4.48
pH=pKa+log([A-/HA])
25% deprotonated tells us that A- is .25 and that the rest (75% is protonated) thats .75.
4 = pKa + log \frac{.25}{.75}
4 - log \frac{.25}{.75} = pKa
4.48=pKa
Answer:
I wanna say c
Explanation:
It makes the most sense, because non-metals don't produce an electrical current, and therefore aren't conductors.
Answer:
-1.05 V
Explanation:
A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.
Oxidation half equation;
Sn(s) ------> Sn^2+(aq) + 2e
Reduction half equation:
Mn^2+(aq) + 2e ----> Mn(s)
Cell voltage= E°cathode - E°anode
E°cathode= -1.19V
E°anode= -0.14 V
Cell voltage= -1.19 V - (-0.14V)
Cell voltage= -1.05 V
A. Adding a catalyst. A catalyst by definition is a compound that speeds up reaction rates.