Calculate the enthalpy change that occurs when 1.00 kg of acetone condenses at its boiling point (329.4 K). The standard enthalpy of vaporization of acetone is 29.1 kJ · mol-1 .
1.) −501 kJ
2.) −29.1 kJ
3.) +501 kJ
4.) +29.1 kJ
5.) −2.91 × 104 kJ
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Ответ:
Option 1 is correct.
ΔH for the condensation of 1 kg of Acetone is - 501 KJ.
Explanation:
Molar mass of acetone = 58.08 g/mol
Number of moles of acetone that condenses = mass/molar mass = 1000/58.08 = 17.22 moles
1 mole of Acetone vaporizer with an enthalpy change of 29.1 KJ.
1 mole of Acetone will condense with an enthalpy change of - 29.1 KJ (since condensation is the exact reverse of vaporization)
That is, vaporization is given by,
Acetone(l) > Acetone(g) ΔH = 29.1 KJ
Then condensation will be given by,
Acetone(g) > Acetone(l) ΔH = -29.1 KJ
So,
1 mole of Acetone will condense with an enthalpy change of - 29.1 KJ
17.22 moles of Acetone will condense with an enthalpy change of 17.22 × - 29.1 KJ = - 501 KJ.
The first option is correct.
Ответ:
1.) -501 kJ.
Explanation:
The enthalpy of vaporization is defined as the amount of energy that must be added to a liquid substance, to transform a quantity of that substance into a gas.
ΔHvap = 29.1 kJ/mol
Δq = n * ΔHvap
Where,
n = number of moles
ΔHvap = standard enthalpy of vaporization
Δq = enthalpy change of vaporization
Molar mass of acetone, C3H6O = (3*12) + (6*1) + (1*16)
= 58 g/mol
Number of moles = mass/molar mass
= 1000/58
= 17.2412 mol.
Δq = 17.2412 * 29.1
= +501.7 kJ.
Δq = 501 kJ is required to be added to evaporate it so Δq = - 501 kJ is required to condense it
Ответ:
37 electrons
Explanation: Rb atom has 37 electrons and 37 protons. when 1 electron is removed number of protons remains 37 but number of electrons is 36.36 negative charges is balanced by 36 positive charges. one extra proton is present so 1 positive charge