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Question
a coin contains 0.253 moles of silver (ag). how many silver (ag) atoms are in the coin?
(there is a calculation diagram and some options below: 6.02×10²³ atoms ag, 1 mole ag, 0.253 mole ag, 1.52×10²³ atoms ag, 6.02×10²³ g ag, 1 g ag, 0.253 atoms ag, 1.52×10²³ mol ag)
Step1: Recall Avogadro's number
Avogadro's number is \( 6.02 \times 10^{23} \) atoms per mole, which means 1 mole of any substance contains \( 6.02 \times 10^{23} \) particles (atoms, molecules, etc.).
Step2: Set up the conversion
We have 0.253 moles of Ag. To find the number of atoms, we multiply the number of moles by Avogadro's number. The formula is: Number of atoms = Moles × Avogadro's number.
So, \( 0.253 \, \text{mol Ag} \times \frac{6.02 \times 10^{23} \, \text{atoms Ag}}{1 \, \text{mol Ag}} \)
Step3: Calculate the result
Multiply 0.253 by \( 6.02 \times 10^{23} \).
\( 0.253 \times 6.02 \times 10^{23} = (0.253 \times 6.02) \times 10^{23} \)
\( 0.253 \times 6.02 \approx 1.52306 \)
So, the number of atoms is approximately \( 1.52 \times 10^{23} \) atoms Ag.
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The number of silver atoms in the coin is \( \boldsymbol{1.52 \times 10^{23}} \) (when calculated as \( 0.253 \, \text{mol} \times 6.02 \times 10^{23} \, \text{atoms/mol} \)).