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fgiga [73]
3 years ago
15

The Lewis structure for a chlorate ion, ClO3-, should show ____ single bond(s), ____ double bond(s), and ____ lone pair(s).

Chemistry
2 answers:
Yuliya22 [10]3 years ago
7 0

<u>Answer:</u> A molecule of ClO_3^- has 1 single bond, 2 double bonds and 8 lone pairs.

<u>Explanation:</u>

Lewis-dot structure is defined as the structure which represents the number of valence electrons around the atoms. The electrons are represented as dots.

This structure helps in the determination of bonding electrons and non-bonding electrons.

For the given compound ClO_3^-

As we know that chlorine has '7' valence electrons and oxygen has '6' valence electron.  The negative charge gets add up in the total number of valence electrons.

Total number of valence electrons in ClO_3^- = 7 + 3(6) + 1 = 26

In ClO_3^-, two electrons of 2 oxygen atoms combine with 2  electrons of chlorine and 1 electron of one oxygen atom combines with 1 electron of chlorine atom.

A lone pair remains on the central atom which is chlorine.

Hence, a molecule of ClO_3^- has 1 single bond, 2 double bonds and 8 lone pairs.

eduard3 years ago
3 0
1 single bond, 2 double bond 2 lone pairs
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pH at equivalence point is 8.52

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HCOOH+NaOH\rightarrow HCOO^{-}Na^{+}+H_{2}O

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So, moles of NaOH used to reach equivalence point equal to number of moles HCOO^{-} produced at equivalence point.

As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.

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At equivalence point, pH depends upon hydrolysis of HCOO^{-}. So, we have to construct an ICE table.

HCOO^{-}+H_{2}O\rightleftharpoons HCOOH+OH^{-}

I: 0.1940                                   0                 0

C: -x                                          +x               +x

E: 0.1940-x                                x                  x

So, \frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}

species inside third bracket represent equilibrium concentrations

So, \frac{x^{2}}{0.1940-x}=5.56\times 10^{-11}

or,x^{2}+(5.56\times 10^{-11}\times x)-(1.079\times 10^{-11})=0

So, x=\frac{-(5.56\times 10^{-11})+\sqrt{(5.56\times 10^{-11})^{2}+(4\times 1.079\times 10^{-11})}}{2}

So, x=3.285\times 10^{-6}M

So, pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52

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