Q: A 25.00-mL sample of a household cleaning solution
A 25.00-mL sample of a household cleaning solution was diluted to 250.0 mL in a volumetric flask. A 50.00-mL aliquot of this solution required 43.04 mL of 0.1776 M HCl to reach a bromocresol green end...
See AnswerQ: A 0.1401-g sample of a purified carbonate was
A 0.1401-g sample of a purified carbonate was dissolved in 50.00 mL of 0.1140 M HCl and boiled to eliminate CO2 Back-titration of the excess HCl required 24.21 mL of
See AnswerQ: The aluminum in a 2.200-g sample of impure
The aluminum in a 2.200-g sample of impure ammonium aluminumsulfate was precipitated with aqueous ammonia as the hydrous Al2O3 xH2O. The precipitate was filtered and ignited at 1000°C to give anhydro...
See AnswerQ: The efficiency of a particular catalyst is highly dependent on its zirconium
The efficiency of a particular catalyst is highly dependent on its zirconium content. The starting material for this preparation is received in batches that assay between 68% and 84% ZrCl4. Routine an...
See AnswerQ: In the determination of lead in a paint sample, it is
In the determination of lead in a paint sample, it is known that the sampling variance is 10 ppm while the measurement variance is 4 ppm. Two different sampling schemes are under consideration: Scheme...
See AnswerQ: Calculate the pH of water at 25°C and 50°
Calculate the pH of water at 25°C and 50°C. The values for pKw at these temperatures are 13.99 and 13.26, respectively.
See AnswerQ: A dilute solution of an unknown weak acid required a 28.
A dilute solution of an unknown weak acid required a 28.94-mL titration with 0.1062 M NaOH to reach a phenolphthalein end point. The titrated solution was evaporated to dryness. Calculate the equivale...
See AnswerQ: A 3.00-L sample of urban air was bubbled
A 3.00-L sample of urban air was bubbled through a solution containing 50.0 mL of 00116MBaOH2 , which caused the CO2 in the sample to precipitate as BaCO3 The excess base was back-titrated to a...
See AnswerQ: Air was bubbled at a rate of 30.0 L/
Air was bubbled at a rate of 30.0 L/min through a trap containing 75 mL of 1% H2O2 H2O2 SO2 H2SO4 After 10.0 minutes, the H2SO4 was titrated with 10.95 mL of 0.00242 M NaOH. Calculate concentrat...
See AnswerQ: The digestion of a 0.1417-g sample of a
The digestion of a 0.1417-g sample of a phosphorus-containing compound in a mixture of HNO3 and H2SO4 resulted in the formation of CO2 ,H2O, and HPO34 Addition of ammonium molybdate yielded a solid...
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