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3 years ago

One mole of nitrogen and one mole of neon are combined in a closed container at STP.How big is the container?V = __________ L

Chemistry

1 answer:

cestrela7 [59]3 years ago

5 0

Answer : The volume of container will be 44.8 L

Explanation :

At STP condition,

The temperature and pressure are 273 K and 1 atm respectively.

As we know that at STP, 1 mole of substance occupies 22.4 L volume of gas.

As per question,

1 mole of nitrogen gas occupies 22.4 L volume

and,

1 mole of neon gas also occupies 22.4 L volume

Thus, total volume will be:

Total volume of container = 22.4 + 22.4 = 44.8 L

Hence, the volume of container will be 44.8 L

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Answer:

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Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

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$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

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$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

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Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

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