Concentration of Cu in unknown (penny) = (from calibration curve) Mass of Cu in sample (penny) = Recall that the volume of the solution obtained by dissolving the penny is 250.00 mL (0.25000 L). To calculate the mass of Cu in the penny, you will first need to calculate the moles of Cu. This is done by multiplying the concentration of Cu (in moles/L) by the volume (0.25000 L). Finally, multiplying the moles of Cu by the molar mass of Cu (in g/mole) yields the mass of Cu in the original sample. %3D Percent Cu in penny =

The acid-base indicator HIn undergoes the following reaction in dilute aqueous solution: HIncolor1H++Incolor2 The following absorbance data were obtained for a 5.00 I0-4 M solution of HIn in 0.1 M NaOH and 0.1 M HC1. Measurements were made at wavelengths of 485 nm and 625 nm with 1.00-cm cells. 0.1 M NaOH A485 = 0.075 A625 = 0.904 0.1 M HC1 A485 = 0.487 A625 = 0.181 In the NaOH solution, essentially all of the indicator is present as In-; in the acidic solution, it is essentially all in the form of HIn. (a) Calculate molar absorptivities for In- and HIn at 485 and 625 nm. (b) Calculate the acid dissociation constant for the indicator ¡fa pH 5.00 buffer containing a small amount of the indicator exhibits an absorbance of 0.567 at 485 nm and 0.395 at 625 nm (1.00-cm cells). (c) What is the pH of a solution containing a small amount of the indicator that exhibits an absorbance of0.492 at 485 nm and 0.245 at 635 nm (1.00-cm cells)? (d) A 25.00-mL aliquot of a solution of purified weak organic acid HX required exactly 24.20 mL of a standard solution of a strong base to reach a phenolphthalein end point. When exactly 12.10 mL of the base was added to a second 25.00-mL aliquot of the acid, which contained a small amount of the Indicator under consideration, the absorbance was found to be 0.333 at 485 nm and 0.655 at 625 nm (1.00-cmcells). Calculate the pH of the solution and Ka for the weak acid. (e) What would be the absorbance of a solution at 485 and 625 nm (1.50-cm cells) that was 2.00 10-4 M in the indicator and was buffered to a pH of 6.000?

Mass spectrometry is an extremely versatile detection system for GC. However, interfacing an HPLC system to a mass spectrometer is a much more difficult task. Describe the major reasons why it is more difficult to combine HPLC with mass spectrometry than it is to combine GC with mass spectrometry.

A 10.00 g sample containing an analyte was transferred to a 250 mL volumetric flask and diluted to volume. When a 10.00 mL aliquot of the resulting solution was diluted to 25.00 mL it was found to give a signal of 0.235 (arbitrary units). A second 10.00 mL aliquot was spiked with 10.00 mL of a 1.00 ppm standard solution of the analyte and diluted to 25.00 mL. The signal for the spiked sample was found to be 0.502. Calculate the weight percent of analyte in the original sample.