A 100 mL of 0.125 M CH₂COOH solution was shaken with 1.2 g charcoal for 5 min. After filtration, 10 mL sample was titrated with 0.50 M NaOH. If 2.1 mL NaOH were needed to reach the end point, calculate the mass of acetic acid adsorbed per gram charcoal. Select one: 2.1g 0.17 g 0.14 g 0.10 g
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- 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.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.Standardization of EDTA was done using MgSO4 standard wherein a 50-mL aliquot of solution obtained from 0.480 g MgSO4 in 500 mL needed 39.4 mL of the EDTA solution to reach the endpoint. Determine how many milligrams of CaCO3 will react per mL of this EDTA solution.
- A 100 mL of 0.375 M CH3COOH solution was shaken with 1.0 g charcoal for 5 min. After filtration, 10 mL sample was titrated with 0.50 M NaOH. If 6.8 mL NaOH were needed to reach the end point, calculate the mass of acetic acid adsorbed per gram charcoal. < Select one: O 0.21 g 0.31 g 0.17 g 0.0035 gPreparation and Standardization of KMnO4 solution Experimental data Complete the table below. Trial 1 0.2001 g Trial 2 0.2065 g Trial 3 Weight of sodium oxalate (Na2C2O4, MM= 134 g/mol) Titration data 0.2050 g Final reading Initial reading 29.86 mL 0.00 mL 30.66 mL 30.52 mL 0.00 mL 0.00 mL Total vol. of KMNO4 used Computed Molarity of KMNO4 solution Mean Molarity Computed Normality of KMNO4 Mean Normality of KMNO4 solution Reaction Involved: Calculations:SO3 + H20 by titration with permanganate in acidic pH: peroxide was diluted to 25 ml and analyzed The sample required 84.2 mL of 0.0175 M permanganate to reach the concentration of hydrogen peroxide in the original sample? 04+H202 →02 + Mn2+ NT COPY NRSITY What is the GE
- 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 CO, 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 311 mL of the standard acid. What is/are the component/s of the sample? O Cannot be determined O NAHCO3 only O NAOH and Na,CO3 O NazCO3 only O NAHCO, and Na2CO3An 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 CO, 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. What is/are the component/s of the sample? O NazCO3 only NaHCO3 only O NAHCO3 and Na>CO3 NaOH and NazCO3 O Cannot be determined1. A 1.000 g sample containing Na,C,0, (MM=134 mg/mmol) is titrated with 40.00 mL of 0.0200 M KMNO, in acid solution. Calculate fhe percentage of Na,C,0, in the sample. 5C20,2- + 2MNO, + 16H* analyte Rxn: 2MN2+ + 10 CÓ, + 8H,0 titrant
- 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?A 500.0-mg sample of waste ground coffee beans used as fertilizer was digested and analyzed for nitrogen using the Kjeldahl method. After digestion, the distilled ammonia was collected in 100 mL of 0.5500 M boric acid. This solution required 19.60 mL of 0.5880 M HCI for titration to the methyl red end point. Calculate the %N in the fertilizer. Prelim Rxn: NH3 + H3BO3 --> NH4: H2BO3 Titration Rxn: NH : H2BO3 + HCI --> H3BO3 CI-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