Answer: 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point
Explanation:
Latent heat of vaporization is the amount of heat required to convert 1 mole of liquid to gas at atmospheric pressure.
Amount of heat required to vaporize 1 mole of lead = 177.7 kJ
Molar mass of lead = 207.2 g
Mass of lead given = 1.31 kg = 1310 g (1kg=1000g)
Heat required to vaporize 207.2 of lead = 177.7 kJ
Thus Heat required to vaporize 1310 g of lead =
Thus 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point
Answer:
the temperature and pressure can greatly affect the volume of a substance. especially with gases. as with mass increasing and decreasing the amount of material also increases and decreases the volume of the substance.
All chemical equations show that the total mass of the reaction stays the same.
If an object's velocity is changing, it's either experiencing acceleration or deceleration.
Answer:
Protons=7, Neutrons=6, Electrons=6
Explanation:
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