Full explanation please ObservationDriving force(s).(DEMO)Mix sodium hydroxide andsodium chlorideppt (rust)iron (III) hydroxide1.iron (III) chloride3 NaOH(aq)3 NaClag)KNO3Patassinn+FeCla(aq)Fe(OH)3(s)Mix lead (II) nitrate andPh wacto Example:a) molecular equation3 NaOH(aq) + FeCl3{aq) → Fe(OH)3(6) + 3 NaCl(ag)b) total ionic equationc) net ionic equation1st reaction:a) molecular equationb) total ionic equationc) net ionic equation2nd reaction: (3a) molecular equationb) total ionic equationc) net ionic equation4.

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Principles of Instrumental Analysis

7th Edition

ISBN:9781305577213

Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

ChapterA1: Evaluation Of Analytical Data

Section: Chapter Questions

Problem A1.22QAP

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0.7050-g of pure KHP (FW = 204.2) was dissolved and titrated with 35.00mL of NaOH solution. The excess NaOH was backtitrated with 5.00mL of HCl solution. In a separate titration it was found that 30.00mL of NaOH will react with 28.00mL of HCl. Find the molarity of NaOH and HCl. Show all the solutions involved and round-off final answers to two decimal places except for Molarity. (Molar concentrations should be in four decimal places)
and Employees CH STATE GE 90 ENDULERSE IS FOREVER "ON-Tame Name Jodi Rayal Molarity = wt. x1000 3. Calculate the massof Potassium hydrogen phthalate (KHP) to prepare a 250.0 mL solution of 0.1000 M KHP solution. Mol ut. x Voluse the 2500×1000 204.22250 mass Initial Buret Reading Final Buret Reading Vol NaOH Added Moles of NaOH Moles of HCI Volume of unknown HCI solution Concentration of HCI Average concentration of HCI of KHP = 5.11g 4. Complete the table below for the following neutralization reaction of an unknown concentration of hydrochloric acid with 0.100 M sodium hydroxide: Mu spate solution Trial 1 0 11.15 го data and calculate the concentration 15 Trial 2 11.15 22.15 Trial 3 22.15 33.25 abyh 067 LANDINUKASHU TROWTHCENTISSATE ARAD Ma That par މއކނި REMURNAR MISAMARALISERDRY Tile
TOPIC: GRAVIMETRY SHOW THE SOLUTION The mercury in a 0.7152-g sample was precipitated with an excess of paraperiodic acid, H5IO6, according to the following reactions:5 Hg+2 + 2 H5IO6 ---> Hg5(IO6)2(s) + 10 H+The precipitate was filtered, washed free of precipitating agent, dried and found to weigh 0.3408-g. Calculate the percentage of Hg2Cl2 in the sample. Molar Masses: Hg5(IO6)2 = 1448.75 Hg2Cl2 = 472.09 Answer: 38.82% Hg2Cl2 An iron ore was analyzed by dissolving a 1.1324-g sample in concentrated HCl. The resulting solution was diluted with water, and the iron (III) was precipitated as the hydrous oxide Fe2O3·xH20 by the addition of NH3. After filtration and washing, the residue was ignited at a high temperature to give 0.5394 g of pure Fe2O3. Calculate (a) % Fe, and (b) % Fe3O4 in the sampleMolar Masses: Fe2O3 = 159.69 Fe = 55.847 Fe3O4 = 231.54 Answer: 33.32% Fe and 46.04% Fe3O4
When a Vitamin C (ascorbic acid; MM = 176.12 g mol-1) tablet is crushed, dissolved and titrated with 0.0340 M KIO3(aq) to a purple/blue endpoint (given by a starch indicator), the volume of KIO3 used is 29.80 mL. If 60 mg of ascorbic acid is the recommended dietary allowance (i.e., 100% of the RDA), then what is the % RDA for the Vitamin C in the tablet? KIO3(aq) + 5 KI + 6 H+ → 3 I2(aq) + 3 H2O I2 (aq) + ascorbic acid → 2 I- + dehydroascorbic acid
1,5419 g of magnetite (Fe3O4) ore; in concentrated HCL to form a mixture of Fe + 2 and Fe+3it's unraveling. By adding HNO3 to it, all Fe+2 s are upgraded to Fe+3. And the addition of FE+3 s NH3,with Fe (OH)3, precipitating into. The sediment is weighed as 0.8525 g in the form of Fe2O3 after the necessary operations. Calculate the percentage of Fe3O4 in the sample.
The mercury in a 0.8142-g sample was precipitated with an excess of paraperiodic acid, H5IO6: 5 Hg 2+ + 2 H5IO6 ---> Hg5(IO6)2 (s) + 10 H + The precipitate (MW = 1448.8 g/mol) was filtered, washed free of precipitating agent, dried, and weighed, and 0.4114 g was recovered. Calculate the a) % Hg (200.6 g/mol) b) % Hg2Cl2 (472.1 g/mol)
Calculate the weight of pure sodium carbonate that is necessary to prepare 2.806 L of 0.223 N Na2CO3 (105.99 g/mol) from the primary-standard solid. Assume the solution is to be used for titrations in which the reaction is: CO32- + 2H+ ----> H2O + CO2 Express you answers in 3 decimal places
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?..
What volume of 0.10 moldm-3 sulphuric acid would be required to neutralise a mixture of 1.06 g of anhydrous sodium carbonate and 4.00 g of sodium hydroxide? (H2SO4 + 2NaOH ---> Na2SO4 + 2H2O) and (H2SO4 + Na2CO3 ® ---> Na2SO4+ H2O + CO2)
You collected the following data from a titration experiment using a 0.129M standardized NaOH solution to titrate a 26.55 mL solution with an unknown Molarity concentration (M) of sulfuric acid (H2SO4). Initial Burette Reading (mL) Final Burette Reading (mL) Vol Delivered (mL) Trial 1 0.44 19.69 ?? Trial 2 0.18 17.2 Trial 3 0.50 19.94 For just Trial 1, determine the amount of NaOH delivered for the titration with appropriate significant digits. Do not include units.
A sample of pure sodium oxalate weighing 0.1050 g is ignited [Na2C2O4 (s) --> Na2CO3 (s) + CO (g)] and the resulting product requires 15.00 mL of a solution of H2SO4 for complete neutralization. What is the normality of the acid? MM of Na2CO3 = 106.0 g/molMM of Na2C2O4 = 134.0 g/molMM of CO = 28.01 g/molMM of H2SO4 = 98.08 g/mol
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: 16

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Example: a) molecular equation 3 N2OH(aqg) + FeCl3{aq) → Fe(OH)3(5) + 3 NaCl(aq) b) total ionic equation c) net ionic equation