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Znceg cnemicaNals) Oz (&}NazOls)adaEeWhat mass of Na in grams requlrcd to fully react with 10 Erams of 0z?0258 28.7 €0 1.25 80 4080 57.48 8Question 13Cons...

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Znceg cnemicaNals) Oz (&}NazOls)adaEeWhat mass of Na in grams requlrcd to fully react with 10 Erams of 0z?0258 28.7 €0 1.25 80 4080 57.48 8Question 13Consider thc following thrce statements Which of the following Is/arc truc?The specific heat capacity of 10 grams of Fc double that of 5 Grams of Fe: Upon gaining 10 of heat, substancc with lower : specifc heat capacity will have grcatcr change temperature than _ substance of equa Mass with higher specific heat capacity: Specific heat capacit

Znceg cnemica Nals) Oz (&} NazOls) adaEe What mass of Na in grams requlrcd to fully react with 10 Erams of 0z? 0258 28.7 € 0 1.25 8 0 408 0 57.48 8 Question 13 Consider thc following thrce statements Which of the following Is/arc truc? The specific heat capacity of 10 grams of Fc double that of 5 Grams of Fe: Upon gaining 10 of heat, substancc with lower : specifc heat capacity will have grcatcr change temperature than _ substance of equa Mass with higher specific heat capacity: Specific heat capacity measures the total amount of heat stored in substance only 0 iand only only hand only 0 Wonly 8 pts



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When 25.0 $\mathrm{mL}$ of 1.0 $\mathrm{M}_{2} \mathrm{SO}_{4}$ is added to 50.0 $\mathrm{mL}$ of 1.0 $\mathrm{MNaOH}$ at $25.0^{\circ} \mathrm{C}$ in a calorimeter, the temperature of the aqueous solution increases to $33.9^{\circ} \mathrm{C}$ . Assuming that the specific heat of the solution is $4.18 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right),$ that its density is
1.00 $\mathrm{g} / \mathrm{mL}$ , and that the calorimeter itself absorbs a negligible amount of heat, calculate $\Delta H$ in kilojoules for the reaction.
$$
\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)
$$

First for party. We could say that the mass of the copper multiplied by the specific heat of copper multiplied by the absolute value of the change in temperature of the copper, plus the mass of the unknown substance times the specific heat of the unknown substance times the change, the absolute value of the change in temperature of the unknown substance. Equaling the mass of the cup times, the specific heat of aluminum, plus the massive water times, the specific heat of water multiplied by the change in temperature of the water, and so solving for the specific heat of the unknown substance. This would be equaling massive the cup specific heat of aluminum, plus massive water, justice of heat of water times the change and the change in temperature of the water minus the mass of the copper specific heat of copper times the absolute value of the change in temperature of the copper, divided by the mass of the unknown substance times, the absolute value of the change in temperature of the unknown substance, and so we can solve this would be equaling. We can say first one over 70 grams times 80 degree Celsius and then we can say 100 grams multiplied by 900 then this would be Jules per kilogram per degree Celsius plus 250 grams multiplied by 4186. And this is me, Jules per kilogram per degree. Celsius multiplied by 10 degrees Celsius minus. And then this would be 50 grammes multiplied by 387 Jules per kilogram per degree. Celsius multiplied by 60 degrees Celsius. And we find that this is equaling 1.82 times 10 to the third Jules per kilogram per degree Celsius. So this would be our final answer for the specific heat of that unknown substance. Four part be looking at the table. This could be beryllium, so other possibilities may exist. However, the material could B beryllium and then for part C, Let's explain. Well, we can say that the material can also be and alloy. So essentially two analogy simply to metals together, Um, two different metals together. So a mix, uh, or a material that isn't listed. That is the end of the solution. Thank you for watching

When using a calorie meter, not only does the water in the calorie meter absorb some heat but often so does the calorie meter itself. So if we were to take a hot object such as iron and place it into a calorie meter, the water in the calorie meter would absorb the heat from the hot iron. And so would the calorie emitter itself, the walls of the calorie meter, the thermometer and so forth. So when performing calculations, we would say that the heat lost, so will make it negative by the hot iron will be equal to the heat gained by the water in the calorie meter. Plus the heat gained by the calorie meter itself. As it warms up, the heat lost by the iron will be equal to its specific heat capacity provided a 0.444, multiplied by its mass provided at 93.3, multiplied by its change in temperature. Final temperature 19.68 minus the initial temperature 65.58. That then will be equal to the Heat gained by the water, which will be the equal to the specific heat of water. Multiplied by the mass of water, multiplied by the change in temperature of water, the final temperature being the same 19.68, but its initial temperature was different at 16.95. Then the heat of the calorie meter, because we don't know the specific heat capacity will just say that the mass times the specific heat is equal to a generic heat capacity. So it'll be the heat capacity of the calorie emitter multiplied by the change in temperature of the calorie emitter, which will be the same as that of the water because the water was in the calorie emitter. Now we've got a single algebraic expression with one unknown. All we need to do is isolate the heat capacity of the calorie meter And in doing so, we get 383 jewels per degree C.

