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Travka [436]
3 years ago
9

Which equation represents an oxidation-reduction reaction?

Chemistry
1 answer:
bazaltina [42]3 years ago
3 0
In an oxidation-reduction reaction there is an exchange of electrons.


The exchange of electrons implies change in the oxidation states: at least one element increases its oxidation number while other reduces it.


By simple ispection you can predict that in the equation b. there is a change in oxidation states of Cl and Mn.


Now you can check it:


Equation             4H   Cl     +    Mn  O2     ->    Mn  Cl2   + 2H2 O     + Cl2    

oxidation sates   1+  1-             4+   2-              2+   1-         1+   2-         0


The oxidation state of Cl in HCl is 1-  and it changed to 0 in Cl2


The oxidation state of Mn in MnO2 is 4+ and it changed to 2+ in MnCl2 


Answer b.   
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V_{initial} = 752\:mL
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converting to Kelvin
TK = TC + 273
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By the first Law of Charles and Gay-Lussac, we have: 
\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }

Solving:
\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }
\frac{ 752 }{ 298.0 } = \frac{ V_{f} }{ 323.0 }
Product of extremes equals product of means:
298.0* V_{f} = 752*323.0
298.0 V_{f} = 242896
V_{f} = \frac{242896}{298.0}
\boxed{\boxed{V_{f} \approx 815.08\:mL}}\end{array}}\qquad\quad\checkmark
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OlgaM077 [116]

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Electrochemistry - Equilibrium
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Answer:

Explanation:

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n = 2

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