Answer:
1.52 L
Explanation:
P1V1T2=V2P2T1
V2=V1T2/T1
Fill in with given values then solve
We can check this by knowing that V and T at constant P have a proportional relationship. Hence, this is correct.
- Hope that helped! Please let me know if you need further explanation.
When converted to a household measurement, 9 kilograms is approximately equal to a
Answer:
The answer is 18.12KJ is required to vaporise 48.7 g of dichloromethane at its boiling point
Explanation:
To solve the above question we have the given variable as follows
ΔHvap = heat of vaporisation of dichloromethane per mole = 31.6KJ/mole
However since the heat of vaporisation is the heat to vaporise one mole of dichloromethane, then, for 48.7 grams of dichloromethane, we have.
The number of moles of dichloromethane present = 48.7/84.93 = 0.573 moles
Therefore, the amount of heat required to vaporise 48.7 grams of dichloromethane at its boiling point is 31.6KJ/mole×0.573moles =18.12KJ