Question #27f47

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Question #27f47

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Answer 1 · 2015-01-17T19:32:28

The answer is $1.66$ $1.66$ $cm$ $"cm"$

So, start with the fact that you have $2.6 \cdot 10^{23}$ $2.6 * 10^23$ atoms of TI, which will allow you to determine how many moles you actually have

$2.6 \cdot 10^{23}$ $2.6 * 10^23$ $atoms \cdot \frac{1 mole Ti}{6.022 \cdot 10^{23} atoms} = 0.43$ $"atoms" * ("1 mole Ti")/("6.022" * 10^23 "atoms") = 0.43$ $moles$ $"moles"$

From here, use $Ti$ $"Ti"$ 's molar mass to find out how many grams you have.

$0.43 moles \cdot \frac{47.9 grams}{1 mole Ti} = 20.6 grams$ $"0.43 moles" * ("47.9 grams")/("1 mole Ti") = "20.6 grams"$

Use the given density to determine the volume you have

$ρ = \frac{m}{V} \Rightarrow V = \frac{m}{ρ} = \frac{20.6 g}{{4.50 g/cm}^{3}} = {4.58 cm}^{3}$ $rho = m/V => V = m/(rho) = ("20.6 g")/("4.50 g/cm"^3) = "4.58 cm"^3$

Since you're dealing with a cube, which has a volume of $V_{cube} = {edge}^{3}$ $V_("cube") = "edge"^3$ , your edge length will be

$edge = V^{\frac{1}{3}} = 1.66 cm$ $"edge" = V^(1/3) = "1.66 cm"$

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Question #27f47

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