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0:08 PM Mon Sep 26 K 148.33 g Mg(NO3)2 X STARTING AMOUNT 8.33 g Mg(NO₂)₂2 × 45.0 g Mg(NO₂)2 What volume in 0.320 mol Mg(NO3)2 45.0 g Mg(NO3)2 0.320 mol Mg(NO3)2 = 0.948 L ADD FACTOR 86.32 141 = 0.948 mol Mg(NO3)2 mL Question 13 of 17 L of a 0.320 M Mg(NO₂)2 solution contains 45.0 g of Mg(NO3)2 ? DELETE A 210.34 6.022 x 1023 ... 148.33 0.0971 g Mg(NO3)2 L 1.63 1 L ANSWER 0.320 0.948 0.303 45.0 M Mg(NO3)2 RESET 3 0.167 14.4 g Mg(NO3)2/mol Submit

0:08 PM Mon Sep 26 K 148.33 g Mg(NO3)2 X STARTING AMOUNT 8.33 g Mg(NO₂)₂2 × 45.0 g Mg(NO₂)2 What volume in 0.320 mol Mg(NO3)2 45.0 g Mg(NO3)2 0.320 mol Mg(NO3)2 = 0.948 L ADD FACTOR 86.32 141 = 0.948 mol Mg(NO3)2 mL Question 13 of 17 L of a 0.320 M Mg(NO₂)2 solution contains 45.0 g of Mg(NO3)2 ? DELETE A 210.34 6.022 x 1023 ... 148.33 0.0971 g Mg(NO3)2 L 1.63 1 L ANSWER 0.320 0.948 0.303 45.0 M Mg(NO3)2 RESET 3 0.167 14.4 g Mg(NO3)2/mol Submit

Chemistry
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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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10:08 PM Mon Sep 26 < 148.33 g Mg(NO3)2 STARTING AMOUNT 148.33 g Mg(NO3)2 X X 45.0 0.320 45.0 g Mg(NO3)2 0.320 mol Mg(NO3)2 What volume in g Mg(NO3)2 mol Mg(NO3)2 = 0.948 L ADD FACTOR x( ) 86.32 141 mol Mg(NO3)2 0.948 mL Question 13 of 17 L of a 0.320 M Mg(NO3)2 solution contains 45.0 g of Mg(NO3)2? DELETE f 210.34 6.022 x 1023 ... 148.33 0.0971 L g Mg(NO3)2 1.63 1 L ANSWER 0.320 0.948 0.303 45.0 RESET 3 0.167 14.4 M Mg(NO3)2 g Mg(NO3)2/mol all 23% Submit
Transcribed Image Text:10:08 PM Mon Sep 26 < 148.33 g Mg(NO3)2 STARTING AMOUNT 148.33 g Mg(NO3)2 X X 45.0 0.320 45.0 g Mg(NO3)2 0.320 mol Mg(NO3)2 What volume in g Mg(NO3)2 mol Mg(NO3)2 = 0.948 L ADD FACTOR x( ) 86.32 141 mol Mg(NO3)2 0.948 mL Question 13 of 17 L of a 0.320 M Mg(NO3)2 solution contains 45.0 g of Mg(NO3)2? DELETE f 210.34 6.022 x 1023 ... 148.33 0.0971 L g Mg(NO3)2 1.63 1 L ANSWER 0.320 0.948 0.303 45.0 RESET 3 0.167 14.4 M Mg(NO3)2 g Mg(NO3)2/mol all 23% Submit
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Here we have to determine the volume of Mg(NO3)2 solution in L having  concentration 0.320M that contains 45 g of magnesium nitrate. 

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