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Unformatted text preview: The mean lifetime of lightbulbs produced at a certain factory is 14,0 0 hours. The standard deviation for the lifetime distribution is 1,00 hours the key number). Assume that losers are those who purchased lightbulbs lasting less than 10,0 0 hours and winners those who purchased lightbulbs lasting longer than 18,0 0 hours. Calculate the percentage of the customers who are winners and losers. Tables will be given! Exam # 1 Problem #2: ANSWER KEYS The normalized Gaussian curve is obtained when the variable x is substituted by a variable z :  = x z After the variable transformation, the normalized Gaussian curve is a Gaussian curve for = 0 and = 1. x _ Table 1. Cumulative distribution function of the normalized Gaussian distribution.  z  P( ,z ) P(0, z )  z  P( ,z ) P(0, z ) 0 0.5000 0.0000 0.1 0.4602 0.0398 2.1 0.0179 0.4821 0.2 0.4207 0.0793 2.2 0.0139 0.4861 0.3 0.3821 0.1179 2.3 0.0107 0.4893 0.4 0.3446 0.1554 2.4 0.0082 0.4918 0.5 0.3085 0.1915 2.5 0.0062 0.4938 0.6 0.2742 0.2258 2.6 0.0047 0.4953 0.7 0.2420 0.2580 2.7 0.0035 0.4965 0.8 0.2119 0.2881 2.8 0.0026 0.4974 0.9 0.1841 0.3159 3.9 0.0019 0.4981 1.0 0.1587 0.3413 3.0 0.001350 0.498650 1.1 0.1357 0.3643 3.1 0.000968 0.499032 1.2 0.1151 0.3849 3.2 0.000687 0.499313 1.3 0.0968 0.4032 3.3 0.000483 0.499517 1.4 0.0808 0.4192 3.4 0.000337 0.499663 1.5 0.0668 0.4332 3.5 0.000233 0.499767 1.6 0.0548 0.4452 3.6 0.000159 0.499841 1.7 0.0446 0.4554 3.7 0.000096 0.499904 1.8 0.0359 0.4641 3.8 0.000072 0.499928 1.9 0.0287 0.4713 3.9 0.000048 0.499952 2.0 0.0227 0.4773 4.0 0.000032 0.499968 2.9 Problem #6 Consider titration a 0.100 M solution of maleic acid (cis butenedioic acid) with 0.100 M solution of NaOH. K A1 =1.23 102 and K A2 =4.66 107 . (a) Plot an approximate pH vs. f curve and define the f. (b) What is the majority species derived from maleic acid at f = 2.5? (c) Calculate the value of the overall K a . 0.5 1 1.5 2 2.5 3 2 4 6 8 10 12 pH f Titration curve of the 0.100 M solution of a diprotic maleic acid (cisbutenedioic acid) with 0.100 M solution of NaOH The acid has very different values of K a1 and K a2 : K A1 =1.23 102 K A2 =4.66 107 ; Two endpoints are clearly visible! Accepted approximations, maleic acid buffers (simple): Before first equivalence C4H2O4H2 = ( C4H2O4H) + H +, K A1 ; H 2 A = AH + H + pH = pK A1 + log[C4H2O4H2/[(C4H2O4H)] Before second equivalence (C4H2O4H) = (C4H2O4)2 + H+, KA2; HA = A 2 + H + pH = pK A2 + log[(C4H2O4)2]/[(C4H2O4H)] Advanced: The (C4H2O4H) , amphiprotic, may also: (C4H2O4H) + H2O = C4H2O4H2 + OH , K B related to K A2 ; Leads to the equation given by Hadi, (Harris: page 190, eq. 910). Chemistry 222 Fall 2010 STATISTICS, ANALYSIS OF EXPERIMENTAL ERROR (4) Practical advice: 1.We should retain one more digit for the average and for the standard deviation than was present in the original data....
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This note was uploaded on 12/28/2010 for the course CHEMISTRY 222 taught by Professor Wieckowski during the Spring '10 term at University of Illinois at Urbana–Champaign.
 Spring '10
 WIECKOWSKI

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