Our professor gave us an equation to solve for an unknown X metal.But whenever w

Question

Our professor gave us an equation to solve for an unknown X metal.But whenever we solve it, we dont get a valid AMU that applies.
Heres the equation, and below it will be how we solved:
You are given a small bar of an unknown metal X. You find thedensity of the metal is 10.5 g/cm^3. Another experiment tells youthe edge-length of the face centered cubic unit cell of thismetal has length of 4.08x10^-10 m. What is the identity of themetal?
We start with: 10.5g ( 1x10^-24)(6.02x10^23amu)( 1A    )(4.09x10^-10m) = and we get 25.852 cm^3 ( 1A )(   1 g )(1x10^-10m)
then when we plug it into the density, it doesntwork. Please help. Thanks.



Heres the equation, and below it will be how we solved:
You are given a small bar of an unknown metal X. You find thedensity of the metal is 10.5 g/cm^3. Another experiment tells youthe edge-length of the face centered cubic unit cell of thismetal has length of 4.08x10^-10 m. What is the identity of themetal?
We start with: 10.5g ( 1x10^-24)(6.02x10^23amu)( 1A    )(4.09x10^-10m) = and we get 25.852 cm^3 ( 1A )(   1 g )(1x10^-10m)
then when we plug it into the density, it doesntwork. Please help. Thanks.

Explanation / Answer

You find the density of the metal is 10.5 g/cm^3. Anotherexperiment tells you the edge-length of the face centered cubicunit cell of this metal has length of 4.08x10^-10 m. What isthe identity of the metal? Assume one unit cell..........what is the volume? (4.08x10^-10 m)3 (100 cm/1m)3 = 6.79E-23cm3 The mass of this unit cell is then calculated: 6.79E-23 cm3 x 10.5 g/cm3 = 7.13E-22g This unit volume contains 4 atoms per cell. (7.13E-22 g/4 atoms)(6.022E23 atoms/mol)= 107g/mol......element is likely Pd or Ag The mass of this unit cell is then calculated: 6.79E-23 cm3 x 10.5 g/cm3 = 7.13E-22g This unit volume contains 4 atoms per cell. (7.13E-22 g/4 atoms)(6.022E23 atoms/mol)= 107g/mol......element is likely Pd or Ag

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