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Ch.6 - Thermochemistry

Chapter 6, Problem 102

Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the equation: CO2(s)¡CO2( g) ◀ When carbon dioxide sublimes, the gaseous cO2 is cold enough to cause water vapor in the air to condense, forming fog. When dry ice is added to warm water, heat from the water causes the dry ice to sublime more quickly. The evaporating carbon dioxide produces a dense fog often used to create special effects. In a simple dry ice fog machine, dry ice is added to warm water in a Styrofoam cooler. The dry ice produces fog until it evaporates away, or until the water gets too cold to sublime the dry ice quickly enough. Suppose that a small Styrofoam cooler holds 15.0 L of water heated to 85 °C. Use standard enthalpies of formation to calculate the change in enthalpy for dry ice sublimation, and calculate the mass of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reaches 25 °C. Assume no heat loss to the surroundings. (The ΔH °f for CO2(s) is -427.4 kJ/mol.)

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Hey everyone, we're told that solid carbon dioxide, more commonly known as dry ice, is directly converted to gas when heated instead of melting first, adding warm water to dry ice is enough to promote this process, calculate the mass of dry ice added to 25 liters of water at 70 degrees Celsius. Given that the dry eyes completely sublime when the water reaches 20 degrees Celsius assume no heat is lost to the surroundings and our heat of formation for carbon dioxide in its solid form is negative 427.4 kg jules Permal first we need to recognize that we have a sublimation reaction. So we have solid carbon dioxide and this is going to form into its gaseous form. Next we can calculate the standard entropy of our reaction and we can do so by taking the heat of formation of our products and subtract our heat of formation of our reactant, plugging in our values and looking it up in our textbooks, we find that the heat of formation of our gasses carbon dioxide is negative 393.5 kg joules per mole. And we're going to go ahead and subtract that negative 427.4 kg joules per mole which was provided to us in our questions. Then this gets us to a total of 33.9 kg joules per mole. Now let's go ahead and calculate for the mole of carbon dioxide, we can do so by recalling that the heat absorbed is equal to the negative heat released. So when we plug in our formulas, we have n times our standard and they'll be of our reaction. And this is going to be in terms of carbon dioxide and this will be equivalent to the negative mass times of specific heat times our change in temperature. And this will be in terms of our water, solving for n of our carbon dioxide. We get the negative of our mass times specific heat times a change in temperature of our water. All divided by the standard entropy of our reaction. Now to calculate the mass of our water, we can go ahead and do so by taking our density times are volume. So this is equal to one g over millimeter. And we want to convert Our male leaders into leaders. So we know that we have 10 to 3rd male leaders per one leader. And we're going to multiply this by 25l which was given to us in our questions them. Now when we calculate this out and cancel out all our units, we end up with a value Of 25,000 g of water. Now we can go ahead and solve for our mole of carbon dioxide. So we have the negative of 25,000 g times the specific heat of water which is 4.184 jewels over grams times degrees Celsius Times are change in temperature which is 20°C -70°C which was told to us in our question stem. And this will be all divided By 33.9 kill a jules Permal. And we need to convert those killer jewels into jewels and we can do so by using dimensional analysis. So we know that we have one killer jewel per attend to the third jewels. So when we calculate this out and cancel out all of our units, We end up with 154. mole of carbon dioxide. Now that we saw for the mole of carbon dioxide, we want to solve for the grams of carbon dioxide. So again, we're going to use dimensional analysis and we have 154.28 mole of carbon dioxide. And using carbon dioxide solar mass, we know that one mole of carbon dioxide contains 44 g of carbon dioxide. This will get us the value of 6788. g. And we can go ahead and convert this into kilograms. Again, with dimensional analysis and we know that we have 10 to 30 g per one kg. So our final answer is going to be 6. kilograms of carbon dioxide. Now, I hope that made sense. And let us know if you have any questions
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Textbook Question

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A 25.5-g aluminum block is warmed to 65.4 °C and plunged into an insulated beaker containing 55.2 g water initially at 22.2 °C. The aluminum and the water are allowed to come to thermal equilibrium. Assuming that no heat is lost, what is the final temperature of the water and aluminum?

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If 50.0 mL of ethanol (density = 0.789 g>mL) initially at 7.0 °C is mixed with 50.0 mL of water (density = 1.0 g/mL) initially at 28.4 °C in an insulated beaker, and assuming that no heat is lost, what is the final temperature of the mixture?

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Textbook Question

Palmitic acid (C16H32O2) is a dietary fat found in beef and butter. The caloric content of palmitic acid is typical of fats in general. Write a balanced equation for the complete combustion of palmitic acid and calculate the standard enthalpy of combustion. What is the caloric content of palmitic acid in Cal/g? The standard enthalpy of formation of palmitic acid is -208 kJ/mol and that of sucrose is -2226.1 kJ/mol. [Use H2O(l) in the balanced chemical equations because the metabolism of these compounds produces liquid water.]

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