Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bn%5Ctimes%20H_f_%7Bproducts%7D%5D-%5Bn%5Ctimes%20H_f_%7Breactants%7D%5D)
Putting the values we get :
![\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B8%5Ctimes%20H_f_%7BCO_2%7D%2B10%5Ctimes%20H_f_%7BH_2O%7D%5D-%5B2%5Ctimes%20H_f_%7BC_4H_%7B10%7D%2B13%5Ctimes%20H_f_%7BO_2%7D%7D%5D)
![\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%288%5Ctimes%20-393.5%29%2B%2810%5Ctimes%20-241.82%29%5D-%5B%282%5Ctimes%20-125.7%29%2B%2813%5Ctimes%200%29%5D)
2 moles of butane releases heat = 5314.8 kJ
1 mole of butane release heat = 
Thus enthalpy of combustion per mole of butane is -2657.4 kJ
In this question, you are given the gasoline density (0.749g/ml) and volume of the gasoline (19.2 gallons). You are asked the mass of the gasoline in pounds. Then you need to change the grams into pounds and the ml into gallons. The calculation would be:
mass of gasoline= density * volume
mass of gasoline= 0.749g/ml * (1 pound/453.592grams) * 3785.41ml/gallon * 19.2 gallon= 120 pounds
Hydrophobic molecules tend to be nonpolar molecules that group together to form micelles rather than be exposed to water. Hydrophobic molecules typically dissolve in nonpolar solvents (e.g., organic solvents).
The strong Base with a pH of 12 is reduced by 4 units upon being added with solution Y. If you added a strong acid to the strong base, all ions are present in the solution, yes? So every OH- is neutralised by every H+ for example, meaning the resultant pH should be 7. The resultant pH is only 8 however, so solution Y must be a <em>weak acid </em>only!
