Butane c4 h10 ( = –125.7), combusts in the presence of oxygen to form co2 (g) (hf = –393.5 kj/mol), and h2o(g) (hf = –241.82) in the reaction: what is the enthalpy of combustion, per mole, of butane? use

Following is the balanced reaction for combustion of butane,

2 C4H10   +      13 O2       →      8 CO2   +     10H2O

Given, Heat of formation of C4H10 =  = -125.7 kJ/mol

Heat of formation of water =  = -241.82 kJ/mol

Heat of formation of CO2 =  = -393.5 kJ/mol

Thus, enthalpy of combustion = ΔH = ∑Hproducts - ∑Hreactants

                                              = 8 + 10 - 2 

                                              = (-393.5X8) + (-241.82X10) - (-125.7X2)

                                              = -5314.8 kJ/mol

This energy would result in combustion of 2 moles of butane

∴ For 1 mole, energy required for combustion of butane = -5314.8/2 = -2657.4 kJ/mol

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.