$2.55 \times 10^{24}$ $2.55xx10^24$ titanium atoms are cast into a cube that has a mass of $202.5 \cdot g$ $202.5g$ . If the density of titanium metal is $ρ = 4.60 \cdot g \cdot c m^{- 3}$ $rho=4.60g*cm^-3$ , what is the edge length of the cube?

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$2.55 \times 10^{24}$ $2.55xx10^24$ titanium atoms are cast into a cube that has a mass of $202.5 \cdot g$ $202.5g$ . If the density of titanium metal is $ρ = 4.60 \cdot g \cdot c m^{- 3}$ $rho=4.60g*cm^-3$ , what is the edge length of the cube?

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Answer 1 · 2016-09-22T19:20:42

The length of the metal cube $=$ $=$ $^{3} \sqrt{45.0 \cdot c m^{3}}$ $""^3sqrt(45.0*cm^3)$ $=$ $=$ $? ? c m$ $??cm$

Explanation:

If there are $2.55 \times 10^{24}$ $2.55xx10^24$ $titanium atoms$ $"titanium atoms"$ , there are $\frac{2.55 \times 10^{24}}{6.022 \times 10^{23} \cdot m o l^{- 1}}$ $(2.55xx10^24)/(6.022xx10^23*mol^-1)$ $=$ $=$ $4.23 \cdot m o l$ $4.23*mol$ $titanium atoms$ $"titanium atoms"$ .

And this has a mass of $4.23 \cdot m o l \times 47.87 \cdot g \cdot m o l^{- 1}$ $4.23*molxx47.87*g*mol^-1$ $=$ $=$ $202.5 \cdot g$ $202.5*g$ .

Now we have a cube that has a mass of $202.5 \cdot g$ $202.5*g$ . But we know its $density$ $"density"$ , which you have kindly provided, $ρ$ $rho$ $=$ $=$ $4.50 \cdot g \cdot c m^{- 3}$ $4.50*g*cm^-3$ .

$ρ$ $rho$ $=$ $=$ $\frac{Mass}{Volume}$ $"Mass"/"Volume"$ . Thus $Volume$ $"Volume"$ $=$ $=$ $\frac{Mass}{ρ}$ $"Mass"/rho$ $=$ $=$ $\frac{202.5 \cdot g}{4.50 \cdot g \cdot c m^{- 3}}$ $(202.5*g)/(4.50*g*cm^-3)$ $=$ $=$ $45.0 \cdot c m^{3}$ $45.0*cm^3$

And thus the side-length of the metal cube $=$ $=$ $^{3} \sqrt{45.0 \cdot c m^{3}}$ $""^3sqrt(45.0*cm^3)$ $=$ $=$ $? ? c m$ $??cm$ $≅$ $~=$ $3.5 \cdot c m$ $3.5*cm$

Don't worry about making a mistake in the question; this happens. Mind you, I hope I have got your question right!

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$2.55 \times 10^{24}$ $2.55xx10^24$ titanium atoms are cast into a cube that has a mass of $202.5 \cdot g$ $202.5g$ . If the density of titanium metal is $ρ = 4.60 \cdot g \cdot c m^{- 3}$ $rho=4.60g*cm^-3$ , what is the edge length of the cube?

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Explanation:

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2.55×10242.55xx10^24 titanium atoms are cast into a cube that has a mass of 202.5⋅g202.5*g. If the density of titanium metal is ρ=4.60⋅g⋅cm−3rho=4.60*g*cm^-3, what is the edge length of the cube?

1 Answer

Explanation:

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$2.55 \times 10^{24}$ $2.55xx10^24$ titanium atoms are cast into a cube that has a mass of $202.5 \cdot g$ $202.5g$ . If the density of titanium metal is $ρ = 4.60 \cdot g \cdot c m^{- 3}$ $rho=4.60g*cm^-3$ , what is the edge length of the cube?