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
<span>What would you use if you needed to determine the density of an object?
</span>Balance and Graduate Cylinder
Hope This Helps! <span>\(★ω★)/</span>
Answer:
45th answer is pure and properties 42 is periodictable
A. Formation of a gas. You can tell a chemical reaction has occurred through: formation of a precipitate, formation of a gas (also known as bubbles) unexpected color change, unexpected odor change, temperature change.
There are 0.454 kg in 1 lb
Cross multiply
0.454 kg = 1 lb
X kg. = 50 lb
50* 0.454 = 22.7 kg
Answer : 22.7 kg in 50 lbs