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If the 3.21 $\mathrm{g}$ of $\mathrm{NH}_{4} \mathrm{NO}_{3}$ in Example 5.6 were dissolved in 100.0 $\mathrm{g}$ of water under the same conditions, how much would the temperature change? Explain your answer.

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Chemistry 101

Chapter 5

Thermochemistry

Lallé N.

October 30, 2020

Don't we add the mass of NH4NO3?

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So this problem is asking us to look at example 5.6 and so instead of example 5.6 where originally we dissolved 3.21 grams of ammonium nitrate in 50 grams of water. It's asking how much with a temperature change in ah, 100 grams of water. So if I cross it out So now our new condition is ah, 100 grams of water. So in the original problem, we found that the amount of energy absorbed by dissolving 3.21 grams of ammonium nitrate was one killer Jule positive in that situation. So if we do the same thing again so we know that one kill a jewel of heat is going to be absorbed by this reaction, we could do our MSL t equation. So one killer, Jule, we know that our new masses ah 100 grands arses the heat is gonna be 4.184 and then delta T is what we're solving for. So be careful because our certificate value is given in jewels and this value is killer drool. So we're gonna convert that real quick into 1000 because there are 1000 jewels in one killer, Jule ah, 100 4.184 tells a T So if I saw her delta T from this expression Aiken divide by for 18.4 it's that's lacrosse. Those out for 18.4. So 1000 divided by 4 18 0.4 is gonna give us two 0.39 degrees Celsius. So our temperature change is going to equal negative 239 points, 2.39 degrees Celsius. And that's because all the heat that is lost from our water is going into our reaction. And so therefore, this number should be negative that the temperature is going down. Thanks for watching.

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