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
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
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 :)
To find the density of the magnesium atom, we need to calculate its volume and then divide the mass by the volume.
Given information:
Radius of the magnesium atom (r) = 160 pm = 160 x 10^-12 cm
Mass of the magnesium atom (m) = 4.04 x 10^-23 g
First, let's calculate the volume of the magnesium atom using the formula for the volume of a sphere:
Volume = (4/3) * pi * r^3
Plugging in the values:
Volume = (4/3) * 3.14 * (160 x 10^-12)^3
Calculating the volume:
Volume ≈ (4/3) * 3.14 * 4.096 x 10^-33 cm^3
Volume ≈ 1.726 x 10^-32 cm^3
Now, we can calculate the density by dividing the mass of the atom by its volume:
Density = mass / volume
Plugging in the values:
Density = (4.04 x 10^-23 g) / (1.726 x 10^-32 cm^3)
Calculating the density:
Density ≈ 2.34 x 10^9 g/cm^3
Therefore, the density of the magnesium atom is approximately 2.34 x 10^9 grams per cubic centimeter.
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 !