Answer:
ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.
Explanation:
Heat water; mechanical (the movement of a turbine is based off of mechanical energy, not chemical or potential).
Answer:
The equation to show the the correct form to show the standard molar enthalpy of formation:
Explanation:
The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.
Given, that 1 mole of 
gas and 1 mole of 
liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.
Divide the equation by 2.
The equation to show the the correct form to show the standard molar enthalpy of formation:
The study of chemistry helps us understand the nature of the world around us. Chemistry is always developing to keep up with any phenomenon that appears in nature.
Therefore, scientists and chemists are always developing new technologies. However, chemists must very careful when developing these new technologies. They should consider any bad chemical reactions that might occur and also chemicals that harmful to either the individuals or the environment.
It is 0.5474 but you can put 0.5 hope this helps
