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Get Full Access to General Chemistry: Principles And Modern Applications - 10 Edition - Chapter 16 - Problem 101
Get Full Access to General Chemistry: Principles And Modern Applications - 10 Edition - Chapter 16 - Problem 101

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# In Example 16-7, rather than use the quadratic formula to

ISBN: 9780132064521 175

## Solution for problem 101 Chapter 16

General Chemistry: Principles and Modern Applications | 10th Edition

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Problem 101

In Example 16-7, rather than use the quadratic formula to solve the quadratic equation, we could have proceeded in the following way. Substitute the value yielded by our failed assumption into the denominator of the quadratic equation; that is, use as the value of and solve for a new value of x. Use this second value of x to re-evaluate Solve the simple quadratic equation for a third value of x, and so on. After three or four trials, you will find that the value of x no longer changes. This is 3CH3NH24 = 10.00250 - second value of x2. 3CH3NH24: 3CH3NH24 10.00250 - 0.00102 x = 0.0010 the answer you are seeking. (a) Complete the calculation of the pH of 0.00250 M by this method, and show that the result is the same as that obtained by using the quadratic formula. (b) Use this method to determine the pH of 0.500 M

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Rachel Ferrell CHEM 1030 2/15/16 Chapter 4 cont.: Ionization Energy: • =minimum energy required to remove an electron from an atom in the gas phase • result is an ion= a chemical species with a net charge o cation= positive charge o anion= negative charge • Na(g)→Na (g) + e ­‐ o Ionization energy of Na= 495.8 KJ/mol o 1 IE of Na→corresponds to the removal of the most loosely held electron in a valence shell • in general→ As Zeff increases, IE also incr

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