DELFIA® (Dissociation Enhanced Lanthanoid Fluorescence Immunoassay) is a time resolved fluorescence (TRF) assay. Based on the attached figure, the time in us in which the fluorescent signal of the analyte should be read is: Laser pulse 200 400 600 800 1 000 Time (us) Fluorescence intensity

What is ln(A) be with reference to ln(A) vs. time graph? What would 1/A be with reference to 1/A vs. time. graph? Trial 1   Time (s) Absorbance  0 0.600 30 0.522 60 0.454 90 0.392 120 0.344 150 0.300 180 0.261

Given the following size frequency for a dust: Size Interval (um) 7-17.5 17.5-21 21-25 25-28 28-30 30-33 33-36 36-41 41-49 49-70 % by Number 10 10 10 10 10 10 10 10 10 10 a. Plot the cumulative frequency distributions (in %) of the number, surface area, and mass on linear graph paper assuming all particles are spheres with Pp 1.6 g cm ³. = b. Is this a log-normally distributed dust?

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please quickly  ,within 1hour thanks ! If it takes longer than an hour, i don't need the answer (1) List the detectors ofHPLC. Describe the advantage and disadvantage for two wideused detectors

Please find the absorbance and plot te calibration curve

The students conducted the assay for LDH activity using the two serum samples. Each cuvette (path length 1 cm) contained 3ml of a suitable assay buffer (including pyruvate as substrate). 20 microlitres of either serum was added to the cuvette and the absorbance values immediately recorded at the optimum wavelength for a period of 5 minutes (absorbance readings taken every 30 seconds).   Protein concentration of serum sample (mg/ml) Change in absorbance at optimum wavelength per minute Control serum (C) 8 -0.04 Diseased serum (D) 7.8 -0.6 1c. Using the molar absorption coefficient of NADH as 6220 M-1 cm-1, and by application of the Beer-Lambert law, estimate the enzyme activity in the two samples (C and D). Express activity as moles per second.  1d. Estimate the specific activity of the two samples (moles per second per microgram).

Three factors were studied: reactor length, carrier flow rate, and sample volume, with the high and low values summarized in the following table. factor high (+1) level low (–1) level X: reactor length 1.3 cm 2.0 cm Y: carrier flow rate 1.6 mL/min 2.2 mL/min Z: sample volume 100 μL 150μL   The authors determined the optimum response based on the greatest sensitivity, as determined by the change in potential for the potentiometric detector. The following table summarizes their optimization results.   run X* Y* Z* ΔE (mV) 1 –1 –1 –1 37.45 2 +1 –1 –1 31.70 3 –1 +1 –1 32.10 4 +1 +1 –1 27.30 5 –1 –1 +1 39.85 6 +1 –1 +1 32.85 7 –1 +1 +1 35.00 8 +1 +1 +1 32.15 Which factor has the greatest effect on sensitivity (change in potential at the detector)? Which set of conditions optimizes the detection sensitivity?

Analyte in an unknown gave a signal of 9.3 mV.  When 100 mL of 0.211 M standard was added to 100.0 mL of unknown, the signal increased to 16.7 mV.  Find the concentration of the original unknown solution.

A solution of a dye was analyzed by spectrophotometry, and the following calibration data were collected using a 1 cm cuvette. *graph in attachemnt*   Using the calibration data above for the dye solution, what is the dye concentration in a solution with an absorbance (A) = 0.52 if it was measured in a 1 cm cuvette?   Group of answer choices 3.0 x 10-6 M 6.0 x 104 M 2.0 x 10-6 M Not enough information is provided.