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Author Question: When 20.0 mL of a 0.250 M (NH4)2S solution is added to 150.0 mL of a solution of Cu(NO3)2, a CuS ... (Read 42 times)

mmmmm

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Question 1

What volume (mL) of a 0.3428 M HCl(aq) solution is required to completely neutralize 23.55 mL of a 0.2350 M Ba(OH)2(aq) solution?
◦ 55.34 mL
◦ 11.07 mL
◦ 16.14 mL
◦ 32.29 mL
◦ 47.10 mL

Question 2

When 20.0 mL of a 0.250 M (NH4)2S solution is added to 150.0 mL of a solution of Cu(NO3)2, a CuS precipitate forms.  The precipitate is then filtered from the solution, dried, and weighed.  If the recovered CuS is found to have a mass of 0.3491 g, what was the concentration of copper ions in the original Cu(NO3)2 solution?
◦ 3.65 × 10–3 M
◦ 1.22 × 10–2 M
◦ 3.33 × 10–2 M
◦ 4.87 × 10–2 M
◦ 2.43 × 10–2 M


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Marked as best answer by mmmmm on Nov 5, 2023

Jim457

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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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mmmmm

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Reply 2 on: Nov 5, 2023
Gracias!


mcabuhat

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Reply 3 on: Yesterday
Wow, this really help

 

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