You wish to undertake a 100 step countercurrent extraction of solutes A & B with distribution coefficients D= 1.5 & Da - 0.6. Suppose the volumes of mobile and stationary are equal. If fractions 28-40 and 50-68 are each pooled find the % recovery and purity of each solute. AA 79.8% recovery of 99.7% pure A B: 82.0% recovery of 86.6% pure B B. A: 72.7 % recovery of 97.0 % pure A B: 89.0 % recovery of 98.9 % pure B OCA 82.6% recovery of 92.3% pure A B: 82.9% recovery of 91.0% pure B O D.A 92.6% recovery of 99.5% pure A 8: 67.0% recovery of 100 % pure B
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- 17- The percentage of protein in meat products is determined by multiplying the %N as determined by the kjeldahl method by the arbitrary factor 6.25. A sample of processed meat scrap weighing 2.000 g is digested with concd H2SO4 and Hg (catalyst) until the N present has been converted to NH&HSO.. This is treated with excess NaOH, and the liberated NH3 is caught in 50.0 mL pipetful H2SO, (1.000 ml 0.01860g NazO). The excess acid requires 28.80 mL NaOH (1.000 mL = 0.1266 g KHP). Calculate % protein in the meat scrap.A 15.00 g sample containing mixed alkali and other inert components was dissolved and diluted to 300 mL with water. A 20 ml, aliquot was titrated with 5.02 mL of 0.5352 M HCI to reach PHP endpoint. Another 20 mL aliquot was titrated to the BCG endpoint, using up 18.87 mL of titrant in the process. Identify the alkali components and their percent weight.Electroly Concentr X Chapter Partial P henrys la x My Q 'index.html Sodium hydroxide is a white, crystalline basic salt that readily dissociates in water to produce a strong base NaOH(s) – - Na* (aq) + OH (aq) According to the CRC Handbook of Chemistry and Physics, sodium hydroxide (NaOH; m.m. = 39.997 g mol-1; d = 2.13 g cm 3) has an aqueous solubility of 100.0g/100g at 20 °C. Below is a table of aqueous sodium hydroxide densities (in kg L-1) at varying concentrations (wt%) and temperatures (from handymath). Conc. (wt%) 0 °C 10 °C 20 °C 40 °C 60 °C 80 °C 100 °C 1.0124 1.0107 • 1.0095 1.0033 0.9941 0.9824 0.9693 2 1.0244 1.0220 1.0207 1.0139 1.0045 0.9929 0.9797 4. 1.0482 1.0444 1.0428 1.0352 1.0254 1.0139 1.0009 8 1.0943 1.0889 1.0869 1.0780 1.0676 1.0560 1.0432 12 1.1399 1.1333 1.1309 1.1210 1.1101 1.0983 1.0855 16 1.1849 1.1776 1.1751 1.1645 1.1531 1.1408 1.1277 20 1.2296 1.2218 1.2191 1.2079 1.1960 1.1833 1.1700 24 1.2741 1.2658 1.2629 1.2512 1.2388 1.2259 1.2124 28 1.3182 1.3094…
- Q2- A 30.0 mL H2C2O4 acidified solution, was treated with 25.0 mL of 0.102 M KMnOg solution. The reaction: 2MNO4- + 5 H2C204+ 6H → 2MN2+ +10CO2 + 8H20 The excess permanganate required 8.34 mL of 0.053 M Fe2* solution, the reaction Mno“ + 5Fe?* + 8H+ g Mn* +5Fe+ + 4H20 Find the molarity of the H2C2O4 solutionAssume that 1.00 mL (1.02 g) of crude product mixture is obtained from the reactionbefore the washing steps, and the distribution constant of the product mixture in brine isK = (Cmixture/Cbrine) = 50.0. Assuming that after mixing the product mixture volume isstill 1.00 mL (i.e., the volume lost is small), and the volume of the first brine wash layeris 2.00 mL. What mass (grams) of product mixture is lost (dissolves) in the first brinewash? (See Technique 13.2, p 53.) Show work.(3) In a titration, the % analyte is computed from the following equation: Vol (titrant) x M (titrant) x MW (analyte) % analyte = x 100 Weight (sample) Calculate the absolute error in the % analyte from the following data. Vol (titrant)= 38.04 t 0.02 ml M (titrant) MW (analyte) = 74.116 t 0.005 mg/mmol Wt (sample) = 800.0 ± 0.2 mg = 0.1137 ± 0.0003 mmol/ml %3!
