The <span>molecule, when added to a solution, would give the solution the highest pH is KOH. This is because when KOH is in the solution, it will give a very strong basic properties due to the -OH and the K which is highly electropositive.</span>
The heat of a chemical reaction (change of enthalý) may be calcualted as the heat of formation of the products less the heat of formation of the reactants.
Then, for the given equation, the total heat of the combustion of pentnae is:
<span>ΔH°f = 5* ΔH°f CO2(g) + 6ΔH°f H2O(g) - ΔH°f C5H12(g) - 8ΔH°f O2(g)
</span>The standard heat of formation of O2(g) is zero because that is its natural state.
Now you can replace in the equation the values given:
<span>ΔH° = 5*(-393.5 kJ/mol) + 6*(-241.8kj/mol) - (-35.1kJ/mol) - 0
ΔH° = -3,383.2 kJ/mol
</span>
Then, the heat of the combustion of pentane is -3,383.2 kj/mol
Answer:
Same number of protrons, and different number of neutrons :)
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