What is the final volume, in milliliters, when 5.80 mL of 25.0 %(m/v) NaOH solution is diluted to give a 3.50 %(m/v) NaOH soluti
1 answer:
Answer:
41.43 mL
Explanation:
First we <u>calculate the mass of NaOH in the original solution</u>:
- 25.0 % m/v means that in 100 mL of solution, there are 25 grams of NaOH.
- This means that in 5.80 mL, we would have (5.80 * 25/100) <u>1.45 grams of NaOH.</u>
Then we <u>calculate the volume of the diluted solution</u>, using the grams of NaOH (that remain the same throughout the dilution process):
- 1.45 g NaOH *
= 41.43 mL
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The question is incomplete, here is the complete question:
When silver nitrate reacts with copper, copper(II) nitrate and silver are produced. The balanced equation for this reaction is:
Suppose 6 moles of silver nitrate react. The reaction consumes___ moles of copper. The reaction produces __ moles of copper(II) nitrate and __ moles of silver.
<u>Answer:</u> The amount of copper metal reacted is 3 moles, amount of copper (II) nitrate produced is 3 moles and amount of silver metal produced is 6 moles.
<u>Explanation:</u>
We are given:
Moles of silver nitrate = 6 moles
For the given chemical reaction:
- <u>For copper metal:</u>
By Stoichiometry of the reaction:
2 moles of silver nitrate reacts with 1 mole of copper metal
So, 6 moles of silver nitrate will react with = of copper metal
Moles of copper reacted = 3 moles
- <u>For copper(II) nitrate:</u>
By Stoichiometry of the reaction:
2 moles of silver nitrate produces 1 mole of copper(II) nitrate
So, 6 moles of silver nitrate will produce = of copper(II) nitrate
Moles of copper(II) nitrate produced = 3 moles
- <u>For silver metal:</u>
By Stoichiometry of the reaction:
2 moles of silver nitrate produces 2 moles of silver metal
So, 6 moles of silver nitrate will produce = of silver metal
Moles of silver metal produced = 3 moles
Hence, the amount of copper metal reacted is 3 moles, amount of copper (II) nitrate produced is 3 moles and amount of silver metal produced is 6 moles.