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Author Question: A solution is prepared by adding 1.60 g of solid NaCl to 50.0 mL of 0.100 mol L-1 CaCl2. What is the ... (Read 74 times)

schs14

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on: Jul 8, 2021

Question 1

A FeCl3 solution is 0.175 mol L-1. How many mL of a 0.175 mol L-1 FeCl3 solution are needed to make 550 mL of a solution that is 0.300 mol L-1 in Cl- ion?
◦ 0.943 mL
◦ 314 mL
◦ 943 mL
◦ 0.314 mL

Question 2

A solution is prepared by adding 1.60 g of solid NaCl to 50.0 mL of 0.100 mol L-1 CaCl2. What is the molarity of chloride ion in the final solution? Assume that the volume of the final solution is 50.0 mL.
◦ 0.747 mol L-1
◦ 0.647 mol L-1
◦ 0.132 mol L-1
◦ 0.232 mol L-1
◦ 0.547 mol L-1
General Chemistry: Principles and Modern Applications ¦ Petrucci, Herring, Madura, Bissonnette ¦ 11th Edition
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Marked as best answer by schs14 on Jul 8, 2021

jackie

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Reply #1 on: Jul 8, 2021
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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schs14

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Reply 2 on: Jul 8, 2021
Wow, this really help

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Reply 3 on: Yesterday
Thanks for the timely response, appreciate it

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