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Dima020 [189]
2 years ago
14

Converts 800kg into ng

Chemistry
1 answer:
elixir [45]2 years ago
5 0

Answer:

yes

Explanation:

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The actions described in the passage are best understood in the context of which of the following?
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I have had this question on a test before!! Hope this helps
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What is a forensic paleontologist?<br>dum it down for me​
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3 years ago
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
Lera25 [3.4K]

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

= 310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}

= 305.97 atm

The temperature T = 415 K

The rate R = 0.0821 L.atm/mol.K

molar mass of O₂  gas = 32 g/mol

∴

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

d_{ideal} = 287.37 g/L

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;

R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol

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

∴

d_{Van \ der \ Waal} = \dfrac{32}{0.1152}

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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the answer is


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