3 years ago

I need helppppppp plssssss

Chemistry

1 answer:

Irina-Kira [14]3 years ago

4 0

Answer:

<h3>1)20</h3><h3>2)13</h3><h3>3)82</h3><h3>4)56</h3><h3>5)26</h3><h3>6)1</h3><h3>hope it helps</h3><h3>plzzz mark itas brainliest...</h3><h3> </h3><h3 />

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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

Rudiy27

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

Explanation :

Formula used :

$\Delta S=\frac{\Delta H_{freezing}}{T_f}$

where,

$\Delta S$ = change in entropy

$\Delta H_{fus}$ = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

$\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol$

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

$\Delta S=\frac{\Delta H_{freezing}}{T_m}$

$\Delta S=\frac{-3170J/mol}{175.4K}$

$\Delta S=-18.07J/mol.K$

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

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