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209 Chemistry Chapter 4: Homework #4 answers 4 (a) Strategy: what's the easiest thing to keep track of? The most complicated item; that is the starting material SO2 Cl2 , as any changes in the amounts of that stuff require that I rebalance 3 of the 4 product molecules. HI I will worry about last, as it looks like it is the source of H atoms that appear in everything on the right except the I2 ; atoms of I, because they are really just spectators to the rest of the reaction I'll worry about last when we have to count up how many spectators there are at this reaction. So we'll start assuming that there is a single unit of the SO2 Cl2 , as see how much trouble that gets us in; as a first guess, I'll write down SO2 Cl2 + HI H2 S + 2 H2 O + 2 HCl + I2 where represents a coefficient to be determined later. Like, now; counting up the H atoms on the right side, I've got 2 in the H2 S molecule, 4 in the two H2 O molecules, and 2 more in the 2 HCl molecules, for a total of 8; that means that I need 8 units of HI (my only source for H) and that leaves me with 8 units of I, or 4 I2 molecules, as passengers along for the ride. Ending with SO2 Cl2 + 8 HI H2 S + 2 H2 O + 2 HCl + 4 I2 (b) Strategy: Count what we have no other source for first; that would be the FeTiO3 ; one unit of which must yield one unit of each of the two product species. So start with FeTiO3 + H2 SO4 + H2 O FeSO4 7 H2 O + TiOSO4 Next, we need enough of the SO4 units to go around; that looks like two on the right, so we need two on the left, giving FeTiO3 + 2 H2 SO4 + H2 O FeSO4 7 H2 O + TiOSO4 Now it's time to ask whether we have atom balances that the added water is meant to fix up; the Fe, Ti and S atoms are all in balance on both sides of the equation; on the product side we have a total of 14 H atoms and 16 O atoms. So far the reactants contribute 3 + 2 4 = 11 O atoms, and 4 H atoms; so we need 10 more H atoms and 5 more O atoms on the reactant side which conveniently enough is provided by exactly 5 water molecules, leading to FeTiO3 + 2 H2 SO4 + 5 H2 O FeSO4 7 H2 O + TiOSO4 (d) This one is a simple combustion problem; C, H and S burn in O to form CO2 , H2 O and SO2 . Simpler if we rewrite the first molecule as C14 H14 S2 ; then C14 H14 S2 + O2 CO2 + H2 O + SO2 Sequentially, there is only one source of C, and one product containing C; so all 14 C units need to be found in 14 CO2 units; similarly, one source of H, and one product; so all 14 H units must appear in 7 H2 O units, and one source of S (2 units), all of which shows up in SO2 (2 units). Counting the O units, we require 14 2 + 7 1 + 2 2 = 39 units of O, or 39 units of O2 , and a balanced equation 2 C14 H14 S2 + 39 O2 14CO2 + 7H2 O + 2SO2 2 1 16 The reaction we are studying is 3 Fe(s) + 4 H2 O(g)-Fe3 O4 (s) + 4H2 (g) (a) 42.7 g of Fe contains (42.7 g) (1 mole Fe/55.847 g Fe) = 0.7646 moles of Fe. (We keep the extra significant figure until the end of the calculation.) For each 3 moles of Fe we produce 4 moles of hydrogen gas; so 0.7647 moles Fe (4 moles H2 /3 moles Fe)= 1.019 1.02 moles of H2 gas is produced if we completely use up the Fe provided as reactant. (b) Step 1; convert to moles. 63.5 g Fe (1 mole Fe/55.847 g Fe) = 1.137 moles of Fe. (We keep the extra significant figure until the end of the calculation.) For each 3 moles of Fe we require 4 moles of steam; so 1.137 moles Fe (4 moles H2 O/3 moles Fe)= 1.516 1.52 moles of H2 O gas is required to completely use up the Fe provided as reactant. (c) 7.36 moles of H2 gas are produced. Our balanced equation states that we form 1 mole Fe3 O4 for every 4 moles of H2 gas produced; so that we expect 1 7.36 = 1.84 moles of 4 Fe3 O4 . Each mole of the product has mass 3 55.847 + 4 15.9994 = 231.5836 grams; 1.84 moles has mass 1.84 231.5836 = 426.031024 426 grams of Fe3 O4 . 50 9.13 g of KI corresponds to 9.13 g KI 1 mole KI/166.002 g KI = 0.0550 moles KI (to 3 significant digits). Our balanced chemical equation claims that 2 moles of KMnO4 are sufficient to react with 10 moles of KI; so we need 0.0110 moles KMnO4 to take care of the KI in this problem. Our solution is 0.0797 moles KMnO4 /liter; in order to get 0.0110 moles, we require 0.0110 moles KMnO4 /(0.0797 moles KMnO4 /liter solution) = 0.138 liters of solution, or 138 milliliters of solution. 