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4 years ago

PLEASE NO GUESSES OR INCORRECT ANSWERS

1 answer:

3 0

Diffusion is the movement of any chemical from one place to another.
Osmosis is the movement of water across a partially permeable membrane.

I hope this helps!

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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

I need help with problem 103 all of the questions.

AfilCa [17]

A. is decomposition so HCL = H2 + Cl2

not balanced cause hcl needs 2

2HCL = H2 + Cl2

balanced

b. Br2 + Al-i = AlBr3 + I2 single rep.

not balanced since br need 3 so watch carefully cause many changes needed

3Br2 + Al-i = AlBr3 + I2 not right is unbalanced so make it 2

3Br2 + Al-i = 2AlBr3 + I2 now left Al is unbal. so make 2 there

3Br2 + 2Ali = 2AlBr3 + I2

Balanced

C. Na + S = Na2S synthesis reaction is not bal. left Na needs 2

2Na + S = Na2S balanced.

4 0

3 years ago

Escribe 4 unidades de temperatura

olchik [2.2K]

Answer:

esta nublado

Explanation:

3 0

3 years ago

Neutral grain spirit is essentially the same as

Dennis_Churaev [7]

The neutral grain spirit is essentially known as rectified spirits or ethyl alcohol.

<h3>Meaning of neutral grain spirit</h3>

neutral grain spirit can be defined as raw material that is very concentrated with alcohol used to produce strong alcoholic products.

neutral grain spirit as the name implies is neutral in nature that is it can be blended with any other alcoholic product.

In conclusion, The neutral grain spirit is essentially known as rectified spirits or ethyl alcohol.

Learn more about ethyl alcohol: brainly.com/question/1049383

#SPJ11

5 0

2 years ago

What does this mean: A saturated solution may be quite dilute because some substances have a very low solubility. I don't unders

atroni [7]

Answer:

Here's how I would explain it.

Explanation:

Think of it this way.

When you mix solutions of silver nitrate and sodium chloride, you get an immediate precipitate of silver chloride. The equation is

Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)

Now, take some AgCl and stir it vigorously with water.

You won't see much happening, because the AgCl is has such a low solubility. Not much of it will go into solution. And yet, a small amount of it does dissolve until the solution is saturated.

The concentration of AgCl in the saturated solution is

about 0.000 01 mol·L⁻¹.

I hope you will agree that this is a dilute solution even though it is saturated with AgCl.

8 0

4 years ago