Answer:
there are no valence electrons left over, so the molecule has four bond pairs and no lone pairs.
Explanation:
222 grams of calcium chloride is produced.
<h3><u>Explanation</u>:</h3>
The mole concept and the chemical equation are very much closely related with each other. In the chemical reaction, the compounds or elements in both sides are balanced according to the number of atoms of each side of the reaction. So from there we can easily find the amount of reactant reacts to produce desired product.
Here we can see that 2 moles of sodium chloride produces 1 mole of calcium chloride.
So, 4 moles of sodium chloride will produce 2 moles of calcium chloride.
Now, atomic weight of calcium =40.
Atomic weight of chlorine =35.5.
So,the molecular weight of calcium chloride = 
=111.
It means, 1 mole of calcium chloride weighs 111 grams.
So 2 moles of calcium chloride weighs 
grams = 222 grams.
<span>The particles are far apart from each other.</span>
Answer:
Net ionic equation:
Zn²⁺(aq) + 2OH⁻(aq) → Zn(OH)₂(s)
Explanation:
Chemical equation:
ZnCl₂ + KOH → KCl + Zn(OH)₂
Balanced chemical equation:
ZnCl₂ + 2KOH → 2KCl +Zn(OH)₂
Ionic equation;
Zn²⁺(aq) + 2Cl⁻(aq) + 2K⁺(aq) + 2OH⁻(aq) → 2K⁺(aq) + 2Cl⁻(aq) +Zn(OH)₂(s)
Net ionic equation:
Zn²⁺(aq) + 2OH⁻(aq) → Zn(OH)₂(s)
The K⁺ and Cl⁻ are spectator ions that's why these are not written in net ionic equation. The Zn(OH)₂ can not be splitted into ions because it is present in solid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.
The anode is the electrode where the oxidation occurs.
Cathode is the electrode where the reducction occurs.
Equations:
Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V
The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.
Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72
Answer: 2.72 V
