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A $3.53$ - $\mathrm{g}$ sample of ammonium nitrate $\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)$ was added to $80.0 \mathrm{mL}$ of water in a constantpressure calorimeter of negligible heat capacity. As a result, the temperature of the water decreased from $21.6^{\circ} \mathrm{C}$ to $18.1^{\circ} \mathrm{C}$. Calculate the heat of solution $\left(\Delta H_{\text {soln }}\right)$ of ammonium nitrate.

$\Delta H_{s o l n}=26625 \frac{J}{m o l}=26.625 \frac{k J}{m o l}$

Chemistry 101

Chapter 6

Thermochemistry

University of Maryland - University College

University of Toronto

Lectures

05:27

In chemistry, a chemical reaction is a process that leads to the transformation of one set of chemical substances to another. Both reactants and products are involved in the chemical reactions.

06:42

In chemistry, energy is what is required to bring about a chemical reaction. The total energy of a system is the sum of the potential energy of its constituent particles and the kinetic energy of these particles. Chemical energy, also called bond energy, is the potential energy stored in the chemical bonds of a substance. Chemical energy is released when a bond is broken during chemical reactions.

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So for this problem, we want to find the heat of solution when some sample of ammonium nitrate is added to water so really quickly to clarify. The seed of solution is the amount of heat that is released per mole of whatever is put into the waters in this case and be promoted ammonium nitrate. So to solve this problem, we're gonna take advantage of that heat of solution equal to math times, a specific heat capacity, times a change in temperature. And this is all from the perspective of water. So first we want to find the mess. We know that there is 80 milliliters of water and that's an information that's given to us in the problem. And we also know that because the density of water is 01 mil, leader of water is one gram of water, so 88 milliliters corresponds to 80 Gramps. We also know that the specific heat capacity water is 4.18 for very useful constant. And we also know that the change in temperature can you calculate because we know what the temperatures are. The final temperature is 18.1 degrees Celsius and the initial temperature is 21.6 degrees Celsius. So when you plugged it all in, you will get that. The amount of heat that it released in the solution is negative 1171.52 jewels. Or he could simply abbreviate this as negative 1.2 killer jewels just to abide by 66 So now that we have the amount of heat released, we want to find how many moles of ammonia night we were actually used so that we can calculate the moulder heat of solution. So we also know that there was 3.53 grams ammonium nitrate, so I'll have to do is divide by the molar mass of ammonium nitrate. When you calculate the molar mass of ammonium nitrate, you'll get that the molar masses 80 point zero for three grams. When you multiply that all out, you will live that there is 0.44 moles of ammonium nitrate. It's now all we need to do is take the heat released, divided by the moles, and that the final answer it will be negative. 26.6 to 5. Cure Jewell's for more or because the final answer specifically asked for the heat of solution. So we know that we have to reverse the sign because the ammonium nitrate is the surroundings in this problem and the water is the system. We just changed it to a positive, and that is the final answer.

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