Answer:
volume of the bubble just before it reaches the surface is 5.71 cm³
Explanation:
given data
depth h = 36 m
volume v2 = 1.22 cm³ = 1.22 × 
m³
temperature bottom t2 = 5.9°C = 278.9 K
temperature top t1 = 16.0°C = 289 K
to find out
what is the volume of the bubble just before it reaches the surface
solution
we know at top atmospheric pressure is about P1 = 
Pa
so pressure at bottom P2 = pressure at top + ρ×g×h
here ρ is density and h is height and g is 9.8 m/s²
so
pressure at bottom P2 = + 1000 × 9.8 ×36
pressure at bottom P2 =4.52 × Pa
so from gas law
here p is pressure and v is volume and t is temperature
so put here value and find v1
V1 = 5.71 cm³
volume of the bubble just before it reaches the surface is 5.71 cm³
Answer:
C the particle must be somewhere.
Explanation:
This is because normalization of wave function means the maximum probability of finding a particle in a region is 1. And a Wave function describes the probability of finding a particle in region. Also Since it is a probability distribution, its integral over all space must be 1, explaining that the probability that the particle is somewhere and thus it must integrate to 1, meaning it must be it must be normalizable
Answer:
Vrms = 291 m/s
Explanation:
The root mean square velocity or vrms is the square root of the average square velocity and is. vrms=√3RTM. Where M is equal to the molar mass of the molecule in kg/mol.
Temperature = 365 K
Root mean square velocity = ?
molar mass of oxygen = 16 g/mol.
But xygen gas (O2) is comprised of two oxygen atoms bonded together. Therefore:
molar mass of O2 = 2 x 16
molar mass of O2 = 32 g/mol
Convert this to kg/mol:
molar mass of O2 = 32 g/mol x 1 kg/1000 g
molar mass of O2 = 3.2 x 10-2 kg/mol
Molar mass of Oxygen = 3.2 x 10-2 kg/mol
Vrms = √[3(8.3145 (kg·m2/sec2)/K·mol)(365 K)/3.2 x 10-2 kg/mol]
Vrms = 291 m/s
