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atroni [7]
3 years ago
8

What is the total amount of heat released when 94.0 g water at 80.0 °C cools to form ice at −30.0 °C?

Chemistry
1 answer:
Wewaii [24]3 years ago
6 0

Answer:

The total amount of heat released  is  68.7 kJ

Explanation:

Given that:

mass of water = 94.0 g

moles of water = 94 / 18.02 = 5.216

80⁰C   ------>  0⁰C  -------->    -30⁰C

Q1 = m Cp dT

      = 94 x 4.184 x (0 - 80)

     = -31463.68 J

     = -31.43 kJ

Q2 = 6.01 x 10^3 x 5.216

    = - 31348.16 J

   = -31.35 kJ

Q3 = - 94 x 2.09 x 30

    = - 5893.8 J

   = -5.894 kJ

Total heat = Q1 + Q2 + Q3  = -31.43 kJ  + (-31.35 kJ  ) + (-5.894 kJ )  = -68.7 kJ

Total heat released = -68.7 kJ

Note that the "negative sign" simply indicates heat released, therefore no need to put it in the answer.

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Answer:

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Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

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\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}

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\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}

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