What volume of 10.08 m hno3 is required to prepare 458 ml of a 1.6 m hno3 solution? answer in units of ml?

This is a dilution question. A portion of the concentrated solution is taken and mixed with water to prepare a diluted solution.
the equation used when dealing with diluting solutions is as follows;
c1v1 = c2v2 
where c1 is the concentration and v1 is the volume of the concentrated solution.
c2 is the concentration and v2 is the volume of the diluted solution to be prepared. 
substituting the values in the equation;
10.08 M  x V = 458 mL x 1.6 M
V = 72.70 mL
a volume of 72.70 mL is taken from 10.08 concentrated nitric acid solution and diluted with distilled water to make a volume of 458 mL which will have a concentration of 1.6 M.

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.