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Specific Heat Capacity (C) The energy transferred as heat that is required to raise the temperature of gram of a substance by kelvin_m x Cp xATq= heat lost or gaine...

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Specific Heat Capacity (C) The energy transferred as heat that is required to raise the temperature of gram of a substance by kelvin_m x Cp xATq= heat lost or gained m= mass of solution (grams) the Specific Heat Capacity of compound (Jllg x 'C)) Trnal - Teul AHsolution moles of saltIf 1,.25 g of ammonium nitrate (NHANOs) is dissolved in 25.0 mL of water, the following data are collected: starting temperature is 25 final temperature after dissolving is 21.9 "C. Calculate the enthalpy of

Specific Heat Capacity (C) The energy transferred as heat that is required to raise the temperature of gram of a substance by kelvin_ m x Cp xAT q= heat lost or gained m= mass of solution (grams) the Specific Heat Capacity of compound (Jllg x 'C)) Trnal - Teul AHsolution moles of salt If 1,.25 g of ammonium nitrate (NHANOs) is dissolved in 25.0 mL of water, the following data are collected: starting temperature is 25 final temperature after dissolving is 21.9 "C. Calculate the enthalpy of solution in Joules You may assume density of 00 glmL for the solution and you may assume that the specific heat is that of water (this is a dilute solution) and is 4.18 Ji(g*"C) Is this an endothermic or exothermic reaction? If 3.8 g of calcium chloride (CaClz) is dissolved in 100.0 mL of water; Ihe following data are collected: starting temperature is 25.8 'C final temperature after dissolving is 29.9 'C Calculate the enthalpy of solution in Joules. You may assume density of 1.00 glmL for the solution and you may assume that the specific heat is that of water (this is dilute solution) and is 4.18 JI(g* C) Is this an endolhermic Or exothermic reaclion?



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Enthalpy A 100 .-g sample of water is placed in an insulated container and allowed to come to room temperature at $21^{\circ} \mathrm{C}$. To heat the water sample to $41^{\circ} \mathrm{C}$, how much heat must you add to it? Consider the hypothetical reaction, $$ 2 \mathrm{X}(a q)+\mathrm{Y}(l) \longrightarrow \mathrm{X}_{2} \mathrm{Y}(a q) $$ being run in an insulated container that contains $100 . \mathrm{g}$ of solution. If the temperature of the solution changes from $21^{\circ} \mathrm{C}$ to $31^{\circ} \mathrm{C}$, how much heat does the chemical reaction produce? How does this answer compare with that in part a? (You can assume that this solution is so dilute that it has the same heat capacity as pure water.) c If you wanted the temperature of $100 . \mathrm{g}$ of this solution to increase from $21^{\circ} \mathrm{C}$ to $51^{\circ} \mathrm{C},$ how much heat would you have to add to it? (Try to answer this question without using a formula.) d If you had added 0.02 mol of $X$ and 0.01 mol of $Y$ to form the solution in part $\mathrm{b}$, how many moles of $\mathrm{X}$ and $Y$ would you need to bring about the temperature change described in part c. Judging on the basis of your answers so far, what is the enthalpy of the reaction $2 \mathrm{X}(a q)+\mathrm{Y}(l) \longrightarrow \mathrm{X}_{2} \mathrm{Y}(a q) ?$

Let's consider the reaction of copper sulfate and potassium hydroxide for part A before mixing how many grams of copper are present in the copper sulfate solution. So we'll calculate the molds of copper sulfate mhm, which be the polarity times. The volume clarity is point or 1.0 Moller Yeah, times the volume, but the volume is 50 mL and 15 millimeters in leaders to be 500 leaders, and then the moles would therefore be 0.500 moles. So this would be the moles of copper sulfate 500 moles. And we want to do yeah, massive copper. So in order to calculate the mass of copper, um, and one more Let's see what. So for just one more of copper, it's a public two plus and one more of copper. Two plus has a mass of 63.54 g, and this would be equal to 3.18 grams of copper. Two plus present. Um, Before mixing for part B, we are asked to predict the identity of the precipitate. So we have copper sulfate to react calcium hydroxide to produce potassium sulfate, which is a clear and copper hydroxide. Okay, solid to this here, Would you have precipitates which would be copper hydroxide to balance? This will put a two in front of the college on the left hand side for part C right to complete and not Ionic equation for the reaction occurs when these tools solutions are mixed. So yeah, the complete ionic equation. Yeah, would be the U two plus, Yeah, that's a four to minus u k plus mhm two inch minus mhm. You pick. Plus, yeah, was a 14 minus. You know, we two solid then. Based on this, the non ionic equation eliminating Spectators eliminate the K plus in the 14 minus so ornate ionic equation to be copper two. Plus, you work with this use of precipitated copper hydroxide and part D, whereas to calculate the Delta H for the reaction upon mixing so Delta eat is equal to Q, which is equal to M c Delta T Yeah, this vehicle to the masses, 100 grounds yeah, and cc capacity for 24 fuels. For Graham Calvin Delta T here is negative 6.2 Calvin and this would work out to negative two points six humor, jewels. This would be the value for Delta age for this reaction

