An experimenter created a solution with a concentration of 20.0%(w/w) from a solid active compound and the solution had a density of 1.4g/mL. From this solution, 10µL was withdrawn and mixed w/ 190µL of reagent & water. It manifested a concentration of 5x10 M compound. a. What is the concentration (in molarity) of the 20%w/w solution? b. Why was the concentration of the solution expressed as %w/w initially?
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- A 100 g soil containing 20% smectitie mineral matter was mixed with 150 mL solution containing 0.10 M Pb. CEC of pure smectitie is 110 emol(+)/kg. CEC of the soil (cmol(+)/kg) is:A 25.00mL wastewater sample was analyzed for its Mg2+ content using a standard gravimetric method. the sample was diluted to 3.00L and an 11.00mL aliquot was treated to precipitate magnesium as MgNH4PO4.6H2O using (NH4)2HPO4 as the precipitating agent. the precipitate was then filtered, washed, dried, and ignited resulting in a 0.1325mg Mg2P2O7 residue. How much Mg (in ppm) is present in the original sampleAspirin 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?..
- To a 0.5 g sample of impure NaCl are added 0.784 g of pure AgNO3 (MW = 169.88) crystal. This is in excess of the amount required to ppt. All the Cl as AgCl. After dissolving and filtering out the pptd. AgCl, the filtrate requires 25.50 ml of 0.28M KCNS. What is the percentage of NaCl (MW=58.44) in the sample?An analyst weighed 0.25045 g of the Buffalo River Sediment reference material.(RM # 8704) to determine Pb levels. The material was digested in an acid medium toBring it to a solution which was made up to the 30.2450 mL mark with distilled water. If thePb concentration in the control certificate of analysis is (150 ± 17) mg / kg, and theanalyst recovered 95%, what was the concentration of Pb in the solution?A empty evaporating dish weighs 30.56g. You place 20.0mL of an NaCl solution in the dish plus the solution weighs 53.31g. You evaporatorate the solution to dryness ant find that the evaporatoring dish plus the residue weighs 32.25g. What is the w/v%? What is the w/w%? What is the molarity?
- An impure sample of calcium carbonate with a mass of 7.95 g was reacted with 50.00 cm3 of 1.00 mol dm hydrochloric acid (an excess). The resulling solution was transferred to a volumetric flask and titrated with 11.10cm3 of 0.300 mol dm-3 sodium hydroxide solution. Determine the percentage purity by mass of the calcium carbonate sample.CaCO3 + HCl -> CaCl2 + H2O + CO2 HCl + NaOH -> NaCl +H2O a. Determine how many moles of hydrochloric acid were used.b. Determine how many moles of excess HCI was titratedc. Determine how much in moles calcium carbonate present in the sample.d. Calculate the mass of calcium carbonate presente. Determine the percentwge calcium carbonate is in the sample.A 50 mL sample solution containing 8-hydroxyquinoline (MW: 145) was analyzed by adding 25 ml, 0.1 M KBrO3, excess KBr and acidified. The mixture was left for 10 minutes in dark place. After this time KI in excess was added followed by titration with 27.9 mL, 0.05 M thiosulfate standard solution. Write balance equations? What is the percent w/v 8-hydroxyquinoline in sample?a) What mass of Fe(NH4)2(SO4)2•2H2O(s) is required to prepare a 500 mL solution containing 100 ppm (m:v) in Fe? b) Using this stock solution, what aliquot must be used to prepare calibration solutions, 100 mL volume, of the following concentrations: 0.100 ppm, 0.500 ppm, 2.00 ppm, 4.00 ppm, and 7.00 ppm.
- The standard addition method is used to analyze a sample of a river water for mercury. Solution A is made by pipetting 5.00 mL of undiluted sample in to a 10 mL volumetric flask and filling to the mark with DI water. Solution B is made by pipetting 5.00 mL of undiluted sample and 3.00 mL of 15.0 ppb of Hg standard into same 10.0 mL volumetric flask and filling to the mark with DI. Solution A and B are analyzed using atomic absorption spectroscopy and give a percent transmittance values of 56 % and 33 % respectively (not blank corrected). A blank has a transmittance of 96%. What is the corrected absorbance of both solution A and B? A. Solution A: 0.123 Solution B: 0.463 B. Solution A: 0.463 Solution B: 0.234 C. Solution A: 0.123 Solution B: 0.234 D. Solution A: 0.234 Solution B: 0.46320 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 claimThe %purity of a powdered crude sample of Na2CO3 containing only inert impurities is to be determined by reacting 225.0 mg of the crude sample to 10.0 mL of 3.00 M HCl solution, and bubbling the resulting CO2(g) product in water that is at exactly 29 °C. After the reaction has completed, the level of the liquid inside the eudiometer rests 4.30 cm above the water level in the beaker. The graduation on the eudiometer indicates that the trapped gas is 44.37 mL. The experiment was done under a barometric pressure of 755.2 torr. a. How many moles of CO2 were collected? b. What is the percent purity of the sample? Round off to the nearest whole number