Answer : The correct option is, (b) +115 J/mol.K
Explanation :
Formula used :

where,
= change in entropy
= change in enthalpy of vaporization = 40.5 kJ/mol
= boiling point temperature = 352 K
Now put all the given values in the above formula, we get:



Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K
Answer:
<h3>The answer is option D</h3>
Explanation:
To find the speed given the kinetic energy and mass we use the formula

where
m is the mass
v is the speed
From the question
KE = 15 J
m = 2.3 kg
We have

We have the final answer as
<h3>3.6 m/s</h3>
Hope this helps you
Answer: Li is the reducing agentg and O is the oxidizing agent.
Explanation:
1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.
2) The given reaction is:
4Li(s) + O₂ (g) → 2 Li₂O(s)
3) Determine the oxidation states of each atom:
Li(s): oxidation state = 0 (since it is alone)
O₂ (g): oxidation state = 0 (since it is alone)
Li in Li₂O (s) +1
O in Li₂O -2
That because 2× (+1) - 2 = 0.
4) Determine the changes:
Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.
O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.
The reaction would be Exothermic