Question
The radius of a magnesium atom is 160 pm & its mass is 4.04 X 10^-23 g. What is the density of the atom in grams per cubic centimeter?
Assume the atom is a sphere with:
Volume= (4/3)*pi*r^3
Assume the atom is a sphere with:
Volume= (4/3)*pi*r^3
Answers
well I think it is 173 pm but oh well
r = 160 * 10^-12 m * 10^2 cm/m
= 160 *10^-10 cm = 1.6 *10^-9 cm
vol = (4/3)pi (1.6)^3 * 10^-27 cm^3
= 17.2*10^-27 cm^3
so
4.04*10^-23/(17.2 * 10^-27)
= .235 * 10^4 g/cm^3
= 2350 g/cm^3
check my arithmetic !
r = 160 * 10^-12 m * 10^2 cm/m
= 160 *10^-10 cm = 1.6 *10^-9 cm
vol = (4/3)pi (1.6)^3 * 10^-27 cm^3
= 17.2*10^-27 cm^3
so
4.04*10^-23/(17.2 * 10^-27)
= .235 * 10^4 g/cm^3
= 2350 g/cm^3
check my arithmetic !
well I think it is 173 pm but oh well
r = 160 * 10^-12 m * 10^2 cm/m
= 160 *10^-10 cm = 1.6 *10^-8 cm
vol = (4/3)pi (1.6)^3 * 10^-24 cm^3
= 17.2*10^-24 cm^3
so
4.04*10^-23/(17.2 * 10^-24)
= .235 * 10^1 g/cm^3
= 2.350 g/cm^3
more reasonable
r = 160 * 10^-12 m * 10^2 cm/m
= 160 *10^-10 cm = 1.6 *10^-8 cm
vol = (4/3)pi (1.6)^3 * 10^-24 cm^3
= 17.2*10^-24 cm^3
so
4.04*10^-23/(17.2 * 10^-24)
= .235 * 10^1 g/cm^3
= 2.350 g/cm^3
more reasonable
I did a double take for 2350. Hg, Au, U etc not even close to that.
I had a hunch it might not be quite that heavy :)
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