How manyatoms of mercury are in 1.4214E+1 ml sample

5.787 x 10²³ atoms.

Explanation:

Firstly, we should calculate the mass of this sample. we have the relation (m = d x V), where m is the mass of the element, d is the density of the element (d of Hg = 13.56 g/cm³) and V is the volume of the element (V = 14.214 ml). m = d x V = (13.56 g/cm³) (14.214 ml) = 192.74 g. Now, we can calculate the number of moles of Hg in this sample: n = mass / atomic mass = (192.74 g) / (200.59 g/mole) = 0.96 mole. It is known that 1.0 mole of the element contains Avogadro's number of atoms (6.023 x 10²³).

Using cross multiplication:

1.0 mole of Hg → 6.023 x 10²³ atoms

0.96 mole of Hg →? atoms

The atoms of Hg in 14.214 ml sample is = (0.96 mole x 6.023 x 10²³ atoms) / (1.0 mole) = 5.787 x 10²³ atoms.