Answer: Concentration of 
in the equilibrium mixture is 0.31 M
Explanation:
Equilibrium concentration of 
= 0.729 M
The given balanced equilibrium reaction is,
Initial conc. x 0 0
At eqm. conc. (x-2y) M (y) M (3y) M
The expression for equilibrium constant for this reaction will be:
3y = 0.729 M
y = 0.243 M
![K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5By%5D%5Ctimes%20%5B3y%5D%5E3%7D%7B%5Bx-2y%5D%5E2%7D)
Now put all the given values in this expression, we get :
concentration of in the equilibrium mixture = 
Thus concentration of in the equilibrium mixture is 0.31 M
Answer:
Iron is oxidized while chlorine is reduced.
Explanation:
The oxidation reduction reactions are called redox reaction. These reactions are take place by gaining or losing the electrons and oxidation state of elements are changed.
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Consider the following reaction:
2FeCl₂ + Cl₂ → 2FeCl₃
in this reaction the oxidation state of iron is increased from +2 to +3. That's why iron get oxidized and it is reducing agent because it reduced the chlorine. The chlorine is reduced from -2 to -3 and it is oxidizing agent because it oxidized the iron.
2Fe⁺²Cl₂⁻²
2Fe⁺³Cl₃⁻³
The iron atom gives it three electrons to three atoms of chlorine and gain positive charge while chlorine atom accept the electron and form anion.
Answer:
The First choice is correct
Explanation:
That is the closest example of what is shown
The first step to answering this item is to convert the given temperatures in °F to °C through the equation,
°C = (°F - 32)(5/9)
initial temperature: 72°F
°C = (72 - 32)(5/9) = 22.22°C
final temperature: 145°F
°C = (145 - 32)(5/9) = 62.78°C
Substituting to the equation,
H = mcpdT
H = (43 g)(0.903 J/g°C)(62.78 - 22.22)
H = 1574.82 J
<em>Answer: 1574.82 J</em>
