Answer:
P = 14.1 atm
Explanation:
Given data:
Mass of methane = 64 g
pressure exerted by water vapors = ?
Volume of engine = 24.0 L
Temperature = 515 K
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O + energy
Number of moles of methane:
Number of moles = mass / molar mass
number of moles = 64 g/ 16 g/mol
Number of moles = 4 mol
Now we will compare the moles of water vapors and methane.
CH₄ : H₂O
1 : 2
4 : 2/1×4 = 8 mol
Pressure of water vapors:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
P = 8 mol × 0.0821 atm.L/mol.K× 515 K / 24.0 L
P = 338.25 atm.L/ / 24.0 L
P = 14.1 atm
The mass of pentane the student should weigh out is
The density of pentane is 0.626 gcm-3
To calculate the mass of pentane following expression is used,
(Density is defined as the mass divide by volume)
Density = mass / volume
mass of pentane = Density of pentane * Volume of pentane
mass of pentane = 0.626 gcm-3 * 45.0 mL
= 28.17 g
Here the unit of mass of pentane is g,
However the unit of density is gcm-3 and unit of volume is mL i.e. cm3
Hence, Mass = gcm-3 * cm3
Mass = g
The mass of pentane the student should weigh out is 28.17g
Learn more about Density on
brainly.com/question/1354972
#SPJ1
Answer:
Explanation:
The photo is not there :o