Answer:
Half-reactions:
Cr³⁺ + 1e⁻ → Cr²⁺; Zn → Zn²⁺ + 2e⁻
Net ionic equation:
2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺
Explanation:
The Cr³⁺ is reduced to Cr²⁺:
<h3>
Cr³⁺ + 1e⁻ → Cr²⁺ -Half-reaction 1-</h3>
Zn is oxidized to Zn²⁺:
<h3>
Zn → Zn²⁺ + 2e⁻ -Half-reaction 2-</h3>
Twice the reduction of Cr:
2Cr³⁺ + 2e⁻ → 2Cr²⁺
Now this reaction + Oxidation of Zn:
2Cr³⁺ + 2e⁻ + Zn → 2Cr²⁺ + Zn²⁺ + 2e⁻
<h3>2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺ - Net ionic equation</h3>
Answer:
True
Explanation:
The sodium atom does become a sodium ion with a charge of +1.
Iodine electron configuration is:
1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 5S^2 4d^10 5P^5
when Krypton is the noble gas in the row above iodine in the periodic table,
we can change 1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 by the symbol
[Kr] of Krypton.
So we can write the electron configuration of Iodine:
[Kr] 5S^2 4d^10 5P^5
Answer : The concentration of ![[Fe(SCN)]^{2+}](https://tex.z-dn.net/?f=%5BFe%28SCN%29%5D%5E%7B2%2B%7D)
is, 
Explanation :
When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is 
and 
is excess reagent.
First we have to calculate the moles of KSCN.
Moles of KSCN = Moles of 
= Moles of = 
Now we have to calculate the concentration of
![\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D)
Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L
![\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%3D%5Cfrac%7B1.08%5Ctimes%2010%5E%7B-5%7Dmol%7D%7B0.025L%7D%3D4.32%5Ctimes%2010%5E%7B-4%7DM)
Thus, the concentration of is,
