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Solved: Calculating Numbers of Molecules and Atoms from

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

Solution for problem 2PE Chapter 3.12SE

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 2PE

Calculating Numbers of Molecules and Atoms from Mass (a) How many glucose molecules are in 5.23 g of C6H12O6? (b) How many oxygen atoms are in this sample? (a) How many nitric acid molecules are in 4.20 g of HNO3? (b) How many O atoms are in this sample?

Step-by-Step Solution:

Solution 2PE To convert from molecules to grams, it is necessary to first convert the number of molecules of a substance by dividing by Avogadro's number to find the number of moles, and then multiply the number of moles by the molar mass of this substance. Avogadro's number is given as 6.022 x 10^23. Step 1 Glucose its molecular formula is C6H12O6. The formula weight of glucose C6H12O6.=180.1559 g/mol. We know that From avogadro's law 1 mol of C6H12O6=6.023*10 .molecules of C6H12O6. So ,molecules C6H12O6=5.23g of C6H12O6.*(1 mol C6H12O6)/180.1559)*(6.023*10 molecules C6H12O6/1 mol C6H12O6). 22 =1.74 *10 molecules C6H12O6. 22 So there are 1.74*10 molecules C6H12O6 present in 5.23 g of C6H12O6. Step 2 There are six oxygen atoms in each C6H12O6 molecule . So ,atoms O=(1.74*10 molecules C6H12O6)*(3 atoms O/1 molecule C6H12O6) 22 =5.22*10 atoms ‘O’ 23 So, there are 1.20*10 atoms of oxygen present in this sample .

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Chapter 3.12SE, Problem 2PE is Solved
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Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

This full solution covers the following key subjects: Molecules, atoms, sample, mass, hno. This expansive textbook survival guide covers 82 chapters, and 9464 solutions. The answer to “Calculating Numbers of Molecules and Atoms from Mass (a) How many glucose molecules are in 5.23 g of C6H12O6? (b) How many oxygen atoms are in this sample? (a) How many nitric acid molecules are in 4.20 g of HNO3? (b) How many O atoms are in this sample?” is broken down into a number of easy to follow steps, and 49 words. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3. Since the solution to 2PE from 3.12SE chapter was answered, more than 433 students have viewed the full step-by-step answer. Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247. The full step-by-step solution to problem: 2PE from chapter: 3.12SE was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM.

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