Answer : The concentration of
at equilibrium is 0 M.
Solution : Given,
Concentration of
= 0.0200 M
Concentration of
= 1.00 M
The given equilibrium reaction is,
![Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)](https://tex.z-dn.net/?f=Fe%5E%7B3%2B%7D%28aq%29%2B3C_2O_4%5E%7B2-%7D%28aq%29%5Crightleftharpoons%20%5BFe%28C_2O_4%29_3%5D%5E%7B3-%7D%28aq%29)
Initially conc. 0.02 1.00 0
At eqm. (0.02-x) (1.00-3x) x
The expression of
will be,
![K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5B%5BFe%28C_2O_4%29_3%5D%5E%7B3-%7D%5D%7D%7B%5BC_2O_4%5E%7B2-%7D%5D%5E3%5BFe%5E%7B3%2B%7D%5D%7D)

By solving the term, we get:

Concentration of
at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M
Therefore, the concentration of
at equilibrium is 0 M.
Answer:
A
Explanation:
Hope it helps
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Answer:
Cyclobutane is a cycloalkane and organic compound with the formula (CH2)4.
Explanation:
The enthalpy change : -196.2 kJ/mol
<h3>Further explanation </h3>
The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The value of ° H ° can be calculated from the change in enthalpy of standard formation:
∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
Reaction
2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)
∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂
