The relative displacement (r/ro) of bovine serum albumin was observed as a function of time: t (s) 700 3580 4540 5020 r/r. 1.0129 1.0679 1.0871 1.0965 Given w = 6 260 /s, find the sedimentation coefficient. Assuming v = 7.34x 10 3 m3 kg and p = 9.93 x 102 kg/m3, diffusivity = 6.97 x 10-11 m2/s at 25 °C, determine the molar %3D %3D mass of the sample in kg/mol.
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- Two components in an HPLC separation have retention times thatdiffer by 22 s. The first peak elutes in 10.5 min and the peakwidths are approximately equal. Use a spreadsheet to find theminimum number of theoretical plates needed to achieve thefollowing resolution, Rs, values: 0.50, 0.75, 0.90, 1.0, 1.10, 1.25,1.50, 1.75, 2.0, and 2.5. How would the results change if peak 2were twice as broad as peak 1?The following volumes of a solution containing 1.10 ppm Zn2+ were pipetted into separatory funnels each containing 5.00 mL of an unknown zinc solution: 0.00, 1.00, 4.00, 8.00 and 12.00. Each was extracted with three 5 mL aliquots of CCl4 containing an excess of 8-hydroxyquinoline. The extracts were then diluted to 25.0 mL and their fluorescence measured with a fluorometer. The results for the standard addition calibration were as follows: Volume Std. Zn2+ / mL [Zn2+] / ppm Fluorometer Reading 0.00 6.12 1.00 7.41 4.00 11.16 8.00 15.68 12 20.64 a) Determine the concentration on Zn2+ ions in each of the standard solutions coming from the coming from the 1.10 ppm Zn2+ solutions, Please show your workingIn coulometric karl Fisher water analysis, 25.0 mL of pure “dry” methanol required 4.23 C to generate enough I2 to react with residual H2O in methanol. A suspension of 0.8476 g of finely ground polymeric material in 25.0 mL of the same “dry” methanol required 63.16 C. Find the weight percent of water in the polymer.
- Two substances A and B have retention times of 16.40 and 17.63 min respectively on 30.0 cm column. An unsuccessful species passed over the column in 1.30 min. The widths (at the base) of the peaks ofA and B are respectively 1.11 min and 1.21 min: Calculate: a) The Resolution Rs of the column. b) The average number (Nav of theoretical plates in the column c) The equivalent height (H) d) length of column required to achieve a resolution of 1.5, e) time required to elute substance B on the longer column.A solution of Tryptophan has an absorbance at 280 nm of 0.54 in a 0.5 cm length cuvette. Given the absorbance coefficient of trp is 6.4 x 103 Lmol-1cm-1. What is the concentration of solution?A spectrophotometric method for the quantitative determination of the concentration ofPb2+ in blood yields an Ssamp of 0.133 for a 1 mL sample of blood that has been diluted to 6 mL. A second sample is spiked with 1 µL of a 1467 ppb Pb2+ standard and diluted to 6 mL, yielding an Sspike of 0.491. Determine the concentration of Pb2+ in the original sample of blood.
- Calculate the amount of phycocyanin in Sample 1 in mg where A620 = 0.193 and A650 = 0.095, taking into account the dilution factor of 100 ul, and the total volume of extract 45ml. Note your answer to 2 decimal places.Using the absorbance of [FeSCN]+2 at the wavelength of maximum absorption which you obtained in part B, and the diameter of the tube, obtained in part A,calculate the molar extinction coefficient. given info: Concentration of Fe3+ stock solution, C1 = 0.001 M Volume of stock Fe3+ solution pipetted, V1 = 25mL Concentration of diluted stock solution, C2 = 0.0004 M Volume of Fe2+ solution and KSCN (upon mixing) = 10mL Weight of KSCN = 1g The reaction taking place is given as: The reaction taking place is given as: F e 3+ (aq) + SC N − (aq) ⇔Fe [SCN] 2+ (aq) the concentration of [FeSCN]2+ after the mixing of Fe3+ and SCN- is 2.5 x 10-3 M. Moles of [FeSCN]2+ = 2.5 x 10-5 moles Diameter:1.2cm The wavelength at which maximum absorption takes place is = 480 nmA 5.00-mL sample of blood was treated with trichloroacetic acid to precipitate proteins. After centrifugation, the resulting solution was brought to pH 3 and extracted with two 5-mL portions of methyl isobutyl ketone containing the lead-complexing agent APCD. The extract was aspirated directly into an air/acetylene flame and yielded an absorbance of 0.527 at 283.3 nm. Five-milliliter aliquots of standard solutions containing 0.400 and 0.600 ppm of lead were treated in the same way and yielded absorbances of 0.396 and 0.599. Find the concentration of lead in the sample in ppm assuming that Beer’s law is followed.
- 0.9583 g of sample was dissolved in 80 mL of xylene and extracted with three 15 mL quantities of water. The combined aqueous extracts were diluted to 50.0 mL and then shaken with 10 mL toluene. The toluene was then discarded. The absorbance of the aqueous solution when measured in a 1cm path length cell was 0.134 at 278 nm.Calculate the free chlorampenicol content of the sample in ppm, given A(1%, 1cm) = 298 at 278 nm.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 molar absorptivities of compounds X and Y were measured with pure samples of each: ε (M-1 cm-1) λ (nm) X Y 272 16,440 3,870 327 3,990 6,420 A mixture of compounds X and Y in a 1.00-cm cell had an absorbance of 0.957 at 272 nm and 0.559 at 327 nm. Find the concentrations of Y in the mixture. 4.43 x 10-5 M 3.03 x 10-6 M 5.95 x 10-5 M 3.56 x 10-4 M