Asked by Caroline
The coinage metals -- copper, silver, and gold -- crystallize in a cubic closest packed structure. Use the density of gold (19.3 g/cm3) and its molar mass (197 g/mol) to calculate an approximate atomic radius for gold.
Work So Far:
{[1/(19.3x197)]x.74}/6.022e23
=3.2319e18
[(3.2319e18x3)/4pi]^(1/3)
=917190
Work So Far:
{[1/(19.3x197)]x.74}/6.022e23
=3.2319e18
[(3.2319e18x3)/4pi]^(1/3)
=917190
Answers
Answered by
DrBob222
I don't follow everything you've done.
There are 4 atoms to the face centered unit cell.
mass unit cell = (197 x 4/6.02E23) = ?grams.
volume unit cell = mass/density = ? cc
cubed root V = ? = a cm
Then a(2)^1/2 = 4r and solve for r = ? in cm. I obtained 1.44E-8 cm = 1.44 Angstroms. The internet gives 144.2 pm as the radius.
There are 4 atoms to the face centered unit cell.
mass unit cell = (197 x 4/6.02E23) = ?grams.
volume unit cell = mass/density = ? cc
cubed root V = ? = a cm
Then a(2)^1/2 = 4r and solve for r = ? in cm. I obtained 1.44E-8 cm = 1.44 Angstroms. The internet gives 144.2 pm as the radius.
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