When E° cell is an electrochemical cell which comprises of two half cells.
So,
when we have the balanced equation of this half cell :
Al3+(aq) + 3e- → Al(s) and E°1 = -1.66 V
and we have also this balanced equation of this half cell :
Ag+(aq) + e- → Ag(s) and E°2 = 0.8 V
so, we can get E° in Al(s) + 3Ag (aq) → Al3+(aq) + 3Ag(s)
when E° = E°2 - E°1
∴E° =0.8 - (-1.66)
= 2.46 V
∴ the correct answer is 2.46 V
Answer:
I think that it it correctly balanced that is my opinion and, because the way it is set up, that the answer will tell you weather or not it is correctly balanced or not.
Answer:
The amount of energy required to break the ionic bonds in CaF2.
Answer:
A
-1440J
Explanation:
Hello,
This question requires us to calculate the work done on a object to move it from point A to point B
Data
Mass = 60kg
Initial velocity (V1) = 8.0m/s
Final velocity (V2) = 4.0m/s
Workdone on an object is equal to force applied on the object to move it through a particular distance.
Work done = force × distance
Force (F) = mass × acceleration
Distance = s
F = Ma
Work done = M× a × s
But a = velocity (v) / time (t)
Work done = mvs / t
But velocity = distance/ time
Work done = mv × v/
Work done = mv²
Work done = ½mv²
Workdone = ½M(V2² - V1²)
Workdone = ½ × 60 (4² - 8²)
Work done = 30 × (16 - 64)
Workdone = 30 × (-48)
Work done = -1440J
Work done = -1.44kJ
The workdone on the object is equal to -1.44kJ
Answer : It take time for the concentration to become 0.180 mol/L will be, 277.8 s
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 
t = time taken = ?
[A] = concentration of substance after time 't' = 0.180 mol/L
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 0.360 mol/L
Putting values in above equation, we get:
Hence, it take time for the concentration to become 0.180 mol/L will be, 277.8 s
