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jeka94
2 years ago
11

Carl has an empty 35.0-liter tank that he wants to fill with 18.3 moles of oxygen gas. The gas in the tank will have a temperatu

re of 25°C. If the air surrounding the tank is at standard pressure, what will be the gauge pressure of the oxygen in the tank in atmospheres?
Chemistry
1 answer:
Anna11 [10]2 years ago
5 0

The gauge pressure of the oxygen in the tank in atmospheres is 11.46 atm.

<h3>What is gauge pressure?</h3>

The gauge pressure is given as the pressure relative to the ambient atmospheric pressure. The gauge pressure can be calculated as:

Gauge pressure = System pressure - Atmospheric pressure

The system pressure of the 35-liter tank with 18.3 moles oxygen is given as:

Pressure = moles * Rydberg constant * Temperature / Volume

Pressure = 18.3 moles * 0.08 L.atm.mol⁻¹.K⁻¹ * 298 K / 35 L

Pressure = 12.46 atm

The standard pressure is given as 1 atm.

Thus, the gauge pressure of the oxygen in the tank can be:

Gauge pressure = System pressure - Atmospheric pressure

Gauge pressure = 12.46 atm - 1 atm

Gauge pressure = 11.46 atm

The gauge pressure of the oxygen in the tank in atmospheres is 11.46 atm.

Learn more about gauge pressure, here:

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Answer:The 1st and 2nd reactions are the example of oxidation -reduction.

Explanation:

Oxidation is basically when a species  loses electrons and  reduction is basically when the species gains  electrons.

A reaction is known as an oxidation -reduction reaction only if  oxidation and reduction simultaneously occur in the reaction. It basically means if a species is getting oxidized in the reaction then the other species present in the system must be reduced in the reaction.

Oxidation-reduction reactions are also known as redox reactions.

In the 1st reaction the oxidation state of Na in reactant is 0 and in products is +1 hence Na is oxidized and the oxidation state of   chlorine is  0 in reactants and  in products is  -1 so chlorine is reduced. Hence Na is oxidized and Cl is reduced so the reaction is a example of oxidation-reduction.

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In the second reaction the oxidation state of Na in reactant is 0 and in products is +1 hence Na is oxidized and the oxidation state of Cu is +1 in reactant and 0 in products so Cu is reduced. Hence Na is oxidized and Cu is reduced so the reaction is an example of oxidation-reduction.

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In the third reaction the oxidation state of Na changes from +1 to +1 and that of Cu also changes from +1 to +1. So there is no change in oxidation state of the species present in reactants and products. Hence this reaction is not an example of oxidation and reduction.

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Which element has the greatest number of valence electrons available for bonding? selenium boron calcium chlorine
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Explanation :

The given elements are, Se, B, Ca, Cl

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The general configuration shows the number of valence electrons are present in the element.

The number of valence electrons present = 2 + 4 = 6

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Atomic number = 5

Group number = 13

General Electronic configuration of the group number 13 elements is, ns^2np^1.

The number of valence electrons present = 2 + 1 = 3

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General Electronic configuration of the group number 2 elements is, ns^2.

The number of valence electrons present = 2

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Atomic number = 17

Group number = 17

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The number of valence electrons present = 2 + 5 = 7

From this we conclude that the chlorine is the element which has the greatest number of valence electrons available for bonding.

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