7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c. The temperature of the water is observed to rise by 2.316c. (You may assume all the heat released by the reaction is absorbed by the water, and none by the calorimeter itself.) Calculate the standard heat of formation of Compound x at 25c. Be sure your answer has a unit symbol, if necessary, and round it to 3 significant digits.

Explanation:

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In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

We plug in the mass of water, temperature change and specific heat to obtain:

Now, this enthalpy of reaction corresponds to the combustion of propyne:

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

Now, we solve for the enthalpy of formation of C3H4 as shown below:

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

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the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

it is b that is the answer.