A KCC student titrated 100.00 mL of a NaOH solution using 55.00 mL of a 0.1000M H2SO4 solution. How many moles of acid were used ? This is part 1 of a titration problem H2SO4 + 2 NaOH NazSO4 + 2 H20 A. 0.0055 mol B. 0.0100 mol C. 0.100 mol D. 0.550 mol E. 5.50 mol
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- Solution Absorbance mg/ml aspirin Standard solution - 1.6 mg/mL A 0.638 0.08 mg/mL B 0.504 0.064 mg/mL C 0.376 0.048 mg/mL D 0.259 0.032 mg/mL E 0.126 0.016 mg/mL A = -log T where T = %T ÷ 100 Construct a callibration curve using the above data. Absorbance should be on the vertical axis and "mg/mL of acetylsalicylic acid" on the horizontal axis. The line should go through the origin. Using the data provided, the graph you have generated, and the procedure that was used to generate the solutions which were examined by spectroscopy, calculate the amount of acetylsalicylic acid per tablet. Commercial tablet 1 labelled as 100 mg enteric coated Absorbance = 0.16 Commercial tablet 2 labelled as 300 mg Absorbance = 0.45 Student prepared tablet from practical 5 Absorbance = 0.19 Using the data provided, the graph you have generated, and the procedure that was used…A mixture containing only KCl and NaBr is analyzed by the Mohr Method. A 0.3172-g sample is dissolved in 50 mL water and titrated to the Ag2CrO4 endpoint, requiring 36.85 mL of 0.1120 M AgNO3. A blank titration requires 0.71 mL of titrant to reach the same endpoint. Report the % (w/w) KCl and NaBr in the sample. [Ans . 84.41 % (w /w); 17.59 % (w /w)]The calibration curve shown below was used to analyze an unknown protein solution. What is the concentration of the unknown solution, if the absorbance of the unknown is 0.505? Answer in ug/mL. 1.2 y%3D0.005x+0.061 R=0.992 0.8 0.6 0.4 0.2 50 100 150 200 250 Concentration ug/ml Absorbance
- DATA Determine the Molar concentration of your dilutions using the formula M₁V₁ = M₂V₂. The initial molarity is 0.0001M (1x10-4M). M1 V1 1.56E-06 500uL Concentrations Tube 1 1.56E-06 Tube 2 Absorbance Tube 1 Tube 2 M2 Tube 3 Tube 3 V2 1000uL Tube 4 Tube 4 Tube 5 Tube 5 Tube 6 Tube 7 Tube 6 Tube 7 Tube 8 Tube 8 Tube 9 Tube 9 Tube 10 Tube 11 Tube 10 Tube 11 10A 5% dextrose in 1/2 normal saline (D5 1/2 NS) solution is commonly administered to patients needing post-operative IV fluids. How could you prepare 500 mL of this solution? Dextrose is 5% (m/v) and the NaCl is 0.45% (m/v).In a titration experiment, potassium iodate serves as the primary standard. Potassium iodate (KIO3) - with 99.4% purity, weight 0.1498 g in 100 mL solution Find the molarity of the titrant given the following data. Trial 1 2 3 Aliquot of standard 10 10 10 Net volume of titrant 33.30 33.50 33.20 Molarity X1 X2 X3
- In a neutral solution, the [H*] is : * 1.0x10-14M zero 1.0×107M O 1.0x10-7Mcan you show how to calculate the amount of two components for the required volume of a buffer with the receipes? Prepare enough 10X stock solution § Prepare 1LProportions§ 0.25M tris§ 1.92M glycine§ pH 8.3§ 1% (w/v) SDSInstructions to prepare 100mL of 10X running buffer1. Weigh 3.02g of Tris base and 14.42g of glycine into ~90mL of ddH2O in a clean beaker and stir todissolve.2. Check the pH with a calibrated pH metre once dissolved.§ The pH of the buffer should be 8.3. pH adjustment is not required if between 8.1-8.5.3. Add 5mL 20% (w/v) SDS stock solution (a liquid) to make a 1% (w/v) solution.4. Top the volume to 100mL by adding 5mL ddH2O from a graduated cylinder.Viscosity measurements were done on native form of Protein XYZ and on 4 denaturants. The crude Protein XYZ was diluted to produce 5% w/v. Using the Ostwald viscometer, the following data were obtained: time, sec time, sec time, sec time, sec Denaturant Blank (Native) Native Blank (Denaturant) Denatured pH= 2.00 70 95 70 101 Temp (90 °C) 70 98 70 150 0.1M Urea 72 97 88 110 DTT 71 96 90 170 a. Calculate and tabulate the reduced viscosities (in mL/g) of the native and denatured Protein XYZ. b. Determine two most effective denaturants. c. What can be inferred with the two most effective denaturants? Dont reject if you dont know the answer!
- As a technician in a large pharmaceutical research firm, you need to produce 200. mL of 1.00 M potassium phosphate buffer solution of pH = 6.96. The pKa of H2 PO4 is 7.21. %3D You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution? Express your answer to three significant digits with the appropriate units. > View Available Hint(s) HẢ Volume of KH2PO4 needed = Valıxe Units SubmitYou dissolve a 3.435 gram sample of potato in acid to determine the levels of solanine (the reason why green potatoes are poisonous) and diluted it to a total volume of 500.0 mL. You have also created a standard curve as outlined below. Solve for the ug/gram concentration of solanine in the potato sample. ppm Solanine Std. Abs 0.213 5 0.315 10 0.423 15 0.532 Unknown sample 0.314 Your Answer:Four lab groups completed the experiment listed in the lab diluting lul of FCF dye into 1ml of water. The A625 values are listed below. For brilliant blue FCF, ɛ =97000 M-1cm–1 MW = 792.8 g/mol %3D Group # A625 % solution (in original stock sample) Percent ror (True value of Stock=0.2%) 1 0.244 0.199 0.5 0.209 0.170 15 3 0.306 0.250 25 4 0.187 0.153 0.235 a. Which group was more accurate? b. What are possible reasons group 3 had a much higher Absorbance than the others? c. What are possible reasons group 4 had a much lower Absorbance than the others?