My answer:
13 group of the periodic table represented by boron, aluminum and gallium subgroup. It includes gallium, indium, thallium. Typical steper oxidation in the subset gallium 3 is explained by the presence of (n-1)d^10 E-configuration.
Aluminium oxidation degree has +3 an electronic configuration of noble gases S^2P^6
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Answer:
The standard change in free energy for the reaction = - 437.5 kj/mole
Explanation:
The standard change in free energy for the reaction:
4 KClO₃ (s) → 3 KClO₄(s) + KCl(s)
Given that ΔGf(KClO3(s)) = -290.9 kJ/mol;
ΔGf(KClO4(s)) = -300.4 kJ/mol;
ΔGf(KCl(s)) = -409 kJ/mol
According to Hess's law
ΔGr (Free energy change of reaction)= ∑(Product free energy - reactant free energy)
⇒ ΔGr⁰ = {3 x (-300.4) + (-409)} - {3 x (- 290.9)}
= - 901.2 - 409 + 872.7
= - 437.5 kj/mole
Answer:
-250.3kJ
Explanation:
Based in the reactions and using -<em>Hess's law-</em>:
(1) P₄(s) + 6 Cl₂(g) → 4PCl₃(g) ΔH₁ = -4439kJ
(2) 4PCl₅(g) → P₄(s) + 10Cl₂ ΔH₂ = 3438kJ
The sum of (1) + (2) is:
4PCl₅(g) → 4PCl₃(g) + 4 Cl₂ ΔH = -4439kJ + 3438kJ = -1001kJ
Dividing this reaction in 4:
PCl₅(g) → PCl₃(g) + Cl₂ ΔH = -1001kJ / 4 = <em>-250.3kJ</em>
The concentration of [H3O⁺]=2.86 x 10⁻⁶ M
<h3>Further explanation</h3>
In general, the weak acid ionization reaction
HA (aq) ---> H⁺ (aq) + A⁻ (aq)
Ka's value
![\large {\boxed {\bold {Ka \: = \: \frac {[H ^ +] [A ^ -]} {[HA]}}}}](https://tex.z-dn.net/?f=%5Clarge%20%7B%5Cboxed%20%7B%5Cbold%20%7BKa%20%5C%3A%20%3D%20%5C%3A%20%5Cfrac%20%7B%5BH%20%5E%20%2B%5D%20%5BA%20%5E%20-%5D%7D%20%7B%5BHA%5D%7D%7D%7D%7D)
Reaction
HC₂H₃O₂ (aq) + H₂O (l) ⇔ (aq) + H₃O⁺ (aq) Ka = 1.8 x 10⁻⁵
![\tt Ka=\dfrac{[C_2H_3O^{2-}[H_3O^+]]}{[HC_2H_3O_2]}}\\\\1.8\times 10^{-5}=\dfrac{0.22\times [H_3O^+]}{0.035}](https://tex.z-dn.net/?f=%5Ctt%20Ka%3D%5Cdfrac%7B%5BC_2H_3O%5E%7B2-%7D%5BH_3O%5E%2B%5D%5D%7D%7B%5BHC_2H_3O_2%5D%7D%7D%5C%5C%5C%5C1.8%5Ctimes%2010%5E%7B-5%7D%3D%5Cdfrac%7B0.22%5Ctimes%20%5BH_3O%5E%2B%5D%7D%7B0.035%7D)
[H₃O⁺]=2.86 x 10⁻⁶ M
