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
1 answer:
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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