How many molecules are in 3cacl2
1 answer:
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Answer : The correct option is, +91 kJ/mole
Solution :
The balanced cell reaction will be,
Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.
First we have to calculate the standard electrode potential of the cell.
Now we have to calculate the standard Gibbs free energy.
Formula used :
where,
= standard Gibbs free energy = ?
n = number of electrons = 2
F = Faraday constant = 96500 C/mole
= standard e.m.f of cell = -0.47 V
Now put all the given values in this formula, we get the Gibbs free energy.
Therefore, the standard Gibbs free energy is +91 kJ/mole
Answer:
1. Hydrogen Iodide
2. 6 molecules of Hydrogen Iodide
3. Iodine is the limiting reagent
Explanation:
The image of the illustration in the question has been attached:
1. The illustration represents a mixture of hydrogen ( light blue ) and iodine ( purple )
H₂ + I₂ ---> 2HI
This forms hydrogen iodide.
2. In the given illustration, 6 product molecules of Hydrogen Iodide. This is indicated in the box on the right side of the illustration.
3. The limiting reagent is the reactant that determines how much of the products are made. It is the substance that is totally consumed when the chemical reaction is completed. In the box on the right side of the illustration, you will see that hydrogen which is indicated by blue is in excess. The limiting reagent is the one that is completely consumed which is the iodine.
