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What is an ion? A cation? An anion?

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 19E Chapter 2

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 19E

Problem 19E

What is an ion? A cation? An anion?

Step-by-Step Solution:
Step 1 of 3

Solution : Gold is a chemical element with symbol Au. It is one of the least reactive chemical elements. Gold often occurs naturally in its elemental (native ) form, i.e in purest form. The consumption of gold produced in the world is about 50% in jewelry, 40% in investments, and 10% in industry. Given, Radius of a gold ( Au ) atom = 1.35 Å (a) Express this distance in nanometers (nm) and in picometers (pm). First let’s see how much is 1Å, 1 nm and 1pm in meters. 1 Å= 10 m.0 -9 1nm = 10 m. 1pm = 10 -12m. The radius of Au in nanometers is : -1 = 1.35×10 × 1nm = 0.135nm Therefore, the radius of Au - 1.35Å in nanometers(nm) is 0.135nm. The radius of Au in picometers is : = 1.35 ×1/10 pm = 135 pm. Therefore, the radius of Au - 1.35Å in picometers(nm) is 135 pm. (b) ow many gold atoms would have to be lined up to span 1.0 mm To find the number of gold atoms required to line up to span 1.0mm, we need to find the istance covered by one gold atom. Therefore, the distance covered by one gold atom is equal to its diameter. Given, radius of gold a tom = 1.35Å. Therefore, diameter will be twice the radius = 2× 1.35Å = 2.7Å. Distance covered by one gold atom in mm is : = [1m = 10 mm] = 2.7Å×10 mm.7 The number of gold atoms required to line up to span 1.0 mm is : = 3.70 ×10 Au atoms. 6 Therefore, the n umber of gold atoms to line up to span 1.0mm is 3 .70 ×10 Au atoms. (c) If the atom is assumed to be a sphere, what is the volume in cm3 of a single Au atom Volume of a sphere is given by formula : Where, r = radius of the atom. So, we calculate the volume by substituting the values in the formula : = = 1.031×10 cm . 3 3 37 3 Therefore, the volume of gold atom in cm is 1.031×10 cm .

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Chapter 2, Problem 19E is Solved
Step 3 of 3

Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247. This full solution covers the following key subjects: atom, gold, atomic, picometers, distance. This expansive textbook survival guide covers 82 chapters, and 9454 solutions. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3. Since the solution to 19E from 2 chapter was answered, more than 477 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 19E from chapter: 2 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM. The answer to “What is an ion? A cation? An anion?” is broken down into a number of easy to follow steps, and 8 words.

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