62 First things first; we need a balanced chemical equation. Hypochlorite corresponds to the OCl- anion; the cation is Ca2+ so we need two hypochlorites/formula unit. So the net chemical reaction should look something like Ca(OCl)2 (s) + 4HCl(aq) 2Cl2 (g) + 2H2 O(l) + CaCl2 (s) (a) 50.0 grams of calcium hypochlorite corresponds to how many moles? The molecular weight of the calcium hypochlorite is 40.08+2 (35.4527+15.9994) = 142.9822 g/mole our 50.0 grams therefore corresponds to 50.0/142.9822 = 0.350 moles of calcium hypochlorite. 275 milliliters of 6.0 M HCl correspond to 0.275liters 6.0moles/liter = 1.65 moles of HCl. Our balanced chemical equation predicts that we will need 4 moles of the acid every for mole of the hypochlorite salt; that suggests that the entire sample of calcium hypochlorite will be used up while 1.4 moles of the HCl is converted to products. We can get 2 moles of Cl2 per mole of hypochlorite; so we expect to get no more than 0.700 moles of Cl2 which weighs 35.4527 2 = 70.9054 grams/mole; so we would predict that up to 49.63 grams of Cl2 gas might be evolved. (b) As stated above, we have an excess of 0.25 moles of HCl; that would correspond to 0.25 moles 36.4527 g/mole = 9.1132 9.1 grams excess HCl 66 Before we can discuss the yields, we need to figure out how many moles of reactants and products are represented in this problem. We have 0.10 liters of nitrobenzene, with density 1.20 g/milliter, corresponding to 120. grams of a material with molecular weight 6 12.011 + 5 1.00794 + 1 14.0067 + 2 15.9994 = 123.112 g/mole; therefore we have 0.9747 moles nitrobenzene to start off. We have 0.30 liters of triethylene glycol, with density 1.12 g/milliliter, corresponding to 336 grams of triethylene glycol (perhaps one too many sig figs, we'll keep it for now). Triethylene glycol has molecular weight 6 12.011+14 1.00794+4 15.9994 = 150.175 g/mole; so we have 2.24 moles of the triethylene glycol. Azobenzene is our product; we have 55 grams of the final product, and its molecular weight is 12 12.011+10 1.00794+2 14.0067 = 182.225 g/mole; corresponding to 0.302 moles of product. 2 (a) The theoretical yield requires that we convert from the amount of the limiting reactant to the amount of product assuming 100% yield. Our balanced chemical equation tells us that we need twice as many moles of the triethylene glycol as of the nitrobenzene; above we found that we have slightly less than one mole of the nitrobenzene, and somewhat more than two moles of the triethylene glycol. Thus our limiting reagent is the nitrobenzene; each mole of nitrobenzene yields 1 mole of the product in the ideal reaction. Thus our 2 0.9747 moles of the nitrobenzene should yield 0.9747/2 = 0.4874 moles of product; call it 0.487 after adjusting the sig figs. (Note that the two sig figs available in the triethylene glycol is irrelevant, as we expect it is present in excess!) (b) As described above, the actual yield is 0.302 moles of the azobenzene product. (c) Percent yield is simply 100% actual yield/theoretical yield, or 100% 0.302/0.486 = 61.9%. 104 Melamine was the additive found in some dog foods this summer as a "protein extender". Tests for protein in foods often are really only tests for the density of N atoms, and melamine is sufficiently rich in N so that a simple analysis suggests that there is more protein in the food than there really is. . . . (a) First we need a balanced chemical equation; we'll take it in two steps, as the problem gives hints as to what we should look for in the process. Step 1 would appear to be (unbalanced): CO(NH2 )2 HNCO+???? where a simple analysis of the atom balance suggests that what is missing on the product side is one nitrogen and 3 hydrogen atoms. Conveniently enough, that corresponds to the ammonia molecule which we are told is one of the by-products, leading us to: CO(NH2 )2 HNCO + NH3 as the first step. The second step (unbalanced) would therefore seem to be: HNCO C3 N3 (NH2 )3 + CO2 Focus first on the N