Масс-спектр10050050ИК-Спектр100-сивность,N 6.Пропускание, %506978871004000Спектр ЯМР ¹Н (300 МГц)Гц8. Контрольные задачи21 05 20 D.EХИ (газ-реагент-метан)1001771 179 50194m/z150230 240m/z250260OM300020001000Волновое число, см-¹SSГц14371591Гц4.03.53.0Спектры ЯМР 13C и DEPT (75.5 МГц)170 160 150 140 130 120 110 100 902.5802.07026.04 2Гцмимими1.560 504030251М.Д.FarLEABeaприттМ.Д.

Note : DBE = C8H15O2Br = 8- (15+1/2)+1 = 9-8 = 1. Carbonyl present. From DEPT six CH2 groups and one…

Масс-спектр 887 100 69- thil 50 0 50 ИК-Спектр 100 Пропускание, % 100 m/z 4000 Спектр ЯМР Н (300 МГц) Гц N6. 8. Контрольные задачи 1431 -2939 2870 3000 1591 150 Гц 4.0 3.5 3.0 Спектры ЯМР ¹3C и DEPT (75.5 МГц) 170 160 150 140 130 120 110 100 чт 0.5 2.0-0.E 100 1771(1 50 2.5 Волновое число, см-¹ Гц 90 179 194 2000 ХИ (газ-реагент-метан) 230 240 250 260 m/z 1000 Гц мимими 2.0 1.5 80 70 60 50 40 251 -253 30 м.д. -22 ------- м.д.

Масс-спектр 887 100 69- thil 50 0 50 ИК-Спектр 100 Пропускание, % 100 m/z 4000 Спектр ЯМР Н (300 МГц) Гц N6. 8. Контрольные задачи 1431 -2939 2870 3000 1591 150 Гц 4.0 3.5 3.0 Спектры ЯМР ¹3C и DEPT (75.5 МГц) 170 160 150 140 130 120 110 100 чт 0.5 2.0-0.E 100 1771(1 50 2.5 Волновое число, см-¹ Гц 90 179 194 2000 ХИ (газ-реагент-метан) 230 240 250 260 m/z 1000 Гц мимими 2.0 1.5 80 70 60 50 40 251 -253 30 м.д. -22 ------- м.д.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Chemistry (i) For T = 298K ΔG° = -8.314 x 298 x ln (2.83x10-3) = 14.524 KJ/mol (ii) For T = 308.15K ΔG° = -8.314 x 308.15 x ln (9.619x10-3) = 11.891 KJ/mol (iii) For T = 318.15K ΔG° = -8.314 x 318.15 x ln (3.14x10-2) = 9.147 KJ/mol (iv) For T = 328.15K ΔG° = -8.314 x 328.15 x ln (8.82x10-2) = 6.621 KJ/mol (v) For T = 338.15K ΔG° = -8.314 x 338.15 x ln (2.42x10-1) = 3.990 KJ/mol Based on the measured ΔG° values, is this equilibrium spontaneous at room temperature? Which factor, entropy, or enthalpy, has the greater impact on spontaneity in this case? Explain your answers.
1-Pentanol to 1-bromopentane Chemicals: - 60ml Conc. Sulfuric Acid - 100ml Saturated Sodium bicarbonate - 65ml 1-Pentanol - 78g sodium bromide - Distilled water - 58.42g 1-Bromopentane 1-Pentanol Sodium Bromide Sulfuric Acid 1-Bromopentane Formula C5H12O NaBr H2SO4 C5H11Br MW (g/mol) 88.15 102.894 98.078 151.04 Density (g/mL) 0.811 3.21 1.84 1.218 Boiling point (*C) 138 1,396 337 130 NaBr(aq) + H2SO4(aq) -> NaHSO4(aq) + HBr(aq) CH3(CH2)4OH(aq) + H+ Br- (aq) CH3(CH2)4OH2 (aq) + Br-(aq) CH3(CH2)4OH2 (aq) + Br-(aq)  CH3(CH2)4Br(aq) + H2O(aq) How do I calculate the percent yield and identify the limiting reagent?
Produce Water mass “X” is known to be a mixture of water masses “1”, “2”, and “3”. Using the data plotted below, calculate the composition of “X” (i.e. determine the fraction from each of the 3 water masses that makes up water mass X). Watermass#1– SF6 =25 S=33.5 Watermass#2– SF6 =75 S=32 Watermass#3– SF6 =50 S=36.5 Water mass X – SF6 =60 S= 34
10.5mL (mw 106) of benzaldehyde and 2.9g(3.63mL, mw 40) of acetone react with 5g of NaOH and 25mL of Ethanol to form 6.7g of Dibenzalacetone. What is the theoretical, actual and percent yield?
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There's 1 drink (and you are asked to determine the glucose concentration in the drink in the units of g/100mL. (Why these units? Well, once you have the concentrations in g/100mL you will be able to compare your values with the nutritional values given on the drink bottles’ labels). The sample of the drink was diluted 1/100 (i.e. by a factor of 100). This was an essential step in the method because, without it, the machine used to analyse the glucose concentration (spectrophotometer) would have given an error as the concentration would have been too high for accurate detection. What this means for you is that the dilution factor will need to be taken into consideration in your calculations (remember the aim is to calculate the concentration in the original drink and not in the diluted drink). You measured the concentration of their diluted drink using the spectrophotometer and their results were provided to them in the units mM (millimolar). Glucose Concentration in mM of drink =…
There's 1 drink (and you are asked to determine the glucose concentration in the drink in the units of g/100mL. (Why these units? Well, once you have the concentrations in g/100mL you will be able to compare your values with the nutritional values given on the drink bottles’ labels). The sample of the drink was diluted 1/100 (i.e. by a factor of 100). This was an essential step in the method because, without it, the machine used to analyse the glucose concentration (spectrophotometer) would have given an error as the concentration would have been too high for accurate detection. What this means for you is that the dilution factor will need to be taken into consideration in your calculations (remember the aim is to calculate the concentration in the original drink and not in the diluted drink). You measured the concentration of their diluted drink using the spectrophotometer and their results were provided to them in the units mM (millimolar). Glucose Concentration in mM of drink =…
Show all steps leading to the final answer po. Here’s a pdf file in accordance with the topic po: https://drive.google.com/file/d/1_FnDtXCrFKSol3RNWIG_9tNQ7IxgxD6t/view?usp=drivesdk
4. Ksp=4.1x10-36 for Pb3(AsO4)2Calculate Eofor Pb3(AsO4)2(s) + 6e-  3Pb(s) + 2AsO43-Pb2+ +2e-  Pb(s) Eo= -0.126 V
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