Answer:
-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
When given the symbol, you can look on the periodic table of elements to find what the atomic number is, which is the number in the corner next to the symbol. The number of protons in the element is the atomic number. The number of neutrons is the mass number minus the number of protons.
Hope it helps:)
It is a single replacement reaction. Li and Mg switch places.
Answer:
depth of the liquid and nature of liquid affects the pressure due to liquid contained in a vessel
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