Enter your answer in the provided box. Hydrogen sulfide decomposes according to the following reaction, for which Kc = 9.30 Ã 10â8 at 700Â°C: 2 H2S(g) â 2 H2(g) + S2(g) If 0.55 mol of H2S is placed in a 3.0âL container, what is the equilibrium concentration of H2(g) at 700Â°C? ? M

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13 Dec 2019

Enter your answer in the provided box.

Hydrogen sulfide decomposes according to the following reaction, for which

K_c = 9.30 Ã 10^â8 at 700Â°C:

2 H₂S(g) â 2 H₂(g) + S₂(g)

If 0.55 mol of H₂S is placed in a 3.0âL container, what is the equilibrium concentration of H₂(g) at 700Â°C?

? M

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Related questions

Hydrogen sulfide decomposes according to the following reaction, for which K_c= 9.30x10^-8 at 700 degrees Celsius.

2H₂S_(g) → 2H_2(g) + S_2(g)

If 0.29 moles of H₂S is placed in a 3.0-L container, What is the equilibrium concentration of H_2(g) at 700 degrees Celsius?

carminecamel917

Hydrogen sulfide decomposes according to the following reaction forwhich Kc=9.3*10^-8 at 700 degrees celcius2H2S(g)<--->2H2(g)+S2(g) if 0.41 mol of H2S is placed in a 3L container, what is the equilibrium eoncentration of H2(g) at 700degrees celcius

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