If you diluted 120 mL of 1,049 M NaCl to 1,920 mL, what would be the new concentration of NaCl? Don't put units into your
answer.

65.56 M

Explanation:

Using the formula as follows:

C1V1 = C2V2

Where;

C1 = initial concentration (M)

C2 = final concentration (M)

V1 = initial volume (L)

V2 = final volume (L)

According to the information in this question,

C1 = 1,049 M

C2 = ?

V1 = 120 mL

V2 = 1,920 mL

Using C1V1 = C2V2

1049 × 120 = C2 × 1920

125880 = 1920C2

C2 = 125880 ÷ 1920

C2 = 65.56 M

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.