Answer:
A. 72.34mol/min
B. 76.0%
Explanation:
A.
We start by converting to molar flow rate. Using density and molecular weight of hexane
= 1.59L/min x 0.659g/cm³ x 1000cm³/L x 1/86.17
= 988.5/86.17
= 11.47mol/min
n1 = n2+n3
n1 = n2 + 11.47mol/min
We have a balance on hexane
n1y1C6H14 = n2y2C6H14 + n3y3C6H14
n1(0.18) = n2(0.05) + 11.47(1.00)
To get n2
(n2+11.47mol/min)0.18 = n2(0.05) + 11.47mol/min(1.00)
0.18n2 + 2.0646 = 0.05n2 + 11.47mol/min
0.18n2-0.05n2 = 11.47-2.0646
= 0.13n2 = 9.4054
n2 = 9.4054/0.13
n2 = 72.34 mol/min
This value is the flow rate of gas that is leaving the system.
B.
n1 = n2 + 11.47mol/min
72.34mol/min + 11.47mol/min
= 83.81 mol/min
Amount of hexane entering condenser
0.18(83.81)
= 15.1 mol/min
Then the percentage condensed =
11.47/15.1
= 7.59
~7.6
7.6x100
= 76.0%
Therefore the answers are a.) 72.34mol/min b.) 76.0%
Please refer to the attachment .
Answer:
864 KN
Explanation:
Atmospheric pressure is defined as the force per unit area exerted against a surface by the weight of the air above that surface.
Please kindly check attachment for the step by step solution of the given problem.
Answer:
1) a. Customers requiring AC electric power transmission for powering remote devices which may include a subsea transmission system where power is distributed to subsea devices
b. Customers that utilize commutator-type motors
2) The primary voltage is 150 volts
Explanation:
The parameters given are;
Number of primary winding, 
= 50 turns
Number of secondary winding, 
= 150 turns
Voltage in secondary winding, 
= 450 volts
Voltage in primary winding = 
The relation between , , and is as follows;
Which gives;
From which we have;
The primary voltage = 150 volts.
Answer:
D
Explanation:
You don't want to crash into a construction vehicle.
Answer:
7 bits
Explanation:
Given
Instruction Set = 110 operation
Memory unit = 32 bits per word.
We get the required bits by using the following formula
2^n = 110
But 110 is not a factor of 2.
So, we pick the nearest decimal number greater than 110 that is a power of 2.
The number is 128
2^n = 110 becomes
2^n = 128
2^n = 2^7 ---- 2 cancels out
So,
n = 7
Hence, the required number of bits needed for the opcode is 7 bits
