Following reaction is involved in above system
HOCl(aq) ↔ H+(aq) + OCl-<span>(aq)
</span>OCl-(aq) + H2O(l) ↔ HOCl(aq) + OH-<span>(aq)
</span>
Now, if the system is obeys 1st order kinetics we have
K = [OCl-][H+<span>]/[HOCl] ............. (1)
</span>∴ [HOCl-] / [OCl-] = [H+] (1 / 3.0 * 10-8<span>) ............. (2)
</span>
and now considering that system is obeying 2nd order kinetics, we have
K = [OH-][HOCl-] / [OCl-] ................. (3<span>)
</span>Subs 2 in 3 we get
K = [OH-][H+] (1 / 3.0 * 10-8<span>)
</span>we know that, [OH-][H+] = 10<span>-14
</span>∴K = 3.3 * 10<span>-7
</span>
Thus, correct answer is e i.e none of these
Answer:
Difference in the potential energy of the reactants and products
Explanation:
The products have a lower potential energy than the reactants, and the sign of ΔH is negative. In an endothermic reaction, energy is absorbed. The products have a higher potential energy than the reactants, and the sign of ΔH is positive.
Answer:
B:the second answer
Explanation:
Because the all have leaves
Answer:
Correct choice are C and D (they are both, the same).
Explanation:
Cathode is the positive(+) electrode where a reduction occurs.
Reduction is the chemical reaction where the oxidation state is reduced.
2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag+(aq) + H2O (l)
A. 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-
B. 2Ag (s) → 2Ag+ (aq) + 2e-
C. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)
D. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)
C or D, are ok. They are the same equation.
Oxygen from ground state reduce the oxidation state from 0 to -2
