In 250 mL of volumetric flask add 0.975875 grams of and dissolve it in the 250 mL of water.
Given:
The solid of calcium fluoride.
To prepare:
The 250 mL solution of 0.100 M of fluoride ions from solid calcium fluoride.
Method:
Molarity of the fluoride ion solution needed = M = 0.100 M
The volume of the fluoride ion solution needed = V = 250 mL
The moles of fluoride ion needed = n
According to the definition of molarity:
Moles of fluoride ion = 0.025 mol
We know that solid calcium fluoride dissolves in water to give calcium ions and fluoride ions.
According to reaction, 2 moles of fluoride ions are obtained from 1 mole of calcium fluoride, then 0.025 moles of fluoride ions will be obtained from:
Moles of calcium fluoride = 0.0125 mol
Mass of calcium fluoride needed to prepare the solution :
Preparation:
- Weight 0.975875 grams of calcium fluoride
- Add weighed calcium fluoride to a volumetric flask of the labeled volume of 250 mL.
- Now add a small amount of water to dissolve the calcium fluoride completely.
- After this add more water up to the mark of the volumetric flask of volume 250 mL.
Learn more about molarity of solution ere:
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I'm assuming that you are asking a general question because you did not include an example.
The limiting reagent is the item in the reactants (reagents) that will run out first. This is because it limits what the reaction can produce, essentially causing the leftover elements/compounds to just sit there.
2SO₂ + O₂ = 2SO₃
n(O₂)=1 mol
n(SO₂)=2n(O₂)
n(SO₂)=2 mol
Answer: sorry I cant help you I need the same answer
Explanation: