Interphase is not a phase of mitosis.
Answer :
The equilibrium concentration of CO is, 0.016 M
The equilibrium concentration of Cl₂ is, 0.034 M
The equilibrium concentration of COCl₂ is, 0.139 M
Explanation :
The given chemical reaction is:

Initial conc. 0.1550 0.173 0
At eqm. (0.1550-x) (0.173-x) x
As we are given:

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)
Now put all the given values in this expression, we get:

x = 0.139 and x = 0.193
We are neglecting value of x = 0.193 because equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.139
The equilibrium concentration of CO = (0.1550-x) = (0.1550-0.139) = 0.016 M
The equilibrium concentration of Cl₂ = (0.173-x) = (0.173-0.139) = 0.034 M
The equilibrium concentration of COCl₂ = x = 0.139 M
Answer:
12 g of choloracetic acid
Explanation:
The buffer equilibrium is:
HCH₂ClCO₂ ⇄ CH₂ClCO₂⁻ + H⁺
pka= -log ka =
Ka: 1,3x10⁻³ = [CH₂ClCO₂⁻] [H⁺] / [HCH₂ClCO₂]
By Henderson-Hasselbalch equation:
pH = pka + log₁₀ [A⁻] / [HA]
3,01 = 2,89 + log₁₀ [A⁻] / [HA]
1,318 = [A⁻] / [HA]
As molar concentration of chloroacetic acid (HA) is 0,20M
[A⁻] = 0,26 mol/L
The volume is 500 mL ≡ 0,5 L
0,26mol/L × 0,5 L = 0,13 moles of chloroacetic acid. In grams:
0,13 mol × (94,5g / 1mol) = <em>12 g of choloracetic acid</em>
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I hope it helps!