Exactly 5.00-mL aliquots of a solution containing phenobarbital were measured in

Question

Exactly 5.00-mL aliquots of a solution containing phenobarbital were measured into 50.00-mL volumetric flasks and made basic with KOH, 0, 0.1, 0.15, 0.2, 0 25 and 0.3 mL volumes of 2.0 mu g/mL of standard solution of phenobarbital were then introduced into each flask and the mixture was diluted to volume of 50 mL. The fluorescence of each of these solutions was measured with a fluorometer, which gave values of 1.300, 3.26, 4.80, 6.41. 802, and 956, respectively. Plot the data. Using the plot from calculate the concentration of phenoharbital in the unknown. Find the concentration of phenobarbital from the equation in (c) Calculate Signal to Noise Ratio for the audio signal series (50, 25, 22, 35, 28) The following data were collected for fluorescence measurements on a noisy system: 0.101.0134. 0.109, 0.122, 0. 43, 0, 105 Assuming the noise random. Calculate the signal-to-noise ratio. How many measurements must be done to have S/N = 20. Calculate the frequency in Hertz, wave number, energy(J) and kJ/mol for electromagnetic radiation 330 nm. 2400 nm. Data is collected for a chemical analysis Blank: Sample: 0.002 0, 038 0.003 0.032 0.001 0.034 0.005 0.030 0.001 0.039 Calculate the signal to noise ratio for the following measurement Suppose the absorbance signal of sample for 0.2 ppm Pb. What would be the limit of detection (LOD) and limit of quantitation (LOQ) for Pb. Calculate the standard deviation of the following set of measured values: 3.15.3.21.3.18.3.30, 3.25. 313, 3.24, 341. 313. 3.42, 3.19 If this is measurement from a chemical system, what is the precision? Calculate the signal to noise ratio of this measurements

Explanation / Answer

Ans 4- (a) WAVELENGTH = 330 Nanometers

frequency = C /WAVELENGTH

C= 3*10^8

Frequency = 9.09*10^14 Hz

Energy = hC/ wavelength

h= plancks constant = 6.626*10^-34 J.s

Energy = 6.095*10^-19 J

=6.095*10^-16 KJ

wave number = K= 1/ wavelength

K=3.03*10^6 m-1

Same for part (b)

(b)- Wavelength = 2400 nm

Frequency = 1.249*10^14 Hz

Energy = 8.276 * 10^ -20 J

= 8.276*10^-17 KJ

Wave number = 416,666.6 m-1

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