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Suppose that 4.0 L sample B is extracted and analyzed using liquid-liquid extraction. Sample B is a monomer qin H2O & ether. If the sample is combined with 50 mL ether, sample B's concentration in ether was 8x10-3 M.
Find the concentration of sample B in the initial 4.0 L sample if its KD is 500.
Note:
KD= {B}ether / {B}aq
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- 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: 0.463Am 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…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 ?
- Use the information in the pictures if needed and solve for the unknown D. please type the answers because it is much easier for me to understand that way. Unknown D To the unknown mixture, 3 drops of 6 M HCl was added and no precipitate formed. To the liquid that was left, 3 drops of 0.10M (NH4)2SO4 was added and a precipitate formed. The precipitate was isolated by centrifuging and the liquid from the top decanted for further testing. To the decanted liquid, the solution was made basic and a few drops of (NH4)2C2O4 was added. No precipitate formed. The liquid was tested for Mg2+ and a bright blue jelly ppt formed. To the unknown, a separate NH4+ test was done and the litmus paper used to test the pH turned blue. To the unknown, a flame test was performed and the color emitted was a light purple-red. What ions are in the unknown? (Identify unknown ions and provide a brief summary) DO NOT SIMPLY RESTATE THE ABOVE OBSERVATIONS.The buret was filled with 0.100 M HCl solution. Then was transferred in a 25.0 mL of saturated calcium hydroxide solution (2g of calcium hydroxide per 100 ml of water) in two separate E-flasks. Then 2 drops of phenolphthalein was added to each flask Titration data for the determination of solubility and Ksp of calcium hydroxide: Trial 2: Final Buret reading (ml)-26.10; Initial Buret reading (ml)- 19.80; Temperature (Celcius)- 25 Voume of HCl used: Trial 2- 6.30mL 1. Compute for the moles of H+ used and the moles of OH- present. moles of H+ used = (concentration of HCl) × (volume of HCl used)moles of OH- = moles of H+ used 2. Construct an ICE table for the reaction.3. Calculate the molar solubility (in mol/L) of OH- and Ca2+.4. Determine the solubility of Ca(OH)2 in g/L. (MM of Ca(OH)2 = 74.096 g/mol). 5. Calculate the Ksp of Ca(OH)26. Compute for the percent error of the experimental value for Ksp of Ca(OH)2 with the literature…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+ Is this an example of total analysis technique or concentration technique? Explain.
- An experiment to observe the effect of ionic strength on the solubility of Ca(OH)2 was conducted. Ca(OH)2 was added to 250 mL 0.01 M KCl, stirred until equilibrium. This was then filtered, and 50 mL of the filtrate was measured out. The 50 mL filtrate was then titrated with standardized 0.13 M HCl solution until phenolphtalein endpoint (11 mL). How do I solve for the solubility s of Ca(OH)2? Please provide an explanation of the process (do I multiply ionic strength μ to the solubility [where s = [Ca(OH)2) = [Ca2+] = 1/2[OH-]?)An experiment to observe the effect of ionic strength on the solubility of Ca(OH)2 was conducted. Ca(OH)2 was added to 250 mL 0.01 M KCl, stirred until equilibrium. This was then filtered, and 50 mL of the filtrate was measured out. The 50 mL filtrate was then titrated with standardized 0.13 M HCl solution until phenolphtalein endpoint (11 mL). How do I solve for the solubility s of Ca(OH)2? Please provide an explanation of the process (do I multiply ionic strength μ to the solubility [where s = [Ca(OH)2) = [Ca2+] = 1/2[OH-]?) I would like to know how the diverse ion effect would affect the solubility of Ca(OH)2. How does the ionic strength μ of KCl affect this?Potassium acid phthalate, KHC8H4O4 ( MM=204.2 ), reacts with sodium hydroxide on a 1:1 molar basis. A sample of the acid weighing 0.4823 g was titrated with NaOH, requiring 24.35 mL to reach the endpoint. Calculate the molar concentration of the sodium hydroxide solution.
- Transfer 3-4 drops or a pinch of the test compound (Aniline) to 3 ml of the solvents, 5% NaOH and 5% HCL Shake the mixture thoroughly. The time required for the solute to dissolve in the solvent should not be more than 2 minutes. Indicate with (+) or (-) if the test compound is soluble or insoluble in the solvent. Is Aniline solube or insolube in 5% NaOH? _______ Is Aniline soluble or insolube in 5% HCL? _______30.0 mL of 0.400 M Zn²⁺ is mixed with 100.0 mL of 0.200 M F⁻ (Ksp ZnF₂ = 0.030). In the mixture you would observeConsider the titration illustrated in Figure 8.] Anhydrous sodium carbonate (Na₂CO₃) is a primary standard. When 0.364 grams of the substance is placed in a conical flask, then 20.00 cm³ of sulphuric acid (H₂SO₄) solution is required to reach the end point of the titration. What is the concentration in mol·dm⁻³ of the sulphuric acid solution? [Give the answer to 3 decimal places. Do not type in the unit. Use a decimal point.] *