the acid‐base and conjugate acid‐base pairs. __________ ___ __________ ___ ___________ ___
8 Explain why the equilibrium arrows were used instead of the single direction reaction arrow as in the previous problem.
9 Consider a 55mL sample of 0.085M NH3 at 25°C, calculate the pH of the solution. Calculate the pOH of the solution after adding 85mL of 0.012M NH4Cl. NH3/NH4 pKb 4.744 Before: After: 10 Rank the following acids in order of decreasing acidity without the aid of a table. Use 1‐4 HCl HF HI HBr 11 Explain your ranking: ©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
For the following two aqueous samples, calculate the initial mass of compound. 12 531mL sample of carbonic acid H2CO3, Ka 4.3 x 10‐7 with a pH of 6.14. 13 92mL sample of hydroxylamine NH2OH, Kb 1.1 x 10‐8 with a pH of 8.2 14 Consider a sample of 0.27M H2SeO4, a strong polyprotic acid with a very small pKa1 value and a pKa2 value of 0.973. Write the acid‐base reaction for the two deprotonations for the compound and their corresponding equilibrium acidity quotient. Make sure to indicate the correct direction of reaction, using either single arrows or equilibrium arrows. Ka1 Ka2 15 Calculate the pH for this solution. ©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
16 Consider an aqueous solution of 0.34M . Calculate the concentration for solute species: , , , 0 , 1.3 · 10 | 7.1 · 10 17 In the previous problem, the concentration of H2S and the concentrations of the solute species following the first deprotonation reaction barely changed after undergoing the second deprotonation reaction. Explain how this result may have been predicted without calculations. Calculate the pH for the following buffers at 25°C 18 A solution of 0.78M and 0.52M / : 3.37 19 A solution of 0.64M and 0.42M / : 7.21 ©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
20 Consider an aqueous solution of 0.85M . What volume of 0.92M would be necessary to prepare a buffer solution with a pH of 3.24? / : 2.00 . 21 Balance the following skeletal equation of the reaction between permanganate ions and sulfide ions within a basic solution using half‐reactions. Oxidation Half‐Reaction: Reduction Half‐Reaction: Combined Reaction: ©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
22 Balance the following skeletal equation of the reaction between hydrazine and chlorine ions within an acidic solution using half‐reactions. Oxidation Half‐Reaction: Reduction Half‐Reaction: Combined Reaction: 23 Calculate the equilibrium constant, K, for the following reaction at 25°C: 2 1.69 2 2 1 25° / 0.025693 24 A silver electrode in 0.032M Ag NO3 is connected to a hydrogen electrode in which the pressure is 3.5 bar. If the cell potential is 0.081V at 25°C, what is the pH of the electrolyte? | 0.80 , | 0 ©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
1 Complete the dissociation reaction of the strong acid, HNO3 Nitric Acid . Identify the acid‐base and conjugate acid‐base pairs. Conjugate Acid Base Conjugate Acid Base 2 Explain why the usage of the single reaction arrow instead of the equilibrium arrows is properly justified. The particular acid is identified as “strong” which means that it will completely dissociate into its ions when in water. Thus the reaction theoretically goes to completion changing all reactants into the products. 3 Consider a 25mL sample of 0.052M HNO3 at 25°C, calculate the pOH of the solution. Calculate the pH of the solution after diluting it with 50mL of pure water H2O at 25°C. Before: Since the reaction essentially goes to completion, we can assume that the moles of H3O and NO3‐ are equal to the moles of HNO3. Since the volume did not change, the concentrations are also the same, thus: 1.0 10 1.0 10 1.92 10 log 12.7 After: Again, we can assume that the moles of H3O is the same as the moles of HNO3, but we have to take into account the new volume due to the addition of water. 25 0.0520 0.0013 1000 0.0013 0.0173 25 50 1000 1.76 log Circle the most appropriate choice 4 The Kb of HNO3 is: 5 The pKa of HNO3 is: 6 Explain your responses: Kb is the basicity constant, but since we’re dealing with an acid, we can only derive the Ka from the reaction which is:
Since the reaction essentially goes to completion, large. Using the following relationship we can derive Kb ©2009 Stephen Kok UC Berkeley Undergraduate
0, thus Ka would be very
General Chemistry 1A Spring 2009 Professor Nitsche
If Ka is very large, then Kb would be very small since Kw stays constant as long as the temperature does not change. The pKa is simply – log where we found Ka to be very large. The logarithm function log takes on values from essentially negative infinity to 0 when we consider the values of x between 0 and 1 normal range of K values . Due to the negative sign, the negative values become positive values, where the smallest the x value, the larger its logarithmic value will be. The inverse applies as well, so since we have a large Ka value, then the logarithmic value will be very small. 7 Complete the dissociation reaction of weak base, NH3 ammonia , in water. Identify the acid‐base and conjugate acid‐base pairs. Conjugate Base Acid Conjugate Acid Base 8 Explain why the equilibrium arrows were used instead of the single direction reaction arrow as in the previous problem. Unlike strong solutions, weak solutions don’t completely dissociate into its ions, in that there is an equilibrium formed between the molecular form of the base and its product ions. Thus the double arrows indicate a dynamic reaction that is occurring at a particular ratio.
9 Consider a 55mL sample of 0.085M NH3 at 25°C, calculate the pH of the solution. Calculate the pOH of the solution after adding 85mL of 0.012M NH4Cl. Dissociates completely Before: Since we’re dealing with a weak base, the reaction will not go to completion, and thus prompts the usage of the “ICE” method. Reaction Initial Molarity 0.085 ‐ 0 0 Change in Molarity ‐x ‐ x x Equilibrium Molarity
0.085
‐
x
x
Note: Water will not contribute anything because it involves the equilibrium constant Now to explain this chart. In the first row, “Initial Molarity”, you will put the initial concentrations of the various compounds, where in this case, you only started with a solution of 0.085M NH3. In “Change of Molarity”, you will indicate how the reactants and products will change following the reaction. Since we do not know the exact amount at the time, we let the value x stand for the amount of change. Since we know that NH3 will reactant to a small extent, we denote its change by –x, for the drop in its concentration following the reaction. Since this reaction is already balanced, we see that for every 1 mole of NH3 that reacts, there will be 1 mole of NH4 and OH‐ produced. Thus we mark their changes by x, since the amount of NH3 that reacts will indicate the amount of NH4 and ©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
OH‐ produced. Finally, like a mathematical addition table, we simply add down the columns. This gives us the “Equilibrium Molarity” which will directly tie into the equilibrium basicity constant as follows: Since we know the equilibrium concentration, we simply replace them in the constant along with the Kb given in the problem indirectly. log 10 1.80 10 1.80 10
0.085 0.085 Now this is very important, especially if you want to avoid using the quadratic equation properly. The value x indicates how much of the reactants will be consumed and how much of the products will be produced. The extent to which this occurs depends on the value of K. In this case, since we’re dealing with the basicity constant, it will deal with the magnitude size of Kb. If Kb is large, that means the reaction essentially goes to completion and x is very large and cannot be ignored. If Kb is small, then there will be barely any products produced in proportion to the reactants. This would mean that x would be very small and would be possible to ignore in terms of quick calculations. Now here’s a good rule of thumb to follow based on the value of K. 1 · 10 , 1 · 10 , Basically, if the K value has a power equal to or greater than ‐3, then you should be using the quadratic equation because the value of x will be pretty significant at that point. So since our Kb value is only at the ‐5th power, we can safely ignore the –x from the denominator you cannot ignore the x’s on the numerator because you’re solving for them . 1.80 10
0.085 0.085 1.80 10 0.085 0.00124 14 14 log 0.00124 2.91 log 14 2.91 11.1 After: This problem is similar to the one above, except only with a slight twist. Since we’ll be using the “ICE” table to compare concentrations again, we’ll need to calculate the new concentrations following the addition of the NH4Cl solution, which will dissociate into its ions NH4 and Cl‐. We will only worry about the NH4 ions since the chloride ions do not show up in the net ionic equation and thus will play no role in our calculations. Solving for new concentrations 55 0.085 0.004675 1000 0.004675 0.0334 55 85 1000 85 0.012 0.00102 1000
©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
0.00102 55 85 1000
0.00729
Now applying the new values to the “ICE” table: Reaction Initial Molarity Change in Molarity Equilibrium Molarity
0.0334 ‐x 0.0334
‐ ‐ ‐
0.00729 x 0.00729
0 x
x
Notice that there is an initial concentration of NH4 ions due to the addition of NH4Cl. Besides this point, the calculation is practically the same. 0.00729 1.80 10 0.0334 Again, we can ignore x values that will make the calculations more difficult since the K value is small enough. Remember you shouldn’t remove significant x values or else that’ll just make your calculation obsolete. Generally aim for those x values that are being added or subtracted from initial values. 1.80 10 0.0334 0.00729 1.80 10 8.247 10 0.00729 0.0334 log log 4.08 If you want to check your value, you can calculate the pH of this solution, which is 9.91, and compare it to the solution before the NH4Cl solution was added. From 11.1, you see a drop in pH, which makes sense since by adding NH4Cl, you added NH4 ions, which means by le Chatelier’s principle, the reaction is pushed back towards the reactants side, and would mean the decrease of OH‐ ions which would mean a lower pH. 10 Rank the following acids in order of decreasing acidity without the aid of a table. Use 1‐4 HCl HF HI HBr 4 2 3 1
11 Explain your ranking: The strength of an acid depends on the polarity of the molecule, in which the more electronegative the anion is, the stronger the acid becomes. This is because the hydrogen atom becomes more positively charged, which allows for greater attraction to the water molecules to form the hydronium ion. However, if all the acids have anions in the same group, then the acids would become stronger as you went down the group. This is because the bonding between the hydrogen atom and the element would become weaker as the element becomes more massive. A weaker bond would mean that it’ll be easier for the molecule to dissociate, and thus would be by definition, a stronger acid.
