A chemist needs to create a series of standard Cu²+ (aq) solutions for an absorbance experiment. For the first standard, he uses a pipet to transfer 25.00 mL of a 2.49 M Cu²+ (aq) stock solution to a 250.0 mL volumetric flask, and he adds enough water to dilute to the mark. He then uses a second pipet to transfer 10.00 mL of that solution to a 50.00 mL volumetric flask, and he adds enough water to dilute to the mark. Calculate the concentration of the Cu²+ (aq) solution in the 50.00 mL. volumetric flask. concentration: M

A chemist needs to create a series of standard Cu²+ (aq) solutions for an absorbance experiment. For the first standard, he uses a pipet to transfer 25.00 mL of a 2.49 M Cu²+ (aq) stock solution to a 250.0 mL volumetric flask, and he adds enough water to dilute to the mark. He then uses a second pipet to transfer 10.00 mL of that solution to a 50.00 mL volumetric flask, and he adds enough water to dilute to the mark. Calculate the concentration of the Cu²+ (aq) solution in the 50.00 mL. volumetric flask. concentration: M

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter17: Electrochemistry

Section: Chapter Questions

Problem 80AP

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A 0.2500-g of dry CaCO3 was completely dissolved into a solution of 5-mL conc. HCl and 50-mL distilled water in an erlemeyer flask. The resulting solution was boiled for 5 minutes to expel CO2 from the reaction, and then it was cooled to room temperature and the solution was transferred in a 500-mL volumetric flask filled to volume by rinsing the erlenmeyer flask and combining all the washings to fill the mark. A 5.0-mL aliquot of the CaCl2 solution was taken from the orignal solution and was transferred into an erlenmeyer flask with 10-mL NH3 buffer, 20-mL distilled water, and 5 gtts of EBT. The solution consumed 4.96 mL of an EDTA solution. MW: CaCO3 = 100.1 Compute for the N EDTA solution.   0.0053 N   0.0531 N   0.5313 N   None of the choices
Malic acid (C4H6O5) is a dicarboxylic acid that naturally occurs in the production of bignay wine. Maria was tasked to find out the acid content in a 200.0 mL wine sample by titrating it with a standardized solution of KOH. He was able to establish the working concentration of the KOH solution by using 39.12 mL of it to titrate a primary standard of KHP that weighed 3.021 g. The grams and %w/w of Malic acid present in the 200.0 mL wine sample were then determined by titrating a 20.00 mL aliquot using 48.95 mL of the standardized KOH solution. C4H6O5 (aq) + 2KOH (aq) ↔ K2C4H4O5 (aq) + 2H2O (l) Assume the density of the wine sample is 1.000 g/mL MM of KHP = 204.22 g/mol MM of C4H6O5 = 134.0874 g/mol ?    What is the molar concentration of the prepared KOH solution?…
Malic acid (C4H6O5) is a dicarboxylic acid that naturally occurs in the production of bignay wine. Maria was tasked to find out the acid content in a 200.0 mL wine sample by titrating it with a standardized solution of KOH. He was able to establish the working concentration of the KOH solution by using 39.12 mL of it to titrate a primary standard of KHP that weighed 3.021 g. The grams and %w/w of Malic acid present in the 200.0 mL wine sample were then determined by titrating a 20.00 mL aliquot using 48.95 mL of the standardized KOH solution. C4H6O5 (aq) + 2KOH (aq) ↔ K2C4H4O5 (aq) + 2H2O (l) Assume the density of the wine sample is 1.000 g/mL MM of KHP = 204.22 g/mol MM of C4H6O5 = 134.0874 g/mol                                                                                                                         What is the molar concentration of the prepared KOH solution?…
1. Calculate the equivalent weight of KMnO₄ in the following reaction: MnO₄⁻ + CN⁻ + H₂O ⟶ MnO₂ (s) + CNO⁻ + OH⁻) 2. Calculate the normality and molarity of a 15.0% (by weight) H₂SO₄ solution with a specific gravity of 1.090. * 3. To what volume should 100 mL of the above solution be diluted to prepare 1.50 N solution?
Tartaric acid, H2C4H4O6, is a diprotic acid that naturally occurs in the production of wine. Jose was tasked to find out the acid content in a 100.0 mL wine sample by titrating it with a standardized solution of NaOH. He was able to establish the working concentration of the NaOH solution by using 35.21 mL of it to titrate a primary standard of KHP that weighed 3.001 g. The grams and %w/w of Tartaric acid present in the 100.0 mL wine sample were then determined by titrating a 25.00 mL aliquot using 43.56 mL of the standardized NaOH solution. H2C4H4O6 (aq) + 2NaOH (aq) ↔ Na2C4H4O6 (aq) + 2H2O (l) Assume that the density of the wine sample is 1.000 g/mL MM of KHP = 204.22 g/mol MM of H2C4H4O6 = 150.087 g/mol What is the molar concentration of the prepared NaOH 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?
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 exact concentration of the KOH titrant in molarity? What is the balanced chemical equation between the analyte and the titrant? What is the color transition (color X ? color Y) expected throughout the course of titration?
Please answer this NEATLY, COMPLETELY, and CORRECTLY for an UPVOTE. Ni(NO3)2From the stock solution (1.0 M), determine the volume needed to prepare 0.00, 0.04, 0.08, 0.12 and 0.16 M Ni(NO3)2 solution in 50 mL volumetric flask. Obtain the calculated amounts, place in corresponding volumetric flasks and dilute to mark.
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When standardizing two liters of a potassium permanganate solution, approximately 0.0100 M, with a primary standard solution of sodium oxalate, it was planned to use between 30.00 and 45.00 ml of the titrated reagent. In which mass range (in grams) the primary standard should be weighed?
Fifty food supplement tablets with a total mass of 66.393 g were crushed into powder to analyse the content of FeSO4·7H2O. 2.988 g of the powder were dissolved in HNO3 and heated to convert all the iron to Fe3+. Addition of aqueous NH3 caused quantitative .precipitation of Fe2O3· X H2O, which was ignited to give 0.0640 g of Fe2O3. Determine the average weight of FeSO4·7H2O in each tablet.
A quantitative analysis for ethanol, C2H6O, can be accomplished by a redox back titration.Ethanol is oxidized to acetic acid, C2H4O2, using excess dichromate, Cr2O72–, which is reduced toCr3+. The excess dichromate is titrated with Fe2+, giving Cr3+ and Fe3+ as products. In a typicalanalysis, a 5.00-mL sample of a brandy is diluted to 500 mL in a volumetric flask. A 10.00-mLsample is taken and the ethanol is removed by distillation and collected in 50.00 mL of anacidified solution of 0.0200 M K2Cr2O7. A back titration of the unreacted Cr2O72–requires 21.48mL of 0.1014 M Fe2+. Calculate the %w/v ethanol in the brandy