What is the concentration of nitrate ions in a 0.225 msr(no3)2 solution?

We are given that there are 2 moles of NO3 from the 0.225 M Sr(NO3)2 solution. Therefore we simply use stoichiometry to solve this problem.

Concentration NO3 = 0.225 M Sr(NO3)2 * (2 moles NO3 / 1 mole Sr(NO3)2)

Concentration NO3 = 0.45 M

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.