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17 March, 04:41

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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  1. 17 March, 04:44
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    Answer: 1.72cm

    Explanation:

    First let us calculate the mass of titanium that contain 2.86*10^23 atoms ...

    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
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