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
Answer:Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). If hydrogen gas is added to this system at equilibrium, will the reaction shift towards reactants
Answer:
C. The lowest-energy electron configuration of an atom has the maximum number of unpaired electrons, all of which have the same spin, in degenerate orbitals.
Explanation:
The Hund's rule is used to place the electrons in the orbitals is it states that:
1. Every orbital in a sublevel is singly occupied before any orbital is doubly occupied;
2. All of the electrons in singly occupied orbitals have the same spin.
So, the electrons first seek to fill the orbitals with the same energy (degenerate orbitals) before paring with electrons in a half-filled orbital. Orbitals doubly occupied have greater energy, so the lowest-energy electron configuration of an atom has the maximum number of unpaired electrons, and for the second statement, they have the same spin.
The other alternatives are correct, but they're not observed by the Hund's rule.
<span>4 Al + 3 O2 → 2 Al2O3
(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2
0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.
((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =
10.1 g O2 left over</span><span>
</span>
