A 54.4 -mL dilute solution of acid at 23.95 °C is mixed with 50.1 mL of a dilute solution of base, also at the same temperature as the acid, in a coffee-cup calorimeter. After the reaction occurs, the temperature of the resulting mixture is 20.84°C. The density of the final solution is 1.03 g/mL. Calculate the amount of heat evolved. Assume the specific heat of the solution is 4.184 J/g•°C. The heat capacity of the calorimeter is 23.9 J/°C.
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First, we need to find the total mass of the final solution. Since the density is given as 1.03 g/mL, we can multiply this by the total volume of the solution (54.4 mL + 50 mL) to find the mass: Total mass = (54.4 mL + 50 mL) * 1.03 g/mL = 104.4 mL * 1.03 g/mL = Show more…
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In a coffee-cup calorimeter, $150.0 \mathrm{~mL}$ of $0.50 \mathrm{M} \mathrm{HCl}$ is added to $50.0 \mathrm{~mL}$ of $1.00 \mathrm{M} \mathrm{NaOH}$ to make $200.0 \mathrm{~g}$ solution at an initial temperature of $48.2^{\circ} \mathrm{C}$. If the enthalpy of neutralization for the reaction between a strong acid and a strong base is $-56 \mathrm{~kJ} / \mathrm{mol}$, calculate the final temperature of the calorimeter contents. Assume the specific heat capacity of the solution is $4.184 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}$ and assume no heat loss to the surroundings.
A coffee-cup calorimeter is used to determine the heat of reaction for the acid-base neutralization $\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}) \longrightarrow$ $$ \mathrm{NaCH}_{3} \mathrm{COO}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) $$ When we add $20.00 \mathrm{~mL}$ of $0.625 \mathrm{M} \mathrm{NaOH}$ at $21.400^{\circ} \mathrm{C}$ to $30.00 \mathrm{~mL}$ of $0.500 \mathrm{MCH}_{3} \mathrm{COOH}$ already in the calorimeter at the same temperature, the resulting temperature is observed to be $24.347^{\circ} \mathrm{C}$. The heat capacity of the calorimeter has previously been determined to be $27.8 \mathrm{~J} /{ }^{\circ} \mathrm{C}$. Assume that the specific heat of the mixture is the same as that of water, $4.184 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C},$ and that the density of the mixture is $1.02 \mathrm{~g} / \mathrm{mL}$. (a) Calculate the amount of heat given off in the reaction. (b) Determine $\Delta H$ for the reaction under the conditions of the experiment.
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