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209 Chemistry Chapter 3: Homework #3 answers 14 P4 units have molecular weight 4 30.9738 = 123.8952. The density of the P4 unit is given as 1.823 g/cm3 and we are given a cylinder 6.50 cm in length and 1.22 cm in diameter. The volume of a cylinder is given by d2 V = r2 h = h 4 2 3 for our sample V = 3.14159 0.61 6.50 cm = 7.598 cm3 ; and given the density d = 1.823 we know that we have a total mass of P4 m = V d = 7.598 cm3 1.823 g = 13.852 g cm3 (a) Given the molecular weight calculated above, that corresponds to n moles of P4 given by n = 13.852 grams sample 1 mole = 0.1118 moles P4 123.8952 g (b) This one is easy; one mole contains NA = 6.0221 1023 molecules; so 0.1118 moles contains 0.1118 6.0221 1023 = 6.733 1022 P4 molecules. 18 Given the compound Ge[S(CH2 )4 CH3 ]4 , we are asked to figure out what we have. First, let's start with a picture of the molecule, or at least an artist's conception of what might be the molecule: CH3 CH2 CH2 CH2 CH2 S | CH3 CH2 CH2 CH2 CH2 S - Ge - SCH2 CH2 CH2 CH2 CH3 | SCH2 CH2 CH2 CH2 CH3 (a) How many atoms are found in one formula unit? Well, there are four equivalent groupings of atoms, each containing 1 S atom, 4 CH2 groups, and a single CH3 group. By my count that corresponds to 1 + 4 3 + 4 = 17 atoms in each supgrouping; there are four of them, plus the central Ge atom, fo I count a total of 69 atoms per formula unit. (b) How many C and/or H atoms? They are found in CH2 and CH3 groups; there are 4 4 of the former, and 4 1 of the latter. So the total count of carbon atoms is 20; the total count of hydrogen atoms is 44; and the ratio of C atoms to H atoms is 20:44; or in simplified numbers, 5:11. (c) The ratio by mass of Ge to S; each Ge weighs 72.61 (on average, assuming that we have a natural abundance distribution of isotopes and that there are sufficient atoms present so that the average reasonably represents any distribution; and there are 4 S atoms per Ge, each of which weighs (again, on average) 32.066. So the ratio of Ge mass to S mass is 72.61 : (4 32.066) which is about 0.5661:1. (d) 1 mole of the compound contains 4 moles of S atoms, each of which contains 32.066 g of sulfur atoms. So 1 mole of the compound contains 4 moles S atoms 32.066 g/mole = 128.264 g sulfur 1 (e) We are given 33.10 grams of sample; we should calculate how many moles of sample we have which requires that we find the formula weight of the compound. That requires that we do the detailed countings of atoms; that's done above; and the formula weight is therefore 1 72.61 + 4 32.066 + 20 12.011 + 44 1.00794 = 485.444. 33.10 grams contains 33.10 g 1 mole/485.444 g = 6.8185 10-2 moles of our compound; each mole contains 20 C atoms, so the number of C atoms in our 33.10 gram sample is 6.8185 10-2 moles 20 atoms C/mole 6.0221 1023 = 8.212 1023 atoms of C 32 The analyzed % by mass of MSG is 13.6% Na, 35.5% C, 4.8% H, 8.3% N, and 37.8% O. We require the empirical formula for this material; that requires that we translate from mass units to mole units. Choose a representative sample of, say, 100 g MSG (this could be done for any size sample, we will ultimately ignore the absolute quantities anyway) and find 13.6 g Na 1 mole Na = 0.5916 moles Na 22.9898 g Na 1 mole C = 2.9566 moles C 12.011 g C 1 mole H = 4.7622 moles H 4.8 g H 1.00794 g H 1 mole Na = 0.5926 moles N 8.3 g N 14.0067 g N 1 mole Na = 2.3626 moles O 37.8 g O 15.9994 g O We wish to find an empirical formula; let's work on the expectation that there are one unit of both Na and N in the empirical formula, so the other atoms are found in quantities that are likely to be integer multiples of the number of moles of Na found in our sample; for example, if there is a single Na atom in our compound represented by the 0.5916 moles calculated above, then the 2.9566 moles of C in our sample represent 2.9566/0.5916 = 5.00 moles of C; similarly, we have 8.05 moles of H and 3.99 moles of O. We're not quite done; remember that (1) there are only 2 or 3 significant figures in the measurements provided, and (2) there is a preference strong for integer numbers of atoms in the empirical formula (or, perhaps, halves or thirds indicating that the formula contains multiple atoms of what we thought was the smallest contributor. So our empirical formula is likely to be NaC4 H8 NO4 ; the ordering of the atoms is largely irrelevant. 35.5 g C 46 1.3020 g of thiophene is burned, and yields 2.7224 g CO2 ; 0.5575 g H2 O, and 0.9915 g SO2 . Each molecule of CO2 represents one C atom that was found in thiophene; similarly, each molecule of SO2 represents one S atom in thiophene. Each molecule of H2 O found represents two H atoms that were originally in the thiophene; after we have accounted for all the atoms of C, H and S the remaining mass in the thiophene sample must represent O atoms that found their way to one of our products. So let's count moles; first, of C in the thiophene: 2.7224 g CO2 0.5575 g H2 O 1 mole CO2 1 mole C = 6.186 10-2 moles C 44.0103 g CO2 1 mole CO2 2 mole H 1 mole H2 O = 6.189 10-2 moles H 18.01528 g H2 O 1 mole H2 O 2 1 mole SO2 1 mole S = 1.548 10-2 moles S 64.0648 g SO2 1 mole SO2 