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
<h2>Answer:</h2>
The correct answer is option C which is, "Electrons in the orbit closest to the nucleus have the least amount of energy".
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Explanation:</h3>
- There are different orbitals around the nucleus on which the electrons moves around the nucleus.
- These orbitals have a specific energy, due to which they are known as energy levels.
- The energy level near to the nucleus has least amount of the energy and the energy of the orbitals increase as the distance of the orbitals increase to the nucleus.
Answer:
Explanation:
Hello,
In this case, we apply the Gay-Lussac's law which allows us to understand the pressure-temperature behavior as a directly proportional relationship:
Thus, we solve for the final pressure P2 to obtain it as shown below:
Hence, we notice that the temperature doubles as well as the pressure.
Best regards.
Answer:
The half-life varies depending on the isotope.
Half-lives range from fractions of a second to billions of years.
The half-life of a particular isotope is constant.
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