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Mashcka [7]
3 years ago
8

What are the respective concentrations (moles/liter) of k+ and po43- afforded by dissolving 0.800 mol k3po4 in water and dilutin

g to 1.63l?
Chemistry
1 answer:
Kitty [74]3 years ago
8 0

<em>Answer: </em>

  •                      Concentration of K+  = 1.47 M
  •                      Concentration of Po4∧-2 = 0.4908 M

<em>Given Data:</em>

   No. of moles of K3PO4 = 0.800 mole

  Molarity of K3PO4 =  0.800÷1.63 = 0.4908

<em>Chemical equations:</em>

        K3PO4 ⇔3 K+   +  PO4∧-2

<em>Solution: </em>

         K3PO4 : K+                                                K3PO4 : PO4∧-2

            1  :    3                                                            1      :    1

        0.4908 = 0.4908 × 3 = 1.47 M                   0.4908 = 0.4908 × 1 = 0.4908 M

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                     ⇒ -\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}  -----------------------------(1)

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from equation (1) we can write

                                   \frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]

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∴ So the rate of formation of SO₃ [\frac{d[SO_{3}] }{dt}] = 7.28 x 10⁻³ M/s

7 0
4 years ago
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