A titanium cube contains 2.86×10^23 atoms. What is the edge length of the cube? The density of titanium is 4.50 g/cm^3. (The volume of a cube is V=l^3.)
Express the length in centimeters to three significant figures.

1mole of a substance contains 6.02x10^23 atoms as we have come to understand from Avogadro's hypothesis. Therefore, 1mole of titanium also contains 6.02x10^23 atoms.

1mole of titanium = 48g

1mole (i.e 48g) of titanium contains 6.02x10^23 atoms,

Therefore Xg of titanium will contain 2.86×10^23 atoms i.e

Xg of titanium = (48x2.86×10^23)/6.02x10^23 = 22.8g

Next, we must find the volume of titanium

Density of titanium = 4.5g/cm^3

Mass of titanium = 22.8g

Volume =?

Density = Mass /volume

Volume = Mass /Density

Volume = 22.8/4.5

Volume = 5.07cm^3

Now we can find the edge length:

Volume = 5.07cm^3

Length =?

V = L^3

L = cube root (V)

L = cube root (5.07)

L = 1.72cm

A titanium cube contains 2.86×10^23 atoms. What is the edge length of the cube? The density of titanium is 4.50 g/cm^3. (The volume of a cube is V=l^3.) Express the length in centimeters to three significant figures.

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A titanium cube contains 2.86×10^23 atoms. What is the edge length of the cube? The density of titanium is 4.50 g/cm^3. (The volume of a cube is V=l^3.)
Express the length in centimeters to three significant figures.