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The Equilibrium Constant for the Formation of Calcium Carbonate from the Ions

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The equilibrium constant for the formation of calcium carbonate from the ions in solution is 2.2 108 according to the reaction: Ca2+(aq) + CO32-(aq) The equilibrium constant for the formation of calcium carbonate from the ions in solution is 2.2 <font face= symbol ></font> 10<sup>8</sup> according to the reaction: Ca<sup>2+</sup>(aq)  + CO<sub>3</sub><sup>2</sup><sup>-</sup>(aq)    CaCO<sub>3</sub>(s)  What is the value of the equilibrium constant for the reverse of this reaction? A)  the same, 2.2 <font face= symbol ></font> 10<sup>8</sup> B)  -2.2 <font face= symbol ></font> 10<sup>8</sup> C)  2.2 <font face= symbol ></font> 10<sup>-</sup><sup>8</sup> D)  4.5 <font face= symbol ></font> 10<sup>-</sup><sup>9</sup> E)  4.5 <font face= symbol ></font> 10<sup>9</sup> CaCO3(s)
What is the value of the equilibrium constant for the reverse of this reaction?


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