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dlinn [17]
3 years ago
6

The liquid-phase reaction follows an elementary rate law and is carried out isothermally in a flow system. The concentrations of

the A and B feed streams are 2 M before mixing. The volumetric flow rate of each stream is 5 dm3/min, and the entering temperature is 300 K. The streams are mixed immediately before entering. Two reactors are available. One is a gray, 200.0-dm^3 CSTR that can be heated to 77C or cooled to 0C, and the other is a white, 800.0-dm^3 PFR operated at 300 K that cannot be heated or cooled but can be painted red or black. Note that k = 0.07 dm^3 /mol/min at 300K and E =20 kcal/mol.
Required:
a. Which reactor and what conditions do you recommend? Explain the reason for your choice.
b. How long would it take to achieve 90% conversion in a 200-dm^3 batch reactor with after mixing at a temperature of 77C?
c. What would your answer to part (b) be if the reactor were cooled to 0C?
Chemistry
1 answer:
Vera_Pavlovna [14]3 years ago
5 0

e kajajdjae no squo viboes  we eso doadks Answer:ipao han meExplanation:no

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Dr. I. M. A. Brightguy adds 0.1727 g of an unknown gas to a 125-mL flask. If Dr. B finds the pressure to be 736 torr at 20.0°C,
AlladinOne [14]

Answer:

The gas that Dr. Brightguy added was O₂

Explanation:

Ideal Gases Law to solve this:

P . V = n . R . T

Firstly, let's convert 736 Torr in atm

736 Torr is atmospheric pressure = 1 atm

20°C = 273 + 20 = 293 T°K

125 mL = 0.125L

0.125 L . 1 atm = n . 0.082 L.atm / mol.K . 293K

(0.125L .1atm) / (0.082 mol.K /L.atm . 293K) = n

5.20x10⁻³ mol = n

mass / mol = molar mass

0.1727 g / 5.20x10⁻³ mol = 33.2 g/m

This molar mass corresponds nearly to O₂

7 0
3 years ago
Why a dipole develops in a molecule
Debora [2.8K]

Explanation:

When the covalent bonds in a molecule are polarized so that one portion of the molecule experiences a positive charge and the other portion of the molecule experiences a negative charge. This separation of opposite charges creates an electric dipole.

7 0
2 years ago
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The volume of 1.20 lol of gas at 61.3 kPa and 25.0 degree Celsius is
aniked [119]

Answer:

V = 48.5 L

Explanation:

Converting °C to K and kPa to atm

T = 25.0°C + 273.15 = 298.15 K

P = 61.3 kPa × (1 atm / 101.325 kPa) = 0.60498 atm

Calculating the volume of gas

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V = (1.20 mol)(0.082057 L•atm/mol•K)(298.15 K) / 0.60498 atm

V = 48.5 L

4 0
2 years ago
How many grams of kclo3 can be dissolved in 100 grams of water at 30 degrees Celsius
kondaur [170]
<span>The solubility of KClO</span>₃ : ( 10.1 / 100 g water ) at 30ºC

10.1 g ------------ 100 g ( H₂O )
    ? g ------------- 100 g ( H₂O )

Mass of KClO₃ :

100 * 10.1 / 100

1010 / 100 = 10.1 g of KClO₃

hope this helps!
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3 years ago
Metals tend to ________ electrons and form _______ ions.
masya89 [10]
<span>Metals tend to lose electrons and form electro-positive ions / cations.</span>
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