When you are presented with additional details, it can sometimes be a little more difficult to determine which information is necessary and how molarity should be applied. Using both a solution map and dimensional analysis can help clarify which information and conversion factors are necessary to determine the desired value.The following dimensional analysis setup could be used to determine the theoretical mass of AlBr₃(s) (molecular mass = 266.69 g/mol) produced based on reacting 86.9 g of a 0.048 mol/L solution of Br₂(l) (density = 1036 g/L) with excess Al(s) as described in the following equation:\[ 3\text{Br}_2(l) + 2\text{Al}(s) \rightarrow 2\text{AlBr}_3(s) \]Complete the dimensional analysis for calculating the mass of the product by placing the values of each conversion factor according to whether they should appear in the numerator or denominator when calculating the mass of AlBr₃(s) produced from a sample of Br₂(l).**Options:**- 0.048 mol Br₂(l) - 1 L solution- 266.69 g AlBr₃(s)- 1 mol AlBr₃(s)- 1 L solution- 1036 g Br₂(l)- 86.9 g Br₂(l)- 3 mol Br₂(l)- 2 mol AlBr₃(s)**Diagram:**- Group 1- Group 2- Group 4- Group 3Each group represents a section for placing the correct conversion factor from the options provided, aligned with completing the dimensional analysis process.Drag the appropriate values to their respective targets.

Bromine reacts to bromine to form aluminum bromide. The equation for the balanced chemical reaction is given below: The stoichiometry of the balanced chemical equation states that three moles of bromine react with two moles of aluminium to form two moles of aluminium bromide. Calculation of moles of bromine: The molarity of bromine is = 0.048 mol/L mass of bromine solution = 86.9 g The density of the bromine solution is = 1036 g/mol The density of a substance is defined as a mass of a unit volume of the substance. The density of a substance is calculated by using the following formula: by rearranging the above equation: The mole of bromine calculated as follows: The mole of product: We know that three moles of bromine form two moles of aluminium bromide.Calculation of mass of Albr3: We know that the molar mass of AlBr3 is =266.69 g/molThe mass of the product is:You can put the given data as follows: The equation gives in the same order as required in the solution.

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Chemistry

When you are presented with additional details, it can sometimes be a little more difficult to determine which information is necessary and how molarity should be applied. Using both a solution map and dimensional analysis can help clarify which information and conversion factors are necessary to determine the desired value. The following dimensional analysis setup could be used to determine the theoretical mass of AlBr3 (s) (molecular mass = 266.69 g/mol ) produced based on reacting 86.9 g of a 0.048 mol/L solution of Br2 (1) (density = 1036 g/L ) with excess Al(s) as described in the following equation: 3Br2 (1) + 2Al(s) → 2AIBr3 (s) Complete the dimensional analysis for calculating the mass of the product by placing the values of each conversion factor according to whether they should appear in the numerator or denominator when calculating the mass of AlBr3 (s) produced from a sample of Br2 (1). Drag the appropriate values to their respective targets. • View Available Hint(s) Reset Help 0.048 mol Br2(1) 1L solution 266.69 g AlBr3(s) 1 mol AlBr3(s) 1L solution 1036 g Br2(1) 86.9 g Br2(1) 3 mol Br2(1) 2 mol AlBr3(s) Group 1 Group 2 Group 4 Group 3 Group 1 Group 1 Group 2 Group 4 Group 3

When you are presented with additional details, it can sometimes be a little more difficult to determine which information is necessary and how molarity should be applied. Using both a solution map and dimensional analysis can help clarify which information and conversion factors are necessary to determine the desired value. The following dimensional analysis setup could be used to determine the theoretical mass of AlBr3 (s) (molecular mass = 266.69 g/mol ) produced based on reacting 86.9 g of a 0.048 mol/L solution of Br2 (1) (density = 1036 g/L ) with excess Al(s) as described in the following equation: 3Br2 (1) + 2Al(s) → 2AIBr3 (s) Complete the dimensional analysis for calculating the mass of the product by placing the values of each conversion factor according to whether they should appear in the numerator or denominator when calculating the mass of AlBr3 (s) produced from a sample of Br2 (1). Drag the appropriate values to their respective targets. • View Available Hint(s) Reset Help 0.048 mol Br2(1) 1L solution 266.69 g AlBr3(s) 1 mol AlBr3(s) 1L solution 1036 g Br2(1) 86.9 g Br2(1) 3 mol Br2(1) 2 mol AlBr3(s) Group 1 Group 2 Group 4 Group 3 Group 1 Group 1 Group 2 Group 4 Group 3

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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