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- Using basic conditions, MnO4- can be used as titrant for the analysis of Mn2+, with both the analyte and the titrant ending up as MnO2. In the analysis of a mineral sample for manganese, a 0.5165-g sample is dissolved, and the manganese is reduced to Mn2+. The solution is made basic and titrated with 0.03358 M KMnO4, requiring 34.88 mL to reach the endpoint. Calculate the %w/w Mn in the mineral sample. Answer: % Mn =When I was a boy, I watched Uncle Wilbur measure the iron content of runoff from his banana ranch. He acidified a 25.0-mL sample with HNO3 and treated it with excess KSCN to form a red complex. (KSCN itself is colorless.) He then diluted the solution to 100.0 mL and put it in a variablepathlength cell. For comparison, he treated a 10.0-mL reference sample of 6.80 3 1024 M Fe31 with HNO3 and KSCN and diluted it to 50.0 mL. The reference was placed in a cell with a 1.00-cm pathlength. Runoff had the same absorbance as the reference when the pathlength of the runoff cell was 2.48 cm. What was the concentration of iron in Uncle Wilbur’s runoff ?A 20 ml aliquot of malonic acid solution was treated with 10.0 ml of 0.25M Ce4+ leading to the reaction CH2(COOH)2 + 6Ce4+ + 2H2O ® HCOOH + 2CO2 + 6Ce3+ + 6H+ After standing for 10 minutes at 60oC, the solution was cooled and the x’ss Ce4+ was titrated with 0.1M Fe2+, requiring 14.4 ml to reach the ferroin end point. Calculate the M of the malonic in the sample.
- Caustic potash that has been exposed to air is found on analysis to contain 90.00% KOH, 2.38% K2CO3 and 7.62% H2O. What weight of residue will be obtained if 1.00 g of this sample is added to 46.00 mL of 1.00 N HCl and the resulting solution, after neutralization with 1.070N KOH is evaporated to dryness?It is required to know the concentration of an aqueous solution of H2SO4 that appeared in the laboratoryChemistry III and it's unlabeled. To this end, a student of analytical chemistry carried out the followingProcedure: He took 5.00 mL of a fresh and standardized solution of 0.525M NaOH and brought them to a250.0 mL balloon to be completed with distilled water. Subsequently, he poured 15.00 mL of the solutionH2SO4 of unknown concentration in an Erlenmeyer flask and added 2 drops of phenolphthalein.Using a burette filled with the last NaOH solution, he noticed that when adding 39.40 mL of the hydroxidethe Erlenmeyer solution reached a faint but permanent pink. With the above dataDetermine the concentration and pH of the H2SO4 solution.The sulfate in a247.1 mg sample was precipitated as BasO4 by addition of 25.00 mL of 0.03992 M BaCl2. The precipitate was removed by filtration and the remaining BaCl2 consumed 36.09 mL of 0.0217 M EDTA for titration to the Camalgite endpoint. Calculate the % SO3 in the sample.
- ) What amperage is required to plate out 0.250 mol Cr from a Cr3+ solution in a period of 8.00 h?The fat in a 1.821 g sample of potato chips is extracted with supercritical CO2. After extraction, the residue weighs 1.139 g. What is the fat content (% w/w) of the potato chips?A 0.64 g sample containing KCl ( mw = 74.6 ) is dissolved in 50mL of water and titrated to the Ag2CrO4 end point, requiring 26.2 mL of 0.15 M AgNO3 .The %w/w KCl in the sample is ?
- From the titration of 50.00 mL of FeSO4 0.2500 mol L-1 with a solution of K2Cr2O7 0.1000 mol L-1 under acidic conditions, in which [H+] = 1.00 mol L-1, calculate the Potential (E) of the system for adding the following volumes of K2Cr2O7 solution: i) 5.00 mL ii) volume at the stoichiometric point (PE) iii) 22.00 mLA 1.250 g sample of cheese was subjected to KKjeldahl analysis to determine the amount of protein. the sample was digested and the nitrogen is oxidized to NH4+ and was then converted to NH3 with NaOH., and distilled into a collection flask containing 50.00 mL of .1050 M HCl. The excess HCl is back titrated with .1175M NaOH, this required 21.65 mL to reach the bromothymol blue endpoint. What is the reaction for the back titration for this analysis and the %N of cheese sample.A 1.000 g sample containing chlorides, iodides and inert materials was treated with dilute nitric acid followed by AgNO3. A precipitate of AgCl (143.32) and AgI (234.77) was produced and weighs 0.9238 g. On heating in a current of Cl2, the AgI is converted to AgCl, and the resulting product weighs 0.7238 g. Find the percentage of a) NaI (149.89) and b) NaCl (58.44) in the sample
