What volume of a 3.5 m hcl is required to completely neutralize 50.0 ml of a 2.0 m naoh

NaOH:
V=50mL=0,05L
Cm = 2M

n = Cm*V = 0,05L * 2M = 0,1mol

HCl         +    NaOH ⇒ NaCl + H₂O
1mol        :     1mol
0,1mol     :     0,1mol

HCl:

n = 0,1mol
Cm = 3,5M

        n             0,1mol
V= =          ≈ 0,029dm³ = 29mL of 3,5M HCl
       Cm           3,5 mol/dm³

Bis the

Answer the air near the ground cools to its dew point