Answer:
The heat capacity of the bomb calorimeter is 7.58 J/°C.
Explanation:
First, we will calculate energy released on combustion:
= enthalpy change = -3267.5 kJ/mol
q = heat energy released
n = number of moles benzene=
q = -90.0657 kJ = -90,065.7 J
Now we calculate the heat gained by the calorimeter let it be Q.
Q = -q= -(-90,065.7 J) = 90,065.7 J (conservation of energy)
where,
Q = heat gained by calorimeter
c = specific heat capacity of calorimeter =?
= final temperature =
= initial temperature =
Now put all the given values in the above formula, we get:
The heat capacity of the bomb calorimeter is 7.58 J/°C.
Compounds that are gases at room temperature are covalent bonds (nonmetal + nonmetal).
I am assuming that the problem ask for the pressure in
the system. To be able to calculate this, we first assume that the system acts
like an ideal gas, then we can use the ideal gas equation to find for pressure
P.
P V = n R T
where,
P = Pressure (unknown)
V = 0.17 m^3
n = moles of lng / methane
R = gas constant = 8.314 Pa m^3 / mol K
T = 200 K
We find for the moles of lng. Molar mass of methane = 16
kg / kmol
n = 55 kg / 16 kg / kmol
n = 3.44 kmol CH4 = 3440 mol
Substituting all the values to the ideal gas equation:
P = 3440 mol * (8.314 Pa m^3 / mol K) * 200 K / 0.17 m^3
P = 33,647,247 Pa
<span>P = 33.6 MPa</span>
Alpha and beta particles, I saw this somewhere, but I can't remember where