Agold atom is a sphere with a diameter of 272 pm and a mass of 3.27 x 10-13 ng. calculate the density of one gold atom (in g/cm3). note the vsphere = 4/3 π r3.

Answer is: density is 3.88 g/cm³.

r(Au) = 272 pm = 272·10⁻¹² m.

r(Au) = 2.72·10⁻¹⁰ m · 10²cm/m.

r(Au) = 2.72·10⁻⁸ cm; diameter of gold atom.

V(Au) = 4/3 · (2.72·10⁻⁸ cm)³ · 3.14.

V(Au) = 8.43·10⁻²³ cm³; volume of gold atom.

m(Au) = 3.27·10⁻¹³ ng · 10⁻⁹ g/ng.

m(Au) = 3.27·10⁻²² g; mass of gold atom.

d(Au) = m(Au) ÷ V(Au).

d(Au) = 3.27·10⁻²² g ÷ 8.43·10⁻²³ cm³.

d(Au) = 3.88 g/cm³; density of one gold atom.

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.