CHEM 335 - Exam 3 Questions and Answers Latest Update Already Passed

CHEM 335 - Exam 3 Questions and Answers Latest Update Already Passed Theoretical Yield Equation - Answers = (actual yield / percentage yield) x 100 Reaction Yield Equation - Answers = (experimental grams / theoretical grams) x 100 or = (experimental moles / theoretical moles) x 100 Mole Fraction Equation - Answers - Xsolute/solvent = moles of solute/solvent / total moles of solution - Xsolute + Xsolvent = 1 Mole Percentage Equation - Answers - Xsolute/solvent = (moles of solute/solvent / total moles of solution) x 100 - Xsolute + Xsolvent = 100% Mass Percentage Equation - Answers = (mass of component / total mass) x 100 Percent Recovery Equation - Answers = (actual (isolated) amount / expected amount) x 100 Proportionality Example - Answers @ 100 degree C → 7.5g/100mL @ 20 degree c → 0.2g/100mL Max grams that would dissolve 80mL of water @ 100 degree C: - 7.5g = 100mL - ? = 80mL -- (7.5g x 80mL) / 100mL = 6.0g Common Spectroscopic Analytical Methods - Answers (1) X-Ray Absorption Spectroscopy (2) UV-Vis Absorption Spectroscopy (3) IR Absorption Spectroscopy (4) Microwave Absorption Spectroscopy (5) Electron Spin Resonance Spectroscopy (6) Nuclear Magnetic Resonance Spectroscopy Absorption Spectroscopy - Answers - involves passing a beam of electromagnetic radiation through a sample - some radiation is absorbed (attenuated) by the sample, and the rest is transmitted (passed through the sample) How is absorbance/transmittance measured? - Answers - as a funct