Write a step-by-step procedure for this serial dilution. The stock concentration of the blue dye standard is 1.00 mg/mL. Serial dilution series from the stock standard with the following final concentrations: 500 ug/mL, 250 ug/mL, 125 ug/mL, 62.5 ug/mL
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Write a step-by-step procedure for this serial dilution.
- The stock concentration of the blue dye standard is 1.00 mg/mL.
-
Serial dilution series from the stock standard with the following final concentrations: 500 ug/mL, 250 ug/mL, 125 ug/mL, 62.5 ug/mL. and a blank that is 0ug/ml
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- Bay Water Titration The concentration of Cl- in ocean water is about 500-600 mM. The baywater is diluted by a factor of 12.5 using a 20.00 mL volumetric pipet and a 250 mL volumetric flask. Using a 15.00 mL volumetric pipet, 15.00 mL is transferred into three clean Erlenmeyer flasks. 10 mL of 1% dextrin solution, 20 mL of de-ionized water, and 3-4 drops of indicator are added and titrated each with the AgNO3 solution. From the procedure, find: Dilution factor Volume of diluted bay water Then calculate: The [Cl-] of diluted bay water The [Cl-] of bay water Consider this information: [AgNO3] = 0.04043177 trial # Veq 1 13.91 2 13.73 3 13.9 4 13.86 5 13.87 6 13.84 average 13.85167 [Cl-]Bay = ([AgNO3](Veq(ave))/Vdil bay water pipeted) (Vflask/Vbay water)Exactly 10.00-mL aliquots of a solution containing phenobarbital were measured into 50.00-mL volumetric flasks and made basic with KOH. The following volumes of a standard solution of phenobarbital containing 2.000μg/mL of phenobarbital were then introduced into each flask and the mixture was diluted to volume: 0.000, 0.500, 1.00, 1.50, 2.00 mL. The fluorescence of each of these solutions was measured with a fluorimeter, which gave values of 3.26, 4.80, 6.41, 8.02, 9.56, respectively. a. plot the data. b. derive a least squares equation for the data plotted in (a). c. find the concentration of phenobarbital from the equation in (b). d. calculate a standard deviation for the concentration obtained in (c).You want to make a solution of the primary standard sodium oxalate. It has a molecular mass of 133.999 g/mol (no error). Your analytical balance has an error of 0.2 mg. You add some weighing paper to the balance and it reads a mass of 0.23626 g. You then add sodium oxalate onto the weighing paper until the mass registers as 2.8539 g. You carefully transfer this to a 250.0 ml volumetric flask with an error of 0.1 ml and dissolve the sodium oxalate in a sufficient volume of water. What is the concentration of the oxalic acid (in M) and what is the absolute error (also in M).
- Help me pleaseYou are preparing a 100.0 mL standard solution needed for titration analysis. The NaOH (40 g/mol) pellets are measured to be 0.5341 g in a balance with acceptable uncertainty of +0.001 g. Mass of the pellets is recorded after taring the container. The pellets are put in a beaker and added with 50 ml distilled water measured by a graduated cylinder. The solution is then quantitatively transferred to a 100-mL volumetric flask with uncertainty of ±0.08 mL and diluted to a 100-mL mark. What is the concentration (M) of the solution and estimate its uncertainty by propagation? Assumption: There is no uncertainty in molar mass. Express your answer as C+/- u M.Beaker 0.00200 M Fe(NO3)3, mL 0.00200 M NaSCN, mL total volume, mL 1 3.000 2.000 10.00 2 3.000 3.000 10.00 3 3.000 4.000 10.00 4 3.000 5.000 10.00 5 (blank) 3.000 0.000 10.00 In solutions 1-4 you are adding 3.00ml of 0.00200M Fe3+ to some SCN- solution and dilutiing to a final volume of 10.00ml. What is the new concentration of the Fe3+ due to dilution? Your answer should have 3 sig figs.
- Show your complete solution and round-off your answers to four decimal places. A 0.1093-g sample of impure Na2CO3 was analyzed by the Volhard method. After adding 50.00 mL of 0.06911 N AgNO3, the sample was back titrated with 0.05781 N KSCN, requiring 27.36 mL to reach the end point. Calculate the % purity of the Na2CO3 sample. Answer = _% Na2CO3A student was given a stock ascorbic acid solution of 2.323 mg/L. Following the directions in the lab, the student obtained the data below. Calculate the concentration of ascorbic acid in the Standard (in microgram/L, ug -). Report your answer with one place after the decimal. initial final initial volume final volume volume volume color reagent color reagent stock stock solution solution solution solution Standard 1.35 mL 6.64 mL 5.12 mL 9.32 mLBased on the analysis of a vinegar solution via titration with 0.1 M NaOH. Results are as follows: Trial 1 Volume of vinegar solution used (mL) 5.00 Final burette reading (mL) 44.20 Initial burette reading (mL) 0.00 Determine the percentage mass of acetic acid in vinegar for Trial 1. The density of vinegar is 1.01 g/mL and molecular mass for acetic acid is 60 g/mol.
- TITRIMETRIC DATA SAMPLE: CANE VINEGAR % acidity in label: 4.5% %purity of KHP: 99.80% FORMULA WEIGHT of KHP: 204.22 g/mol STANDARDIZATION OF NaOH SOLUTION TRIAL 1 TRIAL 2 TRIAL 3 Weight of KHP, g 0.1012 0.1004 0.09987 Initial volume of NaOH, mL 5.00 10.01 Final volume of NaOH, mL 4.95 9.99 14.93 Molarity of NaOH Average Molarity of NaOH ANALYSIS OF ACETIC ACID IN A VINEGAR SAMPLE TRIAL 1 TRIAL 2 TRIAL 3 Volume of vinegar, mL Initial volume of NaOH, mL 1.00 1.00 1.00 14.98 22.90 30.87 Final volume of NaOH, mL 22.84 30.77 38.75 Molarity of acetic acid Average molarity of acetic acid ANALYSIS OF CARBONIC ACID IN A SODA SAMPLE TRIAL 1 TRIAL 2 TRIAL 3 Volume of soda, mL 20.0 20.0 20.0 Initial volume of NaOH, mL 20.20 22.05 23.93 Final volume of NaOH, mL 22.03 23.89 25.8 Molarity of carbonic acid Average molarity of carbonic acidStandard Sodium Chloride Weighed mass of NaCI g in 250.0 mL Purity of NaCl = 100.03 % Molar concentration of NaCl using weighed mass and purity mol L-1 Titrations of standard NaCI (25.00 ml aliquots) with the provided AGNO3 Precise Titration # Volume of AGNO3 required (mL) 1 2 3.A solution of HCl was titrated against sodium carbonate. What is the normality of acid for Trial 3 in the given data? T1 T2 0.3562 0.3479 0.3042 Weight (g) Initial V (mL) 0.80 1.60 0.40 Final V (mL) 35.20 36.70 39.80 Vol HCI used (ml) 35.10 39.40 N of HC (eq/L) 0.1954 0.1870 Average N of HCl (eq/L)
