A substituted cyclohexane has an axial form that is 1.14 kcal/mol higher in energy than the equatorial form. Given this information, what fraction of the molecules (expressed as a percent to the nearest ones) are in the equatorial form at any instant in a sample of this substituted cyclohexane that is at equilibrium? Assume the temperature is 25 °C, and assume that only the axial and equatorial forms are possible. Recall that R equals 0.001987 kcal/KLaTeX: \cdot⋅mol.

Hbr dissociates to form h+ and br-ca(oh)2 dissociates to form ca2+ and 2oh-the h+ and oh- react to form waterassuming the reactants are dissolved in water, the net reaction is h+ + oh- —> h2oif you’re asked to show all components of the reaction, the complete reaction is 2hbr + ca(oh)2 —> cabr2 + 2h2o

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.

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