A 200-gram sample of a radioactive nuclide is left to decay for 60 hours. At the end of 60 hours, only 25 grams of the sample is
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The heat released in the combustion of lactic acid is absorbed by the
calorimeter and in the decomposition of the lactic acid.
ΔH°f of lactic acid is approximately <u>-716.2 kJ</u>
Reasons:
Known parameters are;
Mass of the lactic acid = 3.93 grams
Heat capacity of the bomb calorimeter = 10.80 kJ·K⁻¹
Change in temperature of the calorimeter, ΔT = 5.34 K
ΔHrxn = ΔErxn
ΔH°f of H₂O(l) = -285.8 kJ·mol⁻¹
ΔH°f of CO₂(g) = -393.5 kJ·mol⁻¹
The chemical equation for the reaction is presented as follows;
- C₃H₆O₃ + 2O₂ → 3CO₂ + 3H₂O
The heat of the reaction = 10.80 kJ·K⁻¹ × 5.34 K = 57.672 kJ
Molar mass of C₃H₆O₃ = 90.07 g/mol
Number of moles of C₃H₆O₃ = = 0.043633 moles
Number of moles of CO₂ produced = 3 × 0.043633 moles = 0.130899 moles
Heat produced = 0.130899 mole × -285.8 kJ·mol⁻¹ = -37.4109342 kJ
Moles of H₂O produced = 0.130899 moles
Heat produced = 0.130899 mole × -393.5 ≈ -51.51 kJ
Therefore, we have;
Heat absorbed by the lactic acid = ΔH°f of H₂O + ΔH°f of CO₂ + Heat absorbed by the calorimeter
Which gives;
Heat absorbed by lactic acid = -37.4109342 kJ - 51.51 kJ + 57.672 kJ ≈ -31.249 kJ
The heat absorbed by the lactic acid ≈ -31.249 kJ
ΔH°f of C₃H₆O₃ ≈ -716.2 kJ
Heat of formation of lactic acid ≈ <u>-716.2 kJ</u>.
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