How many milliliters of 0.0050 n koh are required to neutralize 53 ml of 0.0050 n h2so4 ?

Answer is: 53 milliliters of KOH are required.

Balanced chemical reaction:

2KOH(aq) + H₂SO₄(aq) → 2H₂O(l) + K₂SO₄(aq).

N(KOH) = 0.0050 N; normality of potassium hydroxide solution.

V(H₂SO₄) = 53 mL; volume of sulfuric acid.

N(H₂SO₄) = 0.0050 N; normality of sulfuric acid.

Na·Va  = Nb·Vb.

0.0050 N · 53 mL = 0.0050 N · V(KOH).

V(KOH) = 53 mL; volume of potassium hydroxide solution.

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.