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 number of moles of 
reacted was 0.229
Explanation:
According to ideal gas equation:
P = pressure of gas = Total pressure - vapor pressure of water = (749 - 23.8) mm Hg= 725.2 mm Hg = 0.954 atm (760mmHg= 1atm)
V = Volume of gas = 5.87 L
n = number of moles = ?
R = gas constant =
T =temperature =
According to stoichiometry:
1 mole of 
is produced by = 1 mole of
Thus 0.229 moles of is produced by = 
moles of
Thus number of moles of reacted was 0.229
The <span>modern-day quantum model of the atom known as the wave mechanical model depicts that the </span>electron was a three dimensional wave circling the nucleus in a whole number of
wavelengths allowing the waveform to repeat itself as a stable standing wave
(it can absorb energy form a nearby source which is oscillating at a proper
frequency) representing the energy levels of Bohr model.
If it has a positive charge it is a cation if negative it is an anion .
I attached a chart that will help you know the charges of the elements
