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- use the following chromatogram and table as your data. VWD: Sgrad A 284 nm Retention Time 40- -40 20 20 5 10 15 20 25 30 35 Minuteo VWD: Signal A, 284 nm Retention Area Identity Time (min) 7.867 3214697 Vitamin A 8.493 15687 19.410 140354 19.710 64346 20.803 884315 Vitamin E 21.640 36745 22.100 63054 8.403 L98 ÞI 20.803 ; 84年 81Q24) Consider the following data obtained from the size exclusion chromatography of polypropylene, which was made from propylene of molecular weight 42.08 g/mol. S.No. Number of Molecular Weight Molecules Range (g/mol) 5 8,000 - 16,000 16 16,000 - 24.000 3. 24 24,000 - 32,000 4 28 32,000 - 40,000 5 20 40,000 - 48,000 48,000 – 56,000 From the above data, determine a) Number average molecular weight (M). b) Weight average molecular weight ( Mw). c) Degree of polymerization (DP)1) Study the chromatograph (below) of a mixture of Compounds A and B, run on the GCs in the teaching labs at CU Boulder. Compound A has the shorter retention time. STAAT 61 1.11 227 RT TYPE AREA XXXX XXXX XXXX AREAS 0.009 55874 44.117 ARIHT 0.61 XX XX XX 1.11 2.27 XX XX XX TOTAL AREA=XX MUL FACTOR=XX 1. What is the retention time of compound A? Compound B? 2. Which compound is present in a larger amount? 3. Which compound has the lower boiling point? 4. What would happen to the retention times of compounds A and B if the column temperature were raised? 5. You suspect that compound B is octane. What can you do to provide supporting evidence for this hypothesis?
- 16:24 5 ê ☺ I o The glucose concentrations in a patient with mildly elovated glucose lovels were determined by a spectrophotometric analytical method and wore recorded in mot, as tolows 700 775 7eg 774 T77 T80 and 758 Caculato the standard devialion OA 114 OR 135 OC124 O. 129 OL 222 Add a caption... > Status (Custom) +Which one of the following chromatographic techniques is useful in the identification of organic molecules having molecular weights more than 20,000? 1) reverse phase chromatography 2) normal phase chromatography 3) high pressure liquid chromatography 4) gel permeation chromatography透过率/% 3081 20 3012 26 2985 23 2919 17 2861 14 2710 60 2443 81 100 12.1 Which compound is consistent with the spectrum shown below? 薄膜法 2008 1500 1000 500 */cm-¹ OH 4000 1982 8B 1848 81 1645 45 1463 29 1423 17 1348 49 1314 49 3000 1233 44 1181 74 1115 18 1028 4 993 4 918 7 888 64 645 41 556 47 441 84 a b C d OH
- ef Correlate the structures of the standard food dyes to their relative position in the paper chromatogram. Refer to the structures shown below. H3C. *Na O O HO. O CH3 Allura Red AC (pigment in red dye) 01510 106 FO O Nat O Nat O Nat FO O Na+ 0=8=0 HO 0=S=0 Na Na+ Tartrazine (pigment in yellow dye) Brilliant Blue FCF (pigment in blue dye) What are the identities of the unknowns in paper chromatography of food dyes? How did you arrive at this conclusion? Differentiate normal phase chromatography from reverse phase chromatography. From which of the two governs the experiment you did in thin layer chromatography of plant pigments? How do you say so?% Abundance 100- 60 80- 40 20 0 D) Consider the following spectrum. Identify the structure of the compound. 0 C6H12 15 20 29 40 43 m/z 56 60 69 84 80 100Which one of the following compounds is consistent with the mass spectrum below? 100 80 Relative Intensity ő 20- B SDBS: 0-fttm MS-NU-0502 10 20 30 A) (CH3CH₂CH2)2CH C) (CH³CH₂CH₂)2O 40 CH3CH₂CHOHCH₂CH₂CH3 D (CH3CH₂CH₂)2NH 50 60 m/z 70 80 90 100 110
- Polarities of analyte functional group increase in the order of hydrocarbon < ethers < esters < ketones < aldehydes < amides < amines < alcohols Arrange the order of elution [from shortest to longest retention time] for the following six compounds from an HPLC C18 column. C2H5OH CH3COCH3 CH3COOH CH3COOC2H5 C2H50C2H5 (СН3)2NH O A. C2H5OH < CH3COOH < (CH3)2NH < CH3COCH3 < CH3COOC2H513. Electrospray time-of-flight mass spectrometry was used to analyze the eluate from a high-performance liquid chromatography separation. The mass spectrum of one chromatographic peak, containing a protein of unknown molecular mass, displays MH peaks at m/z = 5 654.208, 5 277.326, 4 947.590, 4 656.613, 4 397.990, and 4 166.576. Find the average molecular mass (M) of the neutral protein and its standard deviation.Given the following GC-MS spectra. A. Draw structure of compound. B. Label base peak and molecular ion peak. Benzene 100 MS-NU-0075 80 - 60 40 20 - anone pluylyplıu 45 0. 10 15 20 25 30 35 40 50 55 60 65 70 75 80 m/z Relative IntensitySEE MORE QUESTIONS