What is the molarity of each ion in a solution prepared by dissolving 0.520 g of na2so4, 1.186 g of na3po4, and 0.223 g of li2so4 in water and diluting to a volume of 100.00 ml?

First  calculate  the  number  of moles  of  each  ion  present
for  Na2So4  is
0.52/142=0.0037moles
Na+  =0.0037  x2 =0.0074moles
SO4^-2  =0.0027moes

for Na3PO4=  1.186/164=0.0072 moles
Na+   =0.0072  x3=0.0216moles
po4  ions =  0.0072  moles

for LiSo4 =0.223/110=0.002moles
Li  ions=0.002 x2  =0.004mole
SO4  ions =0.002moles
total moles  for
Na  ions  =0.0216  +0.0074 =0.029moles
forSO4 ions =0.0037 +0.002=0.0057moles
molarity is  therefore
Na  ion=0.029/0.1=0.29M
SO4  ions=0.0057/0.1=0.057M
Li  ions =0.004/0.1=0.04M
PO4  ions=0.0072/0.1=0.072M

There is one s orbital, and there are three p orbitals, five d orbitals, and seven f orbitals. thus, the answer is b. do not confuse the number of orbitals in a subshell with the number of electrons the subshell can hold. each orbital can hold two electrons, so the capacity of an nf subshell is 14 electrons.