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Alika [10]
1 year ago
12

Clo3− draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. Show t

he formal charges of all nonhydrogen atoms.
Chemistry
1 answer:
Karo-lina-s [1.5K]1 year ago
3 0

The formal charges of all nonhydrogen atoms are -1.

Solution:-

<u>O 7-4 = 3 O Double bond on one H 5-4 = 1</u>

O-Cl-O 6-7 = -1x4 = -4 N 5-4=1 H-N-H 1-1=0

O 3-4= -1 O O 6-7 = -1(2)=-2 H 1-0=+1

<u>6-6 = 0 1-2 = -1</u>

It will percentage its last valence electron thru a single bond to the terminal oxygen atom. This is in agreement with carbon and hydrogen atoms that each need to form 4 and 1 covalent bonds respectively. because the terminal oxygen atom best has a single covalent bond, it'll have a proper rate of -1.

According to the lewis structure of SO2, The critical atom is sulfur and it is bonded with 2 oxygen atoms thru a double bond. each oxygen atom acquires 2 lone pairs of electrons and the primary sulfur atom has 1 lone pair of electrons.

Learn more about Nonhydrogen atoms here:-brainly.com/question/2822744

#SPJ4

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At a certain temperature the value of the equilibrium constant, Kc, is 1.27 for the reaction. 2As (s) + 3H2 (g) ⇌ 2AsH3 (g) What
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Answer:

0.887

Explanation:

Hello,

In this case, the law of mass action for the first reaction turns out:

Kc=\frac{[AsH_3]^2}{[As]^2[H_2] ^3}=1.27

Now, for the second reaction is:

Kc=\frac{[As][H_2] ^{3/2}}{[AsH_3]}

Therefore, by applying square root for the first reaction, one obtains:

\sqrt{Kc} =\sqrt{\frac{[AsH_3]^2}{[As]^2[H_2] ^3}} =\sqrt{1.27}

\frac{\sqrt{[AsH_3]^2} }{\sqrt{[As]^2} \sqrt{[H_2] ^3} } =\sqrt{1.27}

\sqrt{1.27}=\frac{[AsH_3]}{[As][H_2] ^{3/2}}

Finally, since Kc is asked for the inverse reaction, one modifies the previous equation as:

Kc'=\frac{1}{\sqrt{1.27} }=\frac{[As][H_2] ^{3/2}}{[AsH_3]}=0.887

Best regards.

8 0
3 years ago
Give 2 reasons that tea is a mixture or solutions​
cricket20 [7]

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Tea is a solution of compounds in water, so it is not chemically pure.

Explanation:

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stealth61 [152]

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