A 0.223 mole sample of gas is held at 33.0 C and 2.00 atm, What's the volume of the gas? R = 0.0821 L atm / mol K answer soon il
1 answer:
Answer:
The volume of the gas is 2.80 L.
Explanation:
An ideal gas is a theoretical gas that is considered to be made up of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
The Pressure (P) of a gas on the walls of the container that contains it, the Volume (V) it occupies, the Temperature (T) at which it is located and the amount of substance it contains (number of moles, n) are related from the equation known as Equation of State of Ideal Gases:
P*V = n*R*T
where R is the constant of ideal gases.
In this case:
- P= 2 atm
- V= ?
- n=0.223 moles
- R= 0.0821
- T=33 °C= 306 °K (being O°C= 273°K)
Replacing:
2 atm* V= 0.223 moles*0.0821 * 306 K
Solving:
V= 2.80 L
<u><em>The volume of the gas is 2.80 L.</em></u>
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Answer:
a): not necessarily due to London Dispersion Forces and dipole-dipole interactions.
b): not necessarily due to London Dispersion Forces.
Explanation:
There are three major types of intermolecular interaction:
- Hydrogen bonding between molecules with H-O, H-N, or H-F bonds and molecules with lone pairs.
- Dipole-dipole interactions between all molecules.
- London dispersion forces between all molecules.
The melting point of a substance is a result of all three forces, combined.
Note that the more electrons in each molecule, the stronger the London Dispersion Force. Generally, that means the more atoms in each molecule, the stronger the London dispersion force. The strength of London dispersion force between large molecules can be surprisingly strong.
For example, (water) molecules are capable of hydrogen bonding. The melting point of
at
is around
. That's considerably high when compared to other three-atom molecules.
In comparison, the higher alkane hexadecane (, straight-chain) isn't capable of hydrogen bonding. However, under a similar pressure, hexadecane melts at around
above the melting point of water. The reason is that with such a large number of atoms (and hence electrons) per molecule, the London dispersion force between hexadecane molecules could well be stronger than that the hydrogen bonding between water molecules.
Similarly, the dipole moments in HCl (due to the highly-polar H-Cl bonds) are much stronger than those in hexadecane (due to the C-H bonds.) However, the boiling point of hexadecane under standard conditions is much higher (at around than that of HCl.