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Glauber's salt, sodium sulfate decahydrate ( $\mathrm{Na}_{2} \mathrm{SO}_{4}$. $10 \mathrm{H}_{2} \mathrm{O}$ ), undergoes a phase transition (that is, melting or freezing) at a convenient temperature of about $32^{\circ} \mathrm{C}$ : $$\begin{aligned}\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4} & \cdot 10 \mathrm{H}_{2} \mathrm{O}(l) \\\Delta H^{\circ} &=74.4 \mathrm{kJ} / \mathrm{mol} \end{aligned}$$As a result, this compound is used to regulate the temperature in homes. It is placed in plastic bags in the ceiling of a room. During the day, the endothermic melting process absorbs heat from the surroundings, cooling the room. At night, it gives off heat as it freezes. Calculate the mass of Glauber's salt in kilograms needed to lower the temperature of air in a room by $8.2^{\circ} \mathrm{C}$ at 1.0 atm. The dimensions of the room are $2.80 \mathrm{m} \times 10.6 \mathrm{m} \times 17.2 \mathrm{m},$ the specificheat of air is $1.2 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C},$ and the molar mass of air may be taken as $29.0 \mathrm{g} / \mathrm{mol}$.

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$8.2^{\circ} \mathrm{C}=28.16 \mathrm{kg}$

Chemistry 101

Chapter 6

Thermochemistry

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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.

06:34

Glauber's salt, sodiu…

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Glauber'$ salt; sodiu…

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The lattice enthalpy of so…

01:50

Molten sodium chloride is …

01:11

Instant cold packs used to…

Okay, This is a very giant problem. We're gonna be looking a glow bird saw Glauber salt, which is sodium sulphate, Decca hydrate. Oops. And a to S 04 And Decca hydrate. Okay. And this undergoes a phase transition at temperatures of about 32 degrees Celsius. This is given it can melt or freeze at about 32 degrees Celsius. So what happens on this is this is a solid, and it can turn into in a two s 4.10 h, 20 l. So there's an S, and there's a now and our delta H for this is 74 0.4 killer jewels promote. Okay, check and see if my cameras on it is so we use this compound, or this compound can be used to regulate the temperature in homes. Um, it's put in plastic bags and put up in the ceiling. And during the day, it absorbs heat from the surroundings. So it cools the room because it sort absorbs heat, and at night it gives off heat as it reads freezes. Our job is to find the mass, and I'm just going to call it Glauber. Salt in kilograms needed to lower. Um, what are we lowering here? Lowering the air temp by 8.2 degrees Celsius at 1.0 atmosphere. Pressure. The rumor trying Teoh The dimensions of the room are 2.80 meters by 10.6 meters by 17.2 meters. Those air, the room dimensions the specific heat of air. Specific heat of air is 1.2 Jules over Graham C degrees and the molar mass of air. I'm going to call it Moeller Mass of air. We're going to take to be 29.0 grams per mole. Okay, so I think that's all my givens. Um, one other thing will be using the molar mass of the globe. Er, salt is I think it was 300 I'll have to find it. Here It was 320 ish, I think. Um, I where did I write that down? Hang on. Just a moment here. So many things. 3 22.3 grams per mole. Okay, That should be about everything. And given. Now, let's make a plan. We're going to find the volume of the room. That's super easy after we have the volume of the room. We're gonna find malls using PV nerd. And once we have moles and that will be of air, I should. Moles of air will go from Moors of Air two massive air using that Mueller Mass. And then we'll use Q equals M Cat. I actually was gonna write em Cat EMC, Delta T and will be fun solving for Q. And that'll be the heat to be removed from the air. And then we will apply. Um, we're going to go to the conservation of energy, and we know that Q plus the air or Q Air plus Q. Salt for us is going to be equal zero. And then we're going to do our cue for the salt. We're gonna call that end times the heat of fusion for the salt, which we were given, and the molar mass of the salt love abides massive salt divided by the molar mass of the Salt Times heat effusion for the salt, which we have blah, blah, blah. Anyway, I'll go through that more detail, but anyway, I think we are ready to start. Let's start with Step one by volume of the room, and that's very simple because we're just going to take 2.80 meters times 10.6 meters times 17.2 meters and we're going to get, um And that was going to give me meters cubed. So I think I'm just gonna do that'll give me my meters cubed. And then I know that there's 1000 leaders in one meter cubed, so I'll just go ahead and take care of that one fell swoop. So the volume of the room will be 5.10 times. Tend to the fifth Leaders. That's my volume. Okay, step two, I'll switch colors here. Step two. We were going to use PV equals NRT to solve for n so n is going to equal p 1.0 atmospheres. V, which is 5.10 times tend to the fifth leaders are will be 0.8 to 1. L A t m overcame all and I forgot my temperature. My temperature was 32 degrees Celsius. So 32 plus 2 73 is 305 Okay, do your math on this and end of air, this be moles of air will equal 2.4 times 10 to the fourth Moles. That's moles of air. Okay, now we're going to convert moles of air to massive air. And we'll do that by using the molar mass that was given. This is air again, and I believe it was 29.0 grams per mole. And that's 5.9 times 10 to the fifth grams of air. Okay, so we're making progress now it's time to use Q equals M See Delta T and Mass. We just figured out and wrote down as by 0.9 times 10 to the fifth grams. See, we were given as 1.2 j over G C degrees C degrees and our delta T waas 8.2 degrees were losing that heat. So do our math here, and that's going to give us negative 5.8 times 10 to the six. Jules, let's go ahead and just get that into killer jewels. And that's the heat that we need to remove from the room. Okay, And one more think about step here. I think I'm gonna go ahead and start this one on another page. Make sure my cameras don't, and it is Okay, so now we know Here's some things that we know. We know that the cue of the air, plus the cue of the salt, has to equal zero. That's what we're aiming for from here. We know. Weaken substitute Q of air plus instead of Q. We're gonna go Cayugas n times the heat of fusion and this is for the salt equals zero in Q of air plus now n we can take the mass of salt divided by the Moeller Mass of salt because that's in times my heat diffusion equals zero. And then let's solve for Let's solve this for mass of salt because that's what we're looking for here. So if we're solving this for massive salt, that's gonna equal negative que for the air, which we calculate it times the molar mass molar mass of the salt divided by my heat of fusion. There. Now what we have to do is substitute and solve into this. Let's do it. Mass of salt equals we have a negative of Q waas. Um, negative. 5.8 times 10 to the third killer jewels. My molar mass of my Salt Waas 3 22 0.3 grams per mole. And last but not least, this is way back from my chemical equation back on page one, killer jewels per mall and let me go back. That's, um I don't even see where he got it. Oh, I've got one more page back, That's why. Boy, did I use that many pages on here. I shouldn't wasted your time with this. It's this number right there. See if I can get back faster. No. Okay, so now we have to do is solve this. So the massive assault Put these numbers into your calculator and you are going to get 2.5 times 10 to the fourth grams we were asked to solve in kilograms, so that would be 2.5 times. Tend to the first kilograms or 25 kilograms. You? That was a big one.

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