Yes u can help I need to see th worksheet to help tho
Answer:
a) P = 1240 lb/ft^2
b) P = 1040 lb/ft^2
c) P = 1270 lb/ft^2
Explanation:
Given:
- P_a = 2216.2 lb/ft^2
- β = 0.00357 R/ft
- g = 32.174 ft/s^2
- T_a = 518.7 R
- R = 1716 ft-lb / slug-R
- γ = 0.07647 lb/ft^3
- h = 14,110 ft
Find:
(a) Determine the pressure at this elevation using the standard atmosphere equation.
(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.
(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.
Solution:
- The standard atmospheric equation is expressed as:
P = P_a* ( 1 - βh/T_a)^(g / R*β)
(g / R*β) = 32.174 / 1716*0.0035 = 5.252
P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252
P = 1240 lb/ft^2
- The air density method which is expressed as:
P = P_a - γ*h
P = 2116.2 - 0.07647*14,110
P = 1040 lb/ft^2
- Using constant temperature ideal gas approximation:
P = P_a* e^ ( -g*h / R*T_a )
P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )
P = 1270 lb/ft^2
Answer:
the molecular formula for the gas is NO₂
Explanation:
since it contains
Nitrogen = n → 30.45%
Oxygen = o → 69.55%
and 30.45%+69.55% = 100% , then the gas only contains nitrogen and oxygen
Also we know that the proportion of oxygen over nitrogen is
proportion of oxygen over nitrogen = moles of oxygen / moles of nitrogen
since
moles = mass / molecular weight
then for a sample of 100 gr of the unknown gas
mass of oxygen = 69.55%*100 gr = 69.55 gr
mass of Nitrogen = 30.45%*100 gr = 30.45 gr
proportion of oxygen over nitrogen = (mass of oxygen/ molecular weight)/(mass of nitrogen / molecular weight of nitrogen ) = (69.55 gr/ 16 gr/mol) /( 30.45 gr /14 gr/mol) = 1.998 mol of O/ mol of N≈ 2 mol of O/ mol of N
therefore there are 2 atoms of oxygen per atom of nitrogen
thus the molecular formula for the gas is:
NO₂
Pretty sure it’s halogens , or groups 14-17
