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How much heat is produced when 100 $\mathrm{mL}$ of 0.250 $\mathrm{M} \mathrm{HCl}(\text { density }, 1.00 \mathrm{g} / \mathrm{mL})$ and 200 $\mathrm{mL}$ of 0.150 $\mathrm{M} \mathrm{NaOH}$ (density, 1.00 $\mathrm{g} / \mathrm{mL} )$ are mixed?$\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H_{298}^{\circ}=-58 \mathrm{kJ}$If both solutions are at the same temperature and the heat capacity of the products is $4.19 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C},$ how much will the temperature increase? What assumption did you make in your calculation?

$\Delta T=1.15^{\circ} \mathrm{C}$

02:35

Aadit S.

Chemistry 101

Chapter 5

Thermochemistry

Rice University

Brown University

University of Toronto

Lectures

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start with their neutralization reaction. We have hcl queers. That's any wage. Take this to produce. And a ceo the nucleus to a liquid Question. We're told that we have 100 ml of 2, 5 or more H. C. O. And the density is one g per middle leader. Therefore this would be 100 g And we have 200 ml of .150 Mueller. And the O. H density is also one g from the leader. Therefore this is 200 g of salt for the moles of each. C L c 10 3 25 0 moller convert the volume Two leaders. Look at 2.0 to 50 moles for the hcl and moles have any ohh c times three 4150 moller 0.0 point 200 leaders. That works out to point 0300 moles. Uh this is the least amount. So this is the limiting. Okay, key captain. So our empathy here, the amount of heat released is equal to. Mhm. Uh this is negative 58 kilo joules per mole mm hmm. Times .0250 moles. This would give us -145 killer jewels. Therefore 1.45 killer jewels of heat is released. And based on this, what is the temperature change? Q. Is equal the M. C. Delta T. 145 Killer Jewels. The mass is the mass of the hcl in any O. H. Which is 300 g. See is there heat capacity here? Which is uh Now this is in terms of grams here, jules here. So let's back this up here. Uh killed jewels including 10,000 tools. 300 g Heat capacity here is a 4.19 mhm jules per gram degrees Celsius and delta T. Is our unknown solving us for delta T. We find the change in temperature is equal to 1.15 degrees Celsius. Let's state the assumptions that were made here. Yeah. First assumption the amount of heat absorbed or released. Yeah. Yeah. Yeah. By the substance or solutions. Yeah, depends upon the specific heat. Yeah, temperature change. Yeah. And mouse. Mhm. Second the calorie meter does not absorb any heat. Mhm. Yeah. From the reaction. Yeah. And there is no heat exchange. Yeah. Oh. Mhm. Mhm. Between the calorie meter and its surroundings. Yeah. Mhm. Mhm. And third the standard entropy of formation is not different. Yeah. Mhm. Yeah. Mhm. Yeah. Mhm. Mhm. Than the heat of combustion. Mhm. Yeah. Mhm. Or heat produced for the reaction. Mhm. Yeah. Mm. Which occurs at constant pressure. Yeah. Come before nothing. Mhm. Okay. Therefore the pressure of the whole process. Mhm. Mhm. Okay. Yeah. Yeah. Yeah. Mhm. Yeah. Is assumed to be constant. Okay. Mhm. Mhm.

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