<h3>1</h3>
Species shown in bold are precipitates.
- Ca(NO₃)₂ + 2 KOH → Ca(OH)₂ + 2 KNO₃
- Ca(NO₃)₂ + Na₂C₂O₄ → CaC₂O₄ + 2 NaNO₃
- Cu(NO₃)₂ + 2 KI → CuI₂ + 2 KI
- Cu(NO₃)₂ + 2 KOH → Cu(OH)₂ + 2 KNO₃
- Cu(NO₃)₂ + Na₂C₂O₄ → CuC₂O₄ + 2 NaNO₃
- Ni(NO₃)₂ + 2 KOH → Ni(OH)₂ + 2 KNO₃
- Ni(NO₃)₂ + Na₂C₂O₄ → NiC₂O₄ + 2 NaNO₃
- Zn(NO₃)₂ + 2 KOH → Zn(OH)₂ + 2 KNO₃
- Zn(NO₃)₂ + Na₂C₂O₄ → ZnC₂O₄ + 2 NaNO₃
<h3>2</h3>
A double replacement reaction takes place only if it reduces in the concentration of ions in the solution. For example, the reaction between Ca(NO₃)₂ and KOH produces Ca(OH)₂. Ca(OH)₂ barely dissolves. The reaction has removed Ca²⁺ and OH⁻ ions from the solution.
Some of the reactions lead to neither precipitates nor gases. They will not take place since they are not energetically favored.
<h3>3</h3>
Compare the first and last row:
Both Ca(NO₃)₂ and Zn(NO₃)₂ react with KOH. However, between the two precipitates formed, Ca(OH)₂ is more soluble than Zn(OH)₂.
As a result, add the same amount of KOH to two Ca(NO₃)₂ and Zn(NO₃)₂ of equal concentration. The solution that end up with more precipitate shall belong to Zn(NO₃)₂.
<h3>4</h3>
Compare the second and third row:
Cu(NO₃)₂ reacts with KI, but Ni(NO₃)₂ does not. Thus, add equal amount of KI to the two unknowns. The solution that forms precipitate shall belong to Cu(NO₃)₂.
Answer:<span> a) the process for the first ionization energy
</span>
<span>Every time you take a electron you're requiring more and more energy. Expelling the first one
will require less energy than expelling the second and the second will
require less than the third, and so on.
When you take the first one, the atom becomes positive and with that the negative forces of the electron will be more attracted to the positive
charge. The more electrons that are lost, the
more positive this ion will become, causing it to be more difficult to separate the
electrons from the atom.
</span>
Answer:
0.3793 M
Explanation:
The unknown metal is zinc. So the equation of the reaction is;
Zn(s) + Cu^2+(aq) -------> Zn^2+(aq) + Cu(s)
From Nernst equation;
E = E° - 0.0592/n log Q
[Cu2+] = 0.050179 M
n = 2
[Zn^2+] = ?
E = 1.074 V
E° = 0.34 - (-0.76) = 1.1 V
Substituting values;
1.074 = 1.1 - 0.0592/2 log [Zn^2+]/0.050179
1.074 - 1.1 = - 0.0592/2 log [Zn^2+]/0.050179
-0.026 = -0.0296 log [Zn^2+]/0.050179
-0.026/-0.0296 = log [Zn^2+]/0.050179
0.8784 =log [Zn^2+]/0.050179
Antilog(0.8784) = [Zn^2+]/0.050179
7.558 = [Zn^2+]/0.050179
[Zn^2+] = 7.558 * 0.050179
[Zn^2+] = 0.3793 M
