A 50.00 mL solution containing NaBr was treated with excessive AgNO3 to precipitate 0.214 6g of AgBr (FM 187.772). What was the Molarity of NaBr in the solution?
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A 50.00 mL solution containing NaBr was treated with excessive AgNO3 to precipitate 0.214 6g of AgBr (FM 187.772). What was the Molarity of NaBr in the solution?
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- Zhongli is a speleologist tasked to analyze the CaCO₃ content of a limestone stalactite. A 5.0000-g sample was dissolved in 25.00 mL of 1.350 M HCl, it was then heated to expel any CO₂ formed. The excess HCl was titrated to a phenolphthalein end point, it used 37.50 mL of 0.1200 M NaOH. How many moles of HCl was added initially to digest the limestone sample? How many moles of CaCO₃ is present in the limestone sample? What is the purity of the limestone in terms of %w/w CaCO₃?A weight of 0.50 g was taken impure container containing sodium carbonate and bicarbonate. Dissolved in water and then crushed with hydrochloric acid (0.1 N), the burette reading game was at the endpoint of phenolphthalein of 10.5 ml and at the end point of the orange methylation point 30.1 ml. The percentage of sodium carbonate was in ................. knowing that the weights are: Na: 23, C: 12, O: 16Zhongli is a speleologist tasked to analyze the CaCO₃ content of a limestone stalactite. A 5.0000-g sample was dissolved in 25.00 mL of 1.350 M HCl, it was then heated to expel any CO₂ formed. The excess HCl was titrated to a phenolphthalein end point, it used 37.50 mL of 0.1200 M NaOH. A. How many moles of HCl was added initially to digest the limestone sample? B. How many moles of CaCO₃ is present in the limestone sample? C. What is the purity of the limestone in terms of %w/w CaCO₃?
- In order to form Ba(IO3)2, 500 mL of 0.5000 M Ba(NO3)2 was mixed with 500 mL of 0.0500 M NaIO3. Ksp= 1.57x10^-9How many millimoles of Ba(NO3)2 are needed to completely react with NaIO3?What is the limiting reagent?What is the excess reagent?A commercial vinegar was analyzed by titration to determine the percent acetic acid. Briefly, 10.00 mL of vinegar sample was diluted to 100. mL solution in volumetric flask. A 25.00 mL aliquot from the diluted vinegar required 25.55 mL of 0.1005 M NaOH to reach the phenolphthalein endpoint. Which is the correct equation between the analyte and titrant reaction? CH3COOH + NaOH → NaCH3COO + H2O 2CH3COOH + NaOH → NaCH3COO + H2O C20H14O4 + NaOH → NaC20H14O4 + H2O CH3COOH + 2NaOH → NaCH3COO + H2OThe salt content was extracted from a 12.0000g junk food sample. The extract was diluted to 100.00mL. From this solution, 15.00mL was taken and required 23.75mL of a 0.08943M AgNO solution to reach the endpoint. What is the percentage by mass of salt as NaCI (58.45 g/mol) in the junk food sample? (Answer: 6.897% NaCl)
- 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.0.5 g mixture of (0.1 N KCI and 4 points KBr) is required to 58.03 mL of 0.1 N AGNO3 to precipitate both bromide and chloride as AgCl and AgBr. Calculate the weight percentage for each of the AgCl and AgBr in a mixture if you know that the molecular weight of KCI = 74.5 and of KBr = 119 * %3DA 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.
- In one trial determination of the concentration of NaOCl in the diluted bleach sample, 10.00 mL of the diluted bleach solution required 23.32 mL of 0.0195 M Na2S2O3 titrant to reach the endpoint. What is the molar concentration of NaOCl in the diluted bleach solution?The solubility of borax, which is made up of sodium tetraborate (Na2B4O5(OH)4 8H2O), was analyzed. The dissolution of borax is: Na2B4O5(OH)4 • 8H2O(s) ⇌ 2 Na+(aq) + B4O5(OH)42–(aq) + 8 H2O(l) A 50 mL saturated solution was prepared. After filtration of solution, 5 mL aliquot was transferred to a flask and titrated using 0.432 M HCl. The endpoint was found to be 4.73 mL of the titrant. Tetraborate anion (B4O5(OH)42-) is a weak base which reacts with HCl like the following reaction: B4O5(OH)42–(aq) + 2 H+(aq) + 3 H2O(l) ⇌ 4 H3BO3(aq) What is Ksp expression for the dissolution? What is the tetraborate ions concentration in the filtrate? What is the molar solubility and Ksp of borax if the titration was done at room temperature (298 K)?Aspirin powder = 0.8110g MW of Aspirin = 180g.mol-1 Volume of 0.5N HCl consumed in back titration = 23.50mL Volume of 0.5N HCl consumed in blank titration = 44.50mL Percent purity (USP/NF) = Aspirin tablets contain NLT 90.0% and NMT 110.0% of the labeled amount of aspirin (C9H8O4) What is the calculated weight (in grams) of pure aspirin?..