Answer:
E° = 2.46 V
Explanation:
We have a voltaic cell of aluminum and silver in their respective solutions. To know where will take place the reduction (cathode) and where will take place the oxidation (anode), we need to compare the standard reduction potentials.
E°(Al³⁺/Al) = -1.66 V
E°(Ag⁺/Ag) = +0.80 V
Since Ag⁺/Ag has the highest reduction potential that is where reduction will take place while oxidation will take place in the Al electrode.
Reduction (cathode): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
Oxidation (anode): Al(s) ⇒ Al³⁺(aq) + 3 e⁻
The standard potential of the cell (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.
E° = E°red,cathode - E°red,anode = 0.80 V - (-1.66V) = 2.46 V
Answer: 0.4mL of the stock solution is needed
Explanation:
This is simply a dilution analysis. From the question, the following were obtained:
C1 = 25 mg/mL = 25x10^-3g/mL
V1 =?
C2 = 25 ug/mL = 25x10^-6g/mL
V2 = 400ml
Applying the dilution formula, we have:
C1V1 = C2V2
25x10^-3 x V1 = 25x10^-6 x 400
Divide both side by 25x10^-3, we have:
V1 = (25x10^-6 x 400) / 25x10^-3
V1 = 0.4mL
Therefore, 0.4mL of the stock solution need.
Willow is right because the model of atom is not weak. The model of the atom has not changed. It’s just due to new discoveries due to which atom’s model is changing. The particles were always present in the atom but we were not able to discover it.
Answer:
Mine is doing good since I dont have any school today it makes it better. Im listening to scary stories on YT right now which always makes my day better so Im having a really good day today!
What about you? Why isn't your day doing so great?
Answer:
To answer this question we need the illustration
Explanation:
You have left the question incomplete since we do not have the illustration making this impossible to answer
