The equilibrium constant for the formation of calcium carbonate from the ions in solution is 2.2 $\times$ 10 8 according to the reaction: What is the value of the equilibrium constant for the reverse of this reaction?

The Answer of The equilibrium constant for the formation of calcium carbonate from the ions in solution is 2.2 $\times$ 10 8 according to the reaction: What is the value of the equilibrium constant for the reverse of this reaction?

The equilibrium constant for the formation of calcium carbonate from the ions in solution is 2.2 $\times$ 10 8 according to the reaction: What is the value of the equilibrium constant for the reverse of this reaction?

The Answer of The equilibrium constant for the formation of calcium carbonate from the ions in solution is 2.2 $\times$ 10 8 according to the reaction: What is the value of the equilibrium constant for the reverse of this reaction?

Solved

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

Question 39

Multiple Choice

The equilibrium constant for the formation of calcium carbonate from the ions in solution is 2.2 $\times$ 10⁸ according to the reaction: $The equilibrium constant for the formation of calcium carbonate from the ions in solution is 2.2 \times 108 according to the reaction: What is the value of the equilibrium constant for the reverse of this reaction? A) the same, 2.2 \times 108 B) (-2.2 \times 108) C) 2.2 \times 10-8 D) 4.5 \times 10-9 E) 4.5 \times 109$ $The equilibrium constant for the formation of calcium carbonate from the ions in solution is 2.2 \times 108 according to the reaction: What is the value of the equilibrium constant for the reverse of this reaction? A) the same, 2.2 \times 108 B) (-2.2 \times 108) C) 2.2 \times 10-8 D) 4.5 \times 10-9 E) 4.5 \times 109$ $The equilibrium constant for the formation of calcium carbonate from the ions in solution is 2.2 \times 108 according to the reaction: What is the value of the equilibrium constant for the reverse of this reaction? A) the same, 2.2 \times 108 B) (-2.2 \times 108) C) 2.2 \times 10-8 D) 4.5 \times 10-9 E) 4.5 \times 109$ What is the value of the equilibrium constant for the reverse of this reaction?