Carbon dioxide and helium which has a higher van der waals force of attraction between molecules
1 answer:
Answer:
Carbon dioxide
Explanation:
Neither helium nor carbon dioxide has a molecular dipole, so their strongest van der Waals attractive forces are London forces.
Helium is a small spherical atom with only a two electrons, so its atoms have quite weak attractions to each other.
CO₂ is a large linear molecule. It has more electrons than helium, so the attractive forces are greater. Furthermore, the molecules can align themselves compactly side-by-side and maximize the attractions (see below).
For example. CO₂ becomes a solid at -78 °C, but helium must be cooled to -272 °C to make it freeze (that's just 1 °C above absolute zero).
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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: