Chem 209 final exam - Study guides, Class notes & Summaries

Chemistry - ️️the science dealing with the composition of matter and the changes in composition that matter undergoes Macroscopic - ️️the overall view Microscopic - ️️looking deeper into things Scientific Method - ️️involves making observations, formulating hypotheses, and performing experiments Hypothesis - ️️a tentative explanation of certain facts that provides a basis for further experimentation Theory - ️️an explanation of the general principles of certain ph...

CHEM 102 Winter 19 Final Exam (C) Questions and Answers Potentially useful data: x  1 atm = 760 torr = 760 mm Hg 2 a [A]t = – kt + [A]o ΔG° = ΔH° – T ΔS° ln[A]t = – kt + ln[A]o ΔG = ΔG° + RT ln Q 1/[A]t = kt +1/[A]o ΔG° = – RT ln K First order half-life: t1/2 = ln2 / k ΔG° = – nFE° pH  pKa  log base acid  E  E  R T ln Q n F R = 8.314 J mol-1 K-1 F = 96500 C/mol e- 1. Please choose the letter “C” as your answe...

CHEM 102 Winter 19 Final Exam (A) Questions and Answers Potentially useful data: x  1 atm = 760 torr = 760 mm Hg 2 a [A]t = – kt + [A]o ∆G° = ∆H° – T ∆S° ln[A]t = – kt + ln[A]o ∆G = ∆G° + RT ln Q 1/[A]t = kt +1/[A]o ∆G° = – RT ln K First order half-life: t1/2 = ln2 / k ∆G° = – nFE° pH  pKa  log base acid  E  E  R T ln Q n F R = 8.314 J mol-1 K-1 F = 96500 C/mol e- 1. Please choose the letter “A” as you...

CHEM 102 Winter 19 Final Exam (A) Questions and Answers Potentially useful data: x  1 atm = 760 torr = 760 mm Hg 2 a [A]t = – kt + [A]o ∆G° = ∆H° – T ∆S° ln[A]t = – kt + ln[A]o ∆G = ∆G° + RT ln Q 1/[A]t = kt +1/[A]o ∆G° = – RT ln K First order half-life: t1/2 = ln2 / k ∆G° = – nFE° pH  pKa  log base acid  E  E  R T ln Q n F R = 8.314 J mol-1 K-1 F = 96500 C/mol e- 1. Please choose the letter “A” as you...

CHEM 102 Winter 19 Final Exam (C) Questions and Answers Potentially useful data: x  1 atm = 760 torr = 760 mm Hg 2 a [A]t = – kt + [A]o ΔG° = ΔH° – T ΔS° ln[A]t = – kt + ln[A]o ΔG = ΔG° + RT ln Q 1/[A]t = kt +1/[A]o ΔG° = – RT ln K First order half-life: t1/2 = ln2 / k ΔG° = – nFE° pH  pKa  log base acid  E  E  R T ln Q n F R = 8.314 J mol-1 K-1 F = 96500 C/mol e- 1. Please choose the letter “C” as your answe...