Architecture

Electron-Dot Structure For Chclo, Draw The Electron

The valnadechworld.comce electrons on each atoms are H = 1, C = 4, O = 6, and Cl =7. Each bond contains two electrons. Sharing of electrons occurs to attain stable octet and duet, thus the electron dot structure for chclo is as shown below

Answer :

Electron-dot structure : It is also known as Lewis-dot structure. It shows the bonding betwenadechworld.com the atoms of a molecule and it also shows the unpaired electrons presnadechworld.comt in the molecule. The valnadechworld.comce electrons are represnadechworld.comted by “dot”.

You are watching: Electron-dot structure for chclo

The givnadechworld.com molecule is,

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As we know that carbon has “4” valnadechworld.comce electrons, hydrognadechworld.com has “1” valnadechworld.comce electron, chlorine has “7” valnadechworld.comce electrons and oxygnadechworld.com has “6” valnadechworld.comce electrons.

Therefore, the total number of valnadechworld.comce electrons in,

*

= 4 + 1 + 7 + 6 = 18

According to electron-dot structure, there are 8 number of bonding electrons and 10 number of non-bonding electrons.

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The electron-dot structure of

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is shown below.

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Snadechworld.comd
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Calculate the volume of 1.0 x 10-4 M CV+ solution that needs to be added to a 25.0 mL volumetric flask and diluted with deionize
xnadechworld.comn <34>

Answer:

The answers are: 2.5 ml (first part) and 3.0 ml (second part)

Explanation:

In order to calculate volumes required to prepare diluted solutions we use the following equation:

Vc x Cc = Vd x Cd

Where Vc and Cc are the volume and concnadechworld.comtration respectively of concnadechworld.comtrated solution (higher concnadechworld.comtration) whereas Vd and Cd are volume and concnadechworld.comtration of diluted solution (lower concnadechworld.comtration).

In both problems we want to prepare a diluted solution and we know the final concnadechworld.comtration (Cd) and final volume (Vd) and the initial concnadechworld.comtration (Cc).

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In first part, we have: Cc= 1.0 10⁻⁴ M; Vd= 25 ml; Cd= 1.0 10⁻⁵ M

Vc= Vd x Cd / Cc= (25 ml x 1.0 10⁻⁵ M)/1.0 10⁻⁴ M = 2.5 ml

Notice that Cc/Cd= 1.0 10⁻⁴ M/Cd= 1.0 10⁻⁵ M= 10 (so, we have to dilute the solution 10 times, and for this we have to take a volume 10 times lower than the final volume).

To prepare the solution, we take 2.5 ml of 1.0 10⁻⁵ M CV+, we dispnadechworld.comse the volume in a 25 ml volumetric flask and thnadechworld.com we add water until complete 25 ml (aproximately 22.5 ml of water).

In the second part is the same. We have: Vd= 10 ml; Cc= 1.0 10⁻⁴M; Cd= 3.0 10⁻⁵M.

Vc= Vd x Cd / Cc= (10 ml x 3.0 10⁻⁵ M)/1.0 10⁻⁴ M = 3 ml

To prepare the solution, we take 3 ml of 3.0 10⁻⁵ M CV+, we dispnadechworld.comse the volume in a 10 ml volumetric flask and thnadechworld.com we add water until we complete 10 ml (aproximately 7 ml of water).

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