Answer: The concentrations of 
at equilibrium is 0.023 M
Explanation:
Moles of = 
Volume of solution = 1 L
Initial concentration of = 
The given balanced equilibrium reaction is,
Initial conc. 0.14 M 0 M 0M
At eqm. conc. (0.14-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO%5D%5Ctimes%20%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Now put all the given values in this expression, we get :
By solving the term 'x', we get :
x = 0.023 M
Thus, the concentrations of at equilibrium is 0.023 M
Answer: There is one way to write it but i’ll also provide an unbalanced equation and a balanced one.
Explanation:
Unbalanced : Ba (aq) + Cl2 (aq)—-> BaCl (aq)
Balanced : 2Ba (aq) + Cl2 (aq)—> 2BaCl(aq)
Answer:
648.5 mL
Explanation:
Here we will assume that the pressure of the gas is constant, since it is not given or specified.
Therefore, we can use Charle's law, which states that:
"For an ideal gas kept at constant pressure, the volume of the gas is proportional to its absolute temperature"
Mathematically:
where
V is the volume of the gas
T is its absolute temperature
The equation can be rewritten as
where in this problem we have:
is the initial volume of the gas
is the initial temperature
is the final temperature
Solving for V2, we find the final volume of the gas:
Missing question:
Chemical reaction: H₂ <span>+ 2ICl → 2HCl + I</span>₂.
t₁ = 5 s.
t₂ = 15 s.
c₁ = 1,11 M = 1,11 mol/L.
c₂ = 1,83 mol/L.
rate of formation = Δc ÷ Δt.
rate of formation = (c₂ - c₁) ÷ (t₂ - t₁).
rate of formation = (1,83 mol/L - 1,11 mol/L) ÷ (15 s - 5 s).
rate of formation = 0,72 mol/L ÷ 10 s.
rate of formation = 0,072 mol/L·s.
