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.
The answer is B.
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I believe the answer is a compound because compounds can be separated when put together while mixtures can not.
Answer:
it gets reduced from a +3 oxidation to a 0.
Explanation:
the decomposition of iron oxide to elemenTal can be represented by the following equation: iron oxide (Fe203), the oxidation state of iron is +3 while that of oxygen is -2. therefore, the above reaction is a redox (reduction oxidation reaction)
Answer:- The pressure of ethanol would be 109 mmHg.
Solution:- This problem is based on Clausius clapeyron equation--
Given, 
= 63.5 + 273 = 336.5 K
= 34.9 + 273 = 307.9 K
= 400 mmHg
= ?
= 39.3 kJ/mol = 39300 J/mol
R = 8.314 J/mol.K
Let's plug in the values in the equation and do the calculations.
= 1.30
On taking anti ln to both sides...
= 
= 3.67
= 400/3.67
= 109 mmHg
