A sample of Melbourne drinking water was injected into an ion chromatograph. The anions were separated isocratically via ion-exchange using 4 mM NAOH as eluent and detected by conductivity after chemical suppression. The system void volume (t) occurred at 1.25 minutes, while chloride and sulfate were eluted at 3.12 and 6.62 minutes respectively. The eluent concentration was then increased to 10 mM The new retention time for chloride is:
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- Benzene and toluene, when analysed isothermally on a 25 m × 0.25 mm i.d. capillary GC column, were found to have retention times of 8.56 and 8.67 min, respectively. An unretained solvent passed through the column in 4.10 min in the same analysis. The peak widths at half height for benzene and toluene were 4.2 and 4.3 seconds, respectively. Calculate: a) The capacity factor for each of these compounds b) The selectivity factor for the separation c) The average number of plates for the column d) The resolution between benzene and toluene. Are they completely separated under these conditions? e) What is the plate height (H) for this column? What can you say about the efficiency of this column? Explain f) The dimensions of a column with the same phase ratio required to obtain a resolution of 1.5 under otherwise identical conditionsPesticide residues were found in a lot of oranges; the compound identified was the epoxide of heptachlor (EH). To quantify the amount of this pesticide in oranges, 20 grams were taken from fruit peels and mixed with methylene chloride to extract it. The concentration of heptachlor epoxide in the sample was determined by gas chromatography coupled to a detector mass (GC-MS); and calibration was performed using anthracene (A) as internal standard. For In this analysis, the anthracene had a concentration of 350 ppm and the same volume was added to each standard. The table below shows the concentrations of the standard and the ratio of the areas.and the sample is: 0.1080.Determine the concentration of heptachlor epoxide in the sample.in HPLC, what does it mean to equilibrate the system. Let us assume that the system is currently running 100%B. You have told them to run the first standard of caffeine using Reverse Phase chromatography with a C18 column, 30% A (Water/0.1%TFA) and 70% B (MeOH/0.1%TFA).What should I equilibrate the system to and for how long?
- A molecular exclusion column has a diameter of 7.8 mm and a length of 30 cm. The solid portion of the particles occupies 20% of the volume, the pores occupy 40%, and the volume between particles occupies 40%. (a) At what volume would totally excluded molecules be expected to emerge? (b) At what volume would the smallest molecules be expected? (c) A mixture of polymers of various molecular masses is eluted between 23 and 27 mL. What does this imply about the retention mechanism for these solutes on the column?Internal standard. A mixture containing 12.8 mM analyte (X) and 44.4 mM standard (S) gave chromatographic peak areas of 306 for X and 511 for S. A second solution containing an unknown quantity of X plus 55.5 mM S had peak areas of 251 for X and 563 for S. Find [X] in the second solution.In ion-exclusion chromatography, ions are separated from nonelectrolytes (uncharged molecules) by an ion-exchange column. Nonelectrolytes penetrate the stationary phase, whereas ions with charge of the same sign as that of the stationary phase are repelled by the stationary phase. Because electrolytes have access to less of the column volume, they are eluted before nonelectrolytes. A mixture of trichloroacetic acid (TCA, pKa 5 20.5), dichloroacetic acid (DCA, pKa 5 1.1), and monochloroacetic acid (MCA, pKa 5 2.86) was separated by passage through a cation-exchange resin eluted with 0.01 M HCl. The order of elution was TCA , DCA , MCA. Explain why the three acids are separated and the order of elution.
- In a High Performance Liquid Chromatography (HPLC) analysis, toluene exhibits a capacityfactor of 3.5 on a 15.0 cm column. The unretained solute is eluted at 1.33 min (tm). a) Determine the retention time tR of toluene. b) Calculate the number of theoretical plates and the plate height that would produceatoluene peak eluting at 12.83 min with a width at the base of 18.4 sA new method is proposed to simultaneously detectand separatefour foodadditives by reversed-phase liquid chromatography. The data were obtained using theliquid chromatographic column with a flow rate of 0.5 mL per min and with the followingparameters: Pore size 100 A Particle Size 3.5 Internal diameter 4.6 mm Length 150 mm The food additives found in the sample resulted in the following data: ( see image for table) a.Calculate the number of plates from each peak. b.Calculate the plate height for the column. c.Retention factor for each peak.A 0.0200 gram blood sample was decomposed by a microwave digestion technique followed by dilution to 100.0 mL in a volumetric flask. Aliquots of the sample solution were treated with a lead complexing reagent and water as follows: Solution 1: 10.0 ml blood sample + 20.0 mL complexing agent + 30.0 mL H20. Solution 2: 10.0 ml blood sample + 20.0 mL complexing agent + 26.0 mL H20 + 4.00 mL of 78 ppb Pb2+ standard. The resulting solutions were analyzed by UV/Vis at 375 nm. Absorbance for solution 1 = 0.155 and for solution 2 = 0.216. Calculate the concentration of lead (ppb) in the original sample.
- Alfred was tasked to synthesize at least 250.0 ppm of his novel gold nanoparticles (AuNP’s) to be able to perform experiments about the efficacy of AuNP for viral detection. He decided to check whether he still needed to synthesize more or not using a UV Vis spectrophotometer. Fortunately, he already prepared a 100.0 ppm AuNP standard solution beforehand. He got a different aliquot of the standard, added stabilizing solutions, and diluted them to 10.0 mL for the determination of absorbance. Data he obtained is shown below. Table 1. Calibration curve data for AuNP v of 100.0 ppm AuNP standard, mL Absorbance 0.00 0.200 0.50 0.450 1.00 0.510 1.50 0.600 2.00 0.680 3.00 0.810 4.00 0.950 After obtaining the calibration curve, he got a 3.00-mL aliquot of the pre-prepared solution, added stabilizing solution, and diluted it to 10 mL. He got 10mL of the final solution and determined its absorbance to be 0.650. The absorbance of the blank…The following data give the recovery of bromide from spiked samples of vegetable matter, measuredusing a gas–liquid chromatographic method. The same amount of bromide was added to each specimen(Roughan, J.A., Roughan, P.A. and Wilkins, J.P.G., 1983, Analyst, 108: 742).Tomato: 777 790 759 790 770 758 764 ug/gCucumber: 782 773 778 765 789 797 782 ug/g(a) Test whether the recoveries from the two vegetables have variances which differ significantly.(b) Test whether the mean recovery rates differ significantly.Provide a brief explanation of one specific application on High Performance Liquid Chromatography (HPLC). Incorporate the specific application name (e.g., Detection and Quantification of Mycotoxins), provide aconcise overview of sample preparation methods, outline instrumental parameters and conditions utilized, and summarize the outcomes and findings achieved through this analytical approach.