Answer:
A cold front is a boundary or zone that separates two air masses. The cooler more dense mass replaces the warmer.
hope this helps :)
Explanation:
Atom is the
most basic unit of matter. They are the general term used to describe pieces of
matter. But a different kind of atom makes up an element. A combination of atom of different elements
creates a molecule. For example, you have water; it has a chemical formula of
H2O. If you separate H from O, you have two atoms. Two atoms for hydrogen and
one atom for oxygen. However, they are of different elements. And when you
combine them to form H2O, you create a molecule.
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Answer : The molal freezing point depression constant of liquid X is, ![4.12^oC/m](https://tex.z-dn.net/?f=4.12%5EoC%2Fm)
Explanation : Given,
Mass of urea (solute) = 5.90 g
Mass of liquid X (solvent) = 450 g = 0.450 kg
Molar mass of urea = 60 g/mole
Formula used :
![\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}}{\text{Molar mass of urea}\times \text{Mass of liquid X Kg}}](https://tex.z-dn.net/?f=%5CDelta%20T_f%3Di%5Ctimes%20K_f%5Ctimes%20m%5C%5C%5C%5CT%5Eo-T_s%3Di%5Ctimes%20K_f%5Ctimes%5Cfrac%7B%5Ctext%7BMass%20of%20urea%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20urea%7D%5Ctimes%20%5Ctext%7BMass%20of%20liquid%20X%20Kg%7D%7D)
where,
= change in freezing point
= freezing point of solution = ![-0.5^oC](https://tex.z-dn.net/?f=-0.5%5EoC)
= freezing point of liquid X = ![0.4^oC](https://tex.z-dn.net/?f=0.4%5EoC)
i = Van't Hoff factor = 1 (for non-electrolyte)
= Molal-freezing-point-depression constant = ?
m = molality
Now put all the given values in this formula, we get
![0.4^oC-(-0.5^oC)=1\times K_f\times \frac{5.90g}{60g/mol\times 0.450kg}](https://tex.z-dn.net/?f=0.4%5EoC-%28-0.5%5EoC%29%3D1%5Ctimes%20K_f%5Ctimes%20%5Cfrac%7B5.90g%7D%7B60g%2Fmol%5Ctimes%200.450kg%7D)
![K_f=4.12^oC/m](https://tex.z-dn.net/?f=K_f%3D4.12%5EoC%2Fm)
Therefore, the molal freezing point depression constant of liquid X is, ![4.12^oC/m](https://tex.z-dn.net/?f=4.12%5EoC%2Fm)
D. Electrons
Atoms can share their electrons in order to create bonds with other atoms
You would want to make sure that you have controlled the variables properly, and if you determine that you did then you would repeat the experiment to be sure of the results.