PV=nRT n=PV/RT
R=.0821 atm x L
Mol x K
T=318K
P=.55atm
V=37.4L
n= (.55atm) (37.4L) /
.0821atm x L (318k)
mol x K
n= .79 mol of sulfur dioxide
Answer:
Q = 3,534.4 lbm/s = 212,062 lbm/min
Explanation:
Mass flowrate of discharge or leakage mass flowrate (Q) is given as
Q = AC₀√(2ρgP)
A = Cross sectional Area of leakage = (πD²/4) = (π×0.7²)/4
A = 0.385 ft²
C₀ = discharge coefficient = 0.98 (For maximum discharge flow rate, the flow is turbulent with discharge coefficient within 1% of 0.98)
ρ = density of butane at 76°F = 35.771 lbm/ft³
g = acceleration due to gravity = 32.2 lbm.ft/lbf.s²
P = Gauge Pressure in the tank = (absolute pressure) - (external pressure) = 19 - 1 = 18 atm = 38091.9 lbf/ft²
Q = AC₀√(2ρgP)
Q = (0.385)(0.98)√(2×35.771×32.2×38091.9)
Q = 3,534.4 lbm/s = 212,062 lbm/min
Hope this Helps!!!
Answer: between 6.5 and 8.5
Explanation:
As shown in my science book it appears in the ph scale
Answer:
6.68 x 10^-4
Explanation:
131g ÷ 261.337g/mol = 0.5012685 moles
0.5012685 moles ÷ 750.0 liters =
0.5012685÷ 750.0=0.000668358
6.68 x 10^-4
First we will calculate the number of moles of Iron:
, where n is the number of moles, m is the mass of iron in the reaction and M is the Atomic weight.
moles of Iron.
The same number of moles of Oxygen will take part in the reaction.
So
where 32 is the Atomical Weight of Oxygen (16 x 2).
=>
g
