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lara31 [8.8K]
2 years ago
13

What is the density of a board dimensions are 5.54 cm x 10.6 cm X 199 cm and whose mass is 28.6kg

Chemistry
1 answer:
exis [7]2 years ago
4 0

Answer:

d=2.44\ g/cm^3

Explanation:

Given that,

The dimensions of the board is 5.54 cm x 10.6 cm X 199 cm.

The mass of the board is 28.6 kg.

Since, 1 kg = 1000 grams

28.6 kg = 28.6 × 1000 g = 28600 grams

Density = mass/volume

So,

d=\dfrac{28600\ g}{(5.54\times 10.6\times 199)\ cm^3}\\\\d=2.44\ g/cm^3

So, the density of a board is 2.44\ g/cm^3.

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Calculate the density of O2(g) at 415 K and 310 bar using the ideal gas and the van der Waals equations of state. Use a numerica
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Answer:

Explanation:

From the given information:

The density of O₂ gas = d_{ideal} = \dfrac{P\times M}{RT}

here:

P = pressure of the O₂ gas = 310 bar

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= 305.97 atm

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d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}

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To find the density using the Van der Waal equation

Recall that:

the Van der Waal constant for O₂ is:

a = 1.382 bar. L²/mol²    &

b = 0.0319  L/mol

The initial step is to determine the volume = Vm

The Van der Waal equation can be represented as:

P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}

where;

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Replacing our values into the above equation, we have:

310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}

310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}

310V^3 -44.389V^2+1.382V-0.044=0

After solving;

V = 0.1152 L

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d_{Van \ der \ Waal} = 277.77  g/L

We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.

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