©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
For the following two aqueous samples at 25°C, calculate the initial mass of compound. 12 52mL sample of carbonic acid H2CO3, Ka 4.3 x 10‐7 with a pH of 2.64. It’s always good to start out with the chemical equation to guide you through the process.
First off, we can interpret the pH value in terms of the concentration of hydronium ions log 10 Assuming that this reaction acts stoichiometrically, that is for every molecule of H2CO3 consumed, a molecule of HCO3‐ and a molecule of H3O are formed: Hence, shifting the equilibrium acidity quotient around: 10 . 12.2 4.3 10 Now comes simply calculations to find the mass of the compound 1 52 62 12.2 | 39.3 1 1000 1 13 92mL sample of hydroxylamine NH2OH, Kb 1.1 x 10‐8 with a pH of 10.2 Starting with the chemical equation Interpret pH in terms of
14
log
10
. 10 1.1 10
2.28 1
1 1000
92
|
33 1
6.92
2.28
14 Consider a sample of 0.27M H2SeO4, a strong polyprotic acid with a very small pKa1 value and a pKa2 value of 1.92. Write the acid‐base reaction for the two deprotonations for the compound and their corresponding equilibrium acidity quotient. Make sure to indicate the correct direction of reaction, using either single arrows or equilibrium arrows.
©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
15 Calculate the pH for this solution. For polyprotic solutions, calculations of pH values will generally require 2 or more steps since each deprotonation reaction must be taken account of. In this case, our first calculation is quite simply since the first reaction goes essentially to equilibrium due to the high pKa1 value. To more clearly clarify this point: 1 log 10 10 , 0 : 0 1 lim ∞ 10 Or in other words, Ka1 is very large. What this means is that the concentration of the products essentially overwhelm the concentration of the reactants. As shown the equilibrium quotient. If Ka1 is very large, then must be very small and/or are very large. Basically this just proves that the reaction essentially goes to completion and the assumption that all the reactants went into products is true. Hence: 0.27 We now have the concentration of hydronium ions from the first reaction, however, the conjugate base that formed , can act as an acid for a second deprotonation reaction where more hydronium ions would be formed. Referring back to the second reaction: 10
1.2 · 10
For this reaction, we must determine the equilibrium concentration of using the “ICE” method, since this reaction does not go to completion. It reaches a dynamic equilibrium between the concentration of products to the concentration of reactants related by the Ka2 value. Hence, setting up the table:
Reaction Initial Molarity Change in Molarity Equilibrium Molarity
0.27 ‐x 0.27
‐ ‐ ‐
0 x
0.27 x 0.27
The initial concentration of the acid of interest, , and the hydronium ions comes from the calculations before in which the products of the previous reactions supplies relevant compounds to the second reaction. Moving on to the equilibrium quotient: 0.27 1.2 · 10 0.27 Again, we have the unknown x values which we are solving for. However! This time, our K value is significantly large, which means we shouldn’t assume that the value of x is ©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
negligible. Hence we’ll multiply out and use the quadratic equation: 0.27 0.27 0.27 1.2 · 10 0.27 1.2 · 10 0.27 0.27 0.27 0.27 1.2 · 10 1.2 · 10 0 0.282 0.00324 0 1, 0.282, 0.00324 4 √ 2 0.011, 0.293 0.27 0.281 log 0.551 To give you a general idea of the difference if you assumed x to be negligible: 0.27 0.27 1.2 · 10 1.2 · 10 1.2 · 10 0.012 0.27 0.27 0.012 0.011 In this particular problem, the difference was only slight, however, you should be careful in terms of assumptions for problems since the answer may be significantly different if you fail to recognize situations in which assumptions should not be used. Time‐Saving Strategy: You should only use this strategy if you’re confident with assumptions. Given a polyprotic solution which you need to calculate the pH for, if the Ka2 and Ka3, if there is one, are small enough to allow the quadratic equation bypass, you can generally roughly estimate the pH based on the first deprotonation reaction since the following reactions will only give a slightly higher or lower value depending on whether it’s a polyprotic acid or base . However, if it’s a free‐response problem, you should show all your steps.
