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Volgvan
3 years ago
13

Why do real gases not behave exactly like ideal gases?

Chemistry
2 answers:
Korvikt [17]3 years ago
5 0

The theory assumes that collisions between gas molecules and the walls of a container are perfectly elastic, gas particles do not have any volume, and there are no repulsive or attractive forces between molecules .

BigorU [14]3 years ago
4 0

At high pressure and low temperature, real gas deviates from ideal gas. Compressibility factor is used to measure the deviation of real gas from ideal gas.

The compressibility factor is "n" which equals PV/RT.This is equal to one in case of ideal gas and this condition does not hold good for real gas.


At high pressures, the molecules in the gas tend to become closer, resulting in greater intermolecular forces. Internal molecular force holds these molecules together,prevents movement and collision and thus reduces the pressure of the gas much below than that of an ideal gas.


At low temperatures, the kinetic energy of the gas molecules also becomes less thus this results in less collision of the gas molecules and thus finally this results in the decrease in pressure of the gas molecules which would be much lower than that of an ideal gas.


For an ideal gas, the translational kinetic energy is 3/2kT. This condition does not hold good for a real gas.

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Answer:

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3 years ago
Calculate the solubility of carbon dioxide in water at an atmospheric pressure of 0.400 atm (a typical value at high altitude).A
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Answer:

The molar solubility of carbon dioxide gas is 4.662\times 10^{-6} M.

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{CO_2}=K_H\times p_{liquid}

where,

K_H = Henry's constant = 3.50\times 10^{-2}mol/L.atm

p_{CO_2} = partial pressure of carbonated drink

p_{CO_2}=p\times \chi_{CO_2}

where = p = Total pressure = 0.400 atm

\chi_{CO_2} = mole fraction of CO_2=3.33\times 10^{-4}

p_{CO_2}=0.400 atm\times 3.33\times 10^{-4} =0.0001332 atm

Putting values in above equation, we get:

C_{CO_2}=3.5\times 10^{-2}mol/L.atm\times 0.0001332  atm\\\\C_{CO_2}=4.662\times 10^{-6} M

Hence, the molar solubility of carbon dioxide gas is 4.662\times 10^{-6} M.

8 0
3 years ago
What do you know about Relative Dating
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When a 17.7 mL sample of a 0.368 M aqueous hypochlorous acid solution is titrated with a 0.301 M aqueous barium hydroxide soluti
nordsb [41]

Answer:

pH = 12.98

Explanation:

Step 1: Data given

Volume of aqueous hypochlorous acid solution = 17.7 mL = 0.0177 L

Molarity of aqueous hypochlorous acid solution = 0.368 M

Molarity of aqueous barium hydroxide solution = 0.301 M

Volume of aqueous barium hydroxide solution = 16.2 mL = 0.0162 L

Step 2: The balanced equation

2HCl + Ba(OH)2 → BaCl2 + 2H2O

Step 3: Calculate moles

Moles = molarity * volume

Moles HCl = 0368 M * 0.0177 L

Moles HCl = 0.0065136 moles

Moles Ba(OH)2 = 0.301 M * 0.0162 L

Moles Ba(OH)2 = 0.0048762 moles

Step 4: Calculate the limiting reactant

For 2 moles HCl we need 1 mol Ba(OH)2 to produce 1 mol BaCl2 and 2 moles H2O

HCl is the limiting reactant. It will completely be consumed 0.0065136 moles. Ba(OH)2 is in excess. There will react 0.0065136/2 = 0.0032568‬ moles. There will remain 0.0048762 moles - 0.0032568‬  = 0.0016194 moles

Step 5: Calculate molarity Ba(OH)2

Molarity Ba(OH)2 = moles / volume

Molarity Ba(OH)2 = 0.0016194 moles / 0.0339 L

Molarity Ba(OH)2 = 0.04777 M

Step 6: Calculate [OH-]

Ba(OH)2 → Ba^2+ + 2OH-

For Ba(OH)2 we have 2* [OH-]

[OH-] = 2*0.04777 = 0.09554 M

Step 7: Calculate pOH

pOH = -log[OH-]

pOH = -log(0.09554)

pOH = 1.02

Step 8: Calculate pH

pH = 14 - 1.02

pH = 12.98

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goldfiish [28.3K]
<span> There are 97.2 mol of O in 10.8 mol of Fe(NO3)3</span>
7 0
3 years ago
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