content, as it is found in only one of the products; we need 6 nitrogen atoms per mole melamine, and the only source is in HNCO so we need 6 moles of the input reagent per mole of melamine. So we guess 6 HNCO C3 N3 (NH2 )3 + CO2 and count atoms; the left hand side has 6 each H, N, C and O; the products so far have 6 H's, 6 N's, 3 C's and no O's. So we can get balanced by adding 3 units of CO2 , leading to (in final form) 6 HNCO 6 C3 N3 (NH2 )3 + 3 CO2 or, for the overall reaction neglecting the intermediate step: 6 CO(NH2 )2 6 NH3 + C3 N3 (NH2 )3 + 3 CO2 (b) 100.0 kg of urea is used as the input. Our balanced chemical equation suggests that 6 moles of urea results in 1 mole of melamine; to find the mass of melamine created we need to (i) convert from kg of urea to moles, (ii) use the stoichiometric coefficients to find the theoretical yield in moles of melamine; (iii) convert from moles to kg, and (iv) take note that the actual yield is only 84% of the theoretical yield. In order:: 3 i. One mole urea has molecular weight 12.011 + 15.9994 + 2 14.0067 + 4 1.00794 = 60.0556. So our 100 kg of urea corresponds to 100 kg 1000 g/kg 1 mole/60.0556 grams = 1665.1 moles. ii. Our balanced chemical equation tells us that 6 moles of reactant yield one mole of the desired product. So our theoretical yield of melamine in moles, is 1665.1/6 = 277.52 moles. iii. Molecular weight of melamine is: 3 12.011 + 6 14.0067 + 6 1.00794 = 126.12 g/mole. So our theoretical yield is 277.52 moles 126.12 g/mole = 35001 g = 35.001 kg. iv. The actual yield is only 84%. So our yield in kg is 35.001 0.84 = 29.4 kg, though you could easily argue that I could only really specify that to be 29 kg. 4
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Path: Cornell >> CHEM >> 2090 Fall, 2007
Description: Chemistry 209 Chapter 8: Homework #5 answers 6 This is most easily done by referring to Figure 8-3, which orders the different wavelengths (or frequencies) of the electromagnetic spectrum, from high energy at the left to low energy at the right end...
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: Prefinal for AMS310 SOLUTIONS Z /2 I-1 E 2 n 2.58 2 * 15 0.086 I-2 E I-3 (a) I-4 (d) I-5 E I-6 (b) I-7 (a) I-8 (d) I-9 Y Z /2 n 2.58 * 60 10 15 .48 t s /2 n 5.841 * 1.44 2 4.2 2 X 2 *10 20 I-10 (a) I-11 P( X I-12 P( X 3) F (3)...
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: AMS310.01 Practice quiz 1 1. If there are 10 people in a class and 8 will pass. How many sets of 8 people are there? 2. P(A) = 0.8 , P(B) = 0.4 , P(A B)= 0.3. P(A B) =_. 3. Suppose we toss a single die and the result is the number of dots observed. T...
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: AMS310.01 PREQUIZ 2 Practice questions for quiz 2. _ 1. P(A B)= 0.2 P(A)=0.5 P(B)= 0.6, P(B|A)= _2. P(B|A) =0.5 P(A)=0.6 P(B)=0.8 P(A B)= _ _ 3. Given S= {1,2,3,4,5,6} the outcomes of the toss of 1 die. Let A={we get less than 5 dots} and let B={we ...
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: preQuiz 4 AMS310 Quiz will be on Monday October 13 _1. Suppose f(x) = 1/6 for 0< X<6 and f(x) = 0 otherwise. Find the mean value of x. Suppose that Z has standard normal density (Z~N(0,1). The p.d.f. of Z is 2 1 e x / 2 for x f(z) = . 2 Use Table ...
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: Practice questions for Quiz 6 to be given on October 27 _1. Suppose that X has mean deviation of Y equals_ =5 and =10 and Y= 6- 4X. The standard _2. Suppose that X1, X2 and X3 have mean =5 and 2= 4 and Cov(Xj, Xk) =0 for j k . Then the variance of...
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: Prequiz 7. PRACTICE QUESTIONS FOR QUIZ TODAY . NOVEMBER 2, 2003 1-2 A box of \"1 lb.\" weights must have a mean weight of 1 lb. with a standard deviation of 0.04 lb. _1. We take a sample of n=100 weights and calculate the average of these. We let X den...
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: Prequiz 8 1-2 We are interested in estimating the mean score on the test, we sample n=36 individuals and compute an average of 80.1 and a sample estimate of variance equal to 1.44 i.e., n=36, x = 80.1 and s2 = 1.44 so s=1.2. _1.Compute a point estima...
