A pure silver ring contains 0.0134 mmol (millimol) Ag How many silver atoms does it contain?
Step 1 of 3
Here, we have to calculate the number of atoms present in 0.0134 mmol of Ag ring.
1 millimole (mmol) = 0.001 mol
1 mol of Ag ring =6.02 x 1023 Ag atoms
The conversion Factors = and 6.02 x 1023 Ag atoms/ 1mol
A pure silver ring contains 0.0134 mmol (millimol) Ag
= 0.0134 mmol x (0.001 mol/1.0 mmol) x (6.02 x 1023 atoms/ 1.0 mol)
= 0.000080668 x 1023 atoms
= 8.066 x 1018 atoms
Thus, A pure silver ring (0.0134 mmol (millimol))contains 8.066 x 1018 atoms of Ag.
Textbook: Introductory Chemistry
Author: Nivaldo J Tro
This full solution covers the following key subjects: silver, mmol, contains, millimol, contain. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. The full step-by-step solution to problem: 33P from chapter: 6 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. The answer to “A pure silver ring contains 0.0134 mmol (millimol) Ag How many silver atoms does it contain?” is broken down into a number of easy to follow steps, and 16 words. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. Since the solution to 33P from 6 chapter was answered, more than 1163 students have viewed the full step-by-step answer.