Answer: 0.52V
Explanation:
Ecell = Ecell(standard) - [(0.0592 logQ)/n]
Q = product of the quotient
n = no of electrons transferred = 2
Ecell = 0.63 - [(0.0592*Log(1 / 2.0 * 10-4) / 2]
Ecell = 0.63 - 0.0194
Ecell = 0.5205V
Answer:
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Explanation:
<u>Step 1: </u>Data given
The solution contains 0.036 M Cu2+ and 0.044 M Fe2+
Ksp (CuS) = 1.3 × 10-36
Ksp (FeS) = 6.3 × 10-18
Step 2: Calculate precipitate
CuS → Cu^2+ + S^2- Ksp= 1.3*10^-36
FeS → Fe^2+ + S^2- Ksp= 6.3*10^-18
Calculate the minimum of amount needed to form precipitates:
Q=Ksp
<u>For copper</u> we have: Ksp=[Cu2+]*[S2-]
Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]
[S2-]= 3.61*10^-35 M
<u>For Iron</u> we have: Ksp=[Fe2+]*[S2-]
Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]
[S2-]= 1.43*10^-16 M
CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Answer:
1.heat a pan of water with just a little bit of water,have a boil
2.chosse ure salt
3.stir in has much salt has u can than take the pan off the heat
4.pour the mix into a glass jar
5.tie a string to an objeet that can lay accross the top and put just the string in ure mix
Explanation oh and look at it everyday hope that helps
is the Calcium
because the calcium in group two which group two has 2 electron and also calcium and Potassium 4s sub level group potassium has 1 electron. :)