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Quiz: Lab 9 Module 4 - Act 1: \( N \) Inbox - iscott12@asu.edu - Ari Home - Google Drive edu/courses/241987/quizzes/1955758/take rofessional... Online Proofing Gall... Portfolio | Capital Gr... Adobe Acrobat College Arizona Move Professional Associa... Question 9 1 pts Probably half of the chemists employed around the world are involved in synthesis of some sort, making new molecules or other materials from other, usually simpler ones. This is where new materials for pharmaceuticals and drugs come from, where new electronic materials come from, where new catalysts come from, where energy materials come from, where specialized chemicals such as fertilizers pesticides, dyes and so on come from, and where man-made biological materials come from, to name just a few! The chemical yield in any synthesis is a critical issue. A low yield wastes chemicals, usually produces unwanted waste chemicals and wastes energy trying to make that chemical. You didn't determine the yield of formation of the Prussian blue in the calorimetry experiment, although you could have if you had filtered it and dried it. In this reaction you started with 20 mL of 0.45 M ferrous sulfate, which is equivalent to 2.5 g of \( \mathrm{FeSO}_{4} \cdot \mathrm{7H}_{2} \mathrm{O} \) In this reaction you started with 20 mL of 0.3 M ferrous sulfate, which is equivalent to 2.0 g of \( \mathbf{K}_{\mathbf{3}}\left[\mathbf{F e}(\mathbf{C N})_{\mathbf{6}}\right] \) Fe is 20\% of the mass of \( \mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O} \) \( \mathrm{Fe}(\mathrm{CN})_{6} \) is \( 61 \% \) of the mass of \( \mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \) According to the balanced equation, 20\% of the mass of the \( \mathrm{FeSO}_{4} . \mathrm{7H}_{2} \mathrm{O} \) and 61\% of the mass of the \( \mathbf{K}_{\mathbf{3}}\left[\mathbf{F e}(\mathbf{C N})_{6}\right] \) should have been in the Prussian blue product. Assume you isolated 1.45 g off Prussian blue by filtering and then drying, what would the yield have been? Use your gapped notes as a guide to solve this problem. ? The \% Yield = 24.8\% ? The \% Yield =48.9\% ? The \% Yield = 71.1\% ? The \% Yield = 84.3\% Question 10 1 pts Search 

          Quiz: Lab 9 Module 4 - Act 1: \( N \)
Inbox - iscott12@asu.edu - Ari
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Question 9

1 pts

Probably half of the chemists employed around the world are involved in synthesis of some sort, making new molecules or other materials from other, usually simpler ones. This is where new materials for pharmaceuticals and drugs come from, where new electronic materials come from, where new catalysts come from, where energy materials come from, where specialized chemicals such as fertilizers pesticides, dyes and so on come from, and where man-made biological materials come from, to name just a few!
The chemical yield in any synthesis is a critical issue. A low yield wastes chemicals, usually produces unwanted waste chemicals and wastes energy trying to make that chemical.
You didn't determine the yield of formation of the Prussian blue in the calorimetry experiment, although you could have if you had filtered it and dried it.
In this reaction you started with 20 mL of 0.45 M ferrous sulfate, which is equivalent to 2.5 g of \( \mathrm{FeSO}_{4} \cdot \mathrm{7H}_{2} \mathrm{O} \)

In this reaction you started with 20 mL of 0.3 M ferrous sulfate, which is equivalent to 2.0 g of \( \mathbf{K}_{\mathbf{3}}\left[\mathbf{F e}(\mathbf{C N})_{\mathbf{6}}\right] \)

Fe is 20\% of the mass of \( \mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O} \)
\( \mathrm{Fe}(\mathrm{CN})_{6} \) is \( 61 \% \) of the mass of \( \mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \)
According to the balanced equation, 20\% of the mass of the \( \mathrm{FeSO}_{4} . \mathrm{7H}_{2} \mathrm{O} \) and 61\% of the mass of the \( \mathbf{K}_{\mathbf{3}}\left[\mathbf{F e}(\mathbf{C N})_{6}\right] \) should have been in the Prussian blue product. Assume you isolated 1.45 g off Prussian blue by filtering and then drying, what would the yield have been?

