A 25.0-mL sample of 0.050 M barium nitrate solution was mixed with 25.0 mL of 0.050 M sodium sulfate solution labeled with radioactive sulfur-35. The activity of the initial sodium sulfate solution was 1.22⨉10 6 Bq/mL. After the resultant precipitate was removed by filtration, the remaining filtrate was found to have an activity of 250 Bq/mL. (a) Write a balanced chemical equation for the reaction that occurred.

Welcome back everyone in this example, A solution was made by mixing 50 mL of 500.0 25 molar potassium sulfate labeled with radioactive sulfur 36 to 50 mL of 0.0 25 molar barium nitrate. The potassium sulfate solution had an initial activity of 1.26 times 10 to the six power becquerels per mil leader. The leftover fill trait was discovered to have an activity of 2 75 becquerels per mil. A leader sorry. After the resulting precipitate was removed by filtration. What is the balanced chemical equation for the reaction? So beginning by writing out our formulas, we have potassium sulfate, which we should recall is written out as K two S. 04. And we're mixing it with our barium nitrate, which we should recall. Barium is B. A. N. Nitrate. We recall Paypal atomic ion N. 03 where it gets the charge of barium with a subscript of two here. So we should recognize that we have our potassium kati, on which we recall has a plus one charge because potassium is located in group one A. On our periodic table we have the poly atomic ion sulfate which carries a two minus charge. And here these charges have crossed to form our below re agent potassium sulfate and in the same sense, we have barium located in group two A. Of our periodic table carrying a two plus charge, whereas our poly atomic ions nitrate carries a minus one charge. Where we can see these charges crossed to form our second re agent barium nitrate. We want to recognize that this type of reaction is known as a double displacement reaction or double replacement where one of our products should be a solid precipitate. And we will only know the states of our products if we look to our Celje bility rules which we must have memorized, but we can also refer to them in our textbooks or online. So right now recall that in a double displacement reaction, we have the combination of these two components with these two components to produce a product here. So two different products. So right now, based on what we would combine here as products, our first product would be the production of barium sulfate. So be A. S. 04 where both of our charges cross here and they would cancel out each other since they both carry a two plus and then a two minus charge. Where we want to recall our Celje bility rules for sulfates. And we would recall that all sulfates are soluble except those of calcium, strontium, barium and silver and lead. And in this case we have a sulfate bonded to our barium catalon. And so therefore this would be an insoluble sulfate. Which means that this would be a solid precipitate product. So we would give it the label S because it's a solid precipitate. Now for our second product, as we stated, we have the combination of what we have underlined here in blue or purple. So for our second product we would form potassium nitrate. So K N 03 where we want to recall our Celje bility rules for nitrates and we would recall that all nitrates are soluble. So we would have an acquis product because it's soluble in water. And as we can see our charges plus one for potassium and -1 for nitrate crossed to give us these or this neutral charge on our Product, potassium Nitrate, 03. So right now we have our products listed out with their phases where each of our reactant, we want to recognize our Aquarius since they mix in solution. Our next step is to make sure that things are balanced here. So we need to add in any coefficients by comparing our reactive side to our product side. So listing out our atoms, we have potassium sulfur, oxygen, barium and nitrate to worry about on both sides of our equation. And beginning with what we do have on our reactant side, at least we have two moles of our potassium atom counting our sulfur. We have one mole of sulfur. So sorry, this should say one here, going to our oxygen, we have a count of three times two, which gives us six plus four, which would give us 10 oxygen's. And for our very um we have a total of just one and then for nitrogen that is also being multiplied by r subscript of to hear from our charge from barium. So we would have to nitrogen on the reactant side Moving to our product side, we have accounts of just one potassium atom as far as our sulfur. We also just have one sulfur as far as our oxygen. We have four here and then three here. So that would give us nine or sorry that would give us seven here. So seven oxygen's because we would have four plus three, then we have barium where we just have a count of one on our product side. And then for nitrogen we have account of also one on our product side. So clearly we have some balancing to do and right now to balance things out, let's focus with our potassium where if we place a coefficient of two in front of our potassium nitrate on the product side. That would now give us two potassium atoms. We would now have two nitrogen atoms and now we would have a total of three times 24 hour oxygen which would give us six plus four which would give us 10. So we would be able to correct our oxygen to now 10 total. And so as we can see everything checks out as balanced with these corrections meaning we now have a balanced equation and so for our final answer, we have this entire equation with the subscript for our final balanced chemical reaction here. I hope everything I reviewed was clear. If you have any questions leave them down below and I will see everyone in the next practice video.

Ch.21 - Nuclear Chemistry

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Brown 14th Edition

Ch.21 - Nuclear Chemistry

Problem 92a

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Chapter 21, Problem 92a

A 25.0-mL sample of 0.050 M barium nitrate solution was mixed with 25.0 mL of 0.050 M sodium sulfate solution labeled with radioactive sulfur-35. The activity of the initial sodium sulfate solution was 1.22⨉10⁶ Bq/mL. After the resultant precipitate was removed by filtration, the remaining filtrate was found to have an activity of 250 Bq/mL. (a) Write a balanced chemical equation for the reaction that occurred.

