Percentage O: 69.54
Percentage N: 30.46
Dividing each percentage by the components molecular weight:
Oxygen: 69.54 / 16 = 4.35
Nitrogen: 30.46 / 14 = 2.18
Ratio of nitrogen to oxygen = 1 : 2
Empircal formula: NO₂
Mass of empirical formula = 46
Repeat units = 92 / 46 = 2
Molecular formula = N₂O₄
<u>Answer:</u> The standard electrode potential of the cell is 0.62 V.
<u>Explanation:</u>
For the given chemical equation:
![3Cu(s)+2NO_3^-(aq)+8H^+(aq.)\rightarrow 3Cu^{2+}(aq.)+2NO(g)+4H_2O(l)](https://tex.z-dn.net/?f=3Cu%28s%29%2B2NO_3%5E-%28aq%29%2B8H%5E%2B%28aq.%29%5Crightarrow%203Cu%5E%7B2%2B%7D%28aq.%29%2B2NO%28g%29%2B4H_2O%28l%29)
The substance having highest positive
potential will always get reduced and will undergo reduction reaction.
<u>Oxidation half reaction:</u>
( × 3 )
<u>Reduction half reaction:</u>
( × 2 )
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the
of the reaction, we use the equation:
![E^o_{cell}=E^o_{cathode}-E^o_{anode}](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3DE%5Eo_%7Bcathode%7D-E%5Eo_%7Banode%7D)
![E^o_{cell}=0.96-(0.34)=0.62V](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3D0.96-%280.34%29%3D0.62V)
Hence, the standard electrode potential of the cell is 0.62 V.
Answer:
A. The equilibrium shifts right to produce more SO3 molecules.
Explanation:
Answer:
Chemical Change
Explanation:
Because it dissolves with the help of saliva , then into stomach and excreted in a different form
I don't think the introduction of a catalyst to a reaction would affect its enthalpy. The reason why I think this is that enthalpy is a state function meaning it does not matter what pathway is taken, the end result will always be the same. A catalyst only changes the reaction mechanism and the starting point and ending point should not change. If you look at a energy diagram of a reaction with and without a catalyst, you should see that the only thing that changed was the activation energy required to make the reaction happen. Not the enthalpy.