Answer:
pH = 2.059
Explanation:
At the Cathode:
The reduction reaction is:
At the anode:
At oxidation reaction is:
The overall equation for the reaction is:
The overall cell potential is:
Using the formula for the Nernst equation:
where;
E = 0.66
(Zn^2+)=0.22 M
Then
![0.66 =0.76- ( \dfrac{0.0591}{2})log \bigg ( \dfrac{[Zn^{2+} ] PH_2}{[H^+]^2} \bigg )](https://tex.z-dn.net/?f=0.66%20%3D0.76-%20%28%20%5Cdfrac%7B0.0591%7D%7B2%7D%29log%20%5Cbigg%20%28%20%5Cdfrac%7B%5BZn%5E%7B2%2B%7D%20%5D%20PH_2%7D%7B%5BH%5E%2B%5D%5E2%7D%20%5Cbigg%20%29)
![0.66 =0.76- 0.02955 * log \bigg ( \dfrac{0.22*0.87}{[H^+]^2} \bigg )](https://tex.z-dn.net/?f=0.66%20%3D0.76-%200.02955%20%2A%20log%20%5Cbigg%20%28%20%5Cdfrac%7B0.22%2A0.87%7D%7B%5BH%5E%2B%5D%5E2%7D%20%5Cbigg%20%29)
3.4 = log ( 0.1914) - 2 log [H⁺]
3.4 = -0.7180 - 2 log [H⁺]
3.4 + 0.7180 = - 2 log [H⁺]
4.118 = - 2 log [H⁺]
pH = log [H⁺] = 4.118/2
pH = 2.059
<span>Answer: 0.094%
</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />
<span>Only the ionization of the formic acid is the important part.
</span><span />
<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />
<span>2) Mass balance:
</span><span />
<span> HCOOH(aq) HCOO⁻(aq) H⁺(aq).
Start 0.311 0.189
Reaction - x +x +x
Final 0.311 - x 0.189 + x x
3) Acid constant equation:
</span><span />
<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />
<span>= (0.189 + x )x / (0.311 - x) = 0.000177
4) Solve the equation:
You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />
<span>With that approximation the equation to solve becomes:
</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M
5) With that number, the percent of ionization (alfa) is:
</span><span />
<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
<span></span><span />
Answer:
Option d. is correct
Explanation:
Nitromethane is an organic compound that can be used as a fuel booster in rockets. Its chemical formula is 
.
This compound is produced as a result of reaction of sodium chloroacetate with sodium nitrite in aqueous solution.
The electron-dot structure of a nitromethane molecule shows 
charge on the nitrogen.
Option d. is correct