Okay. So this is a comparison question asked us to compare the Qantas of the heat required for each following repairs. And he asked us also to explain why we Choose one rather than another. So the first pair is Evaporating one g of water. So evaporating ground grams water compared with sublime one g of the water. Yeah. Okay. So. All right. Um First, what we compare with the evaporation and supply. So for evaporating, we know it's from a liquid phase to the gas face. From the stop line, it's from the solid face to the gas face. And we know that the energy of the liquid phase is larger. So we can actually decompose the soy face to gas phase. Two solid face to the liquid phase and then to the gas face. You can see this part are the same no matter the temperatures as long as the temperature remains the same before and after transition. The value should be seen. However, for the stop line, there's an extra part from the solid to liquid and from the solid liquid actually will absorb heat. So we can justify here the energy required for evaporation for the evaporation. The liquid actually just simple equals to the there's a easy however, the energy required for sublime the water equals two delta age. If you're in the heat of urine plus the heat of the vaporization. So definitely sublime one g of water will actually absorb more heat. So this one will actually absorb more heat and second pairs condensing .5 g of water. So condemns 0.5 g. It was the water 60 degrees C. And compared was a melting .5g with water 0° C. So condensing is from a gas face to the liquid face and melting is from the soilless to the um liquid face. Of course the sign is different. But I think the question asked us to compare the absolute values. So we don't have to care about the temperatures here because the question assumes during the condensation or melting, the temperature does not change. So then it will not influence the heat. So then it becomes to compare the energy of the vaporization and the energy of the heat of the few in and normally for the few in for the melting. That the uh the interactions you uh break up much less than the interactions. Um you have to break up during the gasification, which is evaporation. So definitely there will be uh relations that the delta H. V. Is much much larger than delta HF. Because the interactions you have to break up during this to process are very different for the vaporization. You have like two totally break interactions between the molecules. Why in the future? And you don't have to do that. So we know that the heat of the vocalization is much larger. So which means condense the heat that you used to uh heat transfer. That you used to condense the water is larger. So this one is larger. And questions see, it says vaporization 50 g of liquid water into and colon Set out 200°C. Who vaporizing? And compared with the mm water, placed Both water. So the third question is vaporizing 50g of water in 200 degrees C. Or vaporize e uh 100 g of water um Placed into an open at 100°C. So which one will actually um consume more heat. So let's say so 50 g of water and It's actually at 100 And you place into 200 the greasy And another is 100 grams of water. And you placed 100° C. And it's also 100°C. So off it would transfer into the gas face and which one will actually mm consume more heat. It's actually does not matter. As long as the temperature does not change. That's shouldn't change the vaporization, the energy of big presentation. So the energy is directly linked to the mass wherever has largest mass that will absorb more heat. So the 2nd 100 g of water will actually absorb more heat. So the answer is the second and this is the answer.

In this question. We have given an aluminum calorie meter and it contains water and we have all that both are in thermal equilibrium. Now we put two metallic blocks in this calorie meter and we have given the data for one metallic blog. But we have that the unknown data for the other metallic blog And we have told that the entire system is stabilized at the final temperature of 20° has yes. So first of all we have to find the specific heat of the unknown simple. So I'm going to write the given data for the institution. So we have given that mass of the That is mass of aluminum accelerometer amaze you want to add 100 g or I can say 0.1 kg. We have given the mass of water that is M W Yes to 50 g Where I can see 0.25 kg. And we have given the mass of the that is copper block that is EMC is given to us. There is 0.05 kg and we have given the mass of the unknown block that is 0.07 kg. Now we have given the value of initial temperature that is 10° service. Yes. And this is for calorie meter and we have given the initial temperature of the cooper block. That is I like to see you Corporate block. It is given to us that is 80°C and final temperature is given to us that is granted degrees health. Yes, we have do you want that initial temperature of the unknown Blog. That is 100° cells. Yes. And we have given the value of No, these are the given data. Now I can see that from the energy balance. It can be written as energy lost. That is minus of two lost will be cause to energy again. That is cute. Again, this is not minus here. Dc energy lost. So and the losses returnees. That is this energy is given by the that is mass of upper block into c of corporate blog. And this is the initial temperature minus final temperature for corporate blog plus this is mass of unknown block into specifically hit of a non block and this is the initial minority final for the unknown block. And the energy gain is by the one is water and one is the aluminum calorie meter. So this is a mass of the water into the specific heat of water. And final temperature minus initial temperature of the water plus the mass of the aluminum into specific it of aluminum into the final minus city initial. Now put all the data in this equation. So this is 0.05 And this is 387 into this is yeah, 18 minus 20 plus. Yeah, They said you're a .07 into this we have to find So this is specific it of unknown into disease 100 minus 20. And for water we have given the math and this is 0.25 into this is 4186 and the temperature differences, That is 20-10 plus for aluminum. My cancer disease, 0.1 into 900. And this is also 20 minus stand No. After holding this we get the value of C. Unknown and this will comes out to be 1.82 and 2 10 to the power three jewel per kg degree Celsius. So this is the specific heat of the unknown block. And this is our answer for the first part of the problem. Now, in the second part, we have to test dad, what is the identification of the unknown material? So from this data, we can say that it might be a beryllium. So this is the answer of our second part of the problem from this data, we can told that it might be beryllium. Now, in the third part of the problem, we have to find that tell that. Can we identify the possible material? So we can say that it might be unknown alloy or not the staring table. But these are the s. four of our given fusion. Thank you


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