[xe] 6s2 4f4 is the noble gas configuration for which element

Neodymium

Explanation:

The given element is:

          [Xe] 6s² 4f⁴

To solve this problem, we need to find the number of electrons in the element given;

  Xe  has 54 electrons already

  6s² 4f⁴  where:

                 letters are the orbitals

                 6 and 4 are the energy levels

                superscript are the number of electrons

  So,  2 + 4  = 6

 Total number of electrons  = 54 + 6  = 60

The element is Neodymium

for every 1 mole of Cu3(PO4)2, there are 2 moles of phosphorus. We also know that 1 mole of Cu3(PO4)2 has 6.022 x 10^23 atoms (Avogadro's number
8.90 moles Cu3(PO4)22 moles P1 mol Cu3(PO4)21 mol Cu3(PO4)26.022 x 10^23 atoms
(53.5958 x 10^23)/2 = 26.7979 x 10^23 atoms of P