A titanium cube contains 2.86×1023 atoms. what is the edge length l of the cube? the density of titanium is 4.50 g/cm3 . (the volume of a cube is v=l3.)

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When mass  Ti = density * volume  and when moles Ti = mass Ti/molar mass Ti                                   ∴ Volume = 2.86 x 10^23 atom * ( 1 mol Ti / 6.022 x 10^23) * (47.867 g Ti / 1 mol Ti) *(1Cm3 / 4.5 g Ti )                 = 5.05 Cm^3 when we assume that the sample of Ti is a cube: and we assume the length = X ∴ V = X^3 ∴X^3 = 5.05  ∴X = ∛5.05      = 1.7 Cm


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