Question #bfcbb

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Question #bfcbb

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Answer 1 · 2017-05-19T00:02:30

$3.8 \cdot 10^{- 21} g$ $3.8 * 10^(-21)"g"$

Explanation:

Copper has a molar mass of ${63.546 g mol}^{- 1}$ $"63.546 g mol"^(-1)$ , which means that $1$ $1$ mole of copper has a mass of $63.546 g$ $"63.546 g"$ .

Now, in order to have $1$ $1$ mole of copper, you need to have $6.022 \cdot 10^{23}$ $6.022 * 10^(23)$ atoms of copper $\to$ $->$ this is known as Avogadro's constant.

So, right from the start, you can say that if $1$ $1$ mole of copper has a mass of $63.546 g$ $"63.546 g"$ and $1$ $1$ mole of copper contains $6.022 \cdot 10^{23}$ $6.022 * 10^(23)$ atoms of copper, then

$6.022 \cdot 10^{23} . atoms Cu \to 63.546 g$ $6.022 * 10^(23)color(white)(.)"atoms Cu " -> " 63.546 g"$

All you have to do now is to use this as a conversion factor to calculate the mass of $6$ $6$ atoms of copper

$6 color(red)(cancel(color(black)("atoms Cu"))) * "63.546 g"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Cu")))) = color(darkgreen)(ul(color(black)(3.8 * 10^(-21)color(white)(.)"g")))$

I'll leave the answer rounded to two sig figs.

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Question #bfcbb

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