B. As particles travel in straight lines, their paths sometimes meet, and then they bounce apart with no gain or loss of energy.
Explanation:
The best statement that describes the collision of gas particles according to the kinetic-molecular theory is that as particles travel in straight lines, their paths sometimes meet and then they bounce apart with no gain or loss of energy.
- The kinetic molecular theory is used to explain the forces between molecules and their energy.
One of the postulate suggests that, when molecules collide with each other, or with the wall of the container, there is no loss or gain of energy.
- Molecules are independent of one another and that forces of attraction and repulsion between molecules are negligible.
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Answer: The density of Ammonia is 0.648 g/l
Explanation:
Density = Mass/ Volume
Mass of one mole of Ammonia (NH3) = 17.031g
Volume =?
Using the ideal gas law we can determine the volume.
PV = nRT
P = 0.913 atm, V= ?, n = 1, R = 0.08206 L.atm/K, and T= 293K
Make V the subject of the formular, we then have;
V= nRT/ P = 1 mol x 0.08206 L.atm/ K.mol x 293 / 0.913 atm
V = 24.04358/ 0.913 = 26.3L
Having gotten the value of Volume in this question, we then go back to solve for density.
Density = Mass/ Volume
17.031g/ 26.3L = 0.64756 ≈ 0.648 g/l
Answer:
2.1 × 10⁻¹ M
2.0 × 10⁻¹ m
Explanation:
Molarity
The molar mass of aniline (solute) is 93.13 g/mol. The moles corresponding to 3.9 g are:
3.9 g × (1 mol/93.13 g) = 0.042 mol
The volume of the solution is 200 mL (0.200 L). The molarity of aniline is:
M = 0.042 mol/0.200 L = 0.21 M = 2.1 × 10⁻¹ M
Molality
The moles of solute are 0.042 mol.
The density of the solvent is 1.05 g/mL. The mass corresponding to 200 mL is:
200 mL × 1.05 g/mL = 210 g = 0.210 kg
The molality of aniline is:
m = 0.042 mol/0.210 kg = 0.20 m = 2.0 × 10⁻¹ m
I don't know but I would pick b.2.covalent bonds
Answer:
The new temperature is 373 K
Explanation:
Step 1: Data given
Volume air = 5000 mL = 5.0 L
Temperature = 223K
New volume = 8.36 L
Step 2: Calculate the new temperature
V1/T1 = V2/T2
⇒V1 = the initial volume = 5.0 L
⇒T1 = the initial temperature = 223 K
⇒V2 = the new volume = 8.36 L
⇒T2 = the new temperature
5.0/223 = 8.36 /T2
T2 = 373 K
The new temperature is 373 K
