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2 years ago

A 0.465 g sample of an unknown compound occupies 245 mL at 298 K and 1.22 atm. What is the molar mass of the unknown compound?

1 answer:

5 0

Using the Ideal Gas Law
PV= nRT
PV = (m/Mr) RT The molar mass of the unknown compound is : 38.0 g/mol

Mr = mRT / PV = 0.465 g x 0.08206 L * atm * K⁻¹ mol ⁻¹ x 298 K / 1.22 atm x 0.245 L = 38.0 g / mol

everything else but G / Mol are cancelled out.

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Answer:

Kc of reaction is 20.

Explanation:

The two proteins are X and Y.

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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:

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

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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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3 years ago

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Answer:

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<h2>Hey there! :) </h2>

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