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A 100.0 mL sample of 1.020 M HCl is mixed with a 50.0 mL sample of 2.040 M NaOH in a Styrofoam cup. If both solutions were initially at 24.53°C, and the enthalpy of the neutralization reaction is −57 kJ/mole of H2O formed, what is the final temperature of the mixture? Assume that the solution has a density of 1.00 g/mL and a specific heat of 4.184 J/g°C, and that the Styrofoam cup has an insignificant heat capacity.

Respuesta :

Answer:

[tex]t_2 = 33.793[/tex]

Explanation:

Given data:

0.1 L HCl × 1.020 mol/L = 0.102 mol HCL

0.05 L NaOH × 2.040 mol/L = 0.102 mol NaOH

NaOH + HCl \rightarrow  H_2O + NaCl

0.102 mol HCl /1 mol HCl × 1 mol H2O = 0.102 mol H2O

0.102 mol H2O / 1 mol H2O × 57000 J = 5814 J

[tex]V(t ) = V_1+V_2 = 100 + 50 = 150 mL [/tex]

M(t)=150*1=150 g

we know that heat energy is given as

[tex]Q = mc(t_2 - t_1)[/tex]

where c is specific heat

total energy released is = 57 \times 0.102 = 5.814 kJ

[tex]5814 = 150 \times 4.184 \times (t_2 - 24.53)[/tex]

[tex]t_2 = 33.793[/tex]

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