Answer:
Freezing point is -2.81°C
Explanation:
34g/342gmol^-1 = 0.0994mol
n = m/mr
Molarity= 0.994/ 0.66 = 1.51M
◇T = -i × m ×Kf
Where ◇T is freezing depression
i= Vant Hoff factor
m = molarity
Kf = freezing content = 1.
860kgmol^-1
◇T =-1 × 1.51 × 1.860 = - 2.81°C
Answer: The equilibrium constant for the given reaction is 0.0421.
Explanation:

Concentration of
= 0.0095 M
Concentration of
= 0.020 M
Concentration of
= 0.020 M
The expression of the equilibrium constant is given as:
![K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}=\frac{0.020 M\times 0.020 M}{0.0095 M}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D%3D%5Cfrac%7B0.020%20M%5Ctimes%200.020%20M%7D%7B0.0095%20M%7D)
(An equilibrium constant is an unit less constant)
The equilibrium constant for the given reaction is 0.0421.
Answer:
Scandium(III) fluoride, ScF3, is an ionic compound. It is slightly soluble in water but dissolves in the presence of excess fluoride to form the ScF63− anion.
hope it will help you......
Answer is: this is an example of an Arrhenius acid.
An Arrhenius acid is a
substance that dissociates in water to form hydrogen ions or protons (H⁺).
For example hydrochloric acid: HCl(aq) → H⁺(aq) + Cl⁻(aq).
An Arrhenius base is a
substance that dissociates in water to form hydroxide ions (OH⁻<span>).
In this example lithium hydroxide is an Arrhenius base:</span>
LiOH(aq) → Li⁺(aq) + OH⁻(aq).