- 20 aspirin tablets labeled 80mg were dissolved in 100mL of 90% ethanol. A 10mL aliquot was taken and was used for assay. The analyte followed usual process and was treated with 50mL of 0.1000N NaH and was titrated with 35mL 0.1050N H2O4 until the solution achieved completion. Calculate the % content of the total aspirin capsules and the actual label claimSolution: Kso=1.41-10-11 The resistance of a saturated solution of PbSO4 (very insoluble salt) at 291K is of 1.7x106 0, being the resistivity of pure water 7.1x105 Ocm. 1. Determine the solubility product of PBSO4 at 291K knowing that Am° (SO4 2)=136 0lcm?mol-1 and Am° (Pb+2)=122 o-lcm?mol-1, 2. How many centimeters will the ion move Pb+2 in one second when applying a potential difference of 25V between two electrodes 1 cm apart?Mganga is doing a routine analysis in a copper plating factory. He is regularly making up solutions of verypure CuSO4.5H2O. He suspect that a batch of a reagent used to make up a stock solution of CuSO4.5H2Ois impure and it may contains some CuCl2. He asked you to determine the percentage impurity as follows:Dissolve1.500g of a mixture and make up to 1.00L mark.150mL of this solution was treated with 20.00mLof 0.100M AgNO3 to remove 98.6% of chloride as AgCl precipitate.
- The concentration of a NAOH solution is determined in a single experiment by titrating a weighed sample of pure, dried KHP (MW = 204.229) with the NaOH. The data are: Weight of KHP = (11.6723 ± 0.0001) - (10.8364 ± 0.0001) g = (32.68 ± 0.02) – (1.24 ± 0.02) mL %3D Volume NaOH Where the standard deviations are o, values. (a) What is the molarity of the NaOH? (b) Based on the propagation-of-error formula, what is the standard deviation of the molarity,on ?In a solution containing 25ml (NH4)2S2O8 with a total volume = 100 ml =0.1 liters , the solution composition are shown below.Note; the solution require 25ml of 0.2M S2O8^-2.SOLUTION Kl KNO3 EDTA Na2S2O3 STARCH50ml 23ml 1 drop 1 ml 10 dropsExperiment; initial (S2O8-2) =0.05M ; INITIAL (l-)=0.10 Mml of S2O3^-2 added Time in minutes and seconds for color change Cumulative time in seconds Total moles of S2O8^-2 consumed1 1:43 2.0 x 10^-42 1:20 4.0 x 10^-43 1:13 6.0 x 10 ^-44 1.17 8.0 x 10 ^-45 1.24 10 x 10 ^-41a) Find the cumulative time in secondsb) for the runs ,plot mole of S2O8^2- REACTION vs. time in seconds2) Draw a straight line through ( origin ) the points and calculate the slope3) divide the value of the slope by the total volume of ( 0.1 L) to get the rate in units of M/s4) find the rate5 )calculate x and y and k100kg of a solution of acetic acid and H₂O containing 30% acetic acid by weight is to be extracted three times with isopropyl ether using 40kg of isopropyl ether in each stage. Determine the weight of acetic acid extracted and the percentage recovery of acetic acid. The following data are available. Concentrations of acetic acid in the leaving streams are Isopropyl ether rich phase 11.7% Ist Stage IInd Stage IIIrd Stage 9.5% 7.8% [BALU, K.; RAMALINGAM, S.; DEEBIKA, B. AND SATHYAMOORTHI, K. (2014)] H₂O rich phase 25.8% 22.7% 20%