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grandymaker [24]
3 years ago
14

A mass of 2 kg of acetylene is in a 0.045 m3 rigid container at a pressure of 4.3 MPa. Use the generalized charts to estimate th

e temperature.
Chemistry
1 answer:
sattari [20]3 years ago
5 0

Answer:

361 K

Explanation:

From the tables:

P_r=4.3/6.14=0.7,T_c=308.3\ K, R=0.2193\ kJ/kgK\\\\Volume (V)=0.045m^3, mass(m)=2kg\\\\v=V/m=0.045/2=0.0225\ m^3/kg

Given that:

v=\frac{ZRT}{P}

From the charts, at P_r=0.7,Z_g=0.59,T_r=0.94

v_g=\frac{0.59*0.3193*0.94*308.3}{4300}=0.0127\ m^3/kg\ which\ is\ small\\ \\At\ T_r=1,Z=0.7,v=\frac{0.7*0.3193*1*308.3}{4300}=0.016\ m^3/kg\\\\At\ T_r=1.2,Z=0.86,v=\frac{0.86*0.3193*1.2*308.3}{4300}=0.0236\ m^3/kg

Interpolating, we get T_r=1.17, T = 361 K

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The density of no2 in a 4.50 l tank at 760.0 torr and 25.0 °c is ________ g/l.
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Answer:

A

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Therefore, from the chemical equation, we have that:


\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} +  \Delta H^\circ_f \text{ H$_2$O}  \right]   -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}

In conclusion, our answer is A.

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