A 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? w1/2 = What would be the width at half-height of the peak (in seconds) if the plate height were 43.0 µm? W12 =
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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 13.8 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 S
- 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?A 50 mL aliquot of river water is treated with a conditioning solution, which is stirred and recorded at 500 nm. The equipment reading indicated 40.5% Transmittance with 1.5 cm optical path cuvettes. Calculate the ppm of sulfate (PM 96 g / mol) in river water if a standard of 3.0x10-5 mol / L of sulfate, under the same conditions, gave 55% Transmittance in cuvettes with 1.5 cm of passage. optical.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 photon of light has a wavelength of 1967 nm. Calculate the energy for 1 photon of light. h = 6.626 x 10-34 J*s, c = 3.00 x 10° m/s, Na = 6.02 x 1023 %3D 1.525 x 1014, 60,840 J 1.011 x 10-34, 1.011 x 10-19,Find the [FeSCN^2+]eq ε=4.0×10^3 M-1 cm^-1 at 447 nmThe 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.
- Drinking water with negligible lead content was fortified with 2.28 µg Pb(II)/L. Triplicate measurements yielded a mean of 1.90 µg Pb(II)/L and a standard deviation of 0.2 µg Pb(II)/L. Find the percent recovery of the spike. 100 percent recovery: % IncorrectMany 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!Indicate the relationship that exists between the transmittance expressed as a percentage, T(%), and the absorbance, A.(1). A = - log [1/T(%)] (2). A + log T(%) = 2 (3). A = [log T(%)] / 100 (4). A = 100 log T(%)