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Over [174]
3 years ago
14

The molar volume of oxygen,O2, is 3.90 dm3 mol-1 at 10.0 bar and 200 degree centigrade. Assuming that the expansion may be trunc

ated after the second term, calculate the second virial coefficient of oxygen at this temperature.
Chemistry
1 answer:
schepotkina [342]3 years ago
4 0

Answer:

B = - 0.0326 dm³/mol

Explanation:

virial eq until second term:

  • PVm = RT [ 1 + B/Vm ]

∴ P = 10 bar * (atm/ 1.01325 bar) = 9.869 atm

∴ T = 200°C = 473 K

∴ Vm = 3.90 dm³/mol

∴ R = 0.08206 dm³.atm/K.mol

⇒ PVm / RT = 1 + B/Vm

⇒ ((9.869 atm)*(3.90 dm³/mol)) / ((0.08206 dm³.atm/mol.K)*(473K)) = 1 + B/Vm

⇒ 0.99164 = 1 + B/Vm

⇒ B/Vm = - 8.357 E-3

⇒ B = (3.90 dm³/mol)*( - 8.357 E-3 )

⇒ B = - 0.0326 dm³/mol

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vova2212 [387]

Answer:

44.2 L

Explanation:

Use Charles Law:

\frac{V1}{T1} =\frac{V2}{T2}

We have all the values except for V₂; this is what we're solving for. Input the values:

\frac{56 L}{346K} =\frac{V2}{273K}   -  make sure that your temperature is in Kelvin

From here, we need to get V₂ by itself. To do this, multiply by 273 on both sides:

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2 years ago
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Mrac [35]

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Explanation:

7 0
3 years ago
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ANTONII [103]

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3 years ago
The chemical formula of ordinary sugar is C12H22O11. Calculate the mass of 7.35 mol of sugar.
ch4aika [34]

First, you need to find the mass of 1 mol of sugar. Mass, or molar mass, can simply be found by adding the masses of the individual elements. These are given to you on the periodic table.

C_{12}H_{22}O_{11}

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Add all of the pieces together.

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342.297 g/mol x 7.35 mol =  2,515.88 grams

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