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tatyana61 [14]
3 years ago
11

What best describes why a liquid needs a container when a solid does not?

Chemistry
2 answers:
kvv77 [185]3 years ago
7 0
Because if you have a liquid then you need a glass to keep it together and when it is a solid it is already together so you don't need to do anything
Tanya [424]3 years ago
5 0
Someone deleted my comment XD
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anyanavicka [17]
0.066-0.0069= 0.0591
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2 years ago
Role and significance of guassion or normal distribution?
kkurt [141]
Normal my guy!!!!!!!!!
5 0
3 years ago
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Distinguish between a solution in general and an aqueous solution
Sveta_85 [38]

Answer:

  • What distinguish a solution in general from an aqueous solution is the solvent. A solution in general may contain any solvent, which may be solid, liquid or gas, while an aqueous solution is formed with water as solvent.

Explanation:

A solution in general is a homogeneous mixture in which a substance, named solute, is dissolved, in other substance, name solvent.

Solutions may be in solid, liquid or gas state. There are many kind of solvents. Usually, in a lab you work with liquid solutions. Some liquid solvents are: ethanol, glycerin, hexane, benzene, and water, among many others.

Aqueous solution is a solution where the solvent is water. Of course, the solute may be any one: NaCl, sugar, ethanol, an acid, a base, a salt.

What distinguish a solution in general and an aqueous solution is the solvent.

3 0
3 years ago
What is the relationship between a mole and Avogadro number
docker41 [41]
A mole contains Avogadro’s number of particles of a substance.
4 0
3 years ago
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What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given tha
sladkih [1.3K]

Answer:

The mean free path = 2.16*10^-6 m

Explanation:

<u>Given:</u>

Pressure of gas P = 100 kPa

Temperature T = 300 K

collision cross section, σ = 2.0*10^-20 m2

Boltzmann constant, k = 1.38*10^-23 J/K

<u>To determine:</u>

The mean free path, λ

<u>Calculation:</u>

The mean free path is related to the collision cross section by the following equation:

\lambda =\frac{1}{n\sigma }------(1)

where n = number density

n = \frac{P}{kT}-----(2)

Substituting for P, k and T in equation (2) gives:

n = \frac{100,000 Pa}{1.38*10^{-23} J/K*300K} =2.42*10^{25}\  m^{3}

Next, substituting for n and σ in equation (1) gives:

\lambda =\frac{1}{2.42*10^{25}m^{-3}* .0*10^{-20}m^{2}}=2.1*10^{-6}m

6 0
3 years ago
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