contestada

Commercial cold packs consist of solid ammonium nitrate and water. NH 4NO 3( s) absorbs 25.69 kJ of heat per mole dissolved in water. In a coffee-cup calorimeter, 3.40 g NH 4NO 3( s) is dissolved in 100.0 g of water at 21.0 °C. What is the final temperature of the solution? Assume that the solution has a specific heat capacity of 4.18 J/g•K.

Respuesta :

okpalawalter8

Answer:

The final temperature is [tex]T_F = 18.39^oC[/tex]

Explanation:

Generally number of moles is mathematically represented as

               [tex]n = \frac{mass}{Molar \ Mass}[/tex]

For the given compound NH 4NO 3( s)

      The molar mass would be = [tex]2 M_{N} + 4M_{H} + 3 M_{0}[/tex]      

  Now substituting 14 for [tex]M_{N}[/tex](molar mass of Nitrogen) , 1 for [tex]M_{H}[/tex] molar mass of hydrogen, 16 for [tex]M_{0}[/tex] molar mass of oxygen

    The molar mass of the compound  [tex]=14 *2 + 4*1 + 16*3[/tex]

                                                               [tex]= 80 g/mole[/tex]

  The mass of the compound dissolve is 3,40 g

Substituting into the above equation

                 [tex]n =\frac{3.4}{80}[/tex]

                    [tex]= 0.0425 \ moles[/tex]

Given in the question that  1 mole of NH 4NO 3( s)  absorbs 25.69kJ

                       then 0.0425 moles would absorb  x

 Now stating this mathematically

                  1 → 25.69 kJ

             0.0425 → x

cross-multiplying and making x the subject

               [tex]x=0.0425 * 25.69*10^3[/tex]

                  [tex]=1091.83J[/tex]

Generally the heat energy absorbed  is mathematically denoted as

             [tex]Q = mc_p \Delta T[/tex]

substituting  100.0 g  for m (mass of water) , 4.18 J /g.K for [tex]c_p[/tex](Specific heat capacity)

Making [tex]\Delta T[/tex] the subject

         [tex]\Delta T = \frac{Q}{m c_p}[/tex]

               [tex]= \frac{1091.83}{100 * 4.18}[/tex]

               [tex]= 2.61^o[/tex]C

The initial temperature is 21.0°C

Since is been absorbed by NH 4NO 3 when it dissolves in water in mean that water losses energy so it temperature would reduce

       Therefore final temperature is

                    [tex]T_F = 21 -2.61[/tex]

                   [tex]T_F = 18.39^oC[/tex]

           

               

                             

Eduard22sly

The final temperature of the solution is 18.4 °C

We'll begin by calculating the number of mole in 3.4 g of NH₄NO₃.

Mass of NH₄NO₃ = 3.4 g

Molar mass of NH₄NO₃ = 14 + (4×1) + 14 + (16×3) = 80 g/mol

Mole of NH₄NO₃ =?

Mole = mass / molar mass

Mole of NH₄NO₃ = 3.4 / 80

Mole of NH₄NO₃ = 0.0425 mole

  • Next, we shall determine the heat absorbed by 0.0425 mole of NH₄NO₃

1 mole of NH₄NO₃ = 25.69 KJ = 25690 J

Therefore,

0.0425 mole of NH₄NO₃ = 0.0425 × 25690

0.0425 mole of NH₄NO₃ = 1091.825 J

  • Finally, we shall determine the final temperature of the solution.

Mass (M) = 100 g

Initial temperature of water (T₁) = 21 °C

Specific heat capacity (C) = 4.18 J/gºC

Heat released (Q) = –1091.825 J

Final temperature (T₂) =?

Q = MC(T₂ – T₁)

–1091.825 = 100 × 4.18 (T₂ – 21)

–1091.825 = 418 (T₂ – 21)

Clear the bracket

–1091.825 = 418T₂ – 8778

Collect like terms

–1091.825 + 8778 = 418T₂

7686.175 = 418T₂

Divide both side by 418

T₂ = 7686.175 / 418

T₂ = 18.4 °C

Therefore, the final temperature of the solution is 18.4 °C

Learn more: https://brainly.com/question/18250635

Otras preguntas