A solution of an unknown carbohydrate gave the following optical readings when analyzed by a polarimeter +3.24, +3.15, +3.30, +3.20, 3.21, +3.19, +3.17, +3.23, +3.20, +3.25. 1. (a) Calculate the sample mean.
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- Thimerosal Tincture USP contains 0.1% w/v thimerosal and 50% v/v ethyl alcohol. If the cap is left off of a 15-mL bottle of the tincture, and the ethyl alcohol evaporates leaving a final volume of 9.5 mL, what is the concentration of thimerosal in the evaporated solution expressed as a ratio strength?9. 3 μL of a 45 mM stock solution of a substrate is added to 8 mL of water. Calculate the following values. The substrate molecular weight is 125 g/mol. Substrate volume in mL Dilution factor Substrate number of moles Substrate concentration for the diluted solution in mM Substrate concentration for the diluted solution in mole/L Substrate concentration for the diluted solution in mg/mLAn unknown mixture is known to contain only Ba(OH)2 (MW=171.34 g/mole) and NaOH (MW=40.0 g/mole). If the mixture is known to contain 45% by mass NaOH, and 8.0 grams of the mixture is dissolved completely in 50.0 ml of solution, answer the following. c).If 10.0 ml of a 0.2 M solution of Na2SO4 was added to the 50.0 ml solution, what would be the final concentration of Na+ in solution.
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