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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V = ( 2.44 x 10⁻² ) x ( 1.4 x 10⁻³ ) x ( 8.4 x 10⁻³ )
V = 2.9 x 10⁻⁷ m³
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Dilution<span> is when you decrease the concentration of a </span>solution<span> by adding a solvent. As a result, if you want to </span>dilute<span> salt water, just add water. ... Add more solute until it quits dissolving. That point at which a solute quits dissolving is the point at which it's </span>saturated<span>.</span>
Answer:
Empirical formula:
Fe₂O
Subscript of iron in empirical formula is 2.
Explanation:
Empirical formula:
It is the simplest formula that gives the smallest whole number ratio between the atoms of elements present in a compound.
Given data:
Mass of iron = 87.47 g
Mass of oxygen = 12.53 g
Empirical formula = ?
Solution:
Number of gram atoms of iron = 87.47 /55.845 =1.57 grams atoms
Number of gram atoms of oxygen = 12.53 / 16 = 0.78 grams atoms
Atomic ratio:
Fe : O
1.57/0.78 : 0.78/0.78
2 : 1
Fe : O = 2 : 1
Empirical formula:
Fe₂O
Subscript of iron in empirical formula is 2.
Answer: The concentration of Silver Nitrate in solution # 4 is 3.5x10-6 M
Explanation:
When calculating the concentration of a dilute solution, we use the equation:
C1V1 = C2V2
Where C1 = concentration of the known solution
V1= amount of know solution
C2= new concentration
V2= capacity of the flask
Keep in mind that solution #2 is not related to the preparation of solutions #3 y #4, so the concentration of #2 is irrelevant.
For calculating the concentration of solution #4 , we first calculate the concentration of the solution #3, that comes from the solution #1.
- Calculating concentration solution #3
- C1V1 = C3V3
- C3= (0.0002442 M * 10mL) / 250mL
- C3= 9.8X10-6M
2. Calculating concentration solution #4
- C3V3 = C4V4
- C4= (9.8X10-6M *18mL) / 50mL
- C4= 3.5X10-6 M