Calculate δh∘f (in kilojoules per mole) for benzene, c6h6, from the following data: 2c6h6(l) + 15o2(g)→12co2(g)+6h2o(l) δh∘=−6534.0 kjδh∘f co2=−393.5 kj/molδh∘f h2o=−285.8 kj/mol express the enthalpy change in kilojoules per mole to three significant digits.

48.6 kJ/mol

Explanation:

The balanced chemical reaction is,

The expression for enthalpy change is,

where,

n = number of moles

(as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

Therefore, the enthalpy change for benzene is 48.6 kJ/mol

answer; protein;

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.