How many silver atoms are there in 3.84 g of silver?

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How many silver atoms are there in 3.84 g of silver?

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Answer 1 · 2016-07-10T12:07:16

We know that in $126.94 \cdot g$ $126.94*g$ of silver there are $N_{A}, i.e. Avogadro's number of silver atoms, 6.022 \times 10^{23} individual silver atoms$ $N_A", i.e. Avogadro's number of silver atoms, "6.022xx10^23" individual silver atoms"$ .

Explanation:

So, we work out the quotient, and multiply by Avogadro's number,:

$\frac{3.84 \cdot g}{126.94 \cdot g \cdot m o l^{- 1}} \times 6.022 \times 10^{23} \cdot m o l^{- 1}$ $(3.84*g)/(126.94*g*mol^-1)xx6.022xx10^23*mol^-1$

$=$ $=$ $? ? silver atoms$ $??" silver atoms"$

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How many silver atoms are there in 3.84 g of silver?

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