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An 0.1250 g sample of methanol was burnt in a bomb calorimeter containing 100.0 mL of water. The temperature rose from 17.33 to 23.73° C. Calculate ∆ H for the combustion reaction.
\[\text{First find the heat of water, Q,} \\ \\ \text{They have given you} \\ \\ \text{V=100mL=100g} \\ \\ \text{ΔT=6.4°C}\] \[\text{Using the formula} \\ \\ \text{Q=mcΔT} \\ \\ Q=100(4.184)(23.73-17.33) \\ \\ Q=2677.76J\] \[\text{Now, Use this equation \to find ΔH for reaction} \\ \\ \Delta H = \frac{Q}{n} \\ \\ \text{The reaction requires 'n' which is number of moles of substance,} \\ \\ \text{We can find that by using the mass given,} \\ \\ n=0.125g~of~CH_3OH \times \frac{1~mol~of~CH_3OH}{32~g~of~CH_3OH}=3.91 \times 10^{-3}moles~of~CH_3OH\] \[\text{Using back the equation,} \\ \\ \Delta H=\frac{Q}{n}=\frac{2677.76J}{3.91 \times 10^{-3}mol}=684849J/mol~~~or~~~685kJ/mol\]