The following is a standard solution of phosphorus (P) data measured by a spectrophotometer. ppm P Absorbansi 2150 0.053 4260 0.110 6400 0.173 8520 0.220 A sample of 2.1 g NA2HPO4 is dissolved and diluted to a volume of 100 mL. The absorbance of the diluted solution was measured with a spectrophotometer and obtained an absorbance of 0.128. What is the percentage of phosphorus in the sample?
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- The protein concentration of a protein isolate was determined using the Bradford assay. Five dilution of bovine serum albumin (BSA) were prepared from a 5.00 mg/mL BSA stock solution. To these dilutions, 100 µl of Bradford reagent were added. After 5 minutes, the absorbance was taken at 595 nm. Standard # BSA, in mg/mL Volume of Bradford reagent, A595 in mL 1 5.0 0.000 1.00 5.0 0.134 3 2.00 5.0 0.265 4 3.00 5.0 0.388 4.00 5.0 0.497 The supernatant dilution was prepared by mixing 49 µL of the protein isolate with 16 µL water. The absorbance was taken 5 minutes after addition of 100 mL Bradford reagent. Calculate the protein concentration (in mg/mL) of the original isolate if the absorbance at 595 nm was 0.413. Note: Final answer format must be x.xxx (three decimal places). Round off only in the final answer. Do not round off in the middle of calculation.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 MTo test a spectrophotometer’s accuracy, a solution of 60.06 ppm K2Cr2O7 is prepared and analyzed. This solution has an expected absorbance of 0.641 at 350.0 nm. Several aliquots of the solution produce the following absorbance values. 0.640 0.638 0.640 0.639 0.640 0.639 0.638 is there any significant difference between the experimental mean and the expected value at a 99.9% confidence level? Question 21 options: Yes because tcal<tcritical Yes because tcal>tcritical No because tcal<tcritical No because tcal>tcritical
- A 10.00 mL of natural water sample containing Ni2+ was pipetted into a volumetric flask and diluted to 50.00 mL with pure water. In the second 10.00 mL of natural water sample transferred into a volumetric flask, exactly 4.00 mL of a Ni2+ solution with a concentration of 5.99 mg/L was added and then diluted to 50.00 mL with pure water. The measured absorbance is A1= 0.436 and A2 = 0.663. What is the Ni2+ concentration in unit of mg/L in the natural water sample?A student was given a stock aluminum(III) solution with a concentration of 5.000 parts per million (ppm). (ppm are defined as mg/L for dilute aqueous solutions.) The student prepared five 100 mL standard solutions and an unknown as described in the procedure section of the experiment. The absorbance of each solution was read at 565 nm, using the blank to set zero absorbance. The results are tabulated below. Solution Volume of Al(III) stock solution used (mL) Absorbance 1 8.00 0.701 2 6.00 0.548 3 4.00 0.378 4 2.00 0.208 5 1.00 0.123 Unknown --- 0.534 3. What is your best estimate of the Al(III) concentration of the unknown sample in the cuvet based upon where its absorbance falls on the standard curve?The 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.
- What are the differences between systematic and random errors and how do they effect accuracy and precision? In what circumstances would you use standard addition (versus a normal calibration curve) to determine the amount of an analyte in a sample? A urine sample, containing analyte Z is analysed by the standard addition method where 5 mL of the original sample was mixed with increasing amounts of a Z standard and each solution diluted to a volume of 50 mL prior to analysis. A plot of the final concentration of the standard in each of the 50 mL samples (x axis) versus The measured signal from the analysis of each 50 mL sample (on y axis) produced a straight line with the general equation: y = 44.72x + 4.06 what was the final concentration of Z in the 50 mL standard addition sample? what was the initial concentration of Z in the original urine sample?3ASSAY Calibration solutions Calibration solutions of naproxen in the range 5 – 25 μg/mL were prepared. Sample preparation 20 tablets weighing 12.3819 g were crushed to a fine powder. A portion of the powder (145.4 mg) was shaken with approximately 150 mL of acetic acid (0.05 M) for 5 min and then made up to volume in a 250 mL volumetric flask (stock solution). Approximately 50 mL of the stock solution was filtered and a 25 mL aliquot was diluted to 100 mL in a volumetric flask. 10 mL of the resulting solution was further diluted to 100 mL with acetic acid (0.05 M). Analysis The standards and sample solutions were analyzed by HPLC under the following conditions: Column: octadecylsilyl (ODS), 4.6 mmx150 mm, mobile phase: acetonitrile : 0.05 M acetic acid (85:15), flow rate: 1 ml/min, UV detection at 243 nm. Results: A calibration curve of concentration versus peak area was constructed for the standard solutions and gave the straight-line equation: y = 3555.6x + 85, r = 0.9999 The area…
- From the following data, calculate the concentration of the analyte in the sample read at 700 nm: Absorbance of unknown sample = 0.807 Absorbance of a 130 mg/dl standard = 0.234 Do not answer in image format. Maintain accuracy and quality in your answer. Answer completely.After determining the absorbance of several standards of known concentration, the trend line for a calibration curve plotting absorbance (y-axis) against concentration (uM, x-axis) is determined to be y=4.244x+0.0000. The absorbance of a solution of unknown concentration is determined to be 0.604. Calculate the concentration of the unknown solution in microM. Give answer to 3 decimal placesTable 2: Absorbance Values of Standards and Unknowns. Sample Blank Standard Solution 1 Standard Solution 2 Standard Solution 3 Standard Solution 4 Standard Solution 5 Water Sample 1 Water Sample 2 Phosphate Concentration (in ppm) Y17 0 ppm 0.02 ppm 0.04 ppm 0.08 ppm 0.16 ppm 0.32 ppm Freeman Lake 0.001 0.325 0.292 0.413 0.315 0.039 0.054 0.049 Absorbance at 2= 690 nm Calculations: 1. Construct a phosphate standard curve in Excel by plotting concentration (in ppm) on your x-axis and Absorbance (unitless) on your y-axis for your known solutions. Label the axes on the graph and provide the curve with a title. Use a linear trendline to generate a best fit line to your data. Label the graph with the equation and the R" value. Insert your labeled graph in the space below. X