This question is describing the following chemical reaction at equilibrium:

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

Thus, by recalling the Van't Hoff's equation, we can write:

Hence, we solve for the enthalpy change as follows:

Finally, we plug in the numbers to obtain:
![\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7B-8.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%20%2Aln%280.25%2F9%29%7D%7B%5B%5Cfrac%7B1%7D%7B%2875%2B273.15%29K%7D%20-%5Cfrac%7B1%7D%7B%2825%2B273.15%29K%7D%20%5D%20%7D%20%5C%5C%5C%5C%5C%5C%5CDelta%20H%3D4%2C785.1%5Cfrac%7BJ%7D%7Bmol%7D)
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Answer:
the discovery will always need to be tested more where if you're investigating you should already be in a more controlled environment
The answer is D.timber good luck
Answer:
1.94
Explanation:
moles = mass/Mr so you have 3.889 moles of aluminium, but because the ratio in the equation is 2:1 you need to halve it.
2Ca(OH)2(aq) + 2FeCl3(aq) on the dead locs