<em>i think the answer is b </em>
Answer:
339kJ
Explanation:
Given parameters:
Mass of steam = 150g = 0.15kg
Initial temperature of steam = 100°C
Final temperature of water = 100°C
Unknown:
Quantity of heat that must be removed to condense the steam = ?
Solution:
The heat involved here is a latent heat because there is no change temperature. The process is just a phase change.
H = mL
m is the mass
L is the latent heat of vaporization = 2,260 kJ/kg
Insert the parameters and solve;
H = 0.15kg x 2,260 kJ/kg
H = 339kJ
Answer:
The volume on the tank is 6, 20 L
Explanation:
We use the formula PV=nRT. We convert the units of pressure in kPa into atm and temperature in Celsius into Kelvin:
0°C=273K
101,325kPa---1 atm
275kPa --------x=(275kPax 1 atm)/101,325kPa= 2,71 atm
PV=nRT --> V=nRT/P
V= 0,750 mol x 0,082 l atm /K mol x 273 K/ 2, 71 atm= <em>6, 20 L</em>
Answer:
Only Harry and Jena
Explanation:
Under federal regulations, an UST is any one or a combination of tanks such that the volume of an accumulation of regulated substances is 10% or more beneath the surface of the ground.
Any UST system holding a mixture of hazardous waste and other regulated substances are also are not covered by federal regulations regarding USTs.
Farm or residential tank of capacity more than 11 gallons used for storing motor fuel is covered by federal regulations regarding USTs.
According to the given question,
Only Harry and Jena are covered by federal regulations regarding USTs.
