The density of silver is 10.5g/cm^3. What is the mass (in kg) of a cube of silver that measures 0.62m on each side?

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First lets calculate the volume of the silver cube

The length of a side = 0.62 m

Volume of the cube = (0.62 m)^3

= 0.238328 m^3

The density of silver is given as, 10.5g/cm^3

lets convert that into kg/m^3

density of silver = (10.5 g/cm^3)x(10^6 cm^3/m^3) x (0.001 kg/g)

= 10500 kg/ m^3

Hence the weight of the silver cube ,

=> 0.238328 m^3 x 10500 kg/ m^3

**=> 2502.444 kg**

You need to remember what is the formula of density such that:

rho = m/V

m expresses the mass of silver

V expresses the volume of cube of silver

Notice that the problem provides the density of silver but it does not provide the volume.

Though the problem provides the length of side of cube, hence you may evaluate the volume such that:

V = (0.62)^3 => V = 0.23828 m^3

Notice that the units of measure of density are g/cm^3, hence you need to convert to Kg/m^3

rho = 10.5*10^(-3)/(10^(-6)) Kg/m^3

rho = 10.5*10^3 = 10500 Kg/m^3

You need to substitute 10500 for rho and 0.23828 for V in formula of density such that:

10500 = m/0.23828 => m = 0.23828*10500 = 2502.444 Kg

**Hence, evaluating the mass in Kg of cube of silver yields m = 2502.444 Kg.**

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