This would appear to represent a molecule with the empirical formula C4 H4 SOn ; where n is as yet unknown; and we have 1.548 10-2 moles of the molecule. If 1.3020 g thiophene represents 1.548 10-2 moles of thiophene, then the formula weight of the molecule would appear to be 1.3020 g/1.548 10-2 = 84.13 g/mol; so far, our molecular fragment weighs 4 12.0115 + 4 1.00794 + 32.066 = 84.14 g/mol; therefore, there is no oxygen to worry about! Thiophene has the empirical formula C4 H4 S. 0.9915 g SO2 52 (a) Net charge on the unit is -2; each O atom is assigned an oxidation state of -2, so the total assigned to the 3 oxygen atoms is -6; that leaves the S with an oxidation state of +4 (as the sum of all the oxidation states gives the net charge; +4 + (-6) = -2). (b) Net charge is -2; each O atom is assigned an oxidation state of -2, so the two S atoms have to total +4, and each S atom is assigned an oxidation state +2. (c) Net charge is -2; each O atom is assigned an oxidation state of -2, so the two S atoms have to total +14 between them, and each S atom is assigned an oxidation state +7. (d) Net charge is -1; each O atom is assigned an oxidation state of -2, and the H atom is normally assigned +1 (unless attached to a metal atom). So the S atom must account for +6 units of charge, which is its oxidation state. (e) Net charge is -2; each O atom is assigned an oxidation state of -2, so the four S atoms have to total +10 between them, and each S atom is assigned an oxidation state +2.5. A little peculiar, we'll suggest what this might mean at some later date. 80 Note that we must calculate the Cu content by difference; it isn't measured directly in our experiment. We need to calculate the moles of each of the other metals found in the various salts, and convert those mole quantities back to masses so that we can account for the total of 1.1713 g of sample. SnO2 has formula weight 150.7088; 0.245 grams of the tin salt (stannic oxide, or tin (IV) oxide) represent 1.626 10-3 moles of SnO2 , which represents 1.626 10-3 moles 118.710 g/mole = 0.1930 g Sn metal. PbSO4 has formula weight 303.28; 0.115 grams of the lead salt (plumbous sulphate, or lead (II) sulphate) represent 0.115 g 1 mole/303.28 g PbSO4 = 3.792 10-4 moles of PbSO4 , which represents 3.792 10-4 moles 207.2 g/mole = 0.07857 g Pb metal. Zn2 P2 O7 has formula weight 304.7234 g. Our 0.246 g sample of the salt corresponds to 0.246 g 1 mole Zn salt/304.7234 g Zn salt = 8.073 10-4 moles; that represents 8.073 10-4 moles 2 moles Zn/mole salt 65.39 g/mole Zn = 0.1056 g Zn. The total mass of metal accounted for so far is 0.1930 + 0.07857 + 0.1056 = 0.37717 g; the last digit is probably not reliable at any level, so call it 0.3772 g. That lets us account for the Cu which represents 1.1713 - 0.3772 = 0.7941 g Cu; in percentages, that corresponds to: 0.7941 100% = 67.80% Cu 1.1713 0.1930 100% = 16.48% Sn 1.1713 0.07857 100% = 6.71% Pb 1.1713 0.1056 100% = 9.02% Zn 1.1713 3
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hmwk4
Path: Cornell >> CHEM >> 2090 Fall, 2007
Description: Chemistry 209 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...
hmwk5ans
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...
prefinal_solutions
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)...
preQUIZ1
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...
preQUIZ2
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 ...
preQUIZ4
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 ...
preQUIZ6
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...
preQUIZ7
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...
prequiz8
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...
PRETEST1
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
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...
pretest2
Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: Pretest 2 Pretest 2 Practice questions for Test 2. [I. Short answer questions. All questions on Quiz 1,2,3 , 4 and 4a. (mainly Quiz 3 , Quiz 4 and QUIZ 4a).] 1. Suppose a discrete variable, D has approximately normal distribution. In order to determ...
pretest3
Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: PRETEST 3 for Test 3. Test 3 will be on Monday April 28. Questions and answers. Solutions will be posted with corresponding homework solutions. I. Multiple choice/short answer questions: Study all Prequiz questions (with emphasis on Prequizzes 5 and ...
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Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: Pretest 3 II-3 II-3. Prove: If we accept Ho: = o at the level then o is on the (1)100% confidence interval for . (P247) Ans) 1. (1-)100% confidence interval for (given on p228) is : x t / 2 s s x t / 2 n n (a) 2. 1) Critical region of the...
pretest3_solutions
Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: Solutions to PRETEST 3. TEST 3 will be given on Wednesday December 3 I-1. We decrease the width of the confidence interval by (1) increasing sample size(2) decreasing the confidence value and (3) decreasing the standard deviation of the variable. So ...
quiz2 with soln
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...
