High School

If the reaction

\[ \text{Na}_2\text{CO}_3(aq) + \text{CaCl}_2(aq) \rightarrow \text{CaCO}_3(s) + 2\text{NaCl}(aq) \]

releases 39.4 kJ of energy, how many kilocalories does it release?

(1 cal = 4.184 J)

(Round off the answer to 2 decimal places)

Answer :

Final answer:

The reaction Na2CO3(aq) + CaCl2(aq) -> CaCO3(s) + 2NaCl(aq) releases 39.4 kJ of energy, which is approximately 9.41 kcal when converted using the conversion factor of 1 cal = 4.184 J.

Explanation:

The reaction mentioned Na2CO3(aq) + CaCl2(aq) -> CaCO3(s) + 2NaCl(aq) releases a specified amount of energy, 39.4 kJ. To convert this energy from kilojoules to kilocalories, you would use the energy equivalence between these two units. This equivalence is that 1 cal = 4.184 J. Converting kJ (kilojoules) to J (joules), we multiply by 1000, so 39.4 kJ = 39400 J. Thus, to get the energy in kilocalories, we divide the joules by the conversion factor (4.184 J/cal). So, the reaction releases approximately 9.41 kcal of energy.

Learn more about Energy Conversion here:

https://brainly.com/question/20458806

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