Answer:
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: [tex]x=\frac{c*b}{a}[/tex]
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?
[tex]energy=\frac{94.15 grams*0.826 kJ}{1 grams}[/tex]
energy= 77.7679 kJ
The heat energy required to boil 94.15 g of ethanol is 77.7679 kJ.
The heat energy in joules required to boil the ethanol is 77.77 kJ.
The given parameters;
The heat energy in joules required to boil the ethanol is calculated as follows;
[tex]Q = m\times \Delta H_{vap}[/tex]
Q = 94.15 x 0.826
Q = 77.77 kJ
Thus, the heat energy in joules required to boil the ethanol is 77.77 kJ.
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