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vichka [17]
3 years ago
5

Which is the correct Lewis structure for carbon tetrabromide (CBr4), in which a central carbon atom is bound to four atoms of th

e group 7A element bromine?
A



B



C



D

Chemistry
2 answers:
AveGali [126]3 years ago
8 0
The answer is b hope I helped
artcher [175]3 years ago
7 0

Answer:

B

Explanation:

Hello,

In this case, since carbon is in group IVA, it has four valence electrons so it needs four bonds to attain the octet. However, as carbon and bromine are bonded, carbon needs four bromines as each bromine provides only one electron for them to attain the octet as bormine has seven valence electrons for which it is in group VIIA. Therefore, the Lewis structure showing the four bonds and the seven free electrons of each bromine is B.

Best regards.

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The question is incomplete, here is the complete question:

By titration, 15.0 mL of 0.1008 M sodium hydroxide is needed to neutralize a 0.2053-g sample of an organic acid. What is the molar mass of the acid if it is monoprotic.

<u>Answer:</u> The molar mass of monoprotic acid is 135.96 g/mol

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

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Molarity of NaOH solution = 0.1008 M

Volume of solution = 15.0 mL

Putting values in above equation, we get:

0.1008M=\frac{\text{Moles of NaOH}\times 1000}{15.0}\\\\\text{Moles of NaOH}=\frac{(0.1008\times 15.0)}{1000}=0.00151mol

As, the acid is monoprotic, it contains 1 hydrogen ion

1 mole of OH^- ion of NaOH neutralizes 1 mole of H^+ ion of monoprotic acid

So, 0.00151 moles of OH^- ion of NaOH will neutralize \frac{1}{1}\times 0.00151=0.00151mol of H^+ ion of monoprotic acid

<u>In monoprotic acid:</u>

1 mole of H^+ ion is released by 1 mole of monoprotic acid

So, 0.00151 moles of H^+ ion will be released by \frac{1}{1}\times 0.00151=0.00151mol of monoprotic acid

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of monoprotic acid = 0.00151 mole

Given mass of monoprotic acid = 0.2053 g

Putting values in above equation, we get:

0.00151mol=\frac{0.2053g}{\text{Molar mass of monoprotic acid}}\\\\\text{Molar mass of monoprotic acid}=\frac{0.2053g}{0.00151mol}=135.96g/mol

Hence, the molar mass of monoprotic acid is 135.96 g/mol

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