QUIZ4 with solution
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 / ...
quiz5 with soln
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 =...
Quiz6 with soln
Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: Quiz 6 October 27 Print your name here: _ ID #_ (use all UPPER CASE letters) _1. Suppose that X has mean deviation of Y equals_ =2 and =1 and Y= 3- 2X. The standard _2. Suppose that X1 and X2 have mean =5 and 2= 4 and Cov(Xj, Xk) =0 for j k . Then...
Quiz7 with soln
Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: Quiz 7. NOVEMBER 2, 2003 PRINT YOUR NAME HERE_ STUDENT ID _ 1-2 We are told that the mean diameter of the \"1mm\" bolts produced by the ABC Bolt Factory equals 1mm and the standard deviation equals 0.1 mm., i.e., = 1mm =0.1mm. _C__1. We take a sample o...
Syllabus
Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: Syllabus There are slight changes in schedule (December) in PINK SURVEY OF PROBABILITY AND STATISTICS FALL SEMESTER, 2003 FIRST DAY PACKET for AMS 310.01 MONDAY AND WEDNESDAY 5:20 PM to 6:40 PM Harriman 112 All information for this class, can be do...
Table1_Binomial
Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: ...
Table2_Poisson Distribution function
Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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Table3_std Normal distribution execution
Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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Table4_T
Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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Table5_Chi square distribution
Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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Table6_F distribution function
Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
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Table8_operating charac curves for tests abt means
Path: SUNY Stony Brook >> AMS >> 310.01 Fall, 2003
Description: ...
Test1_makeup_sols
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
Description: Econ 304 Sonoma State University Fall 2007 Dr. Robert Eyler Suggested Answers to Homework #3 Please answer the questions fully and draw graphs or use equations when needed. Show all your work and remember that these are practice for the midterm quest...
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Path: Cornell >> ECON >> 3010 Fall, 2007
Description: Econ 301 F06 ANSWERS PROBLEM SET 2 Wissink 1. Critically evaluate the following statements and explain why or in what way are they true, false, or uncertain. a. Over the set of people in New York, the binary relation \"is the half-brother of\" is tran...
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Path: Cal Poly Pomona >> ME >> 214 Fall, 2006
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2,01 piecewiseCubic example
Path: UBC >> CPSC >> 303 Winter, 2007
Description: Numerical Approximation and Discretization CPSC 303, Term 2, Winter 20072008 Piecewise Cubic Example Consider interpolating the points i xi hi yi 0 0.00 0.10 1.0000 1 0.10 0.15 0.2857 2 0.25 0.25 0.1379 3 0.50 0.50 0.0741 4 1.00 0.0385 Table 1: Da...
1.09 the definition of numerical analysis
Path: UBC >> CPSC >> 303 Winter, 2007
Description: The following essay appeared in the November, 1992 issue of SIAM News and the March, 1993 issue of the Bulletin of the Institute for Mathematics and Applications.] THE DEFINITION OF NUMERICAL ANALYSIS Lloyd N. Trefethen Dept. of Computer Science Cor...
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Path: Cal Poly Pomona >> ME >> 214 Fall, 2006
Description: ...
Chapter 13
Path: Cal Poly Pomona >> ME >> 214 Fall, 2006
Description: ...
Solutions Manual - Engineering Mechanics Statics 10E (Hibbeler)
Path: UMKC >> ME >> 275 Spring, 2008
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Vector Mechanics for engineers Statics 7th - Cap 05
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
Description: Maya Grinberg Comm 201 Actuate Outline Title: Take a CHILL PILL! General Subject: Leisure time Specific Subject: College Students Need to Manage their Stress and Can do so by Engaging in More Leisure Activities. Purpose: To entice my audience into pl...
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Path: Stanford >> CS >> 145 Fall, 2007
Description: Thrift: Scalable Cross-Language Services Implementation Mark Slee, Aditya Agarwal and Marc Kwiatkowski Facebook, 156 University Ave, Palo Alto, CA {mcslee,aditya,marc}@facebook.com Abstract Thrift is a software library and set of code-generation too...
exam2
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...
exam3
Path: Allegheny >> ECON >> 300 Spring, 2008
Description: ECONOMICS 300 FINAL EXAM SPRING 2004 PLEDGE: _ NAME: _ 1. (3) In a model of oligopoly: a. There are many firms that are price takers. b. There are many firms that offer heterogeneous product. c. One firm that is a price setter. d. Several firms that...
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Path: Allegheny >> ECON >> 300 Spring, 2008
Description: ECONOMICS 300 FIRST MIDTERM SPRING 2004 PLEDGE:_ NAME:_ Multiple Choice (3 points each) (48 points total) 1. When a consumer is consuming a bundle on her budget constraint then a. She must be maximizing her utility. b. She must be exhausting her inco...
midterm2version1s07
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...
midterm 2 review questions
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...
final review key
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...
final review problems
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...
Income and Substition Effects
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 ...
