How much heat (in kJ) is needed to evaporate 2.9 moles of CCl4 (liquid) at its standard boiling point of 309 K? ΔS° vap
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How much heat (in kJ) is needed to evaporate 2.9 moles of CCl4 (liquid) at its standard boiling point of 309 K? ΔS° vap
How much heat (in kJ) is needed to evaporate 2.9 moles of CCl4(liquid) at its standard boiling point of 309 K? ΔS° vap for CCl 4(liquid) = 93.9 J/(mol K).