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DanielleElmas [232]
3 years ago
10

The volume of 160. g of CO initially at 273 K and 1.00 bar increases by a factor of two in different processes. Take CP,m to be

constant at the value 29.14 Jmol−1K−1 and assume ideal gas behavior. The temperature of the surroundings is 273 K.
A) Calculate ΔSsurroundings in an adiabatic reversible expansion.

B) Calculate ΔS in an adiabatic reversible expansion.

C) Calculate ΔStotal in an adiabatic reversible expansion.

D) Calculate ΔSsurroundings in an expansion against Pexternal = 0.

E) Calculate ΔS in an expansion against Pexternal = 0.

F) Calculate ΔStotal in an expansion against Pexternal = 0.

G) Calculate ΔSsurroundings in an isothermal reversible expansion.

H) Calculate ΔS in an isothermal reversible expansion.

I) Calculate ΔStotal in an isothermal reversible expansion.

Determine what processes are spontaneous.

1) adiabatic reversible expansion 2) expansion against Pexternal =0 3)isothermal reversible expansion
Chemistry
1 answer:
mel-nik [20]3 years ago
5 0

Answer:

Explanation:

w eFedfweF edf SEDFAsFGFSDSFG BAEWRDA G

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Upon combustion, a 1.3109 g sample of a compound containing only carbon, hydrogen, and oxygen produces 3.2007 g<img src="https:/
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Answer:

The answer to your question is:   C₃H₃O  This is my answer.

Explanation:

Data

Sample = 1.3109 g

CxHyOz

CO₂ = 3.2007 g

H₂O = 1.3102 g

Empirical formula = ?

MW CO2 = 44 g

MW H2O = 18 g

For Carbon

                                     44 g -------------------- 12 g

                                     3.2007 g ------------    x

                                      x = (3.2007 x 12) / 44

                                      x = 0.8729 g of Carbon

                                     12 g of C --------------  1 mol

                                     0.8729 g --------------  x

                                     x = (0.8729 x 1) / 12

                                     x = 0.0727 mol of Carbon

For Hydrogen

                                 18 g ---------------------- 1 g

                              1.3102 g -------------------  x

                                   x = (1.3102 x 1) / 18

                                  x = 0.0727 g of Hydrogen                                      

                                  1 g ------------------------ 1 mol

                                  0.0727g ----------------  x

                                  x = (0.0727 x 1)/1

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For oxygen

                    g of Oxygen = g of sample - g of Carbon - g of hydrogen

                    g of Oxygen = 1.3109 - 0.8709 - 0.0727

                    g of Oxygen = 0.3673

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                                  0.3673 g ---------------------   x

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                                   x = 0.0230 mol of Oxygen

Divide by the lowest number of moles

Carbon              0.0727 / 0.023  = 3.1  ≈ 3

Hydrogen         0.0727 / 0.023 = 3.1  ≈ 3

Oxygen             0.0230 / 0.023 = 1

                                        C₃H₃O

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