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faltersainse [42]
3 years ago
10

ethylene and steam at 320 celsius and 1 atm are fed to a reaction process as an equimolar mixture. the process produces ethanol

which leaves the process as liquid at 25 celcius. what is the heat transfer associated with the process per mole of ethanol produced

Chemistry
1 answer:
Charra [1.4K]3 years ago
6 0

Answer:

Total amount of heat transfer, H_total = 115653.232 J/mol

Explanation:

Full explanation and step by step instructions is attached

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Anika [276]
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3 years ago
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A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit
Evgesh-ka [11]

Answer:

The answer to your question is   P2 = 0.78 atm

Explanation:

Data

Temperature 1 = T1 = 263°K                 Temperature 2 = T2 = 298°K

Volume 1 = V1 = 24 L                             Volume 2 = V2 = 35 L

Pressure 1 = P1 = 1                                  Pressure 2 = P2 = ?

Process

1.- To solve this problem use the Combined gas law

                          P1V1/T1 = P2V2/T2

-Solve for P2

                           P2 = P1V1T2 / T1V2

-Substitution

                          P2 = (1)(24)(298) / (263)(35)

-Simplification

                          P2 = 7152 / 9205

-Result

                          P2 = 0.777

   or                    P2 = 0.78 atm

5 0
3 years ago
Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form. calculate [In-}/{HIn} for the fol
zvonat [6]

Answer:

1.58x10⁻⁵

2.51x10⁻⁸

0.0126

63.10

Explanation:

Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[In-]/[HIn]

pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,

i) pH = 4.9

4.9 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = - 4.8

[In-]/[HIn] = 10^{-4.8}

[In-]/[HIn] = 1.58x10⁻⁵

ii) pH = 2.1

2.1 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -7.6

[In-]/[HIn] = 10^{-7.6}

[In-]/[HIn] = 2.51x10⁻⁸

iii) pH = 7.8

7.8 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -1.9

[In-]/[HIn] = 10^{-1.9}

[In-]/[HIn] = 0.0126

iv) pH = 11.5

11.5 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = 1.8

[In-]/[HIn] = 10^{1.8}

[In-]/[HIn] = 63.10

6 0
3 years ago
Given similar concentrations, the stronger acid corresponds to the lower pH. Comment on the relative strengths of the acids H3PO
andreyandreev [35.5K]

Answer:

H3PO4 is stronger than H2PO4- because

H3PO4 dissociation constant is 6.9×10^-3

H2PO4^- dissociation constant is 6.2×10^-8

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2 years ago
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D. due to the the water it will bring sand with the water there for us is D.
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2 years ago
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