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
NaCl and H2S will experience dipole-dipole interaction because they are permanently polarized.
Answer:
<u>The temperature difference is</u> 
Explanation:
The formula that is to used is :
Δ
Δ
<em>where ,</em>
- <em>Δ
is the heat supplied in calories = 300cal</em> - <em>
is the mass of water taken = m (assumed)</em> - <em>Δ is the change in temperature</em>
- <em>
is the specific heat of water = 
</em>
ΔT :
