It is moldm^-3s^-1
^ means raise to power
Answer: In gases the particles move rapidly in all directions, frequently colliding with each other and the side of the container. With an increase in temperature, the particles gain kinetic energy and move faster. The actual average speed of the particles depends on their mass as well as the temperature – heavier particles move more slowly than lighter ones at the same temperature. The oxygen and nitrogen molecules in air at normal room temperature are moving rapidly at between 300 to 400 metres per second. Unlike collisions between macroscopic objects, collisions between particles are perfectly elastic with no loss of kinetic energy.
Explanation: This is very different to most other collisions where some kinetic energy is transformed into other forms such as heat and sound. It is the perfectly elastic nature of the collisions that enables the gas particles to continue rebounding after each collision with no loss of speed. Particles are still subject to gravity and hit the bottom of a container with greater force than the top, and giving gases weight. Hope this helps with your problem! Byeeee :DDD
Answer:
265 mL is the new volume for the gas
Explanation:
We decompose the Ideal Gases Law in order to find the answer of this question: P . V = n . R . T
We can propose the formula for the 2 situations, where n remains constant.
R refers to 0.082 L.atm/mol.K which is physic constant.
We convert the temperature to Absolute value:
67.5°C + 273 = 340.5 K
80°C + 273 = 353 K
We convert the volume to L → 242.2 mL . 1 L/1000 mL = 0.2422 L
We convert the pressure values to atm:
882 Torr . 1 atm/ 760 Torr = 1.16 atm
840 Torr . 1atm / 760 Torr = 1.10 atm
P₁. V₁ / T₁ = P₂ . V₂ / T₂ → Let's replace data:
1.16 atm . 0.2422L / 340.5K = 1.10 atm . V₂ / 353 K
(1.16 atm . 0.2422L / 340.5K) . 353K = 1.10 atm . V₂
V₂ = 0.291 L.atm / 1.10 atm → 0.2647 L ≅ 265 mL
i think it is 8. I might be wrong.
Answer:
P_2 =0.51 atm
Explanation:
Given that:
Volume (V1) = 2.50 L
Temperature (T1) = 298 K
Volume (V2) = 4.50 L
at standard temperature and pressure;
Pressure (P1) = 1 atm
Temperature (T2) = 273 K
Pressure P2 = ??
Using combined gas law:
