Warm air can “hold” more water vapor than cool air because as the air warms its molecules move farther apart, making room for more molecules. This leads to the idea that as air cools its molecules move closer together, “squeezing” out water vapor.
An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction UNLESS acted upon by an unbalanced force.
Answer:
Explanation:
1)
Given data:
Initial volume of balloon = 0.8 L
Initial temperature = 12°C ( 12+273= 285 K)
Final temperature = 300°C (300+273 = 573 K)
Final volume = ?
Solution:
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 0.8 L .573 K / 285 K
V₂ = 458.4 L / 285
V₂ = 1.61 L
2)
Initial pressure = 204 kpa
Initial temperature = 29°C ( 29 + 273 = 302 K)
Final temperature = ?
Final pressure = 300 kpa
Solution:
P₁/T₁ = P₂/T₂
T₂ = T₁P₂/P₁
T₂ = 302 K . 300 kpa / 204 kpa
T₂ = 90600 K/ 204
T₂ = 444.12 K
3)
Given data:
Initial volume = 14 L
Initial pressure = 2.1 atm
Initial temperature = 100 K
Final temperature = 450 K
Final volume = ?
Final pressure = 1.2 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 2.1 atm × 14 L × 450 K / 100 K × 1.2 atm
V₂ = 13230 L / 120
V₂ = 110.25 L
Answer:
trigonal planar
Explanation:
The molecule SO3 is of the type AX3. The molecule is symmetrical and non polar.
There are three regions of electron density in the molecule. This corresponds to a trigonal planar geometry. This means that the three oxygen atoms are arranged at the corners of a triangle. The bond angle is 120 degrees.
