1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)
=> p1 = 2 p2
Which is easy to demonstrate using ideal gas equation:
p1 = nRT/V = 2.0 mol * RT / 1 liter
p2 = nRT/V = 1.0 mol * RT / 1 liter
=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2
2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.
So, the pressure in both chambers (which form one same vessel) is:
p = nRT/V = 3.0 mol * RT / 2liter
which compared to the initial pressure in chamber 1, p1, is:
p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1
So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.
You can also see how the pressure in chamber 2 changes:
p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
Answer:
PART A: The LDF occurs between all molecules. Dispersion forces result from shifting electron clouds, which cause weak, temporary dipole.
PART B: Dipole dipole operates only between polar molecules. This is when two polar molecules get near each other and the positively charged portion of the molecule is attracted to the negatively charged portion of another molecule.
PART C: Dipole dipole and in some cases hydrogen bonding operate between the hydrogen atom of a polar bond and a nearby small electronegative atom. Only if the atom bonded to it were F, O or N it would be hydrogen bonding. Otherwise it is dipole dipole.
Answer:
Most common insulation materials work by slowing conductive heat flow and--to a lesser extent--convective heat flow. Radiant barriers and reflective insulation systems work by reducing radiant heat gain. To be effective, the reflective surface must face an air space.
Explanation:
To be effective, the reflective surface must face an air space.
<h3>
Answer:</h3>
The root mean square speeds of O₂ and UF₆ is 513m/s and 155 m/s respectively.
<h3>
Solution and Explanation:</h3>
- To find how fast molecules or particles of gases move at a particular temperature, the root mean square speed is calculated.
- Root mean square speed of a gas is calculated by using the formula;
Where R is the molar gas constant, T is the temperature and M is the molar mass of gas in Kg.
<h3>Step 1: Root mean square speed from O₂</h3>
Molar mass of Oxygen is 32.0 g/mol or 0.032 kg/mol
Temperature = 65 degrees Celsius or 338 K
Molar gas constant = 8.3145 J/k.mol
<h3>
Step 2: Root mean square speed of UF₆ </h3>
The molar mass of UF₆ is 352 g/mol or 0.352 kg/mol
Therefore; the root mean square speeds of O₂ and UF₆ is 513m/s and 155 m/s respectively.
The last set which is n=4 l=3 m=3 is a valid set
