If it requires 38.5 milliliters of 0.85 molar nitric acid to neutralize 20.0 milliliters of barium hydroxide, solve for the molarity of barium hydroxide. show all of the work used to solve this problem. (4 points) unbalanced equation: ba(oh)2 + hno3 yields ba(no3)2 + h2o

M of Ba(OH)₂ = 0.818 M.

Explanation:

Ba(OH)₂ + 2HNO₃ → Ba(NO₃)₂ + 2H₂O,

It is clear that every 1.0 mol of Ba(OH)₂ needs 2 mol of HNO₃ to be neutralized completely.

∴ (nMV) of Ba(OH)₂ = (nMV) for HNO₃.

where, n is the no. of producible H⁺ or OH⁻ of the acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

For Ba(OH)₂:

n = 2, M = ??? M, V = 20.0 mL.

For HNO₃:

n = 1, M = 0.85 M, V = 38.5 mL.

∴ M of Ba(OH)₂ = (nMV) of HNO₃ / (MV) for Ba(OH)₂ = (1)(0.85 M)(38.5 L)/(2)(20.0 mL) = 0.818 M.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

a galvanic cell( aka voltaic cell) and electrolytic cell

explanation:

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