A mixture of 0.007263 mol of Cly, 0.04728 mol of H20, 0.05105 mol of HC, and 0.03562 mol of Oz is placed in a 1.0-L stee

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answerhappygod
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A mixture of 0.007263 mol of Cly, 0.04728 mol of H20, 0.05105 mol of HC, and 0.03562 mol of Oz is placed in a 1.0-L stee

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A Mixture Of 0 007263 Mol Of Cly 0 04728 Mol Of H20 0 05105 Mol Of Hc And 0 03562 Mol Of Oz Is Placed In A 1 0 L Stee 1
A Mixture Of 0 007263 Mol Of Cly 0 04728 Mol Of H20 0 05105 Mol Of Hc And 0 03562 Mol Of Oz Is Placed In A 1 0 L Stee 1 (12.93 KiB) Viewed 37 times
A Mixture Of 0 007263 Mol Of Cly 0 04728 Mol Of H20 0 05105 Mol Of Hc And 0 03562 Mol Of Oz Is Placed In A 1 0 L Stee 2
A Mixture Of 0 007263 Mol Of Cly 0 04728 Mol Of H20 0 05105 Mol Of Hc And 0 03562 Mol Of Oz Is Placed In A 1 0 L Stee 2 (6.39 KiB) Viewed 37 times
A mixture of 0.007263 mol of Cly, 0.04728 mol of H20, 0.05105 mol of HC, and 0.03562 mol of Oz is placed in a 1.0-L steel pressure vessel 1744 K The following equilibrium is established: 2 Ca(a) + 2 Hg (9)=4 HCMg) + LOA (9) At equilibrium 0.004023 mol of Oz is found in the reaction mixture (a) Calculate the equilibrium partial pressures of Cz, H0, HC, and o PoglCl₂) Po(H20) = Peg(HCI) = PoolO₂)

PeqCHCI) Peg(0) - (b) Calculate Ky for this reaction Kp-
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