Answer:
44Kj
Explanation:
These are the equations for the reaction described in the question,
Vaporization which can be defined as transition of substance from liquid phase to vapor
H2(g)+ 1/2 O2(g) ------>H2O(g). Δ H
-241.8kj -------eqn(1)
H2(g)+ 1/2 O2(g) ------>H2O(l).
Δ H =285.8kj ---------eqn(2)
But from the second equation we can see that it moves from gas to liquid, we we rewrite the equation for vaporization of water as
H2O(l) ------>>H2O(g)---------------eqn(3)
But the equation from eqn(2) the eqn does go with vaporization so we can re- write as
H2O ------> H2(g)+ 1/2 O2(g)
Δ H= 285.8kj ---------------eqn(4)
To find Delta h of the vaporization of water at these conditions, we sum up eqn(1) and eqn(4)
Δ H=285.8kj +(-241.8kj)= 44kj
<span>HofO3 - [H of O(g) + H of O2] = deltaH
HofO3 = -107.2kJ/mol + 249.8 + 0 (Heat of formn.of O2 is 0
= +142.6kJ/mol </span>
Water<span> is the basic element of nature. It covers 70% of the </span>earth's<span> surface. ... Also known as hydrologic </span>cycle<span>, the </span>water cycle<span> is a phenomenon where </span>water<span> moves</span>through<span> the three phases (gas, liquid and solid) over the four spheres (atmosphere, lithosphere, </span>hydrosphere<span> and biosphere) and completes a full </span><span>cycle</span>
The answer to this question would be: decrease
If the process is releasing heat to the surrounding, that means it should be an exothermic process. In the exothermic process, the process enthalpy difference will be minus because some of the energy is released as heat. That means the enthalpy of the system will be decreased.
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