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here is a Quizlet I have for that
Answer:
El potencial celular estándar,
is +0.46 V
Explanation:
Las reacciones de media célula son;
Media reacción del ánodo Cu²⁺ + 2e⁻ ↔ Cu, E ° = 0.34 V
Media reacción catódica 2Ag + 2e⁻ ⁻ 2Ag, E ° = 0.80 V
Sin embargo tenemos para hierro Fe²⁺ + 2e⁻ ↔ Fe, E ° -0.44 V
y Fe³⁺ + e⁻ ↔ Fe²⁺, E ° = 0.77 V
que es más alta que la del cobre presente, por lo tanto, el cobre se oxidará en el ánodo
Por lo tanto, en el ánodo, tendremos
Cu → Cu²⁺ + 2e⁻ (E ° = -0.34 V)
En el cátodo
2Ag + 2e⁻ → 2Ag (E ° = 0.80 V)
El potencial celular estándar,
= +0.46 V
Answer:
5.64×10²³ atoms C
Explanation:
Convert moles of H to moles of C:
2.81 mol H × (2 mol C / 6 mol H) = 0.937 mol C
Convert moles of C to atoms of C:
0.937 mol C × (6.02×10²³ atoms C / mol C) = 5.64×10²³ atoms C
Answer:
A reaction is non-spontaneous at any temperature when the Gibbs free energy > 0.
Explanation:
There is a state function, that determines if a reaction is sponaneous or non spontaneous:
ΔG = Gibbs free energy
A reaction is non spontaneous when it does require energy to produce that reaction. It will be spontaneous, when the reaction does not require energy to be occured.
The formula is: ΔG = ΔH - T.ΔS
ΔH → Enthalpy → Energy gained or realeased as heat.
ΔH < 0 → <em>Exothermic reaction. Spontaneity is favored
</em>
T → Temperature
ΔS → Entropy → Degree of disorder of a system.
When the system has a considered disorder ΔS > 0, disorder increases.
When the system is more ordered, ΔS < 0, disorder decreases.
The reaction will be non spontaneous if, the enthalpy is positive (endothermic reaction) and the ΔS < 0 (disorder decreases). It will not occur if we do not give energy.
ΔG < 0 → Spontaneous reaction
ΔG > 0 → Non spontaneous reaction
ΔG = 0 → System in equilibrium
Answer:
The bombarding particle is a Proton
Explanation:
A Nuclear transmutation reaction occurs when radioactive element decay, usually converting them from one element/isotope into another element. Transmutation is the process which causes decay, generally, alpha or beta.
¹⁶₈O(P,alpha) ¹³₇N, can be written as
¹⁶₈O + x goes to ¹³₇N + ⁴₂He
Where x can be anything, balancing the equation in order to give us the correct amount of proton number and nucleus number
16 + x = 13 + 4
x = 17 – 16 = 1, Hence we can say that x = ¹₁P
<u>¹⁶₈O + ¹₁P goes to ¹³₇N + ⁴₂He</u>
Here we can clearly see the bombarding particle is ¹₁P (proton). The ejected particle being ⁴₂He which is also known as an alpha particle