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valina [46]
2 years ago
15

Which ion is formed when an atom of fluorine (f) gains one electron? f–1 f–2 f 1 f 2

Chemistry
1 answer:
gregori [183]2 years ago
6 0

Answer:

If a fluorine atom gains an electron, it becomes a fluoride ion with an electric charge of -1.

Explanation:

hope this helps

pls mark brainliest

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After distilling your crude methyl benzoate, you set aside 4.83 grams of the purified ester. You then prepare the grignard reage
Ber [7]

Answer:

95.6 %

Explanation:

For this question, we will have <u>2 reactions</u>, the formation of the <u>grignard reagent</u> and the <u>formation of the alcohol</u>. The first step then is the calculation of the <u>maximum amount</u> of the grignard reagent. For this, we have to convert the grams to moles and check the smallest value. To do this we have to take into account the <u>following conversion ratios</u>:

Molar mass of Mg = 24 g/mol

Molar mass of phenylmagnesium bromide (C_6H_5Br)= 157 g/mol

Density of bromobenzene= 1.5 g/mL

Molar ratio between Mg and  C_6H_5Br= 1:1

2.3~g~Mg\frac{1~mol~Mg}{24~g~Mg}=0.096~mol~Mg

9.45~mL~\frac{1g}{1.5mL}\frac{1~mol~C_6H_5Br}{157~g}=0.0402~mol~C_6H_5Br

The smallest value is the mol of bromobenzene therefore <u>0.0402 mol</u> of phenylmagnesium bromide would be produced.

The next step is repite the same steps for the reaction of <u>formation of the alcohol</u>. Therefore we have to find the moles of methyl benzoate, so:

Molar mass of methyl benzoate: 136.14 g/mol

4.83~g~\frac{1~mol}{136.14~g}=0.35~mol

The we have to <u>divide by the coefficient</u> of each reactive in the balance reaction. So:

\frac{0.35~mol}{1}=0.35

\frac{0.0402~mol}{2}=0.0201

Therefore the <u>limiting reagent</u> would be the phenylmagnesium bromide. Now, the <u>molar ratio</u> between the phenylmagnesium bromide and triphenyl carbinol is <u>2:1</u>, so the amount of alcohol produced is 0.0201 mol triphenyl carbinol. The next step is the conversion from mol to <u>grams of triphenyl carbinol</u>:

Molar mass of triphenyl carbinol= 260.33 g/mol

0.0201~mol\frac{260.33~g}{1~mol}=5.23~g~triphenyl carbinol

Finally, we have to <u>divide</u> the obtanied solid by the calculated one:

Percentage=\frac{5}{5.23}*100=95.6\%

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