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Katena32 [7]
3 years ago
14

A chemist takes 50-gram sample of sulfur powder that has a melting point of 115.2 °C. What is the melting point of a 100-gram sa

mple of sulfur powder?
115.2 °C since melting point is an extensive property

115.2 °C since melting point is an intensive property

Double of 115.2 °C since melting point is an intensive property

Double of 115.2 °C since temperature is an extensive property
Chemistry
1 answer:
Viktor [21]3 years ago
7 0

Answer:

115.2 °C since melting point is an intensive property  

Step-by-step explanation:

The melting point of a substance does not depend on how much you have.

For example, the melting point of water is 0 °C, whether it is an ice cube from the refrigerator or in the frozen pond outside.

The freezing point of a substance is an <em>intensive property</em>.

Thus, the melting point of 100 g of sulfur is 115.2 °C because melting point in an intensive property.

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10 m3 of carbon dioxide is originally at a temperature of 50 °C and pressure of 10 kPa. Determine the new density and volume of
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Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and 7.2m^3 respectively.

Explanation :

First we have to calculate the new or final volume of carbon dioxide gas.

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 10 kPa

P_2 = final pressure of gas = 15 kPa

V_1 = initial volume of gas = 10m^3

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 50^oC=273+50=323K

T_2 = final temperature of gas = 75^oC=273+75=348K

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

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V_2=7.2m^3

The new volume of carbon dioxide gas is 7.2m^3

Now we have to calculate the new density of carbon dioxide gas.

PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}

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\rho_2=\frac{P_2M}{RT_2}

where,

P_2 = new pressure of gas = 15 kPa

T_2 = new temperature of gas = 75^oC=273+75=348K

M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

\rho = new density

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

\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}

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The correct answer between all the choices given is the third choice or letter C. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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Explanation:

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