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- The first goal is to make the oxalic acid standard solution. You measure 1.5232 g of oxalic acid on an analytical balance, add it to a 250-mL volumetric flask and add deionized H2O to a final volume of 250.0 mL. Molar mass of H2C2O4•2H2O = 126.07 g/mol Mass of H2C2O4•2H2O = 1.5232g Volume of H2C2O4•2H2O solution = 250.0mL What is the Molarity H2C2O4 standard solution ?A chemist receive different mixtures for analysis with the statement that it contain NaOH, NaHCO3 , Na2CO3 or compatible mixtures of these substances together with the inert material. From the data given, identify the respective materials and calculate the percentage of each component. 1.000g samples and 0.2500 N HCl were used in all cases. Sample 1 With phenolphthalein as the indicator, 24.32 ml of HCl was used. A duplicate sample required 48.64 ml HCl using methyl orange as the indicator. Sample 2. With phenolphthalein as the indicator it uses 28.2 ml of HCl to make it colorless and added with methyl orange indicator and uses 11.3 ml of HCl to reach the end point.A STOCK SOLUTION containing 0.1581 g/L K2CrO4 was prepared.In order to make the CALIBRATION STANDARD, 5 ml of the STOCK was transferredinto a 50ml volumetric flask and then diluted with an appropriate solvent.Calculate:(a) The ppm of K2CrO4 in the CALIBRATION STANDARD.(b) The molarity of K2CrO4 in the CALIBRATION STANDARD. (c) Calculate the molar absorptivity of K2CrO4 (at 371.0 nm). Assume that Beer's Law isobeyed over this concentration range.At 371.0 nm, this CALIBRATION STANDARD in a cell of path length 1.00 cm gave a %T of 59.752.
- 3. Analysis of a mixture consisting of NaOH + Na2CO3 + inert matter gives the following data: 10.00 g. Its aqueous solution is diluted to 250.0 mL and two separate 25.00 mL sample portions are titrated. With one portion, an end point with phenolphthalein is obtained in cold solution, with 44.52 mL of 0.5000 N HCI. The other portion requires 46.53 mL of the acid for an end point with methyl orange. Calculate the percentage composition of the original sample.Detailed calculations on how to prepare the solutions listed below Solution: 50 mL 1 M oxalic acid Oxalic acid Mw = 90.03 g mol-1 Oxalic acid purity = 98% Solution: 10 mL 3 M sulfuric acid Sulfuric acid Mw = 98.079 g mol-1 Sulfuric acid density = 1.84 g mL-1 Sulfuric acid purity = 98% Solution: 10 mL saturated potassium oxalate potassium oxalate (monohydrate used) Mw = 184.23 g mol-1 Solubility of potassium oxalate in water at 25°C = 360 mg mL-1 Solution :20 mL 3% hydrogen peroxide Hydrogen peroxide is commercially available as a 32% solution.Am I calculating this correctly? Single 10 mL Extraction and Determination of Distribution Coefficient 1) Calculation of initial amount of benzoic acid in grams Volume of benzoic acid solution: 50.00 mL Molarity of benzoic acid solution: 0.0205 M 50.00 mL x 1 L/1000 mL L x 0.0205 mol/L x 122.122 g/mol = 0.12517505 g ≈ 0.125 g of benzoic acid A2. Calculation of amount of benzoic acid (g) remaining in aqueous solution after extraction with 10 mL of methylene chloride BA + NaOH --> sodium benzoate + H2O BA = 0.0205 M NaOH = 0.0189 M NaOH volume used = 17.4 mL Moles of NaOH: 0.0189 mol/L x 17.4 mL x 1L/1000 mL = 0.00032886 moles of NaOH BA:NaOH = 1:1, therefore there is 0.00032886 moles of BA remaining in aqueous phase 0.00032886 moles BA x 122.122 g/mol = 0.04016106 g 0.12517505 g BA (original) - 0.04016106 g BA (remaining) = 0.08158944 g BA in 10 mL of dichloromethane If everything is done correctly, how would I calculate of the amount of benzoic acid (g) extracted into the…
- 1. Why is there a need to correct conductivity of the sample solution?2. Cite three possible sources of error in the experiment and discuss their effects on the computed ionization constant of the acetic acid. (See details below----- literature value- Ki, acetic acid= 1.75 x 10^-5). 3. What is the effect of dilution on the fraction of the acetic acid ionized?The first goal is to make the oxalic acid standard solution. You measure 1.5232 g of oxalic acid on an analytical balance, add it to a 250-mL volumetric flask and add deionized H2O to a final volume of 250.0 mL. Molar mass of H2C2O4•2H2O = 126.07 g/mol Mass of H2C2O4•2H2O = 1.5232g what is the Number of moles of H2C2O4•2H2O?The thiourea in a 1.455 g sample of organic material was extracted into a dilute sulfuric acid solution and titrated with 37.31 mL of 0.009372 M Hg2+ via reaction: 4(NH2)2CS + Hg2+ →[(NH2)2CS]4 Hg2+ P.S. Answer only the last two letters of the following questions. (Only C and D) a. Is this an example of total analysis technique or concentration technique? Explain. b. Calculate the percent (NH2)2CS ( 76.12 g/mol) in the sample. c. What is classification of the analysis based on the amount of sample and amount of analytes present? Explain. d. If the true value is 10.00%, calculate the absolute and relative error.
- A 500ml solution of NaOH was made using 2g of NaOH(s) Three trials of titration were made with using KHP(s) as the acid dissolved with about 25ml of deionized water and 4 drops of phenolphthalein indicator. Slowly adding the NaOH solution until the clear solution had turned pink which would give us our end point and allow us to find the NaOH molarity by equivalence point. Trial 1: - 0.484g of KHP were used - initial volume of buret containing NaOH solution was 0.0 ml - final volume of buret was 23.60ml Trial 2: - 0.485g of KHP were used - initial volume was 0.0ml - final volume was 24.00ml Trial 3: - 0.486g of KHP - initial volume was 0.0ml - final volume was 23.80ml The molarity of NaOH was found by using the moles of KHP(as at equivalence, both solutions are balanced in moles) divided by the total volume of NaOH used to neutralize the solution. Giving us 0.100M for trial 1, 0.099M for trial 2, and 0.100M for trial 3. Make a rough sketch of a titration curve that…If an instrument gives a response of 1240 for a standard containing 8 ppm of a substance, how much if this substance is in a sample that gives a response of 1705? Are any assumptions needed?The standard addition method is used to analyze a sample of a river water for mercury. Solution A is made by pipetting 5.00 mL of undiluted sample in to a 10 mL volumetric flask and filling to the mark with DI water. Solution B is made by pipetting 5.00 mL of undiluted sample and 3.00 mL of 15.0 ppb of Hg standard into same 10.0 mL volumetric flask and filling to the mark with DI. Solution A and B are analyzed using atomic absorption spectroscopy and give a percent transmittance values of 56 % and 33 % respectively (not blank corrected). A blank has a transmittance of 96%. What is the corrected absorbance of both solution A and B? A. Solution A: 0.123 Solution B: 0.463 B. Solution A: 0.463 Solution B: 0.234 C. Solution A: 0.123 Solution B: 0.234 D. Solution A: 0.234 Solution B: 0.463
