Answer:
Zn(s) → Zn⁺²(aq) + 2e⁻
Explanation:
Let us consider the complete redox reaction:
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
This is a redox reaction because, both oxidation and reduction is simultaneously taking place.
- Oxidation (loss of electrons or increase in the oxidation state of entity)
- Reduction (gain of electrons or decrease in the oxidation state of the entity)
- An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet configuration. An octet configuration is that of outer shell configuration of noble gas.
Here Zn(s) is undergoing oxidation from OS 0 to +2
And H in HCl (aq) is undergoing reduction from OS +1 to 0.
Therefore, for this reaction;
Oxidation Half equation is:
Zn(s) → Zn⁺²(aq) + 2e⁻
Reduction Half equation is:
2H⁺ + 2e⁻ → H₂(g)
If we analyze the situation analytically, there are situations or states. Then, we are also given with 2 values of pressure and 1 value of volume. Lastly, temperature was set as constant. Thus, this means we use the Boyle's Law.
P₁V₁ = P₂V₂
Let's find V₂.
(1 atm)(1.72 L) = (35 atm)(V₂)
Solving for V₂,
<em>V₂ = 0.049 L</em>
Answer:
C. Yes, because they have a definite composition.
Explanation:
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Answer:
The answer to your question is:
Explanation:
See the attachment
