Ethanol, ch.0, is common beverage alcohol. at its boiling point of 78.5 °c, the enthalpy of vaporization of ethanol is 38.6 kj/mol. how much heat is required to vaporize 250 g of ethanol at 78.5 °c? ans = 209.8 kj

209.8 kilo Joules heat is required to vaporize 250 g of ethanol at 78.5 °C.

Explanation:

Mass of an ethanol = 250 g

Molar mass of ethanol = 46 g/mol

Moles of an ethanol =

Enthalpy of vaporization of ethanol =

Heat required to vaporize 250 g of ethanol at 78.5 °C : Q

Q = 209.8 kilo Joules

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

distance, luminosity, brightness, radius, chemical composition, and temperature