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Simora [160]
3 years ago
5

Calculate the percent dissociation of crotonic acid (C,H,CO,H) in a 0.63 mM aqueous solution of the stuff.

Chemistry
1 answer:
Sauron [17]3 years ago
5 0

Answer:

16%

Explanation:

Crotonic acid : C₃H₅CO₂H

C₃H₅CO₂H  ⇄ C₃H₅CO₂⁻ +  H⁺

   C                      O                O

where : C = C ( 1 - ∝ ) ,   O = C∝

also:  Ka = C∝² / ( 1 - ∝ )  ---- ( 1 )

<em>From Alexa data resource : </em>

Pka = 4.69 ,  [  Ka = 2.04 * 10^-5 = C∝² / ( 1 - ∝ ) ]

back to equation 1

2.04 * 10^-5 = [ ( 0.63 * 10^-3 ) * ∝² / ( 1 - ∝ ) ]  ----- ( 2 )

∴ ∝² / ( 1 - ∝ ) = 3.24 * 10⁻²

Resolving equation above

∝ = 0.1645 =  16.45%

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Answer:

Molecular formula = P₄O₆

Explanation:

P(s)         +        O₂(g)------------------------------------⇒ PₓOₙ (g)

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10g                        7.77g                                          17.77g  (gramme ratio)

The molecular mass of Phosphorus  (P) = 31g/mole

The molecular mass of Oxygen atom (O) = 16g/mole

Mole ratio is given by:

P              :             O

10/31                    7.77/16

0.3226       :         0.4856                Mole ratio---------------------------- (1)

Divide (1)  through by 0.3226

 1                 :        1.5-------------------------------------------- (2)

From  (2), the empirical formula for Phosphorus oxide :

Empirical formula = P₁O₁.₅

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Since the molecular formula is a multiple of the empirical formula we have

Molecular formula = (PO₁.₅)ₙ----------------------------------- (3)

Since we are given the molecular mass of the oxide formed, we have:

(PO₁.₅)ₙ = 220-----------------------------(4)

[31 + (16 x 1.5)] x n = 220

[31 + 24]n = 220

55n =220

n = 4

Substituting into (3), we have :

Molecular formula = (PO₁.₅)₄

                               = P₄O₆

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<span>CH</span>₃<span>CH</span>₂<span>COOH + H</span>₂<span>O </span>↔ <span> CH</span>₃<span>CH</span>₂<span>COO</span>⁻<span> + H</span>₃<span>O</span>⁺<span> 
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