The dissociation of molecular iodine into iodine atoms is represented as I2(g) â 2I(g) At 1000 K, the equilibrium constant Kc for the reaction is 3.80 Ã 10â5. Suppose you start with 0.0454 mol of I2 in a 2.28âL flask at 1000 K. What are the concentrations of the gases at equilibrium?

What is the equilibrium concentration of I2? M

What is the equilibrium concentration of I? M

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Keith LeannonLv2

13 Dec 2019

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Pleeaseee Helppp I've tried it on my own but I can't get past plugging it into the quadratic formula

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The dissociation of molecular iodine into iodine atoms is represented as

I₂(g) â 2I(g)

At 1000 K, the equilibrium constant K_c for the reaction is 3.80 Ã 10^â5. Suppose you start with 0.0455 mol of I₂ in a 2.25âL flask at 1000 K. What are the concentrations of the gases at equilibrium?

What is the equilibrium concentration of I₂?

M

What is the equilibrium concentration of I?

The dissociation of molecular iodine into iodine atoms isrepresented as
I2 (g)-------> 2I (g)
<-------
At 1000 K, the equilibrium constant Kc for the reaction is 3.80 x10^-5. Suppose you start with 0.0457 mol of I2 in a 2.27-L flask at1000 K. What are the concentrations of the gases atequilibrium?

What is the equilibrium concentration of I2?
__________ M
What is the equilibrium concentration of I?
__________M