4 Part B Dimensional analysis simply refers to the inclusion of units in an equation. After setting up the solution map, dimensional analysis can be used to set up the conversion factors that lead to the desired values within this map. The correct setup of an equation can be verified by checking if the result will only have the desired units after unit cancellation. 3NaCl(aq) + Al(NO3)3 (aq) → AlCl3 (s) + 3NaNO3(aq) For the described reaction, you have a 0.380 L solution of 0.680 M NaCl(aq). Using the solution map from Part A as a guide, complete the following dimensional analysis that allows you to calculate the moles of AlCl3 (s) that can be formed by placing the values of each conversion factor according to whether they should appear in the numerator or denominator Drag the appropriate values to their respective targets. ▸ View Available Hint(s) Submit 0.186 mol AlCl3 (s) 1.676 mol AlCl3 (s) 0.086 mol AlCl3(s) 1 mol AlCl3(s) 3 mol NaCl(aq) 0.775 mol AlCl3 (s) 1 L solution 0.380 L solution 0.680 mol NaCl(aq) Group 1 Group 2 Group 1 Group 2 Group 1 Group 2 Reset Help
4 Part B Dimensional analysis simply refers to the inclusion of units in an equation. After setting up the solution map, dimensional analysis can be used to set up the conversion factors that lead to the desired values within this map. The correct setup of an equation can be verified by checking if the result will only have the desired units after unit cancellation. 3NaCl(aq) + Al(NO3)3 (aq) → AlCl3 (s) + 3NaNO3(aq) For the described reaction, you have a 0.380 L solution of 0.680 M NaCl(aq). Using the solution map from Part A as a guide, complete the following dimensional analysis that allows you to calculate the moles of AlCl3 (s) that can be formed by placing the values of each conversion factor according to whether they should appear in the numerator or denominator Drag the appropriate values to their respective targets. ▸ View Available Hint(s) Submit 0.186 mol AlCl3 (s) 1.676 mol AlCl3 (s) 0.086 mol AlCl3(s) 1 mol AlCl3(s) 3 mol NaCl(aq) 0.775 mol AlCl3 (s) 1 L solution 0.380 L solution 0.680 mol NaCl(aq) Group 1 Group 2 Group 1 Group 2 Group 1 Group 2 Reset Help
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Part B
Dimensional analysis simply refers to the inclusion of units in an equation. After setting up the solution map, dimensional analysis can be used to set up the conversion factors that lead to the desired values
within this map. The correct setup of an equation can be verified by checking if the result will only have the desired units after unit cancellation.
3NaCl(aq) + Al(NO3)3 (aq) → AlCl3 (s) + 3NaNO3(aq)
For the described reaction, you have a 0.380 L solution of 0.680 M NaCl(aq). Using the solution map from Part A as a guide, complete the following dimensional analysis that allows you to calculate the
moles of AlCl3 (s) that can be formed by placing the values of each conversion factor according to whether they should appear in the numerator or denominator
Drag the appropriate values to their respective targets.
▸ View Available Hint(s)
Submit
0.186 mol AlCl3 (s)
1.676 mol AlCl3 (s)
0.086 mol AlCl3(s)
1 mol AlCl3(s)
3 mol NaCl(aq)
0.775 mol AlCl3 (s)
1 L solution
0.380 L solution
0.680 mol NaCl(aq)
Group 1
Group 2
Group 1
Group 2
Group 1
Group 2
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