Heroin, a tertiary amine, is a weak base because it contains a nitrogen atom with a lone pair of electrons. Therefore, it can be titrated with a mineral acid, e.g., HCI. Heroin has a molecular mass equal to 369.4176 g/mol. The crime laboratory took 75.000 mg of the mysterious white powder and dissolved it in one hundred mL of water. The resulting aqueous solution was titrated to the equivalence point with 5.000 mL of 0.02500 M HCI solution. Using these data, the percentages of heroin in the sugar/heroin mixture is O 30.12% O 21.90% 43.80% O 60.24%

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
icon
Related questions
icon
Concept explainers
Question
A8
A mysterious white powder is found at a crime scene. A simple chemical analysis
concludes that the powder is a mixture of sucrose and heroin (I).
I
Transcribed Image Text:A mysterious white powder is found at a crime scene. A simple chemical analysis concludes that the powder is a mixture of sucrose and heroin (I). I
Heroin, a tertiary amine, is a weak base because it contains a nitrogen atom with a
lone pair of electrons. Therefore, it can be titrated with a mineral acid, e.g., HCI.
Heroin has a molecular mass equal to 369.4176 g/mol.
The crime laboratory took 75.000 mg of the mysterious white powder and
dissolved it in one hundred mL of water. The resulting aqueous solution was
titrated to the equivalence point with 5.000 mL of 0.02500 M HCl solution.
Using these data, the percentages of heroin in the sugar/heroin mixture is
O 30.12%
O 21.90%
O 43.80%
O 60.24%
Transcribed Image Text:Heroin, a tertiary amine, is a weak base because it contains a nitrogen atom with a lone pair of electrons. Therefore, it can be titrated with a mineral acid, e.g., HCI. Heroin has a molecular mass equal to 369.4176 g/mol. The crime laboratory took 75.000 mg of the mysterious white powder and dissolved it in one hundred mL of water. The resulting aqueous solution was titrated to the equivalence point with 5.000 mL of 0.02500 M HCl solution. Using these data, the percentages of heroin in the sugar/heroin mixture is O 30.12% O 21.90% O 43.80% O 60.24%
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps

Blurred answer
Knowledge Booster
Ionic Equilibrium
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY