1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
vladimir1956 [14]
3 years ago
10

Question NO LINKS!

Chemistry
1 answer:
yuradex [85]3 years ago
3 0

Answer:

Climate of Georgia. Maritime tropical air masses dominate the climate in summer, but in other seasons continental polar air masses are not uncommon.

Explanation:

You might be interested in
Which of the following will form a solution? (Select all that apply.)
SSSSS [86.1K]

Answer:

sand and water

Explanation:

when u mix these both it will become like a paste

4 0
3 years ago
An ___is usually a homologous structure that is shared by all organisms in a group~science
Paraphin [41]
A shared derived characteristics is usually a homologous structure, such as a backbone, that is shared by all organisms in a group.
8 0
3 years ago
A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

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:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

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:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

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.

8 0
3 years ago
Read 2 more answers
How many grams of N2 are in 44.8L at STP?
lutik1710 [3]

Answer: 10 i think

Explanation:

3 0
3 years ago
Câu 12. Hiện tượng nào sau đây không phải là sự nóng chảy?
Alik [6]
Mình nghĩ C là đáp án đúng
5 0
3 years ago
Other questions:
  • Why is calgon conditioning better than phosphate conditioning?
    6·1 answer
  • PLEASE HELP ME BUT FAST!
    9·2 answers
  • Write a balanced half-reaction describing the oxidation of solid chromium to aqueous chromium(IV) cations.
    5·1 answer
  • 10) A hydrobromic acid (HBr) solution has a molar concentration of 0.0085 M. Calculate the
    9·1 answer
  • Assuming equal concentrations of conjugate base and acid, which one of the following mixtures is suitable for making a buffer so
    13·1 answer
  • A chemist performed a reaction in which two chlorine atoms form a compound with a magnesium atom through an ionic bond. Which st
    9·1 answer
  • Which statement is true of covalent bonds?
    9·1 answer
  • How many grams of vanadium, V, would contain the same number of atoms as 32.066 g of sulfur, S?
    8·1 answer
  • A 2.50 L balloon is filled with water at 2.27 atm. If the balloon is squeezed into a 0.80 L beaker and does NOT burst, what is t
    13·2 answers
  • If you place a weighing boat (0.229 g) on the balance, hit the tare/zero button, weigh 0.678 g of NaCl, then remove everything f
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!