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: AMS310 PRETEST 1 practice questions for Test 1 I. SHORT ANSWER/MULTIPLE CHOICE QUESTIONS. 1. ALL QUIZ QUESTIONS Given f(x)= 0.1x for x=1,2,3, 4. 2 (a) What is the value of x? (b) What is the value of x? 3. Given X has Poisson probability distribution...
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: AMS310.01 QUIZ 2 September 17 PRINT YOUR NAME HERE_ USE ALL UPPER CASE LETTERS. UNDERLINE YOUR LAST NAME STUDENT ID NUMBER_ _ 1. P(A B)= 0.15 P(A)=0.5 P(B)= 0.2, P(B|A)= _2. P(B|A) =0.8 P(A)=0.2 P(B)=0.6 P(A B)= _ _ 3. Given S= {1,2,3,4,5,6} the o...
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: Quiz 4 AMS310.01 Monday October 13 Print your name here_ _1. Suppose f(x) = 3 for 0< X<1/3 and f(x) = 0 otherwise. Find the mean value of x. Hint = 3xdx = 3x2/2. Suppose that Z has standard normal density (Z~N(0,1). The p.d.f. of Z is 2 1 e x / ...
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: QUIZ 5 AMS310.01 Monday, October 20, 2003 Print your name here_ _1. Suppose that X has mean Y equals_ =10 and =50 and Y= (X-10)/50. The mean of _2. Suppose that X is a binomial variable with n=81 and p=0.5. 81 x0.5 x0.5 = 4.5. Then =np= 40.5 and =...
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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Table4_T
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: AMS310.03 TEST 1-A FORM Fall 2003 1. PRINT YOUR NAME HERE_ USE ONLY UPPER CASE LETTERS. UNDERLINE YOUR LAST NAME TWICE. 2. STUDENT ID NUMBER 3. CHECK TO MAKE SURE THAT YOUR TEST HAS 7 PAGES INCLUDING THIS ONE. 4. SHOW YOUR WORK FOR ALL QUESTIONS I...
test2A_sol
Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: AMS310 TEST 2 Fall 2003 FORM A 1. PRINT YOUR NAME HERE_ USE ONLY UPPER CASE LETTERS. UNDERLINE YOUR LAST NAME TWICE. 2. Write your student ID number here_ 3. CHECK TO MAKE SURE THAT YOUR TEST HAS 8 PAGES INCLUDING THIS ONE. 4. SHOW YOUR WORK FO...
test2B_sol
Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: AMS310 TEST 2 Fall 2003 FORM B 1. PRINT YOUR NAME HERE_ USE ONLY UPPER CASE LETTERS. UNDERLINE YOUR LAST NAME TWICE. 2. Write your student ID number here_ 3. CHECK TO MAKE SURE THAT YOUR TEST HAS 8 PAGES INCLUDING THIS ONE. 4. SHOW YOUR WORK F...
test3_sol
Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: ...
anshw330407
Path: Sonoma >> ECON >> 304 Fall, 2007
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Path: Cornell >> ECON >> 3010 Fall, 2007
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Path: Cal Poly Pomona >> ME >> 214 Fall, 2006
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Path: UBC >> CPSC >> 303 Winter, 2007
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Path: UMKC >> ME >> 275 Spring, 2008
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Path: UMKC >> ME >> 275 Spring, 2008
Description: PROBLEM 5.1 Locate the centroid of the plane area shown. SOLUTION A, in 2 x , in. y , in. xA, in 3 yA, in 3 1 2 8 6 = 48 16 12 = 192 -4 9 6 -192 432 1152 1584 8 1536 1344 240 xA 1344 in 3 = A 240 in 2 yA 1584 in 3 = A 240 in 2 Th...
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Path: Cornell >> COMM >> 2010 Fall, 2007
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Path: Allegheny >> ECON >> 300 Spring, 2008
Description: ECONOMICS 300 EXAM 2 SPRING 2004 PLEDGE:_ NAME:_ Use this information to answer questions 1-4. Suppose that the supply curve for news stories concerning Michael Jackson is perfectly elastic at the price of $10. The demand curve is downward sloping an...
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Path: Allegheny >> ECON >> 300 Spring, 2008
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Path: Allegheny >> ECON >> 300 Spring, 2008
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Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: Econ 101 Principles of Microeconomics Korinna K. Hansen Student Name: Section #: TA Name: Second Midterm Examination Spring 2007 Version 1 DO NOT BEGIN WORKING UNTIL THE INSTRUCTOR TELLS YOU TO DO SO. READ THESE INSTRUCTIONS FIRST. You have 50 m...
