Naf final exam - Study guides, Class notes & Summaries

CISSP CBK Review Final Exam CISSP CBK Review Page 1 1. A risk is the likelihood of a threat source taking advantage of a vulnerability to an information system. Risks left over after implementing safeguards is known as: A. Leftover risks. B. Residual risks. C. Remaining risks. D. Exposures. 2. Copyright provides what form of protection: A. Protects an author’s right to distribute his/her works. B. Protects information that provides a competitive advantage. C. Protects the right of a...

CHEM 102 – Winter 18 Final Exam Review Practice Problems 1) What is the name of the compound below? A) 2-ethyl-2-butene B) 2-ethylbutane C) 3-methyl-2-pentene D) 3-methyl-3-pentene E) 2,2-ethyl-2-butene 2) Proteins are biopolymers formed via multiple condensation coupling of which two monomers? A) ester and diamine B) diamine and dicarboxylic acid C) dialcohol and amide D) dialcohol and diamine E) alcohol and carboxylic acid 3) Which of the following substances wou...

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...