contestada

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 normal boiling point for ccl4. the specific heat of ccl4(l is its heat of fusion is and its heat of vaporization is

Respuesta :

BlueSky06

The solution would be like this for this specific problem:

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

 (29.82 kJ / mole)(0.163 moles) = 4.86 kJ 

total heat = 1.11 kJ + 4.86 kJ = 5.97 kJ 

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

Otras preguntas