Answer:Oxygen,Carbon dioxide,Nitrogen
Explanation:
Answer: Partial pressures are 0.6 MPa for nitrogen gas and 0.4 MPa for carbon dioxide.
Explanation: <u>Dalton's</u> <u>Law</u> <u>of</u> <u>Partial</u> <u>Pressure</u> states when there is a mixture of gases the total pressure is the sum of the pressure of each individual gas:
The proportion of each individual gas in the total pressure is expressed in terms of <u>mole</u> <u>fraction</u>:
= moles of a gas / total number moles of gas
The rigid tank has total pressure of 1MPa.
molar mass = 14g/mol
mass in the tank = 2000g
number of moles in the tank: = 142.85mols
molar mass = 44g/mol
mass in the tank = 4000g
number of moles in the tank: = 90.91mols
Total number of moles: 142.85 + 90.91 = 233.76 mols
To calculate partial pressure:
For Nitrogen gas:
= 0.6
For Carbon Dioxide:
0.4
Partial pressures for N₂ and CO₂ in a rigid tank are 0.6MPa and 0.4MPa, respectively.
Answer:
abrir candados y abrir puertas
Explanation:
Answer:critical stress= 20.23 MPa
Explanation:
Since there was an internal crack, we will divide the length of the internal crack by 2
Length of internal crack, a = 0.7mm,
Half length = 0.7mm/2= 0.35mm changing to meters becomes
0.35/ 1000= 0.35 x 10 ^-3m
The formulae for critical stress is calculated using
σC = (2Eγs /πa) ¹/₂
σC = critical stress=?
Given
E= Modulus of Elasticity= 225GPa =225 x 10 ^ 9 N/m²
γs= Specific surface energy = 1.0 J/m2 = 1.0 N/m
a= Half Length of crack=0.35 x 10 ^-3m
σC= (2 x 225 x 10 ^ 9 N/m² x 1.0 N/m /π x 0.35 x 10 ^-3m)¹/₂
=(4.5 x 10^11/π x 0.35 x 10 ^-3)¹/₂
=(4.0920 x10 ^14)¹/₂
σC=20.23 x10^6 N/m² = 20.23 MPa
Answer:
We want to determine the location after 10mins
Explanation:
The release of diborane is continuous
It is release at a rate of 500lb per 20mins
Then, let find the rate in mg/s
1 lb = 453592.37 mg
So, the mass rate Q is
Q = 500lb / 20mins
Q = 500 × 453592.37 mg / 20 × 60sec
Q = 188,996.82 mg/s
Given that mass concentration of
m~ = 5mg/m³
Then,
Rate of volume is
V~ = Q / m~
V~ = 188,996.82 / 5
V~ = 37,799.364 m³/s
The wind speed is
V = 5mph
Let convert to m/s
1 mph = 0.447 m/s
Then, 5mph = 2.235 m/s
From Pasquill Gilford, the cloud atmosphere characteristic is class A
To know the location, we will divide the velocity by heat rate
X = V / Q
X = 2.235 / 188,996.82 mg/s
X = 2.235 / 0.188996kg/s
X = 11.83 m / kg
The location is 0.000001183 m per mg of diborane