The combustion of propane is represented by the following chemical equation. C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(l) The standard enthalpies of formation ( ΔH∘f ) for C3H8(g), CO2(g), and H2O(l) are −103.8 kJ/mol, −393.5 kJ/mol, and −285.8 kJ/mol respectively. What is the enthalpy of combustion for propane at 25 °C and 1 atm?

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

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.