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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:

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Charlene conserves mass by collecting "the gas and dust released by the burning charcoal", and Lee does not . . . so the answer is . .

<u><em> . . . He did not collect the gas and dust released by the burning charcoal.</em></u>
3 0
3 years ago
You wish to prepare a tape-casting slip containing 50 vol% Al2O3 and 50 vol% polyvinyl butyral (PVB) binder. If the density of A
belka [17]

<u>Answer:</u> The mass of PVB required to produce 1000 grams of tape is 213.4 grams

<u>Explanation:</u>

To calculate the mass of aluminium oxide, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}       .......(1)

  • <u>For Al_2O_3</u>

We are given:

50% (v/v) of Al_2O_3

This means that 50 mL of aluminium oxide is present in 100 mL of tape

Calculating the mass of aluminium oxide by using equation 1:

Density of aluminium oxide = 3.98 g/cm^3

Volume of aluminium oxide = 50mL=50cm^3     (Conversion factor:  1mL=1cm^3 )

Putting values in equation 1, we get:

3.98g/cm^3=\frac{\text{Mass of aluminium oxide}}{50cm^3}\\\\\text{Mass of aluminium oxide}=(3.98g/cm^3\times 50cm^3)=199g

Mass of aluminium oxide = 199 g

  • <u>For PVB:</u>

We are given:

50% (v/v) of PVB

This means that 50 mL of PVB is present in 100 mL of tape

Calculating the mass of PVB by using equation 1:

Density of PVB = 1.08 g/cm^3

Volume of PVB = 50mL=50cm^3

Putting values in equation 1, we get:

1.08g/cm^3=\frac{\text{Mass of PVB}}{50cm^3}\\\\\text{Mass of PVB}=(1.08g/cm^3\times 50cm^3)=54g

Mass of PVB = 54 g

Mass of tape = Mass of aluminium oxide + mass of PVB

Mass of tape = [199 + 54] g = 253 g

To calculate the mass of PVB required to produce 1000 g of tape, we use unitary method:

When 253 grams of tape is made, the mass of PVB required is 54 g

So, when 1000 grams of tape is made, the mass of PVB required will be = \frac{54}{253}\times 1000=213.4g

Hence, the mass of PVB required to produce 1000 grams of tape is 213.4 grams

4 0
3 years ago
The light traveled from _____________________ to __________________. (What transparent materials?)
sergiy2304 [10]

Explanation:

High density medium to low density medium.

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2 years ago
Imagine a moving boat making waves in the middle of a pool of water. Also, imagine two observers, one behind the boat and one in
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3 0
3 years ago
What happens when a metal is reduced. Give an example of a metal being reduced
AlexFokin [52]

Answer:

The oxidation number of the metal decreases

2 Al  + Fe₂O₃ → Al₂O₃ + 2 FeO

The metal element iron, is reduced from Fe⁺³ in Fe₂O₃ to Fe⁺² in FeO

Explanation:

When an element gains electron, the element becomes reduced, hence when a metal is reduced, the metal gains electrons, which reduces the oxidation number of the metal

An example of a metal being reduced is;

2 Al  + Fe₂O₃ → Al₂O₃ + 2 FeO

In the above reaction, the iron (III) oxide is reduced to iron (II) oxide by aluminium metal.

6 0
3 years ago
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