Answer:
The correct answer is option a.
Explanation:
Equilibrium concentration cadmium ions = ![[Cd^{2+}]=0.0585 M](https://tex.z-dn.net/?f=%5BCd%5E%7B2%2B%7D%5D%3D0.0585%20M)
Equilibrium concentration fluoride ions = ![[F^{-}]=0.117 M](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%3D0.117%20M)
Molar solubility is the maximum concentration of salt present in water in ionic form beyond that no more salt will exist in its ionic form and will settle down in bottom of the solution.
The molar solubility of the solid cadmium fluoride = 0.0585 M
..[1]
Due to addition of sodium fluoride will increase concentration of fluoride in the solution.And due to common ion effect the equilibrium will shift in backward direction in [1], that is precipitation of more cadmium fluoride.
Hence, decrease in solubility will be observed.
Its C, Outermost Electrons.
x=1
Simplify both sides of the equation and then isolate the variable
Answer:
337.22 K
Explanation:
Given that:
P₁ = 1 atm
T₁ = 350 K
P₂ = 0.639 atm
T₂ = ??? (unknown)
R(rate constant) = 8.34 J k⁻¹ mol⁻¹
Using Clausius-Clapeyron equation, we can determine the final boiling point of the process.
Clausius-Clapeyron equation can be written as:
![In\frac{P_2}{P_1}=\frac{\delta H_{vap}}{R}[\frac{T_2-T_1}{T_2T_1}]](https://tex.z-dn.net/?f=In%5Cfrac%7BP_2%7D%7BP_1%7D%3D%5Cfrac%7B%5Cdelta%20H_%7Bvap%7D%7D%7BR%7D%5B%5Cfrac%7BT_2-T_1%7D%7BT_2T_1%7D%5D)
Substituting our values given; we have:
![In\frac{0.639}{1}=(\frac{34.4*10^3J/mol}{8.314 J K^{-1}mol^{-1}})[\frac{T_2-350}{350T_2}]](https://tex.z-dn.net/?f=In%5Cfrac%7B0.639%7D%7B1%7D%3D%28%5Cfrac%7B34.4%2A10%5E3J%2Fmol%7D%7B8.314%20J%20K%5E%7B-1%7Dmol%5E%7B-1%7D%7D%29%5B%5Cfrac%7BT_2-350%7D%7B350T_2%7D%5D)
![In({0.639})=(\frac{34.4*10^3}{8.314K^{-1}})[\frac{T_2-350}{350T_2}]](https://tex.z-dn.net/?f=In%28%7B0.639%7D%29%3D%28%5Cfrac%7B34.4%2A10%5E3%7D%7B8.314K%5E%7B-1%7D%7D%29%5B%5Cfrac%7BT_2-350%7D%7B350T_2%7D%5D)
![- 0.4479 = 41317.599 [\frac{T_2-350}{350T_2} ]K](https://tex.z-dn.net/?f=-%200.4479%20%3D%2041317.599%20%5B%5Cfrac%7BT_2-350%7D%7B350T_2%7D%20%5DK)
![-\frac{0.4479}{4137.599} = [\frac{T_2-350}{350T_2} ]](https://tex.z-dn.net/?f=-%5Cfrac%7B0.4479%7D%7B4137.599%7D%20%3D%20%5B%5Cfrac%7BT_2-350%7D%7B350T_2%7D%20%5D)
![- 1.0825118*10^{-4} = [\frac{T_2-350}{350T_2} ]](https://tex.z-dn.net/?f=-%201.0825118%2A10%5E%7B-4%7D%20%3D%20%5B%5Cfrac%7BT_2-350%7D%7B350T_2%7D%20%5D)
∴ the boiling point of CH3COOC2H5 when the external pressure is 0.639 atm is <u>337.22</u> K.
This problem is asking for the dissolution reaction of barium fluoride, both the equilibrium and Ksp expressions in terms of concentrations and x and its molar solubility in water. Thus, answers shown below:
<h3>Solubility product</h3>
In chemistry, when a solid is dissolved in water, one must take into account the fact that not necessarily its 100 % will be able to break into ions and thus undergo dissolution.
In such a way, and specially for sparingly soluble solids, one ought to write the dissolution reaction at equilibrium as shown below for the given barium fluoride:
Next, we can write its equilibrium expression according to the law of mass action, which also demands us to omit any solid and refer it to the solubility product constant (Ksp):
![Ksp=[Ba^{2+}][F^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BBa%5E%7B2%2B%7D%5D%5BF%5E-%5D%5E2)
Afterwards, one can insert the reaction extent, x, as it stands for the molar solubility of this solid in water, taking into account the coefficients balancing the reaction:
Finally, we solve for the x as the molar solubility of barium fluoride as shown below:
![2.5x10^{-5}=(x)(2x)^2\\\\2.5x10^{-5}=4x^3\\\\x=\sqrt[3]{\frac{2.5x10^{-5}}{4} } \\\\x=0.0184M](https://tex.z-dn.net/?f=2.5x10%5E%7B-5%7D%3D%28x%29%282x%29%5E2%5C%5C%5C%5C2.5x10%5E%7B-5%7D%3D4x%5E3%5C%5C%5C%5Cx%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B2.5x10%5E%7B-5%7D%7D%7B4%7D%20%7D%20%5C%5C%5C%5Cx%3D0.0184M)
Learn more about chemical equilibrium: brainly.com/question/26453983
