Assign all the Protons in HNMR Normalized Intensity1.00.90.80.745HH3OHや1·CH₂D0.60.50.40.30.20.107.317.307.30-7.190.000.957.5TE7.0No(npm)(Hz)Height= MHHJ0.01 0.116.5-5.230.290.17ㅂ6.05.55.04.50.00 0.00 0.00 1.06 0.01ㅂ |0.004.03.5Chemical Shift (ppm)-3.0.010.0113.02.52.01.451.43 1.440.00 1.561.51.0

Step 1: For methyl group (CH₃) protons, the chemical shift peak was observed as a doublet at 1.43 to…

Answered: Normalized Intensity 1.0 0.9 0.8 0.7 4…

Chemistry for Engineering Students

3rd Edition

ISBN:9781285199023

Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter1: Introduction To Chemistry

Section: Chapter Questions

Problem 1.77PAE

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….
What is Eºcell in volts? (C) || (aq) | 12 (s) || Cl₂ (g) | CI (aq) | (C) I₂ (s) + 2 e² → 2 I¯ (aq) Cl₂ (g) + 2 e → 2 C1 (aq) O +0.824 O -0.824 O -1.896 O +1.648 O -1.648 O +1.896 Eº = +0.536 V Eº = +1.360 V
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In a survey of 1000 large corporations, 250 said that, given a choice between a job candidate who smokes and an equally qualified nonsmoker, the nonsmoker would get the job (USA Today).(a) Let p represent the proportion of all corporations preferring a nonsmoking candidate. Find a point estimate for p.(b) Find a 0.95 confidence interval for p.(c) As a news writer, how would you report the survey results regarding the proportion of corporations that hire the equally qualified nonsmoker? What is the margin of error based on a 95% confidence interval?
6. Using dimensional analysis, show how 0.082057 Latm mol. K Significant figures note: the last digit on both values has been rounded. NOTE: 1 J = 1 N-m = 1 kg.m², and 1 Pa = 1 N m² is equal to 8.3145 or 1 kg m.s mol - K
•0= AG° + RT * In (K) %3D • -AG° = RT * In (K) %3D • AG° = -RT * In (K) %3D AG° = -0.616 * In (K)... %3D
Phase change properties of pure substances Cu C6H12 C (C6H5CH₂)₂0 (CH3CH₂)20 CH3CH₂OH (CH₂OH)2 NH₂COH Au C C6H14 H₂ copper cyclohexane diamond dibenzyl ether diethyl ether ethanol ethylene glycol formamide gold graphite hexane hydrogen 1084.62 6.7 4440 1.8 -116.22 -114.14 -13.0 2.57 1064.18 4489 -95.27 -259.16 2560 80.7 298 34.4 78.24 197.5 217 2836 3825 68.72 -252.879 0.385 1.841 0.51 2.369 2.438 2.394 2.389 0.129 0.709 2.27 14.304 13.26 2.68 -- 7.19 4.931 9.96 8.44 12.55 117.4 13.08 0.12 29.97 45.6 26.52 38.56 50.5 60.2 324 28.85 0.9 280.3 193.7 242 446 -- 234.4 -240.212 40.2 35.9 61.7 80 29.9 12.69
2HF (g) = H₂(g) + F2 (g), Ke=1.0 x 10 -95 ibrium mixture contains mostly reactants. Desktop 100 2NO(g) + O₂(g) = 2NO₂(g), Kc = 6.0 × 10¹3 Address CO(g) + Cl₂ (g) = COC12 (g), Kc = 5.0 × 10-⁹ Equilibrium mixture contains mostly products. 8:18 PM ^ Tuesday 6/21/2022 ENG IN
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In the technique of high performance liquid chromatography (HPLC) the peak area response for a particular analyte is directly proportional to concentration Analgesic tablets typically consist of aspirin and caffeine as the active ingredients, the remainder made up of inert binding material. The concentrations of aspirin and caffeine may be routinely monitored using HPLC, whereby the peak areas obtained from a solution of the tablet are compared with those obtained from standard solutions of the analytes. In an analysis to determine the concentration of caffeine in such a tablet, the following data were recorded: Solutions Analysed: Tablet Sample: Mass of tablet (dissolved in 250 cm methano): 0.523 g Caffeine Standard: Mass of caffeine (diessolved in 500 cm methanol) 0.105 g HPLC results: Solution Peak Area (catfeine) 1. Tablet Sample 325 2. Caffeine Std. 865 The concentration of caffeine in the tablet (expressed as a mass (w/wl percentage) was Oa 1.89 Ob 13.4 O 3.77 Od. 267 Oe 7.54 You…
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