Answer : The equilibrium concentration of CO in the reaction is, 
Explanation :
The given chemical reaction is:
The expression for equilibrium constant is:
![K_c=\frac{[COCl_2]}{[CO][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCOCl_2%5D%7D%7B%5BCO%5D%5BCl_2%5D%7D)
As we are given:
Concentration of 
at equilibrium = Concentration of 
So,
![K_c=\frac{[Cl_2]}{[CO][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCl_2%5D%7D%7B%5BCO%5D%5BCl_2%5D%7D)
![K_c=\frac{1}{[CO]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B1%7D%7B%5BCO%5D%7D)
![1.2\times 10^3=\frac{1}{[CO]}](https://tex.z-dn.net/?f=1.2%5Ctimes%2010%5E3%3D%5Cfrac%7B1%7D%7B%5BCO%5D%7D)
![[CO]=8.3\times 10^{-4}M](https://tex.z-dn.net/?f=%5BCO%5D%3D8.3%5Ctimes%2010%5E%7B-4%7DM)
Therefore, the equilibrium concentration of CO in the reaction is,
Answer:
28.0mL of the 0.0500M NaOH solution
Explanation:
<em>0.126g of lactic acid diluted to 250mL. Titrated with 0.0500M NaOH solution.</em>
<em />
The reaction of lactic acid, H₃C-CH(OH)-COOH (Molar mass: 90.08g/mol) with NaOH is:
H₃C-CH(OH)-COOH + NaOH → H₃C-CH(OH)-COO⁻ + Na⁺ + H₂O
<em>Where 1 mole of the acid reacts per mole of the base.</em>
<em />
You must know the student will reach equivalence point when moles of lactic acid = moles NaOH.
the student will titrate the 0.126g of H₃C-CH(OH)-COOH. In moles (Using molar mass) are:
0.126g ₓ (1mol / 90.08g) = <em>1.40x10⁻³ moles of H₃C-CH(OH)-COOH</em>
To reach equivalence point, the student must add 1.40x10⁻³ moles of NaOH. These moles comes from:
1.40x10⁻³ moles of NaOH ₓ (1L / 0.0500moles NaOH) = 0.0280L of the 0.0500M NaOH =
<h3>28.0mL of the 0.0500M NaOH solution</h3>
A because it’s more reasonable and reliable plus it the right answer
Answer:
Heterogeneous mixture...is the answer
