Answer:
Explanation:
1)<u><em> Ionization equilibrium equation: given</em></u>
- H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)
2) <em><u>Ionization equilibrium constant, at 25°C, Kw: given</u></em>
<u>3) Stoichiometric mole ratio:</u>
As from the ionization equilibrium equation, as from the fact it is stated, the concentration of both ions, at 25°C, are equal:
- [H₃O⁺(aq)] = [OH⁻(aq)] = 1.0 × 10⁻⁷ M
- ⇒ Kw = [H3O⁺] [OH⁻] = 1.0 × 10⁻⁷ × 1.0 × 10⁻⁷ = 1.0 × 10⁻¹⁴ M
<u><em>4) A solution has a [OH⁻] = 3.4 × 10⁻⁵ M at 25 °C </em></u><em><u>and you need to calculate what the [H₃O⁺(aq)] is.</u></em>
Since the temperature is 25°, yet the value of Kw is the same, andy you can use these conditions:
Then you can substitute the known values and solve for the unknown:
- 1.0 × 10⁻¹⁴ M² = [H₃O⁺] × 3.4 × 10⁻⁵ M
- ⇒ [H₃O⁺] = 1.0 × 10⁻¹⁴ M² / ( 3.4 × 10⁻⁵ M ) = 2.9⁻¹⁰ M
As you see, the increase in the molar concentration of the ion [OH⁻] has caused the decrease in the molar concentration of the ion [H₃O⁺], to keep the equilibrium law valid.
Answer:
C.) That all matter was composed of earth, fire, water and air
Explanation:
Just took the test
Answer:

Explanation:
Hello,
In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

Best regards.