The pNa of a mixture prepared by mixing 200 ml of 0.00063 Mof NaCl and 300 ml of 0.00040 M in Na3PO4 is: O 2.10 O 1.50 O 4.84 O 3.34 3.01 ge Next page
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- må iddake used mh thiosulfat Eanation 4. 10.00 mL of a solution containing calcium iodate was titrated to a starch endpoint with 23.73 mL of 0.02454 M NazS203. What is the [IO3 ] in the initial solution? 5. What is the Ksp of a saturated calcium iodate solution containing 0.0147 M iodate ion? Ksp = [10;!] 944. a)What is the concentration of Na* in % for a solution prepared by mixing 100.0 ml of 0.20 M Nacl and 200.0 ml of 0.200 M Na,SO. b) What is the pNa* for this solution?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
- A protein of molecular weight = 55,000 g/mol and at 1 mg/mL concentration was titrated with Ellman’s reagent. What is the molar concentration of the protein in solution?1.Which of the following correlations about the analyses of iodine and saponificationnumbers is/are incorrect?a. Alcoholic KOH: dissolve KOH to lipid sampleb. Hanus reagent: react with saturated lipid bondc. Reflux set-up: hasten reactiond. Blank titration: determine total mol of KOH2. A triacylglycerol (Sample X) was subjected to saponification analysis. 1.1 grams of thesample was refluxed with 5% alcoholic KOH and the refluxed solution was titrated with 17.2 mLof 0.44 M HCl titrant. The blank solution was titrated with 25.6 mL of the titrant.i. What is the saponification number of the sample? ii. Which of the following could be the identity of the sample? Show solutions pls a. Tripalmitin (MW: 807.3)b. Trilaurin (MW: 639.8)c. Triolein (MW: 885:4)d. Tristearin (MW: 891.48)A 10.00mL sample of alcoholic ethyl acetate was diluted to 100.00 mL. 20.00 mL was aliquoted and mixed with 40.00 mL of 0.04672 M KOH. The resulting mixture was heated for 2 hours. CH3COOC2H5 + OH- → CH3COO- + C2H5OH After cooling, the excess OH was back titrated with 3.41 mL of 0.05042 M H2SO4. Answer the following: Calculate the number of moles of OH- that reacted with ethyl acetate. Calculate the number of moles of ethyl acetate in the 20.00 mL solution. What is the mass of ethyl acetate (FW=88.11 g/mol) in the original 10.00 mL sample?
- An unknown sample containing mixed alkali (NAOH, NaHCO3, or NazCO3) was analyzed using the double flask method. A 250 mg sample was dissolved in 250 mL CO2 free water. A 20.0 mL aliquot of this sample required 11.3 mL of 0.009125 M HCI solution to reach the phenolphthalein end point. Another 29.0 mL aliquot of the sample was titrated 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? O 0.28 mmol Na,CO3 O 1.29 mmol NaOH, 0.97 mmol Na2CO3 O 0.103 mmol NaOH, 0.078 mmol NazCO3 O 1.29 mmol Na2CO3, 0.97 mmol NaHCO3 O Cannot be determinedStandardization of a Solution A 50.00 mL aliquot portion of 0.01020 M KIO3 solution was transferred into a conical flask. After adding 2.0 g KI and 2 mL of 6M HC1, the mixture was titrated with 31.32 mL of the Na2S2O3 solution to the starch end point. Calculate the number of mmoles of IO3-. mmoles 103- How many mmoles of 12 was released after the addition of KI? mmoles I₂ How many mmoles Na2S2O3 reacted with the 12 produced? mmoles Na2S2O3- What is the molarity of the Na2S2O3 solution? M Na2S2O3 Write the half reaction representing the transformation of OC1- (hypochlorite ion) to Cl-(chloride) in acidic medium. Reduction Half Reaction: Write the half reaction representing the transformation of 1- (iodide ion) to 12 (iodine) in acidic medium. Reduction Half Reaction: Analysis of Household Bleaching Agent The composition of a commercially available household bleaching agent is advertised as 5% (w/v) NaOC1. . Calculate the number of mmoles of NaOCl present in 50 mL of the bleaching agent.…The pNa of a mixture prepared by mixing 50 ml of 0.00063 M of NaCl and 300 ml of 0.00040 M in NA3PO4 is: O 1.50 O 2.95 О 3.05 О 3.34 O 4.84
- The thiourea in a 1.455-g sample of organic material was extracted into a dilute H2SO4solution and titrated with 37.31 mL of 0.009372 M Hg2+via the reaction4(NH2)2CS + Hg2+ S [(NH2)2CS]4Hg2+Find the percentage of (NH2)2CS (76.12 g/mol) 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 KMnO416:19 An ore is analyseod for the manganese content by converting the manganese to Mng04 and woighing it a 152 g sample yelds Mng04 woighing 0.126 g, what would be the percent Mngog in the sample? 3 Mng03 a 2Mng04 + % 02 3 mol Mng03 =2 mol Mng04 OA858 % OR5.97 % OC4 95 % O0.4 20 % OE 33.32 % Add a caption... > Status (Custom) +