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Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: Review Questions for Midterm 2 Econ 101, Lecture 4, Spring 2007 Professor Korinna K. Hansen TAs: Jonathan Thornhill, Kyoung Jin Choi, Woo Jin Choi and Jim Lin 1) In a world of two goods X and Y with prices P X and PY, an increase in both prices PX a...
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Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: Econ 101, Final Review Key Spring 2007 1) c 2) a 3) c 4) d 5) b, this is assuming that fixed cost is included in the marginal cost of the first unit. 6) b 7) a 8) d 9) b 10) c 11) b 12) d 13) b 14) c 15) c 16) d 17) a 18) a 19) c 20) b 21) c 22) d 23...
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Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: Review Questions for the Final Exam Econ 101, Lecture 4, Spring 2007 Professor Korinna K. Hansen TAs: Jonathan Thornhill, Kyoung Jin Choi, Woo Jin Choi and Jim Lin Use this graph for the next question. 1) Points A, B, and C in the above Figure indic...
finalexamreviewtopics
Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: Additional Topics for Final Exam Econ 101, Lecture 4, Spring 2007 Professor Korinna K. Hansen In addition to all topics you prepared for Midterms 1 & 2, please prepare the following topics: Monopoly Barriers to entry Revenue curves for the Monopolist...
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Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: Y Income and Substitution Effects: Px increases E\' E2 E1 New budget line Hypothetical budget line Original budget line X Income effect Substitution effect Y Income and Substitution Effects: Px decreases E1 E\' Original budget line E2 Hypothetic...
LearningGuide1
Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: University of Wisconsin Department of Economics Economics 101: Principles of Microeconomics Korinna K. Hansen Learning Guide 1 Due Date: Week of Monday, Jan. 29th, 2007 Reading Assignment: Case & Fair Chapter 2 and Lecture Material. Problem Assignm...
LGuide2
Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: University of Wisconsin Department of Economics Economics 101: Principles of Microeconomics Korinna K. Hansen Learning Guide 2 Due Date: Week of Monday, February 5th, 2007 Reading Assignment: Case & Fair Chapter 3 up to page 59 and Lecture Materia...
LGuide3
Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: University of Wisconsin Department of Economics Economics 101: Principles of Microeconomics Korinna K. Hansen Learning Guide 3 Due Date: Week of Monday, February 12th, 2007 Reading Assignment: Case & Fair Chapter 3 (mostly beyond page 59) and all ...
LGuide4
Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: University of Wisconsin Department of Economics Economics 101: Principles of Microeconomics Korinna K. Hansen Learning Guide 4 Due Date: Week of Monday, February 26th, 2007 Reading Assignment: Case & Fair Chapter 6 up to the middle of page 126. Pl...
midterm1reviewtopics
Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: Review Topics for Midterm 1 Econ 101, Lecture 4, Spring 2007 Professor Korinna K. Hansen Opportunity Cost Absolute Advantage Comparative Advantage Production Possibility Frontier Marginal Rate of Transformation Law of Increasing Opportunity Cost Law ...
Long-run Producer Theory & Input Market Graphs
Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: ...
midterm1 review questions
Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: Review Questions for Midterm 1 Econ 101, Lecture 4, Spring 2007 Professor Korinna K. Hansen TAs: Jonathan Thornhill, Kyoung Jin Choi, Woo Jin Choi and Jim Lin Use the graph below to answer the following question. 1) In the above Figure, which of th...
Midterm1reviewkey
Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: Econ 101: Principles of Microeconomics - Lecture 4 - Spring 2007 Key to the Review problems for First Midterm 1) d 2) d 3) b 4) b 5) a 6) d 7) a 8) b 9) a 10) c 11) d 12) c 13) a 14) c 15) c 16) d 17) c 18) b 19) a 20) b 21) d 22) b 23) b 24) b 25) c...
midterm2version1key
Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: Version 1 1) b 2) b 3) a 4) a 5) b 6) b 7) b 8) b 9) a 10) a 11) b 12) d 13) c 14) d 15) a 16) c 17) a 18) d 19) c 20) b 21) b 22) a 23) b 24) d 25) c 26) d 27) b 28) c 29) d 30) b ...
midterm1version1key
Path: Wisconsin >> ECON >> 101 Spring, 2007
Description: Version 1 1) a 2) b 3) a 4) a 5) a 6) b 7) b 8) a 9) b 10) b 11) b 12) c 13) a 14) d 15) c 16) c 17) b 18) c 19) a 20) b 21) a 22) d 23) d 24) a 25) c 26) b 27) d 28) d 29) b 30) d ...
