Answer:
plqojs then came back and her story about how the ocean had become toper for her own purposes and share the ocean and the mosquitoes and it that way in notebook life that they were never able of English orto make the ocean view the mosquitoes in the first place they would have to make it to. 
 
        
                    
             
        
        
        
The correct answer is B) they operate at a higher efficiency. sorry hope the answers not to late :(
        
                    
             
        
        
        
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
 
        
             
        
        
        
Given:
The density of air = 1.19 g/L at 25°C and atmospheric pressure,
or
density = 1.19 x 10⁻³ kg/L
Volume of air in the room is
V = 12.5*19.5*6.0 = 1462.5 ft³
Note that
1 ft³ = 28.317 L
Therefore
V = (1462.5 ft³)*(28.317 L/ft³) = 4.1414 x 10 ⁴ L
By definition, mass = density*volume.
Therefore, the mass is
(1.19 x 10⁻³ kg/L)*(4.1414 x 10⁴ L) = 49.283 kg
Answer: 49.3 kg (nearest tenth)