Balance the following equation:
KOH + H2SO4 ⇒ K2SO4 + H2O

2KOH(aq) + H2SO4(aq) ⇒ K2SO4(aq) + 2H2O(l)

Explanation:

The reaction is a neutralization reaction since an acid, aqeous H2SO4 reacts completely with an appropriate amount of alkali, aqueous KOH to produce salt, aqueous K2SO4 and liquid water, H2O only.

2KOH(aq) + H2SO4(aq) ⇒ K2SO4(aq) + 2H2O(l)

Alkali + Acid → Salt + Water.

During this reaction, 2 moles of KOH neutralize 1 mole of H2SO4 to yield 1 mole of K2SO4 and 2 moles of H2O.

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.