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vovikov84 [41]
3 years ago
10

AICI3(s)+Ca 3N2(s)-AIN(S)+CaCl2(s)

Chemistry
1 answer:
disa [49]3 years ago
8 0

Answer:

2AlCl3 + Ca3N2 - 2AlN+ 3CaCl2

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How many molecules of N2O gas are in 250 L at STP?
Fittoniya [83]

Answer:

I have no idea all I know is poop.

Explanation:

Because I said so

4 0
2 years ago
Please help due tomorrow
lord [1]

Answer:

synthesis, single-replacement, double replacement, decomposition, or combustion reaction

Explanation:

pretty sure this is right

6 0
3 years ago
What conditions would produce the highest pressure?​
anzhelika [568]

air that is cooling will cause the highest pressure

Explanation:

This is because heat has to escape a certain area and if the heat can't escape then more and more pressure will build up in a area.

3 0
3 years ago
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
Fe(II) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts Fe(II)
vovangra [49]

Answer:

0,051g of O₂

Explanation:

The reaction of precipitation of Fe is:

4Fe(OH)⁺(aq) + 4OH⁻(aq) + O₂(g) + 2H₂O(l) → 4Fe(OH)₃(s)

<em>-Where the Fe(OH)⁺ is Fe(II) and Fe(OH)₃ is Fe(III)-</em>

This reaction is showing you need 1 mol of O₂(g) per 4 moles of Fe(II) for a complete reaction.

85mL= 0,085L of 0,075M Fe(II) are:

0,085L*0,075M = 6,34x10⁻³ moles of Fe(II)

For a complete reaction of 6,34x10⁻³ moles of Fe(II) you need:

6,34x10⁻³ moles of Fe(II)×\frac{1molO_2}{4moles Fe(II)} =

1.59x10⁻³ moles of O₂. In grams:

1.59x10⁻³ moles of O₂×\frac{32g}{1molO_2} = <em>0,051g of O₂</em>

<em></em>

I hope it helps!

7 0
3 years ago
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