Ecell = E°cell - RT/vF * lnQ
R is the gas constant: 8.3145 J/Kmol
T is the temperature in kelvin: 273.15K = 0°C, 25°C = 298.15K
v is the amount of electrons, which in your example seems to be six (I'm not totally sure)
F is the Faradays constant: 96485 J/Vmol (not sure about the mol)
Q is the concentration of products divided by the concentration of reactants, in which we ignore pure solids and liquids: [Mg2+]³ / [Fe3+]²
Standard conditions is 1 mol, at 298.15K and 1 atm
To find E°cell, you have to look up the reduction potensials of Fe3+ and Mg2+, and solve like this:
E°cell = cathode - anode
Cathode is where the reduction happens, so that would be the element that recieves electrons. Anode is where the oxidation happens, so that would be the element that donates electrons. In your example Fe3+ recieves electrons, and should be considered as cathode in the equation above.
When you have found E°cell, you can just solve with the numbers I gave you.
1 mol of anything has 6.02 * 10^23 items.
4 mol has 4 times as much
C: Answer <<<<< =========
Write balanced molecular, ionic, and net ionic equations<span> (NIE) for each of the ... </span>3.Mg(OH)2(aq) +. 2HCl(aq). →. MgCl2(aq). +. 2H2O(l). Ionic Equation: Mg2+(aq). + ... 2-(aq). +. Ca2+(aq). →. Ca(C2O4)(s). 5. 2(NH4)3PO4(aq) +. 3Zn(NO3)2(aq) → ... 2-(aq<span>). →. MgCrO4(s). 9. </span>2FeCl3(aq<span>). +. </span>3Mg<span>(s). →. </span>3MgCl2(aq<span>) +. </span>2Fe<span>(s).</span>
Answer:
hey just where are the rections keep a pic
Explanation:
okay?.........................................
Answer:
Explanation:
Given that:
Half life = 
days
Where, k is rate constant
So,
The rate constant, k = 0.00007 days⁻¹
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given:
5 % is lost which means that 0.05 of is decomposed. So,
![\frac {[A_t]}{[A_0]}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D)
= 1 - 0.05 = 0.95
t = ?
![\frac {[A_t]}{[A_0]}=e^{-k\times t}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%3De%5E%7B-k%5Ctimes%20t%7D)
t = 732.76 days =
