Calculate the heat required to melt 7.35 g of benzene at its normal melting point. heat of fusion (benzene) = 9.92 kj/mol heat = kj

calculate the heat required to vaporize 7.35 g of benzene at its normal boiling point. heat of vaporization (benzene) = 30.7 kj/mol heat = kj

The heat required to melt 7.35 g of benzene at its normal melting point is 934.8 Joules.

The heat required to vaporize 7.35 g of benzene at its normal melting point is 2,893 Joules.

Explanation:

Mass of benzene = 7.35 g

Moles of benzene =

Heat fusion of benzene,

1) Heat required to melt 7.35 g of benzene at its normal melting point = Q

(1 kJ = 1000 J)

2) Heat vaporization of benzene,

Heat required to vaporize 7.35 g of benzene at its normal melting point = Q

(1 kJ = 1000 J)

c.

explanation:

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.