Answer:K subscript e q equals StartFraction StartBracket upper C upper O subscript 2 EndBracket StartBracket upper C a upper O EndBracket over StartBracket upper C a upper C upper O subscript 3 EndBracket EndFraction
Explanation: the answer has it's root in Law of mass action which states that; the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants raised to their respective stoichiometric coefficients.
Explanation:
There are several ways to define acids and bases, but pH and pOH refer to hydrogen ion concentration and hydroxide ion concentration, respectively. The "p" in pH and pOH stands for "negative logarithm of" and is used to make it easier to work with extremely large or small values. pH and pOH are only meaningful when applied to aqueous (water-based) solutions. When water dissociates it yields a hydrogen ion and a hydroxide.
Answer:
The pressure is higher than the atmospheric one, therefore the temperature is less.
Explanation:
When it is closed permanently, the pressure of the pot inside it increases, generating that the atoms and particles of the water are closer together, increasing their kinetic energy, if intermolecular friction and therefore the boiling point is lower, because the water reaches a boil or boil at a lower temperature.
Answer:
2H⁺ + NO₃⁻ + 1e⁻ → NO₂ + H₂O
Explanation:
NO₃⁻ → NO₂
In left side, Nitrogen acts with +5 by oxidation number
In right side, the oxidation number is +4
This is a reduction reaction, because the oxidation number has decreased. So the N has gained electrons.
NO₃⁻ + 1e⁻ → NO₂
In acidic medium, we have to add water, where there are less oxygens to ballance the amount. We have 2 O in left side, and 3 O in right side, so we have to add 1 H₂O on left side.
NO₃⁻ + 1e⁻ → NO₂ + H₂O
Now that oxygens are ballanced, we have to ballance the hydrogens by adding protons in the opposite side
2H⁺ + NO₃⁻ + 1e⁻ → NO₂ + H₂O
<u>Answer:</u> The value of equilibrium constant is 0.997
<u>Explanation:</u>
We are given:
Percent degree of dissociation = 39 %
Degree of dissociation,
= 0.39
Concentration of
, c = 
The given chemical equation follows:

<u>Initial:</u> c -
<u>At Eqllm:</u>

So, equilibrium concentration of ![N_2O_4=c-c\alpha =[1-(1\times 0.39)]=0.61M](https://tex.z-dn.net/?f=N_2O_4%3Dc-c%5Calpha%20%3D%5B1-%281%5Ctimes%200.39%29%5D%3D0.61M)
Equilibrium concentration of ![NO_2=2c\alpha =[2\times 1\times 0.39]=0.78M](https://tex.z-dn.net/?f=NO_2%3D2c%5Calpha%20%3D%5B2%5Ctimes%201%5Ctimes%200.39%5D%3D0.78M)
The expression of
for above equation follows:
![K_{c}=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
Putting values in above equation, we get:

Hence, the value of equilibrium constant is 0.997