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Alborosie
3 years ago
8

Which scientific discipline belongs in the blue box?

Chemistry
2 answers:
Brums [2.3K]3 years ago
7 0

Answer:

chemistry I think.

Explanation:

I don't need a thanks.

Stella [2.4K]3 years ago
7 0

Answer:

chemistry

Explanation:

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Consider two solid blocks, one hot and the other cold, brought into contact in an adiabatic container. After awhile, thermal equ
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Answer: The statement is not correct because the decrease in entropy of the hot solid CANNOT BE equal to the increase in entropy of the cold one

Explanation:

Let us start by stating the second law of thermodynamics and it the second law of thermodynamics states that there is an entity called entropy and entropy will always increase all the time. Also, the second law of thermodynamics states that the change in entropy can never be negative. The second law of thermodynamics can be said to be equal to Change in the transfer of heat, all divided by temperature.

So, the first law of thermodynamics deals with the conservation of energy. But there is nothing like conservation of entropy.

Therefore, the decrease in entropy of the hot solid CANNOT BE equal to the increase in entropy of the cold one because entropy is not a conserved property.

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3 years ago
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the density of silver is 10,5 g/cm^3. what is the mass (in kilograms) of a cube of silver that measure .44 m on each side
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Chromium-58 is a(an)
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Answer:

Element

Explanation:

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Ethanol has a heat of vaporization of 38.56kj/mol and a normal boiling point of 78.4 ∘c.
Elanso [62]

This is an incomplete question, here is a complete question.

Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 °C. What is the vapor pressure of ethanol at 14 °C?

Answer : The vapor pressure of ethanol at 14.0^oC is 5.174\times 10^{-2}atm

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of ethanol at 14.0^oC = ?

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm

T_1 = temperature of ethanol = 14.0^oC=273+14.0=287K

T_2 = normal boiling point of ethanol = 78.4^oC=273+78.4=351.4K

\Delta H_{vap} = heat of vaporization = 38.56 kJ/mole = 38560 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{38560J/mole}{8.314J/K.mole}\times (\frac{1}{287K}-\frac{1}{351.4K})

P_1=5.174\times 10^{-2}atm

Hence, the vapor pressure of ethanol at 14.0^oC is 5.174\times 10^{-2}atm

4 0
3 years ago
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