The heat of vaporization for ethanol is 0.826 kJ/g. Calculate the heat energy in joules required to boil 94.15 g of ethanol.

The heat energy required to boil 94.15 g of ethanol is 77.7679 kJ.

Explanation:

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three.

If the relationship between the magnitudes is direct, the following formula must be followed:

a ⇒ b

c ⇒ x

Then:

It is called "heat of vaporization" to the energy necessary to change 1 gram of substance in solid state, to liquid state, without changing its temperature. In this case, the heat of vaporization of ethanol is 0.826 kJ / g, that is, 0.826 kJ of energy is required to change 1 gram of substance from solid state to liquid state. The thermal energy required to boil 94.15 g of ethanol can be calculated using the following rule of three: if 0.826 kJ is required to change 1 gram of ethanol, to change 94.15 grams, how much energy does it require?

energy= 77.7679 kJ

The heat energy required to boil 94.15 g of ethanol is 77.7679 kJ.

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they are made of gases.

they have a spherical shape.

they make energy through fusion.