100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temperature rises from 21.80 °c to 23.20 °c. assuming the density and specific heat of the resulting solution is 1.00 g/ml and 4.18 j/g ∙ °c, respectfully, calculate how much heat (q) was released. group of answer choices 19.4 kj 598 j 586 j 1.17 kj

Respuesta :

ΔH°rxn = 39013.33 J/mol

           = 39.013 kJ/mol.

What is Calorimeter?

Calorimetry is a branch of science that measures a body's status with regard to temperature in order to look at changes in its physical and chemical composition. The changes could be chemical, like burning, acid-base neutralisation, or physical, such melting or evaporation.

We can calculate the amount of heat (Q) released from the solution using the relation:

Q = m.c.ΔT,

Where, Q is the amount of heat released from the solution (Q = ??? J).

m is the mass of the solution (m of the solution = density of the solution x volume of the solution = (1.0 g/mL)(200 mL)

                                      = 200 g.

c is the specific heat capacity of the solution (c = 4.18 J/g∙°C).

ΔT is the difference in the T (ΔT = final temperature - initial temperature = 23.20 °C - 21.80 °C = 1.4 °C).

∴ Q = m.c.ΔT

     = (200 g)(4.18 J/g∙°C)(1.4 °C) = 1170.4 J.

∵ ΔH°rxn = Qrxn/(no. of moles of AgNO₃).

Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.

M = (no. of moles of AgNO₃)/(Volume of the solution (L)).

∴ no. of moles of AgNO₃ = (M)(Volume of the solution (L))

                                          = (0.3 M)(0.1 L) = 0.03 mol.

∴ ΔH°rxn = Qrxn/(no. of moles of AgNO₃) = (1170.4 J)/(0.03 mol)

                 = 39013.33 J/mol

                 = 39.013 kJ/mol.

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