Answer: depending on the method, 138-139 (imperial) gallons, or 626 L
Explanation:
The wt of water is 4924-3547 lb = 1377 lb = 1377/2.2 kg = 626 kg = 626 L
1gallon = 4.5 L (it does where I come from and who still measures things in imperial anyway?) so I guess that is 139 gallons
or alternatively, I recall from distant childhood, “a pint of water weighs a pound and a quarter” which means 1 gallon = 10 lb, so 1377 lb = ~138 gallons
Explanation:
The given data is as follows.
Moles of propylene = 100 moles, 
= 300 K
= 800 K, 
, 
of propylene = 100 J/mol
Now, we assume the following assumptions:
Since, it is a compression process therefore, work will be done on the system. And, work done will be equal to the heat energy liberating without any friction.
W = 
= 
= 5 MJ
Thus, we can conclude that a minimum of 5 MJ work is required without any friction.
Equation :
PbF2 ------> Pb2+ + 2F-
Solubility product is given as:
Ksp = [Pb2+][F-]^2
let x = [Pb2+]
Ksp = x(2x)^2 = 4(x)^3 = 4(2.08*10^-3)^3
Ksp = 3.60*10^-8
Answer:
<u>1.7 atm</u>
Explanation:
The formula you would want to use it P2=p1v1/v2
Plug in the numbers and solve
(1)(6)/3.5
Let me know if you need any other help!
Use Arrhenius equation:
k = A*exp(-Ea/RT)
We have:
1.35x10^2/s = A*exp(-85600/(8.314*298.15))
or: A = 1.342x10^17/s
It is a piece of cake to calculate:
k = 1.342x10^17*exp(-85600/(8.314*348.15))
= 1.92x10^4/s
