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3 years ago

What are Free Radicals ?

Chemistry

2 answers:

DENIUS [597]3 years ago

4 0

Answer:

an uncharged molecule (typically highly reactive and short-lived) having an unpaired valency electron.

larisa [96]3 years ago

4 0

an uncharged molecule (typically highly reactive and short-lived) having an unpaired valency electron.

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Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

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

$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

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

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

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

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

8 0

3 years ago

What is the standard unit used to measure mass? cubic meters gram liter meter

azamat

Your answer is grams

8 0

3 years ago

Read 2 more answers

How much does a gallon of water WEIGH?

elixir [45]

A gallon of water weighs 8.34 pounds

4 0

3 years ago

DUE TOMORROW!!! 15 POINTS

Lunna [17]

Answer:

A. write balanced chemical equation (including states), for this process.

Explanation:

Almost all hydrocarbon 'burn' reactions involve oxygen; it's by far the most reactive substance in air.

Hydrocarbon combustions always involve

[some hydrocarbon] + oxygen --> carbon dioxide + steam.

C6H6(l) + O2 (g)--> CO2 (g)+ H2O (g)

Balance carbon, six on each side:

C6H6(l) + O2 (g)--> 6CO2 (g)+ H2O (g)

Balance hydrogen, six on each side:

C6H6(l) + O2 (g)--> 6CO2(g) + 3H2O (g)

Now, we have fifteen oxygens on the right and O2 on the left.

Two ways to deal with that. We can use a fraction:

C6H6 (l)+ (15/2)O2 (g)--> 6CO2 (g)+ 3H2O (g)

Or, if you prefer to have whole number coefficients, double everything

to get rid of the fraction:

2C6H6 (l)+ 15O2 (g)--> 12CO2 (g)+ 6H2O (g)

With the SATP states thrown in...

C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(g)

6 0

3 years ago

A mixture of propane and butane is fed into a furnace where it is mixed with air. the furnace exhaust leaves the furnace at 305°

IgorLugansk [536]

In this item, I supposed, that we are determine the molar fraction of oxygen and carbon dioxide in the sample. This can be done by dividing their respective partial pressures by the total pressure of the sample.

O2 : mole fraction = (100.7 mmHg) / (763.00 mmHg) = 0.13

CO2 : mole fraction = (33.57 mmHg) / (763.00 mmHg) = 0.044

Answers: O2 = 0.13
CO2 = 0.044

3 0

2 years ago

What are Free Radicals ?​

What are Free Radicals ?