Answer:
3.43×10¹ mol
Explanation:
Given data:
Initial number of moles = 12.4 mol
Initial volume = 122.8 L
Final number of moles = ?
Final volume = 339.2 L
Solution:
The number of moles and volume are directly proportional to each other at same temperature and pressure.
V₁/n₁ = V₂/n₂
122.8 L/ 12.4 mol = 339.2 L / n₂
n₂ = 339.2 L× 12.4 mol / 122.8 L
n₂ = 4206.08 L.mol /122.8 L
n₂ = 34.3mol
In scientific notation:
3.43×10¹ mol
If Robert has 4 grams of a substance and Jill has 10 grams of the same substance <span>Jill's sample will weigh more than Robert's sample.</span>
Um a rainbow wand others colors ! Hoped I help!
The overall molecule is Polar because the shape of the molecule is Trigonal Pyramidal, which means it has the lone pair electrons. Becuase of the lone pair the pulling is unequal.
H3O+ has 3 polar bonds.
To know if the bonds are polar or nonpolar find the difference of the element's electronegativity charge.
H has electronegativity charge of 2.2, and O has 3.4.
Always subtract the smaller number from the greater one.
So 3.4 - 2.2 = 1.2
If the difference is from 0-0.4 the bond is nonpolar, but if it's from 0.5-1.9 the bond is polar.
So, 1.2 is polar bond. So H3O+ has 3 polar bonds, and the overall molecule is polar too.
A simple way to know if it's polar or nonpolar is to draw the lewis dot structure, and use VSEPR.
Larger molecules experience larger dispersion forces due to more distance of valance of electrons from the nucleus.
<h2>Cause of stronger dispersion force</h2>
Larger and heavier atoms and molecules have stronger dispersion forces than smaller and lighter ones because in a larger atom or molecule, the valence electrons are farther from the nuclei than in a smaller atom or molecule.
They are less tightly held to the nuclear charge present in the nucleus and can easily form temporary dipoles so we can conclude that larger molecules experience larger dispersion forces due to more distance of valance of electrons from the nucleus.
Learn more about London dispersion force here: brainly.com/question/1454795
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