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kompoz [17]
3 years ago
13

In a calorimetry experiment, a salt is dissolved in water within an insulating container. The temperature of the solution (surro

undings) decreases from 25°C to 14°C. Given that the mass of the solution is 120 g, and the specific heat of the solution is 3.18 J/g°C, calculate the enthalpy change (in kJ) for the dissolving process. Be sure to give the correct sign for your answer, and identify the dissolving process as endothermic or exothermic. Group of answer choices
Chemistry
1 answer:
Stels [109]3 years ago
7 0

Answer:

Explanation:

heat released  by the solution

= 120 x 3.18 x ( 25 - 14 )

= 4197.6 J

= 4.1976 kJ

This is the heat gain by the salt

so enthalpy change = +  4.1976 kJ

because there is increase  in enthalpy

It is endothermic process .

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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,

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As we know that:

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

T_f = freezing point temperature = -97.6^oC=273+(-97.6)=175.4K

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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