16 Consider an aqueous solution of 0.34M . Calculate the concentration for solute species: , , , 0 , 1.3 · 10 | 7.1 · 10 First consider the two deprotonation reactions along with their equilibrium quotients.
Since the reaction reaches a dynamic equilibrium, we cannot assume that the reaction goes to completion. Moreover, it will be shown that it is actually the reverse. Using the “ICE” method to solve for equilibrium concentrations. Reaction Initial Molarity 0.34 ‐ 0 0 Change in Molarity ‐x ‐ x x ©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
Equilibrium Molarity
0.27
‐
Setting up appropriate values in the quotient 1.3 · 10 0.27 Since the K value is significantly small, we can assume that the value of x is negligible. 0.27
0.27
1.3 · 10
3.51 · 10
3.51 · 10 0.27 0.27 Note that the concentrations for and are only initial values, this is because they participate in the second deprotonation reaction and their concentration values will change accordingly. Thus brings us to the second “ICE” table for the second reaction. Reaction Initial Molarity ‐ 0 3.51 · 10 3.51 · 10 Change in Molarity ‐y ‐ y y Equilibrium Molarity
3.51 · 10
‐
3.51 · 10
The variable y was used in place of the standard x because they represent different values and should not be mixed together. Hopefully you won’t make such a mistake. Regular procedure: 3.51 · 10 7.1 · 10 3.51 · 10 K value is extremely small, the value of y is definitely negligible. 3.51 · 10 3.51 · 10 7.1 · 10 3.51 · 10 3.51 · 10 7.1 · 10 3.51 · 10 3.51 · 10 3.51 · 10 3.51 · 10 Using the autoprotolysis constant for water to solve for the concentration of hydroxide. 1 · 10 1 · 10 2.85 · 10 0 Equilibrium Concentrations 0.27 2.85 · 10 3.51 · 10 3.51 · 10 7.1 · 10
17 In the previous problem, the concentrations of the solute species following the first deprotonation reaction hardly changed after undergoing the second deprotonation reaction. Explain how this result may have been predicted without calculations. Considering the equilibrium constants for the two deprotonation reactions. 1.3 · 10 | 7.1 · 10 The first constant was already significantly small. To review on equilibrium constants, they’re numerical representations of the extent of reactions. If the K value is very high, that ©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
means the concentration of the products is significantly larger than the concentration of the reactants, and vice‐versa. In this case, we could predict that the concentration of the products would have been much less than the concentration of the product, which explains why the initial concentration of the main polyprotic acidic compound did not change by a significant amount. The same explanation may be given for the concentrations of the compounds after the second deprotonation reaction, in which the reaction barely proceeded, producing barely any of the products. In the case of multiple‐choice problems, such judgments could be made given appropriate conditions such as these, and would also save a decent amount of time.
Calculate the pH for the following buffers at 25°C
18 A solution of 0.78M and 0.52M / : 3.37 Always start out with the chemical equation and the equilibrium coefficient.
log
.
