Answer:
[tex]q_{rxn}=-488.86\frac{kJ}{mol}[/tex]
Explanation:
Hello!
In this case, since this calorimetry problems can by analyzed by figuring out that the heat lost due to the combustion of C6H12O6 is gained by the water and the calorimeter which undergo the mentioned temperature rise, we can write:
[tex]Q_{rxn}=-(m_wC_w+C_c)(T_2-T_1)[/tex]
Thus, by plugging in the given data, we obtain:
[tex]Q_{rxn}=-(950.0g*4.184\frac{J}{g\°C} +933\frac{J}{\°C})(42.3\°C-35.0\°C)\\\\Q_{rxn}=-17.5kJ[/tex]
Next, we compute the moles of C6H12O6 by using its molar mass (180.18 g/mol) as shown below:
[tex]n=6.45g*\frac{1mol}{180.18g} =0.0358mol[/tex]
Thus, the value for q of the reaction turns out:
[tex]q_{rxn}=\frac{Q_{rxn}}{n} \\\\q_{rxn}=\frac{17.5kJ}{0.0358mol} \\\\q_{rxn}=-488.86\frac{kJ}{mol}[/tex]
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Answer: -35,800 J
Explanation:
To solve this problem, first we an equation to relate the heat transfers involved.
qrxn+qwater+qbomb=0
This is is a consequence of the law of conservation of energy. In other words, this is true because heat can neither be created nor destroyed. Therefore, we solve for qrxn as follows.
qrxn=−[qwater+qbomb]
−[(4.184 Jg∘C)(950. g)(42.3∘C−35.0∘C)+(933 J∘C)(42.3∘C–35.0∘C)]
−(29,016 J+6,811 J)
−35,800 J
