For the information given below predict the structure of the unknown compound using the information provided. Annotate each spectra provided to justify your answer.You will also need to calculate the HDI for the proposed structure and provide the chemical formula. Don't forget to draw your predicted structure.Make sure to when annotating each spectra to be clear, detailed, and accurate. TRANSMITTINKEIZI50LODRelative Intensity10080604040Unknown #3Elemental Analysis:MM HA24.78% Carbon, 2.08% Hydrogen, remainder is two moles of a halogen20-Lto4000MS-HW-0847Mass ofMolecular lon=96 m/zMass Spec102030405060708090100110 120m/zIR3000KAVENUMBERI2000+1500109800HPM-00-2161H NMR13C NMR6ST5ppmChemical Formula:200 180 160 140120100CDS-05-47043210ppmmyProposed Structure:1000500HDI Calculation:- 80-696040200

TRANSMITTINKEIZI 50 LOD Relative Intensity 100 80 60 40 40 Unknown #3 Elemental Analysis: MM HA 24.78% Carbon, 2.08% Hydrogen, remainder is two moles of a halogen 20- L to 4000 MS-HW-0847 Mass of Molecular lon= 96 m/z Mass Spec 10 20 30 40 50 60 70 80 90 100 110 120 m/z IR 3000 KAVENUMBERI 2000 + 1500 10 9 8 00 HPM-00-216 1H NMR 13C NMR 6 ST 5 ppm Chemical Formula: 200 180 160 140 120 100 CDS-05-470 4 3 2 1 0 ppm my Proposed Structure: 1000 500 HDI Calculation: - 80 -69 60 40 20 0

TRANSMITTINKEIZI 50 LOD Relative Intensity 100 80 60 40 40 Unknown #3 Elemental Analysis: MM HA 24.78% Carbon, 2.08% Hydrogen, remainder is two moles of a halogen 20- L to 4000 MS-HW-0847 Mass of Molecular lon= 96 m/z Mass Spec 10 20 30 40 50 60 70 80 90 100 110 120 m/z IR 3000 KAVENUMBERI 2000 + 1500 10 9 8 00 HPM-00-216 1H NMR 13C NMR 6 ST 5 ppm Chemical Formula: 200 180 160 140 120 100 CDS-05-470 4 3 2 1 0 ppm my Proposed Structure: 1000 500 HDI Calculation: - 80 -69 60 40 20 0

