To determine the amount of heat needed to evaporate a certain amount of water at its boiling point, we use the equation,
H = m(Hv)
where H is the heat, m is the mass (in grams), and Hv is the latent heat of vaporization. Substituting the known values,
H = (105 g)(540 cal/g)
H = 56700 cal
H = 56.7 kcal
Answer: 56.7 kcal
Answer:
P = 3.23 atm
Explanation:
Given data:
Number of moles of gas = 30 mol
Temperature = 120°C
Volume of gas = 300 L
Pressure of gas = ?
Solution:
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will convert the temperature.
120+273.15 = 393.15 K
P × 300 L = 30 mol × 0.0821 atm.L/mol.K × 393.15 K
P = 968.33 atm. L /300 L
P = 3.23 atm
Answer:
the answer is False I hope this helps
Answer:
I think it's D
Explanation:
I'm so sorry if that's wrong, hope it helps!
