Answer:
E° = 0.65 V
Explanation:
Let's consider the following reductions and their respective standard reduction potentials.
Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:
Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V
The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V
First you would multiply 5,3, and 2 to get a volume of 30. Density is mass over volume so you would then divide 129 by 30. You would get 4.3. The answer would be 4.3 g/cm^3
The experiment would contain 70% U-235 and 30% U-238. This is because some of the nuetrons need to be absorbed, but more of them must cause the fission Too much U-238, however, would cause to <span>many netrons to be absorbed and stall the reaction process</span>