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Author Question: 17.5 mL of a 0.1050 M Na2CO3 solution is added to 46.0 mL of 0.1250 M NaCl. What is the ... (Read 127 times)

aeb093

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on: Nov 5, 2023

Question 1

158 mL of a 0.148M NaCl solution is added to 228 mL of a 0.369M NH4NO3 solution.  The concentration of ammonium ions in the resulting mixture is
◦ 0.157 M
◦ 0.218 M
◦ 0.625 M
◦ 0.369 M
◦ 0 M

Question 2

17.5 mL of a 0.1050 M Na2CO3 solution is added to 46.0 mL of 0.1250 M NaCl.  What is the concentration of sodium ion in the final solution?
◦ 0.205 M
◦ 0.119 M
◦ 0.539 M
◦ 0.148 M
◦ 0.165 M
Chemistry ¦ Chang, Goldsby ¦ 11th Edition
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Marked as best answer by aeb093 on Nov 5, 2023

Jim457

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Reply #1 on: Nov 5, 2023
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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aeb093

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Reply 2 on: Nov 5, 2023
Wow, this really help

LegendaryAnswers

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Reply 3 on: Yesterday
YES! Correct, THANKS for helping me on my review

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