Answer:
Carbon dioxide and water
Explanation:
Photosynthesis is a process used by plants and other organisms to convert light energy into chemical energy that can later be released to fuel the organisms' activities. This chemical energy is stored in carbohydrate molecules, such as sugars, which are synthesized from carbon dioxide and water – hence the name photosynthesis.
Answer : The correct expression for equilibrium constant will be, ![K=[O_2]^5](https://tex.z-dn.net/?f=K%3D%5BO_2%5D%5E5)
Explanation :
Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.
The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.
As we know that the concentrations of pure solids are constant that is they do not change. Thus, they are not included in the equilibrium expression.
The given equilibrium reaction is,

The expression of
will be,
![K=[O_2]^5](https://tex.z-dn.net/?f=K%3D%5BO_2%5D%5E5)
Therefore, the correct expression for equilibrium constant will be, ![K=[O_2]^5](https://tex.z-dn.net/?f=K%3D%5BO_2%5D%5E5)
Solubility product constants are values to describe the saturation of ionic compounds with low solubility. A saturated solution is when there is a dynamic equilibrium between the solute dissolved, the dissociated ions, the undissolved and the compound. It is calculated from the product of the ion concentration in the solution. For barium chromate, the dissociation would be as follows:
BaCrO4 = Ba^2+ + (CrO4)^2-
So, the expression for the solubility product would be:
Ksp = [Ba^2+] [(CrO4)^2-]
we let x = [BaCrO4] = [Ba2+] = [(CrO4)2-] = 2.81x10^-3 g/L ( 1 mol / 253.35 g ) = 1.11x10^-5
Ksp = x(x)
Ksp= x^2
Ksp = (1.11x10^-5)^2
Ksp = 1.23x10^-10
The Ksp of Barium chromate at that same temperature for the solubility would be 1.23x10^-10.
The given chemical reaction is:

The standard heats of formation of
are:
Δ
) = -285.8kJ/mol[/tex]
Δ
) = -187.6 kJ/mol[/tex]
Calculating the change in heat:
Δ
=∑ΔH
-∑ΔH
= [{2 * (-285.8 kJ/mol)} -{2*(-187.6 kJ/mol)}]
= -196.4 kJ/mol
Therefore, the change in enthalpy for the given reaction is -196.4 kJ/mol