First of all, the formula for finding Kelvin is Celsius + 273
Therefore, if we subtract 273, we get the temperature in degrees
120 - 273 = - 153
Therefore, the answer is (1), or -153 degrees Celsius
Hope this helped!! :D
When P1/P2 = C1/C2
and C is the molarity which = moles/volume
so, P1/P2 = [(mass1/mw)/volume] / [(mass2/mw)/volume]
P1/P2 = (mass1/mw)/1.5L / (mass2/mw)/1.5L
so, Mw and 1.5 L will cancel out:
∴P1/P2 = mass1 / mass2
∴ mass 2 = mass1*(P2 / P1)
= 0.278g * (78 bar / 62 bar)
= 0.35 g
∴ the quantity of argon that will dissolve at 78 bar = 0.35 g
Answer:
The right solution according to the question is provided below.
Explanation:
According to the question,
(a)
The initial conditions will be:
DO = 
= 
= 
The initial oxygen defict will be:
Do = 
= 
The initial BOD will be:
Lo = 
= 
= 
(b)
The time reach minimum DO:
tc = ![\frac{1}{(kr-kd)} ln{(\frac{0.76}{0.61} )[1-\frac{1.674(0.76-0.61)}{0.61\times 6.453} ]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%28kr-kd%29%7D%20ln%7B%28%5Cfrac%7B0.76%7D%7B0.61%7D%20%29%5B1-%5Cfrac%7B1.674%280.76-0.61%29%7D%7B0.61%5Ctimes%206.453%7D%20%5D%7D)
= 
By putting the values of log, we get
= 
The distance to reach minimum DO will be:
Xc = 
= 
= 
Answer:
The answer is Option a, that is "−9kJmole,5kJmole".
Explanation:
Please find the complete question in the attached file.
In the question, it uses the catalyst inside a process, which does not modify the process eigenvalues, however, it decreases the active energy with an enthalpy of -9kJmole, and also the power for activating decreases around 13 to 5 kJ mole, that's why the choice a is correct.
You didn’t give us the options???
anyways: Water diffusion is called osmosis. Oxygen is a small molecule and it's nonpolar, so it easily passes through a cell membrane. Carbon dioxide, the byproduct of cell respiration, is small enough to readily diffuse out of a cell. Small uncharged lipid molecules can pass through the lipid innards of the membrane.