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3 years ago

Draw the lewis structure of the hypochlorite ion clo- . include lone pairs

Chemistry

1 answer:

Dmitry_Shevchenko [17]3 years ago

6 0

Solution:- Hypochlorite ion $ClO^-$ has one Cl and one O atom. Cl has 7 valence electrons and O has 6 valence electrons. Since there is one negative charge on the ion,

total valence electrons = 7 + 6 +1 = 14

(note:- if there is negative charge then it is added and if there is positive charge then it is subtracted while calculating the valence electrons)

Both Cl and O atoms wants to complete their octet and so for this we put a single bond between them. Single bond means two electrons, so the remaining electrons would be 14 - 2 = 12

It means 12 electrons will be placed as lone pair of electrons. To complete the octet, we put 6 dots around each of the atom. Oxygen is more electron negative than Cl, so we show the -1 charge for oxygen.

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Consider an exceptionally weak acid, HA, with a Ka = 1x10 -20 . You make a 0.1M solution of the salt Na

Vedmedyk [2.9K]

Answer:

$pH=13$

Explanation:

Hello,

In this case, given the acid, we can suppose a simple dissociation as:

$HA\rightleftharpoons H^+ + A^-$

Which occurs in aqueous phase, therefore, the law of mass action is written by:

$Ka=\frac{[H^+][A^-]}{[HA]}$

That in terms of the change $x$ due to the reaction's extent we can write:

$1x10^{-20}=\frac{x*x}{0.1M-x}$

But we prefer to compute the Kb due to its exceptional weakness:

$Kb=\frac{Kw}{Ka}=\frac{1x10^{-14}}{1x10^{-20}} =1x10^{-6}$

Next, the acid dissociation in the presence of the base we have:

$Kb=\frac{[OH^-][HA]}{[A^-]}=1x10^{6}=\frac{x*x}{0.1-x}$

Whose solution is $x=0.0999M$ which equals the concentration of hydroxyl in the solution, thus we compute the pOH:

$pOH=-log([OH^-])=-log(0.0999)=1$

Finally, since the maximum scale is 14, we can compute the pH by knowing the pOH:

$pH+pOH=14\\\\pH=14-pOH=14-1\\\\pH=13$

Regards.

5 0

3 years ago

Read 2 more answers

HELP!!! The chemical formula is different from the empirical formula in

Lelechka [254]

Answer:

the Ans is H2O2

Explanation:

THANKS.............

3 0

2 years ago

Read 2 more answers

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

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3 years ago

Sickle cell is anemia is a deadly,recessive disease.How is sickle cell anemia inherited

vichka [17]

Autosomal recessive: cystic fibrosis (CF), sickle cell anemia (SC), Tay Sachs disease. Genes are inherited from our biological parents in specific ways. One of the basic patterns of inheritance of our genes is called autosomal recessive inheritance.

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Isotopes percent abundance of antimony

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