What is the new mass/volume (m/v)% of a koh solution that is prepared by diluting 110 ml of a 6% (m/v) koh solution to 330 ml?

We can use the dilution formula where we make diluted solutions from more concentrated solutions
c1v1 = c2v2 
c1 is concentration and v2 is volume of the concentrated solution
c2 is concentration and v2 is volume of the diluted solution to be prepared 
in this case we have been given m/v % which we can use as the concentrations
substituting the values in the equation 
6% x 110 mL = C x 330 mL 
C = 2%

the new m/v % of the diluted solution prepared is 2% (m/v) 

The correct answer is c. orbital shape. "the angular momentum quantum number l describes the shape of an orbital. it corresponds to each orbital type, i.e. (0,1,2,−1)=(s,p,d,f,g,"