Answer:
Ka = 1.39x10⁻⁶
Explanation:
A monoprotic acid, HX, will be in equilibrium in an aqueous medium such as:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
<em>Where Ka is:</em>
Ka = [H⁺] [X⁻] / [HX]
<em>Where [] is the molar concentration in equilibrium of each specie.
</em>
The equilibrium is reached when some HX reacts producing H+ and X-, that is:
[HX] = 1.64M - X
[H⁺] = X
[X⁻] = X
As pH is 2.82 = -log [H⁺]:
[H⁺] = 1.51x10⁻³M:
[HX] = 1.64M - 1.51x10⁻³M = 1.638M
[H⁺] = 1.51x10⁻³M
[X⁻] = 1.51x10⁻³M
And Ka is:
Ka = [1.51x10⁻³M] [1.51x10⁻³M] / [1.638M]
<h3>Ka = 1.39x10⁻⁶</h3>
Answer:
The change in entropy is -1083.112 joules per kilogram-Kelvin.
Explanation:
If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (
), in joules per gram-Kelvin, by the following model:
(1)
Where:
- Mass, in kilograms.
- Specific heat of water, in joules per kilogram-Kelvin.
, 
- Initial and final temperatures of water, in Kelvin.
If we know that 
, 
, 
and 
, then the change in entropy for the entire process is:
The change in entropy is -1083.112 joules per kilogram-Kelvin.
<span>Definition of Metallurgy-The branch of science and technology concerned with the properties of metals and their production and purification.</span>
The answer u are looking for is b
