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.

plants need sunlight, water, and carbon dioxide to make food. plants use the raw materials of water, nutrients they absorb from soil, and carbon dioxide to make sugars the plant can use. energy from the sun and chlorophyll are also needed.

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be