This question is providing the exothermic heat of neutralization of acetic acid in units of kilojoules per mollimole (-49,8 kj/mmol) and asks for the same value but in calories per millimole which results -11,902.5 cal/mmol.
In this case, according to the given problem, it turns out necessary to solve a two-factor conversion in order to convert the kilojoules to joules and finally to calories as shown below:
[tex]-49.8\frac{kJ}{mmol}*\frac{1000J}{1kJ}*\frac{1cal}{4.184J}[/tex]
Thus, we cancel out the kJ and J, to obtain the following result, rounded to one decimal place:
[tex]-11,902.5\frac{cal}{mmol}[/tex]
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