A 200.00 mL sample of a citrus fruit drink is acidified and 10.00 mL of 0.05000 MI2solution is added. After the reaction is complete the excess l2 is titrated with 3.86 mL of0.1200 M Na,S,O3 solution. Calculate the concentration of ascorbic acid, C,HgO, (M.M.= 176.14) in the juice drink sample in mg/100 mL sample.CeHeOs → CeH&Os + 2H + 2eA 32.41 mg CgH3O6/100 mL sample23.64 mg C&H8O6/100 mL sampleC47.28 mg C&H8O6/ 100 mL sampleD400 mg CoHBO6/ 100 mL sample

Answered: Image /qna-images/answer/04ee2cd6-7ac9-4e5c-8f58-156f52a7c372.jpg

Answered: A 200.00 mL sample of a citrus fruit…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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A 0.8159 g sample of HCl was placed into a 50 mL volumetric flask and the sample was thoroughly dissolved in water to make 50 mL of solution. It required 22.07 mL of NaOH to reach the endpoint in the titration. What is the molarity of the NaOH solution? I mg IHE/ 2.022 4 mol 0.81599 Hcl 36.4bg.
A solution is prepared by pipetting 25.0 mL of 0.350 M CuCl2 into a 100.0 mL volumetric flask, diluting to the calibration mark on the flask with water and quantitatively mixing the contents of the flask. What is the molar concentration of Cl- in the solution?
Mark and John, two vinegar enthusiasts, are each tasked to determine the acetic acid content of their respective vinegar concoctions by titration. First, a 1 M-labeled KOH solution was standardized against the KHP (MW = 204.22 g/mol) standard that is 99.4% pure. In the process, 0.540 g KHP was found to require 2.80 mL of the KOH solution to completely react up to the phenolphthalein endpoint. Then, Mark and John both prepared their samples by taking 10.0-mL aliquots of each vinegar and diluting them to 25.0 mL. Using the same titrant and indicator, Mark’s vinegar required 18.60 mL of the standardized titrant to reach the endpoint, while John’s vinegar required 16.50 mL of the same titrant to reach the same endpoint. Question: What is the acetic acid concentration of Mark’s and John’s vinegar in molarity given that the Concentration of KOH is 9386 M?
A pure organic compound has the formula C4H8SOx. A sample of a compound wasdecomposed; the sulfur was converted to sulfate and titrated in a modified version of theadsorption indicator procedure. From the following data, determine the correct formula forthe compound.Given: weight of sample = 12.64 mgBa(ClO4)2 needed for titration = 10.60 mL of 0.0100 M Ba(ClO4)2
3. A 4.912-g sample of a petroleum product was burned in a tube furnace, and the SO2 produced was collected in 3% H₂O₂. Reaction: SO2(g) + H₂O2 → H₂SO4. A 25.00-mL portion of 0.00873 M NaOH was introduced into the solution of H₂SO4, following which the excess base was back-titrated with 15.17 mL of 0.01102 M HCI. Calculate the sulfur concentration in the sample in parts per million.
A 1.00 gram sample of anhydrous cerium sulfate was dissolved in water. Barium nitrate solution was then added until no further precipitation occurred. The precipitate was filtered from the solution then dried and was found to weigh 1.41 g. Determine the formula of the cerium sulfate. Encom calt is hydrated
+2 When excess Na,so, solution was added to 20.0 mL of an unknown solution containing Ca ion, 0.0472 g of Caso, precipitated. What is the molar concentration (molarity) of Ca in the unknown solution? (molar mass of Caso, = 136.14 g/mol A) 1.36 x 102 B) 2.47 x 102 C) 2.10 x 10 D) 1.73 x 10 E) 3.83 x 10 Select one: -2 O a. 2.47 x 10 -2 с. 3.83 х 10 d. 1.73 x 10 -2 е. 2.10 х 10
When 50.0mL of a 0.335M potassium carbonate solution is mixed with 0.045L of a 0.390M nickel(II) chloride solution the following reaction occurs: K2CO3 (aq) + NiCl2 (aq) → a. Complete the reaction equation b. Calculate the mass of precipitate formed c. Calculate the concentration of each ion in the mixture when the reaction is complete
The alcohol content in a 10.0 g sample of blood from a driver required 4.23 mL of 0.07654 M K2Cr2O7 for titration. If the current legal limit of blood alcohol content is 0.08 percent by mass, should the driver be prosecuted for drunken driving? What is the percent alcohol in the driver’s blood? 3 C2H5OH (ethanol) + 2 K2Cr2O7 + 8 H2SO4 → 3 HC2H3O2 + 2 Cr2(SO4)3 + 2 K2SO4 + 11 H2O

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