6-36. Calculate [H ] and pH for the following solutions: (a) 0.010 M HNO 3 (b) 0.035 M KOH (e) (c) 0.030 MHCl

a)2

b)12.5

c)1.5

Explanation:

a)

pH = -log [H^+]

pH = -log [0.01]

pH = 2

b)

pOH = -log[OH^-]

pOH = -log [0.035]

pOH= 1.5

But

pH + pOH = 14

pH = 14 - pOH

pH= 14 - 1.5

pH = 12.5

c)

pH = -log[H^+]

pH= -log[0.030]

pH = 1.5

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.

it is a good conductor

explanation: