What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚c) and 1 atm? report your answer two points past the decimal with the unit j/molk. ∆h˚fus = 3.17 kj/mol.

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

where,

= change in entropy

= change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

= freezing point temperature =

Now put all the given values in the above formula, we get:

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

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~ trueboss72