The topic is about competing for the heat associated when the substance changes in temperature, the amount of it in order to change the temperature of the substance is solved as Q equals M. C. Delta T. Where DELTA T. Is equal to the difference of the T. F. And T. I. Where TF is the final temperature and Tia is the initial temperature. M. Is the mass and C. Is the specific it, which value is determined by knowing what type of material or substance there is. Now let us suppose we are given with water initially are initially the volume of the 250 ml of water is set at 20 degrees Celsius and it is, its temperature is raised to 35 degrees and shoes. We wish to find the queue that is needed in order to cost this change in temperature and that also the heat where the queue that is needed to call it them. We note that water has a density that is usually taken to be equal to one g per ml as such. Even with density and volume result for the masses, the product of density and volume. So density is one g per ml and the volume is 250. In other words, M where the mass is also equal to 250 g for water, The specific heat is equal to 4.184 jews program degree cell shows so again specific it is a property of the material, knowing what type of material or substance that is. You can know what specific property value, but what specific heat capacity value you will use. So now that the salt for the Q. We have M. C. D. F minus T. A. And this is equal to 250 Times 4.184 times 35 minus 20. This gives us the value. Yeah. 15,690, Jews. Yeah. Mhm. Yeah. Next we find the heat that is associated when it's uh Cool back to 20. So in this case RTF will be 20 and R. T. I will be 35. So that is 202,050 Times 4.184 times 20 minus 35. And this is equal to -15619. The negative sign indicates that hit is lost as stop cooling process proceeds. So this is the heat needed to raise the temperature and this is the heat that is needed to call the temperature by the same degree. So I hope everything is clear

There we are continuing on with the topic of heat and energy and energy transfer. And so the first thing we can do here is determine the total mass of the solution, which is the sum of the mass of NH- four n. 03 and water. The mass is equal to 25, g. We have 275g. It's on maths and so Q. P. Is equal to negative of the specific heat capacity. Multiplied by the mass, multiplied by the change in temperature. We plug in our valley's we got 851 4.4 units of jewels. And so we can convert the quantity of energy transferred for 25 g to the end will be changed for bar molo 80 g of NH four N. 03. Where delta H. North seawater negative 8514.4 units of jewels multiplied by 80 g divided by 2 50 g. What we're left with is negative 27.2 units of kilo jewels. So the entropy change for the reaction is negative 27.2 units of kilo jewels. And because the sign of entropy changes, negative energy is released in this reaction

Good day. The topic is about heat. When the substance absorbs or loses heat. This heat can cause or may cause a change in its temperature. And the heat that is associated with the temperature change of the substance is solved as Q. Equals M. C. Delta T. Where Q. Is the heat. And Mr. Mass sees the specific it, which value depends on the type of substance and all that is the change in the temperature which can be solved as the difference in the final temperature and the initial temperature. Now it could also it could also be the the heat that is subscribed and lost by the system, may not cause the change in its temperature but rather on its face. And that heat is quantified as Q equals M. L. Where L. Is the latin heat. And depending on which type of uh face change the substance undergoes L. Can can take different names. So like for example when the substance undergoes vaporization then and here is the latin heat of vaporization. Now let us consider A container that contains 200 g of water which here is a. And in this iniquity, presume with 50 g of ice b. And in it is placed 30 g of water, see Which is temperature is at 90° and shoes. Now, we wish to find the final condition of the system that is after mixing this hotter water. What will happen to the system that this will there be ice or they're all with ice cold milk or if all melts, what would be the final temperature and so on and so forth. Now. Since the temperature of since liquid water A. Is in equilibrium with ice, it follows that the temperature of A. And B. Are both equal to zero degrees self shoes. So let us suppose in this case. All right there like this Sold for the amount of heat 1st that liquid water that is poured into it. So that's Qc. How much how much heat can can it release if its temperature drops from 92 0°C. So that is if The the 30 g of water it's added goes all the way down to the freezing temperature of water then how much heat would be released because that heat would be absorbed by the ice and the water that's already in the container. So that's all that since there's only change in temperature that's involved, then we will use the formula Q. Equals M. C. Delta T. Yeah. In other ways of putting it CTF minus D. A. So this is the heat released by what they're seeing. So Q. C. is equal to the mass of that water that's put in his 30 the specific it of uh of water. It's used to the calorie. That's one calorie program, decrease our shoes. And then we assume the temperature drops to 90 from 90 to 0. So that's zero minus 90. So the heat that can be lost when um 30 g sample Cools down all the way to 0°C is negative, 2700 Kyle. And let us assume that? Let us let us now solve how much ice with milk. If these 2,700°C shoes are 2700 calories is absorbed. So since melting is a process of that is that involves face change. So we will use Q. Equals um And so ice how much of the ice and we will melt If 2700 calories of heat are absorbed. So that will be M. V. Times. Now the latin heat of uh fusion specifically because fusion is retains to melting From our reference for water is equal to 80 Kyle program. Okay so let me write that one. The latitude official is a tickle program for what they're. So let's solve for M. V. So that will be 2700 divided by 80. And this is equal to 33 0.75 g. So what does this say if The water that sport in that at 90°C And it's it goes down from 90 all the way to the freezing point that is losing 2700 calorie. It will cause 33.75 g of ice to melt. And let us recall how much ice is. It is in the Contain. It's 50g. So that means 33.75 out of 50g of the ice will melt remaining. That means the remain the remainder will still remain to be a nice And what's more interesting or what's more what what what another thing that we can interpret out of that is that since not all I smells, it follows that the temperature will remain to be at the freezing temperature of ice, which is 0°C. So to put it, we have, so we say that Only 33.7 g of ice melts, leaving 16.25g and melted so that the final mixture will still contain liquid water and ice and 0°C. So I hope that is clear.


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