Principles of Instrumental Analysis
7th Edition
ISBN: 9781305577213
Author: Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher: Cengage Learning
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Chapter 31, Problem 31.4QAP
Interpretation Introduction
Interpretation:
The cause of low temperature endotherm for pressure to be found at the same temperature but for high temperature they differs is to be evaluated.
Concept introduction: Temperatures for phase transition and the enthalpies associated with the temperature were plotted for benzoic acid. This is done by the use of differential scanning calorimetry and known as thermogram.
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Principles of Instrumental Analysis
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- Apply the Q test to the following data sets to determine whether the outlying result should be retained or rejected at the 95% confidence level. a. 85.10, 84.62, 84.70b. 85.10, 84.62, 84.65, 84.70arrow_forwardanswer properlyarrow_forwardUsing Boltzmann equation to estimate the populations in J=1 and v-1, relative to J-0 and v-0, respectively. the following values were obtained at room temperature for a diatornic molecule = 0.98 NJ-0 Nu-l 8.9 10-7 N. which mean Select one: Oa J-1 has almost same number of molecules as J=0, while v-1 has almost no malecules Ob. Both J=1 and v- 1 have almost no molecules Oc Both J=1 and v=1 have almost same number of molecules as in J-0 and v=0 O d. v=1 has almost same number of molecules as v=0, while J=1 has almost no moleculesarrow_forward
- The force constants for ¹H₂ and 79 Br2 are 575 and 246 N-m¹, respectively. The atomic masses for ¹H₂ and 79Br2 are 1.0078 amu and 78.9183 amu, respectively. ▼ ▼ Part F Calculate the ratio of the vibrational state populations n₁/no for 79Br2 at 1190. K. Express your answer to three significant figures. n₁/no = 0.452 Submit Previous Answers Request Answer ΑΣΦ Part G X Incorrect; Try Again; 5 attempts remaining Calculate the ratio of the vibrational state populations no/no for 79Br2 at 280. K. Express your answer to three significant figures. Submit VE ΑΣΦ n2/mo= 0.9679 PRZ Previous Answers Request Answer ?arrow_forwardMust answer all and properly else downvotearrow_forwardThe answer was incorrect and I don't understand whyarrow_forward
- a) Make A grouped frequency distribution of the data 4,16,32,52,56,62,73,81,83,88,94,100,102,110,138,152,165,173,176,184,186,202,204,207,209,223,229,238,248,249,262,264,270,272,279,281,285,287,293,296,305,309,314,320,321,323,337,353,356,360,362,366,371,373,381,393,395,398,403,407,418,419,426,429,433,434,435,444,446,447,450,451,469,486,508,513,521,522,524,529,533,538,544,551,554,558,560,561,562,572,596,607,609,621,624,652,657,660,677,695,701,702,712,713,718,724,726,740,750,764,765,766,773,778,781,785,786,795,816,857,861,864,873,874,876,885,888,894,899,900,901,912,913,918,920,933,941,944,945,957,961,962,965,969,972,982,983,990,991,996 b) Draw Histogram for the frequency distribution table madearrow_forward6 v=1720 cm-1 v = 1150 cm 1 1. NaNH, 2. CH3l v=3500-3300 cm¹ CI CH₂OH Heat OH 1. Hg(OAc)2 e H₂O 2. NaBH4 v = 3500-3300 cm-¹ HBr No characteristic v tBuOK v=3050 cm¹¹, 1650 cm1 H₂SO4, heat HBr a v = 3050 cm¹¹, 1650 cm¹ 1. BH3 2. NaOH, H₂O₂ v = 3500-3300 cm 1 CrO₂, H₂SO4 v = 1705 cm¹arrow_forwardPlease Explainarrow_forward
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