Please help me %Transmittance501245803056.011656.001626.721595.461576.661447.911317.921278.361176.441150.841075.45999.14943.29919.94764.75813.15695.44638.69436.55-Wavenumbers (cm-1)00s700s00100S0007000800sE000For55-F096570어75-85-F6FO0L105-110-115 Berizophenonne-Starting Material

IR spectroscopy mainly used to determine the functional groups that are present in a molecule In…

Answered: %Transmittance 30 40 50 80 3056.01…

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter25: Instruments For Optical Spectrometry

Section: Chapter Questions

Problem 25.17QAP

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Similar questions

Use the data in Figure 4.8 to estimate the ratio of radiation intensity at 10,000 Å (infrared) to that at 5000 Å (visible) from a blackbody at 5000 K. How will this ratio change with increasing temperature? Explain how this change occurs.
Determine the operational wave length for the data Wavelength (nm) Transmittance (%T) 350 100.0 375 92.2 400 86.6 425 75.3 450 54.6 475 37.6 500 28.2 525 31.0 550 54.6 575 77.2
What is threshold frequency?
Why does fluorescence tend to be the mirror image of absorption?
The drug pentobarbital is a sedative. It was extracted from a urine sample during a particular ultraviolet-visible spectrophotometric forensic analysis procedure. The extraction solvent was evaporated and the residue reconstituted in a small volume of the transparent solvent methanol. The pentobarbital solution thus obtained had a percent transmittance (%T) of 91 at the relevant wavelength of maximum absorbance. If the urine analyte concentration were halved, what would its percent transmittance be? (You may assume that the Beer-Lambert law holds). Can you show all the steps of the solution
A beam of electrons with a speed of 600 m/s passes through double slit with the slits spaced 2 μm apart. If the detector is 2 meters away from the double slit, how far off the center will the m = 1 signal appear?
Potassium dichromate and potassium permanganate have overlapping absorptionspectra in 1M H2SO4 . K2Cr2O7 has an absorption maxima at 440 nm, and KMnO 4 has a band at 545nm (the maximum is actually at 525 nm, but the longer wavelength is generally used whereinterference from K2C2O7 is less). A mixture is analyzed by measuring the absorbance at these twowavelengths with the following results: A 440 =0.405, A 545 =0.712 in a 1-cm cell. The absorbances ofpure solutions of K2Cr2O7 (1x10 -3 M) and KMnO 4 (2.00x10 -4 M) in 1 M H 2 SO 4 , using the same cell gavethe following results: A Cr,440 = 0.374, A Cr,545 = 0.009, A Mn,440 =0.019, A Mn,545 = 0.475. Calculate theconcentrations of dichromate and permanganate in the sample solution.
An absorption measurement with a 1 cm path length yields a reading of 0.004 absorbance units with a noise of 0.001 absorbance units and a mean noise reading of 0 for 10 scans averaged together. Assuming the noise for a single scan does not change if the path length is increased to 6 cm and the number of scans is increase to 100, what is the signal to noise of the new measurement.
What issues experienced with fluorescence detectors are resolved using chemiluminescence detectors?
A 2.0 mm cell was filled with a solution of benzene in a non-absorbing solvent. The concentration of the benzene was 0.010 mol dm−3 and the wavelength of the radiation was 256 nm (where there is a maximum in the absorption). Calculate the molar absorption coefficient of benzene at this wavelength given that the transmission was 48 per cent. What will the transmittance be through a 4.0 mm cell at the same wavelength?
Convert the accompanying transmittance data to absorbances: (a) 25.5% (b) 32.8% (c) 0.085
Why are wavlength maxima different for UV-Vis and Fluorescence?

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