A 275 mL, Sample of\'0.400 M NaCl solution is diluted to 885 ml. What is the fin

Question

A 275 mL, Sample of'0.400 M NaCl solution is diluted to 885 ml. What is the final concentration (M) of NaCl in the solution? You must show your work in order to receive credit. Write your answer in the box provided with 3 significant figures. Aqueous sodium sulfide reacts with aqueous manganese (III) nitrate in a double replacement type reaction. Determine the, limiting reagent and the. theoretical yield of the resulting sulfide salt, in. grams, when 250 ml of 0.40 M aqueous manganese (III) nitrate solution reacts with, 0.300 L of a 0.60 M aqueous sodium sulfide solution. You must show your work in order to receive credit. Write your theoretical yield answer with 2 significant digits in the box provided below.

Explanation / Answer

3)
number of mol of NaCl present = Mi*Vi
= 0.400 M * 0.275 L
= 0.110 mol

final volume = 885 mL = 0.885 L

Final concentration = number of mol / volume
= 0.110 mol / 0.885 L
=0.124 M
Answer: 0.124 M

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