Answer: The standard entropy of vaporization of ethanol is 0.275 J/K
Explanation:
Using Gibbs Helmholtz equation:
For a phase change, the reaction remains in equilibrium, thus 
Given: Temperature = 285.0 K
Putting the values in the equation:
Thus the standard entropy of vaporization of ethanol is 0.275 J/K
Hi there!
• Avogadro's number = 6.023 × 10²³
• No.of molecules in N = 1.806 × 10²² [ Given ]
It's known that :-
Number of molecules = Moles × Avogadro's number
=> 1.806 × 10²² = Mol. × 6.023 × 10²³
=> Mole =
=> Moles = 0.03 mol.
Hence, 0.03 mol. is th' required answer.
~ Hope it helps!
Non metals are electronegative in nature whereas metals are electropositive. A non-metal accepts electron and gains negative charge whereas metals lose eletrons and gains positive charge. Hence correct answer is non-metals. Hope this helps!
Scientist examine fossil evidence during the pre-time yes researching about the things at the pre-Cambrian time might help them to identify what would’ve happened before Paleozoic era
Answer
A. -1305 kJ
Explanation
Given:
Equation: C2H4(g) + 302(g) ---- > 2C02(g) + 2H20(l).
Formation ΔH values:
for C2H4(g) = 52.30 kJ/mol,
for 02(g) = 0 kJ/mol,
for CO2(g) = -393.5 kJ/mol, and
for H20(1) = -285.8kJ/mol.
What to find:
The ΔHrxn from the ΔH of formation for the given reaction.
Step-by-step solution:
ΔHrxn = (Sum of ΔH formation for the product) - (Sum of ΔH formation for the reactants).
ΔHrxn = (ΔH for 2CO2(g) + ΔH for 2H2O(l)) + (ΔH for C2H4(g) + ΔH for 3O2(g))
ΔHrxn = [(2 x -393.5) + (2 x -285.8)] + [52.30 + (3 x 0)]
ΔHrxn =(-787.0 - 571.6) + (52.30 + 0)
ΔHrxn = -1358.6 + 52.30
ΔHrxn = -1306.3 kJ
so the closest answer is A. -1305 kJ
