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Oliga [24]
3 years ago
13

What mass of chlorine gas is contained in a 10.0 L tank at 31.0° C and 3.50 atm?

Chemistry
1 answer:
jeyben [28]3 years ago
8 0
You need to use the ideal gas law (PV=nRT) and solve for n. ((3.50atm•10.0L)/(0.0821(L•atm/mol•K)•304K) = n = 1.40 moles. 1 mole of Cl2 = 70.9 gm/mole. The mass would be 99.43 gm
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I will give brainliest. If you burn the carbon in limited air, the reaction is
Fynjy0 [20]

This reaction is different in that the carbon undergoes an incomplete combustion as opposed to complete combustion where carbon is fully oxidized. A caveat: incomplete combustion products in general can be difficult to predict without sufficient information, as it's not uncommon to obtain a mixture of different products.

Here, we are told that solid carbon is burned in limited air to produce a gas. I am presuming that, in the equation that's given, the "0" represents a blank where you must fill in a chemical symbol. In this case, our equation would be: 2C(s) + O₂(g) → 2CO(g).

There is not enough information here to provide the numerical answers to the two questions. From the words in the question (e.g., "is different" and "this time"), it would seem that this question is an excerpt from a larger or preceding question where specific numbers had been provided or computed.

However, it's possible to make some general observations on how one may go about answering these questions <em>if </em>one had more information.

Since we're to assume that oxygen is the limiting reagent, if one is given the amount of solid carbon (either in mass, moles, or number of atoms), it's possible to determine the moles of CO(g) that's produced since C and CO have an equal stoichiometric ratio. So, for example, if one burns 2 moles of C(s), then 2 moles of CO(g) would be produced.

<em><u>But</u></em>, there is still not enough information to compute the volume of CO gas if this is the line of questioning. We don't know, for instance, the temperature or pressure of the reaction conditions. In fact, the only way it would be possible to answer this would be if you were given beforehand a conversion factor that relates the volume of CO(g) to its quantity (e.g., to assume that one mole of gas occupies <em>x </em>liters).

As for the second question, this would depend on what you know about the quantity of the C(s) reacted and/or the quantity (or volume, from question a) of CO(g) produced. If you can get the number of moles of C(s) reacted or CO(g) produced, the number of moles of O₂(g) used up: It would be half the number of moles of C(s) reacted or half the number of moles of CO(g) produced). <u>Again</u>, it's impossible to determine the volume of O₂(g) using just the information provided here, so I suspect that you must have further information relating gas quantity to volume. As we did with CO(g), the volume of O₂(g) used up can be found using whatever conversion factor you have.

If you have any further information or questions, please feel free to follow up.  

6 0
3 years ago
What is the density of chlorine gas (MM = 71.0glmol) at 1.50 atm and 25.0C
IrinaVladis [17]

Answer: D=4.35g/L

Explanation:

The formula for density is D=\frac{M}{V}. M is mass in grams and V is volume in liters.

Since we are give pressure and temperature, we can use the ideal gas law to find moles/volume. FInding moles/volume would give us the base for density. All we would have to do is convert moles to grams.

Ideal Gas Law: PV=nRT

\frac{n}{V} =\frac{P}{RT}

\frac{n}{V}=\frac{1.50 atm}{(0.08206Latm/Kmol)(25+274.15K)}

\frac{n}{V} =\frac{0.061309mol}{L}

Now that we have moles, we can use molar mass of chlorine gas to find grams.

0.061309mol*\frac{71.0g}{mol} =4.3529g

With our grams, we can find our density.

D=\frac{4.3529g}{L}

We need correct significant figures so our density is:

D=\frac{4.35g}{L}

5 0
3 years ago
Consider the following data on some weak acids and weak bases:
Juli2301 [7.4K]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

8 0
3 years ago
What types of compounds do not dissolve in water?
Nataliya [291]
Oil doesn't dissolve in water, which could be your answer.
8 0
3 years ago
Boron’s chemistry is not typical of its group.(a) Cite three ways in which boron and its compounds differ significantly from the
Mila [183]

Boron’s chemistry is not typical of its group. is group 3A (13) shows the increasing metallic character from Al to Tl.

All Boron compounds are covalent whereas the other elements in group 3A (13) form mostly ionic compounds.

Except for Boron, the other elements of group 3A (13) show increasing metallic character from Al to Tl. But Boron is a metalloid.

Compared to the other elements in group 3A, boron has a lower reactivity in chemical terms (13)

The metalloid boron (B), as well as the metals aluminium (Al), gallium (Ga), indium (In), and thallium, are all part of group 3A (or IIIA) of the periodic table (Tl). In contrast to the other members of Group 3A, the element borax primarily forms covalent connections.

To learn more about group 3A (13) refer the link:

brainly.com/question/5489194

#SPJ4

6 0
1 year ago
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