Answer: This is an oxidation-reduction (redox) reaction:
3 C-II - 12 e- → 3 CII (oxidation)
4 CrVI + 12 e- → 4 CrIII (reduction)
C2H5OH is a reducing agent, K2Cr2O7 is an oxidizing agent.
<span>I think this is due to aerodynamics. Air is the stuff this is all around you. carry your hand in front of your physique which includes your palm dealing with sideways so as that your thumb is on appropriate and your pinkie is dealing with the floor. Swing your hand back and forth.</span>
Answer:
1,321.986 (the first one)
Answer:
II. The reaction is exothermic.
III. The enthalpy term would be different if the water formed was gaseous.
Explanation:
For the reaction:
C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)
The ΔH is -1.37×10³ kJ. As the change in enthalpy is <0,<em> II. The reaction is exothermic.</em>
The ΔH formation of a compound is different if the chemical is in liquid or gaseous phase. For that reason: <em>III. The enthalpy term would be different if the water formed was gaseous.</em>
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I hope it helps!
