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- How many (a) millimoles, (b) milliequivalents, and (c) milliosmoles of calcium gluconate (Ca(C6H11O7)2 - m.w. 430) are represented in 15 mL of a 10% w/v calcium gluconate solution?B. Solid AgCl (Ksp = 1.8×10-10 ) is dissolved in 1 Liter of water , then 0.585 gm of sodium chloride is added to silver chloride .caſcuſate and decide if soſubility of siſver chlſoride will decreased or increased by this addition and why ? (M.wt of NaCl= 58.5).5. A 300.0 mg sample containing Na,CO3, NaHCO3 and NaOH and inert material either alone or in some combination was dissolved and titrated with 0.1000 M HCI the titration required 24.41 mL to reach the phenolphthalein endpoint. And an additional 8.67 mL to reach the methyl red endpoint. Determine the composition of the sample and calculate the percent of each titrated component.
- In how much volume of diluted sulfuric acid should you dissolve a multivitamin tablet of approximately 1700 mg of Vitamin C to back analyze 10.00 mL of that solution, with 25.00 mL of 0.01 M KIO3 and 0.04 M thiosulfate to use approximately 8-10 mL of titrant ? You have 25, 50, 150, 250 and 500 mL flasks.5. A 300.0 mg sample containing Na,CO3, NaHCO, and NaOH and inert material either alone or in some combination was dissolved and titrated with 0.1000 M HCI the titration required 24.41 ml to reach the phenolphthalein endpoint. And an additional 8.67 mL to reach the methyl red endpoint. Determine the composition of the sample and calculate the percent of each titrated component.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?..
- An unknown sample containing mixed alkal (NaOH NaHCO or Na,CO) was analyzed using the double flask method, A 250 mg sample was dissolved in 250 m CO3 free water. A 20.0 mL aliquot of this sample required 11.3 ml. of 0.009125 M HCl solution to reach the phenolphthalein end point. Another 29.0 ml aliquot of the sample was bitrated to the bromocresol green endpoint using 31.1 ml of the standard acid. How many millimoles of the components are there in the original solid sample) 0-928 mmol No. 0.0103 mmol NaOH 0078 m N.CO Cannot be determined 012 mol NaC0, 197 minol NaHCO 129 mmol NaOH, 097ml NajcoA 0.5131-g sample that contains KBr (MM: 119.0023) is dissolved in 50 mL of distilled water. Titrating with 0.04614 M AgNO3 requires 25.13 mL to reach the Mohr end point. A blank titration requires 0.65 mL to reach the same end point. Report the %w/w KBr in the sample.A 0.2028 g sample of primary standard grade Na2C2O4 (M.M. = 134.00) was dissolved in 100 mL of 1 M H2SO4. The solution required 22.42 mL of the KMnO4 solution to reach the phenolphthalein end point. Titration of the blank (100 mL of 1M H2SO4) required 0.02 mL of the KMnO4 solution. What is the molarity of KMnO4
- An 800.0 mg sample containing sulfate was treated with a slight excess of Bac vielding a precipitate that contained 4.3 mg of co-precipitated BaCO3. After i and cooling, the precipitate (BaSO4+ BaCO3) weighed 377.0 mg. Calculate th following:1. What is the molar solubility of PbCl2 in a solution of 0.23 M CaCl2? Ksp = 1.6 ✕ 10-5 for PbCl2. 2. What is the Qsp when 32.0 mL of 6.50 ✕ 10-7 M Na3PO4 are mixed with 41.4 mL of 5.50 ✕ 10-5 M CaCl2? Assume the volumes are additive. Ksp = 2.0 ✕ 10-29. 3. What is the Qsp when 61.0 mL of 1.50 ✕ 10-4 M AgNO3 are mixed with 80.0 mL of 5.20 ✕ 10-3 M CaCl2? Assume the volumes are additive. Ksp = 1.6 ✕ 10-10. 4. What [I-1] is needed to start the precipitation of AgI from a saturated solution of AgCl? Ksp = 2.9 ✕ 10-16 for AgI and Ksp = 1.2 ✕ 10-10 for AgCl.The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.00 mL. A 20.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04672 M KOH:CH3COOC2H5 + OH- → CH3COO- + C2H5OHAfter cooling, the excess OH2 was back-titrated with 3.41 mL of 0.05042 M H2SO4. Calculate theamount of ethyl acetate (88.11 g/mol) in the original sample in grams