Explanation:
The given data is as follows.
Mass flow rate of mixture = 1368 kg/hr
in feed = 40 mole%
This means that 
in feed = (100 - 40)% = 60%
We assume that there are 100 total moles/hr of gas 
in feed stream.
Hence, calculate the total mass flow rate as follows.
40 moles/hr of N_{2}/hr (28 g/mol of ) + 60 moles/hr of 
(2 g/mol of )
= 1120 g/hr + 120 g/hr
= 1240 g/hr
= 
(as 1 kg = 1000 g)
= 1.240 kg/hr
Now, we will calculate mol/hr in the actual feed stream as follows.
= 110322.58 moles/hr
It is given that amount of nitrogen present in the feed stream is 40%. Hence, calculate the flow of into the reactor as follows.
= 44129.03 mol/hr
As 1 mole of nitrogen has 28 g/mol of mass or 0.028 kg.
Therefore, calculate the rate flow of into the reactor as follows.
= 1235.612 kg/hr
Thus, we can conclude that the the feed rate of pure nitrogen to the mixer is 1235.612 kg/hr.
Answer:
Explanation:
because they want to make us aliens
There are 3 major stages in cellular respiration, glycolysis, Krebs cycle, and the electron transport chain.
To solve this question,
let us first calculate how much all the nucleons will weigh when they are apart,
that is:
<span>Mass of 25 protons = 25(1.0073) = 25.1825 amu </span>
Mass of neutrons = (55-25)(1.0087) = 30.261 amu
So, total mass of nucleons = 30.261+25.1825 =
55.4435 amu
<span>Now we subtract the mass of nucleons and mass of the Mn
nucleus:
55.4435 - 54.938 = 0.5055 amu
This difference in mass is what we call as the mass defect of
a nucleus. Now we calculate the binding energy using the formula:</span>
<span> E=mc^2 </span>
<span>But first convert mass defect in units of SI (kg):
Δm = 0.5055 amu = (0.5055) / (6.022x10^26)
<span>Δm = 8.3942x10^-28 kg</span>
Now applying the formula,
E=Δm c^2
E=(8.3942x10^-28)(3x10^8)^2
E=7.55x10^-11 J</span>
Convert energy from Joules
to mev then divide by total number of nucleons (55):
E = 7.55x10^-11 J *
(6.242x10^12 mev / 1 J) / 55 nucleons
<span>E = 8.57 mev / nucleon</span>
Answer:
Imine can be isolated from the reaction mixture as water is continuously removed from the reaction chamber
Explanation:
In this reaction, a non -aqueous solvent is not used (not mentioned in the question). Thus, we can say that there is continuous removal water under suitable reacting conditions and hence the imine formed is left behind.
