Aquantity 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

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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