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KIM [24]
3 years ago
14

How much Sr(OH)2 • 8 H2O (M = 265.76) is needed to prepare 250.0 mL of solution in which [OH–] = 0.100 M?

Chemistry
1 answer:
attashe74 [19]3 years ago
4 0
The required is the amount of Sr(OH)2·8H2O needed to prepare a solution having a concentration of 0.100 M OH-. We start with the concentration of the hydroxide ion to determine the number of moles hydroxide ion for a given volume of 250.0 mL solution.

0.100 mol/L (0.250 L) = 0.4 mol hydroxide ion needed.

We know that, for every mole of Sr(OH)2·<span>8H2O in solution, two hydroxide ions is formed. Thus,

0.4 mol OH- ( 1 mol </span>Sr(OH)2·<span>8H2O / 2 mol OH-) ( 265.76 g / mol) = 53.152 g 

Thus, 53.152 grams of </span>Sr(OH)2·<span>8H2O is needed to make a solution with 0.100 M OH-.</span>
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Answer:

Kp = 0.049

Explanation:

The equilibrium in question is;

2 SO₂ (g)  +  O₂ (g)   ⇄ 2 SO₃ (g)  

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The initial pressures are given, so lets set up the ICE table for the equilibrium:

atm        SO₂         O₂          SO₃

I              3.3        0.79           0

C              -2x           -x          2x

E             3.3 - 2x    0.79 - x    2x

We are told 2x = partial pressure of SO₃ is 0.47 atm at equilibrium, so we can determine the partial pressures of  SO₂ and O₂ as follows:

p SO₂  = 3.3 -0.47 atm = 2.83 atm

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Now we can calculate Kp:

Kp = 0.47² /[ ( 2.83 )² x 0.56 ] = 0.049 ( rounded to 2 significant figures )

Note that we have extra data in this problem we did not need since once we setup the ICE table for the equilibrium we realize we have all the information needed to solve the question.

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