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marishachu [46]
4 years ago
10

Calculate the total quantity of heat required to convert 25.0 g of liquid ccl4(l from 35.0°c to gaseous ccl4 at 76.8°c (the norm

al boiling point for ccl4. the specific heat of ccl4(l is its heat of fusion is and its heat of vaporization is
Chemistry
1 answer:
GREYUIT [131]4 years ago
6 0

The solution would be like this for this specific problem:

<span>25.0 g CCl4 x (1 mole CCl4 / 153.8 g CCl4) = 0.163 moles CCl4 

 </span>(29.82 kJ / mole)(0.163 moles) = 4.86 kJ 

total heat = 1.11 kJ + 4.86 kJ = 5.97 kJ 

<span>I hope this helps and if you have any further questions, please don’t hesitate to ask again. </span>

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The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

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

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