Compare the evaporation of a contained liquid with that of an uncontained liquid.

1 answer:

8 0

Answer: all other conditions equal, the rate evaporation of a contained liquid will be slower than the rate of evaporation of an uncontained liquid.

Justification:

1) The rate of evaporation increases as the surface area of the liquid (relative to the whole content) increases. This is, the greater the surface is the faster the evaporation.

2) That is so because the higher the surface of the liquid the more the number of particles in the liquid that are in contact with the surrounding air and so the more the particles will escape from the liquid to the air (which is what evaporation is).

3) A liquid contained will take the form of the container, so part of the liquid wil remain below the surface, while an uncontained liquid will spread all over the surface and so pratically all the liquid is in contact witht the air surrounding it.

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First, you need to find:
One mole of $CaCl_{2}$ is equivalent to how many grams?

Well, for this you have to look up the periodic table. According to the periodic table:
The atomic mass of Calcium Ca = 40.078 g (See in group 2)
The atomic mass of <span>Chlorine Cl = 35.45 g (See in group 17)
</span>
As there are two atoms of Chlorine present in $CaCl_{2}$ , therefore, the atomic mass of $CaCl_{2}$ would be:

Atomic mass of $CaCl_{2}$ = Atomic mass of Ca + 2 * Atomic mass of Cl

Atomic mass of $CaCl_{2}$ = 40.078 + 2 * 35.45 = 110.978 g

Now,

110.978 g of $CaCl_{2}$ = 1 mole.
75.9 g of $CaCl_{2}$ = $\frac{75.9}{110.978} moles$ = 0.6839 moles.

Hence,
The total number of moles in 75.9g of $CaCl_{2}$ = 0.6839 moles

According to <span>Avogadro's number,
1 mole = 1 * </span> $6.022 * 10^{23}$ molecules
0.6839 moles = 0.6839 * $6.022 * 10^{23}$ molecules = $4.118*10^{23}$ molecules

Ans: Number of molecules in 75.9g of $CaCl_{2}$ = $4.118*10^{23}$ molecules

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