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10 November, 08:14

A quantity of 4.00? 102 ml of 0.600 m hno3 is mixed with 4.00? 102 ml of 0.300 m ba (oh) 2 in a constant-pressure calorimeter of negligible heat capacity. the initial temperature of both solutions is the same at 16.91°c. what is the final temperature of the solution? (assume that the densities and specific heats of the solutions are the same as for water (1.00 g/ml and 4.184 j/g·°c, respectively). use - 56.2 kj/mol for heat of neutralization.) webassign will check your answer for the correct number of significant figures. °c

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  1. 10 November, 08:25
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    The overall reaction formula for this neutralization would be:

    Ba (OH) 2 + 2HNO3 = Ba (NO3) 2 + 2H2O

    The heat of neutralization is based on the number of moles released.

    Calculating for the number of moles of the reactants where moles can be calculated as: moles = molarity * volume

    moles of Ba (OH) 2 = 0.300 M * 0.4 L = 0.12 moles Ba (OH) 2

    moles of HNO3 = 0.600 M * 0.4 L = 0.24 moles HNO3

    In the reaction, 2 moles of HNO3 is needed per 1 mole of Ba (OH) 2, therefore the reactants react completely. The moles of Ba (NO3) 2 = moles of Ba (OH) 2.

    Therefore the heat of neutralization, H is:

    H = ( - 56.2 kJ / mol) * 0.12 mol

    H = - 6.744 kJ

    Converting the Cp to J/°C:

    Cp = (4.184 J/g·°C) (1 g/mL) (400 mL + 400 mL) = 3347.2 J / °C

    Calculating for the final temperature using the formula of enthalphy:

    H = - Cp * (Tf - Ti)

    6,744 J = (3347.2 J / °C) (Tf - 16.91°C)

    Tf = 18.92 °C
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