What volume in milliliters of 0.100 M HCIO3 is required to neutralize
40.0 mL of 0.165 M KOH?

66.0 mL HClO3

Explanation:

M1V1 = M2V2

M1 = 0.100 M HClO3

V1 = ?

M2 = 0.165 M KOH

V2 = 40.0 mL KOH

Solve for V1 --> V1 = M2V2/M1

V1 = (0.165 M)(40.0 mL) / (0.100 M) = 66.0 mL HClO3

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.