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Ksju [112]
3 years ago
13

PLEASE HELP!!!

Chemistry
2 answers:
Ostrovityanka [42]3 years ago
8 0
Solubility is the maximum amount of a substance that will dissolve in a given amount of solvent at a specific temperature. There are two direct factors that affect solubility: temperature and pressure. Temperature affects the solubility of both solids and gases, but pressure only affects the solubility of gases.
yarga [219]3 years ago
4 0

Answer:

Here are some: Gas

1. The reaction is exothermic, so an increase in temperature means that solubility would decrease.

2. The change in solubility cannot be determined from the given information. Increasing pressure increased solubility, but increasing temperature decreases solubility

3. An increase in pressure and an increase in temperature in this reaction results in greater solubility.

4. An increase in pressure results in more gas particles entering the liquid in order to decrease the partial pressure. Therefore, the solubility would increase.

Water

The more polar the compound, the more soluble it is in water

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Because they’re both made up of two substances that are not chemically combined
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This supercontinent was part of what specific early Earth Science theory?
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Explanation:

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A sample of gas is held at constant pressure. Increasing the Kelvin temperature of this gas sample causes the average kinetic en
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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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