Ozone reacts completely with NO, producing NO2 and O2. A 13.0 L vessel is filled with 1.30 mol of NO and 1.30 mol of O3 at 401.0
1 answer:
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Answer:
The mass of the solute and the volume of the solution.
Explanation:
Hello,
In this case, given the formula of molarity:
In such a way, since the moles could not be directly measured, we must measure the mass of the solute and by using its molar mass, one could compute its moles. Moreover, since the solution is composed by the solvent (typically water) and the solute, we consequently must measure the volume of the solution needed for the preparation of such concentration-known solution. In such a way, we can actually prepare the required solution.
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Explanation:
1. In the first column
<u>INCREASE CONCENTRATION</u>: these are the items that needs to be placed in this bin
a. Add drink mix solid to a diluted mixture of drink mix in pure water
b. Add drink mix solution to a diluted mixture of drink mix in pure water
c. Add drink mix solid to pure drink mix solution
d Evaporate water from a diluted mixture of drink mix in pure water
e Evaporate water from pure drink mix solution
2. The items below are what should be placed in this bin,<u>DECREASE CONCENTRATION:</u>
a. Add water to a diluted mixture of drink mix in pure water
b. Add water to pure drink mix solution
3. These items here should be placed here. <u>DOES NOT AFFECT CONCENTRATION:</u>
a. Add pure drink mix solution to pure drink mix solution
b. Drain the pure drink mix solution
c. Drain the diluted mixture
Answer:
ΔG° = -80.9 KJ
Assuming this reaction takes place at room temperature (25 °C):
K=
Explanation:
1) Reduction potentials
First of all one should look up the reduction potentials for the species envolved:
→
E°red=1.61V
→
E°red=0.771V
2) Redox pair
Knowing their reduction pontentials one can determine a redox pair: one species must oxidate while the other is reducing. <u>Remember: the table gives us the reduction potential, so if we want to know the oxidation potential all that has to be done is reverce the equation and change the potencial signal (multiply to -1).</u>
1) reduces while
oxidates
(oxidation) →
E°oxi=-0.771V
(reduction) →
E°red=1.61V
(overall equation) →
E°=Ereduction + Eoxidation= 1.61 v+(-0.771 v) = 0.839v
The cell potential can also be calculated as the cathode potencial minus the anode potential:
E° = E cathode - E anode =1.61 v - 0.771 v=0.839 v
3) Gibbs free energy and Equilibrium constant
ΔG°=-nFE°, where 'n' is the number of electrons involved in the redox equation, in this case n is 1. 'F' is the Faraday constant, whtch is 96500 C. E° is the standard cell potencial.
ΔG°=-nFE°=-1*96500*0.839
ΔG° = - 80963 J = -80.9 KJ
The Nerst equation gives us the relation of chemical equilibrium and Electric potential.
E=E°-
Where 'R' is the molar gas constant (8.314 J/mol)
It's known that in the equilibrium E=0, so the Nerst equation, at equilibrium, becomes:
E°=
Isolating for 'K' gives:
This shows that 'K' is a fuction of temperature. Assuming this reaction takes place at room temperature (25 °C):
K=