## Chemistry (4th Edition)

First, I would add 3.00 mL of that $4.00$ $M$ $HNO_3$ solution to an adequate recipient, and then I would add sufficient water to reach 60.0 mL of solution.
1. Find the necessary volume: $C_1 * V_1 = C_2 * V_2$ $0.200* 60.0= 4.00 * V_2$ $12.0 = 4.00 * V_2$ $V_2 = 3.00$ mL