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Julli [10]
2 years ago
12

A stream consisting of 48.9 mole% benzene (B) and 51.1% toluene (T) is fed at a constant rate to a process unit that produces tw

o product streams, one a vapor and the other a liquid. The vapor flow rate is initially zero and asymptotically approaches one third of the molar flow rate of the feed stream. Throughout this entire period, no material accumulates in the unit. When the vapor flow rate has become constant, the liquid is analyzed and found to be 28.0 mole% benzene.
Required:
How would you expect the vapor flow rate to vary with time?
Chemistry
1 answer:
marshall27 [118]2 years ago
7 0

The vapor flowrate will vary with time by ;  gradually increasing from zero to one half of the molar feed flow rate and then remain constant ( C )

<u>Although the options related to your question is missing attached below are the missing options </u>

<u />

Vapor flow rate is the amount of vapor per unit of time that flows through a specific area at a given time .

Given that the stream consists of 48.9 mole% of benzene and 51.1% toluene fed at a constant rate into a process unit. The vapor flowrate will vary with time because it will gradually increase from zero to 1/2 of the molar feed flow rate before it will remain constant.  

Hence we can conclude that The vapor flowrate will vary with time by  gradually increasing from zero to one half of the molar feed flow rate and then remain constant.

Learn more : brainly.com/question/24232544

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2 years ago
Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 1.000 mg of caffeine
Yuliya22 [10]

Answer:

  • <u>194 g/mol</u>

Explanation:

<u>1) Content of C:</u>

All the C atoms in the 1.000 mg of caffeine will be found in the 1.813 mg of CO₂.

  • Mass of C in 1.813 mg of CO₂

       Since the atomic mass of C is 12.01 g/mol and the molar of of CO₂ is 44.01, there are 12 mg of C in 44 mg of CO₂ and you can set the proporton:

       12.01 mg C / 44.01 mg CO₂ = x / 1.813 g CO₂

        ⇒ x = 1.813 × 12.01 / 44.01 g of C = 0.49475 mg of C

  • Number of moles of C

      number of moles = mass in g / atomic mass = 0.49475×10⁻³ g / 12.01 g/mol = 4.1195×10⁻⁵ moles = 0.041195 milimol

<u>2) Content of H</u>

All the H atoms in the 1.000 mg of caffeine will be found in the 0.4639 mg of H₂O

  • Mass of H in 0.4639 mg of H₂O

       Since the atomic mass of H is 1.008 g/mol and the molar of of H₂O is 18.015 g/mol, there are 2×1.008 mg of H in 18.015 mg of H₂O and you can set the proporton:

       2×1.008 mg H / 18.015 mg H₂O = x / 0.4639 mg H₂O

        ⇒ x = 0.4639 mg H₂O × 2 × 1.008 mg H / 18.015 mg H₂O = 0.051913 mg H

  • Number of moles of H

      number of moles = mass in g / atomic mass = 0.051913 ×10⁻³ g / 1.008 g/mol = 5.1501× 10⁻⁵ moles = 0.051501 milimol

<u>3) Content of N</u>

All the N atoms in the 1.000 mg of caffeine will be found in the 0.2885 mg of N₂

  • Mass of N in 0.2885 mg of N₂ is 0.2885 mg

  • Number of moles of N

      number of moles = mass in g / atomic mass = 0.2885 ×10⁻³ g / 14.007 g/mol = 2.0597× 10⁻⁵ moles = 0.020597 milimol

<u>4) Content of O</u>

The mass of O is calculated by difference:

  • Mass of O = mass of sample - mass of C - mass of H - mass of N

       Mass of O = 1.000 mg - 0.49475 mg C - 0.051913 mg H - 0.2885 mg N

     Mass of O = 0.1648 mg

  • Moles of O =  0.1648 × 10 ⁻³ g / 15.999 g/mol = 1.0303×10⁻⁵ mol = 0.01030 milimol

<u>5) Ratios</u>

Divide every number of mililmoles by the smallest number of milimoles:

  • C:  0.041195 / 0.01030 = 4
  • H: 0.051501 / 0.01030 = 5
  • N: 0.020597 / 0.01030 = 2
  • O: 0.01030 / 0.01030 = 1

  • C: 4
  • H: 5
  • N: 2
  • O: 1

<u>6) Empirical formula:</u>

  • C₄H₅N₂O₁

<u>7) Calculate the approximate mass of the empirical formula:</u>

  • 4 × 12 + 5 × 1 + 2 × 14 + 1 × 16 =  97 g/mol

So, since that number is not between 150 and 200 g/mol, multiply by 2: 97 × 2 = 194, which is between 150 and 200.

Thus, the estimate is 194 g/mol

7 0
3 years ago
Question 6 1.75 pts The following reaction 2H2S(g)=2H2(g)+S2(g), Kc=1.625x10-7 at 800°C is carried out at the same temperature w
kirza4 [7]

Answer:

[S₂] = 1.27×10⁻⁷ M

Explanation:

2 H₂S(g) ⇄ 2 H₂(g) + S₂(g), Kc=1,625x10⁻⁷

The equation of this reaction is:

1,625x10⁻⁷ = \frac{[H_2]^2[S_2]}{[H_{2}S]^2}

The equilibrium concentrations are:

[H₂S] = 0,162 - 2x

[H₂] = 0,184 + 2x

[S₂] = x

Replacing:

1,625x10⁻⁷ = \frac{[0,184+2x]^2[x]}{[0,162-2x]^2}

Solving:

4x³ + 0,736x² + 0,033856x - 4,3x10⁻⁹

x = 1.27×10⁻⁷

Thus, concentration of S₂ is:

<em>[S₂] = 1.27×10⁻⁷ M</em>

4 0
3 years ago
What is the balanced equation for calcium sulfate?​
kicyunya [14]

Answer:

CaSO4

Explanation:

Calcium sulfate (or calcium sulphate) is the inorganic compound with the formula CaSO4 and related hydrates.

8 0
3 years ago
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