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Step 1: Identify the reactants and products. The reactants are barium nitrate (Ba(NO3)2) and sodium sulfate (Na2SO4). The product is barium sulfate (BaSO4), which is a precipitate, and sodium nitrate (NaNO3).

Step 2: Write the unbalanced chemical equation. Ba(NO3)2 + Na2SO4 → BaSO4 + NaNO3.

Step 3: Balance the chemical equation. The balanced chemical equation is: Ba(NO3)2 + Na2SO4 → BaSO4 + 2NaNO3. This is because there are 2 nitrate ions on the left side and only 1 on the right side in the unbalanced equation. To balance it, we need to have 2 nitrate ions on both sides, which is achieved by adding a coefficient of 2 in front of NaNO3.

Step 4: Check the balanced equation. There should be the same number of each type of atom on the left side of the equation as on the right side. In this case, there are 1 Ba, 2 N, 6 O, 2 Na, and 4 O on both sides, so the equation is balanced.

Step 5: Understand the context of the problem. The radioactive sulfur-35 is used to track the sulfate ions. The decrease in activity from the initial sodium sulfate solution to the filtrate indicates that some of the sulfate ions have been removed from the solution, presumably by forming the barium sulfate precipitate.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Precipitation Reactions

Precipitation reactions occur when two soluble salts react in solution to form an insoluble compound, known as a precipitate. In this case, barium nitrate and sodium sulfate react to form barium sulfate, which is insoluble in water. Understanding the solubility rules helps predict which compounds will precipitate during a reaction.

Balancing Chemical Equations

Balancing chemical equations is essential to ensure that the number of atoms of each element is conserved in a reaction. This involves adjusting the coefficients of the reactants and products so that they match. For the reaction between barium nitrate and sodium sulfate, the balanced equation reflects the stoichiometry of the reactants and products involved.

Radioactive Decay and Activity

Radioactive decay refers to the process by which unstable atomic nuclei lose energy by emitting radiation. The activity of a radioactive sample, measured in becquerels (Bq), indicates the number of decay events per second. In this scenario, the initial and final activities of the sodium sulfate solution provide insight into the distribution of the radioactive sulfur after the precipitation reaction.

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Rate of Radioactive Decay

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Related Practice

Open Question

Calculate the mass of octane, C8H18, that must be burned in air to evolve the same quantity of energy as produced by the fusion of 1.0 g of hydrogen in the following fusion reaction: 4 1^1H → 4 2He + 2 0^1e. Assume that all the products of the combustion of octane are in their gas phases. Use data from Exercise 21.50, Appendix C, and the inside covers of the text. The standard enthalpy of formation of octane is -250.1 kJ/mol.

Open Question

Naturally found uranium consists of 99.274% 238U, 0.720% 235U, and 0.006% 234U. As we have seen, 235U is the isotope that can undergo a nuclear chain reaction. Most of the 235U used in the first atomic bomb was obtained by gaseous diffusion of uranium hexafluoride, UF6(g). (a) What is the mass of UF6 in a 30.0-L vessel of UF6 at a pressure of 695 torr at 350 K? (b) What is the mass of 235U in the sample described in part (a)? (c) Now suppose that the UF6 is diffused through a porous barrier and that the change in the ratio of 238U and 235U in the diffused gas can be described by Equation 10.23. What is the mass of 235U in a sample of the diffused gas analogous to that in part (a)? (d) After one more cycle of gaseous diffusion, what is the percentage of 235UF6 in the sample?

Open Question

Charcoal samples from Stonehenge in England were burned in O2, and the resultant CO2 gas bubbled into a solution of Ca(OH)2 (limewater), resulting in the precipitation of CaCO3. The CaCO3 was removed by filtration and dried. A 788-mg sample of the CaCO3 had a radioactivity of 1.5 × 10^-2 Bq due to carbon-14. By comparison, living organisms undergo 15.3 disintegrations per minute per gram of carbon. Using the half-life of carbon-14, 5700 years, calculate the age of the charcoal sample.

Textbook Question

A 25.0-mL sample of 0.050 M barium nitrate solution was mixed with 25.0 mL of 0.050 M sodium sulfate solution labeled with radioactive sulfur-35. The activity of the initial sodium sulfate solution was 1.22 × 10⁶ Bq/mL. After the resultant precipitate was removed by filtration, the remaining filtrate was found to have an activity of 250 Bq/mL. (b) Calculate the K_sp for the precipitate under the conditions of the experiment.

Textbook Question

The two most common isotopes of uranium are 235U and 238U. (d) 238U undergoes radioactive decay to 234Th. How many protons, electrons, and neutrons are gained or lost by the 238U atom during this process?

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