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3 years ago

Draw a lewis structure for hccl3. Show all unshared pairs and the formal charges

Chemistry

1 answer:

klasskru [66]3 years ago

6 0

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

The given molecule is, $HCCl_3$

Carbon has '4' valence electrons, hydrogen has '1' valence electron and chlorine has '7' valence electrons.

Therefore, the total number of valence electrons in $HCCl_3$ = 1(1) + 1(4) + 3(7) = 26

According to electron-dot structure, there are 8 number of bonding electrons and 18 number of non-bonding electrons.

The electron-dot structure of $HCCl_3$ is shown below.

Now we have to calculate the formal charges on carbon, hydrogen and chlorine.

Formula for formal charge :

$\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}$

$\text{Formal charge on carbon}=4-0-\frac{8}{2}=0$

$\text{Formal charge on hydrogen}=1-0-\frac{2}{2}=0$

$\text{Formal charge on chlorine}=7-0-\frac{2}{2}=0$

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Total weight = 1 + 35.45 + 3 * 16 = 84.45

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3 years ago

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Answer:

ΔG < 0

Explanation:

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3 years ago

What is the [OH-] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 m

velikii [3]

Answer: The $[OH^-]$ of a solution is $10^{-12}$ M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

moles of $HCl$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{0.0912g}{36.5g/mol}=0.0025mol$

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.0025\times 1000}{250}=0.01$

pH or pOH is the measure of acidity or alkalinity of a solution.

$HCl\rightarrow H^++Cl^{-}$

According to stoichiometry,

1 mole of $HCl$ gives 1 mole of $H^+$

Thus $0.01$ moles of $HCl$ gives = $\frac{1}{1}\times 0.01=0.01$ moles of $H^+$

Putting in the values:

$[H^+][OH^-]=10^{-14}$

$[0.01][OH^-]=10^{-14}$

$[OH^-]=10^{-12}$

Thus the $[OH^-]$ of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution is $10^{-12}$ M

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