2nd Edition

ISBN: 9780321809261

Author: John E. McMurry, Robert C. Fay

Publisher: Prentice Hall

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Textbook Question

Chapter 14, Problem 14.116CHP

Nicotine (C₁₀H₁₄N₂) can accept two protons because it has two basic N atoms (K_b1 = 1.0 × 10^–6; K_b2 = 1.3 × 10^–11). Calculate the values of K_a for the conjugate acids C₁₀H_I4N₂H⁺ and C₁₀H₁₄N₂H₂²⁺.

Chapter 14, Problem 14.116CHP, Nicotine (C10H14N2) can accept two protons because it has two basic N atoms (Kb1 = 1.0 106; Kb2 =

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Chapter 14, Problem 14.115CHP

Chapter 14, Problem 14.117CHP

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Chapter 14 Solutions

General Chemistry: Atoms First

Ch. 14.1 - Write a balanced equation for the dissociation of...Ch. 14.1 - What is the conjugate acid of each of the...Ch. 14.1 - Prob. 14.3CP Ch. 14.1 - Some of the following ions have been detected...Ch. 14.2 - If you mix equal concentrations of reactants and...Ch. 14.2 - The following pictures represent aqueous solutions...Ch. 14.3 - Prob. 14.7P Ch. 14.4 - Prob. 14.8P Ch. 14.4 - The concentration of OH in a sample of seawater is...Ch. 14.4 - Prob. 14.10P

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Nicotine (C 10 H 14 N 2 ) can accept two protons because it has two basic N atoms ( K b1 = 1.0 × 10 –6 ; K b2 = 1.3 × 10 –11 ). Calculate the values of K a for the conjugate acids C 10 H I4 N 2 H + and C 10 H 14 N 2 H 2 2+ .

Solution Summary: The author explains the value of the acid dissociation constant (K_ a) for conjugated acids of Nicotine.

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