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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<h2><u>Answer:</u></h2>
It wasn't an adjustment in the condition of issue on the grounds that the vitality in the can did not change. Additionally, since this was a physical change, the atoms in the can are as yet similar particles. No synthetic bonds were made or broken. You added enough vitality to make a stage change from strong to fluid.
The main changes recorded which don't include framing or breaking substance bonds would bubble and liquefying. Bubbling and liquefying are physical changes as opposed to synthetic changes, so no new items are shaped.
What is the exoeruent. Searched it up on google and only came up with two search results. None related to chemistry
Answer:
6. d, 7. a
Explanation:
6. Molarity is a number of moles solute in 1 L solution.
7. 1 L solution - 2.5 mol K2CO3
20 L - x mol K2CO3
x =20*2.5/1 = 50 mol K2CO3
Molar mass(KCO3) = M(K) + M(C) + 3M(O)= 39 +12 +3*16= 99 g/mol
99 g/mol *50 mol = 4950 g KCO3 Closest answer is A.
Actually KCO3 does not exist, in reality it should be K2CO3.
