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2 years ago

Predict the sign of ΔS for each process:(c) Cl₂(g) (100°C and 1 atm) → Cl₂(g) (10°C and 1 atm)

Chemistry

1 answer:

dexar [7]2 years ago

7 0

The entropy for the process:

Cu(s) (350°C and 2.5 atm) → Cu(s) (450°C and 2.5 atm) decreases. The sign of entropy is negative.

<h3>What is Entropy? </h3>

Entropy is defined as the measure of the thermal energy of a system per unit temperature which is unavailable for doing useful work. As the work is obtained from ordered molecular motion, therefore the amount of entropy is also gives a measure of the molecular disorder, or randomness, of a system.

<h3>Effect of Temperature</h3>

There is increase in the Entropy as temperature increases. An increase in the temperature results that particles of substance have greater kinetic energy. The fast moving particles have more disorder than particles which are moving slowly at the given lower temperature.

<h3>Effect of Pressure</h3>

The entropy increases with increases in the pressure on the substance.

Since the temperature of given reaction decreases. So, entropy also decreases.

Thus, we concluded that the entropy for the process:

Cu(s) (350°C and 2.5 atm) → Cu(s) (450°C and 2.5 atm) decreases. So, the sign of entropy is negative.

learn more about Entropy:

brainly.com/question/14131507

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A new substance forms with identical properties

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How much energy is required to convert 15.0 g of ice at −106 °C to water vapor at 125 °C? Specific heats are 2.09 J/g K for both

myrzilka [38]

Answer:

49.3 kJ of energy is required

Explanation:

An exercise of calorimetry at its best

First of all, convert the ice to water before melting.

Q = ice mass . C . ΔT

Q = 15 g . 2.09 J/g°C (0° - (-106°C)

15 g . 2.09 J/g°C . 106°C = 3323.1 J

Now we have to melt the ice, to change its state

Q = mass . latent heat of fusion

Q = 15 g . 0.335 kJ/g = 5.025 kJ .1000 = 5025 J

After that, we have liquid water at 0° and the ice has melted completely. We have to release energy to make a temperature change, to 100° (vaporization)

Q = 15g . 4.18 J/g°C (100°C - 0°C)

Q = 6270 J

Water has been vaporizated so we have to calculate, the state change.

Q = mass . latent heat of vap

Q = 15 g. 2.260 kJ/g

Q = 33.9 kJ (.1000) = 33900 J

Finally we have to increase temperature from 100°C to 125°C

Q = 15 g . 2.09 J/g°C . (125°C - 100°C)

Q = 783.75 J

To know how much energy is required to conver 15 g of ice, to water vapor at 125°C, just sum all the heat released.

3323.1 J + 5025 J + 6270 J + 33900 J + 783.75 J = 49301.85 joules.

Notice I have to convert kJ to J in two calcules to make the sum.

49301.85 joules / 1000 = 49.3 kJ

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3 years ago

HOW IS A COMPOUND DIFFERENT FROM A MIXTURE?

valentinak56 [21]

D.It can not be separated by a physical change.

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3 years ago

A tank contains an ideal gas mixture of 5 g of O2 and 8 g of CO2 at 160kPa and specified temperature. If O2 were separated from

erica [24]

Answer:

74 or 74 kPa.

Explanation:

Hello,

In this case, based on the initial information, it is seen that the oxygen and the carbon dioxide form the mixture at 160 kPa, thus, by isolating the oxygen, its pressure will be equal to its initial partial pressure because it gets isolated, hence, we compute its molar fraction as:

$x_{O_2}=\frac{5gO_2*\frac{1molO_2}{32gO_2} }{5gO_2*\frac{1molO_2}{32gO_2} +8gCO_2*\frac{1molCO_2}{44gCO_2} } =0.46$

Therefore, its initial pressure turns out:

$p_{O_2}=160kPa*0.46=73.9kPa$

Such pressure will be the oxygen's pressure once it is isolated. Finally, considering the request, the answer will be just 74 (by rounding to the nearest integer and without units).

Best regards.

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3 years ago