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ivolga24
11 hours ago
13

At the boiling point, the density of the liquid is 809 g/l and that of the gas is 4.566 g/l. how many liters of liquid nitrogen

are produced when 297.0 l of nitrogen gas is liquefied at 77.36 k?
Chemistry
1 answer:
KiRa [994]11 hours ago
6 0

Result: 1.68 L of liquid nitrogen is generated during the gas liquefaction process.

Clarification:

This process involves transforming gaseous nitrogen into its liquid form.

The two states possess distinct densities, thus occupying varying volumes; however, the mass remains constant.

Step 1: Calculate the mass of nitrogen gas

Let’s determine the mass of nitrogen gas associated with 297 L.

The density formula is:

Density = \frac{Mass}{Volume}

With a density of nitrogen gas at 4.566 g/L and a volume of 297 L, we can compute the mass of nitrogen gas as follows:

Using these values yields:

Mass = Density \times Volume

Mass = \frac{4.566g}{L} \times 297L

Mass = 1356g

The mass of nitrogen gas calculates to be 1356 g.

Step 2: Derive the volume of liquid nitrogen from the mass obtained

The mass for liquid nitrogen remains the same.

With the density of liquid nitrogen at 809 g/L, we can substitute this into our formula to find the volume of liquid.

Volume = \frac{Mass}{Density}

Volume = \frac{1356g}{809g/L}

Therefore, the volume of liquid nitrogen is 1.68 L.


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The correct statement is D. The water in Glass A is at a lower temperature compared to Glass B; consequently, the particles in Glass A exhibit reduced movement.

Explanation:

Raising the temperature increases the solubility of solutes.

The experiment indicates that's glass B is at a higher temperature than glass A since the antacid dissolves more quickly in glass B than in glass A. Therefore, glass A must be cooler, leading to slower particle movement compared to glass B.

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Answer:

a-294

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b-The molar mass is expressed in grams/mol, hence convert 1 mg to g, which equals 0.001, and divide it by the molar mass.

c/d-1 mole of any substance consists of 6.023×10^23 (ions, molecules, etc.), therefore we need to find the moles here as

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Butane (c4h10) undergoes combustion in excess oxygen to generate gaseous carbon dioxide and water. given δh°f[c4h10(g)] = –124.7
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The Δ H value for butane (g) is -124.7 kJ/mol.

The Δ H value for CO2 (g) is -393.5 kJ/mol.

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Considering the reaction,

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Now, we will calculate how many moles of butane are in 8.30 grams.

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= 0.143 moles

Therefore, the total energy released during the reaction is given by,

Q = number of moles × ΔH° rxn

= 0.143 × (2658.3)

= 380.14 kJ

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The full question can be found in the image linked to this response.

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