The pH of a solution at 25. 0 °C that contains 2. 95 × 10^-12 m hydronium ions is 13.5.
<h3>What is pH? </h3>
pH is defined as the concentration of the hydrogen bond which is released or gained by the species in the solution which depicts the acidity and basicity of the solution.
<h3>What is pOH? </h3>
pOH is defined as the concentration of the hydronium ion present in solution.
pOH value is inversely proportional to the value of pH.
pH value increases, pOH value decreases and vice versa.
Given,
Total H+ ions = 2.95 ×10^(-12)M
<h3>Calculation of pH</h3>
pH = -log[H+]
By substituting the value of H+ ion in given equation
= log(2.95× 10^(-12) )
= 13.5
Thus we find that the pH of a solution at 25. 0 °C that contains 2. 95 × 10^-12 m hydronium ions is 13.5.
learn more about pH:
brainly.com/question/12942138
#SPJ4
Answer:
The answer to your question is Argon
Explanation:
Electron configuration given 1s² 2s² 2p⁶ 3s² 3p⁶
To find the element whose electron configuration is given, we can do it by two methods.
Number 1. Sum all the exponents the result will give you the atomic number of the element.
2 + 2 + 6 + 2 + 6 = 18
The element with an atomic number of 18 is Argon.
Number 2. Look at the last terms of the electronic configuration
3s² 3p⁶
Number three indicates that this element is in the third period in the periodic table.
Sum the exponents 2 + 6 = 8
Number 8 indicates that this element is the number 8 of that period without considering the transition elements.
The element with these characteristics is Argon.
It is water soluble so is also soluble in aqueous solutions of NaOH or NaHCO3.
Answer:
0.32 M
Explanation:
Step 1: Write the balanced reaction at equilibrium
Ag₂S(s) ⇌ 2 Ag⁺(aq) + S²⁻(aq)
Step 2: Calculate the concentration of Ag⁺ at equilibrium
We will use the formula for the concentration equilibrium constant (Keq), which is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.
Keq = [Ag⁺]² × [S²⁻]
[Ag⁺] = √{Keq / [S²⁻]}
[Ag⁺] = √{2.4 × 10⁻⁴ / 0.0023} = 0.32 M
Answer:
Water will boil at 
.
Explanation:
According to clausius-clapeyron equation for liquid-vapor equilibrium:
![ln(\frac{P_{2}}{P_{1}})=\frac{-\Delta H_{vap}^{0}}{R}[\frac{1}{T_{2}}-\frac{1}{T_{1}}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_%7B2%7D%7D%7BP_%7B1%7D%7D%29%3D%5Cfrac%7B-%5CDelta%20H_%7Bvap%7D%5E%7B0%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D-%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%5D)
where, 
and 
are vapor pressures of liquid at 
(in kelvin) and 
(in kelvin) temperatures respectively.
Here, = 760.0 mm Hg, = 373 K, = 314.0 mm Hg
Plug-in all the given values in the above equation:
![ln(\frac{314.0}{760.0})=\frac{-40.7\times 10^{3}\frac{J}{mol}}{8.314\frac{J}{mol.K}}\times [\frac{1}{T_{2}}-\frac{1}{373K}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B314.0%7D%7B760.0%7D%29%3D%5Cfrac%7B-40.7%5Ctimes%2010%5E%7B3%7D%5Cfrac%7BJ%7D%7Bmol%7D%7D%7B8.314%5Cfrac%7BJ%7D%7Bmol.K%7D%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D-%5Cfrac%7B1%7D%7B373K%7D%5D)
or, 
So, 
Hence, at base camp, water will boil at
