How many grams of Ca(OH)2are required to make 1.5 L of a 0.81 M solution?
1 answer:
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Answer:
2.5 % butane, 42.2 % pentane and 55.3 % hexane
Explanation:
Hello,
In this case, the mass balance for each substance is given by:
Whereas accounts for the fractions at the outlet distillate and
for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:
So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:
The total distillate flow:
And the total bottoms flow:
Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:
Then, the molar fraction of pentane and hexane:
Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.
NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.
Regards.
Answer:
The edge length of a face-centered cubic unit cell is 435.6 pm.
Explanation:
In a face-centered cubic unit cell, each of the eight corners is occupied by one atom and each of the six faces is occupied by a single atom.
Hence, the number of atoms in an FCC unit cell is:
In a face-centered cubic unit cell, to find the edge length we need to use Pythagorean Theorem:
(1)
Where:
a: is the edge length
R: is the radius of each atom = 154 pm
By solving equation (1) for "a" we have:
Therefore, the edge length of a face-centered cubic unit cell is 435.6 pm.
I hope it helps you!