Answer:
8.7 L
Explanation:
T2(V1/T1) = V2
417.15 K(6.2 L/296.45 K) = 8.7 L
Remember to almost always change celcius to kelvin. Also, this is part of Charle's Law (temp and volume are proportional, so if temp increaces so must the volume or vice versa). Lastly, Charle's Law has the formula of V1/T1 = V2/T2. I just rearranged it to go along with your problem. Hence, the T2(V1/T1) = V2
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
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Equilibrium expression is ![Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BH3O%2B%5D%5BHCO3%5E-%5D%7D%7B%5BH2CO3%5D%7D%5C%5C)
<u>Explanation:</u>
Equilibrium expression is denoted by Keq.
Keq is the equilibrium constant that is defined as the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.
Example -
aA + bB = cC + dD
So, Keq = conc of product/ conc of reactant
![Keq = \frac{[C]^c [D]^d}{[A]^a [B]^b}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BC%5D%5Ec%20%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%20%5BB%5D%5Eb%7D)
So from the equation, H₂CO₃+H₂O = H₃O+HCO₃⁻¹
![Keq = \frac{[H3O^+]^1 [HCO3^-]^1}{[H2CO3]^1 [H2O]^1}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BH3O%5E%2B%5D%5E1%20%5BHCO3%5E-%5D%5E1%7D%7B%5BH2CO3%5D%5E1%20%5BH2O%5D%5E1%7D)
The concentration of pure solid and liquid is considered as 1. Therefore, concentration of H2O is 1.
Thus,
![Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BH3O%2B%5D%5BHCO3%5E-%5D%7D%7B%5BH2CO3%5D%7D%5C%5C)
Therefore, Equilibrium expression is ![Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BH3O%2B%5D%5BHCO3%5E-%5D%7D%7B%5BH2CO3%5D%7D%5C%5C)