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telo118 [61]
3 years ago
10

While elements and compounds can both be classified as pure substances because they have distinct properties and composition, __

________ cannot be decomposed further. A) compounds B) elements C) formulas D) molecules
Chemistry
2 answers:
Tresset [83]3 years ago
4 0
Elements as they are fundamental (one single hydrogen atom is still an element
makkiz [27]3 years ago
3 0
Elements or answer b
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Explanation:

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A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide
Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe

Next, the total moles:

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After that, since the process is isobaric, we can compute the work as:

W=P(V_2-V_1)

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3

Thereby, the magnitude and direction of work turn out:

W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).

Regards.

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