<span>This is due to the fact that the air pressure in that certain section of Earth’s atmosphere decreased. As density of gas particles decreases as air pressure decreases. Therefore, density of gas particles and air pressure have a direct relationship. An increase in air pressure would then effect to an increase in gas particles. </span>
Answer:
ΔH°c = -2219.9 kJ
Explanation:
Let's consider the combustion of propane.
C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(l)
We can find the standard enthalpy of the combustion (ΔH°c) using the following expression.
ΔH°c = [3 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l))] - [1 mol × ΔH°f(C₃H₈(g)) + 5 mol × ΔH°f(O₂(g))]
ΔH°c = [3 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol)] - [1 mol × (-103.8 kJ/mol) + 5 mol × (0 kJ/mol)]
ΔH°c = -2219.9 kJ
Answer: <span>9330 j/mol
</span>
The temperature of the gas is 475 ° Celcius which is equal to: 475 +273= 748 °K. The formula for kinetic energy of individual atoms would be
K= 3/2 * kB * T
If kB is 1.38 * 10^-23 J/K and 1 mol is made from 6.02*10^23 molecule, then the kinetic energy of 1 mol CO2 would be:
K= 3/2 * kB * T
K= 3/2 * 1.38 * 10^-23 * 748 * 6.02 *10^23 =9324 J/mol
When heat energy is supplied to a material it can raise the temperature of mass of the material.
Specific heat is the amount of energy required by 1 g of material to raise the temperature by 1 °C.
equation is
H = mcΔt
H - heat energy
m - mass of material
c - specific heat of the material
Δt - change in temperature
substituting the values in the equation
120 J = 10 g x c x 5 °C
c = 2.4 Jg⁻¹°C⁻¹