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter28: Atomic Spectroscopy

Section: Chapter Questions

Problem 28.15QAP

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Sample IdentificationCodeConcentration of M%TA=2-log(%T)Q50004.00 x 10-417.90.75Q50013.20 x 10-425.00.6Q50022.40 x 10-435.70.46Q50031.60 x 10-450.20.3Q50048.000 x 10-570.80.15SampleIdentificationCode%TA=2-log(%T)AMQ0210150143.70.359518560.360.000192Q0210150244.10.355561410.360.00018Q0210150343.80.358525890.360.00017Q0210150444.10.355561410.360.00018Q0210150543.80.358525890.360.00017What was their percent error?43%Does Batch 021015 meet legal requirements?No, because it is not between 2.85 * 10(4) and 3.15 * 10(4)Well #DropsBluedye1234567891012345678910Drops 9Distilled water876543210Concne 0.26tration0.52Test Tube #0.781.041.3Solutions3Concentration (M)2.082.32.6Concentration (ppm)1:1 dilution11.82Starting Dilution21.562:1 dilution0A.Zero standard0Was your calibration curve as linear as you expected?B.Did you experience any â€œdriftâ€ of the resistance readings?C.What is the equation of your best-fit line?D.What commercial drink did you analyze?E.Assuming…
TERNSHITTANCET THE B 5D LD0 Relative Intensity Unknown #2 Elemental Analysis: 55.81% Carbon, 7.03% Hydrogen, 37.17% Oxygen Mass Spec Exchanges with D₂O 100 MS-NW-4945 40 40 80 20- 10 20 Mass of Molecular lon= 86 m/z 50 30 40 50 60 70 80 90 100 m/z IR 110 120 ww D 4000 3000 2000 MAVENUMBER + 1500 1000 500 12 1H NMR 13C NMR 5.32 ppm 5.25 ppm 6.07 ppm PPM H | | 2.90 ppm Chemical Formula: 200 180 160 140 120 100 80 60 40 20 0 CDS-03-015 HDI Calculation: ppm Proposed Structure:
s for E X Bb Error 401 Bb Exp 51 Postlab - Sulfanilami x M Inbox (1,713) - ranjansah94 X obcswebdav/pid-4022217-dt-content-rid-100630455_1/courses/2212-merged-CHEM2125SEC101-CHEM. (D Page view A Read aloud V Draw 1. This experiment is a multistep synthesis. In the first step as described in the textbook, acetanilide (1) reacts with chlorosulfonic acid to give p-acetamidobenzenesulfonyl chloride (2). HN HO HN O=S=0 ČI 2 To obtain pure (2), it is necessary to destroy the excess chlorosulfonic acid with water. Write the equation showing how chlorosulfonic acid reacts with water. Hint: water acts as a nucleophile. hp
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Analytes than can be trapped inside Sephadex G-25 (1,000-5000) Cytochrome C (MW=12384) Aprotinin (MW=6512) Vitamin B12 (MW=1355) Gly-gly-gly (MW=189)
Gb6cd_i-5c7IV_IVquz8U4UO2epq7z4kBnpYZ5iA/edit > 2 ons Help Calibri - - [12] + B 7 U . UA 2 3 ge 2: Problems of Water and Air Quality Questions 1. Why is it recommended to have a carbon monoxide detector on each floor of a home? Et © ª @ E- E 4-B-E- 2. What impact would the increased use of electric vehicles have on environmental damage caused by acid rain? 3. Why is ground level ozone most likely to reach its highest concentrations on hot, sunny days during periods of heavy traffic in large cities? Page 3: More Air Pollution Issues: Questions 4. The COVID-19 pandemic forced many people to work from home for extended periods. During these periods, many cities reported unusually low levels of smog. Suggest an explanation for this observation. Page 4: Water Pollution: Questions 5. Explain how ozone can be both harmful and beneficial. 6. India relies heavily on burning coal to generate electricity. How does India's reliance on coal contribute to the poor air quality experienced in many of…
Ayurveda is a form of medicine practiced in India. In one study, 20% of U.S.- and Indian-manufactured Ayurvedic medicines purchased through the Internet in 2005 contained detectable levels of several toxic elements. Identify as many peaks as you can in the X-ray fluorescence spectrum of the Ayurvedic medicine. Use a question mark (7) for peaks that cannot be identified. Refer to the table of X-ray fluorescence energies as needed. 49 2 000 10.55 1 800 12.63 9.99 11.84 1 600 1 400 1 200 1 000 6.40 800 25.25 9.20 13.84 600 14.20 14 76 3.70 8.73 400 28.48 4.01 7.06 15.17 200 2 3 4 5 6 7 89 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 X-ray energy (kev) X-ray fluorescence spectrum of Ayurvedic medicine [Data provided by P T. Palmer, San Francisco State University. Energy (ke V) Assignment Answer Bank 3.70 KK, KK, Ca K Se K Sc K 401 Detector response (counts/s)
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An atomic absorption method for the determination of the amount of iron present in used jet engine oil was found from pooling 30 triplicate analyses to have a standard deviation s = 3.1 µg Fe/mL. If s is a good estimate of o, calculate the 95 and 99% confidence intervals for the result 21.7 µg Fe/mL if it was based on the following criteria. (a) a single analysis 95% confidence interval= 21.7 + 99% confidence interval = 21.7 + (b) the mean of two analyses 95% confidence interval = 21.7 + 99% confidence interval= (c) the mean of six analyses 21.7 ± 95% confidence interval = 21.7 + 99% confidence interval= 21.7 + μg Fe/mL μg Fe/mL μg Fe/mL μg Fe/mL μg Fe/mL μg Fe/mL
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I need step by step process solving [4.184x156(Tf-295)] + [4.184x85.2(Tf-368)] = 0. because my work is wrong. Thank you!!