10
4.26 10
Setting up an “ICE” table should be natural by now. Set the initial concentrations into the table. Note for the change in molarity, unless you’re very good and judging which way the reaction would like to go, it won’t matter whether you assume that the reaction is going towards the product side losing HNO2 molecules or going towards the reactant side gaining HNO2 molecules . If the direction is wrong, the value of x will simply be negative. Reaction Initial Molarity 0.78 ‐ 0.52 0 Change in Molarity ‐x ‐ x x Equilibrium Molarity
0.78
‐
0.52
0.52 4.26 10 0.78 Since Ka is sufficiently small, we can assume that the value of x is negligible. 0.52 4.26 10 6.39 10 0.78 log 3.19
19 A solution of 0.64M Chemical Equation
and 0.42M
/
:
7.21
Note that you can set up this problem for the deprotonation reaction rather than the protonation reaction and still get the same result, just use Ka instead. log 10 . 6.17 10 ©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
0.64 ‐x
Reaction Initial Molarity Change in Molarity Equilibrium Molarity
‐ ‐
0.64
0.42 x
‐
0.42
0 x
0.64 6.17 10 0.42 is sufficiently small, we can assume that the value of x is negligible. 0.64 6.17 10 4.05 10 0.42 log 7.39 14 6.61
Once again, since
20 Consider an 1.00L aqueous solution of 0.85M . What volume of 0.92M would be necessary to prepare a buffer solution with a pH of 2.21? / : 2.00 . First off, start with the chemical reaction and equilibrium quotient: Let us isolate
since we’re dealing with the pH of the buffer.
log
log
log
log
log log
Fortunately, the above equation will be provided on the exam, so you don’t have to worry about deriving it every time you need to use it. This was just for informational purposes. First off we should solve for the ratio of the concentrations required for the particular buffer pH. 2.21
2.00
log 10
Next, we’ll calculate the moles of 0.85
0.21 .
0.85
1.62
log
present in the solution. 1
0.85
Since concentrations of both compounds will be in the same solution, the volumes cancel ©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
out and gives the following: Solving for the volume of required: 1.38 0.92
1.62
0.85 1.38
1.5
21 Balance the following skeletal equation of the reaction between permanganate ions and sulfide ions within a basic solution using half‐reactions. Oxidation Half‐Reaction: No hydrogen or oxygen atoms to take care of, thus only balance the charges: 2 Reduction Half‐Reaction: Balance oxygen atoms using water molecules
2 Rebalance the hydrogen atoms by adding water to the side requiring additional hydrogen atoms and adding a hydroxide ion on the other side 4 4 2 Cancelling out repeated molecules 2 4 Balancing charges 3 2 4 Combined Reaction: Retrieved half‐reaction, multiply by a coefficient to match their number of electrons. 3 2 2 2 3 4 New Half‐Reactions 3 3 6 4 8 2 6 2 Electrons cancel out and form a balanced redox equation 2 4 8 3 6
22 Balance the following skeletal equation of the reaction between hydrazine and chlorine ions within an acidic solution using half‐reactions. Oxidation Half‐Reaction: First balance out the nitrogen atoms
2 Add water molecules to balance out the oxygen atoms
©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
2 Add hydrogen ions to balance out hydrogen atoms 2 2 Balance out charges 2 2 Reduction Half‐Reaction:
2 8 8
8
Add water molecules to balance out oxygen atoms
3 Add hydrogen atoms to balance out hydrogen atoms 6 3 Balance out charges 6 6 3 Combined Reaction: Retrieved half‐reactions, multiply by coefficient 22 2 8 8 3 6 6 3 New Half‐reactions 4 2 4 16 16 3 3 18 18 9 Combine half‐reactions to form full reaction electrons cancelled out 4
2
3
4
18
16
9
3
Cancel out repeated molecules to form balance equation 2 3 2 4 3 5 23 Calculate the equilibrium constant, K, for the following reaction at 25°C: 2 2 2 1.69 1 25° / 0.025693 As defined earlier in the book Δ ° And recently defined Δ ° ° Hence: ° °
.
1.358 10 With such a large K value, this reaction essentially goes to completion. .
24 A silver electrode in 0.032M Ag NO3 is connected to a hydrogen electrode in which the pressure is 3.5 bar. If the cell potential is 0.081V at 25°C, what is the pH of the electrolyte? 0.80 , 2 0 This problem requires the use of the Nernst equation given as followed: 0.0257
©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche
First to set up the redox half‐reactions 2 2 2 2 2 Balanced full reaction and determine cell potential 2 2 2
0.80 0.00 0
0.80
0.80
0.0257 0.081
0.80 0.719
ln
ln
0.0257 ln 2 0.01285 ln
0.032
3.5
0.003584
55.95 ln 0.03584 2ln 1 55.95 ln 0.03584 2 1.342 10 log 12.87
29.64
©2009 Stephen Kok UC Berkeley Undergraduate
General Chemistry 1A Spring 2009 Professor Nitsche