What temperature kelvin would .97mol of gas be to occupy 26L at 751.2mmHg?
1 answer:
Answer: 322.56 Kelvin
Explanation:
Use the Ideal Gas Law
R is the gas constant
T is the temperature in Kelvins
P is the pressure in atmospheres
V is the volume in liters
n is the number of moles of gas
First, the mm of mercury need to be converted to atmospheres using the conversion factor 1atm = 760 torr.
Now plug everything in
You might be interested in
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>.
Learn more here: