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.

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

indium, and thallium have three electrons i think