Answer:
[tex]\Delta _mH=63J/g[/tex]
Or:
[tex]\Delta _mH=12.4kJ/mol[/tex]
Explanation:
Hello,
In this case, since the required heat to turn solid gold into liquid gold is computed by:
[tex]Q=m\Delta _mH[/tex]
Which is mass and enthalpy of fusion- dependent, we compute the enthalpy of fusion as shown below:
[tex]\Delta _mH=\frac{Q}{m} =\frac{756J}{12g}\\ \\\Delta _mH=63\frac{J}{g}[/tex]
Or in kJ/mol (typical data):
[tex]Delta _mH=63\frac{J}{g}*\frac{1kJ}{1000J} *\frac{197g}{1mol}\\ \\Delta _mH=12.4kJ/mol[/tex]
Best regards.
