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# A pure silver ring contains 0.0134 mmol (millimol) Ag How many silver atoms does it ISBN: 9780321910295 34

## Solution for problem 33P Chapter 6

Introductory Chemistry | 5th Edition

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Problem 33P

Problem 33P

A pure silver ring contains 0.0134 mmol (millimol) Ag How many silver atoms does it contain?

Step-by-Step Solution:
Step 1 of 3

Solution 33P

Here, we have to calculate the number of atoms present in 0.0134 mmol of Ag ring.

We know,

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

Therefore,

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.

Step 2 of 3

Step 3 of 3

##### ISBN: 9780321910295

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.

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A pure silver ring contains 0.0134 mmol (millimol) Ag How many silver atoms does it