<span>así que te está diciendo que</span>
First, consider the steps to heat the sample from 209 K to 367K.
1) Heating in liquid state from 209 K to 239.82 K
2) Vaporaizing at 239.82 K
3) Heating in gaseous state from 239.82 K to 367 K.
Second, calculate the amount of heat required for each step.
1) Liquid heating
Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol
=> number of moles = 12.62 g / 17 g/mol = 0.742 mol
Heat1 = #moles * heat capacity * ΔT
Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J
2) Vaporization
Heat2 = # moles * H vap
Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J
3) Vapor heating
Heat3 = #moles * heat capacity * ΔT
Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J
Third, add up the heats for every steps:
Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J
Fourth, divide the total heat by the heat rate:
Time = 22,466.3 J / (6000.0 J/min) = 3.7 min
Answer: 3.7 min
Explanation:
As the given data is as follows.
Height, H = 150 feet
Heat gain = 30,000 BTU/hr, and Heat loss = 25000 BTU/hr
m = mass of water heated = 700 gallons = 5810 lbs
is the heat capacity of water = 1 BTU/lb 
(given)
= temperature difference = 
Heat energy required to heat 700 gal can be calculated as follows:
Heat Required = 
Thus, water rises till 
.
Answer:
A. 15859.2 L or 15900 L
B. 0.629 mol
Explanation:
At STP, one mole is equal to approximately 22.4 L
L or mL is volume, so you are attempting to solve for L or mL.
A.
708 mol x (22.4 L/1 mol) = 15859.2 L (w/ significant figures included - 15900 L)
B.
(14.1 L) x (1 mole/ 22.4 L) = 0.629 mol.
Also called thermal capacity, is when the amount of heat required to raise the temp. of one mole or one gram of a substance by one degree Celsius without change of phase.
Hope this helps :)
