Answer : The concentration of
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 ![[Fe(SCN)]^{2+}](https://tex.z-dn.net/?f=%5BFe%28SCN%29%5D%5E%7B2%2B%7D)
![\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, 
Answer is: <span>concentration of NOCl is 3.52 M.
</span>
Balanced chemical reaction: 2NOCl(g) ⇄ 2NO(g) + Cl₂<span>(g).
Kc = 8.0.
</span>[NOCl] = 1.00 M; equilibrium concentration.
[NO] = x.
[Cl₂] = x/2; equilibrium concentration of chlorine.<span>
Kc = </span>[Cl₂] ·[NO]² / [NOCl].
8.00 = x/2 · x² / 1.
x³/2 = 8.
x = ∛16.
x = 2.52 M.
co(NOCl) = [NOCl] + x.
co(NOCl) = 1.00 M + 2.52 M.
co(NOCl) = 3.52 M; the initial concentration of NOCl.
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Answer:
Acid are those substances which release H + ions when dissolved in water.
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