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pogonyaev
3 years ago
11

Can a tank of oxygen gas ever be half empty? explain

Chemistry
2 answers:
ikadub [295]3 years ago
3 0
<span>No, it is practically very unlikely situation. Oxygen is a gas and gas has no shape it occupies the shape of the container. moreover, according to the kinetic theory of gases, the particles of gas move in random motion occuping the space. The particles move randomly and keep on colliding with each other. Therefore, it is not correct to say that a tank of oxygen gas can ever be half empty.</span>
MatroZZZ [7]3 years ago
3 0
<span>The answer is yes. The compacted air moves to the sifter bed channels. The strainer bed channel assumes an imperative part, as the gadget expels the nitrogen from the air. After air is first compacted in the concentrator, it is constrained into the primary strainer bed. Oxygen is sent into the item tank.</span>
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What would happen to the annual rates of limestone (CaCO3) precipitation in the ocean if the ocean water were to become much war
stiv31 [10]

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CaCO3 exoskeleton dissolves in acidic water

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4 0
3 years ago
Which of the following reactions have a positive ΔSrxn? Check all that apply.
PolarNik [594]

Answer:

The reactions that have a <em>positive ΔS rxn </em>are the first and the fourth choices:

  • <em>2A(g) + B(s) → 3C(g)</em>

  • <em>2A(g) + 2B(g) → 5C(g)</em>

Explanation:

<em>ΔS rxn </em>is the change of entropy of the chemical reaction.

ΔS rxn = S after reaction - S before reaction.

Therefore, a positive ΔS rxn  means that the entropy after the reaction is greater than the entropy before the reaction.

You may use some assumptions to predict whether a reaction will lead an increase or decrease of the entropy.

First, assume that all the non-shown conditions, such as temperature and pressure, are constant.

Under that assumption, and from the meaning of entropy as a measure of the disorder or randomness of a system you can predict the sign of the change of entropy.

  • <em><u>2A(g) + B(s) → 3C(g)</u></em>

        1)  The solid compounds, B(s) in this case, are very ordered and so they have low entropy.

        2) Gas molecules are highly disordered (scattered), and the greater the number of molecules of the gas the larger the entropy, S).

Hence, since the product side shows 3 gas molecules and the reactant side shows 2 gas molecules and 1 solid molecule, you predict that the products have a larger entropy than the reactants, meaning an increase in entropy: <em>ΔS rxn is positive.</em>

  • <em><u>2A(g) + B(g) → C(g)</u></em>

Using the same reasoning, 3 gas molecules in the  reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

  • <u><em>A(g) + B(g) → C(g)</em></u>

Again, 2 gas molecules in the  reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

  • <u><em>2A(g) + 2B(g) → 5C(g)</em></u>

With the same reasoing, 5 molecules in the product side, lets you predict that will have more entropy than 4 molecules in the reactant side, and, the entropy will increase: <em>ΔS rxn is positive.</em>

6 0
3 years ago
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