The standard heat of formation of a branched alkane is -35 kcal/mol; the standard heat of formation of the unbranched version of alkane (same molecular formula) is -28 kcal/mol. Finally, the standard enthalpy of combustion of the branched alkane is -632 kcal/mol. Given this information, what is the standard enthalpy of combustion of the unbranched alkane in terms of kcal/mol to the nearest ones?

-625 kcal/mol

Explanation:

The method to solve this question is based on Hess´s law of constant heat of summation which allows us to combine the enthalpies of individual reactions for which we know their enthalpy to obtain the enthalpy change for a desired reaction.

We are asked to calculate the standard enthalpy of formation  of combustion of an unbranched alkane :

CnHn+2  unbranched  + O₂  ⇒ CO₂ + H₂O  ΔcHº = ?

where CnH2n+2 is the general formula for alkanes.

and we are given information for

n C+ (2n + n)/2 H₂  ⇒ CnHn+2  unbranched    ΔfHº = -35 kcal/mol     (1)

n C+ (2n + n)/2 H₂  ⇒ CnHn+2  branched        ΔfHº = -28 kcal/mol     (2)

CnHn+2  branched + O₂  ⇒ CO₂ + H₂O          ΔcHº = -632 kcal/mol  (3)

If we reverse  (1)  and add it to the sum (2) and  (3)  we get the desired equation for the combustion of the unbranched alkane:

CnHn+2  unbranched  + O₂  ⇒ CO₂ + H₂O  

Thus

ΔcHº  unbranched = + 35 kcal/mol + (-28 kcal/mol) + (-632 kcal/mol)

                                = -625 kcal/mol

answer: i have one answer

acids are substances that ionize (break off) in an aqueous solution to produce hydrogen (h+) ions

explanation:

hope this and sorry i could only come up with one answer!