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kogti [31]
3 years ago
5

Lab question: When a student chemist transferred the metal to the calorimeter, some water splashed out of the calorimeter. Will

this technique error result in the specific heat of the metal as being reported too high or too low?
Chemistry
1 answer:
Natasha2012 [34]3 years ago
5 0
<span>The temperature change is greater when assuming that the mass is constant and the heat must be greater </span>in the result the heat of the metal as being reported too high.

hope it helps
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N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol
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A

Explanation:

Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:


\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)

Therefore, from the chemical equation, we have that:


\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} +  \Delta H^\circ_f \text{ H$_2$O}  \right]   -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}

In conclusion, our answer is A.

5 0
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