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koban [17]
3 years ago
15

When are atoms considered stable?

Chemistry
2 answers:
Debora [2.8K]3 years ago
7 0

Answer:

<u>Atoms are at their most stable when their outermost energy level is either empty of electrons or filled with electrons. Sodium atoms have 11 electrons. Two of these are in the lowest energy level, eight are in the second energy level and then one electron is in the third energy level.</u>

Explanation:

<em>purple</em><em> </em><em>u</em><em> </em><em /><em> </em>

katrin [286]3 years ago
6 0

Answer:

when they have satisfied the octet rule naturally or through bonding to obtain full valence shells

Explanation:

Generally, most atoms of an element are unstable because they have a void in their electron shell to fill, hence, they need to react with other elements to fulfil this task of octet.

Octet rule states that atoms of elements engage in reactions to form compounds so they can have eight (8) valence electrons in their shell. Noble gases e.g argon, neon etc. are elements that have naturally satisfied this octet rule by possession of 8 valence electrons in their shell. Other elements that do not have this naturally becomes reactive and enter bonding with other atoms to obtain full valence shells.

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How many liters of 3.5 M solution can be made using 23 moles of LiBr *must show work to get credit*
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<u>Answer:</u> 6.57 L of solution can be made.

<u>Explanation:</u>

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}} .....(1)

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Putting values in equation 1, we get:

3.5mol/L=\frac{23mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{23mol}{3.5mol/L}=6.57L

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Answer:

A. write balanced chemical equation (including states), for this process.

Explanation:

Almost all hydrocarbon 'burn' reactions involve oxygen; it's by far the most reactive substance in air.  

Hydrocarbon combustions always involve  

[some hydrocarbon] + oxygen --> carbon dioxide + steam.  

C6H6(l) + O2 (g)--> CO2 (g)+ H2O (g)

Balance carbon, six on each side:  

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Balance hydrogen, six on each side:  

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Now, we have fifteen oxygens on the right and O2 on the left.  

Two ways to deal with that. We can use a fraction:  

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Or, if you prefer to have whole number coefficients, double everything  

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C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(g)

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