Ni2 Study guides, Class notes & Summaries

State University of New York at Binghamton, Department of Chemistry Chemistry 108, Introductory Chemistry II, Exam 3, Version A 1. A reaction will always be spontaneous if: a. H is + & S is + b. H is – & S is – c. H is – & S is + d. H is + & S is – e. H is + & S is 0 2. In the following reaction, which substance is behaving as a reducing agent? Cr2O72(aq) + 3Ni(s) + 14H+(aq)  3Ni2+(aq) + 2 Cr3+(aq) + 7H2O a. Cr2O72 b. Ni c. H+ d. ...

Exam (elaborations) TEST BANK FOR Introductory Quantum Optics 1st Edition By Gerry C and Knight P [Cambridge Solution Manual] Adil Benmoussa Solutions Manual To INTRODUCTORY QUANTUM OPTICS By C. C. Gerry and P. L. Knight May 15, 2005 Table of Contents Table of Contents 3 1 Introduction 7 2 Field Quantization 9 2.1 problem 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.2 problem 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.3 problem 2.3 . . . . . ...

Chemistry Final Exam (Answered) Oxidation Rules Rule 1: The oxidation number of an element in its free state = 0 (Also true for diatomic elements) -Mg=0,H2=0, C=0 Rule 2: The oxidation number of a monatomic (one-atom) ion is the same as the charge on the ion -Mg^2+=+2, Na^+=+1, O^2-=-2 Rule 3: The sum of all oxidation numbers in a neutral compound is zero. The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion. Rule 4: The oxidation numb...

Chemistry Final Exam (Answered) Oxidation Rules Rule 1: The oxidation number of an element in its free state = 0 (Also true for diatomic elements) -Mg=0,H2=0, C=0 Rule 2: The oxidation number of a monatomic (one-atom) ion is the same as the charge on the ion -Mg^2+=+2, Na^+=+1, O^2-=-2 Rule 3: The sum of all oxidation numbers in a neutral compound is zero. The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion. Rule 4: The oxidation numb...

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...

Exam (elaborations) TEST BANK FOR CISCO TECHNICAL SOLUTION SERIES IP Telephony Solution Guide Version 2.1 CHAPTER 1 Introduction to IP Telephony 1-1 Overview 1-1 Organization 1-1 Audience 1-2 Scope 1-2 Revision History 1-2 Related Information 1-3 CHAPTER 2 IP Telephony Architecture Overview 2-1 CHAPTER 3 Planning the IP Telephony Network 3-1 In this Chapter 3-1 Related Information 3-1 Evaluating and Documenting the Existing Data Infrastructure 3-1 LAN/Campus Environment 3-2 WAN En...

CHEM 102 Winter 18 Final Exam (A) 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 “A” as your a...

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...