Use your gapped notes as a guide to solve this problem.
? The \% Yield = 24.8\%
? The \% Yield =48.9\%
? The \% Yield = 71.1\%
? The \% Yield = 84.3\%

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Quiz: Lab 9 Module 4 - Act 1: N Inbox - iscott12@asu.edu - Ari Home - Google Drive edu/courses/241987/quizzes/1955758/take rofessional... Online Proofing Gall... Portfolio | Capital Gr... Adobe Acrobat College Arizona Move Professional Associa... Question 9 1 pts Probably half of the chemists employed around the world are involved in synthesis of some sort, making new molecules or other materials from other, usually simpler ones. This is where new materials for pharmaceuticals and drugs come from, where new electronic materials come from, where new catalysts come from, where energy materials come from, where specialized chemicals such as fertilizers pesticides, dyes and so on come from, and where man-made biological materials come from, to name just a few! The chemical yield in any synthesis is a critical issue. A low yield wastes chemicals, usually produces unwanted waste chemicals and wastes energy trying to make that chemical. You didn't determine the yield of formation of the Prussian blue in the calorimetry experiment, although you could have if you had filtered it and dried it. In this reaction you started with 20 mL of 0.45 M ferrous sulfate, which is equivalent to 2.5 g of FeSO4Β·7H2O In this reaction you started with 20 mL of 0.3 M ferrous sulfate, which is equivalent to 2.0 g of 𝐊3[𝐅 𝐞(𝐂 𝐍)6] Fe is 20% of the mass of FeSO4Β· 7 H2O Fe(CN)6 is 61 % of the mass of K3[Fe(CN)6] According to the balanced equation, 20% of the mass of the FeSO4 . 7H2O and 61% of the mass of the 𝐊3[𝐅 𝐞(𝐂 𝐍)6] should have been in the Prussian blue product. Assume you isolated 1.45 g off Prussian blue by filtering and then drying, what would the yield have been? Use your gapped notes as a guide to solve this problem. ? The % Yield = 24.8% ? The % Yield =48.9% ? The % Yield = 71.1% ? The % Yield = 84.3% Question 10 1 pts Search

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Quiz: Lab 9 Module 4 - Act 1: \( N \) Inbox - iscott12@asu.edu - Ari Home - Google Drive edu/courses/241987/quizzes/1955758/take rofessional... Online Proofing Gall... Portfolio | Capital Gr... Adobe Acrobat College Arizona Move Professional Associa... Question 9 1 pts Probably half of the chemists employed around the world are involved in synthesis of some sort, making new molecules or other materials from other, usually simpler ones. This is where new materials for pharmaceuticals and drugs come from, where new electronic materials come from, where new catalysts come from, where energy materials come from, where specialized chemicals such as fertilizers pesticides, dyes and so on come from, and where man-made biological materials come from, to name just a few! The chemical yield in any synthesis is a critical issue. A low yield wastes chemicals, usually produces unwanted waste chemicals and wastes energy trying to make that chemical. You didn't determine the yield of formation of the Prussian blue in the calorimetry experiment, although you could have if you had filtered it and dried it. In this reaction you started with 20 mL of 0.45 M ferrous sulfate, which is equivalent to 2.5 g of \( \mathrm{FeSO}_{4} \cdot \mathrm{7H}_{2} \mathrm{O} \) In this reaction you started with 20 mL of 0.3 M ferrous sulfate, which is equivalent to 2.0 g of \( \mathbf{K}_{\mathbf{3}}\left[\mathbf{F e}(\mathbf{C N})_{\mathbf{6}}\right] \) Fe is 20\% of the mass of \( \mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O} \) \( \mathrm{Fe}(\mathrm{CN})_{6} \) is \( 61 \% \) of the mass of \( \mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \) According to the balanced equation, 20\% of the mass of the \( \mathrm{FeSO}_{4} . \mathrm{7H}_{2} \mathrm{O} \) and 61\% of the mass of the \( \mathbf{K}_{\mathbf{3}}\left[\mathbf{F e}(\mathbf{C N})_{6}\right] \) should have been in the Prussian blue product. Assume you isolated 1.45 g off Prussian blue by filtering and then drying, what would the yield have been? Use your gapped notes as a guide to solve this problem. β—― The \% Yield = 24.8\% β—― The \% Yield =48.9\% β—― The \% Yield = 71.1\% β—― The \% Yield = 84.3\% Question 10 1 pts Search
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