# Determine the number of oxygen atoms present in 0.698 moles of silver sulfate.?

I got 1.68 x 10^24, but I'm not sure if that's right.

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Silver sulfate is Ag2SO4:

You have:

.698 moles of Ag2SO4 *(6.022x10^23 mlc of Ag2SO4/1 mol of Ag2SO4) * (4 atoms of O/1 mlc of Ag2SO4) = 1.68 x 10^24 atoms of O.

I got the same answer so I think you're right

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• Silver Sulfate, Ag2SO4.

1 mole is Avagadro's number (6.022 x 10^23) molecules. By definition

.698 mole is .698 x 6.022 x 10^23 = 4.20 x 10^23 By math

There are 4 atoms of O per molecule of Ag2SO4. By formula

Therefore, there are (4.20 x 10^23) x 4 atoms = 1.68 x 10^24 atoms. By math

Source(s): Common chemistry knowledge. I had chemistry in high school in 1961. I had to look up Avagadro's Number on Wikipedia. I forgot the exact number over the years but remembered what a mole was.
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• 1mole Ag2SO4-------------4*6.022*10^23atoms

0.698mole-------------4*6.022*0.698*10^23atoms= 1.68 x 10^24

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• 1mole Ag2SO4-------------4*6.022*10^23atoms

0.698mole-------------4*6.022*0.698*10^2… 1.68 x 10^24

There you are now

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