Five uL of a 10 to 1 dilution of a sample were added to 5 mL of Bradford reagent. The absorbance at 595 nm was 0.78 and, according to a standard curve, corresponds to 0.015 mg of protein on the x axis. What is the protein concentration of the original solution? Why?
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Five uL of a 10 to 1 dilution of a sample were added to 5 mL of Bradford reagent. The absorbance at 595 nm was 0.78 and, according to a standard curve, corresponds to 0.015 mg of protein on the x axis. What is the protein concentration of the original solution? Why?
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- Blue Blue dye stock solution Absorbance at 630 nm Calibration curve Part 1: 0.1944 M 0.00697 y = 0.0682x A purple solution is prepared by mixing 5.00 mL of the blue dye solution with 6.75 mL of red dye solution. The calibration curve for each color is provided. Part 2 out of 3 (a) Determine the theoretical molar concentration of each dye in the mixture. [Blueltheoretical = 8.27 x 10-2 [Red] theoretical = 1.16 [Bluelexperimental [Redlexperimental " *10-1 M M x 10 (b) Using the absorbance and calibration curve for each dye, calculate the experimental molar concentration of each dye in the mixture. x 10 Red Red dye stock solution Absorbance at 500 nm Calibration curve 0.2015 M 0.00193 y = 0.0170x M MA student is analyzing a tap water sample for Mg by atomic absorption spectroscopy and has the following table of standards data: Conc. (ppm) Absorbance 0.20 0.40 0.70 1.0 0.131 0.255 0.498 0.641 She prepared 3 dilutions of her tap water (see below). Which dilution is the best to use? Tap water dilution factor Absorbance 1:4 1:20 1:50 1.196 0.390 0.119 the 1:4 dilution is best the the 1:50 dilution is best any value could be used with equal precision the 1:20 dilution is best"[FeSCN2+]eq is calculated using the formula: [FeSCN2+]eq = Aeq/Astd x [FeSCN2+]std where Aeq and Astd are the absorbance values for the equilibrium and standard sample respectively." I am confused about what Aeq and Astd mean. We did an experiment where we measured the absorption of this chemical formation and I got around 0.18. Is that one of the values?
- An unknown amount of a compound with a molecular mass of 270.57 g/mol is dissolved in a 10 mL volumetric flask. A 1.00 mL aliquot of this solution is transferred to a 25 mL volumetric flask, and enough water is added to dilute to the mark. The absorbance of this diluted solution at 355 nm is 0.495 in a 1.000 cm cuvette. The molar absorptivity for this compound at 355 nm is €355 = 6149 M-¹cm-¹. What is the concentration of the compound in the cuvette? concentration: What is the concentration of the compound in the 10 ml flask? concentration: How many milligrams of compound were used to make the 10 mL solution? mass: M M mgA student weighed out 0.150 g of protein powder and dissolved it in 100 mL of water (Solution 1). The student then diluted this solution by transferring 1 mL into a 25 mL flask and diluting with water (Solution 2). Finally, 1 mL of that solution was transferred to a test tube and combined with 4 mL Bradford reagent. The absorbance of the solution in the test tube was 0.11. Assuming that the best fit linear line of the standard curve was y = 0.04144 x + 0.01521 (μ g mL), calculate the percent protein by mass in the original protein powder.Give a clear handwritten answer with explanation
- .Q1. To determine the concentrations (mol/L) of Co(NO3)2 (A) and Cr(NO3)3(B) in an unknown sample, the following representative absorbance data wereobtained.A (mol/L) B (mol/L) 510nm 575nm5×10−1 0 0.714 0.0970 6×10−2 0.298 0.757Unknown Unknown 0.671 0.330Measurements were made in 1.0 cm glass cells.i. Calculate the four molar absorptivities: ∈A(510), ∈A(575), ∈B(510) and ∈B(575).ii. Calculate the molarities of the two salts A and B in the unknown.6. Blue Blue dye stock solution 0.293 M Absorbance at 630 nm 0.00265 Calibration curve y = 0.0833x A solution is prepared by diluting 2.79 mL of the blue dye stock solution to 25.00 mL. The measured absorbance for the prepared solution is listed in the data table. (a) What is the theoretical molar concentration? [Blue]theoretical x 10 |M (b) What is the experimental molar concentration? [Blue]experimental x 10 M (c) What is the percent error? Percent error (blue) = %You obtained the following raw data when setting up a Biuret standard curve: Absorbancy 540nm BSA (mg/ml) 0.158 1 0.210 2 0.260 3 0.305 4 0.360 0.410 0.455 7 0.510 8 0.530 9 0.550 10 0.554 After blanking against a biuret-dH20 sample, the protein concentration of an unknown sample was determined using the same method and an absorbancy of 0.279 was obtained. Set up a standard curve, excluding outliers (experimental and statistical) and determine the protein concentration in the unknown sample in mg / ml (up to 3 significant figures).
- A student weighed out 0.150 g of protein powder and dissolved it in 100 mL of water (Solution 1). The student then diluted this solution by transferring 1 mL into a 25 mL flask and diluting with water (Solution 2). Finally, 1 mL of that solution was transferred to a test tube and combined with 4 mL Bradford reagent. The absorbance of the solution in the test tube was 0.144. Assuming that the best fit linear line of the standard curve was y=0.04144x+0.01521 (μgmL), calculate the percent protein by mass in the original protein powder.A solution is prepared by diluting 2.79 mL of the blue dye stock solution to 25.00 mL. The measured absorbance for the prepared solution is: Blue dye stock solution = 0.293 M Absorbance at 630 nm = 0.00265 Calibration curve y = 0.0833x A.) What is the theoretical molar concentration? B.) What is the experimental molar concentration? C.) What is the percent error?One common way to determine phosphorus in urine is to treat the sample, after removing the protein, with Mo (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue colored species called molybdenum blue. The absorbance of molybdenum blue can be measured at 650nm. A 24-hour urine sample was collected and the patient produced 1122mL in 24 hours. A 1.00mL aliquot of the sample was treated with Mo(VI) and ascorbic acid and diluted to a volume of 50.00mL. A calibration curve was prepared by treating 1.00mL aliquots of phosphate standard solutions in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at 650nm and the following results obtained: Solution Absorbance at 650nm: 1ppm P0.2302ppm P0.4363ppm P0.6384ppm P0.848Urine sample 0.518 How many grams of phosphorus were eliminated per day by the patient?