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yan [13]
2 years ago
11

Nitrogen (N) is much more electronegative than hydrogen (H). Knowing this, consider how the electrons will be shared in ammonia

(NH3). What do you predict about the polarity of ammonia
Chemistry
1 answer:
irakobra [83]2 years ago
7 0

Answer:

Ammonia is a polar molecule

Explanation:

Nitrogen is more electronegative than hydrogen, so it's ability to attract electrons is higher.

This means that shared electrons would be mostly pulled towards the central nitrogen atom.

It is a polar molecule because it has an asymmetrical shape, and unequal charges. Nitrogen has lone pair, and lone pair-bonding pair repel second most (after lone pair-lone pair), this contributes to its polarity

You should also note, that hydrogen bonded with either nitrogen, oxygen or fluorine has the most strongest bond called hydrogen bonding.

Hope this helps!

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Consider the nitration by electrophilic aromatic substitution of salicylamide to iodosalicylamide. Reaction scheme illustrating
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<u>Answer:</u> The percent yield of the reaction is 68.68%.

<u>Explanation:</u>

To calculate the mass of salicylamide, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of salicylamide = 1.06 g/mL

Volume of salicylamide = 3.65 mL

Putting values in above equation, we get:

1.06g/mL=\frac{\text{Mass of salicylamide}}{3.65mL}\\\\\text{Mass of salicylamide}=(1.06g/mL\times 3.65mL)=3.869g

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of salicylamide = 3.869 g

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Putting values in equation 1, we get:

\text{Moles of salicylamide}=\frac{3.869g}{137.14g/mol}=0.0295mol

The chemical equation for the conversion of salicylamide to iodo-salicylamide follows:

\text{salicylamide }+NaI+NaOCl+EtOH\rightarrow \text{iodo-salicylamide }

By Stoichiometry of the reaction:

1 mole of salicylamide produces 1 mole of iodo-salicylamide

So, 0.0295 moles of salicylamide will produce = \frac{1}{1}\times 0.0295=0.0295moles of iodo-salicylamide

Now, calculating the mass of iodo-salicylamide from equation 1, we get:

Molar mass of iodo-salicylamide = 263 g/mol

Moles of iodo-salicylamide = 0.0295 moles

Putting values in equation 1, we get:

0.0295mol=\frac{\text{Mass of iodo-salicylamide}}{263g/mol}\\\\\text{Mass of iodo-salicylamide}=(0.0295mol\times 263g/mol)=7.76g

To calculate the percentage yield of iodo-salicylamide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iodo-salicylamide = 5.33 g

Theoretical yield of iodo-salicylamide = 7.76 g

Putting values in above equation, we get:

\%\text{ yield of iodo-salicylamide}=\frac{5.33g}{7.76g}\times 100\\\\\% \text{yield of iodo-salicylamide}=68.68\%

Hence, the percent yield of the reaction is 68.68%.

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