Methanol, ethanol, and n−propanol are three common alcohols. when 1.00 g of each of these alcohols is burned in air, heat is liberated as indicated. calculate the heats of combustion of these alcohols in kj/mol.

For methanol: Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol: Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol: Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

Explanation:

Given:

Mass of Methanol = 1.0 g

Mass of ethanol = 1.00 g

Mass of n-propanol = 1.00 g

For methanol:

2 CH₃OH + 3 O₂ ----> 2 CO₂ + 4 H₂O, ∆H₀ = -22.6 kJ/g  (negative sign signifies release of heat)

1 g of methanol on combustion gives 22.6 kJ of energy

Calculation of moles of methanol:

Molar mass of methanol = 32.04 g/mol

Thus moles of methanol = 1 g/ (32.04 g/mol) = 0.0312 moles

Hence energy in kJ/mol:

Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol:

C₂H₅OH + 3 O₂ ----> 2 CO₂ + 3 H₂O, ∆H₀ = -29.7 kJ/g  (negative sign signifies release of heat)

1 g of ethanol on combustion gives 29.7 kJ of energy

Calculation of moles of ethanol:

Molar mass of ethanol = 46.07 g/mol

Thus moles of ethanol = 1 g/ (46.07 g/mol) = 0.0217 moles

Hence energy in kJ/mol:

Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol:

2 C₃H₇OH + 9 O₂ ----> 6 CO₂ + 8 H₂O, ∆H₀ = -33.4 kJ/g , (negative sign signifies release of heat)

1 g of methanol on combustion gives 33.4 kJ of energy

Calculation of moles of methanol:

Molar mass of methanol = 60.09 g/mol

Thus moles of methanol = 1 g/ (60.09 g/mol) = 0.0166 moles

Hence energy in kJ/mol:

Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

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.