A peak elutes from an HPLC column 11.4 cm in length in 13.9 min. What would be the width at half-height of the peak (in seconds) if the plate height were 5.22 μm? W1/2 = What would be the width at half-height of the peak (in seconds) if the plate height were 33.0 μm? W1/2 = S
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- A peak elutes from an HPLC column 17.4 cm in length in 11.4 min. What would be the width at half-height of the peak (in seconds) if the plate height were 4.66 µm? W1/2 What would be the width at half-height of the peak (in seconds) if the plate height were 42.0 µm? W 1/2 = S SA peak elutes from an HPLC column 13.8 cm in length in 14.8 min. What would be the width at half-height of the peak (in seconds) if the plate height were 7.75 µm? W 1/2 = What would be the width at half-height of the peak (in seconds) if the plate height were 46.0 μµm? W 1/2 = S SA peak elutes from an HPLC column 13.3 cm in length in 15.3 min. What would be the width at half-height of the peak (in seconds) if the plate height were 8.62 µm? wi/2 S What would be the width at half-height of the peak (in seconds) if the plate height were 43.0 µm? Wi/2 = S
- A typical HPLC column is 15.0 cm long and 4.6 mm in diameter. What is the internal volume of the column? What is the volume of a 5-μm diameter particle (approximate it as a hard sphere)? Because of the way spherical particles pack in a column, the volume that is external to the particles is approximately 40% of the entire column volume, meaning that 60% of the column volume is occupied by the particles. Given this, the total column volume you calculated in part (a), and the volume of a particle you calculated in part (b), calculate the number of particles packed in a 15.0 cm x 4.6 mm columnA peak elutes from an HPLC column 13.8 cm in length in 14.8 min. What would be the width at half-height of the peak (in seconds) if the plate height were 7.75 µm? W 1/2 4.1 W 1/2 Incorrect What would be the width at half-height of the peak (in seconds) if the plate height were 46.0 µm? 24.3 Incorrect S SThe absorbance of a cationic iron(II) sample solution was measured in a spectrophotometer, but the instrument returned an error because the absorbance was too high. The sample was then diluted by using a pipette to take 100.0 μLμL of the sample and injecting it into a cuvette already containing 2.00 mLmL of water (total volume is 2.00 mLmL + 100.0 μLμL). The absorbance value of the diluted solution corresponded to a concentration of 7.71×10−6 M M . What was the concentration of the original solution? Express the concentration to three significant figures with the appropriate units.
- The table contains data for a liquid chromatography separation of four compounds, A, B, C, and D, on a 45.0 cm column. Retention Width at the Compound time (min) base (min) A 5.44 0.439 B 13.3 1.09 C 14.2 1.18 D 23.8 1.96 Calculate the number of theoretical plates, N, for each peak. NA = NB =For a concentration technique, Stotal is given as 19.31 ± 0.035, Smb is 0.22 ± 0.008, and kA is 0.154 1 0.007 ppm, where Stotal is the signal, kA is the method's sensitivity for the analyte and Smb is the signal from the method blank. If we want the absolute percent uncertainty of the concentration CA to be 2.6%, which statement is false below? (Hint: refer to sections 4B.1 and example 4.7 in your textbook) We must improve the uncertainty in KA to 10.004 ppm-¹. O Uncertainty in the method's sensitivity dominates the absolute uncertainty. O Improving the signal's uncertainty will improve the absolute uncertainty.A series of standards were analyzed which gave the following results SOLUTION % Transmittance 0 mL of 0.100 mg/mL Hg + 2 mL tinchloride in a total volume of 75 mL 100 1.0 mL of 0.100 mg/mL Hg + 2 mL of tinchloride in a total volume of 75 mL 81.3 2.0 mL of 0.100 mg/mL Hg + 2 mL of tinchloride in a total volume of 75 mL 63.9 3.0 mL of 0.100 mg/mL Hg + 2 mL of tinchloride in a total volume of 75 mL 55 5.0 mL of 0.100 mg/mL Hg + 2 mL of tinchloride in a total volume of 75 mL 35.1 How would you make a Beer-Lambert Plot with this information?
- To determine the molar concentration of a metal ion in a solution of unknown concentration, a student fırst made five standard solutions that contain the metal ion of interest and measured the absorbance of each solution in a spectrophotometer at its Amax- A calibration curve was obtained that had an equation of y = 5.747 x + 0.013 Next, the student pipetted 15.0 mL of the initial solution of unknown concentration into a 100.0 mL volumetric flask, and filled the flask with deionized water to the line. The absorbance of this final diluted solution was found to be A = 0.226 at Amax. The color of the original and diluted solution was blue. What is the molarity of the original solution, as well as an approximate Amax for this metal ion? 2max = 599 nm and concentration is 0.247 M Amax 457 nm and concentration is 0.247 M %3D 2 max = 457 nm and concentration is 0.0371 M 1 max 599 nm and concentration is 0.0371 M Amax = 599 nm and concentration is 0.00557 MMany ions form colored solutions when dissolved in water. For example, aqueous Cu2+ solutions are light blue and aqueous Ni2+ solutions look light green. These metals in particularare often found in stainless steel. Would you expect their presence in your samples to interfere in the spectrophotometric analysis of permanganate at 525 nm? Why or why not? I think the answer is that I would expect the Cu2+ in my samples to interfere but not Ni2+ because the Ni2+ is green, signifying a higher wave length that the spectrometer wouldn't be able to pick up on because it was only set to 525 (blue-ish) where as the green is 530-ish. I just want to make sure thanks!Example: You wish to prepare a calibration curve for the spectrophotometric determination of permanganate. You have a stock 0.100 M solution of KMnO4 and a series of 100 mL volumetric flasks. What volumes of the stock solution will you have to pipet into the flasks to prepare standards of 1.00 x10³, 2.00 x10-³, 5.00 x103, and 10.0 x103 M KMnO4 solutions? Solution: for 1.00 x10-³ 0.1 (M1X V1)conc. = (M2 X V2) dil mmol |x V1 = 1.0 × 10-3 -) x 100 (ml) mL mmoly mL V₁= 1.0 mL stock solution. Take home work for the 2.00 x10-3, 5.00 x10³, and 10.0 x10-³ M.