Answer:- 29.6 moles of carbon.
Solution:- We have been given with 3.7 moles of and asked to calculate the moles of C.
Looking at the formula of the compound, there are 8 carbons in it means 1 mol of he compound has 8 moles of C. So, if we multiply the given moles of the compound by 8 then we get the moles of C.
= 29.6 mol C
Hence. there are 29.6 moles of C in 3.7 moles of .
The energy required to vaporize 155 g of butane at its boiling point: 61,723 kJ
<h3>Further explanation</h3>Enthalpy is the amount of system heat at constant pressure.
The enthalpy is symbolized by H, while the change in enthalpy is the difference between the final enthalpy and the initial enthalpy symbolized by ΔH.
Delta H reaction (ΔH) is the amount of heat change between the system and its environment
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The standard unit is kilojoules (kJ)
The enthalpy change symbol (ΔH) is usually written behind the reaction equation.
Change in Standard Evaporation Enthalpy (ΔH vap) is a change in enthalpy at the evaporation of 1 mol liquid phase to the gas phase at its boiling point and standard pressure.
Examples of water evaporation:
H₂O (l) ---> H₂O (g); ΔH vap = + 44kJ
The enthalpy of evaporation is positive because its energy is needed to break the attraction between molecules in a liquid
relative molecular mass of butane (C₄H₁₀) = 4.12 + 10.1 = 58 gram / mol
tex]\large{\boxed{mole\:=\:\frac{grams}{relative\:molecular\:mass}}}[/tex]
mole = 2,672
Since the heat of vaporization for butane is 23.1 kj / mol, the energy needed to evaporate 2,672 moles of butane is:
23.1 kJ / mol x 2,672 mol = 61,723 kJ
<h3>Learn more</h3>the heat of vaporization
The latent heat of vaporization
Keywords: the heat of vaporization, butane, mole, gram, exothermic, endothermic
