This edition is a specially designed self-study aid to help you get the most from the text and to effectively prepare for tests.Equation #2: relative abundance of 11 3In + relative abundance of 11 5In = 1 The hard part of this problem is coming up with the ... #1 and solve for X. -1.9998(X) + 0(Y) = -0.0841 0.0841 1.9998 = 0.042 To solve for Y, plug the answer for X back into the original equation #2. 0.042 + Y = 1 Y = 1 -0.042 = 0.958 To calculate the percent relative abundance of each isotope, multiply 48 Chapter 3 - Stoichiometry.
|Title||:||Study guide for Chemistry, third edition [by] Steven S. Zumdahl|
|Author||:||Paul B Kelter, Steven S. Zumdahl|
|Publisher||:||D C Heath & Co - 1993-06-01|